Q stringlengths 18 13.7k | A stringlengths 1 16.1k | meta dict |
|---|---|---|
Are neutrons an actual particle? I'm a senior student in high school and we are learning about Particle Physics in Physics. I wanted to ask a question about neutrons. Is there a possibility that neutrons may not even be a particle, just a bond, relationship or pairing between electrons and protons? Can neutrons just be the cancelling out of electrons and protons charges, forming a neutral charge inside the nucleus and not an actual particle? For example, carbon has 6 electrons, 6 protons and 6 neutrons. Could the 6 neutrons and 6 protons cancel each other out forming 6 neutral charges making the atom stable? The protons and electrons would still exist but they just form a stable atom by being neutrally charged. I know this is extremely unlikely and most likely wrong but I really wanted to know if there was an answer to this or is the theory we have have no correct and there is no need for further debate.
| If I understand your thoughts correctly, and you had a typo when you said "Could the 6 neutrons (you wanted to say electrons) and 6 protons cancel each other", you expected a carbon atom to have the same number of protons and neutrons. If you know that there are carbon 13 (which has 7 neutrons) and carbon 14 (which has 8 neutrons), you may think twice.
| {
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If radio signals attenuate when travelling through space, then what kinds of emissions are we looking for when searching for extraterrestreal life? My understanding is that radio waves travel forever, like ripples in a pond, but attenuate with distance. They get mixed with other signals and become cosmic noise.
I'm looking at these answers, which suggest that the TV and radio signals that originate from earth attenuate and are indistinguishable from background noise after some distance.
This makes me ask:
How do we hope to detect signals from alien civilizations, if those signals attenuate as well?
Are there other types of signals, that can cross vast interstellar distances (> 600 light year) and carry data that is recoverable?
| Electromagnetic waves are still the best bet and distances are huge but we’re reasonably good at sending and detecting weak signals, and constantly improving at that job. We are still communicating with the Voyager probes, even if they
have both reached the interstellar region, and signal strengths have energy 20 billions times smaller than needed to run a digital watch.
For direct “communication”, the basic assumption is that an alien technology is better than ours at sending directional signals.
There is also hope that secondary information might reveal the presence of civilisation: we are still in the infancy of exoplanet exploration and again the hope is that the presence of chemicals in the atmosphere of an exoplanet might reveal signs of civilisation.
| {
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Does AdS/CFT correspondence take place only near black holes? Is it related to black holes or is AdS/CFT a separate thing itself?
| While the AdS/CFT correspondence is related to black holes and a tool for studying them, it is a much broader framework. Generally speaking, it concerns a correspondence between a quantum field theory defined on anti de Sitter space and conformal field theory.
Anti de Sitter space is the maximally symmetric, Lorentzian manifold with constant negative scalar curvature, which is to say the Ricci scalar $R = g_{ab}R^{ab} <0 $.
On the other side of the correspondence is a conformal field theory, which is a quantum field theory that has conformal symmetry. This is generated by,
*
*Translations $x \mapsto x^\mu + a^\mu$
*Dilations $x^\mu \mapsto \lambda x^\mu$
*Rotations $x^\mu \mapsto \Lambda^\mu_\nu x^\nu$
*Special conformal transformations which are a combination of an inversion, translation and another inversion, where by inversion we mean $x^\mu \mapsto x^\mu/|x|^2$.
Conformal field theories have a more rigorous formulation in terms of vertex algebras and vertex operator algebras, which are associative algebraic structures you can think of as a vector space with an infinite family of left-multiplications, indexed by an indeterminate.
| {
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What are these geometric patterns on ice? In a winter day, I noticed the water frozen inside a canal in our building. As you see there are very nice geometric patterns formed by the ice, with specific angles. What is the physical interpretation of these patterns? Are the giant cristals? The only physical inputs that I can think of is the shape of the canal: width ~30cm and depth of water ~10cm
| They are crystals of ice which grow outward from nucleation sites, where the freezing process prefers to start. Those crystals expand until they hit an edge of the container, or another crystal as it grows.
This effect is exploited in a process called directional casting, in which a gas turbine blade can be grown as a single crystal from molten superalloy by cleverly initiating the freezing process at one end of the blade so only the crystallite with the desired lattice orientation can take over and grow. The resulting blade has exceptional strength because it has no grain boundaries, and the lattice orientation provides maximum strength.
| {
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Why does general homogenous solution of differential equations modelling circuits die off after a long time? I was reading this answer in Elecronic engineering stack exchange which said that when solving the linear second order differential equation modelling circuits having ac source. We only need to account for particular solution as homogenous part of the differential always dies to zero.
How do we prove mathematically that for all such linear AC circuits the homogenous part of the solution dies to zero at $t\to +\infty$?
Respone to discussion in comments:One of the has comment on this post that it is direct consequence of solution of differential equations. Ok, so for simplicity I assumed a second order differential equation of form:
$$py''+qy' + r = g(t)$$
Now for the homogenous part I do $$ py'' + gy'+r=0$$ , I find that for the solutions of the above equation to die at infinity, I need both roots of the homogenous part of equation to have have negative real part. Now , I know for sure this isn't always mathematically guaranteed, so why so for linear circuits?
| When you apply power to a circuit, there is a transient response (such as when you connect an RC circuit to a battery) followed by a steady-state behavior. I would assume that the homogeneous part refers to the transients.
| {
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$\gamma^5$ rotation of chiral fermion in (1) Peskin&Schroeder, (2) Weinberg, or (3) Srednicki The theta angle due to the chiral gamma^5 rotation of chiral fermion results in the phase alpha(x) that has different + or - sign for
(1) Peskin&Schroeder, (2) Weinberg or (3) Srednicki.
Here
*
*Peskin&Schroeder does a psi' = exp(i alpha(x) gamma5) psi rotation, and it gives a (+) sign F F dual term in (19.79)
*Weinberg book does a psi' = exp(i alpha(x) gamma5) psi rotation, and it gives a (-) sign F F dual term in (23.68)
*Srednicki book does a psi' = exp(-i alpha(x) ) psi rotation, and it gives a (-) sign F F dual term in (94.4)
It looks that (1) and (3) have the same results, while (2) Weinberg gives a different result from others. Does someone know why? Who may possibly make a mistake?
p.s. Pardon me to show their results directly in images below, I believe that it is easier to show them directly so people can see from their books -- which will be more clear than I typed the equations in a partial manner.
Peskin&Schroeder
Weinberg
Srednicki
| The gamma matrices that Peskin & Schroeder work with are $$\gamma^0 = \begin{pmatrix}0 & 1\\ 1 & 0\end{pmatrix},\quad \gamma^i=\begin{pmatrix}0& \sigma^i\\ -\sigma^i & 0\end{pmatrix}\tag{3.25}$$
On the other hand, for Weinberg, the gamma matrices are $$\gamma^0=-i\begin{pmatrix}0 & 1 \\ 1 &0\end{pmatrix},\quad \gamma^i=-i\begin{pmatrix}0 & \sigma^i \\ -\sigma^i & 0\end{pmatrix}\tag{5.4.17}$$
In particular we see that the gamma matrices used by Weinberg, call them $\gamma^\mu_W$, are related to the gamma matrices used by Peskin & Schroeder, call them $\gamma^\mu_{PS}$, by $$\gamma^\mu_W=-i\gamma^\mu_{PS}.$$
Peskin & Schroeder define $\gamma^5_{PS}=i\gamma^0_{PS}\gamma^1_{PS}\gamma^2_{PS}\gamma^3_{PS}$, while Weinberg defines $\gamma^5_W=-i\gamma^0_W\gamma^1_W\gamma^2_W\gamma^3_W$. In that case we can compare the two and we observe that $$\gamma^5_W=-i(-i\gamma^0_{PS})(-i\gamma^1_{PS})(-i\gamma^2_{PS})(-i\gamma^3_{PS})=-i\gamma^0_{PS}\gamma^1_{PS}\gamma^2_{PS}\gamma^3_{PS}=-\gamma^5_{PS}.$$
So because of the different conventions for the gamma matrices we see that what Peskin & Schroeder define as $\gamma^5$ differs from what Weinberg defines by a minus sign. In that case when Weinberg does $\psi' = e^{i\alpha\gamma_W^5}\psi$ he is doing $\psi' = e^{-i\alpha\gamma_{PS}^5}\psi$. So for Weinberg transforming with $\alpha$ is the same as transforming with $-\alpha$ for Peskin & Schroeder. Therefore the two are really consistent. Finally it is immediate to see that Srednick is also consistent with the other two as you said yourself in the question.
| {
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Two bubbles attract or repel? I have this weird thought that bothers me, but unfortunately, I know little physics and I thought I can get help to mentally settle it.
Here we go:
Let's say you have infinite sand or at least a vast quantity of it. And no gravity.
Just sand, tight in a vast container
You manage to inflate two big (relative to the sand grains) balloons inside the sand.
Due to size, the balloons will push the sand to create tension, as the sand is tightly packed.
If you bring the balloons close to each other, will they repel due to this tension or attract? Does the shape matter?
| Let's suppose a big container, but not so big that gravitational effects become relevant. When the baloons are inflated, the sand is forced to be more compact in the neighborhood. But unlike a fluid, the pressure is not evenly distributed to all container. It is like a pile foundation, that compact locally the soil, but not 100m away.
So, if 2 baloons are inflated close, the sand volume between them is more stressed than the average of the container. The consequence is that, if A and B are the points of the initial center of inflation of each one, these points will be a little more distant after the inflation is finished.
| {
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Definition of the probability current I know that the definition of probability current is given by
\begin{equation}
J\sim \psi^*\frac{\partial \psi}{\partial x}-\psi \frac{\partial \psi^*}{\partial x}
\end{equation}
However, some papers are using this definition:
\begin{equation}
J\sim \psi^*\frac{\partial H}{\partial k_x}\psi
\end{equation}
where $H$ is the Hamiltonian of the system and $k_x$ is the wavevector. So, how can I understand that these two are equivalent, and is it possible to derive the second one, I mean where does it come from?
update
this question is also related to this one where I think we might have a close answer.
| Ok, I think I get it and would like to share in case any additional input.
So in general the probability current can be written as
\begin{equation}
J=\frac{-ie\hbar}{2m} [\psi^*(\nabla\psi)-(\nabla\psi^*)\psi]
\end{equation}
knowing that the momentum operator is $\bf p=-i\hbar\nabla$, we get
\begin{equation}
J=\frac{e}{2m} [\psi^*(\bf p\psi)+(\bf p\psi^*)\psi]=\frac{e}{m}Re[\psi^*(p\psi)]
\end{equation}
Now, we can recall that velocity operator $\bf{v}$ can be written with respect to $\bf p$ as ${\bf{v}}=\bf{p}/m$, yeilding
\begin{equation}
J=eRe[\psi^*(\bf v\psi)]
\end{equation}
and since $\bf v\sim\frac{\partial H}{\partial k}$ we get the desired form of probability current
\begin{equation}
J\sim eRe[\psi^*(\bf \frac{\partial H}{\partial k}\psi)]
\end{equation}
| {
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How do charge conjugate fields transform under $SU(2)$ and $SU(3)$? I am trying to derive the gauge transformation for the charge conjugate field of a quark doublet (left handed quark) such that its field $Q$ transforms under $SU(2)$ and $SU(3)$ as:
$SU(2):$ $Q \rightarrow \exp\left[ \frac{i}{2}\theta^{a} \sigma^{a}\right] Q$, where $\sigma^{a}$ are the Pauli matrices and $\theta^{a}$ some group parameter.
$SU(3):$ $ \rightarrow \exp\left[ \frac{i}{2}\alpha^{a} t^{a}\right] Q$, where $\alpha^{a}$ is some group parameter and the $t^{a}$ are $SU(3)$ generators in the fundamental representation, for instance Gell-Mann matrices.
Now I need to find the corresponding transformations for $SU(2)$ and $SU(3)$ for the charge conjugate field of $Q$ defined as: $Q^{c}\equiv i\gamma^{2}\gamma^{0}\bar{Q}^{T}$.
What I started to do is try to find $\delta Q^{c}$ for both cases:
$SU(2):$ $Q^{c}\rightarrow i\gamma^{2}\gamma^{0}(1-\frac{i}{2}\theta^{a}{\sigma^{a}}^{\ast}) Q^{\ast}$, then $\delta Q^{c}=i\gamma^{2}\gamma^{0}(-\frac{i}{2}\theta^{a}{\sigma^{a}}^{\ast})Q^{\ast}$
Now my question is how do I commute ${\sigma^{a}}^{\ast}$ with the gamma matrices here so I can get a $Q^{c}$ on the LHS?
I have the same question for the $SU(3)$ case:
$SU(3):$ $\delta Q^{c}=i\gamma^{2}\gamma^{0}(-\frac{i}{2}\alpha^{a}{t^{a}}^{\ast})Q^{\ast}$. Do the Gell-Mann matrices commute with the gamma matrices here or not?
Thanks a lot and sorry if this is a too much of a beginner kind of question.
| The gamma matrices act on spinor indices while the $SU(2)$ and $SU(3)$ matrices act on group indices, for one is like the others do not exist. Therefore they commute.
| {
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Does the magnetic field produced by a current carrying wire, exert a magnetic force on the wire itself? I have to calculate the pressure on a current carrying wire. Since there is a pressure on the wire, there must be a force on it, which is a magnetic force. Does the magnetic field produced by the wire, exert a magnetic force on the wire itself? If this is true, why?
| The answer is: no.
The 'magnetic force' on the wire is due (indirectly) to magnetic Lorentz forces acting on the moving electrons in it. It is true that there will be attractive magnetic forces between electrons moving in parallel paths at different points in the wire's cross-section (for example between electrons at opposite ends of a diameter). But these forces are equal and opposite, so there will be no resultant force on these electrons taken together, and no resultant force on the wire.
It's a different story when we apply an external magnetic field with a component at right angles to the wire. The electrons will then experience forces in a direction given by $\textbf F = -e \mathbf v \times \mathbf B$, (or by Fleming's left hand rule). [The moving electrons would be forced out of the wire, were it not for 'bonding forces' that stop them from leaving. Strictly, it is these 'bonding' forces (or their Newton's Third Law partners) that the rest of the wire experiences, rather than the magnetic Lorentz forces directly. However the magnitude of the force on the wire can be correctly calculated as the vector sum of the Lorentz forces, which is easily shown to be equal in magnitude to $F=BIL\ \sin \theta$ with the usual notation.]
| {
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Conjugate momentum notation I was reading Peter Mann's Lagrangian & Hamiltonian Dynamics, and I found this equation (page 115):
$$p_i := \frac{\partial L}{\partial \dot{q}^i}$$
where L is the Lagrangian. I understand this is the definition of conjugate momentum, but I wanted to know if there is a particular reason for the momentum index to be a lower index and the coordinate index to be an upper index. Is it simply the author's preference or there is a deeper reason?
| Usually, you would write your Lagrangian in some sort of form like:
$$L = {\dot q}^{i}{\dot q}_{i} = g^{ij}{\dot q}_{i}{\dot q}_{j}$$, because the lagrangian itself is a scalar. Then, if you took a variation with respect to the "downed" version, you'd be left with
$$\frac{\delta L}{\delta {\dot q}_i} = 2 g^{ij}{\dot q}_{j} = 2 {\dot q}^{i}$$
So, variation of a scalar with respect to a "downed" index leaves an "upped" index.
| {
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Can the Hamiltonian be interpreted as the "speed" of unitary evolution? The Schrodinger equation
$$i\hslash \frac{d}{dt} \psi = H \psi$$
means a quantum state $\psi(t)$ evolves unitarily, that is,
$$\psi(t) = \exp(-\frac{i}{\hslash} H t) \psi(0)$$
where $\psi(0)$ is the initial state at time $t = 0$.
Suppose if we scale the energy levels of the Hamiltonian by some factor $0 < \zeta$, let
$$\tilde{H} = \frac{1}{\zeta} H$$
then for the evolution from $\psi(0)$ to a target state $\phi$,
$$\phi \equiv \psi(\tilde t = \zeta t) = \exp(-\frac{i}{\hslash} \frac{1}{\zeta} H \zeta t) \psi(0)$$
where the time $\tilde t$ to reach $\phi$ has to scale contravariantly to complement the change in $H$.
So can it be said the Hamiltonian $H$ is the "speed" of unitary evolution?
| To paraphrase @Qmechanic in more mundane language: Hamiltonian is the generator of unitary evolution, i.e., the rate of infinitesimal change. In this sense it can be called speed.
Note however, that the Hamiltonian appears in the exponent, and we do not have here a physical quantity, the derivative of which we could call speed. In other words, whether we call Hamiltonian speed has to do more with semantics rather than physics or math.
| {
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Why voltage is same across parallel circuit if it is work done per unit charge? Suppose we have the following circuit:
If voltage is work done per unit charge, why voltage is same across each resistor if the charge has to do more work in resistor R2 than in resistor R1?
| The work is the same. Both resistors would require the same work per coulomb to move a test charge through the resistor. One way to get an intuition for why this is true is from a physical picture. This won't be true always, but let's imagine the reason why the resistor $R_2$ has double the resistance is because it's twice as long. What happens is that the electric field generated in $R_2$ is half as strong as the electric field in $R_1$. Consequently, the electric force on a test charge in $R_2$ is also halved. But the work is the same, because charges move twice as far in $R_2$, and $W = Fd$.
