Q stringlengths 18 13.7k | A stringlengths 1 16.1k | meta dict |
|---|---|---|
Solution to Dirac equation with external source The Dirac equation is:
\begin{equation}
\left[i\gamma^{\mu}(\partial_{\mu}-iA_{\mu})-m\right]\psi=0, \tag{1}
\end{equation}
where $A_\mu$ is a gauge field. The solution to this equation is:
\begin{equation}
\psi=\exp\Big(i\int^xA_{\mu}dx^{\mu}\Big)\psi_0, \tag{2}
\end{equation}
where $\psi_0$ is the solution to the Dirac equation in the absence of the gauge field.
Now suppose there is an external field $B_\mu$, which has an $SU(3)$ symmetry such that $B_{\mu} = B_{\mu}^{a}T_{3}^{a}$ where $T_{3}^{a} = \frac{1}{2}\lambda^{a}$ with $\lambda^{a}$being Gell-Mann matrices. Then the Dirac equation becomes
\begin{equation}
\left[i\gamma^{\mu}(\partial_{\mu}-iA_{\mu})-m\right]\psi = B_{\mu}^{a}T_{3}^{a}\psi, \tag{3}
\end{equation}
or
\begin{equation}
\left[i\gamma^{\mu}\left(\partial_{\mu}-iA_{\mu}+ iB_{\mu}^{a}T_{3}^{a}\right)-m\right]\psi = 0. \tag{4}
\end{equation}
Then, in analogy to eq. (2), can we write solution to eq. (4) as
\begin{equation}
\psi=\exp\left[i\int^x \left(A_{\mu}-B_{\mu}^{a}T_{3}^{a}\right)dx^{\mu}\right]\psi_0? \tag{5}
\end{equation}
If not, how should we solve eq. (4)?
| Neither of your claimed solutions are valid in general. In order for an expressions like
\begin{equation}
\psi(x) =\exp\Big(i\int^xA_{\mu}dx^{\mu}\Big)\psi_0,
\end{equation}
to be a valid function of positionm the integral has to be independent of the path chosen. This is only the case if the $F_{\mu\nu}= \partial_\mu A_\nu-\partial_\nu A_\mu$ gauge field is zero.
In general, given an equation such as your Eq (1), you have to work hard to solve it. The case of a uniform magnetic field is the simplest example. In that case there is a solution in terms of associated Legendre polynomials that is related to the solution of the 2 dimensional Harmonic oscillator in polar coordinates. I suggest you work though that example and find it.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/710124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What is the instant velocity? The velocity is the variation rate of the position correct? So does it make sense to talk about velocity without time?
| Instantaneous speed is just the speed something has at some instance in time. Time is still involved, but in an infinitesimal fashion.
You are correct in that if you took a picture of a moving object at some time, the picture would be still and you cannot tell anything about the motion of the object from the photo. Only with multiple photos at different times motion can be measured.
But that does not mean that motion does not exist. This is the classic Zeno paradox.
Instantaneous motion is an intrinsic property of object relevant to many physical calculations, but one that cannot directly measure directly, but rather from the displacement over time and thus actually measuring average speed and not instantaneous speed.
The only way to measure the instantaneous speed of object from a photo is if the object was of known length and a ruler was present in the photo while the body is moving with relativistic speeds. From the length contraction measured once can estimate the instantaneous speed without needed multiple photos.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/710296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 7,
"answer_id": 2
} |
Can't understand a statement about motion From the book where I am studying motion, It says
Motion is a combined property of the object under study and the observer. There is no meaning of rest or motion without the viewer.
I know that, for an object, it can be said that 'it is moving' in one frame of reference, and it can be said that 'it is at rest' in another frame of reference, but the sentence I mentioned above seems somewhat confusing. How can a phenomena be a property of two things? Also, how is it that, when there is no one to see, the topic of motion and rest is irrelevant? I don't know exactly what the second sentence is trying to say, provided that my understanding of the second sentence is wrong. I need assistance.
| It is just saying that all motion is measured relative to something else that could also be moving since there is no absolute reference for position in the universe. Though I think there was supposed to be something special about the CMB.
So it's just saying that you always need something to measure velocity against (velocity, not acceleration).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/710421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
Contradiction in my understanding of wavefunction in finite potential well Most things like to occupy regions of lower potential. So the probability amplitude should be higher in a region of lower potential. I denote the potential by V.
However, we also know that the kinetic energy of a particle is given by E-V - the bigger this E-V difference (the lower the V basically), the bigger the kinetic energy, thus the bigger the velocity of that particle. By having a bigger velocity though, this means that less time is spent by the particle in the region, thus a lower wavefunction amplitude.
How do both of these statements reconcile with one another? By looking at the diagram below, we see that we have a decreasing amplitude as we go across to the left. But surely the particle will want to be in a region of lower potential?
| Imagine a perfectly elastic ball dropped vertically onto a flat surface. The ball heads for the point of lowest potential, ie the ground, but because of conservation of energy it bounces back to its initial position and continues to bounce up and down. It spends most of its time near the top of its bounce, because it is ravelling more slowly there, so the probability of finding it at a given instant is highest at the top of the bounce and lowest at the bottom, even though the bottom is the most attractive place to be in terms of PE.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/710744",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 2
} |
Renormalization Group Flow I am reading a book on effective field theory where the following "renormalization group equation" is given:
Now a quick search on google shows a bunch of interesting pictures of "renormalization group flows":
My question is: what is the relationship between the RG equation and the RG flow? In the picture, what is the "time parameter" of the flow, and what is being "flowed"?
| RG time is usually introduced as a dimensionless parameter like e.g. $$\Lambda\equiv\Lambda_0\mathrm{e}^{-t},$$
where in this convention $t=0$ corresponds to the ultra-violet (UV) at the initial momentum scale $\Lambda_0$ (which might be asymptotically large/infinite) and $t\rightarrow\infty$ corresponds to the infra-red (IR) with $\Lambda\rightarrow0$.
The RG equation in the question encodes Wilson's RG approach of integrating out momentum modes successively. Fluctuating fields are split into high (H) and low (L) according to their momentum. $S^\mathrm{eff}(\Lambda')$ is obtained from $S^\mathrm{eff}(\Lambda)$ at the higher scale $\Lambda$ by integrating out the high momentum modes $\phi_H$ with momenta $k\in(\Lambda',\Lambda]$. Higher momentum modes with $k>\Lambda$ are already integrated out in $S^\mathrm{eff}(\Lambda)$.
The RG flow is the change of (the couplings in) $S^\mathrm{eff}(\Lambda)$ with the RG scale $\Lambda$ or equivalently RG time $t$. For simplicity, practicality or sometimes physical reasons usually the flow of a few couplings (in many cases one coupling: e.g. the running coupling of QED or QCD in their perturbative regimes) is considered. RG flows are depicted or sketched in the typical diagrams one finds when googeling "RG flow". Usually not as functions of $t$ but rather as parametric plots in the space of couplings.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/711150",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Eyes shut, can a passenger tell if they’re facing the front or rear of the train? Suppose you’re a passenger sitting in one of the carriages of a train which is travelling at a high, fairly steady speed. Your eyes are shut and you have no recollection of getting on the train or the direction of the train’s acceleration from stationary. Can you tell whether you’re facing the front or the back of the train?
This isn’t a theoretically perfect environment - there are undulations, bends and bumps in the track. Not a trick question - you cannot ask a fellow passenger!
Edit: This is intentionally lacks rigorous constraints. Do make additional assumptions if it enables a novel answer.
| Yes, because the train predictably hits bumps/gaps in the track.
As you travel on a train, you hear first one set of wheels then the other then the other going from the front of the train to the back hit these imperfections in the track. If you are facing the direction of travel you hear these coming from in front of you. If you are facing away, they come from behind.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/711352",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 9,
"answer_id": 1
} |
How do experimentalists measure the exciton binding energy? The exciton binding energy in semiconductors is determined theoretically by the energetic difference between the fundamental gap and the optical gap or, in other words, as the energetic difference of the fundamental gap and the first exciton peak in an optical spectrum.
My question is related to experiment. How would one obtain this energetic difference or the values of the fundamental and optical gap with good accuracy?
| One can measure light absorption in a semiconductor. Peaks in absorption could appear corresponding to the photon -> exciton transition. They appear bellow optical gap
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/711604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What is a "cycle time" in Allan deviation formula for atomic clock instability? And why does having more independent atoms reduce $\sigma_y(\tau)$? The usual formula for clock instability is given as
$\sigma_y(\tau)\approx\frac{\Delta f}{f_0\sqrt{N}}\sqrt{\frac{T_c}{\tau}}$
First off what do each of these symbols really mean?
What is $T_c$? The literature mentions it is cycle time. What "cycle" is it? Is it the period of oscillation $1/f_0$? But, that can't be right as it is already present in the formula.
Next, as per my intuition, the more independent atoms $N$ you have more scrambled the phase of the clock would be. So I don't understand why it is in the denominator. Same for the averaging time $\tau$. Why is it in the denominator? Wouldn't the phase of a clock made of independent atoms scramble more and more with time?
| Partial answer: Each atom is an independent measurement of the local oscillator frequency so you get more statistics on your frequency estimate each shot. However, you’re right that if different atoms are frequency shifted relative to each other due to, for example, a non zero velocity distribution or a inhomogeneous zeeman shifts due to an inhomogeneous magnetic field, then yes this can impact clock stability. The standard quantum limit for clock performance that you cited assumes there a no such broadening mechanisms like these.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/712068",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Points where electric field is zero when charges are present at vertices of a regular polygon There is a $n$-sided regular polygon with a charge $q$ at each vertex. I know that there are $n$ points, other the center of the polygon, where the electric field is zero. But why is this so? Is there a general way to prove it?
PS: I know some questions related to my question have been asked, but none of them gives me a satisfactory reason why there should be a total of $n+1$ neutral points in space for such a charge distribution.
| Consider a line connecting two adjacent charges on your polygon, and then another line which bisects that one, going in the positive x direction through the center of the polygon (at x = 0). To show that your premise is true, you would need to find one (and only one, if (n) is odd) other point (other than at x = 0) on that line of symmetry, where the sum of the x components of the fields is zero. For a general proof, the distances and components of distances would be a function of (n), and would depend on whether (n) was odd or even.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/712514",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How are these Covariant Derivative Identities found? In David Tong's Gauge Theory notes on page 137 near eq. (3.30) he makes use of the following expressions for the covariant derivative $D_{\mu}$
$$\frac{1}{2}[\gamma^{\mu},\gamma^{\nu}]D_{\mu}D_{\nu}=\frac{1}{4}[\gamma^{\mu},\gamma^{\nu}][D_{\mu},D_{\nu}]\tag{1}$$
and
$$e^{-ikx}e^{D^2}e^{ikx}=e^{(D_{\mu}+ik_{\mu})^2}\tag{2}$$
I'm guessing the first is just a change of dummy indices in the second term of the commutator, but I don't see how the indices are dummy.
The second expression I'm more confused about. It looks like $x^{\mu}$ is acting like a generator of translation in momentum space, but I'm not sure.
| Hint for eq. (2):
$$e^{-ik\cdot x} f(D) e^{ik\cdot x}~=~f\left(e^{-ik\cdot x} D e^{ik\cdot x}\right)$$
and
$$\begin{align} e^{-ik\cdot x} D_{\mu} e^{ik\cdot x}~\stackrel{\text{Hadamard}}{=}&~e^{-ik_{\nu} [x^{\nu},\cdot]} D_{\mu}\cr
~=~~~&D_{\mu}+ik_{\nu} [D_{\mu},x^{\nu}]\cr
~=~~~&D_{\mu}+ik_{\mu},\end{align}$$
where we used Hadamard's formula.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/712683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
In Hamilton-Jacobi theory, how is the new coordinate $Q$ time-independent when Hamilton's principal function separates? Following the notation in Goldstein, the solution to the Hamilton-Jacobi equation is the generating function $S$ for a canonical transformation from old variables $(q,p)$ to new variables $(Q,P)$ where the new Hamiltonian is $K=0$ and the new momentum is the integration constant $P=\alpha$. Therefore, $\dot{Q}=\dot{P}=0$. If the old Hamiltonian is not a function of time, the solution may be expressed as
$$
S(q,\alpha,t)=W(q,\alpha) - \alpha t.\tag{1}
$$
What's confusing me is that, by definition, $S$ is a type 2 generating function where the new variables are constant in time. However, the new coordinate $Q$ from the type 2 generating function above is expressed as
$$Q=\frac{\partial S}{\partial P} = \frac{\partial S}{\partial \alpha} = \frac{\partial W}{\partial \alpha} - t.\tag{2}$$
Therefore giving
$$\dot{Q} = -1 \neq 0.\tag{3}$$
What am I missing here?
| I will assume you are only talking about the case of $1$ degree of freedom (or else, your notation is problematic).
Generally, it is always a good reflex to study a specific case when you have two conflicting arguments. Take for example a free particle with $H = \frac{p^2}{2}$, solving the Hamilton-Jacobi equation, you get:
$$
S = \sqrt{2P}q-Pt
$$
from which
$$
P = \frac{p^2}{2} \\
Q = \frac{q}{p}-t \\
$$
and since
$$\dot p = 0\\
\dot q = p
$$
you have $\dot Q = 0$.
As you can see, your mistake was in the last step
$$
Q = \left(\frac{\partial W}{\partial P}\right)_q-t \\
\dot Q = -1
$$
Remember that $W$ is a function of $q,P$, so there is an implicit time dependence. Using the chain rule, you get the extra terms:
$$
\dot Q = \left(\frac{\partial^2 W}{\partial P^2}\right)_q\dot P+\frac{\partial^2 W}{\partial q\partial P}\dot q-1
$$
The first one is trivial by construction, since $\dot P = 0$, but the second one will cancel the $-1$.
To prove this fortuitous cancellation, the fastest route is to refer to the general proof of the HJE. In this simple case, there is also the following shortcut starting from the HJE:
$$
H\left(\frac{\partial W}{\partial q},q\right) = P
$$
and by applying $\left(\frac{\partial}{\partial P}\right)_q$, you get:
$$
\left(\frac{\partial H}{\partial p}\right)_q\frac{\partial^2 W}{\partial P\partial q} = 1 \\
\dot q \frac{\partial^2 W}{\partial q\partial P} = 1 \\
$$
hence the result. Notice that using $p = \left(\frac{\partial W}{\partial q}\right)_P$ and using $\dot H = 0$, you get from the HJE $\dot P = 0$.
Hope this helps and tell me if you need more details.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/713564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Is the universe made up of particles or fields? QFT describes the electron as an excitation of the electron field. The spin of electrons create magnetic fields. So which came first? How can a particle created from a field then create its own field? Or are some fields not fundamental?
|
QFT describes the electron as an excitation of the electron field.
Quantum field theory fields are completely different than the fields of Newtonian gravity and classical electrodynamics. It is a theoretical model that is very successful in fitting existing data and observation of elementary particles, the electron in your example, and is predictive of new data and observations at the microscopic world of elementary particles and neuclei..
The electric and magnetic and classical electromagnetic fields are part of the theory for macroscopic dimensions and fit the data for these large dimensions.
the spin of electrons create magnetic fields.
It can be shown mathematically that the classical theories emerge from the basic quantum mechanical theories. So the mathematics of the spin of the electrons leads at large dimensions to classical magnetic fields.
so which came first? how can a particle created from a field then create its own field?
It is in the mathematical model of QFT that an electron is created by an operator acting on a qft field. A model. It is the measurements and observations that are fitted by the model that leads to describing electrons as created by the operator on the field. The electrons exist in nature whether one models them or not.
or are some fields not fundamental?
In the hierarchy of models, first come the QFT fields, from which emerge the classical fields, so in that sense, if we call the QFT fields fundamental then the classical are derived. BUT this is the present theoretical model. It could be that in the future new theories would have the QFT model emerging from a more fundamental theory.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/713942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
What is the generally expected or more useful form of supersymmetry, on-shell or off-shell? If the on-shell numbers of degrees of freedom of bosons and fermions match we
have on-shell supersymmetry, if the off-shell numbers match we have off-shell
supersymmetry. When LHC people say they are searching for supersymmetry are they searching for the on-shell or off-shell or both?
In the Introduction to the ADS/CFT Correspondence by Horatiu Nastase it was mentioned that for the Super Yang-Mills theory only the on-shell version is known and there is no fully off-shell formulation with Lorentz invariance. Does that mean the results (not talking about the results which are useful in QCD calculations irrespective of whether supersymmetry is real, talking about Quantum Gravity results related to the Black Hole Information Paradox etc) of ADS/CFT are valid only if the universe has on-shell version of supersymmetry and are invalid if the universe has the off-shell version?
Or is it like we can always (in principle) introduce some auxiliary fields so that we have both on-shell and off-shell supersymmetry?
| On-shell SUSY means the theory is supersymmetric when taking into account equations of motion. Off-shell means that the action is supersymmetric even before writing equations of motion (obviously off-shell SUSY holds on shell as well). Physically, we are interested in on-shell stuff, because physics is ultimately expressed by equations of motion, but off-shell SUSY is still useful because it helps us to write down most general interactions between various supermultiplets.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/714352",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Why I cannot write the time evolution operator $e^{-i(T+V)t}$ as the product of operators $e^{-iTt}e^{-iVt}$ To calculate the wave equation of a time-independent Hamiltonian we use:
$$
\Psi_{i}(r,t)=e^{-iH^{0}t}\psi_{i}(r,0).
$$
We also know that the time-independent Hamiltonian $H^{0}=T+V$ is given to the sum of kinetic and potential energies, in an isolated atom.
However, we cannot write $e^{-iH^{0}t}$ as the product of the operator, $e^{-iTt}e^{-iVt}$. Why is that?
| Let $t$ be a dummy variable to keep track of orders in a series expansion:
$$e^{tA} e^{tB}
=
\Big(1 + tA + \frac{1}{2} A^2 t^2 + \mathcal{O}\left(t^3\right) \Big)
\Big(1 + tB + \frac{1}{2} B^2 t^2 + \mathcal{O}\left(t^3\right) \Big)
= 1 + (A + B)t + \frac{1}{2} (A^2 + 2 AB + B^2) t^2 + \mathcal{O}\left(t^3\right)
\, .
$$
Now look at the series expansion for
$$e^{t(A+B)}
= 1 + (A+B)t + \frac{1}{2} (A + B)^2 t^2 + \mathcal{O}\left( t^3 \right)
= 1 + (A+B)t + \frac{1}{2} \left(A^2 + AB + BA + B^2\right) t^2 + \mathcal{O}\left( t^3 \right)
\, .$$
If $A$ and $B$ are numbers, then $AB + BA = 2AB$ and the series expansions agree. But in quantum mechanics the kinetic and potential energies are operators. Then $$AB + BA = 2AB + [B, A]$$
and generally the commutator $[B, A]$ is nonzero. You can check that
$$\exp (tA) \exp (tB) = \exp t\Big(A + B + \frac{1}{2}[A, B] \Big) + \mathcal{O}\left(t^3 \right) $$
which is the start of the BCH formula mentioned in the other answer. With some more algebra and combinatorics, you can work out the higher order terms, too.
Because all the terms in the BCH formula except $A + B$ involve commutators, the usual formula for the exponential of numbers is recovered when all commutators vanish. As a generalisation of that, a good principle to remember is that the only thing that can "go wrong" when trying to derive an operator version of a formula that is true for numbers, is that the operators can fail to commute. So for commuting operators formulas you are used to for numbers still hold.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/714531",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Infinite conducting plane Let $\pi$ be an infinite conducting plane laying in $z=0$.
the plane is kept in potential of 8 volts, ($\phi(z=0)=8[V]$).
Prove or disprove:
the surface charge density - $\sigma$ is well defined.(there is a single solution to the surface charge density.
What I thought:
You have to solve a Laplace Equation for the potential above the surface and below the surface.
The solution to laplace equation is unique, and therefore the electric field above and below is unique and well defined. According to the equation: $\hat{n}\cdot (\vec{D_2}-\vec{D_1}) = \rho_s$ the surface charge density is also unique and well defined.
| Without knowing what happens at any other point, the solution is indeed non-unique. The potential has to obey Laplace's equation below and above the plane, so
$$\phi''(z) = 0,$$
meaning that
$$\phi(z) = \begin{cases}
A_+ z + B_+ & z>0\\[5pt]
A_- z + B_- & z<0
\end{cases}$$
Note that since the system has translational invariance along the $x$ and $y$ directions, all quantities are only a function of $z$.
Now the only boundary condition given is that $\phi(0) = 8$ V. So $B_+ = B_- = 8$ V, i.e. all potentials of the form
$$\phi(z) = \begin{cases}
A_+ z + 8 \ \mathrm{V} & z\geq0\\[5pt]
A_- z + 8 \ \mathrm{V}& z<0
\end{cases}$$
are valid solutions. The corresponding surface charge density $\rho_s = \epsilon_0 \hat{\mathbf{z}}\cdot \big[\mathbf{E}(0^+) - \mathbf{E}(0^-)\big]$ is
\begin{align}
\rho_s &= -\epsilon_0 \Bigg[\frac{\partial \phi}{\partial z}(0^+)-\frac{\partial \phi}{\partial z}(0^-)\Bigg]\\[5pt]
&= -\epsilon_0 (A_+-A_-),
\end{align}
which is clearly non-unique, as it depends on $A_\pm$. I encourage you to look at the proof of the uniqueness theorem and see what assumption breaks down for this example.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/714941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Issue expanding $\sin \theta$ about $\theta_{eq}$ Quoting a textbook:
$$(m_1 + 2m_2\sin^2\theta)\ddot\theta = m_1\Omega^2\sin\theta\cos\theta - \frac g L (m_1 + m_2)\sin\theta.\tag{10}$$
We can simplify this expression a bit by relating $\frac g L (m_1 + m_2)$ to the equilibrium angle $\theta_{eq}.$
$$(m_1 + 2m_2\bbox[yellow]{\sin^2\theta})\ddot\theta = m_1\Omega^2\sin\theta(\cos\theta - \cos\theta_{eq})\tag{11}.$$
Keeping only the first order term in the smallness, we can replace
$$\sin\theta(\cos\theta - \cos\theta_{eq}) \rightarrow \sin\theta_{eq}(-\sin\theta_{eq}(\theta - \theta_{eq})).\tag{12}$$
This leads to the equation of motion for small oscillations
$$(m_1 + 2m_2\bbox[yellow]{\sin^2\theta_{eq}})\ddot\theta = -m_1\Omega^2\sin^2\theta_{eq}(\theta - \theta_{eq}).\tag{13}$$
Here we are expanding about $\theta_{eq}$ and everything makes sense except the parts highlighted in yellow. How was the jump made for $$\sin^2\theta \approx \sin^2 \theta_{eq}~ ?$$
I know that expanding about $\theta_{eq}$ and keeping only linear terms we get $$\sin\theta \approx \sin\theta_{eq} + \cos\theta_{eq} (\theta - \theta_{eq})$$
logically then $$\sin^2\theta \approx \sin^2\theta_{eq} + 2\sin\theta_{eq} \cos\theta_{eq} (\theta - \theta_{eq}) $$
why is the $\sin\theta_{eq} \cos\theta_{eq} (\theta - \theta_{eq}) $ term dropped? it is linear in $\theta$
| This is because your $\ddot{\theta}$ is already assumed small.
I think the simplest way to deal with this is to introduce an explicit counter $\epsilon$ which keeps track of the "smallness" of various things.
If you are expanding near $\theta_{eq}$, write
$$
\theta=\theta_{eq}+\epsilon \delta\theta
$$
where $\delta\theta=\theta-\theta_{eq}$.
More importantly, $\ddot{\theta}=\epsilon \delta \ddot{\theta}$
With this setup, you can expand in powers of $\epsilon$ and then set $\epsilon=1$ at the very end once you have truncated the series. Thus indeed:
$$
\sin^2(\theta)=\sin^2(\theta_{eq})+2\epsilon \cos\theta_{eq}\delta \theta
$$
but
$$
\sin^2(\theta)\ddot{\theta}\to
(\sin^2(\theta_{eq})+2\epsilon\cos(\theta_{eq}) \delta \theta)\epsilon\delta \ddot{\theta}
= \epsilon \sin^2(\theta_{eq}) \delta \ddot{\theta}
+2\epsilon^2 \cos\theta_{eq}\delta\theta \delta\ddot{\theta}
$$
and you see that there's an extra power of your counter $\epsilon$ that occurs because there's already an $\epsilon$ baked in the substitution $\ddot\theta=\epsilon \delta\ddot{\theta}$.
Thus, the linear term is indeed $\sin^2(\theta_{eq})\ddot{\theta}$ when expressed in terms of your initial variables.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/715750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Is this proof that massless objects cannot be charged? In the realm of pre relativistic physics.
$$\vec{p}=m\vec{v}$$
$$F= \frac{dp}{dt}= m\vec{a}$$
If there exists an electric field in space, the force experienced by it would be $$F= q\vec{E}$$
Applying newtons laws:
$$q\vec{E} = m \vec{a}$$
This an equation stating a relation between the force, and the acceleration of an object.
Given $$m=0$$
It follows that
$$q\vec{E} = 0$$
Since we assumed in the beginning that there exists and electric field.
$$q=0$$
Thus a massless object must have no charge. [Or newtons laws don't work for massless objects even in pre relativistic physics]
My question is: is my reasoning correct? And is there a generalisation of this to the relativistic equations.
| As described in Michael Seifert's answer, the question requires relativity, not Newtonian mechanics. As explained in the comments by Andrew Steane and Michael Seifert, the question also requires quantum mechanics. So we're in the realm of quantum field theory.
Massless charged particles are problematic in QFT, because then we would get pair production at arbitrarily low energies. This would manifest itself in ordinary blackbody radiation, for example, and that's not what we observe. There are various other issues, but this is the most obvious. It's one of the reasons that Penrose's CCC theory, at least in its original version, was totally non-viable.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/715865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 4
} |
How to solve for the trajectory of the center of mass? I'm working on the physics engine component of a game engine I'm building, and I need some guidance with this particular situation.
Consider a square with mass M that is free to translate in the xy plane and free to rotate about any axis perpendicular to the page (Fig. 1)
If a linear impulse J is applied at a point above the center of mass (CM) as shown below, I know there must be some angular impulse (momentary torque) generated since there is a component of J that is perpendicular to the displacement vector from CM. I imagine this angular impulse will tend to rotate the square clockwise.
However, I can also imagine that the CM will also undergo translation since the square is not constrained. How would I go about computing the overall rotational + translational motion of this system?
| If $\mathbf J(x(t),y(t))$ is the external force acting on square between times $t_0, t$, then the total impulse is $\int_{t_0}^{t}\mathbf J(x(u),y(u)))du$. So we get $\int_{t_0}^{t}\mathbf J(x(u),y(u)))du=\int_{t_0}^{t}m\frac{d\mathbf v}{du}(u)du=m\mathbf v (t)-m\mathbf v (t_0)=\mathbf p(t)-\mathbf p (t_0)$, where $m$ is the mass of the square, $\mathbf v$ the velocity and $\mathbf p$ the total linear momentum. Similarly you get that $\int_{t_0}^{t}\mathbf τ du=\mathbf L (t)-\mathbf L (t_0)$, that is the total torque is the change of angular momentum from $t_0$ to $t$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/715966",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Are two states with the same measurement probabilities necessarily equal up to unitary equivalence? Let $\rho$ and $\rho'$ be $n\times n$ density matrices, and suppose that for every observable $A$ and every $\lambda$ in the spectrum of $A$ we have
$$
\text{tr}(\rho P_{\lambda})=\text{tr}(\rho' P_{\lambda}),
$$
where $P_{\lambda}$ is the orthogonal projection onto the $\lambda$-eigenspace of $A$. Does it then necessarily follow that $\rho'=U\rho U^{\dagger}$ for some unitary $U$?
| Let $H$ denote a finite-dimensional complex Hilbert space, $\rho$ and $\rho^\prime$ be two density matrices, i.e. positive semi-definite operators with unit trace and $A$ an arbitrary hermitian operator. Its spectral representation reads
$$ A=\sum\limits_\lambda \lambda(A)\, P_\lambda (A) \quad .$$
If the equality in the question holds for all $A$, then $\rho^\prime = \rho$. To see this, multiply the equality with $\lambda(A)$ and sum over all $\lambda$. We then find
$$\mathrm{Tr}\rho A = \mathrm{Tr}\rho^\prime A \quad, $$
which we can write as
$$\mathrm{Tr} (\rho-\rho^\prime)A = 0 \quad \forall A \quad .$$
Now there are several ways to see that this implies $\rho-\rho^\prime =0$. For example, pick $A:=\rho-\rho^\prime$. Then $\mathrm{Tr} A^2 = 0$. But $A^2 \geq 0$ and thus $A=0$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/716096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
} |
Number of Independent Components of Levi-Civita Christoffel Symbol Can anybody explain why Levi-Civita Christoffel symbol in general $N$ dimensional space have $\frac{N^2(N+1)}{2}$ independent components?
I have read that in $N$-dimensional space, metric tensor has at most $\frac{N(N+1)}{2}$ independent components and for each component we can have $N$ independent christoffels and thus one arrives at the final expression. But I can't understand how these numbers are arising. Kindly elaborate or suggest any book where I can get the detalied discussion on this.
| First, let's talk about the metric tensor; it has $\frac{N(N+1)}{2}$ independent components, because it is a symmetric tensor: $g_{ab}=g_{ba}$. Writing this out in a matrix format:
\begin{align}
[g]&=
\begin{pmatrix}
g_{11}&\dots& g_{1N}\\
\vdots &\ddots&\vdots\\
g_{N1}&\dots & g_{NN}.
\end{pmatrix}
\end{align}
In this format, everything which lies on the main diagonal and the upper triangle are the independent components, because those in the strictly lower triangle are determined by symmetry. So, you just count how many such guys there are; it is easily seen to be $\frac{N(N+1)}{2}$. For example, $N$ independent entries in the first row, $N-1$ in second row, ..., $1$ in the last row. An alternate way of counting them is that there are a total of $N^2$ entries; if we ignore the $N$ of them which are on the main diagonal, that leaves us with $N^2-N$ components; but by symmetry only half are independent. So, the total number of independent components is $N+\frac{N^2-N}{2}=\frac{N(N+1)}{2}$ (which is just another way of saying that $1+2+\dots +N=\frac{N(N+1)}{2}$).
Next, the Christoffel symbols $\Gamma^{a}_{bc}$ have $3$ indices. Each index $a,b,c$ can take on any value between $1$ and $N$. So, a-priori, that's a total of $N^3$ functions. However, if you consider torsion-free connections (which is the case for the Levi-Civita connection, which comes from the metric), then we have symmetry in the lower entries, so $\Gamma^a_{bc}=\Gamma_{cb}$. That reduces the number of components down to $N\cdot\frac{N(N+1)}{2}=\frac{N^2(N+1)}{2}$ components ($N$ choices for $a$, and $\frac{N(N+1)}{2}$ for the lower slots $b,c$, taking symmetry into account).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/716251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
The value of $g$ in free fall motion on earth When we release a heavy body from a height to earth. We get the value of $g=9.8 \ ms^{-2}$. Now, I'm confused about what it means. For example, does it mean that the body's speed increases to $9.8$ every second? Or, does it mean that the speed of the body is $9.8 \ m/s$?
| It means the speed increases by $9.8$ m/s every second.
At the beginning (when you release the body) its speed is $0$.
After $1$ second the speed is $9.8$ m/s, after $2$ seconds the speed is $19.6$ m/s, and so on.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/716376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Solving 3D Kepler Problem substitution goes wrong I'm trying to arrive at the effective potential equation in Kepler Problem using Routh reduction method. We can procede in two ways, either using polar coordinates in the plane where the orbit happens or using spherical coordinates. I'm having trouble with this last one. I'm gonna follow steps taken in this Wikipedia page. Recalling,
\begin{gather*}
\mathcal{L}(r, \dot{r}, \theta, \dot{\theta}, \dot{\phi}) = \frac{1}{2}m \left(\dot{r}^2 + r^2 \dot{\theta}^2 + r^2 \sin^2 (\theta) \dot{\phi}^2\right) - V(r).\tag{1}
\end{gather*}
Because $\phi$ is cyclic, its momentum conjugate is conserved
\begin{gather*}
p_{\phi} = \frac{\partial \mathcal{L}}{\partial \dot{\phi}} = mr^2 \sin^2 (\theta) \dot{\phi} = L_z = cte.\tag{2}
\end{gather*}
Now, consider the Routhian
\begin{gather*}
\mathcal{R}(r, \dot{r}, \theta, \dot{\theta}) = \frac{1}{2} \frac{p_{\phi}^2}{mr^2\sin^2 (\theta)} - \frac{1}{2}m \left(\dot{r}^2 + r^2 \dot{\theta}^2 \right) + V(r).\tag{3}
\end{gather*}
Now consider $\theta$ Lagrange equation, which is equivalent to the conservation of the modulus of the momentum
\begin{align*}
m(2r\dot{r}\dot{\theta} + r^2\ddot{\theta}) &= \frac{p_{\phi}^2\cos(\theta)}{mr^2\sin^3(\theta)} \tag{4} \\
m^2r^4\dot{\theta}^2 + \frac{p_{\phi}^2}{\sin^2(\theta)} &= L^2 = cte.\tag{5}
\end{align*}
However, if I substitute in the Routhian, it does not work properly
\begin{gather*}
\mathcal{R}(r, \dot{r}, \dot{\theta}) = \frac{1}{2} \frac{L^2}{mr^2} - \frac{1}{2}m \left(\dot{r}^2 + 2r^2 \dot{\theta}^2 \right) + V(r).\tag{6}
\end{gather*}
| *
*The underlying reason for OP's flawed argument is, that a premature use of EOMs in the stationary action principle
$$\begin{align} S~=~-&\int\!dt ~R(r,\dot{r};\theta,\dot{\theta}), \cr
-R(r,\dot{r};\theta,\dot{\theta})~=~&\frac{1}{2}m(\dot{r}^2 +r^2\dot{\theta}^2) -\frac{p_{\phi}^2}{2mr^2\sin^2\theta} -V(r),
\end{align}\tag{A}$$
invalidates the variational principle, cf. e.g. this Phys.SE post.
*In OP's case the EOM happens to be the integrated EOM for $\theta$:
$$\begin{align}{\rm const}~\approx~\vec{L}^2~=~&(\vec{r}\times \vec{p})^2~=~r^2\vec{p}_{\perp}^2~=~m^2r^2\dot{\vec{r}}_{\perp}^2\cr
~=~&m^2r^4\left(\dot{\theta}^2+\sin^2\theta ~\dot{\phi}^2\right)~=~p_{\theta}^2 + \frac{p_{\phi}^2}{\sin^2\theta}.\end{align}\tag{B}$$
(An on-shell conserved quantity is always the result of some EOM.)