Ultimately, the reason the voltage drop is the same across both resistors is that total energy is conserved. Any individual charge moving through $R_1$ starts at the positive terminal of the battery and ends at the negative terminal (assuming conventional current). The exact same thing is true of any single charge moving through $R_2$. If the two charges start at the exact same place (the positive terminal), then they must begin their journey with the exact same amount of electric potential energy. If they end at the same place (the negative terminal), then they must end their journey with the exact same electric potential energy. Hence, both charges must lose the same amount of electric PE along their path. But if the wires are ideal, then the only place they can lose their electric PE is inside the resistor that they traveled through.
| {
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What is our time-size in spacetime? Are we spaghetti or flat blobs? In special and general relativity time is treated as a dimension ($ict$, being $t$ a real number). Computations usually revolve around describing world lines, and events (crossings of world lines).
But objects can't physically be time-lines. Future unfolds through projections of quantum states, so the lines would have to have tips. Thinking about tips leaving a literal energy-trails in the time direction sounds bad (You leave a pebble in space for a day, so now you have a light-day*mc more energy in your universe, leaving a longer and longer trail every day)
Lets take a human as an example. An approximation where a set of points are roughly always moving together, are not very much apart (measured in $ct$), are subjected always roughly to the same gravitational potential. The watch on your left hand is usually always ticks the same way as the watch on your right hand. So humans are essentially flat blobs flying through time. How bad is this approximation?
If we're flat blobs, what is our size in the time dimension? Planck's constant/momentum-uncertainty?
A flat-blob also means there's no way you could meet your younger self by using a time machine (unless doing some cloning). Because you're something with near-zero time-size, you can't reach back to a younger-self because you're not a line in time.
| Although in the geometric picture of general relativity time is pictured as a dimension this is not how Einstein viewed the matter. He was against interpreting physics in a purely geometric fashion which began with Minkowski interpreting special relativity with a Lorentz metric. In fact, he saw general relativity as the unification of inertia and gravitation.
| {
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Field strength tensor This question seems very simple however I'm stuck:
given the following:
How can we find those components? And how can we check the antisymmetry of $F$?
The equations are from Nakahara, Geometry, Topology and Physics.
$$
\mathcal A_\mu=A_\mu{}^\alpha T_\alpha \quad \mathcal{F}_{\mu\nu}=F_{\mu\nu}{}^\alpha T_\alpha\tag{10.40}
\\ [T_\alpha, T_\beta]=f_{\alpha\beta}{}^\gamma T_\gamma
$$
$$
\\F_{\mu\nu}{}^\alpha=\partial_\mu A_\nu{}^\alpha-\partial_\nu A_\mu{}^\alpha+f_{\beta\gamma}{}^\alpha A_\mu{}^\beta A_\nu{}^\gamma\tag{10.41}
$$
| The easiest way to check the antisymmetry of $F$ is to write it as
$$
\mathcal{F}_{\mu \nu} = \partial_\mu \mathcal{A}_\nu - \partial_\nu \mathcal{A}_\mu + [ \mathcal{A}_\mu , \mathcal{A}_\nu]
$$
This is manifestly antisymmetric, because the commutator is antisymmetric. Note then that
\begin{align}
[ \mathcal{A}_\mu , \mathcal{A}_\nu] &= [A_\mu^a T_a, A_\nu^b T_b] \\
&= [T_a, T_b] A_\mu^a A_\nu^b \\
&= f^c_{\; ab}T_c A_\mu^a A_\nu^b
\end{align}
which allows us to match the original expression with the one you provided.
In particular, from the definition
$$
[T_a, T_b] = f^c_{\; ab} T_c
$$
we can see that
$$
f^c_{\; ab} = -f^c_{\; ba}.
$$
| {
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Singularity in Robertson Walker metric with flat spatial slices In Sean Carroll's GR book, pg. 76, a special case of the Robertson-Walker metric, where the spatial slices are flat is given by
$$ds^2=-dt^2+a^2(t)[dx^2+dy^2+dz^2].$$
It was said that $t=0 $ represents a true singularity of the geometry (the 'Big Bang') and should be excluded from the manifold. The range of the $t$ coordinate is therefore $0<t<\infty$.
Why is $t=0$ a singularity? What is infinite or undefined when $t=0$ ?
| Carroll mentions that in the solution under consideration, the scale factor $a\rightarrow 0$ as $t\rightarrow 0$. Since the metric must always be non-degenerate, this represents a singular point. Operationally, if the metric is degenerate than the dual metric $g^{\mu\nu}$ becomes undefined.
| {
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Spherical Lens Instead of Parabolic Lens I know that using the paraxial approximation, spherical lenses behave like parabolic lenses.
It seems that there is no reason to use spherical lenses instead of parabolic (because they are used in the same way, and parabolic lenses do not required paraxial approximation) apart from the fact that parabolic lenses are more complicated to make.
*
*Does spherical lenses have more advantages over parabolic lenses?
*Are there any application that required specifically spherical lenses (and not parabolic)?
| Spherical reflectors have the outstanding advantage over parabolic reflectors in that by moving the feedpoint/receiver off-center above the spherical surface, the beam aperture can be steered without having to steer the entire antenna. This principle was used in the design of the radio telescope at Arecibo, PR which recently fell down when several of the cables supporting the steerable feedhorn assembly broke.
Note that because the reflections from a spherical mirror do not come to a single focus, an array of aspheric reflectors in front of the feedhorn is required to correct for this- which the Arecibo system possessed.
| {
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Why voltage is not the same for the capacitors in series?
In this picture, there are two capacitors C1 and C2 joined in series and connected to a battery. We know there are two terminals in a battery, a positive terminal and a negative terminal.
The potentials of the positive and negative terminals are +P and -P respectively. And there potential difference, in other word the voltage of the battery is V.
The plate on the left side of the capacitor C1 is directly connected to the positive terminal of the battery. So its potential will also be +P. Similarly the potential of the plate on the right side of C2 will have a -P potential.
My questions:
A) I am assuming the magnitudes of the potentials of the opposite sides of the battery are same just different in signs. For the right plate of C2 to have a potential -P it takes -q amount of charge. The left plate of C1 also gives off this equal amount of charge and acquires the state +q. Since C1 and C2 are two different capacitors, why gain or loss of same amount of charge would cause them to have the same magnitude of potential?
B) If the left plate of C1 has a charge of +q and a potential +P, then C1's other plate should also have the potential -P as it has -q charge. Because two plates of the capacitor C1 are same in material and geometry. Similarly left plate of C2 has to acquire to the potential +P. If this happens then the voltage of C1 and C2 will be same V and net voltage of the two capacitors will become 2V which is wrong.
If someone could clarify this confusion it would be a big help.
| This is an important point which is commonly overlooked. The crux is that the voltage drop on each capacitor is not necessarily the batteries voltage. Let's work out the ideas step by step.
Firstly, to find the charge on capacitor system let us take equivalent capacitance. When we replace the two with an equivalent one, all voltage will drop on that, hence the voltage of that capacitor will be same as the battery.
Finding equivalent capacitance:
$$ C_{eq} = \frac{C_1 C_2}{C_1 + C_2}$$
Now, we can find the charge, which is basically by saying whole voltage is dropped on the equivalent capacitor:
$$ V \frac{C_1C_2}{C_1 + C_2} = Q$$
Note that charge on equivalent capacitor would be same on each individual capacitor which made it, let us find voltage on each capacitor starting with $C_1$ via the equation $ V = \frac{Q}{C}$
$$ V_1 =\frac{ \frac{VC_1 C_2}{C_1 + C_2}}{C_1} = \frac{VC_2}{C_1 + C_2}$$
Similarly,
$$ V_2 = \frac{VC_1}{C_1 + C_2}$$
We find that each doesn't drop the same amount of voltage, and after charging the charge built up is such that $V_1 +V_2=V$.
P.S: This idea also apply in series resistor, suppose two resistor in series connected to battery. Try applying the 'incorrect logic' to that, you will find the same error.
| {
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What is a simple example illustrating the concept of "commensurate" and "incommensurate" order in condensed matter physics? In a wide range of contexts in condensed matter physics, e.g this paper, the concepts of commensurate and incommensurate orders are invoked to describe particular ordered phases. I think I have some intuition for what this means - that somehow lattice translation symmetry is inconsistent with the structure of the ordered phase - but I would like to refine this notion. A clear formal definition of these terms would also be useful.
| I am very new at this and have seen these two terms come up in various topics within CM physics so the answer depends on what you are talking about. I'll provide an answer in different words than what was stated in the previous answer.
Commensurate means integer multiple of the crystal lattice constant.
Incommensurate means not an integer multiple of the crystal lattice constant.
Near-commensurate is, of course, somewhere in between.
Example:
A commensurate charge density wave has a wavelength that is exactly an integer multiple of the lattice constant.
See the second paragraph in Charge Density Wave Wikipedia
| {
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Newton's Laws of Motion, Pulleys, Rope and tension I was solving some questions to apply my concepts, and I came across the atwood machine and pulley block problems.
Consider the following for example:
The pulley is massless and frictionless, string, too, is ideal.
Why does the book say that the tension in the green string is $2T$ if the tensions in the two wings of the lower string is $T$ and $T$. Like if we see closely the strings only apply the normal force on the pulley how is it equal to two times tension $T$.
Also if the pulley would have been having mass and friction props. (string is mass less but with friction) would the tensions in the lower string be the same throughout? And what about the upper string?
And what happens if everything is non ideal?
| If pulley had mass $M_{pulley}$ and it is in equilibrium in the frame in which you are working, then assume that tension in string is $T_{g}$
$$\sum \vec{F}_{net,pulley}=0$$
$$T_{g}-T-T-M_{pulley}g=0$$
$$T_{g}=M_{pulley}g+2T$$
If pulley is ideal, then $M_{pulley}=0$, so, $T_{g}=2T$.
If there is friction in pulley and strings have mass then scenario becomes quite different, Pulley will rotate with some angular acceleration due to torque about its axis as well as Tension won't be same throughout the string.
| {
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What does $\rm kg m^{-2}$ mean in simple terms? I'm reading a research paper on fish stocking density and they measure the density in terms of $\rm kg m^{-2}$ (i.e. stocking density is 1.27 $\rm kg m^{-2}$). I've googled it but still don't have a good grasp on what a kg m is. Is kg m just an alternate way of writing kg/m or does it mean something else? How does the -2 exponent factor in?
| it is density of kg per unit area.
m^-2= 1/m^2
aka kg/m^2
| {
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What is a simple way to slow down this cart? What can I do to slow the wheels down and make this 4 wheel cart harder to push? I'd prefer a fairly straight forward solution. It will be used for exercise. I am very naïve when it comes to physics.
| We do not know how much slow down you need, but I can propose a scalable method over a wide range.
I expect it may not be practical for your case, but it could be.
You could make the wheels roll on a layer of a sticky and highly viscous fluid.
You could vary the viscosity, but also the layer depth.
Drive through honey...
The fluid could be honey, for example. You can roll on a surface just wet with honey (or something similar), or in a 3 mm layer, or in 30 mm.
The sugar solution can vary from pure water to saturated solution, and could contain sugar crystals to increase viscosity. It is easy to imagine that it has different effects on the movement.
It changes it's viscosity and its saturation point with the temperature.
This scales from no braking with pure water to quite a lot with 20 mm viscous honey.
... or inverted sugar syrup
But if that is not enough, we need a fluid with higher Viscosity: You can use inverted sugar syrup instead of the sugar solution. That is a mixture or solution of glucose and fructose instead of the sucrose used before.
It has, depending on temperature and moisture, a viscosity from like honey to basically solid - still flowing, but too slow to see it.
That covers the scale from "no slow down at all" continuously down to "apparently stuck in place".
| {
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How does the Bohm interpretation/pilot wave theory explain particles getting spontaneously created and destroyed? Pilot wave theory says that classical particles are riding on waves.
Quantum field theory says that particles are the excitations of a field.
Both of these descriptions seem like essentially the same thing, if we equate pilot waves with quantum fields. What are the differences?
Also, the quantum field description allows particles to dissolve back into the field, and also to be spontaneously created out of the field. This has been observed in experiments. How does the pilot wave description explain this?
|
if we equate pilot waves with quantum fields. What are the differences?
The fields of quantum field theory are not associated with a specific particle, they are generic over space and time, like a coordinate system , on which creation and annihilation operators operate to create a given particle. The pilot waves are connected with the given particle. In addition the theory is not Lorenz invariant, whereas field theory is by construction.
Also, the quantum field description allows particles to dissolve back into the field,
this is a misunderstanding, the fields are invariant, nothing can dissolve into them,
and also to be spontaneously created out of the field.
Energy must be supplied to be able to create a pair of particle anti-partcle
This has been observed in experiments.
Only with incoming energy . See this entry.
How does the pilot wave description explain this?
Pair production needs par excelence special relativity , and pilot theory is not good at special relativity. There is still research on this though. If they succeed, the connection with QFT though will not be as simple as you envisage.
| {
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Using Christoffel symbols to derive formulas for div, grad, curl In Sean Carroll's GR book, pg. 1oo, it was said that in flat space, the Christoffel symbols vanish in Cartesian coordinates. However, in curvilinear coordinates, they do not vanish. For example, for plane polar coordinates
$$ds^2=dr^2+r^2 d\theta^2$$
the Christoffel symbols are
$$\Gamma^r_{rr}=\Gamma^r_{r\theta}=\Gamma^{\theta}_{rr}=\Gamma^\theta_{\theta\theta}=0$$
$$\Gamma^r_{\theta\theta}=-r$$
$$\Gamma^\theta_{r\theta}=\frac{1}{r}$$
The author then said that using these and similar expressions, we can derive formulas for the div, grad and curl in curilinear coordinate systems. How exactly is it done?
| In an arbitrary coordinate system on a manifold, the grad, div and curl are defined using covariant derivatives:
$$\nabla\phi=(\nabla_a\phi)\textbf{e}^a$$
$$\nabla \cdot \textbf{v}=\nabla_av^a$$
$$(\text{curl} \textbf{v})_{ab}= \nabla_av_b -\nabla_bv_a$$
where $\phi$ is a scalar field, $\textbf{v}$ is a vector field and $(\text{curl} \textbf{v})_{ab}$ is a rank 2 tensor.
Since covariant derivatives of tensors $\phi$ and $v^a$ can be expressed using partial derivatives and Christoffel symbols, i.e.
$$\nabla_a\phi=\partial_a\phi$$
$$\nabla_a v^a= \partial_av^a +{\Gamma^a}_{ca}v^c$$
we say that the formulas for grad, div and curl in an arbitrary coordinate system can be derived using Chirstoffel symbols.
| {
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Creating a Bouyancy Force Generator How much energy is needed to maintain boiling water? (I'm assuming $1400~\rm W$). With that said, when air is added to an object to decrease its density, does the strength of its buoyant force increase? Can it be enough to create/maintain $1400~\rm W$?
| No.
At a temperature of $100^\circ~\rm C$ and a pressure of $1~\rm atm$, the heat of vaporization is $2257~\rm kJ/kg$. The more power you put in, the faster the water boils. It also generally decreases with higher temperatures and increases with higher pressures.
If this water vapor is used to displace water in an object, then yes, its buoyant force increases. But the work done by this extra buoyant force is always less than the energy required to vaporize the water, so you can't use this as any sort of perpetual motion device. If you put in $1400~\rm W$, whatever you get out will be less than $1400~\rm W$.
| {
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Is the many-worlds interpretation really just an interpretation? Is the many-worlds interpretation just a different interpretation to quantum mechanics or does it contain some different predictions?
In other words, is it possible theoretically to conduct an experiment that checks the many-worlds interpretation?
| Actually, there is one experiment documented that could just prove this. It is based on proving that communication exist between the proposed parallel worlds. They basically isolate a particle in an ion trap. Then they make a quantum measurement on another system (with two discrete outcomes), creating two parallel worlds (branching). Depending on the measurement, the ion is excited (before it decoheres) only from one of the parallel worlds. A detection of this excitation is evidence for the MWI.
The key to this experiment is time and decoherence. The ion must be trapped long enough so that the effects from the parallel worlds can affect it before decoherence.
I will show in section 4 that these single ions are isolated from the environment
to such a degree that the decoherence timescale is on the order of seconds or longer with
existing technical ion-trap equipment. Moreover it is possible to excite these atoms before
they are correlated with the environment to such a degree that complete decoherence took
place. In our example above Silvia1 switches on the microwave emitter long enough to
excite an ion in a trap with a large probability. After that, Silvia2 measures the state
of the ion and finds that it is excited with some finite probability, though Silvia verified
it was in the ground state before the branching took place. From that Silvia2 infers the
existence of Silvia1. In an obvious way Silvia1 and 2 can exchange informations (bit strings
of arbitrary length), e.g. by preparing more than one isolated ion. Single ions in traps can
act as \gateway states" and communication between parallel worlds is possible.
https://cds.cern.ch/record/289177/files/9510007.pdf
So the ion must be isolated from the environment long enough (because any information leak to the environment causes decoherence) so that they can detect any effects from the parallel worlds.
So the answer to your question is that yes, theoretically it is possible to conduct an experiment. The MWI is a unique one, because it is the only one that actually deals with many existing worlds so if this could be proved then it would change the way we think about QM.
| {
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Mesoscopic Bose-Einstein Condensate Bose-Einstein condensates of molecules of a few daltons have been already created, so I was wondering: would making a Bose-Einstein condensate on a system of Quantum Dots, due to their properties, cause the system to display any different effects?
| Bose-einstein condensate is a state of a multiparticle boson system. It is not clear in which sense QDs can be viewed as particles. Otherwise, condensates of cold atoms and excitons are mesoscopic. So are superfluidity and superconductivity - since they are characterized by a macroscopic order parameter.
One quantum-dot-like possibility is the coherence between different superconducting islands, see, e.g., this paper.
Yet, another option is polaritonic condensates, achievable in periodic semiconductor structures where optical modes are coupled to electronic transitions. This certainly has been done with the arrays of quantum dots as well, but I cannot provide an exact reference.
| {
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How do I find the approximate surface area of a chicken? I'm working on building a chicken army and I'm trying to find out how much metal or kevlar (still deciding) I need to make armor for the chickens. this measurement does not need to be exact I'm just trying to get an estimate for how much I will need. You will be spared when my chickens take over the world if you give me a working answer.
| Assume a spherical chicken, of some radius $r$. The surface area of said bird would then be $4\pi r^2$, which would be a good approximation of the amount of metal required to cover it completely.
If you assume $r\approx 30\text{ cm}$, that would come to an area of around $$A \approx 1\text{m}^2/\text{chicken}.$$
If the metal has a thickness $h$, this would be a total volume of roughly $V = 4\pi r^2 h.$ If $h\approx 5\text{mm}$, that means $$V \approx 5 \times 10^{-3} \text{m}^3/\text{chicken}.$$
How much would such an armour weigh? Unless you have some sort of "super-chicken" (though I immediately regret giving you the idea), you would require the weight to not be too substantial. If you are using some sort of metal (say, Aluminium or Steel) its density would be around $\sim 5 \times 10^3 \text{ kg/m}^3$. As a result, the mass of a metallic armour would be around: $$M = \rho \times V\approx 25 \text{ kg}/\text{chicken}.$$
Pretty heavy for the average chicken, I would think. Using Kevlar would reduce the mass by about a factor of $5$ (its density is around $1000 \text{ kg/m}^3$), so that would make it a slightly more bearable $5 \text{ kg}$ per chicken.
P.S: Come the revolution, my house will be the one flying the chicken flag. Inform the troops.
| {
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Understanding a Poynting vector equation I'm reading this section in the Griffiths Introduction to Electrodynamics book.
I trying to understand where equation 9.57 comes from (the middle part of the equation at least; I see where the $cu\;\hat{\mathbf{z}}$ part on the right comes from).
Does it come directly from calculating $\frac{1}{\mu_0}\mathbf{E} \times \mathbf{B}$ for a wave travelling in the $ \hat{\mathbf{z}}$ direction?
| A monochromatic plane wave in $z$-direction has the fields
$$\begin{align}
\mathbf{E}&=E_0\cos(kz-\omega t+\delta)\ \hat{\mathbf{x}} \\
\mathbf{B}&=\frac{1}{c}E_0\cos(kz-\omega t+\delta)\ \hat{\mathbf{y}}
\end{align}$$
I guess you find these a few pages earlier in your book.
Or you can check that these fields actually satisfy all Maxwell's equations.
Plugging these fields $\mathbf{E}$ and $\mathbf{B}$
into the definition of the Poynting vector $\mathbf{S}$ (9.56) you get
(also using $\hat{\mathbf{x}}\times\hat{\mathbf{y}}=\hat{\mathbf{z}}$ and
$\frac{1}{c\mu_0}=c\epsilon_0$)
$$\begin{align}
\mathbf{S} &= \frac{1}{\mu_0}\mathbf{E}\times\mathbf{B} \\
&= \frac{1}{\mu_0} E_0\cos(kz-\omega t+\delta)\ \hat{\mathbf{x}}
\times \frac{1}{c}E_0\cos(kz-\omega t+\delta)\ \hat{\mathbf{y}} \\
&= c\epsilon_0 E_0^2 \cos^2(kz-\omega t+\delta)\ \hat{\mathbf{z}}
\end{align}$$
Here you recognize the term $\epsilon_0 E_0^2 \cos^2(kz-\omega t+\delta)$
as the energy density $u$ from (9.55).
So you finally get
$$\mathbf{S} = c u \ \hat{\mathbf{z}}$$
| {
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Is no acceleration a cause or consequence of no net force? If a body is moving with constant velocity, or is at rest, then the net force on it must be $0$. If the net force on a body is $0$, then it must be moving with constant velocity or must be at rest.
Is $0$ net force a consequence of being at rest or moving with constant velocity or is moving at constant velocity or being at rest a consequence of $0$ net force?
| It may be a matter of opinion, but based on Newton’s first law. I would lean towards the latter.
This is Newton’s first law. It states if a body is at rest or moving at constant speed in a straight line it will remain at rest or continue move in a straight line unless acted upon by a net force. The law begins with the state of motion of a body.
If it said a net force is only considered to exist if there is a change in motion of a body, I might lean towards the former.
The law is due to the inertia of a body which is a property of its mass. I don’t believe the law is based on the property of forces.
Hope this helps.
| {
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Normalization of squeezed vacuum for QFT in the black hole background (Hawking radiation) In the context of BH evaporation, one can represent an initial vacuum as the final state containing particles, following Polchcinski (2016, page 9)
$$|0\rangle_{a}=\mathcal{N} \exp \left(\int_{0}^{\infty} \frac{d \omega}{2 \pi} e^{-\omega / 2 T_{\mathrm{H}}} b_{\omega}^{\dagger} \tilde{b}_{\omega}^{\dagger}\right) $$
where $\mathcal{N}$ is the normalization constant.
On the other hand, one can read in Lectures on quantum gravity edited by Gomberoff, Marolf (Springer) that the condition for normalizability is
the average total number of excitations must be finite. If it is not, the state $\left|0_{i n}\right\rangle$ does not lie in the Fock space built on the the state $\left|0_{\text {out }}\right\rangle$
We know that in the case of BH evaporation the number of excitations is infinite.
My question is: is that state actually normalizable, if yes, how to do this?
|
is that state actually normalizable...?
Yes. To see this, start with the state $U|0\rangle$ with
$$
\newcommand{\ra}{\rangle}
U =
\exp\left(
\int_0^\infty\frac{d\omega}{2\pi}\
f(\omega)\left(b_\omega\tilde b_\omega
- b_\omega^\dagger\tilde b_\omega^\dagger\right)
\right)
\tag{1}
$$
for some real-valued function $f(\omega)$. The operator $U$ is manifestly unitary, so the norm of $U|0\ra$ is the same as the norm of $|0\ra$. Use equation (3.7.50) in ref 1 to get the identity
$$
U|0\ra = A|0\ra
\tag{2}
$$
with
$$
A =
\mathcal{N}
\exp\left(
-\int_0^\infty\frac{d\omega}{2\pi}\
\tanh\big(f(\omega)\big) b_\omega^\dagger\tilde b_\omega^\dagger
\right)
\tag{3}
$$
and
$$
\mathcal{N}
=
\exp\left(
-\int_0^\infty\frac{d\omega}{2\pi}\
\log\Big(\cosh\big(f(\omega)\big)\Big)
\right).
\tag{4}
$$
The function $f(\omega)$ can be chosen so that the state $A|0\ra$ is the state shown in the question, namely
$$
-\tanh\big(f(\omega)\big) = e^{-\omega/2T}.
\tag{5}
$$
The only thing left to check is that (4) is finite. To see that it is, use the fact that for large $\omega$, equation (5) implies
$$
-f(\omega)\approx e^{-\omega/2T}\ll 1,
\tag{6}
$$
which in turn implies
$$
\log\Big(\cosh\big(f(\omega)\big)\Big)
\sim
e^{-\omega/T},
\tag{7}
$$
so the integral (4) converges. If we're concerned that the manipulations leading to the identity (2) might not be well-defined, then we can use a regulator, like taking $\omega$ to be a discrete parameter and deferring the continuum limit until the end of the calculation, after all desired inner products have been calculated.
Reference:
*
*Chapter 3 (Operators and States) in Barnett and Radmore (2002), Methods in Theoretical Quantum Optics (https://copilot.caltech.edu/documents/16975/acprof-9780198563617-chapter-3.pdf)
| {
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Which is the centripetal term here? In spherical coordinates the acceleration can be written as
$$\textbf{a} = \dot{\textbf{v}} = \ddot{r} \hat{\textbf{r}} + \dot{r} ( \dot{θ} \boldsymbol{\hat{\theta}} + \sin θ \dot{\phi} \boldsymbol{\hat{\phi}}) + \dot{r} \dot{θ} \boldsymbol{\hat{\theta}} + r \ddot{\theta} \boldsymbol{\hat{\theta}} + r \dot{θ} ( \cos θ \dot{\phi} \boldsymbol{\hat{\phi}} - \dot{θ} \hat{\textbf{r}} ) + \dot{r} \sin θ \dot{\phi} \boldsymbol{\hat{\phi}} + r \cos θ \dot{θ} \dot{\phi} \boldsymbol{\hat{\phi}} + r \sin θ \ddot{\phi} \boldsymbol{\hat{\phi}} + r \sin θ \dot{\phi} ( - \sin θ \dot{\phi} \hat{\textbf{r}} - \cos θ \dot{\phi} \boldsymbol{\hat{\theta}}) $$
and from this we have the radial component of acceleration
$$\ddot{r} - r \dot{θ}^2 - r \sin^2 θ \dot{\phi}^2$$
Do we call any of the above terms as centripetal acceleration? If so why?
| The two negative terms combine to give a resultant, -r$ω^2$.
| {
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How would velocity of sound, the fundamental frequency and wavelength of sound vary when the temperature of an organ pipe is increased? Here is my approach to this:
neglecting any thermal expansion of the pipe:
By the Laplace formula for the speed of sound,
$V=\sqrt{\frac{\gamma P}{\rho}}$ where P is the pressure, $\gamma$ is the adiabatic constant and $\rho$ is the density of the medium.
Assuming the gas to be an ideal gas, we can use the ideal gas equation.
Hence we have:
$V=\sqrt{\frac{\gamma RT}{M}}$ where R is the gas constant, T is the absolute temperature, and M is the molar mass of air
So, when we increase the temperature, clearly, the velocity would increase as well. Coming to the fundamental frequency, we know
$f_0$ (fundamental frequency) $\alpha$ V (velocity of sound)
Hence, the fundamental frequency would also increase.
But how would we get the variance of the wavelength with temperature? I thought of using the relation
$V=f\lambda$ where V is the velocity of sound, f is the frequency and $\lambda$ is the wavelength.
According to which, the wavelength should also increase but according to my book, that's not right. Why does this happen?
| HINT: Sound waves in an organ pipe are examples of standing waves. The allowed wavelengths of a standing wave on (for example) a string are not arbitrary, but are instead determined by the length of the string.
| {
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Force on the bottom of a tank full of liquid - Hydrostatic Pressure or Gravity
Imagine a tank filled with water that has some height $h$ and at the bottom area $A$ but as it goes up, for example at height $h/2$, it's area is now $A/2 $. What's the correct way to calculate the force at the bottom of the tank? (Let's ignore atmospheric pressure for now)
*
*If I use $W=mg$, we get $F=W=ρVg=ρ(\frac{Ah}{2}+\frac{Ah}{4})g=\frac{3}{4}ρghA$
*If I calculate the hydrostatic pressure at the bottom, it's $p=ρgh$, and then $F=pA=ρghA.$
Which one is the correct one and why?
| I would like to point out that the question you encountered poorly defines how the cross-sectional area is related to the height at which it is being observed.
Is it like:
*
*Area(x) = (x/h)*A, x being the distance from the bottom, h being the total height, A being the area at the bottom.
*The way parsa639 describes it.
Either way im not sure how you calculated the volume of the tank. Could you please explain your method.
I performed an integration to find the volume while assuming area to be a function of x as described in (1.) and got the volume to be (Ah)/2 so the weight would be (ρAhg)/2. Something else that comes into play in such a scenario is the force the sloped walls of the container apply on the liquid. This
conversation covers it well.
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Understanding renormalizability I want to understand when a given theory is renormalizable and how to find renormalizable theories for different dimensions (the latter will become clearer later on).
To do so, we work through an example: the following real scalar field theory in $d$ spacetime dimensions
\begin{equation*}
\mathcal{L} = \frac 1 2 \partial_{\mu} \phi \partial^{\mu} \phi - \frac 1 2 m^2 \phi^2 - \frac{\lambda_3}{3!} \phi^3 - \frac{\lambda_4}{4!} \phi^4 - \frac{\lambda_5}{5!} \phi^5 - \frac{\lambda_6}{6!} \phi^6
\end{equation*}
Let us work with $3 \leq n \leq 6$. We first determine the coupling constants
\begin{align*}
[\mathcal{L}] = d, \quad [\partial_{\mu}]=1, \quad [m] = 1,\quad \Rightarrow \quad& [\phi^n] = \frac 1 2 n(d-2), \quad [\lambda_n] = d- \frac 1 2 n(d-2) \\
&[\phi] = \frac 1 2 (d-2), \quad [\phi^2] = d-2
\end{align*}
I learned that a theory with $[\lambda_n] < 0$ is nonrenormalizable (from M. Srednicki's beautiful book, chapter 18). Hence, from $[\lambda_n] = d- \frac 1 2 n(d-2)$, we get the following condition
\begin{equation} [\lambda_n]<0 \iff n> \frac{2d}{d-2} \tag{*} \end{equation}
From $(*)$ we see that the renormalizability of the theory depends on the powers of $\phi$ and the dimension $d$. In particular, we see that our theory is renormalizable if we work with $d = 1$ and $d=3$. For $d > 3$ our theory becomes nonrenormalizable
Let me ask some questions
*
*How to check whether our theory is renormalizable when $d=2$ or not? Our formula $n> \frac{2d}{d-2}$ breaks down for such a case.
*Let's say we want to find renormalizable theories for higher dimensions. Based on $(*)$, for $d=4$ we see we would need to drop $\phi^5$ and $\phi^6$ terms. For $d=5,6$ we see we would need to drop $\phi^4$, $\phi^5$ and $\phi^6$. For $d=7,8,9,10,..,10^6,...$ (I did not check further than $d=10^6$ but it seems that the relation holds for further dimensions) we see we would need to drop $\phi^3$, $\phi^4$, $\phi^5$ and $\phi^6$. Does this mean that the simplest scalar field theory $\mathcal{L} = \frac 1 2 \partial_{\mu} \phi \partial^{\mu} \phi - \frac 1 2 m^2 \phi^2$ is renormalizable for all dimensions?
PS: Please note that I am aimed at understanding under which conditions is a theory renormalizable and not solving this one in particular.
| Your statement (*) is valid only for $d > 2$ (you have to take care when dividing an inequality by a non-positive number). Based on the original formula for $[\lambda_n]$, renormalizability holds for all $n$ if $d \le 2$.
You are correct that the "simplest" (free) scalar field theory is renormalizable for all $d$.
| {
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Why is the constant of integration zero when solving the Friedmann equations? I'm confused regarding relating the Hubble constant H$_0$ to current age of the universe $t_0$.
I'm looking at solutions for different epochs of the Friedmann equation, which I've got in the general form
\begin{align}
\frac{a}{\sqrt{\Omega_va^4+\Omega_ca^2+\Omega_ma+\Omega_r}}\frac{\mathrm da}{\mathrm dt}=\mathrm H_0.
\end{align}
Take for instance the radiation-only, $\Omega_r=1$,
\begin{align}
a\frac{\mathrm da}{\mathrm dt}=\mathrm H_0,
\end{align}
if I were coming at this from a more naive perspective, I would take my initial condition to be $a(t_0)=1$,
\begin{align}
\int_1^aa'\mathrm da'=\int_{t_0}^t\mathrm H_0\mathrm dt',
\end{align}
giving
\begin{align}
a(t)=\sqrt{1+2\mathrm H_0(t-t_0)}.
\end{align}
This solve the equation, fits the initial condition, and is unit-less, but differs from the usually stated solution
\begin{align}
a(t)=\sqrt{\frac{t}{t_0}}.
\end{align}
I've figured people arrive at this via some method similar to the following. Either (1), use indefinite integration and take the constant to be zero,
\begin{align}
\int a\mathrm da=\int \mathrm H_0\mathrm dt\\
a=\sqrt{2\mathrm H_0t},
\end{align}
then use the assumption that $a(t_0)=1$ to find a relationship between H$_0$ and $t_0$,
\begin{align}
a(t_0)=1=\sqrt{2\mathrm H_0t_0}\\
2\mathrm H_0=\frac{1}{t_0},
\end{align}
giving the usual solution $a(t)=\sqrt{t/t_0}$.
Or (2), use definite integration and take $a(0)=0$ to get the same result.
What confuses me is that I cannot get H$_0=1/2t_0$ (which reduces the equation I get to the usually stated) using the definition H$_0=\frac{\dot a_0}{a_0}$ or $a(t_0)=1$. I'm not convinced that I should just take the integration constant to be zero, and I'm even more not convinced of using $a(0)=0$, as I believe that solutions are not guaranteed to be unique for $a=0$ given the Picard-Lindelof theorem.
Thanks for any help.
| I'm not sure where the confusion is; this seems straightforward.