*Specifically, OP plugs EOM (B) into the Routhian (A) to obtain
$$-R(r,\dot{r};\theta,\dot{\theta})~\approx~\frac{1}{2}m(\dot{r}^2 +\color{red}{2}r^2\dot{\theta}^2) -\frac{\color{red}{\vec{L}^2}}{2mr^2} -V(r)\tag{C}$$
on-shell. However the on-shell Routhian (C) cannot be used to deduce the EOMs.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/716481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Is there a version of the Einstein field equations that uses the Riemann curvature tensor instead of the Ricci curvature tensor? I understand that the Einstein field equation uses the Ricci curvature tensor, Ricci curvature scalar, and stress-energy momentum tensor. But is there a way to form an equation that uses the Riemann curvature tensor and the stress-energy momentum tensor?
| Here is the Lichnerowicz' form of Einstein's equations:
\begin{eqnarray}
R^μ{}_{νρσ;μ}=J_{νρσ},\\
R_{μνρσ;α}+R_{μναρ;σ}+R_{μνσα;ρ}=0,\\
\end{eqnarray}
where the “current” tensor is
$$J_{νρσ}=\left(T_{νσ;ρ}-\frac12 g_{νσ}T_{;ρ}\right)-\left(T_{νρ;σ}-\frac12 g_{νρ}T_{;σ}\right) .$$
The second equation is just (differential) Bianchi identity, while the first could be obtained by combining Bianchi identity with the standard form of Einstein equations.
Compare this system with Maxwell's equations:
\begin{eqnarray}
F^μ{}_{ν;μ}=J_ν,\\
F_{μν;ρ}+F_{ρμ;ν}+F_{νρ;μ}=0.
\end{eqnarray}
Just from the formal analogy between gravitational and electromagnetic equations we could guess that free gravitational field propagates at the speed of light (at least when the linearized approximation is appplicable) and that derivatives of stess-energy tensor serve as sources for that free gravitational field.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/716623",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 2
} |
Double slit experiment: Are electrons interacting with other electrons to create a wave? Assume a double slit experiment with electrons and no observer (light source). Can the wave-like behavior and resulting interference pattern be explained by the single electron that is being shot, doesn't really travel to the detector, but interacts with other electrons in the medium (e.g. air) between source (electron gun) and target (detector), and this creates the wave? I imagine it as if the shot electron is repelled from other electrons and they again repel other electrons and so forth.
Furthermore, in case of a light source acting as an observer. Could they interact electromagnetically with all these electrons, between the source and target, in a way that the electrons are not moving freely and repel each other. But, acts a contiguous block and the shot electron hits the first electron, that transfers the energy to the second electron, then to the next until the last electron hits the detector. Similarly to Newton's cradle?
| Let me first point out that the double-slit experiment is a Gedankenexperiment, designed to illustrate quantum mechanics - not some puzzling observation requiring explanation. What happens in this experiment is a consequence of the wave-like nature of electrons when described by Schrödinger's equation - the math is nearly exactly the same as for the two-slit experiment in optics (i.e., for waves described by Maxwell equations).
Furthermore, descriptions of the experiment usually resort to language like "electrons are emitted one by one", to stress that we have created all the conditions to avoid that two or more electrons are present at the same time. One could even say that the next electron is emitted after the previous one has been detected on the screen. This is just "classical" interference... applied to electrons.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/716763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
When is it admissible to neglect the pressure term to use the Burgers' equation instead of the Navier-Stokes equation? The Burgers equation can be understood as a simplification of the Navier-Stokes equations when the pressure term is neglected:
$$
\frac{\partial u_i}{\partial t}+\ u_j\frac{\partial u_i}{\partial x_j}=-\frac{1}{\rho}\frac{\partial p}{\partial x_i}+\nu\frac{\partial^2u_i}{\partial x_j^2}
\quad\Rightarrow \quad
\frac{\partial u_i}{\partial t}+\ u_j\frac{\partial u_i}{\partial x_j}=\nu\frac{\partial^2u_i}{\partial x_j^2}
$$
Using the Burgers equation instead of the Navier Stokes equation has the advantage that there is an analytical solution for the Burgers equation. This can be obtained using the Hopf-Cole transformation, which reduces the Burgers PDE, which is nonlinear, to the heat equation, which is linear.
However, when is it possible to neglect the pressure term $\boldsymbol{\nabla}p$ in the Navier-Stokes equations? What could be a physical example of such a situation?
| Two relevant examples, the decay of an initially linearly increasing wind and a source of water spreading out in a circular symmetric pattern, are illustrated here. As for neglecting the pressure gradient, one may think it as a toy model of the barotropic NS equations where pressure is a function of density alone , $p(\rho)=\rho^\gamma$ for $\gamma\rightarrow 0$. More interestingly, Burgers' equation also provides a simple model for Kolmogorov's turbulent flow having finite, nonvanishing dissipation as viscosity vanishes and the Reynolds number goes to infinity. In the limit the equation becomes hyperbolic and energy dissipation is concentrated in shock waves.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/716953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
How can we know the maximum of a scalar field after Lorentz transforamtion? Support that we have a field $\phi(x)$. we do the Lorentz transformation for it, namely,
$$
\phi(x) \to \phi'(x)= \phi(\Lambda^{-1}x).
$$
If the field $\phi(x)$ takes the maximum at point $x=a$, where does the field $\phi'(x)$ take the maximum at point x? From the Lorentz transformation, I think that at point $x=\Lambda^{-1} a$, the field $\phi'(x)$ takes the maximum. Is it correct?
| Obviously the transformation $x^{\mu}=\Lambda^{\mu}_{\nu} x^{\nu}$ considering you are transforming four vector from one frame to another. Then the field's mode coordinates will be changed but the field itself will be invariant since proper time are connected by Lorentz transformations.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/717249",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Breaking down of 2nd law of thermodynamics Do you know a scenario where the second law of thermodynamics breaks down?
| There is a really cool example I heard of that was published a few years ago, an instance of it actually being violated in practice, not just in theory.
Very roughly, what they did was construct a thermodynamic analogue to an RLC circuit, where energy oscillates back and forth, instead of entropy uniformly increasing by monotonically increasing uniformity of the system.
The key here is that most of what you learn in statistical mechanics is predicated on slow changes of uniform systems. Kinetics and irreversible processes are much subtler. Bear in mind that most popular science is usually not going to go past the face value statement of a physical law.
https://www.media.uzh.ch/en/Press-Releases/2019/Thermodynamic-Magic.html
DOI: 10.1126/sciadv.aat9953
Edit: To add more clarity, their experiment involved two heat reservoirs connected with a Peltier device that had an inductor connecting the terminals. Heat passing through the Peltier device generates a current, the inductor keeps the current going past the point of equilibrium, causing equilibrium overshoot, so the temperature difference appears like a damped harmonic oscillator instead of monotonically going to zero.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/717412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
Why do we need Legendre transformation for thermodynamic potentials? I get the idea that thermodynamic potentials are introduced because it is not always easy to describe a system's energy as a function of variables like $S,V,N$ as we normally do with internal energy $U=U(S,V,N)$. For example, temperature is much easier to control. This is why Helmoltz free energy is introduced: $F=F(T,V,N)$ having in mind that $T=\left(\frac{\partial U}{\partial S}(S,V,N)\right)_{V,N}$. But now my question is: given that we are able to invert the latter relation and write $S=f(T,V,N)$, couldn't we get away with a function $F=F(T,V,N)$ defined simply as $F=U(f(T,V,N),V,N)$ without using the Legendre transform? Why do we even need the Legendre transform?
| Something that hopefully adds to Salvatore Manfredi D's answer: the goal isn't just to change variables, it's also to build a function that is minimized when equilibrium is reached after a specific process completes.
Start with $U$:
$$dU=T\,dS-P\,dV$$
You can read two things in this equation:
*
*$U$ is a function of $S$ and $V$.
*When equilibrum is reached after a process with $S$ and $V$ constant, $U$ is minimized.
If you build $F$ with a Legendre transformation $F=U-TS$:
$$dF=-S\,dT-P\,dV$$
Similarly, you deduce from this equation that:
*
*$F$ is a function of $T$ and $V$.
*$F$ is minimized after a process with $T$ and $V$ constant.
It's the latter that, I think, could answer your question. Not only did you build a state function with the variables you wanted, but this function has useful properties to study thermodynamic equilibrium.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/717564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Minus sign in Feynman's The reason for antiparticles I was going through Feynman's lecture on "The reason for antiparticles", which can be found here, and I got a little confused early on. His statement of Eq. 3 seems clear to me, from which Eq. 5 follows directly. But then he says that from that an Eq. 8 one can derive the relation depicted in Fig. 3. I'm missing how do you get that minus sign. If you substitute Eq. 5 on the first term of Eq. 8, you get $$1 - 2 \Re(\beta) +\sum P_{\varphi_0 \rightarrow p} =1,$$ but then the $1$'s cancel aout and you get $$ 2 \Re(\beta) =\sum P_{\varphi_0 \rightarrow p},$$ which has the opposite sign as that it the picture. What went wrong? Here I'm using $\beta$ to summarize that long integral he writes down. Is it something subtle like distinguishing between "not doing anything" $P_{\varphi_0 \rightarrow \varphi_0}$ and actually scattering twice but ending up looking as if you did nothing $P_{\varphi_0 \rightarrow p \rightarrow \varphi_0}$?
| The probability of being in some state or another at the end of the interactions is $1$. Equation (5) gives the probability of being in the state $\phi_0$ after the two interactions. Then equation (8) gives the probability of being either in state $\phi_0$ or another state, which is $p$. So equation (8) there is
$$
\text{Prob}_{\phi_0\rightarrow\phi_0}+ \int\text{Prob}_{\phi_0\rightarrow p}= 1\\
1+\int\text{Prob}_{\phi_0\rightarrow\phi_0}\text{via p}+\int\text{Prob}_{\phi_0\rightarrow p}= 1.\\
$$
So to get the mysterious equivalence, rearrange the above to remove the $1$s.
I think the thing that you missed is where it says $\langle\phi_0|\phi_0\rangle=1$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/717654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How can quantum tunneling happen conceptually? I have read in Griffiths' Quantum Mechanics that there is a phenomenon called tunneling, where a particle has some nonzero probability of passing through a potential even if $E < V(x)_{max}$.
What I don't understand about this is how to conceptualize how this can happen. I have read on Wikipedia that tunneling means that objects can "in a sense, borrow energy from their surroundings to cross the wall". How can the object "know" that across the wall there's going to be a lower energy and, thus, the borrowed energy will be restored and not depleted.
| From the viewpoint of the Schrödinger equation, the key relation is
$$\hbar k = \sqrt{2m(E - V)},$$
where $k$ is the wave number ($\psi \propto e^{ikx}$). This holds locally in a spatially varying potential $V$ under the semiclassical WKB approximation. In the classically allowed region, $E > V$ and $k$ is real, i.e., the wave function is oscillatory. As $V$ increases and approaches $E$ (classical turning point), the kinetic energy and momentum go to zero and the oscillations halt. Then when $E < V$ (tunneling region), $k$ is imaginary ($k = i\kappa$) and the wave function decays rather than oscillating ($\psi \propto e^{-\kappa x}$). This spatial decay occurs at a finite rate $\hbar\kappa = \sqrt{2m(V - E)}$, and so some of the wave function reaches the other side as long as the barrier is finite in height and width.
How can the object "know" that across the wall there's going to be a lower energy and, thus, the borrowed energy will be restored and not depleted.
It doesn't need to know. If the barrier extends for infinite distance (e.g., if $V$ continues to grow without bound), the wave function will still decay at the prescribed finite rate, and there will be some nonzero probability of finding the particle far into the classically forbidden region. This probability merely continues to rapidly decrease rather than ever "breaking free".
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/717902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 0
} |
Will quantum events ever occur on a macro-scale rather than a vacuum? Michio kaku says there's a chance we'll wake up on Mars tomorrow https://www.theatlantic.com/science/archive/2018/10/beyond-weird-decoherence-quantum-weirdness-schrodingers-cat/573448/
In this post, it is shown that quantum decoherence in the macro world occurs almost instantly, except in vacuum and at low temperatures. Therefore, in the macro world, we cannot pass through walls. Nor can you suddenly wake up on Mars the next day. But Michio Kaku says that if we wait longer than our cosmic lifetimes, we could get through a wall one day or wake up on Mars the next day. Was Michiokaku right? Wrong? Is it possible that such events are possible because there is a possibility that our body does not interact with all other oxygen or photons and thus exhibits quantum coherence, just as there is a possibility that oxygen particles in a room only gather in the opposite direction? Will quantum events ever occur on a macro-scale rather than a vacuum?
| The rate of a tunneling event in which an object of your mass (or any human's mass) moves from Earth to Mars is not zero, but so small that by the time it has a reasonable chance to occur, you, Mars, the solar system, the galaxy, all visible light and known structures in the observable Universe, and the black holes those structures eventually and inevitably collapsed into, would long since have ceased to exist.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/718042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Why magnetic field doesn't do any work on moving charge? As we know the Lorentz force $F=qv\times B$ never does work on the particle with charge $q$. But it does work on a dipole.
My question is, doesn't a moving charge behaves like a dipole? I mean an electron moving around nucleus behaves like magnetic dipole then why doesn't a moving charge behaves like a magnetic dipole?
| In a particle accelerator, the charges are deflected sideways by magnets and thus forced into a circular path. What happens in the process?
Firstly, permanent magnets could be used for this and they would not lose their magnetic field nor would it weaken. Secondly, the particles emit electromagnetic radiation in a narrow cone forwards and slightly sideways out of the ring.
Source
What happens in the process? We have particles with kinetic energy (previously accelerated by electric fields) and their magnetic dipoles are aligned in the magnetic field. It looks like the particles will respond by emitting photons. And in the process, the particles - as long as they are exposed to the external magnetic field - are forced into a circular-like path. They discontinuously lose kinetic energy (converted into photons) and slow down.
The magnet only serves as a kind of catalyst or like a spring. When the particles pass, the magnetic field of the magnet and the magnetic dipoles of the particles interact. After the passage, the magnetic field of the magnet is just as strong as before the passage.
My question is, doesn't a moving charge behaves like a dipole? A charged particle is a permanent magnetic dipole. The value of the magnetic dipole of the electron is a constant (see NIST).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/718142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Moment of Inertia Tensor and Center of Mass Can the Moment of Inertia Tensor be diagonal in a reference frame where the Center of Mass of the system is not on some of the axes?
| Given the general inertia tensor about the center of mass as
$$ {\rm I}_C = \begin{bmatrix}
I_{xx} & I_{xy} & I_{xz} \\
I_{xy} & I_{yy} & I_{yz} \\
I_{xz} & I_{yz} & I_{zz} \end{bmatrix} $$
and the parallel axis theorem in 3D given the center of mass at a position $(x,y,z)$ from the reference frame origin, the MMOI about the origin is
$$ {\rm I}_0 = {\rm I}_C + m \begin{bmatrix} y^2+z^2 & - x y & x z \\ -x y & x^2+z^2 & -y x \\ -x z & - y z & x^2+y^2 \end{bmatrix} $$
You are asking if the above can be diagonal.
So for example you are looking for cases where
$$ \begin{aligned}
I_{xy} - m\, x y & = 0 \\
I_{xz} - m\, x z & = 0 \\
I_{yz} - m\, y z & = 0 \end{aligned} $$
If the the center of mass coordinates are not on the axes, then $x \neq 0$, $y \neq 0$, and $z \neq 0$.
The solution of the above is
$$ \begin{aligned}
x & = \sqrt{ \frac{I_{x y} I_{x z} }{m\, I_{y z}}} \\
y & = \sqrt{ \frac{I_{x y} I_{y z} }{m\, I_{x z}}} \\
z & = \sqrt{ \frac{I_{x z} I_{y z} }{m\, I_{x y}}} \\
\end{aligned} $$
A second solution exists also with $(-x,-y,-z)$.
Now from the three cross terms $I_{xy}$, $I_{xz}$, $I_{yz}$ not all of them will have the same sign. The problem here, is that not all will be positive. In fact, most of the time one the cross terms is negative. This means their product, and the thing under the square root will be negative, and as a result the location of the reference frame will be a complex number (or a solution won't be available).
So you can construct a theoretical set of cross terms to diagonalize the MMOI tensor, but I think in real life this will never happen. At best, only one of the cross terms will be eliminated.
As an example, I used CAD to construct and arbitrary shape. If the above is correct, then I should be able to find a solution for $(x,y,z)$.
The values I get from CAD are
Quantity
Symbol
Value
Units
Mass
$m$
$2.4543256$
${\rm kg}$
Center of mass
$\boldsymbol{cg}$
$\pmatrix{ -25.108283 \\ -3.8535717 \\ -59.530438}$
${\rm mm}$
MMOI Cross terms
$I_{xy}$
$511.959$
${\rm kg\,mm^2}$
$I_{xz}$
$1314.109$
${\rm kg\,mm^2}$
$I_{yz}$
$-1503.307$
${\rm kg\,mm^2}$
The problem now is that all the terms under the square root are negative, and the reference location should be
$$ \boldsymbol{r} = \pmatrix{ 13.5034 {\rm i} \\15.44755 {\rm i} \\ 39.65117 {\rm i} } \; {\rm mm}$$
When calculating the MMOI from the reference point, I get only one zero non-diagonal term
$${\bf I}_r = \begin{bmatrix}
4771.346 & 1023.918 & 2628.218 \\
1023.918 & 5063.45 & 0 \\
2628.218 & 0 & 1318.564 \end{bmatrix}\;{\rm kg \,mm^2} $$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/718315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Assume a photon has a wave function, what can be said about its Hamiltonian and eigenvalues? I understand a photon does not have an associated wave function, but what if we assumed a photon does in fact have a wave function. How would this look and how could its Hamiltonian be determined?
| In classical mechanics, a particle with 0 mass is not so clear, and this extends to quantum mechanics as well.
In relativity, however, light has a rigorous framework. In relativity, $E \neq \frac{p^2}{2m}$ for a free particle. It's instead:
$$E^2 = (mc^2)^2+(pc)^2$$
and so for light, which has 0 mass:
$$E^2=(pc)^2 \implies E=pc$$
So in some sense, you can define the photon hamiltonian (in 1 dimensional case) as:
$$H = c i \hbar \frac{\partial}{\partial x}$$
So The Schrodinger Equation becomes:
$$i\hbar \frac{\partial \psi}{\partial t} =i \hbar c \frac{\partial\psi}{\partial x} \implies \frac{\partial \psi}{\partial t} = c \frac{\partial \psi}{\partial x}$$
Note that for a complete hamiltonian of a photon, you actually need to learn about relativistic quantum mechanics. This is because even though for a photon:
$$E=pc$$
momentum is still a 3-vector while energy is a number. And the Schrodinger equation would have a gradient which is a vector.