An arbitrary solution to $a\dot a= H_0$ for constant $H_0$ satisfies
$$
a^2(t_1)-a^2(t_2)=2 H_0 (t_1-t_2)\qquad (\ast)$$
for any $t_1,t_2$. For fixed $H_0$, $a^2(t)$ is uniquely specified by the single initial condition at some time $t'$, as 1st-order ODEs usually are. From this you take a square root to obtain $a(t)$, which could get you in trouble if $a(t)$ goes on to change sign, but this is beside the point here.
You could, instead of fixing $H_0$, specify $two$ initial conditions. If these are at times $t_1,t_2$ then clearly
$$H_0=\frac{1}{2(t_1-t_2)} (a^2(t_1)-a^2(t_0))\,.$$
Then specifying the initial conditions
$$
a(0)=0\,,\quad a(t_0)=1
$$
gives $H_0=1/(2t_0)$ and $a^2(t)=t/t_0\implies a(t)=\sqrt{t/t_0}$.
| {
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What do angle have to do with waves? Why do we use trigonometric functions to model waves? I've just learned SHM and I'm not able to understand why "phi" (initial phase) is an angle. Why are angles used in SHM?
| The solution to the differential equation of a harmonic oscillator involves rotation.
If you specify your position and velocity as the real and complex components of a complex number $z$, then the solution of the differential equation is $z k^t$ where $k$ is some complex number, which is just rotation and scaling.
If you specify the position and velocity as components of a vector $x$, then you have a matrix $K$ for which the solution is $K^t x$ where $K$ is a scaled, rotation matrix.
You might be interested in this fantastic video in which he solves the differential equation of a harmonic oscillator in order to motivate matrix exponentiation.
| {
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Standard boost for one-particle states in Weinberg's QFT chapter 2 Weinberg's approach to QFT starts with particles which are not necessarily more fundamental than fields but are known more for certain. A particle of a particular species can have different momenta $p$ in different frames but its internal degrees of freedom (like electron spin) is described by a discrete quantum number $\sigma$, and the one-particle state is denoted by $\Phi_{p,\sigma}$. A standard boost is defined to relate one-particle states with different momenta but the same quantum number of internal degrees of freedom $\Phi_{p,\sigma} = U(L(p))\Phi_{k,\sigma}$, where $k$ is a chosen standard momentum ($(m,0,0,0)$ for particles with mass $m$) and $L(p)$ is a standard boost that takes $k$ to $p$. With the non-compact boost components of the Lorentz group trivialized in this way, essentially particles are defined as finite-dimensional unitary representation of the little group, that is, the stabilizer of $k$.
My question is about the choice of the standard boost $L(p)$. Is there any physical consideration (it doesn't change the quantum number of internal degrees of freedom) that determines $L(p)$? Different choices of $L(p)$ differ by an element in the stabilizer of $p$ (which is $SO(3)$), the standard boost for one choice is not standard in another, hence mixed quantum numbers of internal degrees of freedom, does it matter for physics? Perhaps it's just redefinition of quantum number of internal degrees of freedom?
| I had some difficulty understanding a fair amount of chapter 2 because Weinberg’s book on QFOT (which you appear to be describing) sometimes skips some of the logic (like what are the “sigma’ degrees of freedom). I managed to find a set of videos that helped me a lot - much is filled in to Weinberg’s terse presentation. Here is the link:
https://freevideolectures.com/course/3090/relativistic-quantum-mechanics/22
| {
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Angular momentum commutation relations The operator $L^2$ commutes with each of the operators $L_x$, $L_y$ and $L_z$, yet $L_x$, $L_y$ and $L_z$ do not commute with each other.
From linear algebra, we know that if two hermitian operators commute, they admit complete sets of common/simultaneous eigenfunctions. The way I understand this statement is that the eigenfunctions of both operators are the same. So, if that were the case, that would mean that $L_x$ has the same eigenfunctions as $L^2$. The same goes for $L_y$ and $L_z$. That would mean that $L_x$, $L_y$ and $L_z$ all have the same eigenfunctions, which doesn't seem to be true since they do not commute with each other. How is this resolved?
| Commuting operators do not necessarily share ALL eigenstates, just some set.
An eigenstate shared by $L^2$ and $L_x$ will not be the same as that shared by $L^2$ and $L_y$
| {
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What is light, a wave or a particle or A wave-particle? What is light?
And how do we know that light is an electromagnetic wave?
I asked my teacher and he said that when you place a compass in light's path, the needle of the compass rotates. Which I think is not a valid answer and thats not what actually happens when we place a compass in path of a light.
| Your instructor is wrong, and you are right. Placing a compass needle in a light beam will not cause the needle to turn, the way it certainly will if you place that compass near a magnet or a piece of wire carrying an electrical current. This is because light beams do not generate electrical current flow or magnetic fields in space, to which a compass could possibly react, because the fields associated with light beams are changing far too rapidly and are too weak. The situation is more complicated, and goes something like this:
Light was thought to be distinct from the realm of electricity and magnets until the 1860's when Maxwell wrote down his four equations which described all the (known) interactions between electricity and magnets. Maxwell then noted that his equations contained the possibility of waves consisting of coupled electric and magnetic fields, and that by manipulating his equations he could actually solve for the speed at which those waves would travel through space.
He then discovered that the speed of these "electromagnetic waves" was exactly the speed of light, which was known fairly accurately at that time. This proved that visible light was fundamentally electromagnetic in nature, and it represented just a tiny slice of the whole electromagnetic wavelength spectrum- encompassing radio, microwaves, infrared, visible, ultraviolet, x-ray, and gamma radiation.
But the flow of light, consisting of photons of electromagnetic energy, is not deflected by electric or magnetic fields, and its passage does not create electric or magnetic fields either that a compass could detect.
This story and all the background details I have omitted here will be found in any high-school level physics textbook, of which there are literally hundreds of good examples available. A college-level text will contain all the math and lots more detail.
| {
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Bosonic coherent state normalization In the paper arXiv:1208.3469 (equation (85)) it is stated that the coherent state for two bosons with corresponding annihilation operators $a,b$ can be written as:
$$|\Psi_\lambda \rangle = \sqrt{1-|\lambda|^2} e^{-\lambda a^\dagger b^\dagger}|0\rangle,$$
where $|0\rangle$ is the vacuum.
I tried to check that this state is indeed normalised and tried to derive equation (86) in the paper but failed. My attempt is the following:
\begin{align}
\langle \Psi_\lambda |\Psi_\lambda \rangle
& = (1-|\lambda|^2)\langle 0|e^{-\lambda^\star ab} e^{-\lambda a^\dagger b^\dagger}|0\rangle
\\ & = (1-|\lambda|)^2 \sum_{m,n=0}^\infty \langle 0|a^n (a^\dagger)^m b^m (b^\dagger)^n |0\rangle (-\lambda)^n (-\lambda^\star)^m.
\end{align}
Then we see that only cases with $m=n$ survive and that each term simply gives $(n!)^2$ since e.g. $\langle 0|a^n (a^\dagger)^n|0\rangle = n!$.
What is wrong in my reasoning here?
| You forgot a factor of $1/n!$ in your exponential expansions, and then forgot to use the geometric series. The correct set of steps are:
$$\langle \Psi_\lambda |\Psi_\lambda \rangle = (1-|\lambda|^2)\langle 0|e^{-\lambda^\star ab} e^{-\lambda a^\dagger b^\dagger}|0\rangle = (1-|\lambda|)^2 \sum_{m,n=0}^\infty \frac{\langle 0|a^n (a^\dagger)^m b^m (b^\dagger)^n |0\rangle}{n! m!} (-\lambda)^n (-\lambda^\star)^m.$$
And then as you point out, the expectation values are $\frac{\langle 0|a^n (a^\dagger)^m b^m (b^\dagger)^n |0\rangle} = \delta_{mn} n! m!$, and from there you get
$$
\cdots = (1-|\lambda|)^2 \sum_{m,n=0}^\infty |\lambda|^{2n} = 1.
$$
where the geometric series says that $\sum_{n=0}^\infty r^n = \frac{1}{1-r}$ for $r$ in the unit ball.
| {
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What is the strongest result that has subsequently been shown to just be a statistical glitch? While reading up on the $g-2$ result, I noticed that the authors stopped short of announcing a definitive result because they only had 4.2 sigma of confidence rather than the 5 that is the standard for confirmation. That got me thinking: Has anyone ever published results anything close to 5 sigma and then had it turn out to be a statistical glitch? (As opposed to for example a systemic error in construction or execution of the experiment.)
Edit:
I ran into a mention of a 3.9 sigma result at the LHC that turned out to be a statistical glitch
| There are lots of unexplained >5-sigma results out there, and the only way establish that one of them was a statistical glitch would be to prove that it had no systematic errors. This is impossible. There is a well-known aphorism that "all models are wrong", and all real measurements have inherent systematic errors. We can work hard to keep their effects small, but they can never be eliminated. When a 5-sigma result is reported that later goes away without explanation, it is always far more likely that it was an unknown systematic effect than a statistical fluctuation.
One can, however, make a statistical argument that 5-sigma statistical glitches have been published. Several million scientific articles appear every year and the total number of scientific papers ever published is probably of order $10^8$. Not all scientific papers have experimental results, of course, but many report more than one result, so $10^8$ is a conservative guesstimate for a lower bound on the total number of published scientific measurements. Given that the (2-sided) probability of a 5-sigma fluctuation is about one in two million, one would expect quite a few 5-sigma statistical fluctuations to have appeared in the scientific literature. It is just impossible to distinguish them from the more common 5-sigma unknown systematic errors.
| {
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According to general relativity planets and Sun bend the spacetime (explaining gravity), but does this hold true for smaller objects? According to general relativity planets and the sun bend spacetime, and that is the explanation of gravity. However, does this hold true for smaller objects, like toys, pens, etc.? Do they also bend spacetime?
| As explained in the other answers, yes. Its so hard to tell because, compared to the other forces, gravity is very, very, very, very, very, very, very, very, very, very weak. Only when a whole lot of matter gets together does it become significant.
For example, the Earth is 6,000,000,000,000,000,000,000,000 kg of matter, yet we can overcome its gravity with an almost effortless jump. So can a good magnet using electromagnetism.
We don't know why gravity is so weak, it remains a great mystery.
| {
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Why is there a volumetric dimension to pressure-volume work? This concept was a bit hard to grasp, because I'm bad at seeing the real-world implications of multiplication. Division makes sense, but multiplication is harder for me. Pressure-volume work is measured in $1$ Pa $\times$ m$^3$. It says that $1$ Pa $\times$ m$^3 = 1$ J.
Does that mean that no matter what the volume of the thermodynamic system is, if it exerts $1$ Pa onto its surroundings, its doing $1$ Joule of work? Again, I'm not very learned in science, math and thermodynamics, but this looks like $1$ Pa $\times$ $1$ m$^3 = 1$ J. Is this another way of saying:
$x$ Pa $\times$ $1$ m$^3 = x$ J
OR
$1$ Pa $\times$ $x$ m$^3 = x$ J
Basically saying that either the volumetric quantity or the pressure quantity has a $1/1$ relation with the energy of the work (J), meaning the other quantity is equal to $1$.
|
Basically saying that either the volumetric quantity or the pressure quantity has a 1/1 relation with the energy of the work (J), meaning the other quantity is equal to 1.
The definition of mechanical pressure-volume work is the integral of the external pressure times the differential change in volume of the system $w = -\int p_{ext} dV$. In SI, the units can be Pa m$^3$. The units can also be lbf ft from (lbf/ft$^2$)(ft$^3$) or Torr L or N m from (N/m$^2$)(m$^3$). These are all unit designations for mechanical work.
The expression 1 J = 1 Pa $\times$ 1 m$^3$ is a unit conversion expression. The unit on the left is a derived unit for energy. The units on the right consist of a derived unit for pressure (Pa = 1 N/m$^2$ = kg/m s$^2$) and a fundamental unit for length m$^3$.
The relationship is true regardless of whether it is used to convert the units of mechanical expansion/compression work to energy or whether it is used to convert the $pV$ component in the ideal gas law to an energy to compare with the $nRT$ term (that has the same units) or even whether it is used to convert $pV$ in the definition of enthalpy $H \equiv U + pV$ from pressure times volume to energy units.
The relationship is not saying that one or the other quantity is equal to 1. How can you appreciate this? Consider that all of these conversions give the same value of 1 J.
$$ 0.5 Pa \times 2 m^3$$
$$ 1 Pa \times 1 m^3$$
$$ 2 Pa \times 0.5 m^3$$
The mistake is perhaps to believe that multiplication of units in a unit conversion factor can be "inverted" to get a numerical value of one or the other factors. You cannot and must not go down that path. The better approach in calculations is to take three steps. First, calculate the number using the equation. Second, determine and validate the units on the number by unit analysis through the equation. Finally, convert the calculated units as needed to other forms.
| {
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Why do images not appear inverted when looking directly through a pinhole camera? I understand that the way light takes through a pinhole creates an inverted image on a surface behind the pinhole. I remember this effect from school experiments, it's also described in this wikipedia article. I punctured a piece of paper and looked through it (instead of watching the reflection), the image appeared as normal to me. Why is that? Why doesn't the scene appear upside down when looking through the hole?
| Has the OP just reduced the size of his pupil by adding another aperture (the pinhole) in front of his eye?
A smaller aperture of the will make the scene seem dimmer and if the OP requires corrective lenses their vision will improve as the "stopped down" aperture improves depth of field.
| {
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Hyperfine Splitting What does Hyperfine Splitting exactly mean?
I am having a bit of trouble understanding the concept of hyperfine splitting.
I solved the problem of including hyperfine interaction for the ground state of Hydrogen and I found the eigenvalues. However, I can't understand the physical meaning.
What I think...
Once we include the hyperfine interaction for the ground state of Hydrogen, we find three states with energy $A$ and one state with energy $-3A$. Does it mean that the ground state is not just a single state but is split in four states?
| The ground of hydrogen atom in the Coulumb potential is
$$
E_0 = - 13.6057 eV.
$$
Its wavefunction
$$
\psi(\vec{r}) = \mathbb{N} \exp\left(-\frac{r}{a_B}\right).\,\,\,\text{ where } a_B = 0.529177 \dot A
$$
The degeneracy of the ground state is two electron spin states: $\vert \uparrow \rangle$ , and $\vert \downarrow \rangle$
The above description is our common knowledge about the ground state of $H(1S)$. But the above saying omitted the states of proton in the nuleus of hydrogen. There, the proton has spin $I = \frac{1}{2}$. Taking the proton's spin into account, the degeneracy of $H(1s)$ is $4$:
$$
\vert S_3 I_3 \rangle = \vert \uparrow \uparrow \rangle;\,\vert \uparrow \downarrow \rangle;\,\vert \downarrow \uparrow \rangle;\,\vert \downarrow \downarrow \rangle
$$
Therefore, is we consider the interaction between the electron spin and the proton spin (the hyperfine interaction) into consideration, $S\cdot I$. Using additive of angular momentum
$$
\vert S= \frac{1}{2}\rangle \oplus \vert I =\frac{1}{2}\rangle \Longrightarrow j= 0\,; \,\,\, 1.
$$
The singlet state $j=0$ gives one single state, and the triplet state $j=1$ gives three degenerstate:
\begin{align}
\vert j=0, m \rangle = & \,\, \vert 0, 0 \rangle; \,\,\Longrightarrow \Delta E = -3A\\
\vert j=1, m \rangle = & \,\,\vert 1, 1 \rangle, \,\vert 1, 0 \rangle,\,\vert 1, -1 \rangle\,\,\Longrightarrow \Delta E = +A
\end{align}
The pertubation of hyperfine interaction conserves the total sum of energy, therefore as the triplet states rise an amount of energy $A$ to $13.6057+A$, the singlet lower the energy by $3A$ to $13.657 - 3A$. The value of $A$ is very small, $A \approx 6\times 10^{-6} eV$. The splitting can only by observed by ESR (electron spin resonance, a microwave absorption technique).
| {
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Is it possible to stir the tea so that it cools more slowly? I recently learned that stirring the tea with vertical movements speeds up the cooling process. As a result, I had a question, and whether it is possible to stir it so that it cools down, as slowly as possible. Conditions: the spoon is an aluminum tea spoon, take its shape as a rectangle, the temperature of the tea is 363K, normal environmental conditions.
| There are more factors involved here.
Stirring in a way than causes turbulence does not have the same effect as stirring in circles, creating a vortex. Moreover, heat loss due to evaporation and surface radiation may or may not exceed heat loss due to wall contact, depending on environmental conditions and the choice of material the cup is made of.
In general, convection itself does not cool the tea down at all, unless it can make it flow out of the cup. Convection merely facilitates radiation, evaporation and conduction. In this sense, making the tea less temperature homogeneous by stirring the hotter tea away from the surfaces, may have an insulating effect. Vortex flow being known to concentrate heat in its center, like in a vortex tube, supports the possibility of this. I have however no knowledge of experiments in this respect.
.
| {
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Do all Joukowski aerofoils violate no-penetration condition at trailing edge? In our fluids course we calculated the velocity distribution around a completely symmetric Joukowski aerofoil (as shown below) and used the Kutta condition to ensure that the velocity was not infinite at the trailing edge. However, even after applying this condition, the trailing edge had $v_y = 0$ but $v_x \neq 0$ which violates the no-penetration boundary condition.