To fix this, you need to work with relativistic equations which keeps energy squared and momentum squared:
$$-\hbar^2\frac{\partial^2 \psi^\mu}{\partial t^2}=-\hbar^2 c^2\nabla^2\psi^\mu \implies \frac{1}{c^2} \frac{\partial^2 \psi^\mu}{\partial t^2}=\nabla^2 \psi^\mu$$
Which you can say is Maxwell's equation describing light. But this also doesn't give the complete picture. Because I'm trying to keep the answer in the scope of your question, I cannot answer. For a complete picture, you need QFT. For more information check this.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/718464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If water is nearly as incompressible as ground, why don't divers get injured when they plunge into it? I have read that water (or any other liquid) cannot be compressed like gases and it is nearly as elastic as solid. So why isn’t the impact of diving into water equivalent to that of diving on hard concrete?
|
So why isn't the impact of diving into water equivalent to that of diving on hard concrete?
Water is rather incompressible, but it is not very hard. It deforms rapidly under shear stress, unlike concrete. This greatly prolongs the duration of the deceleration and therefore reduces the impact.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/718786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "38",
"answer_count": 13,
"answer_id": 3
} |
Obtaining the conjugated momentum operator after coordinate transformation I am mostly uncertain if my definition of the momentum operator and the resulting commutator is correct after a simple coordinate transformation. Lets say we have in our first coordinate system the following definitions,
$$
H = -\frac{\hbar^2}{2m}\frac{d^2 }{dx^2} + V(x),\\
p_x = -i\hbar \frac{d}{dx}
$$
Now we introduce the following coordinate $y=m^{1/2}x$.
My first question is do I obtain the correct momentum operator by simply using the coordinate transformation laws? I.e. is the new momentum given by the follwing?
$$
p_y = -i\hbar \frac{dy}{dx}\frac{d}{dy}
$$
which yields in for this particular case
$$
p_y = -i\hbar m^{1/2}\frac{d}{dy}.
$$
Using this momentum operator and the new coordinate in the commutator yields
$$[y,p_y] = i\hbar m^{1/2} $$
which looks weird. Should I have defined the new momentum operator instead as $-i\hbar \frac{d}{dy}$ instead to obtain the common commutation relation?
It also occurred to me that I do not know how to obtain the conjugated momentum operator after a general coordinate transformation, while I have no trouble to do so for the classical quantities using the Lagrangian and $p_y=\frac{\partial L(x(y), \dot x(y))}{\partial \dot y}$, yet $\hat p_y = ?$
Is there a simple prescription to obtain the new momentum operator based on an given invertible coordinate transformation?
| The comments and the preliminary answer from User Feynman_00 helped me to see where I went wrong. My error was assuming that the new momentum operator is the same as old one expressed in the new coordinates.
$$\hat p_y \neq \hat p_x=-i\hbar \frac{d}{dx}=-i\hbar \frac{dy}{dx}\frac{d}{dy} = -i\hbar m^{1/2}\frac{d}{dy}=\hat p_x(y)$$
I carried the error over to the commutator by
$$
i\hbar = [x,p_x] \neq [y, \hat p_x(y)]=i\hbar m^{1/2}
$$
If the classical Lagrangian has the form
$$L_x = \frac{1}{2}m\dot x^2 - V(x)\\
L_y = \frac{1}{2}\dot y^2 - \tilde V(y)
$$
we can derive
$$
p_y = \dot y \\
p_x = m\dot x = m^{1/2} \dot y = m^{1/2}p_y
$$
Assuming that this carries over to the quantum operators, yields the proper result
$$\begin{aligned}
\hat p_y &= m^{-1/2} \hat p_x \\
&= m^{-1/2} \hat p_x(y) = -i\hbar \frac{d}{dy} \Rightarrow \\
[\hat x, \hat p_x] &= [\hat y, \hat p_y] = i\hbar
\end{aligned}$$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/718910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
How does a decelerating universe agree with this version of the Friedmann-Robertson-Walker formula? I was watching a video discussing dark matter, and he presented the following simplified version of the Friedmann-Robertson-Walker equation:
$$ \left(\frac{\dot{a}(t)}{a(t)}\right)^2 = \frac{8\pi G}{3} \rho(t) -\frac{k}{a(t)^2}$$
where $a$ is the scale factor, $\rho$ is the density, $k$ is a constant, and $G$ is the gravitational constant.
He said that if the $k$ factor is more negative than the $G$ factor, the universe expansion would have negative deceleration. However, the L.H.S term is squared so it can never be negative. So, what did he mean by negative?
| There are two Friedmann equations, and they are usually written as:
$$ \frac{\dot{a}^2 + kc^2}{a^2} = \frac{8\pi G \rho + \Lambda c^2}{3} \tag{1}$$
$$ \frac{\ddot{a}}{a} = \frac{-4\pi G}{3}\left(\rho + \frac{3p}{c^2}\right) + \frac{\Lambda c^2}{3} \tag{2} $$
The equation you cite is obtained from equation (1) by assuming the cosmological constant, $\Lambda$, is zero. With this assumption the equation simplifies to:
$$ \frac{\dot{a}^2 + kc^2}{a^2} = \frac{8\pi G \rho}{3} \tag{1a}$$
and subtracting $kc^2/a^2$ from both sides gives your equation (I assume that in your equation the $c^2$ has been subsumed into the constant $k$).
$$ \left(\frac{\dot{a}}{a}\right)^2 = \frac{8\pi G \rho}{3} - \frac{kc^2}{a^2} \tag{1b}$$
But then you say:
He said that if the $k$ factor is more negative than the $G$ factor, the universe expansion would have negative deceleration (my emphasis)
But equation (1b) gives us the velocity $\dot{a}$ not the acceleration (or deceleration) $\ddot{a}$. The value of $\dot{a}$ can be positive or negative because when we take the square root of $(\dot{a}/a)^2$ we get both positive or negative roots, but the right side cannot be negative or the square root would be imaginary.
It's the second equation that gives us the acceleration, but the acceleration is independent of $k$ and indeed if we assume $\Lambda = 0$ the acceleration is always negative anyway.
You have not linked the video so I cannot watch it myself to check, but either you have misunderstood what the video is claiming or the video is simply wrong.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/719178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
In the twin paradox or twins paradox what do the clocks of the twin and the distant star he visits show when he's at the star? In the twins paradox of relativity one twin stays on earth while the other travels to a star ten light years away, and then immediately flies back. Because his rocket travels at just under the speed of light, the entire (including the return) journey takes just under twenty years as measured by clocks on earth, but only a day, say, according to the clocks on the rocket. Furthermore, the twin on earth has physically aged by twenty years, while the traveling twin has aged only a day.
If the clocks at the distant star are synchronized with those on earth, (Einstein synchronized, if that helps, which just means that looking at across the ten light year gap, each would see ten year old light indicating a time that was ten years less than one's own clock shows, a symmetrical situation) what do they say the time is when the twin arrives there and what do the clocks on the rocket say?
Note, "paradox" might be a bit of a misnomer, as it's really just counterintuitive.
| This is an example of the relativity of simultaneity. Let's suppose the distances and speeds are such that the journey takes exactly 10 years in the Earth frame and 12 hours in the ship frame.
When the twin arrives at the star, the clock on the star will read 10 years, while the clock on the spaceship with show 12 hours.
The time dilation effect is entirely symmetrical, so in the traveller's frame, the clock on the distant star has advanced only 12 hours since the start of the traveller's journey. In the traveller's frame the time at the distant star when the traveller started their trip was minus ten years, so from the traveller's perspective the clock on the distant star was out of synch by ten years.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/719318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Bernoulli's equations on a falling (not freefall) bucket of water If a bucket of water with a tiny spout at the bottom (allowing the water to jet out) is falling (not at freefall due to a pulley system), will the air pressure above the water level in the bucket be equal to the water pressure at the spout (that is, atmospheric pressure.) If it is true, when using Bernoulli’s equation to find the velocity of the water jet at the spout, can I simply cancel out the two air pressures since they are equal, and use v_waterjet=√2ah, where a is the relative acceleration of the water, and h is the surface height of the water above the bucket.
| The air pressure above and external to the spout do not change. But, if the bucket is accelerating downwards, the water in the bucket experiences a (fictitious) upward force that counters gravity and lowers the pressure at the bottom, hence lowering the flow rate out the spout. (In free fall the pressure at the bottom would be the same as the external pressure at the top and there would be no flow out the hole.) You can account for this by using a smaller "effective" value for the acceleration of gravity: $g_{eff} = g - a$ where a is the value of the downward acceleration. Also, you need to include a coefficient of discharge to accurately estimate the mass flow rate out the spout to account for the vena contracta effect.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/719748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
On average, how often does any given hydrogen nucleus run into another hydrogen nucleus in the Sun? I think there is a misprint in an article. I will include the link if you don't mind and cut paste the sentence that I think is a misprint. Here is the link
https://www.abc.net.au/science/articles/2012/04/17/3478276.htm#:~:text=The%20power%20output%20of%20the,can%20it%20be%20so%20low%3F
...and here is the sentence in the article"
" On average, any given hydrogen atom will run into another hydrogen atom only once every five billion years."
Overall I think it is an interesting article but maybe I am not grasping what the author has intended with this sentence.
By the way I wasn't specifically looking for this but ran across it by accident while trying to find the rate that fusion occurs in the sun's core and compare that to man made fission or fusion events I haven't finished researching that and then I ran into this and became puzzled...
| If “run into” means takes part in a fusion reaction then this is correct as an order of magnitude approximation. Wikipedia says:
... each proton (on average) takes around 9 billion years to fuse with one another using the PP chain
As noted in comments, if the average lifetime of a lone proton were very much shorter than this then the Sun would have already used up most of its hydrogen. However, we know that the proportion of hydrogen in the core of the Sun is still around $40\%$ and the Sun is only about half way through its main sequence life span.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/719899",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Normal ordered exponential of one-body operators Let $\{a_i\}_{i=1}^N$ be a set of annihilation operators (they are either all bosons, or all fermions) satisfying the canonical commutation or anti-commutation relation. In the book Quantum Theory of Finite Systems by Blaizot and Ripka, Problem 1.6 claims that (summation over repeated indices is implied)
$$
\exp(a^\dagger_i M_{ij} a_j)
= N[\exp(a^\dagger_i (e^M-1)_{ij} a_j)]
\tag{1}
$$
where $M_{ij}$ is an $N \times N$ complex matrix, and $N[A]$ puts creation operators in $A$ to the left, treating all $a^\dagger, a$ in the argument as commuting or anti-commuting numbers. For example, with $\eta = +1$ for bosons, and $-1$ for fermions, we have
$$
N[a_4 a^\dagger_2 a_1 a^\dagger_3]
= \eta^{1 + 2} a^\dagger_2 a^\dagger_3 a_4 a_1
= \eta a^\dagger_2 a^\dagger_3 a_4 a_1
$$
I tried to prove eq. (1) by series expansion and comparing terms, but the expansion soon becomes rather complicated. I would appreciate it if someone can provide an elegant and clean proof.
My current attempt: For fermions the exponential function can be greatly simplified. Below I give a proof for fermions when $N = 1$, so that $M$ reduced to a complex number.
The RHS (right hand side) of eq. (1) now actually means
$$
\begin{align*}
\text{RHS}
&= N[\exp[(e^{M}-1) a^\dagger a]]
\\
&= 1 + \sum_{n=1}^\infty
\frac{(e^{M}-1)^n}{n!}
N\left[(a^\dagger a)^n\right]
\end{align*}
$$
Normal ordering gives:
$$
\begin{align*}
N\left[(a^\dagger a)^n\right]
&= N[a^\dagger a a^\dagger a \cdots a^\dagger a]
\\
&= \eta^{1 + \cdots + (n-1)}
a^{\dagger n} a^n
\\
&= \eta^{n(n-1)/2} a^{\dagger n} a^n
\end{align*}
$$
For fermions, $a^n = 0$ for $n \ge 2$, which is the key to simplify the exponential function:
$$
\begin{align*}
\text{RHS}
&= 1 + (e^M - 1) a^\dagger a
\end{align*}
$$
Meanwhile,
$$
\begin{align*}
\text{LHS}
&= \exp(M a^\dagger a)
= 1 + \sum_{n=1}^\infty
\frac{M^n}{n!} (a^\dagger a)^n
\end{align*}
$$
But with $a a^\dagger = 1 - a^\dagger a$, we notice that
$$
\begin{align*}
(a^\dagger a)^2
&= a^\dagger a a^\dagger a
= a^\dagger (1 - a^\dagger a) a
\\
&= a^\dagger a - \underbrace{
a^{\dagger 2} a^2
}_{= 0} = a^\dagger a
\end{align*}
$$
which further leads to $(a^\dagger a)^n = a^\dagger a$ for any $n \ge 1$. Therefore
$$
\begin{align*}
\text{LHS}
&= 1 + \bigg[
\sum_{n=1}^\infty \frac{M^n}{n!}
\bigg] a^\dagger a
\\
&= 1 + (e^M - 1) a^\dagger a
= \text{RHS}
\end{align*}
$$
But obviously things will be complicated for bosons, since the $a$ operator is no longer nilpotent.
| Let us prove OP's claim for a single bosonic mode:
Proposition:
$$ e^{ta^{\dagger}a}~=~:e^{(e^t-1)a^{\dagger}a}: \qquad t~\in~\mathbb{C}.\tag{A}$$
Sketched proof of eq. (A): Let's call the LHS for $U(t)$ and the RHS for $V(t)$. Both sides can be written as a function of the operator $n=a^{\dagger}a$ without the use of $a$ and $a^{\dagger}$. They satisfy the same first-order ODE:
$$U^{\prime}(t)~=~a^{\dagger}a e^{tn}~=~a^{\dagger} e^{t(n+1)}a~=~e^t a^{\dagger} U(t)a, \tag{B}$$
$$V^{\prime}(t)~=~e^t a^{\dagger} V(t)a, \tag{C}$$
with the same initial condition $U(0)={\bf 1}=V(0)$. Hence they must be equal. $\Box$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/720290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
How conservation of energy looks like in the moving frame? It is obvious that a car accelerates by converting its chemical energy in the fuel to produce kinetic energy to accelerate itself.
If the energy is lost into friction and heat, the car will slow down.
But from the point of view of the car, it always appears to be stationary, which means its kinetic energy is zero. So what is the thing that the fuel actually does in the car frame?
| Energy is not invariant across reference frames. From the car's frame you would correctly calculate the car's kinetic energy to be 0. The earth, on the other hand, would have massive kinetic energy.
The fuel does work against the pistons in the engine. The energy is lost in the same places as before.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/720630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Can a mass on a spring oscillate when it is in contact with a heat bath? I am reading statistical mechanics from Concepts in Thermal Physics, the author states the following after deriving the equipartition theorem.
A mass on a spring has energy $E$ which is given as the sum of two quadratic terms:
$$E = \frac{1}{2}mu^2 + \frac{1}{2}kx^2$$
and by Equipartition theorem we have:
$$ E = 2\cdot\frac{1}{2}k_\mathrm{B} T$$
What troubles me is the following:
How big is this energy? At room temperature, $k_\mathrm{B} \approx 0.025$ eV, which is a tiny energy. This energy isn’t going to set a $10$ kg mass on a stiff spring vibrating very much!
Isn't the energy of the system,
$$E = \sum_{i=1}^n E_i =\sum_{i=1}^{n} (K_i + P_i)$$
where $n$ is the number of particles composing the system? Are the two expressions for energy equal?
Edit
What I can't understand is what are the degrees of freedom of the system. When we have a box filled with gas particles we don't examine the macroscopic motion of the box (or at least this is not subject of statistical mechanics). Why the author considers the macroscopic velocity of the mass on the spring? Shouldn't we take only into account the (solid) particles that constitute the mass on the spring?
Also if there is vibration then some force must have initiate this motion. For example, we must first stretch the string in order to start vibrating.
| Short answer is: Mass will oscillate with an amplitude given by the following expression:
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/721512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
What is virtual photon concept in classical electrodynamics? If we observe a charged particle like an electron passing us at some high speed $u$, then as $u \to c$ the field we observe looks like a superposition of plane waves normal to the trajectory of the electron. The field can be Fourier transformed, and the modes associated with virtual photons. See for example the discussion in chapter 19 of Classical Electricity and Magnetism by Panofsky and Phillips.
Is this virtual photon we talk about in classical electrodynamics the same as virtual photon that is the the force carrier in quantum electrodynamics?
| After searching lots of sources i haven't found a direct answer. But my suggestion is yes. It may not be entirely correct. But when
$r→0$
We need to find a impact parameters which can't be zero because integral contain $1/x$ term that convert $\log(x)$ and $x=0$ is quite annoying. Therefore the lower limit definition required to use uncertainty principles.
$$XP=h$$
$$X=hc/(E_2-E_1)$$
The energy is corresponding to vertual photon we considering. What actually happens is that a vertual photon do exchange between them and colide with one anathor transfering momentum $dp$ or energy $dE$ and scattering result repulsion
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/721941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
What is the meaning of a thermal equilibrium between matter and radiation? I understand that the thermal equilibrium between two bodies means that the two bodies attain the Same temperature. Therefore,there is no flow of a thermal energy between them. However, I don't know how i should understand the meaning of it when it comes to radiation and matter. Does it mean the photon and the matter in the question attain the same temperature? But actually before even Physicists knew about photons, How did they understand the meaning of that? What should I understand from Planck when he said in his paper and i quote "the thermal equilibrium of N identical resonators in the same stationary radiation field" What is the meaning of a stationary radiation field?
| A "Stationary radiation field" means that the intensity and spectrum of the radiation is not changing with time. For the matter and radiation to also be in thermal equilibrium, requires that the radiation has reached an equilibrium state with the matter that both absorbs and emits it. For every photon that is absorbed, another is emitted (at every frequency).
When this state is reached, the radiation field and the matter can be characterized by the same temperature. For the radiation field, it is the temperature of the blackbody Planck function that wholly determines its spectrum. For the matter, it is the temperature that determines the speed distribution and energy level occupancy of its constituents.
Note that it isn't merely that the energy being absorbed by the matter is the same as the energy being received by the matter. It is that this is true in any frequency window.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/722244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
$(a - a^t)e^{\frac{1}{2} a^ta^t} | 0 \rangle = 0$ When talking about the limits of squeezed states, we reach the conclusion that $e^{\frac{1}{2} a^ta^t} | 0 \rangle$ must be an eigenstate for momentum since the uncertainty in momentum becomes zero.
As such I would expect that $(a - a^t)e^{\frac{1}{2} a^ta^t} | 0 \rangle = 0$ if it is the zero eigenstate. However, I am unable to follow through with this calculation:
$$(a - a^t)e^{\frac{1}{2} a^ta^t} | 0 \rangle = ae^{\frac{1}{2} a^ta^t} | 0 \rangle - a^te^{\frac{1}{2} a^ta^t} | 0 \rangle = [a,e^{\frac{1}{2} a^ta^t}] | 0 \rangle - a^te^{\frac{1}{2} a^ta^t} | 0 \rangle = [a,e^{\frac{1}{2} a^ta^t}] | 0 \rangle = [a,a^ta^t]e^{\frac{1}{2}a^ta^t}| 0 \rangle = 2a^te^{\frac{1}{2}a^ta^t}| 0 \rangle $$
Where does this cancel out?
| If I may suggest another approach:
Defining $\left|-1\right\rangle$ as the zero vector,$$\begin{align}(a-a^\dagger)e^{\tfrac12a^ta^t}\left|0\right\rangle&=(a-a^\dagger)\sum_{n\ge0}\frac{\sqrt{(2n)!}}{n!2^n}\left|2n\right\rangle\\&=\sum_{n\ge0}\left(\frac{\sqrt{(2n)!}}{n!2^n}\sqrt{2n}\left|2n-1\right\rangle-\frac{\sqrt{(2n+1)!}}{n!2^n}\left|2n+1\right\rangle\right),\end{align}$$which you can show telescopes to $0$, using$$\frac{\sqrt{(2n)!}}{n!2^n}\sqrt{2n}=\frac{\sqrt{(2n-1)!}}{(n-1)!2^{n-1}}$$for $n\ge1$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/722516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Producing cherenkov radiation using radioactive source I want to produce cherenkov radiation by transpering $\beta$ particles through a dialectric media. To do this , I will use a radioactive decay as a source for the $\beta$ particles, The thing that bother me is that the radiation (the visible spectrum) will be too faint. My question is
:
How can I relate between my radioactive source and the brightness of the radiation (the visible spectrum)?
| Use the Frank-Tamm formula, which describes the energy emitted ($E$) per unit frequency ($\omega$) per unit length ($x$):
$$ \frac{\partial^2E}{\partial x\,\partial \omega}=
\frac{q^2}{4\pi}\mu(\omega)\omega\Big[1-
\frac{c^2}{v^2n^2({\omega})}
\Big],$$
where $\mu(\omega)$ and $n(\omega)$ are the frequency dependent permeability and index of refraction, respectively.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/722849",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Did electric charge not exist in the early universe? I ask this question based on something Don Lincoln said in a video about leptogenesis over at the Fermilab Youtube channel. Video for reference: https://www.youtube.com/watch?v=PsqEcGMjEfo
Here is his exact quote:
... during a time in the universe very shortly after the Big Bang -
far less than a trillionth of a second after the cosmos began ... the
universe was extremely hot - much, much, hotter than we can achieve in
any conceivable particle accelerator.
...there is a ghostly field in the universe called the Higgs field
that gives mass to subatomic particles. The Higgs field depends on the
temperature, which is to say, the energy, of the universe. Above that
temperature, the Higgs field is zero, meaning that all of the familiar
particles have no mass.
...This is kind of hard to imagine, as the universe was way different
than we currently understand, with properties like mass and even
electrical charge not existing yet.
What is he referring to? Did electric charge not exist in the early universe? And does that have something to do with the Higgs field being zero?
| In the earliest part of the big bang, the electromagnetic force had not yet split out from the unification of all the forces because it was still too hot for this to happen. At that point in the process, there were fractionally-charged quarks in existence but they didn't "talk" to one another electromagnetically. Things had to cool off some more before electromagnetism could split off and electrons and positrons could exist, and start to interact electromagnetically. See Stephen Weinberg's book The First Three Minutes for an excellent and detailed account of all this.
I do not know if this history depends at all on the Higgs field.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/723045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Besides traveling at the speed of light, how can we be sure that it is possible to have energy and momentum without mass? How can we be sure that it is possible to have energy and momentum without mass? If something were to continually lose energy, would it not also lose a corresponding amount of mass? I understand that photons are predicted to have no mass because they travel at the speed of light, but is there another reason this is believed?
| You cannot separate the idea of having energy and momentum without mass from the idea of moving at the speed of light. I.e. there is no "besides". The basic equation that relates energy, momentum, and mass is: $$m^2 c^2 = E^2/c^2 -p^2$$ From this equation, in the special case that $p=0$ we get the famous $E=mc^2$. And in the special case that $m=0$ we get the usual photon relationship between momentum and energy $E=pc$. Since the velocity is always given by $$v=\frac{pc^2}{E}$$ for a massless particle we always get $v=c$. There is no way to separate $m=0$ from $v=c$, one implies the other.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/723144",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Can speed be entangled? When reading about quantum entanglement, I see that all variables that are discussed (spin for instance) have discrete states.
Can a continuous variable such as speed be entangled? In the case of annihilation for instance with two particles moving at a certain speed - by measuring the speed of one of them one knows what the other speed is.
In that case, though, I wonder how different this is from the laws of conservation.
| Yes, speed can - and almost always is, if you pick a random particle in the universe - entangled. As are most variables. The corresponding operator for speed is just the magnitude of total momentum operator divided by mass. Positions can be entangled, momenta can be entangled... all it takes for a state to be entangled is that the wave function in (for instance) momentum space cannot be factored into separate products of individual momenta:
$$\psi(p_1, p_2, ... p_n) \neq \psi(p_1) \psi(p_2) ...\psi(p_n)$$
Generally, except in special cases, there is no reason why the state should be of the form above.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/723558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Is the equal sign in "$0 ^\circ \mathrm{C} = 273.15\, \mathrm{K}$" fair? I'm in doubt whether the equal sign in an expression like "$0 ^\circ \mathrm{C} = 273.15 \,\mathrm{K}$" is fair because, normally, if $A=B$, then, say, $2A=2B$, which is hardly applicable to "$0 ^\circ \mathrm{C} = 273.15 \,\mathrm{K}$".
So is it okay to use the "$=$" if we cannot actually perform multiplication or division by the same number on both sides? Which sign would suit better here if the "$=$" does not work? Thanks a lot in advance!
| The equal sign is fair, the multiplication is not.
$^\circ C$ does not have a meaningful meaning of multiplication, for any value. One requirement for multiplication to be meaningful, you basically need a zero that has a philosophical meaning of zero (i.e., that there is no quantity of something). The Kelvin scale satisfies this criteria (at $0K$ there is no physical movement), but the Celsius scale does not ($0^\circ C$ is an arbitrary point).
Now, you can multiply changes in $^\circ C$, because changes by definition have a philosophical zero ("no change" actually is in fact zero). But, interestingly, changes in $^\circ C$ is equivalent to changes in $K$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/723684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 5,
"answer_id": 3
} |
Wave Function Collapse and the Dirac Delta Function When the wave function of a quantum system collapses, the probability of finding it at some specific point is given depends on $||\Psi||^2$:
$$
\int_{\mathbb{R}^3}{d^3 \mathbf x \; ||\Psi||^2} = 1
$$
Could this modulus square, the instant you measure, be thought as the Dirac Delta Function, because all the probability condensates to a single point, and its integral over all $\mathbb{R}^3$ gives 1.
$$
||\Psi||^2 = \delta(\mathbf x)\\
\int_{\mathbb{R}^3}{d^3 \mathbf x \; \delta(\mathbf x)} = 1
$$
If yes, what are the initial conditions the wave equation must have the instant after being collapsed.
The first one shall be this:
$$
\Psi(\mathbf x, t) \\
||\Psi(\mathbf x, 0)||^2 = \delta(\mathbf x)
$$
No? How would you plug this condition onto the Schrödinger Equation?
| I read this question as to be simple and precisely asked. Without knowing what a collapse of the wave function could be like, it understand: there is no longer a wave function when the object represented by the wave function is "measured".
But then the object exists, so it is just "there". It is all about "what actually is a wave function". It is a theoretical construct that can explain why atoms exist, why electrons can tunnel and we can produce EPROMS, and much more.
But the delta function is no function but a distribution and the only fact you know is that integral over defined range is 1.
Let there be a delta distribution. And do the integration from A to B. Will this be a timely process to integrate? Or spacial at no time? Whatever: one instant the integral value will jump from 0 to 1. But that doesn't allow you to stop the integration as you have to go to B to be sure, that the integral value will not change.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/723822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 3
} |
What causes light passing through a hole to change direction? On diagrams showing light passing through a hole, the wave of light appears to change direction when it emerges from the hole.
What causes that change of direction? Is it maybe the walls of the hole imparting a pulling force or the sudden absence of light next to the emerging beam causes the light to spread?
Or maybe light does this all the time and we only notice when we put a wall with a hole in the way.
Please explain this to like I'm a five year old.
| It might be useful to first ask a different question. Namely, why does light not scatter (change direction) when passing through a bulk medium such as glass (assuming no impurities, etc.)? One answer is the Ewald-Oseen extinction theorem which gives a rigorous mathematical account of how light propagates through matter. The atoms in a material absorb the incident light and then re-radiate it in different directions. The extinction theorem guarantees that the light radiated by the atoms will interfere with the incident beam in such a way that the light continues through the medium without changing direction. What happens, then, if a hole is carved out of that medium? The atoms on the edge of the hole are free to radiate light in all directions without this cancellation effect taking place.
One might also ask why atoms do not radiate light in all directions in the case of specular reflection (i.e. reflection off of a smooth surface). Again, the extinction theorem shines light on this issue because Snell's law and the law of reflection are derived from it.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/723976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
How can the energy-momentum tensor influence the metric outside an energy-momentum distribution? The elements in the energy-momentum tensor are determined by the mass-energy-impuls distribution as viewed from an inertial frame.
So, if you see a collection of masses with different momenta you can fill in their values in the MEI-tensor and calculate the metric by applying the Einstein field equations.
My question is hoe that gives you the values of the metric outside of the masses. Outside the masses the components are all zero, so you would expect a flat metric, which obviously is false. Is there a kind of analytical continuation going on?
Let me give an example of what I mean. Take a point in the vacuum around the Earth. Clearly the energy-momentum tensor is zero. How does one calculate the components of the metric at that point? What's different from calculating the metric of a "totally empty" vacuum? You need infinitesimal differences of the metric in the first place to plug in the equations.
| The absence of energy-momentum ma only requires the space to be Ricci flat. In other words it requires $R_{\mu\nu}\equiv {R^{\lambda}}_{\mu \lambda\nu}=0$ and not that ${R^\mu}_{\nu\sigma\tau}=0$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/724115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
How could proper distance today be infinity in a curvature only Universe when the age is finite? So for a curvature only universe, the Friedmann equation becomes
and we get the solution $a(t) = t/to$, and $to = 1/Ho$.
If we calculate the proper distance today we will get
As $z-> infinity$, the proper distance today also approaches infinity.
Typical textbook would say something like, in this universe we could see things infinitely far away.
I understand that when a galaxy emits a light, as the light travels towards us that galaxy is moving away from us, thus the proper distance to that galaxy today would be larger than the distance to that galaxy when that photon was emitted. But how could that galaxy's proper distance be infinity today? The age of this universe is $1/Ho$, finite, how could that galaxy have travelled to infinitely far from us by today?
| When $a(t) = t/t_0$ and $κ=-1/t_0^2$, the FLRW metric
$$ds^2 = dt^2 - a(t)^2 \left( \frac{dr^2}{1-κr^2} + r^2 dΩ^2 \right)$$
with the coordinate substitution
$$\begin{eqnarray} T &=& t\sqrt{1-kr^2} \\ R &=& r\,a(t) \end{eqnarray}$$
becomes
$$ds^2 = dT^2 - dR^2 - R^2 dΩ^2$$
which shows that it's just Minkowski space in odd coordinates. This shouldn't be surprising since the stress-energy tensor is identically zero.
If you fix $t$ and $Ω$, and look at the locus of points at all $r$ in the Minkowski $(T,R)$ coordinates, you'll see that these points form a hyperbola which fits entirely inside the future light cone of the origin ($T=R=0$), even though it's of infinite length. Therefore, it's possible for a test particle starting at the origin (the big bang) to reach any $r$ at a later time without exceeding the speed of light.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/724227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Why does the opposing force differ in when falling on concrete vs on water in spite of Newton's third law? If a person jumps from the first floor of a building and lands on a concrete surface, they will suffer serious injury because of Newton's third law.
If the same person jumps the same distance and lands in swimming pool filled with water, however, then there will not be any serious injury.
The person in both cases lands with same amount of force. Why doesn't water offer the same amount of force in return as concrete?
| Newtons third law says that the $F = ma$, where the acceleration $a$ is the change of velocity per time $\Delta v/ \Delta t$. At the instant you hit the ground, $\Delta v$ is very high if you are falling fast, resulting in a high force. When you fall on the water, $\Delta v$ is less, since you do not totally stop. The reason you are confused is because just before you hit the ground/water, the force is indeed the same, but exactly when you hit the surface, it is different for both cases.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/724438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 9,
"answer_id": 2
} |
If Aristoteles was right and heavier objects falled faster towards the ground how would be Newton's Laws of Motion described? It seems like it would be like:
a(m)=km
and may be a(m1,m2)=K(m1-m2)
Am I doing any sense?
Btw I'm no negationist, nor I'm trying to create a negationist movement here, I just wonder how physics would be If we lived on different physical environment in order to understand better the physics we have now.
| If acceleration due to gravity depended on mass then you have the following conceptual obstacle (first pointed out by either Galileo or Newton, I think).
Two objects each of mass $m$ will fall with a certain acceleration. If you join them together into an object of mass $2m$ then they fall with a different acceleration. But suppose they are joined by a long thin rope that is not under tension, so not applying any force to either object. What is it that changes the acceleration of the objects from the mass $m$ acceleration to the mass $2m$ acceleration ? If we cut the rope in the middle then we have two separate objects again, and their accelerations should change back to the mass $m$ acceleration. But how does cutting the long rope in the middle change the acceleration of either object ?
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/724569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How does pilot wave theory explain "identical particle" interference? Pilot wave theory says that there exist waves in 3D space which carry particles. This explains, say, the double slit experiment.
But this does not explain the behavior of identical particles. According to standard QM, a system of two identical particles will have quantum interference. But this interference does not take place in the real world 3D space, but rather in an abstract space. The wavefunction of two identical particles looks like $\psi (x_1,x_2)$. But the points $(x_1,x_2)$ live in an abstract space, as in, $(x_1,x_2)$ is not to be identified with a location in the real world 3D space. Rather, it is to be identified with a configuration of the system.
So, since the waves in pilot wave theory live in the real world space, how can it explain the wave-like interference in abstract spaces?
| The pilot wave theory is mostly a strategy for denying the implications of quantum theory. The pilot wave theory takes the wavefunction and adds particles on top of it. Pilot wave theorists then have two options. (1) They can deny that the wavefunction is real, in which case it makes no sense that the particles are influenced by it and the theory can't explain single particle interference let alone more complicated issues like EPR correlations. (2) They say the wavefunction is real in which case they still have to deal with the issue of how to understand the fact that it looks approximately like a collection of parallel universes under conditions of decoherence. So pilot wave theory doesn't solve the measurement problem and also adds structure to quantum theory so it has parallel universes and other stuff on top.
Also, contrary to one of the answers given above, pilot wave theory does make predictions different from those of quantum theory for the added particles.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/724699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
What frame of refernce to select in statistical mechanics? Suppose we have a solid particle suspended inside a fluid such as an ideal gas, as shown in the following picture:
Our system is the solid particle and the environment is the gas (which acts as a heat bath). Our frame of reference is attached on the edge of the container (shown with black color, ignore the other origin for the moment).
The energy of the solid particle, composed of $N$ atoms, in this frame of reference is (for sake of simplicity we assume we have a monoatomic solid and neglect the potential energy terms):
$$ E = \sum_{i=1}^N \frac{1}{2}mV_{i}^2 = \underbrace{\frac{1}{2}MV_{\mathrm{cm}}^2}_{\text{KE of the center of mass}} + \underbrace{\sum_{i=1}^N \frac{1}{2}mu_{i}^2}_{\text{KE with respect to center of mass}} $$
Selecting a frame of reference
What is the correct frame of reference to apply statistical mechanics? The center of mass of the system or the one attached on the edge of the container?
If it is the second, then that means that even macroscopic objects such as a rock emerged on a fluid (e.g. sea), they have on average $\frac{3}{2}k_\mathrm{B}T$ energy associated with their center of mass, which means that they move a little bit (because of the very high mass). Is that correct?