The lecture notes alluded to this being due to our 'especially symmetric choice of aerofoil', but I can't find online the velocity distributions of aerofoils that aren't symmetric over the x-axis, but I feel like they too may have a 'non-infinite but still violating no-penetration condition' trailing edge.
My question is, is this violation just an artefact of the Joukowski aerofoils all having a cusp at the trailing edge? Or is it truly just an artefact of this one particular symmetrical Joukowski aerofoil?
| As @alephzero and @Chiral Anomaly said, the Joukowski mapping produces a cusp at the trailing edge of the aerofoil which doesnt have a defined normal to the surface, therefore it's less that the no-penetration condition is violated there, but that a direction normal to the surface from which to apply the no-penetration condition doesn't even exist.
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If a polarized light wave is indistinguishable from its original self after being flipped 180°, why doesn't a photon have a spin of two? The spin of a photon has a counterpart in classical physics, it's polarization, right?
And if you spin a polarized light wave by 180°, (or pi radians), it is now the same as before, correct?
So why isn't the spin of a photon said to be two, rather than one?
EDIT: To Michael Seifert and Anna V: I deeply appreciate the answer(s) and comment(s) on this and other questions of mine, etc. ... But....
HOW does the phase shift of 180° (or 1-pi radians) manifest itself in 'observable(s)'? How can one distinguish a '-i' photon from one that was twisted by 1-pi radians in the opposite way (now i, rather than -i)?
| If you rotate a polarized classical electromagnetic wave by 180° about its axis of propagation, you don't get the same wave back; you get the same wave phase-shifted by half a cycle. You need to make a full rotation of 360° to get the same classical wave solution back, so by your logic, the photon should indeed have spin 1.
This is contrast to a gravitational wave, for which a rotation of 180° about the propagation direction yields exactly the same solution with no phase shift. This manifests on the quantum level through the graviton being a spin-2 particle.
| {
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"timestamp": "2023-03-29T00:00:00",
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What motivated Einstein to formulate general relativity? I never really fully understood what motivated general relativity or why the Newtonian concept of gravity was considered problematic.
One thing I always hear is that it is because it doesn't address what causes two masses at a distance to be attracted. Maybe quantum mechanics aside, nobody seemed to have a problem with two electrons at a distance repelling each other.
One thing I read is that mass is performing double-duty being both a resistance to a change in velocity and a cause of gravitational fields. I can see that being a curiosity but not convinced that it is a problem.
Ok, gravitational field changes propagating instantly might be a reasonable motivation.
Mercury's orbit may be a problem solved by general relativity but I don't see it being the motivation to come up with the theory. One would only know it solves the problem after the theory is fully developed.
I have never studied general relativity at any sort mathematical level and I've always been curious about this.
| Einstein came up with General Relativity because he recognized that Special Relativity was not a good description of the universe (he couldn't make gravitation work sensibly within it).
He came up with Special Relativity because Newtonian physics had failed to accommodate light in any useful or consistent way.
| {
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How can you measure the polarization of a single-photon without destroying it? In Quantum Key Distribution, namely the BB84 protocol, Alice creates single-photons in the horizontal/vertical basis or the (45 degree rotated) diagonal basis which she then sends to Bob over a transmission line.
Now if an eavesdropper Eve that has the ability to measure the polarization of the photons is intercepting this transmission line, she can try to figure out the secret key that Alice is trying to send to Bob. But how can Eve measure the polarization of a single-photon without destroying it?
In videos like this https://youtu.be/uiiaAJ3c6dM?t=171 it looks like the polarization filters are able to measure the polarization of the photons without destroying them. But is that feasible in the real world? I mean if I use a vertical polarization filter to measure a horizontally polarized photon, that photon will be absorbed/destroyed by my filter.
| In general, if you want to know the polarization of light experimentally, you need to make a series of measurements in different bases/directions using a different filter each time. Each measurement indeed will destroy a single photon, so in order to fully know the polarization of light you need to measure an ensemble/collection of photons. I think that this is what the linked video is trying to explain: Eve has to choose one basis in which to measure the single photon (and then create a new photon of the measured polarization to send to Bob). However, if Eve chose the wrong basis in which to measure the photon's polarization, she will send Bob a photon of the wrong polarization. The video describes, however, that Alice can send a classical message to Bob to tell him which basis to measure the photon's polarization in.
| {
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Analysis of a state vector $\,|\psi\rangle\,$ in the basis of eigenvectors of a $4\times 4$ Hamiltonian matrix I have the following matrix
\begin{equation}
A=
\begin{pmatrix}
0 &1 &0 &0 \\
1 & 0 &0 &0\\
0 &0&0&1\\
0&0&1&0
\end{pmatrix}
\end{equation}
The eigenvalues of this matrix is $\left\{1,1,-1,-1\right\}$. We set
\begin{equation}
|\psi\rangle=\alpha|++\rangle + \beta |+-\rangle +\gamma |-+\rangle +\delta|--\rangle
\end{equation}
To determine the constants $\left\{\alpha,\beta,\gamma,\delta\right\}$, we apply the equation to the eigenvalues for the positive energy firstly as
\begin{equation}
\begin{pmatrix}
0 &1 &0 &0 \\
1 & 0 &0 &0\\
0 &0&0&1\\
0&0&1&0
\end{pmatrix}\begin{pmatrix}
\alpha \\
\beta\\
\gamma\\
\delta
\end{pmatrix}=\begin{pmatrix}
\alpha \\
\beta\\
\gamma\\
\delta
\end{pmatrix}
\end{equation}
which implies that
$\beta=\alpha$, $\delta=\gamma$. The normalization condition gives
\begin{equation}
\langle \psi \lvert\psi \rangle=1 \longrightarrow 2\rvert\alpha\rvert^{2} + 2\rvert\gamma\rvert^{2}=1
\end{equation}
My problem is that: I did note know the method used to find the relation between $\alpha$ et $\gamma$ and to do the rest of the calculation.
| You have found the eigenvectors in the 1-eigenspace to be
$$
|\psi \rangle = \alpha
\begin{pmatrix}
1\\
1\\
0\\
0
\end{pmatrix}
+ \gamma
\begin{pmatrix}
0\\
0\\
1\\
1
\end{pmatrix} = \alpha |v_1\rangle + \gamma|v_2\rangle,
$$
where $\alpha$ and $\gamma$ are any complex numbers constrained by $|\alpha|^2 + |\gamma|^2 = \frac{1}{2}$. This means that this eigenspace is two-dimensional, and any linear combination of the basis vectors $|v_1\rangle$ and $|v_2\rangle$ is an eigenstate of your matrix $A$ with eigenvalue $1$. In other words, you are completely free to choose whichever linear combination in the subspace you want, and you have to provide an extra constraint to fix the answer to whatever question you have.
| {
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Why is the half-wave dipole the most used antenna design? When producing em waves using a dipole antenna (of length L), you could theoretically use any L and adjust the frequency of the oscillating voltage to get the desired wavelength. Then why are most antennas half a wavelength long? I'd also like to know why it's useful in the context of receiving em waves. Thanks.
| An antenna, in general, is NOT a resonator (contra @oliver). The purpose of the antenna is to be a direction dependent impedance transformer that matches the wave impedance of the transmission line connecting the transmitter/receiver to space (air) having impedance $\sqrt{\frac{\mu_0}{\epsilon_0}}=377\Omega$ in all direction. In other words, the apparent impedance driving the antenna in the desired direction is matched to the line; waves that are to be in or from undesired direction be completely mismatched (reflected). Being resonant can help with impedance matching be achieved at a given frequency/direction. If wider bandwidth than that of a simple resonant curve is required then different radiators need be employed, for example, a horn that is being an aperture radiator can easier be made wider bandwidth.
It happens that a "half-wave dipole* has about $70\Omega$ input impedance for radiation in its "mid-plane" in which it radiates uniformly, and $70\Omega$ is a convenient number to match to $50\Omega - 75\Omega$ the impedance level of most RF equipment. If the length of the antenna is longer relative to its wavelength then the radiation breaks up into several narrower lobes with two practical consequences, (1) matching gets more difficult (2) a complicated antenna shape is difficult to maintain because it becomes more sensitive to environmental effects.
| {
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What is the proof of $C_{V} = \frac{fR}{2}$? I came across this formula in thermodynamics. Please give me a rigorous proof to this formula. My teacher did not even give any proof neither do any of my books. The formula is :
$C_{V}=\frac{fR}{2}$ where $C_{V}$ is the molar heat capacity at constant volume, $f$ is the total number of degrees of freedom .
$$f=f_{\rm translational}+f_{\rm rotational}+f_{\rm vibrational}$$
And $R$ is the universal gas constant.
| This is the so called equipartition theorem, that says that each degree of freedom contributes with $\frac{k_BT} {2} $ to the average energy, or using mol $\frac{RT} {2} $.
More precisely, each degree of freedom in the Hamiltonian that is quadratic contributes for $\frac{k_BT} {2} $.
If we can write the energy:$ E = \sum_n \alpha_n x^2_n$ like a sum of quadratic degrees of freedom, we can compute the average, using Boltzmann distribution:
$$ \langle E \rangle =\frac{\int E \exp(-\beta E) d^3 x_1...d^3 x_N}{\int \exp(-\beta E) d^3 x_1...d^3 x_N} =
\frac{\int \sum_n \alpha_n x^2_n\exp(-\beta \sum_n \alpha_n x^2_n) d^3 x_1...d^3 x_N} {\int \exp(-\beta \sum_n \alpha_n x^2_n) d^3 x_1...d^3 x_N}
$$
We can factorize, using the property of the exponential and Gaussian integrals:
$$ \frac{\int \alpha x^2\exp(-\beta \alpha x^2) d^3 x} {\int \exp(-\beta \alpha x^2) d^3 x} = \frac{\frac{\alpha \sqrt{\pi} }{2{(\alpha \beta})^{\frac{3}{2} } }} {\frac{\sqrt{\pi}}{{(\alpha \beta})^{\frac{1}{2} } }} = \frac{1} { 2 \beta} $$
Knowing that: $\beta = \frac{1} {k_B T} $, we have:
$$ \langle E \rangle = N f \frac{k_B T} {2} $$
Where $N$ is the number of molecules and $f$ is the number of degrees of freedom per molecule.
Or writing this for a mole of gas: $\langle E \rangle = f \frac{R T} {2}$, where $R$ is the constant of gas.
Now, we know that the heat capacity at constant volume is: $C_V = \frac{d\langle E \rangle}{dT}$ then we have:
$$ C_V = f \frac{R} {2} $$
For a perfect monoatomic gas, where we neglect interactions, we have only the kinetic term:
$$ E = \sum_n^N \frac{p^2_n}{2m}$$
Where the sum is over the number of molecules. Using the equipartition theorem we have: $\langle E \rangle = N(\frac{3}{2})k_B T$ or $\langle E \rangle = \frac{3} {2}RT$. The factor of $3N$ arise from the number of molecules and the fact that we have three dimensional space, then we have three degrees of freedom from the momentum of each molecule. Clearly the degrees of freedom include translational, rotational and vibrational terms, depending on the system.
| {
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Commutation relations inconsistent with constraints In section $9.5$ of Weinberg's Lectures on Quantum Mechanics, he uses an example to explain the clasification of constraints. The Lagrangian for a non-relativistic particle that is constrained to remain on a surface described by
$$f(\vec x)=0\tag{1}$$
can be taken as
$$L=\frac 12 m \dot{\vec x}^2-V(\vec x)+\lambda f(\vec x).\tag{2}$$
Apart from the primary constraint $(1)$ there is also a secondary one, arising from the imposition that $(1)$ is satisfied during the dynamics
$$\dot {\vec x}\cdot\vec \nabla f(\vec x)=0.\tag{3}$$
Then he states that imposing $[x_i,p_j]=i\hbar\delta_{ij}$ would be inconsistent with the constraints $(1)$ and $(3)$ (which reads $\vec{p}\cdot\vec{\nabla}f=0$ in the Hamiltonian formalism). How can I see this inconsistency?
| Would an example suffice? If so, consider the case $f(\vec x) = x_1$. Then (1) says $x_1=0$, which is already inconsistent with the commutation relation, and (3) says $p_1=0$, which is again inconsistent with the commutation relation. If $x_1$ or $p_1$ is zero, then we can't have $[x_1,p_1]\neq 0$.
| {
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Heisenberg uncertainty principle at relativistic velocities Would the Heisenberg uncertainty principle, the energy of h-bar in particular, be affected by the Lorentz factor at relativistic velocities from an external pov?
If a rocket were to speed by at relativistic velocities and if we could see inside it, or if we could view the environment close to a black hole, would quantum effects be more pronounced there?
That is to say, would the magnitude of h-bar increase in a similar way that the Lorentz factor increases the magnitude of momentum and kinetic energy at relativistic velocities?
My thought process stems from H bar = 1.054571817×10−34 joules x sec, yet mass and energy are equivalent. So if the Lorentz factor can intensify mass and energy shouldn't H bar be affected as well?
| Planck constant $h= 6.62607015\times 10^{−34}$ Js (i.e. Joules times second and not Joules per second) in SI units. See here. As such it has the units of action. Although energy transforms as a result of Lorentz transformations, the Planck constant does not. There may be some interesting effects of special relativity (or general relativity) on the uncertainty principles, but as far as we know this does not include a variation in the value of $h$.
| {
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Can a body float in the middle of a fluid? Let's say we have a cubic body of side $a$ and made of a material with density $\rho$ and we measure its immersed height in a fluid of density $\rho_f$ by the variable $y$. Then, its potential energy (and considering a gain of potential due to buoyancy) can be written as:
$V = -Mgy + \frac{\rho_f}{2}a^2y^2g$
To find the system equilibrium points, one can derivate the previous expression in order to y, obtaining:
$\begin{equation}
\frac{\partial V}{\partial z} = 0 \Longleftrightarrow \rho_f a^2gy_{eq} = Mg \Longleftrightarrow y_{eq}=\frac{Mg}{\rho_f a^2 g} = \frac{\rho a^3 g}{\rho_f a^2 g} = \frac{\rho}{\rho_f}a
\end{equation}$
Which leads to something that I don't know how to explain. Having $\rho > \rho_f$, one will obtain that the body floats mid-water. How is this even possible if, theoretically with the equations obtained, there isn't any change on the fluid's density with depth?
| An object can remain at a fixed depth, fully submerged as long as the buoyant force equals the weight. This will generally require monitoring and adjustment. A submarine can adjust the amount of compressed air in a tank which admits outside water. A scuba diver can adjust the amount of air in an inflatable vest.
| {
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Why are constant volume and constant pressure heat capacities basically the same for solids? Are degrees of freedom involved? I knowv that $C_V=\frac{\frac{f}{2} Nk_B}{m}$ and $C_P=\frac{(\frac{f}{2} +1)Nk_B}{m}$. Since for solids their values are very close to each other, I would assume $\frac{f}{2} +1$ is very close to $\frac{f}{2}$. Namely, I thought I would need to have $\frac{f}{2} >> 1$. This would require a high number of degrees of freedom. However, is this the case? We are talking about a solid, of which the particles can move in less directions then, let's say, a gas molecule. Or is the very constraint that makes the number of degrees of freedom get very high? And is this the case or is there another explanation for why $C_V$ and $C_P$ have close values for solids?
| Although changing the temperature of a solid under constant volume conditions can cause extremely high pressures, the corresponding volume changes under constant pressure conditions are very small, so the amount of work that a solid can do while expanding when heated, is also very small. It is this work that makes the difference between $C_P$ and $C_V$, which explains why they are not that much different.
To put it colloquially: by heating a metal block, you can probably lift a whole skyscraper by a 1mm, but you aren't able to design a lift by this principle, even if it should only carry a few people from the lobby to the first floor.
| {
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In special relativity, how do we know that distance doesn't change in the direction perpendicular to velocity? In the theory of special relativity it is said that the distance in the direction of the speed changes by a factor of
$$\gamma=\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$$
How do we know that the distance perpendicular to the velocity doesn't change?
| We can simply derive this from the Lorentz transform. With the $x$ direction parallel and the $y$, $z$ directions perpendicular the Lorentz transform is: $$t'=\gamma \left(t-\frac{vx}{c^2} \right)$$ $$x'=\gamma (x - vt)$$ $$ y' = y$$ $$z' = z$$ For an object with a length $L$ in the $y$ direction at rest in the unprimed frame we have the worldlines of the ends as $y_0 = y$ and $y_L= y+L$ so the distance between the ends is $y_L-y_0=L$.
To obtain the distance in the primed frame we simply transform $y'_0=y=y'$ and $y'_L=y+L=y'+L$ so the distance is $y'_L-y'_0=L$.
Therefore the distance in the perpendicular direction does not contract.
| {
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How to find the falling time of an object when acceleration is not a constant? Let's say we are throwing an object from the surface of the earth, this object reaches 70,000km with initial velocity of $10713 \mathrm{m}/\mathrm{s}$ until it reaches the peak high , the g value at 70,000km is $0.068 \mathrm{m}/\mathrm{s}^2$.
Now my question is: How to calculate the falling time of the object from the peak height (70,000km) until it reaches the ground? Acceleration is not constant here.
| I will only outline an answer for now.