I am giving below a gif from Wikipedia which can help visualizing the process. In this gif the yellow "ball" is a dust particle.
Frame of reference and net velocity
Suppose now that we describe our system based on the origin with blue color, which happens to move relatively to the container (and also to the gas). This means that now $\langle V_\mathrm{cm} \rangle \neq 0$. This is not a proper frame of reference to apply statistical mechanics since if the relative velocity is increased, the temperature of the solid particle will increase which doesn't make sense. Is that also correct?
In summary, when we want to describe our system in statistical mechanics, what frame of reference should we use?
| $V_{cm}=0$ in the box frame. What makes you think the gas is macroscopically moving in the box?
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/724790",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Dispersion equation with variable wavenumber The wave equation
$$u_{tt}=c^2 u_{xx}$$ is known to have a simple wave solution $u(x,t)=Ae^{i(kx-\omega t)}$ where the dispersion equation is simply $c=\omega/k$. Yet, let the wavenumber be a function in $x$, then the independent variable $x$ will appear in the dispersion solution cause the first and second derivatives are functions in $x$ as the following:
$$ \dfrac{\partial{u}}{\partial x} = (ik+ixk_x) e^{i(kx-\omega t)}$$ and $$ \dfrac{\partial^2 u}{\partial x^2} = \left( (ik_x+ik_x+ixk_xx) + (ik+ixk_x)^2 \right) e^{i(kx-\omega t)} .$$
then $$c^2=\dfrac{-\omega^2}{(2ik_x+ixk_xx)+(ik+ixk_x)^2}$$
Did anyone encounter an independent variable as $x$ explicitly in the dispersion relationship as in the terms $ixk_x x$ and $ixk_x$?
| From a mathematical point of view your idea does not make sense. Remember that plane waves do not exist, they are just a tool that physicists introduce to build up a mental representation. Ask a mathematician; he will tell you that to get a dispersion relation, you just need to take the Fourier transform of your wave equation. But (x,k) is the pair of variables involved in the Fourier transform. They can only appear together in: $ e^{i(k.x-\omega t)}$. with k being the Fourier variable independent of x.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/724931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Current loop and the associated magnetic field under mirror reflection Let a circular current loop in the $xy$ plane carries a current in the counterclockwise sense. Therefore, it produces a magnetic field in the $+z$ direction (think of it as an arrow parallel to the $+z$-axis). Now consider reflecting this by placing a mirror in the $yz$ plane. Of course, in the image, the current will flow in the clockwise sense. But what would be the direction of the magnetic field in the mirror image?
Case I Should it remain unchanged i.e. continue to point along the $+z$-axis? If so, is it not a problem that the image does not correspond to a real physical situation (because in the real world, a clockwise current will produce a magnetic field in the $-z$-direction)? If this is really the case, this would imply that parity is violated (which is not true for electromagnetism).
Case II On the other hand, if you think that in the image the magnetic field points in the $-z$-direction, how will you explain that? Because if we think that the magnetic field is represented by an arrow (I mean, a real solid arrow made of metal or something) along $+z$, that arrow cannot be flipped by reflecting it in a mirror in the $yz$ plane. You are not allowed to use the laws of physics of the real world in the reflected world. You have to first know what you will see in the mirror and then decide whether that mirror image corresponds to the real-world situation.
| Some physical quantities that most of us treat as vectors, they are indeed pseudovectors, whose nature can be appreciated when you do reflections as a transformation of coordinates.
Examples of pseudovectors are angular velocity $\omega$, force moment and torque $M$, angular momentum $\Gamma$, and the magnetic field $b$.
As a rule of the thumb, all these vectors appears in formulas relating two vectors and the pseudovectors through a vector product (or a curl operator)
*
*$v = \omega \times r$ in the rigid body rotation;
*$\Gamma = r \times Q$ for the angular momentum of a point particle;
*$M = r \times F$ for the force moment of a point particle;
*$F = q v \times b$ for Lorenz force in absence of electric field;
where position $r$, velocity $v$, force $F$, linear momentum $Q$ are vectors.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/725095",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Einsteins gravity theory, but for static objects https://www.youtube.com/watch?v=XRr1kaXKBsU this is a good video to explain Einsteins gravity.
The video claims that objects always move in a straight line, with an absence of a force of gravity, rather that space is curved, so that the straight line becomes curved.
It demonstrates it well, but there is one catch. If you were to place an object with 0 velocity, that means that it isn't moving, so therefore, it won't be moving through space, which means that, no matter how close that 0 velocity object is to another object, it won't budge.
However, if the aforementioned scenario were performed, the object would start moving, therefore demonstrating the existence of a force. That violates Einstein's theory.
Is there an explanation for this?
| The geodesic equation of general relativity reduces to Newton's Law of Gravity, to a very high approximation, in such a case. Therefore, there is no contradiction. In the language of GR, the object is moving uniformly along a geodesic in spactime and, in the language of Newton, it is accelerating in space. It is simply two different ways of saying the same thing. The only way to really understand this is to follow the math and understand the equations for yourself. But I can't teach GR in real life or in a comment so I'd refer you to Exploring Black Holes by Wheeler.
https://www.eftaylor.com/exploringblackholes/
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/725247",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
What is the current going into a resistor-capacitor parallel circuit and the current coming out of it? I'm looking at the circuit below and know that I1 = I2. Can someone explain why those 2 currents are the same?
| It might appear that because there is a charge storing device, the capacitor, the two current can be different and a difference in the currents results in the charge stored in the capacitor to change.
However the fact is that the net charge on the capacitor is zero in that a change in positive charge on one plate is always equal in magnitude to the change in negative charge on the other plate.
Since there is no net change in the charge stored in the capacitor the two currents must be equal which, of course, you would expect by applying Kirchhoff’s current law to the circuit.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/725405",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Diagonalising the Hamilton operator, why does this magic work? Let the Hamilton operator $H= \omega_1 a_1^\dagger a_1 + \omega_2 a_2^\dagger a_2 + \frac{J}{2} (a_1^\dagger a_2 + a_1 a_2^\dagger)$ be given, of course $a_j$ and $a_j^\dagger$ are the creation and annihilation operators, respectively. This operator can be rewritten as
$$
H=
\begin{pmatrix} a_1^\dagger & a_2^\dagger \end{pmatrix}
\begin{pmatrix}
\omega_1 & J/2\\ J/2& \omega_2
\end{pmatrix}
\begin{pmatrix} a_1 & a_2 \end{pmatrix}
$$
The Eigenvalues of the matrix in the middle, which may be rewritten as
$$
\begin{pmatrix}
\omega_0 - \frac{\Delta}{2\omega_0} & J/2\\ J/2& \omega_0 + \frac{\Delta}{2\omega_0}
\end{pmatrix}
$$
where $\omega_0 = \frac{\omega_1 + \omega_2}{2}$ and $\Delta = \omega_2 - \omega_1$, are
$$
\lambda_{\pm} = \omega_0 \pm \frac{1}{2}\sqrt{J^2 + \Delta^2}.
$$
Question: Could these be the Eigenvalues of that thing $H$ that maps vectors from an infinite dimensional space to others ? Why is that ? How could the diagonalisation of the little $2\times2$-matrix have anything to do with diagonalising $H$, whose true matrix is infinite dimensional ?
Please mention the subject that rigorously explains this and some references about that.
| We are dealing here essentially with a one-particle Hamiltonian - more precisely, a Hamiltonian for many non-interacting particles. The Hamiltonian in question is simply the second quantized version of
$$
H=\omega_1 |1\rangle\langle 1|+ \omega_2 |2\rangle\langle 2| +
\frac{J}{2}\left(|2\rangle\langle 1| + |1\rangle\langle 2|\right),
$$
where the second quantized version is obtained via usual prescription
$$
\hat{\Psi}=a_1|1\rangle + a_2|2\rangle,\\
\hat{\Psi}^\dagger=a_1^\dagger\langle 1| + a_2^\dagger\langle 2|,\\
\mathcal{H}=\Psi^\dagger H\Psi
$$
Thus, we could have diagonalized the original single particle Hamiltonian (which is really a 2-by-2 matrix) and then introduce the second quantization OR we can first perform the second quantization and then diagonalize the Hamiltonian in "the space of creation and annihilation operators" as is done in the OP.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/725688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Andy Weir's Project Hail Mary - Does the travel time to Tau Ceti make sense? I'm an English teacher with a modest science background teaching Science Through Science Fiction this fall, and I need some help with the physics in Andy Weir's Project Hail Mary. Here's one question I have. (Light spoilers for the book.)
The ship is going to Tau Ceti (11.9 light years from Earth). Astrophage (their propulsion system) can achieve avelocity of 0.92c. Weir tells us the time that will pass from Earth's frame of reference during the ship's journey is about 13 years.
I don't know the equations, but my intuition says that would make sense for cruising at 0.92c the whole way, but does it account for acceleration and deceleration? The ship would have to accelerate and decelerate with low enough $g$'s not to kill the human crew, which I believe is at least less than 5 $g$'s (borrowing from Kim Stanley Robinson's Aurora). It seems to me this would significantly extend the travel time. But I haven't seen any science critique call this out, so maybe I'm just that hazy on how the math actually works out.
Does Weir's stated time frame make sense or is it not fully accounting for acceleration?
Edit: Catiffiny reminds me that the book states acceleration is 1.5 g. I'm not sure it states this is constant for the whole trip (including when the crew is asleep), but it does at least imply constancy by not mentioning changes anywhere else.
| The book clearly states
1.5g acceleration outbound from Earth, &
1.5g deceleration approaching Tau Ceti.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/725874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What is the matrix representation of a function of an quantum operator? If we know the matrix representation of a quantum operator, say $J$, will the matrix representation of any function of the operator i.e$f(J)$, same as acting the function on the matrix of the operator?
| It's not clear what you mean by "the function acting on the matrix of the operator."
If you mean
$$f\left(\begin{pmatrix}a & b\\ c & d\end{pmatrix}\right) = \begin{pmatrix}f(a) & f(b) \\ f(c) & f(d)\end{pmatrix}$$
then the answer is no, that is not what it usually means in physics.
The typical meaning is this: if a function has a power-series expansion $f(x) = \sum_n c_n x^n$ then you can define its operation on a matrix as $f(J) = \sum_n c_n J^n$ the power of $J$ are obtained from repeated matrix multiplication. Whether this is really well-defined for every matrix depends on the function.
If you are only interested in matrices that are diagonalizable like $J = PDP^{-1}$ then you can also define $f(J) = P\begin{pmatrix}f(d_1)&&\\&f(d_2)&\\& & \ddots\end{pmatrix}P^{-1}$. As long as $f(d)$ is defined for each of the diagonal entries this is well-defined and has the same result as the previous definition.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/726112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Do Neutrons Have a Charge Radius? The radius of a proton is described as a "charge radius", about 0.84 fm. The neutron is about the same size, 0.8 fm, but has no measureable charge. Is this a contradiction? Are the two radii measured in the same way?
| Netrons have no net charge the same way a neutral hydrogen atom has no net charge. I don't know what the electric charge profile of a neutron looks like, but it's going to be rather complicated. See the parton distribution for hadrons - it describes how much each particle type contributes at each length scale.
At a guess, though, the neutron will have more positive charge at the center and more negative toward its outside. This is because it's more electromagnetically stable when you have one negative valence particle (the up quark) and two negative ones (the down quarks). I would expect it to be similar for protons - more positive charge toward the outside, less toward the center (possibly even going negative at the very center). This is all wild speculation, though, so take it with a fist sized lump of salt.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/726318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
„Orbital“ of an quark Inspired by the idea of the electron orbitals ( probability of finding an electron in an atom) i was wondering what that would look like inside a proton or neutron for quarks. For simplicity consider a meson where we know the location of one quark. What is the orbital - or probability of finding the antiquark around this quark? Sure it must be in the range of about 1 fm, since this is the range of the strong force. Is it correct to postulate therefore that the antiquark orbital is a sphere of 1 fm in size? Next, we know that when quarks/antiquarks are very near together, the strong force gets weaker and diminishes. Can this possibly mean, that the probability of finding the antiquark in our meson goes here to zero? As result i would conclude that the orbtial of an antiquark around a localized quark in a meson look like this: The possibility of finding the antiquark is highest at a distance of 1 fm and then goes to zero as we approach the quark. Also, the probability of finding the antiquark outsite this 1 fm sphere is also zero. Of course things get more complicated when we treat both quark and antiquark in their orbital form so thats why i came up with that example.
| Unlike atomic electrons, the spin-orbit coupling in the strong interation is always non-negligible. If you know the spin and parity $J^P$ for a meson, you can figure out the allowed values for $L$. But in general, even a wavefunction which consists only of valence quarks will allow multiple values of $L$ to end up with an observed $J^P$. For a low-energy example, consider the two-nucleon deuteron, for which the spin $S=1$ is well-defined but whose orbital wavefunction contains a non-trivial mixture of s-wave and d-wave states.
In higher-energy hadron interactions, you must also consider the contributions of the gluons and the sea quarks, and you end up reinventing the parton distribution functions.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/726448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Which direction will the beam of electrons be deflected?
For D.C. sources like the one shown, do we apply Fleming's right hand rule or left hand rule? I derived my answer using Fleming's left hand rule in this case. If I am not mistaken, we use Fleming's right hand rule for A.C. motors only. I got C but the answer is B.
| D.C. source connected to solenoid will produce constant current, generating constant magnetic field (directed up in the solenoid and down outside the solenoid). The right direction of electron velocity equals left orientated current of positively charged particles, so if the current goes to the left anf field points down, the force will act to the page by the rule of vector product (according to Lorents law). You have to point your right hand fingers in to the current direction and make the field has to come in your palm, so the thumb will be looking into the page.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/726917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What work does a microwave oven do? I learned that when energy is transfered it either produces work or it becomes thermal energy (heat).
Work implies a force that acts on an object producing changes in its position.
I'm learning these concepts from Khan Academy, and in this article they say:
A hot cup of coffee has a measurable amount of thermal energy which it
acquired via work done by a microwave oven, which in turn took
electrical energy from the electrical grid.
What work did the microwave oven if all the energy transferred became heat?
What is heat actually? What is so special about it?
Does heat imply movement of atomic particles and hence the microwave oven sucessfully did work on them?
Thanks for your patience.
|
Work implies a force that acts on an object producing changes in its position.
Not necessarily. It is possible to do work on an object and only change its internal energy, not its position. Heating coffee in a microwave is one example; heating a wire by passing a current through it is another; compressing a spring is a third example.
Heat is simply the internal energy that a macroscopic object possesses by virtue of the microscopic motions of its constituent atoms or molecules.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/727193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Lagrangian Mechanics: semi-holonomic constraints By switching to a different set of coordinates, can you make problem with semi holonomic constraints into a problem with holonomic constraints? If so, then when can you do this? I wold like to know if this is possible for all semi-holonomic problems, some semi-holonomic problems or no semi-holonomic problems.
My intuition is that it probably works for some specific cases, but not in general. However, I don't know the reason why. Would be very nice with some sort of proof.
(Constraints on the form: $f=f(q_i,\dot q_i,t)$.)
| *
*A non-holonomic$^1$ constraint is by definition a constraint that is not holonomic,
e.g. on the form $f(q,\dot{q},t)=0$ or an inequality.
*A semi-holonomic constraint
$$ \omega~\equiv~\sum_{j=1}^na_j(q,t)~\mathrm{d}q^j+a_0(q,t)\mathrm{d}t~=~0 $$
is equivalent to a holonomic constraint iff there exist an integrating factor $\lambda(q,t)\neq 0$ and a one-form $\eta$ such that
$$ \lambda\omega+ f\eta~\equiv~\mathrm{d}f ,$$
cf. my related Phys.SE answer here.
--
$^1$ If you are using the 3rd edition of Goldstein, be aware of erratum.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/727314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Synthetic aperture for visible light camera It seems to me a camera array for visible light could be constructed to synthesize a large effective aperture to achieve high angular resolution just as a synthetic aperture radar does. This could be used for example on a satellite with several small cameras mounted on extended poles. However, I was told the present engineering cannot accomplish this. Is this true and if so, why?
| Short answer: Yes... and also no...
Longer answer as to why "yes" and how to turn it into a "no" below:
For wavelengths of visible light our sensors can only collect intensity (amplitude) but the phase information is lost. So just attching several cameras spread out is not an option with current technology.
But there is a way to keep the phase information: we just don't convert to electrical signals before we do the interference. There is a name for that: Interferometry .
We're doing it on earth already for example with ESAs VLT which consists of 4 telescopes with 8.2m mirrors and 4 movable 1.8m mirrors. They have a incredibly precise system of mirrors to combine the images from the 8 telescopes optically to form basically a single large telescope.
In the foreground you can see the tracks on which the auxilary telescopes can move. The auxilary telescopes are the small pods with the round heads in front to the big ones (Image credit ESO/Y.Beletsky )
If we're picky, every telescope with a segmented mirror (like JWST but also many earth bound large telescopes) is some kind of interferometer. But that's another discussion.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/727474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Property of reciprocal lattice I came across the following property of the reciprocal lattices.
Let be $\Lambda$ a Bravais lattice and $\Lambda^*$ its reciprocal lattice; let be $\vec{G} \in \Lambda^*$ and $\vec{G}_0$ the shortest vector in $ \Lambda^*$ that is parallel to $\vec{G}$. Then, the relation between the two vectors is: $\vec{G}=n\vec{G}_0$, where n is integer.