The object is on a suborbital trajectory. Classically this just means the objects or it is a 2 body gravitational orbit like earth and moon except the periapsis is lower altitude then the surface of the earth. Describing orbital characteristics is the same.
Now there is usually no time parameterization of the orbit. One usually has to use an approximation method like newton's method, to get the actual angle (true anomaly) from the circle angle (mean anomaly) for times ellapsed past an epoch (known position). This is due a interesting term which involves sin function, and there is no way to algebraically solve.
So basically, object experience known not necessarily constant acceleration is simple. Object orbiting from gravitational field is not so simple. It gets into transcendental functions, non linear pdes etc. That's classical gravity and differential equations for you.
| {
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How can parallel rays meet at infinity? I found that in every book (till my 12th) it is written that, in concave mirror, when object is at focus, then reflected rays will be parallel and they meet at infinity to form a real image.
But, as we know, parallel rays never meet. Then, does this mean that all books are wrong ? If not, then why?
| The books are just suggesting that as the object distance approaches the focus (from outside of the focus), the image distance will approach infinity.
| {
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Does Einstein's Equivalence Principle ignore time dilation? It seems Einstein's equivalence principle is neglecting time dilation. If an observer is at rest in an inertial reference frame, free of any gravitation, she will experience time flow at the "native" rate of a universe empty of mass and energy. However, an observer in free fall at the surface of the earth will experience time flow at a rate determined by earth's gravitation, which is slower than the native time flow rate. This seems to imply non-equivalence?
| Not only is the equivalence principle not ignoring time dilation, you can actually use the equivalence principle to derive gravitational time dilation.
The issue that you are running into is probably the most common issue in applying the equivalence principle. That is that the equivalence principle is strictly local. It cannot be applied over a region of spacetime which is large enough for tidal effects (spacetime curvature) to be noticed. The comparison of the free faller with some distant clock is a non-local comparison and the equivalence principle makes no claim about the outcome.
| {
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Mathematical proof of charging by induction If we bring a positive charge +Q near a neutral conductor , we know that the surface near the source gets-Q and opposite to it gets +Q, but why do these induced charges have to be equal in magnitude to source charge, why isn't a charge distribution such as -7Q on surface near the source and +7Q on the opposite surface, not possible ? Can we show the result mathematically?
| In the equilibrium situation any free charge inside the conductor has no prefered direction to travel.
If a charge Q is near the conductor, and 7Q of opposing changes at its surface, clearly a free charge would have a prefered direction.
As the are plenty of free charges in a conductor, eventually (and very quick indeed) the charges redistribute in order to reach the equilibrium.
| {
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Distinguishing between static and dynamic pressure in fluids I read about the difference between static and dynamic pressure in fluids in the mechanics textbook (part 2) by BM Sharma. It explained me by using a Pitot tube and an ordinary pipe attached to the same main pipe. I applied Bernoulli's theorem at point $A$ and at the datum line:
$$P_1+(1/2)(\rho)v^2 +0 = P_0 + 0 + (\rho)g(h_s).$$
The heights $h_s, h_d$, the datum line and point $A$ and $B$ are indicated in the image.
Similarly, I applied the equation for point $B$:
$$P_1 + (1/2)(\rho)v^2 + 0 = P_0 + 0 + (\rho)gh_d.$$
Thus, I obtained the result that the heights $h_s$ and $h_d$ are identical.
Please explain where my analysis is wrong. Please explain by applying Bernoulli's theorem at point $A$ and the datum level.
| $$\underline{\textit{Analysis}}$$
Let $P_A=P_0$, $P_B=P_0$ and $P'_A$, $P'_B$ denote the static pressures at the points $A$, $B$, and their vertical projections $A'$, $B'$, on the datum line respectively. Let $v_A=0$, $v_B=0$ and $v'_A$, $v'_B=0$ denote the horizontal velocity components at the points $A$, $B$, and their vertical projections $A'$, $B'$, on the datum line respectively. We assume that the fluid is incompressible.
In order to analyze the static pressure differences between points in the two columns, we apply Pascal's law. In order to analyze the dynamic pressure differences between two points at two points on the streamline coinciding with the datum line, we apply the Bernoulli's equation. The Pascal's law implies that $$P'_A=P_A+\rho g h_s=P_0+\rho g h_s,$$ $$P'_B=P_B+\rho g h_d=P_0+\frac{1}{2}\rho g h_d,$$ so that $$0<\rho g (h_d-h_s)=P'_B-P'_A,$$ which is the quantitative difference between the static and dynamic pressure heads in the flow. The Bernoulli's equation applied to the to the streamline $A'B'$ implies yields $$P_A'+\frac{1}{2}\rho (v'_A)^2=P_B'+\frac{1}{2}\rho (v'_B)^2$$ so that $$0<P'_B-P'_A=\frac{1}{2}\rho((v'_A)^2-(v'_B)^2)=\frac{1}{2}\rho(v'_A)^2,$$ wherein the quantity on the right hand side of the above expression defined as the dynamic pressure of the flow. Therefore the dynamic pressure head is estimated (on neglecting head losses) by measuring the difference $h_d-h_s$ as $$P_\text{dynamic}:=\frac{1}{2}\rho(v'_A)^2=\rho g (h_d-h_s)=P_B-P_A.$$
$$\underline{\textit{Explanation}}$$
The Bernoulli's equation reads $P+\frac{1}{2}\rho v^2 + \rho g h=constant$ at all points along a streamline in a steady, incompressible and inviscid flow. The variable $P$ corresponds to the physical quantity of static pressure at the associated point, while the expressions $\frac{1}{2}\rho v^2, \rho g h$ are referred to as the dynamic pressure and pressure head due to gravity head at the associated point, respectively. The critical error in the analysis presented in the OP is that the Bernoulli's equation is not applied at points along streamlines, but perpendicular to them. Specifically, the Bernoulli's equation cannot be applied to relate the flow properties across the points $A$ and $A'$ as in the OP, since there is no streamline connecting the points $A$ and $A'$ (in others words, there is no flow occurring across those points since the 'flow' or fluid in the column $AA'$ is static).
| {
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Problem understanding what densities of states represent I understand that the one particle partition function for a particle in a box can be written as: $$Z_1 = \sum_{k_x} \sum_{k_y} \sum_{k_z} (2s+1) e^{- \beta \epsilon( \vec k) } $$
My first question is that I'm not sure why we can also write it in the following way: $$Z_1 = \int_0^{\infty} g(\epsilon) e^{-\beta \epsilon} d \epsilon$$
I don't understand what the the quantity $g (\epsilon) d\epsilon $ or the multiplicity of states in energy range $\epsilon + d \epsilon $ means physically. I'm trying to see what happens when I integrate: $$ \int g( \epsilon) d \epsilon $$ For various functions but I'm getting confused. I also don't see how from this function alone I can get expressions for things like the Fermi energy for $N$ spin $1/2$ Fermions in a volume $V$ or the Debeye frequency. Any help clarifying this would be great!
| Maybe it would help to look at a simple finite-dimensional example. Consider the hamiltonian
\begin{align}
H = \begin{pmatrix}
\varepsilon & 0 &0\\
0 & 3\varepsilon &0\\
0 & 0 & 3\varepsilon
\end{pmatrix}
\end{align}
The partition function in the canonical ensemble is
\begin{align}
Z &= \text{Tr}\,e^{-\beta H}\\
&= e^{-\beta \varepsilon} + e^{-3\beta \varepsilon} + e^{-3\beta \varepsilon}\\
&= \sum_{i=1}^3e^{-\beta \varepsilon_i},\\
\end{align}
where the sum is over all energy eigenvalues. However, we could also write this as a sum over distinct energy eigenvalues
\begin{align}
Z &= e^{-\beta \varepsilon} + 2e^{-3\beta \varepsilon}\\
&= \sum_{i=1}^2g(\varepsilon_i) e^{-\beta \varepsilon_i},\\
\end{align}
with $g(\varepsilon_i)$ giving the degeneracy of the $i$th energy eigenvalue, i.e. $g(\varepsilon) = 1$, $g(3\varepsilon) = 2$. This latter expression is the same as your second equation with the integral, just with a finite sum instead of an integral. The integral is taken over all distinct values of $\epsilon$, and the $g(\epsilon)$ factor accounts for the degeneracy of energy eigenvalues at each $\epsilon$.
We need to include the density of states when finding, for example, the mean occupancy
\begin{align}
n = \int f(\epsilon) g(\epsilon)\,d\epsilon
\end{align}
to account for the fact that there may be multiple states available to occupy at each value of $\epsilon$.
The last integral you wrote down $\int g(\epsilon)\,d\epsilon$ just counts the total number of energy eigenstates in some range of energies.
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Why force between charges increases when it moves, instead of decreasing? Imagine two positive charges in a space ship moving with a velocity,v with respect to an observer on earth.
according to the person in the spaceship,the electrostatic force between the charges is $F'=(\frac{1}{4π\epsilon_0})\times \frac{q_1q_2}{r^2} $
But if you are the observer on earth,then the equation becomes $F=\gamma \times F' =\gamma\times(\frac{1}{4π\epsilon_0})\times \frac{q_1q_2}{r^2} $
So,the force is greater in the frame on earth.
But actually shouldn't the force be lesser because with respect to earth the charges are moving and hence have an inward magnetic attractive force which partially cancels the outwards repulsive force?
| A common class of error, in these sorts of problems, is getting the transformation backwards and putting $\gamma$ where you should have $1/\gamma$. That's what's happened to you here.
To see this intuitively, consider a pair of opposite charges instead of same-sign charges. Released from rest, the opposite charges will accelerate towards each other and collide after some finite time. An observer in a different frame will see this collision happen more slowly, due to time dilation.
How the moving observer explains this delayed collision is sensitive to the geometry of the problem. If the line between the charges is perpendicular to the velocity, as in your diagram, the moving observer believes there is a magnetic interaction between the charges which counteracts their electrical attraction. (A classic problem from Griffiths: consider parallel line charges, and find the speed at which the magnetic repulsion exactly cancels the electrical attraction.) If the line between your point charges were parallel to the velocity vector, there would be zero magnetic force between them, but the electrical attraction would have been reduced thanks to the Lorentz boost. At other angles, the boosted electrical force and the magnetic interaction conspire together so that your opposite charges collide after exactly the right dilated amount of time.
| {
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Why is the power of a car limited? As an introduction this is what's written in my textbook:
$Power=\frac{work\ done\ by\ driving\ force}{time}$
$=\frac{Driving\ force×distance}{time}$
(as distance over time = speed)
$= Driving\ force×speed$
Note: If the power is constant and the body is accelerating then the speed is changing thus the driving force must also change to keep the power constant.
And in most of the questions we solved in class we assume that the power (of a car) is constant and the "note" in the text I mentioned explains how it will be constant and I understand that.
What I don't understand is why will it be constant in the first place? Shouldn't the car exert the same force all the time— so the speed will increase increasing the distance— and thus the work done will increase as time passes and the power will keep increasing?
My teacher informed us that we "assume" that the power is constant and that it variates between a certain range.
• Yet why does the power have to be limited to a certain range? Why doesn't it increase forever?
Or at least increase until it comes to its maximum speed due to air resistance?
• Why does speed decrease when the driving force gets bigger even when there's no friction to prevent the car from speeding up forever?
|
Yet why does the power have to be limited to a certain range?
Aside from the power limits of car itself, as stated in @silverrahul answer, power is limited because there is an upper limit to the driving force on the car equal to the maximum possible static friction force between the tire(s) of the drive wheel(s) and the road, which is given by
$$F_{max}=\mu_{s}N$$
where $\mu_s$ is the coefficient of static friction between the tire(s) and the road and $N$ is normal reaction force of the road on the tire(s). Once $F_{max}$ is exceeded, the tires begin to slip preventing acceleration.
Neglecting air resistance, the static friction force is the only external force acting on the car. It acts forward and is equal and opposite to the backward force of the tire on the road due to the torque applied to the wheel(s), per Newton's third law.
Hope this helps.
| {
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If I pull a metal bar for long enough with a constant small force, will it eventually break? Let's say I have a strong metal bar. I pull it apart with a very small constant force -- obviously it doesn't break. However, this would disturb the internal configuration. If I let go, then eventually the internal configuration would return to what it was before I started pulling on the bar. However, if I keep pulling on the bar long enough, would the bar eventually break, no matter how small the force is?
| Lots and none at all.
Clearly not usefully, anyway.
Is it not obvious that if what you Ask were true, every screw or bolt holding any shelf would eventually break?
You might fall back on "if… long enough" meaning that the experiment should continue beyond the corrosion life of the screw (or bolt) and if you do, are there not endless examples of screws holding up shelves for all the hundreds of years shelves have been held up by metal screws rather than wooden pegs? (Come to that, why is the Question restricted to metal as opposed to wood, plastic or anything else?)
How is the Question different from "If I push on a load for long enough with a constant small force, will it eventually move?"
| {
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Is there an intuitive way to view the concept of momentum? Ideas like distance, velocity and force are very intuitive to understand because you can "see" their real-world applications and so one can come to understand them without having any knowledge of their mathematical formulas.
Momentum as it is defined is the product of mass times the velocity. I can see how it came to be mathematically derived via combining Newton's second law and the equation for acceleration (change in velocity over change in time).
However, besides just remembering that p=mv, is there an intuitive way to view what momentum is? Am I supposed to have an intuitive understanding of it? or is that just how every physics student thinks of it?
| Imagine two large iron rolling balls, one larger than the other. Imagine they are moving at the same speed.
Imagine trying and stopping them. Which would be easier? The lighter one, of course. And if the balls are the same size, their speed is what matters.
So basically momentum is the property of the object that makes it easier or harder to stop, i.e. the momentum of a body is what changes the force that must be applied to stop it.
This quantity is transferred during collisions, and of course we know that it is dependent on the mass and velocity ; $p=mv$ as you stated.
| {
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Is friction always opposite to velocity? Let's say an object is sliding on a slope and is the object has a velocity of $(0,0,5)$. The friction would be acting in the opposite direction of motion, being $(0,0,-1)$.
However, gravity is also affecting the object on the slope. The gravity is exerting a force of $(-1,-1,0)$ in the direction of the slope tangent. Should the object also experience friction in the direction of $(1,1,0)$? Or would it only experience friction in the direction of $(0,0,-1)$?
What confuses me is that if friction is only acting in the opposite direction of motion, when an object is standing still on a slope due to friction acting in a direction opposite to gravity, when you add a perpendicular impulse the object will suddenly be sliding downwards as well because friction is no longer acting in the direction opposite to gravity. Is that correct?
| A better phrasing is:
Friction always pulls in the direction that prevents sliding (often called relative motion).
Remember, there are two types of friction:
*
*Kinetic friction when the object slides. To prevent sliding (to stop sliding), kinetic friction pulls exactly opposite to the velocity. Regardless of any forces acting.
*Static friction when the object is stationary but there are forces trying to make it slide. To prevent sliding (to keep the object stationary), static friction pulls opposite to whichever direction the net force pulls.
Gravity is an external force. When standing still on flat ground, gravity doesn't try to initiate sliding, so there is no need for a static friction to act opposite to gravity. When standing on a slope, a component of gravity pulls along with this slope and tries to start sliding down - static friction thus must pull opposite to this, so up the slope. But not if I at the same time push on the object up along the slope - then static friction might have to pull downwards to prevent the object from sliding upwards.
So static friction is not related to gravity in general. Gravity is just one possible force to hold back against. There is no requirement for frictions to act opposite to gravity.
| {
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How to know the true value of the measured quantity from an interferometer phase? For an interferometer, the measured signal will oscillate as a function of the accumulated phase $\phi(x)$ as a sinusoid or cosine, where $x$ is a quantity we are trying to measure from the interference. The signal will have the same properties for all $\phi(x_2)=\phi(x_1)+2n\pi$ where n is an integer, my question regards how would we know from the signal, whether the quantity of interest $x$, is $x_1$ or $x_2$ considering we don't know $n$. Also, how would we be able to differentiate between negative and positive quantities of $x$?
For example, if the signal is due to a rotation $x\rightarrow\Omega$ of the frame of reference and the effect causing the phase accumulation the Sagnac effect $\phi(\Omega)=A\Omega$, where $A$ is just some constant we need not concern ourselves with for now. Then the signal output from the interferometer might go something like $S(\Omega)=C\cos(A\Omega)$, where $C$ is the contrast. Clearly, the signal is the same for $\pm\Omega$ and we can't tell the difference between large $\Omega$ and small $\Omega$ if the amplitude of the signal is the same for those two values of $\Omega$.
| This is a great example why physical theories are necessary a priori for interpretting the results of any experiment. One must first have an idea of the magnitude of the effect they are trying to measure before really being able to ascertain its value. When we are talking about a physical rotation, perhaps all that matters is the final orientation, so the extra $2\pi n$ may not matter. When we are talking about a relative phase between two oscillators, the extra $2\pi n$ does not matter.
But when we are talking about a cumulative effect, the total phase matters! So, if you want to measure a cumulative effect using an interferometer, you must already know within which $2\pi$ window you expect to find anything... otherwise you'll be out of luck. For something like gravitational wave detection, we might know that the signal is always rather small, so we might only expect to use $n=0$. For other cases, without a priori assumptions or knowledge, you are absolutely correct that an interferometer will only give you a definite answer modulo factors of $2\pi$.
| {
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Why 2nd Shell can have 8 electrons? I recently watched this video:https://youtu.be/INYZy6_HaQE
and understood why 1st orbital can have only 2 electrons: According to Pauli's exclusion principle, two electrons cannot have the same quantum states.In the first energy level,all other quantum states are same except for the spin. And spin can have two values $+\frac{1}{2}$and $-\frac{1}{2}$.So,there can be two electrons in the first shell.