I looked for a proof, but I noticed that for example on Ashcroft-Mermin (chapter 5) the proof is left as a problem to the reader. The hint is: suppose the contrary is true (this means "suppose n is not integer", I think) and show that there is a vector parallel to $\vec{G}$ that is shorter than $\vec{G}_0$. Despite the hint, I'm not able to prove this statement. Does anyone please know this proof?
| Suppose $n$ is not an integer. Denote by $[n]$ the integral part of $n$, and $\{n\}=n-[n]$, so $0<\{n\}<1$. Consider $\vec{G}-[n]\vec{G}_0=\{n\}\vec{G}_0$, which is still parallel to $\vec{G}$, and belongs to $\Lambda^*$. But the length of $\{n\}\vec{G}_0$ is strictly smaller than $\vec{G}_0$, so a contradiction.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/727633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Why only harmonics allowed in the Casimir effect? My question is really a request for an intuitive explanation as to why only harmonic frequencies of photons allowed between two conduction plates.
Why do the plates have to be conductive?
And can real photons with unmatching frequencies exist between the two plates in the casimir experiment?
| Plates do not have to be conducting. Various dielectric materials can be used, and the Casimir force on the plates will depend on the characteristics of those materials such as dielectric constants.
But perfectly conducting plates are the simplest toy model, because perfect conductor means electric field is zero in and at the plates.
For two infinitely long perfectly conducting plates separated by distance $a$, relevant frequencies used in the Fourier expansion have to obey the relation (due to Maxwell equations and the mentioned boundary condition)
$$
\frac{\omega_n^2}{c^2} = \frac{n^2\pi^2}{a^2} + k_x^2 + k_y^2.
$$
So we can see allowed frequencies are harmonics only in 1D world, where there is no $k_x,k_y$. In 3D space, any frequency is allowed, due to the fact $k_x,k_y$ can have any real value.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/727803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
What is the difference between the state $| \psi \rangle$ of quantum mechanics and the microscopic state $St(q,p)$ of statistical mechanics? In terms of physical quantities, the $|\psi \rangle$ of quantum mechanics and the microstate $St(q,p)$ of statistical mechanics are both a vector, and the microstate of statistical mechanics can be considered as a vector with a size of $3N$, while $| \psi\rangle$ is also a vector, although it's size is unknown.
When the correspondence between the number of microstates and the volume of the phase space is proved by one-dimensional harmonic oscillators to be $\Omega = \dfrac{\Gamma}{\hbar}$ relational proof, the number of energy eigenvalues is regarded as the number of microstates, then whether the number of $\{St\}$ is the same as the number of $\{|\psi\rangle\}$?
$N$ is the number of particles. $\Omega$ is the number of microstates. $\Gamma$ is the volume of accessible area of phase space.
|
What is the difference between the state $| \psi \rangle$ of quantum mechanics and the microscopic state $St(q,p)$ of statistical mechanics?
Apples and oranges.
The $| \psi \rangle$ of quantum mechanics and the $St(q,p)$ of statistical mechanics use mathematics, vectors , tensors and even complicated functions of special relativity four vectors, but they describe observations through a different mathematical process.
In physics theories, extra axioms are imposed so as to pick up those mathematical solutions which will have correct units and obey experimentally observed laws. These are called laws in classical mechanics and electrodynamics, and postulates and principles in quantum mechanics; the objective is to have a theory for physical observations that fits them and, important , is predictive.
It is true that statistical mechanics relies a lot on probability, but the probability postulate of quantum mechanics describes a probable location for a particle that is biased by the wavefunction, a solution of the appropriate quantum differential equation. The distribution is not statistical, also there is no "number of $| \psi \rangle$, $ψ$ is a continuous mathematically function on which creation and annihilation operators act in order for a particle to appear at (x,y,z,t).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/727961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
If reference frames are equally valid, then why do teachers say the geocentric view is wrong? If all reference frames are valid, then why is the geocentric model taught as "wrong" in schools?
I've checked many websites but none of them clear the issue. Wiki says that in relativity, any object could be regarded as the centre with equal validity. Other websites and answers make a point on the utility of the heliocentric model (simplicity, Occam's razor...) but just because something is not so easy to deal with doesn't mean it is wrong.
Note: I am not asking for evidence that geocentrism is wrong; I am asking for a way to resolve the contradiction (from what I see) between relativity and this "geocentricism is wrong" idea.
| "Geocentric model" is a loaded term. It's not just referring to the the sun rotating around the Earth. If you are only concerned with whether the sun orbits the Earth or whether the Earth orbits the sun it's a wash. It is referring to the model as whole which must explain all the observed phenomena seen. This includes things such as why the retrograde motion of planets where they changes direction in the sky, or the phases of Venus.
Despite all the additional complexity of concentric circles and such that were introduced to the geocentric model it was never able to adequately explain some of these, whereas the heliocentric model did, and with much less complexity.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/728129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 10,
"answer_id": 2
} |
I'm having trouble understanding exactly what $δ$ represents in thermodynamics I know that $δ$ sometimes represents the Dirac delta function but in my book it states "Suppose that equilibrium has been established Then a slight change in the position of the piston should not change the free energy since it is at a minimum that is $δA=0$" but in terms of this what exactly does it mean?
| That $\delta$ is not a Dirac delta, but the symbol for indicating a differential form that is an inexact differential, i.e. a differential whose integration depends on the path of integration,
$\displaystyle \int_{\ell^1_{A\rightarrow B}} \delta f \ne \displaystyle \int_{\ell^2_{A\rightarrow B}} \delta f \qquad$ in general, for different integration paths $\ell^i_{A\rightarrow B}$ with the same extreme points $A$, $B$
and not only on the values of a function (its primitive) at the extreme points of the integration path, like exact differentials
$\displaystyle \int_{\ell_{A\rightarrow B}} d f = f(B) - f(A) \qquad $ for every integration path with extreme points $A$ and $B$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/728326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Relativistic invariants of a classical field in 4D fashion: why the relation between the components of the current density holds? I'm trying to understand how is justified the following relation between the first component of the current density integrated over the volume and the scalar product of the 4-vector current density and a hypersurface orthogonal to the $x^0$ axis
$$\int j^0 dV=\int j^\rho dS_\rho$$
Is introduced in "The Classical Theory of Fields: Volume 2 - Landau"(§14-§29-§32),
where it's written that is possible to write the total charge $\int \rho dV=\int j^0 dV$ in a four dimensional form $\int j^\rho dS_\rho$, where the integral is taken over the entire four-dimensional hyperlane perpendicular to the $x^0$ axis and this mens that the integration is performed over the whole 3D space and that in general the last integral over an arbitrary surface is the sum of the charges whose world lines pass through this surface.
The point that i'm still not getting is how the first component is related to the other 4 in this integral relation. I also precise that i encountered this problem evaluating the Noether's charge under Lorentz transformation of a field and finding that the angular momentum $M^{\mu\nu}$ (the Noether's conserved charge)can be expressed in a Lorentz-covariant form: $$M^{\mu\nu}=\int dV (x^\mu T^{0\nu}-x^\nu T^{0\mu})=\int dS_\rho (x^\mu T^{\rho\nu}-x^\nu T^{\rho\mu})$$ In these last lines, mathematically, the problem is the same becuse i'm not getting how the zero-$\nu$ component integrated over the volume is equal to all the 4 component multiplied for these 4 hypersurfaces.
| The definition of $dS^i$ (Landau §6) is that it is a four-vector equal in magnitude and normal to the hypersurface element; in other words, $dS^i$ is the projection of the hypersurface element, respectively, onto the hyperplanes $x^0=$ const, $x^1=$ const, $x^2=$ const and $x^3=$ const.
By definition, the integral
$\int_{x^0=\text{const.}} j^i dS_i$
is carried out for the hypersurface $x^0=$ const whose projections onto other three components are null, i.e., the only contribution to the above integral is
$\int_{x^0=\text{const.}} j^0 dS_0.$
The integral
$\int j^i dS_i$
over an arbitrary hypersurface is, of course, cannot be written only as integrations over $dS_0$; in this case, the integral describes the sum of all charges whose world lines pass through this hypersurface (Landau §28).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/728469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solution of Einstein's field equation for local energy density If there is a uniform positive energy density present in a patch of space-time, what would be the metric describing the patch? a de-Sitter patch? What would be the gravitational potential felt by a mass in that region?
| Under the approximation of spherical symmetry, the FLRW metric describes any homogeneous patch. Note that in general such a patch would not remain static for the same reason that the universe should not be static. The Friedmann equations precisely describe how the patch would evolve.
The gravitational potential is a Newtonian idea, but we can generalize it to this picture pretty easily. Suppose the patch has energy density $\rho(t)$ and pressure $p(t)$. For a spherical region of radius $r$ inside our patch, the second Friedmann equation says that
$$
\frac{d^2r}{dt^2} = -\frac{4\pi G}{3}[\rho(t)+3p(t)]r.
$$
That must also be the acceleration of a test mass at radius $r$ with respect to the (arbitrary) origin. But this is just a harmonic oscillator with potential
$$
\Phi(r) = \frac{2\pi G}{3}[\rho(t)+3p(t)] r^2.
$$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/728608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Why does particle leave circular motion after string slacks? If a particle is attached to a string and made to move in a vertical circle with initial velocity of $\sqrt{4gl}$ $m/s$ where l is the length of string, at some angle (approx $131°$ with the initial position), the string slacks and the particle leaves the circular path and undergoes projectile motion. Why does this phenomenon occur even though the component of weight can provide centripetal acceleration?
If we throw the particle at $\sqrt{2gl}$ $m/s$ , the velocity and tension both become 0 when the string is in horizontal direction. How can we know whether the particle will now oscillate or leave the circular path?
| Following is not a full answer but a comment to help OP figure out the answer on their own.
Gravity does not provide centripetal acceleration to the particle. This is because centripetal acceleration always points towards a definite point while gravity always points towards a fixed direction.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/728852",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Does a solar panel encased in glass give off less power? The speed of light in glass is about 2/3 c. If we encase a solar panel in glass, does it give off less power because of this reduced speed of light? (and ignoring other things like the reflectivity of glass, etc.)
| No, there would be no change in the amount of generated power.
The situation you describe would not change the rate that photons are arriving at the surface of the solar cell. Assuming we aren't worried about angular dependence or changes in reflection, photons would be converted into usable power at the same rate.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/729017",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Is the normal force the reaction force? There is a box on the surface of earth. The earth exerts a force to the box (black arrow). The box exerts a reaction force to the earth(brown arrow). But this reaction force is exerted to the earth not the box, so where is the normal force? If the reaction is the normal force then why they put the arrow starting in the box upwards and not starting in earth? (second diagram).
| There are two different third-law force pairs. One pair is the downward gravitational force on the box and the upward gravitational force on the earth. The other is the upward normal force on the box and the downward normal force on the earth.
In equilibrium, the gravitational and normal force are equal and opposite to each other, but that's due to the first law (no net force in equilibrium), not the third law.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/729179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Does the quantum mechanical wave function work as a charge distribution? The quantum mechanical wave function is traditionally interpreted as a probability distribution of the particles position. It is possible to interact with an electron in a conductor, without collapsing the wave function, which can be distributed over the entire conductor. Can the wave function be understood as a charge distribution in this case? And if so, wouldn’t that mean that the particle actually takes the shape of the wave function?
| Schrödinger initially believed that the wave function is related to actual charge distribution density, rather than to probability density. Some objections were raised to such interpretation. One of the objections is based on wave packet dispersion. I offered some modification of the interpretation (Entropy 2022, 24(2), 261), where it is assumed that the wave function is related to coarse-grained charge density, and one-particle wave functions are modeled as plasma-like collections of a large number of particles and antiparticles. This modification seems to be immune to the problem of wave packet dispersion.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/729300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Can a Matrix Have EigenBras? I only need a yes or no, but I cannot find anything online. I know a matrix $A$ can have eigenkets found by using.
$$A\psi=\lambda\psi.$$
However, I was wondering if my matrix $A$ was Hermitian, could I just apply the Hermitian conjugate to the equation and find eigenbras for the system too?
| Look for left eigenvectors in linear algebra,
$v^H A = \lambda v^H$.
If the matrix is Hermitian, left and right eigenvectors are the same (just evaluate the conjugate transpose of the expression above to get $A^H v = \lambda^* v$; then you can prove that $\lambda$ is real as well).
Then ket goes to Right-eigenvectors, bra goes to Left-eigenvectors.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/729611",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Two solutions for a collision which satisfy both conservation of momentum and energy. Which is correct? For the following problem, there are two sets of solution which appear to be equally valid:
v_car = 15.7 m/s, v_truck = 12.9 m/s and v_car = 12.3 m/s, v_truck = 15.2 m/s. How can I know which set of solutions is correct in such cases?
"A 1000-kg car traveling with an x component of velocity of +20 m/s collides head-on with a 1500-kg light truck traveling with an x component of velocity of +10 m/s.If 10% of the system's kinetic energy is converted to internal energy during the collision, what are the magnitudes of the final speeds of the car and truck?"
| To approach a more general form of this problem, let's skip the first step of your problem and just suppose you have already calculated:
Total momentum
$P' = m_1 v'_1 + m_2 v'_2$
Total energy
$E' = (1/2) m_1 v_1^{'2} + (1/2) m_2 v_2^{'2}$
Let $f = m_1/m_2$, and let's do our work in the center of mass frame where P=0. And instead of conserving energy E, I'll use a convenient constant C which is proportional to the center-of-mass energy, to save some factors of 2 and $m_2$. For ease, the velocities $v_1$ and $v_2$ represent center-of-mass velocities.
Momentum:
$f v_1 + v_2 = 0$, or $v_2 = -f v_1$
Energy:
$A = f v_1^2 + v_2^2 = f v_1^2 + f^2 v_1^2$
Now $v_1 = \pm \sqrt{\frac{A}{f+f^{2}}}$.
In the center-of-mass frame, it's obvious which root to choose based on the principle that @knzhou described in his comment, which is a little more general than it sounds! You choose the opposite sign of the initial velocity in the center-of-mass frame.
And then of course to solve your actual problem, you add back the center of mass velocity.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/729822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Why do we observe particles, not quantum fields? My understanding is that, in the context of quantum field theory, particles arise as a computational tool. We perform an expansion in the path integral in some parameter. The terms in these expansions correspond to Feynman diagrams which can be interpreted as interactions between particles.
My question: if particles are just a computational tool, why does it seem like particles really exist, and not quantum fields. For example, it seems like chemical reactions really do occur through the exchange of discrete particles called electrons. Are states with definite particle number somehow more probable, more easily observed, or some kind of classical limit?
| The theorists have mathematical rationalizations, but the reality (as Bohr pointed out) is that if your experiment senses fields you observe fields, while if it senses particles, you observe particles. Every radio detects the electromagnetic field, not photons. Waves in any field may be observed to diffract so long as their wavelength is accessible to a diffraction structure. On the other hand, if your experiment tracks particles, you'll observe particles.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/729950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "35",
"answer_count": 5,
"answer_id": 4
} |
Torus shaped event horizon Is there a solution to GR field equations for a rotating black hole that has a torus shaped event horizon? If so, when a craft flies through the torus, it can pass through a ring singularity while remaining outside of the event horizon. Would it enter a strange region of repulsive gravity and closed timelike curves? How far from the black hole would this region extend?
| There is a result in four-dimensional General Relativity that forbids such situations. Namely, the Hawking's Theorem on the Topology of Black Holes. It is given, e.g., on the classic book by Hawking and Ellis, The Large Scale Structure of Space-Time, which I quote:
Proposition 9.3.2
Each connected component in $J^+(\mathscr{I}^{-},\overline{\mathscr{M}})$ of the horizon $\partial \mathscr{B}(\tau)$ in a stationary regular predictable space is homeomorphic to a two-sphere.
Hence, at least under this proposition's conditions, you can't have a toroidal black hole. Notice that we are assuming the black hole is stationary, for example, and, implicitly, that we are working with four-dimensional GR. The dimension of spacetime is extremely relevant: in five-dimensions, you do get black rings.
If you're interested, I believe the original reference due to Hawking is Comm. Math. Phys. 25, 152–166 (1972). DOI: 10.1007/BF01877517.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/730094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 2,
"answer_id": 1
} |
Do theoretical particle physicists really believe the particles they work on do not exist? Inspired by this article in The Guardian.
In private, many physicists admit they do not believe the particles they are paid to search for exist – they do it because their colleagues are doing it
This makes no sense to me - there are surely better things to work on than stuff one believes to be wrong.
Is it really the case that many theoretical particle physicists don't believe the particles they work on exist?
| She is referring to particles in speculative theories like Garrett Lisi's E8, various string theories, GUT, and SUSY, etc. The experimental physicists and engineers sometimes have theories but mostly they are testing others weird ideas. Sabine makes good arguments that it's probably a waste of money and resources. It's unfortunate, but there's no particular reason to assume that the grand unified theory is something we can confirm or even could confirm. So, experimental physicists are like drunks looking for their keys around the lamp post because that is where the light is shining brightest.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/730232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
How high and when the bead will be on a spinning parabolically bent wire? Imagine a wire, bend it parabolically with equation $ 4ay = x^2 $, and the vertex as usually at the origin. It is now rotated with angular velocity, $\omega$, on an axis, passing in its plane through the vertex. A bead, of mass $m$, inserted in the wire, just slightly displaced from the vertex. Then calculate the max height it attains and at what time, would it?
This isn't a homework question just came out of curiosity!
What I did-
Normal $N$ is perpendicularly up from the bead, and the weight, $mg$ is vertically downwards, and the centerifugal force, $F_c$ is horizontally right, $$ F_c = \frac {mv^2}{R} = m \omega ^2 R = m \omega ^2 x $$
where x is the x coord of the coordinate plane.