I need a similar explanation for 2nd shell
In other words, why the 2nd shell can have 8 electrons?
I know that there are other quantum states. I am requesting to explain how many values each quantum state can take.For example spin can have two values.Likewise,how many values can the angular momentum can have?
How all these lead to the 8 electrons in the second shell?
(Don't make it simple,you can use maths.Iam really on to this)
|
In other words, why the 2nd shell can have 8 electrons?
I do not know just how deep and detailed of an answer you are looking for.
In very simple, not deep at all terms, 2nd shell can have 8 electrons, because 2nd shell has 4 orbitals, and each orbital can have 2 electrons.
1st shell has only 1 orbital, hence 1st shell can have 2 electrons
| {
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Do electromagnetic fields also have thermal fluctuations? As far as I know electromagnetic fields have vacuum fluctuations. But can electromagnetic fields also have thermal fluctuations and can they induce current in coils? If the answers are positive. How does one model a fluctuating electromagnetic field inside the core of coil and its induced current. Could one model this as an ideal coil with a parallel fluctuating current source?
| Not only can this happen, but what you are describing is essentially why you hear static on your radio or see it on your TV.
Thermal radiation (eg from the sun) follows a blackbody spectrum. Thermal radiation can interact with a resonant circuit consisting of an inductor (coil), as well as a capacitor, tuned to amplify a certain frequency. Since the phase of the thermal radiation is random, any signal coming out of this circuit will be noisy. You can model this as a random current in the circuit.
| {
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Does universal speed limit of information contradict the ability of a particle to pick a trajectory using Principle of Least Action? I'm doing some self reading on Lagrangian Mechanics and Special Relavivity. The following are two statements that seem to be taken as absolute fundamentals and yet I'm unable to reconcile one with the other.
*
*Principle of Least action states that the particle's trajectory under the influence of a potential is determined by minimising the action that is $\delta S =0$.
*The universal speed limit of information propagation is $c$.
Question: Consider a potential that exists throughout space (like gravity perhaps). Now, I give a particle $x,p$ and set it motion. As per principle of least action the motion is now fully deterministic since the Lagrangian $L$ is known. We arrive at this claim by assuming that the particle is able to "calculate" the trajectory such that $\delta S = 0$ using Principle of Least Action.
This is where my confusion is. The first principle is a global statement whereas the second principle is a local statement.
What is the modern day physics answer to this?
I hope I could get my questions across! Look forward to your valuable opinions!
| There are typically an infinite number of solutions to Lagrange equations, corresponding to one starting point and an infinity of end points. If the system starts out on one trajectory, its end point is determined. The system does not "know" in advance what its endpoint is, but the physicist who has advance knowledge of the beginning trajectory and the Lagrangian (which "knows" everything about what's between the starting point and all possible trajectories) can predict the rest of the trajectory and the endpoint.
| {
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Why don't evanescent waves give rise to electromagnetic waves? I'm reading about evanescent waves for the first time. I understand that even thought no electromagnetic wave is transmitted across the boundary, an electric field is transmitted which decays exponentially into the material.
As far as I understand this is still an oscillating electric field. Can anyone explain, therefore, why it doesn't generate a magnetic field and give rise to an electromagnetic wave?
Or is there an electromagnetic wave moving parallel to the boundary but people just talk about the E field?
| An evanescent wave is a superposition of the incident (penetrated) wave and the medium reaction to this incident wave (a radiated everywhere wave). The interference is "destructive" in the medium direction and "constructive in the opposite direction.
You must consider all sources of a wave to understand its behaviour.
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Expressing acceleration in terms of velocity and derivative of velocity with respect to position we know that
$$a = \dfrac{dv}{dt}$$
dividing numerator and denominator by $dx$, we get $$a=v\dfrac{dv}{dx}$$ provided that $dx$ is not equal to zero or instantaneous velocity not equal to zero
when I questioned my teacher that this formula implies instantaneous velocity should not be zero or their should be no turnaround points then why we use this formula for deriving equations of particle's position , velocity which is performing SHM, but got no satisfactory answer.
what is wrong in my argument and what are the conditions under which above mentioned equation is not true?
| What is wrong is assuming that dv/dx is finite when v=0. Try this for motion with uniform acceleration. For example, for object thrown upwards with some initial velocity. At the top turning point the expression fir dv/dx tends to infinity.
| {
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Coaxial Cable acts as differentiator In order to determine the characteristic impedance of a coaxial cable, I put a resistance on the one end and connected the other end of the cable to a signal generator and an oscilloscope.
This worked fine when using a sine signal as I just had to find the resistance, for which the returning signal was (almost) gone. However, I also wanted to try this with a square signal.
The result was first a positive sharp peak decreasing similar to an exponential function followed by a negative peak of the same form.
Why does this happen?
My first thought was a differentiator since the sine signal kept its form and the sides of the square signal resulted in sharp peaks with the constant middle part being changed to $0$ voltage.
After some reading I found out, that an RL-circuit can be looked at as an differentiator. So my assumption is that the coaxial cable's inductance makes the experiment a RL-circuit.
However, where do you draw the inductor into the circuit? My first guess would be to just add it in series with R:
Where V is the oscilloscope and the red parts are the coaxial cable.
This also produces the next question: I am measuring the voltage over R and the inductor while for the differentiator the voltage is taken over the inductor only:
I only made a sketch of the seen image on the oscilloscope:
The coaxial cable was $40\; \mathrm{m}$ long.
| The dominant characteristic of a piece of coax is its parallel capacitance per foot of length. This effect is significant and must be included in a model of the cable along with its series inductance (small) and series DC resistance (very small) per foot.
| {
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Velocity is relative, which means acceleration is relative, which further implies that forces are relative as well So how would we know whether a force truly exists or not. I can be accelerating a car my 5 meters per second squared but another car accelerating with the same acceleration would think that my car is at rest relative to them. So is there any force on the car? Or are forces just relative and their existence just depends on our reference frame?
|
I can be accelerating a car my 5 meters per second squared but another car accelerating with the same acceleration would think that my car is at rest relative to them
That is not true.
When we consider just velocity, it’s simple because there are no forces involved.
But in accelerated motion we have to consider pseudo forces. You both are being pushed in this case, due to the pseudo force. That force is same for both of you, but that doesn’t mean you won’t notice a force on the other.
Say, you and your friend are moving in your cars with same acceleration. But your friend has a pendulum in his car that is hanging. Due to pseudo force, the pendulum is pushed back, and it stays like that because of the constant force on it backwards.
Like this:
You look over, and notice that pendulum is acting weird. That’s how you prove, that even though both are experiencing same forces, they don’t cancel out. Instead both experience it.
| {
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Why $\frac{1}{E-H_0+i\eta}|n\rangle \stackrel{?}{=} \frac{1}{E-E_n+i\eta}|n\rangle$? A Hamiltonian $H_0$ is diagonalized in $\{|n\rangle\}$ i.e. $H_0|n\rangle=E_n|n\rangle$. Why can we write
$$\frac{1}{E-H_0+i\eta}|n\rangle \stackrel{?}{=} \frac{1}{E-E_n+i\eta}|n\rangle$$
$H_0$ is in denominator, how come we can apply $H_0$ directly on $|n\rangle$? Are there any situations where this is not true?
The actual problem I am struggling with (It would be great if anybody can give me some direction to proceed): I have a Bogoliubov (4-by-4) Hamiltonian $H_B$ which is diagonalized by an anti-unitary matrix $T$ i.e. $T^+\sigma T=\sigma$ where $\sigma=diag(1,1,-1,-1)$. And $T^+H_BT=E_k$ where $E_k$ is diagonal matrix with eigenvalues of $H_B$. Due to Bogoliubov form $H_B$ obey $H_B|T_n\rangle = (\sigma_{nn}E_{nn})\sigma|T_n\rangle$. I wonder if I can write expression
$$
\frac{1}{E-H_B+i\eta}|T_n\rangle \stackrel{?}{=} \frac{1}{E-\sigma_{nn}E_{nn}+i\eta}\sigma|T_n\rangle
$$
| Whenever you see weird manipulations of operators, like $\sqrt{\mathcal{O}}$, $\frac{1}{\mathcal{O}}$, or $\log{\mathcal{O}}$, they are always defined by just doing those operations to the eigenvalues. Now, if the operators aren't diagonalizable in an eigenbasis, then it is more complicated to define these operations, if they can be defined at all.
Let me just say that in the case of inverses, though, you're literally just inverting the matrix. So $\frac{1}{H - E I} = (H - E I)^{-1}$.
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Is there an alternative to radio waves that can go through metallic objects? Radar can pass through materials such as paper, wood, glass, brick, and concrete, but it reflects off of metal. Is there an alternative to radar that can pass through metal substances? If not, is it likely that we will ever find such a wave?
| As pointed out in another answer, ultrasound is an option for detection of objects behind sheet metal. Actually developing such a system is very difficult, of course.
The rear doors of the Tesla model X are wing doors (referred to by Tesla as 'falcon wing doors'). These wing doors are moved by electric motors, controlled via software interface. That means the doors have to be equipped with sensors to detect whether some obstacle is present. Elon Musk, the CEO of Tesla, describes that he was keen to avoid to have a detector/sensor visible on the outside of the door panel; he was keen to achieve a very clean look.
The Tesla engineers succeeded in developing ultrasonic sensing that operates through the aluminum door panel.
As attested in video footage: the model Y wing doors detect the presence of any obstacle, and avoid touching the detected obstacle when the user gives a command to open the doors.
Examples of what counts as obstacle: a person standing next to the door, another car, a low ceiling of the garage.
| {
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Physical interpretation of relativistic synchrotron radiation When an electron approaches the speed of light, the emission pattern is sharply collimated forward, in the direction of motion. I can see how this is mathematically true from taking the relativistic limit of the angular distribution of the radiated energy (as is done on the Wikipedia page about synchrotron radiation).
What I am wondering is how to think about this phenomenon from a physical point of view. How can both the particle and the radiation move in the forward direction? Why isn't the collimated beam pointing in the opposite, i.e. the ''backwards'', direction instead?
| The Bremsstrahlung diagram is used for a decelerating electron.
It could equally apply to an accelerating one, (instead of a nucleus, just the accelerating field). Feynman diagrams are in the center of mass system of the input and output particles
The collinearity comes because the whole diagram should be Lorentz transformed to the lab system where the electron has very high energy
Here u is the velocity of the particle in the CM frame. This function is always strongly peaked in the forward direction unless u≈c and cosθC≈−1.
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Kinematic behavior of a flat, a closed, an open universe
according to An Introduction to Modern Astrophysics, 1263p, there is explanation about evolution of scale factor.
“ For the early universe($R<1$) there is little difference among the kinematic behaviors of a flat, a closed, an open universe because the early universe was essentially flat.”
what does it mean? i don’t know relationship between ‘kinematic behaviors’ and ‘early universe flat’
add)
Help me please
| This answer is based on the Figure 5 you added to your question. I believe that the concept of "kinetic behavior" in this context means the mathematics of the expanding universe based on the cosmological model related to Figure 5. The "early flat universe" refers to the similarity between the three assumed models (open, flat and closed) in Figure 5 on the left side where R approaches zero. Note that near R = 0, both the open and closed models look very similar to the flat model.
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Why is there no kinetic term in the Hamiltonian of the Ising model? I am used to the Hamiltonian formalism in the context of (quantum) field theory, where as far as I can remember it always has the form of a kinetic term + a potential term. For me the absence of kinetic terms means a theory without dynamics. In Wikipedia the Hamiltonian of the Ising Model reads:
$$H(\sigma) = \sum_{i,j} J_{ij} \sigma_i \sigma_j + \sum_j h_j \sigma_j\,, \tag{1}$$
where the first term corresponds to interactions and where the sum runs over nearest neighbors, so I suppose $i \neq j$. The second term corresponds to an external potential.
Why is there no kinetic term? Can the system evolve in time, e.g. by seeing the 2d Ising Model as a (Euclidean) (1+1)d model? How should I picture this Hamiltonian as a system in my head?
| In classical statistical mechanics (as opposed to quantum stat-phys), the dynamical (kinetic energy) part of the Hamiltonian decouples from the interaction (potential) part, since both momentum and position are independent. The same is true for classical spin systems: their thermodynamics is independent from the dynamics, which has to be added by hand.
Depending on the choice of dynamics (an example of which is Glauder dynamics), the dynamics exponents at the thermal phase transition changes, see the review by Hohenberg and Halperin.
Of course, classical spin models are microscopically coming from a corresponding quantum model. However, it is well known that thermal phase transitions of quantum systems are described by static classical models (here meaning described by the classical Ising model without dynamics). How to find the classical dynamics of a quantum spin model at a thermal transition is still quite an open question, since computing the dynamics of a quantum system at a phase transition is very hard.
| {
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Shouldn’t the Lorentz force acting on an electron increase it’s tangential speed? I know some of you might be tempted to just answer “No, the Lorentz force acting on a particle is in every point perpendicular to the velocity vector” but please help me with my doubt.
Let’s say we have an electron that for some undefined reason is already moving in a circle and thus has a magnetic moment m. Then it also has an initial tangential velocity which magnitude is v, and again, it has a centripetal acceleration with magnitude a1 = v^2/R.
If we now allow a magnetic field B perpendicular to the surface area within the circumference drawn by the trajectory of the electron, we should notice a Lorentz force immediately acting on the electron and pointing in the same direction of the initial basic centripetal force that used to mantain the uniform circular motion when no magnetic field was applied.
As a result, the electron will now experience a total centripetal force given by the sum of the Lorentz force and the initial force.
One could write (a2 = ma1 - evB)/m. At the beginning of the experiment though, the total centripetal acceleration was just a = v^2/R.
The magnitude of the tangential velocity must have changed then if the radius R stayed the same (imagine in fact, that the electron was circulating into a single coil).
Now, this phenomena doesn’t happen in real life (right?), so how does physics justify such a paradox? Increased velocity should also increase the magnitude of the Lorentz force, which again would increase the velocity of the electron and so on in an infinite loop. Free infinite current basically. What is the right math behind everything?
Thanks in advance for your patience.
| If the radius is going to stay the same as you require (which is fine), the centripetal force needs to stay the same. The initial force that was the agent of that centripetal force will change if some force is added or subtracted in that direction.
If (for simplicity) a string was supplying $F_C$, then you add a magnetic force in that's acting in the radial direction, $F_C$ will stay the same while $F_{string}$ decreases by exactly $F_B$. Initially, $F_C = F_{string1}$ , then $F_C = F_{string2} + F_B $.
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Where does the kinetic energy of a spaceship in flight go, when it is stopped by firing thrusters in the opposite direction? Where does the kinetic energy of a spaceship in flight go, when it is stopped by firing thrusters in the opposite direction?
If you stopp a driving car, the kinetic energy is converted to thermal energy in the brakes, tires, air.
But where does the kinetic energy of a spaceship go, when it is stopped by firing thrusters in the opposite direction of travel.
I do understand, that the energy expenditure of the braking process is the same, as the energy expenditure that was needed for the acceleration in the first place...
But this makes it even more weird... sort of? Where does the kinetic energy go? Why can it not be reused, like in a hybrid/electric car? Or ist there just no way yet?
| The kinetic energy of the spacecraft is carried off by the exhaust gas from the rocket engines that are being fired to stop its motion.
| {
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In a semiconductor, how exactly do electrons move into the conduction band? From what I know, electrons that are excited can move into the conduction band, but how and what causes electrons to be excited? Heat/light seem to make sense as to something that can 'excite' the electron, but I am not completely sure.
Does the location of the electron matter at all, or can an electron at the lowest energy of the valence band be excited to the conduction band (even if there is an acceptor defect above it, could it possibly skip that and move straight to conduction)?
| There are many processes that may excite electrons into the conduction band:
*
*light (even if we do not illuminate the semicodnuctor, there is usually some background radiation)
*strong electric field (Zener tunneling)
*collisions with other electrons, in which they may exchange energy
*scattering particles, such as neutrons against the semiconductor
and others.
If nothing special is done in terms of forcing excitations, it is usually background radiation that causes the excitation. In statistical mechanics terms the semiconductor can be thought of as a subsystem in thermal equilibrium with its surroundings, which enables us to use the Boltzmann/Fermi distribution without thinking of the exact processes that cause the thermal equilibrium (which are usually neglected in statistical physics, but whose existence is always implied).
Let me note that heat is not the same as light, but rather a general measure of energy exchange. In this case it is indeed light (meaning generally electromagnetic radiation, not necessarily visible spectrum).
The probability of transition from this or that place in the valence band is governed by the energy conservation and the selection rules (i.e., the magnitude of the matrix element for the specific type of excitation mechanism).
| {
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Does this research paper prove that warp drives are impossible? Does this preprint prove that warp drives are impossible?
J. Santiago, S. Schuster and M. Visser, "Generic warp drives violate the null energy condition"
It states that the NEC (Null Energy Condition) is violated in this paper and many others: E. W. Lentz, "Breaking the Warp Barrier: Hyper-Fast Solitons in Einstein-Maxwell-Plasma Theory"
| It proves that warp drives violate an energy condition that many physicists think is reasonable. The universe of course has no obligation to be reasonable.
The NEC states roughly that the energy embodied in tension or negative pressure cannot be larger than the density of mass-energy somewhere. That is very extreme: no known material can do it, and were such things possible many other crazy things would be possible. Yet the energy conditions have not always worked so well. But few physicists would bet they can exist for real.
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Earths electric charge: is it neutral? Is the whole earth electric neutral? I know about the negative charge of the earths surface and the positive charge of the ionosphere, but i wonder if the whole earth (including its atmosphere) has an equivalent number of electrons and protons.
It could be measured by some space probe with devices to measure electric fields. Was this ever done?
| No, I won't say that.You answered it yourself the atmosphere is constantly being irradiated by ionizing radiation and some of the charged particles might leave the earth's atmosphere and hence would make the earth a charged object at least on a very small scale.
This question has been asked before, you can find a more detailed answer here.
The net charge of Earth
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How do you differentiate this differential equation? I have to differentiate this equation (Gravitational force between N-Bodies)
$\begin{align}
\frac{d^2}{dt^2}\vec{r_i}(t)=G
\sum_{k=1}^{n}
\frac
{m_k(\vec{r}_k(t)-\vec{r}_i(t))}
{\lvert\vec{r}_k(t)-\vec{r}_k(t)\rvert^3}
\end{align}$
where $\vec{r_{i/k}}(t)$ is the position of a body in 3D space and $m_{i/k}$ is its mass
How would you calculate $\frac{d^3}{dt^3}\vec{r_i}(t), \frac{d^4}{dt^4}\vec{r_i}(t) ...$?
Edit: I know that solving for $\vec{r_i}(t)$ is very complicated but is that also the case for the third derivative of $\vec{r_i}(t)$? I'm asking because since $\vec{r_i}(0)$ and $\frac{d}{dt}\vec{r_i}(0)$ are given (and therefore $\frac{d^2}{dt^2}\vec{r_i}(0)$ is also known) one could make a Taylor series with $\frac{d^3}{dt^3}\vec{r_i}(t), \frac{d^4}{dt^4}\vec{r_i}(t)$ and so on
| The right hand side of your equation is an explicit function of $r_{i}$ and the $r_{k}$s. One just needs to apply the chain and product rules from introductory calculus. The expression will be super-ugly, and will have explicit $\dot r$'s in it, but I don't see what difficulty you're having with the computation.
Perhaps the only complication is that your denominator is really:
$$\left[\left({\vec r}_{i} - {\vec r}_{k}\right)\cdot\left({\vec r}_{i} - {\vec r}_{k}\right)\right]^{3/2}$$
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Why does the intensity of the bright fringes decrease as we move away from the central maxima in Young's Double Slit Experiment? I studied that in Young's Double Slit Experiment the variation of intensity ($I$) of the fringes on the screen with respect to the phase difference ($Φ$) is given by :
$I = 4I_{0} \cos^{2}\frac{Φ}{2}$
$I_{0}$ is the intensity of light coming from each slit.
At maximas or constructive interference, $Φ = nλ$, where $n$ is any whole number and hence we get $I = 4I_{0}$ Below I have given the image of an interference pattern from a laser beam passing through double slit. As you can see as we move away from the central maxima, the intensity decreases and eventually it becomes zero. But how is this possible? According to our equation, the intensity of the centre all the bright fringes should be $4I_{0}$ and hence we should get equal brightness in all the maximas. But why does the intensity decrease and become zero at some point? Shouldn't the interference pattern extend upto infinity and there should be equal brightness at all the maximas? Please explain. I am so confused.
| In your formula, $I_0$ is the intensity of wave from either one of the sources, at the point of consideration. Now, as we move further from Central Bright Fringe, $I_0$ decreases too, varying as $I \propto \frac 1r$, if we consider line source (and hence cylindrical wavefront). Hence bright fringes become dimmer.
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Non-constant wavelength of particles Is it possible for a particle's wavelength to change with respect to time? I'm thinking of a massless particle like a photon, but as it evolves through time its wavelength changes. I'm aware photons wavelengths do change from the expansion of spacetime, but I'm thinking on a smaller scale.
EDIT
Is it possible to have a particle that has its wave length change in time? So imagine a particle that is created with a wave length of say 500nm then it decays to infinity or something as it evolves through time.
| The photon wavelength may vary for a photon propagating in a non-uniform medium - with variable $n(\vec{r})$, the frequency being constant.
| {
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Since the speed of light is constant and also the speed limit; would you, in your reference frame, have no upper bound on your speed? Let us imagine you are in a vacuum and after having maintained a speed of 0 km/s (standing still) you accelerate to 297,000 km/s (99%). You know this is now your speed because you have a speedometer telling you so. You then decide to maintain that speed for a while.
With the speed of light is always ~300,000 km/s faster than you, what is preventing you from (again in your reference frame) increasing your speed, as shown by a speedometer, an arbitrary amount faster than ~300,000 km/s? After all, the speed of light will always be always faster.
I feel like length contraction even backs this since it will make your space wheels tinier. You're essentially scaled down and your tiny wheels would have to rotate many more times to go the distance just 1 rotation would have taken you with your non-contracted length. This then would cause the speedometer to relay speeds faster than the speed of light.
|
You know this is now your speed because you have a speedometer telling you so.
This is precisely where you hit a (metaphorical) roadblock.
A speedometer must use something outside of your reference frame to measure your speed, as speed inside your frame is either 0 or meaningless (take your pick).
It's measuring the speed of your space wheels.
Your space wheels will never spin faster than the speed of light (where the speed of your wheels is the linear velocity at the edge of the wheel).
It's measuring your current work being put in and converting it to a speed.
Then you are measuring kinetic energy and not speed. There is no upper limit on the kinetic energy an object can have! However when you solve backwards for speed infinite kinetic energy leads to your speed approaching (but never exceeding) c.
$$ KE = mc^2 (\gamma - 1) \\
\gamma = \frac{1}{\sqrt{1 - (v/c)^2}} = \frac{KE}{mc^2} + 1 \\
\sqrt{1 - (v/c)^2} = \frac{1}{\frac{KE}{mc^2} + 1} \\
1 - (v/c)^2 = \left( \frac{1}{\frac{KE}{mc^2} + 1} \right)^2 \\
v = c \cdot \sqrt{ 1 - \left( \frac{1}{\frac{KE}{mc^2} + 1} \right)^2} \\
$$
If you plug in $\infty$ for KE you should see that you recover $c$.
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An algebra step in the Quantum Partition Function for the Harmonic Oscillator On page 183 of Altland Simons, we are told:
$$ \prod_{n = 1}^{\infty} \Big[ \Big( \frac{2\pi n}{\beta} \Big)^2 + \omega^2 \Big]^{-1} \sim \prod_{n = 1}^{\infty} \Big[ 1 + \Big( \frac{\beta \omega}{2\pi n} \Big)^2 \Big]^{-1} \sim \frac{1}{\sinh(\beta \omega / 2)}. $$
How do we get the first relation?
This physics SE question suggests that it is a multiplication and division by $$\prod_{n = 1}^{\infty} (\beta / 2\pi n)^2 ,$$ but I'm confused how multiplication and division by the same factor switches the $\beta, n$ from denominator/numerator to numerator/denominator:
$$\Big[ \Big( \frac{2\pi n}{\beta} \Big)^2 + \omega^2 \Big]^{-1} \times \frac{(\beta / 2\pi n)^2}{(\beta / 2\pi n)^2} \sim \Big[ 1 + \Big( \frac{\beta \omega}{2\pi n} \Big)^2 \Big]^{-1}.$$
I understand that one answer in the linked post addresses this in a different way (zeta function regularization) and why we can multiply by $\prod_{n = 1}^{\infty} (\beta / 2\pi n)^2 $. I'm stuck on the step before that -- why would multiplication and division of that product help in the first place?
| They are dropping an $\omega$-independent factor. Such factors are often dropped as they have no physical effect.
| {
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Wave propagation through conductors During wave propagation through conductor, why is the current density out of phase with electric field ?
When the oscillating electric field applies a driving force on the free electrons, they will start oscillating. Then the current density should be in phase with electric field , isn't it ?
| The effect is identical to that of a harmonic oscillator. The electric field accelerates the electrons similar to how the spring accelerates the mass at the end. In the harmonic oscillator, too, the velocity is the highest when the force is lowest, i.e. when the displacement is zero.
If I have to guess at the origin of your confusion, you are trying to apply Ohms law $J=\sigma E$, but this applies in a different situation than what you describe. The reason is as follows. Electrons cannot move freely though the conductor, but they will collide with the atoms, meaning they cannot accelerate indefinitely. How often they collide gives the value of $\sigma$ in Ohms law. If you create an electric field that oscillates very rapidly, the electrons will not accelerate to high speeds, and will not travel for great lengths, so these collision effects will not be large. In this case, the behaviour will be similar to the harmonic oscillator, rather than to what Ohms law describes.
| {
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What happens to entropy when half the particles are removed? Curiosity question. What happens to entropy in the following situation?
A gas fills an entire container and is in equilibrium. Suddenly all particles are removed from half the container. As such, there are now 1/2 the original number of particles, but all occupy one side of the container.
What happens to entropy? Does it rise or fall?
| The question is ambiguous, since talking about entropy we have to say entropy of what we discuss. In most cases "what" is omitted, since the answer is obvious, but not here. More specifically, assuming that initially we had $2N$ molecules:
*
*One can discuss the entropy of the $2N$ molecules after half of them was removed. Obviously, it will depend on the process of removal, which is non-quasistatic. We can confidently say that this entropy has increased.
*One could also discuss the entropy of the remaining $N$ particles. The entropy of this new system might well be less than that of the original one, but it is not the same system. Obviously, its entropy will further increase as the molecules spread over the container.
Let me also note that the discussion above and the question assumed that we are dealing with an isolated container. One could also formulate a more realistic problem of a subsystem being able to exchange particles with the reservoir, in which case the entropy difference between $2N$ and $N$ particles can be calculated using the grand canonical ensemble.
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Why is $kx−ωt$ a constant for a travelling wave? In my physics textbook, there is a statement like this:
The motion of a fixed phase point on a progressive wave is given by
$kx−ωt$= a constant.
What does this mean? Why is it a constant?
Does fixed phase point mean that it is a particle of the medium always at the same height? Then why are we considering its motion? Are they talking about different points? Or are they talking about a point on the wave itself moving in its direction of propagation?
| This relation pretty much defines a travelling wave:
$$
\mathbf{k}\mathbf{x}-\omega t=const \Rightarrow \mathbf{x}=\mathbf{v}t+\mathbf{x}_0,
$$
i.e., the wave front (or the surface pf constant phase) of a wave $v(\mathbf{k}\mathbf{x}-\omega t)$ moves with a constant velocity (where $v(\phi)$ is an arbitrary shape).
| {
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What is the force pair for the normal force? Clarification on Newton's 3rd Law In the process of trying to wrap my head around Newton's 3rd law I've come across 2 definitive statements.
*
*Forces must occur in pairs
*Forces must act on different bodies
This is confusing to me when applied to the classic box on a flat plane scenario (assuming the flat plane is Earth). I've been taught in school that the present forces are like this:
So accordingly I always assumed that the gravitational force of the box and the normal force were pairs. However, after watching a few videos explaining the concept, namely this one, the impression I was left with is that the paired forces are FBE and FEB.
However, I'm aware that the normal force still exists, but if it can't exist without a paired force, what would be its paired force?
Additionally, can a force be paired with multiple forces?
| I find that Newton's 3rd law is often written in a way that makes it easy to get confused.
Here's how I like to write it:
If object A exerts a force on object B, then object B exerts a force on object A that has the same strength but opposite direction.
This makes it clear that you will never ever find two forces from a pair acting on the same object. The force of gravity on a ball and the normal force on a ball can't be a Third Law pair because they both on act on the same object, and that's not how Third Law Pair works.
If you want to find the a force's pair partner you need to look at another object!
For example, if you want to know the pair partner of the normal force you have to ask "which object is causing this normal force?" If a book is feeling a normal force from a table, then the pair partner is the force that the book exerts on the table.
| {
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How to measure the speed of an electric current? We all know that the definition of a current is the amount of charge flowing per second, that is often expressed by the equation $i=dq/dt$. But is it possible to measure the speed of an electric current in m/s? And also how can we measure such speed?
| Let's say that you have a wire with cross section $S$. Denote by $v$ the average speed of charge carriers, whose density is $n$. Assuming also that each charge carrier has charge $e$, it is clear that
\begin{equation}
J=e v n S
\end{equation}
From here you can find $v$
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How do you understand stall in terms of Newtonian mechanics (i.e. without Bernoulli's Principle)? Okay, so. I understand how a wing generates life using Newtownian mechanics, to wit: the air molecules crash into wing, which is at an angle to the air molecules. As a result, the air molecules are deflected downward and, by conservation of momentum, the wind must now have an upward component of momentum.
So why would it be that having too great of an angle of attack would cause stall? I know the typical explanation is that the fluid flow over the top of the wing separates from the wing, but I fail to see how that would keep the air molecules crashing into the bottom of the wing from creating lift.
If y'all can forgive me a hokey MS Paint drawing:
| The way to think about is in term of the ratio of lift and drag.
Wing design is optimized for good lift/drag ratio. (Provided, of course, that the pilot maintains the optimal angle of attack for the wings and the speed of the aircraft.)
Aircrafts that are designed for aerial acrobatics have flat wings, giving the pilot freedom to fly the aircraft upside down, along a horizontal line.
The lift/drag ratio of acrobatics aircrafts is less good, but an acrobatics aircraft doesn't need to have distance capability.
So let's take the case of an aircraft that has completely flat wings. With a sufficiently strong engine that aircraft will fly. Here's the rub: with flat wings there is an optimal angle of attack too.
Stalling is that the lift/drag ratio becomes so bad that you keep losing speed.
To see that there must be an optimum: exaggerate the angle of attack to the point where the amount of drag exceeds the amount of lift. Clearly the aircraft will lose control at that point.
So: even when the wings are flat there is still an optimal angle of attack
Going back to optimized wings:
The purpose of the wing shape is to facilitate a good rejoining of the air that passes underneath and the air that flows over the wing. The better those two flows rejoin, the less turbulence. The less turbulence, the less drag.
The better the two flows rejoin, the better the mileage of the aircraft.
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Why the mass of initial particle has to be greater than the sum of masses of final particles? Suppose we have a decay of a rest particle $A$ into other particles $a_1,...,a_n$
\begin{equation}
A \rightarrow a_1+a_2+\cdots+a_n
\end{equation}
It is always stated that in order to this particle decay to be possible, the mass of the initial particle $M_A$ has to be greater than the sum of masses of final particles, i.e,
\begin{equation}
M_A > M_{a_1}+\cdots+M_{a_n}=\sum_{k=1}^n M_{a_k}
\end{equation}
How can we proof this mathematically from physical principles?
I have tried using fourmomentum conservation, but I can't get to the inequality.
| At the most fundamental level [in terms of the structure of special relativity],
the reason that
"the parent mass is greater than the sum of the daughter masses" (always the invariant masses)
is the same as
"the Clock Effect (which I will describe analogous to the above by: the proper time of the inertial path from O to Z is
greater than the sum of the proper times along a piecewise-inertial path from O to Z via an intermediate event not on OZ)".
And these are ultimately due to the "reversed triangle inequality", which involves the law of "cosines".
\begin{align}
\tilde C &= \tilde A + \tilde B\\
\tilde C\cdot \tilde C &= \tilde A\cdot \tilde A + \tilde B\cdot\tilde B + 2\tilde A\cdot \tilde B\\
C^2 &= A^2 + B^2 + 2AB \mbox{("cosine of angle between"})\\
\end{align}
[In special relativity, the time-dilation (the gamma $\gamma$) factor plays the role of [hyperbolic] cosine, as adjacent/hypotenuse, where hypotenuse is the triangle side that is opposite the "right angle" (the corner where the perpendicular sides meet).]
Next, note the algebraic fact $A^2+B^2+2AB=(A+B)^2$.
How the magnitude $C$ compares to the sum of magnitudes $(A+B)$
depends on the metric geometry,
which is essentially tied to the nature of $\mbox{("cosine of angle between"})$.
[For relativity, we restrict to future-timelike vectors.]
*
*In Euclidean space, $\mbox{("cosine of angle between"})\le 1$.
So, $C^2 = A^2 + B^2 + 2AB \mbox{("cosine of angle between"}) \le (A+B)^2$ ["triangle inequality"].
*In Minkowski spacetime, $\mbox{("cosine of angle between"})=\cosh\theta=\gamma\ge 1$.
So, $C^2 = A^2 + B^2 + 2AB \mbox{("cosine of angle between"}) \ge (A+B)^2$ ["reversed triangle inequality"... hence the Clock Effect and the Mass Effect (where the vector equation is conservation-of-total-4-momentum in energy-momentum space)].
*In Galilean spacetime, it can be argued that its $\mbox{("cosine of angle between"})=1$.
So, $C^2 = A^2 + B^2 + 2AB \mbox{("cosine of angle between"}) = (A+B)^2$ ["no triangle inequality"... hence no Clock Effect and no Mass Effect].
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"answer_id": 1
} |
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