If $\theta$ is from horizontal then, $$N cos \theta = m \omega ^2 x $$ $$ N sin \theta = mg $$ $$ tan \theta = \frac {g}{\omega ^2 x}$$ After differentiating the curve, $$ \frac {dy}{dx} = \frac {x}{2a} = m = tan (90 -\theta) $$
And $$ tan (90 -\theta) = \frac{\omega ^2 x}{g} = \frac {x}{2a}$$But the prob is the x gets cancelled and I get a wrong relation between $g$ and $\omega$. Am I missing something? Is there any alternate way to do this?
| Your equations are correct and it can only mean that at certain $\omega^2=g/2a$ the body can be stabilized for any x on the wire and for any $\omega^2\neq g/2a$ the solution will not be stable and probably bead will fall to the center for $\omega^2<g/2a$ and will fly away for the $\omega^2>g/2a$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/730354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Finding width of a slit in diffraction $500nm$ light is illuminating an aperture of width
$$d_1=0.2mm$$ The diffracted light from this first aperture then illuminates a second aperture $1m$ away. What is the smallest width, $d_2$, of the second aperture that will allow most of the diffracted light cone to pass through it?
I have tried to solve it, even checking my notes for the lecture
I could not find information to help to solve this problem. I have tried to utilise one formula as
however 2.5mm is not the answer. How can I proceed? Thank you.
| The sketches in your solution do not match the text of the problem. The way I understood the text, it is describing what is known as Fraunhofer diffraction. Check the corresponding wikipedia article. Especially look at the first example: https://en.wikipedia.org/wiki/Fraunhofer_diffraction#Examples . It has the solution to your question: most of the light is contained within the angle $\alpha = 2\lambda/W$, in your case $\lambda=0.5\mu m$ and $W=200\mu m$. Then the size of the central band at a distance $z=1m$ is given by $d_f = \alpha z=2\lambda/W\approx 5mm$. This is your answer: At a distance $z=1m$ the most diffracted intensity will get through a slit $d_f=5mm$ wide.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/730463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Since water can evaporate at any temperature above absolute 0, would the water cycle still be possible even without the Sun? Since water can evaporate at any temperature (or even sublimate at less than 0°C, although at a very low rate: Why does water not evaporate in below 0 degrees?), could there be a water cycle if there was no Sun? Would evaporation or sublimation of water occur with no external source of heat (even if extremely slowly),that then could fall as precipitations (in form of ice, snow, water...)?
| To have a water cycle you need water to evaporate and condense on the surface of the water at different rates. Without an outside source of energy it would reach equilibrium and there wouldn't be excess humidity condensing into clouds readily.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/730897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
At which electric field or voltage does field emission occur? Imagine our environment is a vacuum. At which value of electric field or voltage, does field emission occur? I just want to know what is the maximum electric field we are allowed, to place between two electrodes, without field emission phenomenon. So, you can choose any metal that you wish in order to mention this maximum value.
| Well, I am a new master circuit student and I do know so much about physics.
The definition of work function is here:
In solid-state physics, the work function (sometimes spelled workfunction) is the minimum thermodynamic work (i.e., energy) needed to remove an electron from a solid to a point in the vacuum immediately outside the solid surface.
And also, according to field electron emission:
Field electron emission, also known as field emission (FE) and electron field emission, is emission of electrons induced by an electrostatic field. The most common context is field emission from a solid surface into a vacuum.
So, we can say that field emission occurs when the electron can absorb enough energy to overcome the work function the metal.
So according to my research, I can say that based on Work functions of elements the maximum work functions of materials is 5.93eV (and I do not know a material with a higher work function is discovered or not). So, the voltage required to remove an electron from metal to a point in vacuum will be equal to:
5.93/(1.6e-19)=3.7e19 volts
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/731087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Complex Grassmann Dirac Functional - How do we integrate over it? I'm following the Book of Brian Hatfield (Quantum Field Theory of point particles and Strings), p.192 here: For real Grassmann numbers (and Functionals thereof):
If $\Phi[\psi]$ is a functional, and $\psi(x)$ is a Grassmann-valued function, we demand that $$\int \mathcal{D}\psi \delta[\psi - \xi] \Phi[\psi] = \Phi[\xi]$$ (this is equation 9.67) , and one option to do this is to let (equation 9.66) $$\delta[\psi - \xi] = \prod_x (\psi(x)-\xi(x)).$$
The complex case of the delta function is NOT treated in the book, and I want to deduce how the mentioned relations would turn out for that case. Here, $\psi$ now has two components ($\psi = \frac{1}{2} \psi_1 + i \psi_2$) - Which makes me wonder: How does the fundamental relation turn out? For complex $\psi$:
\begin{align}
\int \mathcal{D}\psi \delta[\psi - \xi] \Phi[\psi] = \Phi[\xi]
\end{align}
or
\begin{align}
\int \mathcal{D} \psi \int \mathcal{D} \psi^* \delta[\psi - \xi] \Phi[\psi] = \Phi[\xi]?
\end{align}
The first version works, but only if I assume that $\delta[\psi - \xi] = \prod\limits_x (\psi(x) - \xi(x))$ and $\delta[\psi-\xi] = \delta[\psi-\xi]^*$, and those exclude each other.
In either case, what would be a realization of the $\delta$ functional? Would it still be
\begin{align}
\delta[\psi - \xi] = \prod_x (\psi(x) - \xi(x))?
\end{align}
| *
*Be aware that integration over a complex Grassmann-odd variable $\psi$ has 2 different notations in the literature:
$\int \!\mathrm{d}^2\psi$ and $\int\! \mathrm{d}\psi~\mathrm{d}\psi^{\ast} $. This is similar to the standard notations for a Grassmann-even complex integration.
*Similarly, a complex Grassmann-odd Dirac delta distribution can denote just the holomorphic part or also include the anti-holomorphic part, depending on conventions.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/731486",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Understanding page 141 of Blundell’s Concepts in thermal physics On this page (in the second edition), there is a figure containing two states A and B of a system:
There are two paths between A and B: one is an irreversible change, and the other is a reversible change. However, on the same page it says that entropy stays the same for a reversible change, and that entropy increases for an irreversible change. The first seems to imply that the entropy of A and B are the same; and the latter seems to imply that the entropy of B is larger than the entropy of A. This is a contradiction.
Am I missing something, or is this a mistake in the book?
|
on the same page it says that entropy stays the same for a reversible change,
yes, it says that
and that entropy increases for an irreversible change
yes, it says that
and the latter seems to imply that the entropy of B is larger than the entropy of A.
No, this does not follow from what the authors say.
They say that the heat along the reversible path and the the heat along the irreversible path satisfy the Clausius equation:
$$\int_A^B \frac{dQ}{T} \leq \int_A^B \frac{dQ_\text{rev}}{T} \tag{14.7}$$ It then says that since $dS = dQ_\text{rev}/{T}$ we must have
$$dS = \frac{dQ_\text{rev}}{T} \geq \frac{dQ}{T} \tag{14.8}$$
It follows that for an adiabatic process ($dQ=0$) we must have
$$dS \geq 0 \tag{14.9}$$
which is a statement of the second law.
The result does not imply that $\Delta S_{A,B}$ along the irreversible path is larger than $\Delta S_{AB}$ along the reversible path, if by $\Delta S$ we are referring to the entropy change of the system. The statement would be correct if we are referring to the entropy change of the universe.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/731814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Equivalence between small distance and high energy I see in a lot of particle physics literature statements along the lines of: 'This is valid for high energies (small distances)'. Exactly what do we mean by small distances in this case? From QFT, QM and other physics courses, the connection between small distance and high energy makes intuitive sense, but I am wondering if there is a concrete connection that authors have in mind?
| In HEP, where (in units of c=1) momenta are much larger than masses, so they are tantamount to energies, the fundamental scale relation of QM, $[x,p]=i\hbar$, comes to bear. The de Broglie relation ${\lambda\over 2\pi}= \hbar/p$ of the particles/waves involved relates distances to momenta/energies inversely.
Consequently, "small distance" is a synonym for "high energy" or momentum. So, analogously to diffractive crystallography, one probes ever smaller distance scales with ever higher enery/momentum scattering probes.
In HEP natural units, (ℏ=1) this is codified by the duality of the units for distance or energy,
such as the radius of a proton being 4 GeV$^{-1}$ actually meaning
$$
4 \mathrm {GeV}^{-1} (\hbar c) = 8\cdot 10^{-16}{\mathrm m}~.
$$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/731971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Electromagnetic radiation reflected normally causing interference to itself Is it possible to have a body reflect almost perfectly any radiation falling normally with the insident ray so that the reflected ray interferes with the incident and hopefully reduce or cancel it out ?
| High-quality optical lenses are often coated to reduce reflections. A thin film of a transparent material with a different index of refraction is coated onto the lens, with a thickness of a quarter of the wavelength of the light of interest. Incoming light passes through the coating and bounces off the other side, returning to the top surface half a wavelength out of phase. This partially cancels the reflection from the top surface. The result is that more light gets through the lens, giving a brighter image with better contrast, and less is reflected away or bounced around inside the instrument.
It only works perfectly at one specific wavelength. For optical equipment designed to work with the full range of visible light, such as binoculars, this is usually in the green part of the spectrum, where the eye is most sensitive and in the centre of the visible range, to give the best overall performance. This is why such lenses often appear to have a bluish tinge - light is preferentially reflected at the ends of the spectrum.
Conversely, dielectric mirrors can be made with transparent coatings with a thickness designed to reinforce the reflection. For specific, narrow wavelengths they can reflect 99.999% of the light.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/732282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Calculating heat removed by cooling system using output fluid temperature I am trying to calculate the amount of energy removed by a cooling system for some medical research. I'm a little out of my depth with the physics calculations.
I have fluid flowing through a variable temperature object (an organ being heated) at a known unchanging flow rate. The fluid input temperature is constant. A thermocouple on output side measures the temperature of the fluid leaving the object at high frequency.
So I have the mass of water, the change in temperature of the water (and obviously it's specific heat capacity). I do not have the average temperature of the water at the end of the experiment.
I think I can calculate the amount of energy removed by taking the area under the curve of the time temperature graph but I'm not sure what the units on the x-axis should be. Would it be mass of water?
Would taking the average temperature of the output water be equally accurate given the flow rate is unchanging? The temperature rises and falls several times.
Thanks so much
| The heat that is transferred from the organ to the water can be calculated as the difference between the enthalapy of the inlet and outlet water. Within time $dt$ this heat is
$$
dQ = \dot M C_p (T_\text{out} - T_\text{in}) dt
$$
where $\dot M$ is the flow rate of the cooling water and $C_P$ is its heat capacity. We need to integrate with respect to time:
$$
Q = \int_0^t \dot M C_p (T_\text{out} - T_\text{in}) dt
$$
Only $T_\text{out}$ is a function of time:
$$
Q = \dot M C_p \int_0^t T_\text{out} dt - \dot M C_p T_\text{in} t
$$
Therefore we have to calculate the integral
$$
\int_0^t T_\text{out} dt
$$
where $T_\text{out} = T_\text{out}(t)$ is the outlet temperature as a function of time.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/732679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Thermodynamics : an empty container surrounded by an atmospheric gas I have a small exercices of thermodynamics if anyone can help me :
We start with an empty container of volume $V$. The walls of this container are adiabatic and will not change over time surrounded by a gas of pressure $P_0$ and of temperature $T_0$.
*
*Define and give the characteristics of the system we'll study
*Write the internal energy inside the box using the characteristic of the exterior gas when the equilibrium is reached
*What is the internal energy if the box is closed when the mechanical equilibrium is reached but not the thermal one ?
I know the system is the volume of the box + the matter inside the atmosphere which will go inside the box.
Idk how to characterize it and how I can write just the internal energy and not its variations.
Edit : by applying the first principle in open system, supposing that work and thermal energy are equal to 0 :
$u_2 = P_0(v+v_0) - P_f \times v$
with $m$ the mass of the matter of the system, $v = \frac{V}{m}$, $v_0$ the volume occupied by the gas outside the box at initial time.
I think this relation is wrong because next question ask for internal energy when the box is closed when you reach mechanical equilibrium but not thermal one (my relation doesn't need temperature at all ??!!).
| Since the container is empty, the pressure inside the container is 0. There is no material/mass inside the system whose thermodynamical quantities can be considered. Please check the question again.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/732997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Dimensional Analysis, How to determine the right order for the power relation? I came across this question in Solved Problems in Classical Mechanics by O.L. de Lange and J. Pierrus. Question 2.12 is as follows:
Use dimensional analysis to determine the dependence of the period $T$ of a simple
pendulum on its mass $m$, weight $w$, length $\ell$ and arc-length of swing, $s$.
Solution:
$$
T=k m^\alpha w^\beta \ell^\gamma s^\delta .
$$
Hence
$$
M^0 L^0 T^1=M^\alpha\left(M L T^{-2}\right)^\beta L^\gamma L^\delta,
$$
and so
$$
\alpha+\beta=0, \quad \beta+\gamma+\delta=0, \quad-2 \beta=1 .
$$
These yield $\alpha=-\beta=\frac{1}{2}$ and $\gamma=\frac{1}{2}-\delta$. Consequently, becomes
$$
T=k \sqrt{\frac{m \ell}{w}}\left(\frac{s}{\ell}\right)^\delta,
$$
where $\delta$ is an undetermined number.
Question
My question is how do I know that $\delta$ is the "correct" undetermined? I could have rearranged so $\delta=\frac{1}{2}-\gamma$ and obtained:
$T=k \sqrt{\frac{m s}{w}}\left(\frac{\ell}{s}\right)^\gamma$,
This seems wrong to me as the equation for a pendulum is: $T=k \sqrt{\frac{m \ell}{w}}$.
what am I misunderstanding?
| There's no way to know, since you're starting with four parameters and only 3 equations to solve them. Thus, both $\gamma$ and $\delta$ can be thought as undetermined numbers.
Nevertheless, the results you'll obtain will be identical with any of those two choices, since from a dimensional point of view you have powers of the quantity $\dfrac{l}{s}$ or $\dfrac{s}{l}$, which are scalar quantities.
In general, just from the dimensional analysis it's possible to show the following relations:
$$
T=k\sqrt{\dfrac{ml}{w}} \chi\left(\dfrac{s}{l}\right)\\
T=k\sqrt{\dfrac{ms}{w}} \chi'\left(\dfrac{l}{s}\right)
$$
Where $\chi$ and $\chi'$ are undetermined functions. The two expressions are nevertheless the exact same expression, maybe one function will be easier to compute than the other one but at the end the result for T is uniquely defined.
This seems wrong to me as the equation for a pendulum is: $T=k\sqrt{\dfrac{mℓ}{w}}$. what am I misunderstanding?
It seems wrong because you know a priori the answer, but just looking at the dimensional analysis you cannot state which one of the two expression is correct. They both are, the real challenge is to find an analytic expression for the two functions $\chi$ and $\chi'$ so that at the end they explain the same physics.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/733081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Center of Gravity derivation question I need a sanity check, please.
When determining the center-of-gravity of a lamina described by $f(x)$, we know that by definition,
$\bar{x} M = \sum_{i=1}^{N} m_i \tilde{x_i}$
where $\tilde{x_i}$ is the location of the centroid of strip located at $x_i$
Assuming uniform density and thickness, this becomes
$\bar{x} A = \sum_{i=1}^{N} A_i \tilde{x_i}$
or
$\bar{x} = \frac{\sum_{i=1}^{N} A_i \tilde{x_i}}{\sum_{i=1}^{N} A_i}$
Let us consider a stip with width $\Delta x$ at a distance $x_i$ from y-axis
The area of the strip is $\Delta x \times f(x_i)$
The centroid of the strip, $\tilde{x_i}$ will be at a distance $x_i + \Delta x/2$ from y axis
Hence
$\bar{x} = \frac{\sum_{i=1}^{N} (x_i + \Delta x/2)(\Delta x f(x_i)) }{\sum_{i=1}^{N}\Delta x f(x_i)} = \frac{\sum_{i=1}^{N} (x_i \Delta x f(x_i)) + ( \frac{\Delta x^2}{2} f(x_i)) }{\sum_{i=1}^{N}\Delta x f(x_i)}$
Now, to arrive at the famous equation.
$\bar{x} = \frac{\sum_{i=1}^{N} x_i f(x_i)}{\sum_{i=1}^{N} f(x_i)}$
the term
$\frac{\Delta x^2}{2} f(x_i)$ is eliminated by some means.
Is this reasoning correct?
Is it because it is safe to assume $\Delta x^2$ is negligible?
Thanks so much
| to obtain the center of mass use those equations
$$y_{CM}=\frac{\iint\,y\,dx\,dy}{A}\\
y_{CM}=\frac{1}{A}\int_{x_1}^{x_2}\left(\int_0^{f(x)}y\,dy\right)\,dx$$
$$x_{CM}=\frac{\iint\,y\,dx\,dy}{A}\\
x_{CM}=\frac{1}{A}\int_{x_1}^{x_2}\left(\int_0^{f(x)}dy\right)\,x\,dx$$
where $~A~$ is the area under the curve $~f(x)~$
$$A=\iint\,dx\,dy=\int_{x_1}^{x_2}\,f(x)\,dx$$
$\Rightarrow$
$$y_{CM}=\frac 12\frac{\int_{x_1}^{x_2}f(x)^2\,dx}{\int_{x_1}^{x_2}\,f(x)\,dx}$$
$$x_{CM}=\frac{\int_{x_1}^{x_2}f(x)\,x\,dx}{\int_{x_1}^{x_2}\,f(x)\,dx}$$
with this function
$$f(x)=a\,(x-x^2)$$
and for example with $~a=2~,x_2=0.4~,x_1=0.2~$ you obtain the results
$~x_{CM}=0.306~,y_{CM}=0.209$
Remark
of course you can obtain analytical solution but I didn't put it , because it is too long
the Parable equation is:
$$f(x)=a\,x^2+b\,x+c$$
for your parable you obtain 3 equations for the 3 unknowns $~a,b,c$
$$f(0)=0\\
f(x_f)=0\\
\frac{df}{dx}\Bigg|_{x=\frac{x_f}{2}}=0$$
$\Rightarrow$
$$a\mapsto \pm a\quad, b=-a\,x_f\quad ,c=0$$
with $~x_f=1,f(x)=a\,(x-x^2)$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/733288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.