Q stringlengths 18 13.7k | A stringlengths 1 16.1k | meta dict |
|---|---|---|
Evaluating the Ricci tensor effectively If given a metric of the form $$ds^2=\alpha^2(dr^2+r^2d\theta^2)$$ where $\alpha=\alpha(r)$, then can one immediately conclude that $$R_{\theta\theta}=r^2R_{rr}$$ where $R_{ab}$ is the Ricci tensor, without doing any explicit calculations? I can show that this is true by the long-winded way of computing both explicitly, but it would seem that there may be a more elegant way?
| I'm not sure what OP exactly is requesting, but OP's equation follows e.g. from the general fact that for an arbitrary 2D surface, the Ricci tensor
$$ R_{\mu\nu} ~\propto~g_{\mu\nu} $$
is always proportional to the metric tensor $g_{\mu\nu}$. This is basically a consequence of that in 2D the Riemann curvature tensor is complete determined by the scalar curvature.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/64139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Motivation for Wigner phase space distribution Most sources say that Wigner distribution acts like a joint phase-space distribution in quantum mechanics and this is justified by the formula
$$\int_{\mathbb{R}^6}w(x,p)a(x,p)dxdp= \langle \psi|\hat{a}\psi\rangle$$
as this allows us to compute the expectation value for the operator $\hat{a}$ corresponding to the physical quantity $a$. However, without the knowledge of this formula, how did Wigner come up with this definition:
$$w(x,p)=\frac{1}{(2\pi)^3}\int_{\mathbb{R}^3}\psi\left(x-\frac{v}{2}\right)\psi^*\left(x+\frac{v}{2}\right)e^{iv \cdot p}dv \quad ?$$
I would be greatly indebted for any mathematical motivation that anyone could provide. Similarly, I also wonder at the mathematical motivation for the Weyl quantization.
| I want to add an interesting detail to the excellent answer by Qmechanic which I recently learned from Cosmas Zachos, here.
Royer showed in his 1977 paper, that the Wigner function $w$ is given by the expectation value of the parity operator $\Pi_{\mathbf{r}\mathbf{p}}$ which inverts the phase space around the point $\mathbf{z}=(\mathbf{r},\mathbf{p})$, i.e., he showed that
$$
w(\mathbf{r},\mathbf{p}) \propto \langle \psi| \Pi_{\mathbf{r}\mathbf{p}}| \psi \rangle ,
$$
with (in one dimension for simplicity)
$$
\begin{align}
\Pi_{rp} &= \int\text{d} s~ e^{-2ips/\hbar} ~| r-s\rangle \langle r+s |
\\
&=
\int\text{d} k~ e^{-2ikr/\hbar} ~| p+k\rangle \langle p-k |,
\end{align}
$$
where $|r\rangle$ and $|p\rangle$ are eigenstates of position $\hat{r}$ and momentum $\hat{p}$. He proved that this operator has the property
$$
\begin{align}
\Pi_{rp} (\hat{r} -r) \Pi_{rp} & = -(\hat{r} -r) \\
\Pi_{rp} (\hat{p} -p) \Pi_{rp} & = -(\hat{p} -p) .
\end{align}
$$
As Royer states himself:
[The Wigner function] is proportional to the overlap of $\psi$ with its mirror image about $r,p$, which is clearly a measure of how much $\psi$ is "centered" around $r,p$.
Although this is not an answer to the question as Wigner himself probably did not know about this, it still adds some physical depth to the problem and could serve as a modern motivation.
Something which you can immediately understand based on this interpretation is how the Wigner function is bounded and what these bounds represent. Royer showed that if $\psi$ is antisymmetric about $r,p$, the Wigner function attains its lower bound.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/64216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 2,
"answer_id": 1
} |
Row of pivoted magnets and energy scale This question is about a system involving a horizontal row of length L of equally spaced pivotable magnets, each with a pole at either end. These magnets will often be referred to as units.
So each unit when rotated causes it's neighbours to rotate in the opposite direction to itself.
When the first unit is quickly rotated 1/4 of a turn and fixed in place rotationally, each will likely turn less than the previous at first, taking longer to complete the full 1/4 of a turn. They will all eventually rotate the full 1/4 as the far end of the line is still not fixed.
.
Turning the first magnet will require energy. For a line of 1000, while this will be more energy than that for a line of for example 2, it will be less than 1000 times the energy required for a line of 1. This is because:
*
*Momentum means that every unit doesn't need to be rotated the full distance to turn the first unit, it is not as if they are connected together by rods, the first can be turned all the way before the others have had time to rotate much at all.
*More distant magnets contribute less force, so most units are contributing very little force to the first unit directly.
Another way to think about this is that the line of separated magnets forms a compound magnet that will result in more force than each component magnet has individually, but not as much as 1000 times more, for example if metal becomes attached to an end it will take less than 1000 times the force of an individual magnet to remove.
.
Coils around each unit slow rotation but produce N electricity after 1/4 of a turn, however long the 1/4 turn takes.
The amount of electricity produced appears to be L * N, for an input power that is increment by a diminishing amount for the growth of L, an example of this sort of growth could be sqrt(L).
.
Question:
This has the potential not to add up energy wise for long lines, what's going on?
| $\def\vm{{\bf m}}$If the magnets are free to rotate they will find the lowest energy state.
This occurs when all the magnets are aligned head to tail along the line.
$$\begin{equation*}
\rightarrow\rightarrow\rightarrow\cdots\rightarrow\tag{1}
\end{equation*}$$
If now the magnet at one end is rotated a quarter turn and then fixed, it does not follow that all the other magnets will align themselves in an alternating pattern.
$$\begin{equation*}
\uparrow\downarrow\uparrow\cdots\uparrow\tag{2}
\end{equation*}$$
In fact, the magnets will align themselves something like this
$$\begin{equation*}
\uparrow\searrow\rightarrow\cdots\rightarrow\tag{3}
\end{equation*}$$
since the potential energy of this configuration is lower than that of (2).
Some details
The potential energy of the magnetic dipole-dipole interaction between nearest neighbors $i$ and $i+1$ is of the form
$$H_{i,i+1} = -J(3(\vm_i\cdot \hat x)(\vm_{i+1}\cdot \hat x)-\vm_i\cdot \vm_{i+1})$$
where we take $\hat x$ to be the unit vector pointing from dipole $i$ to $i+1$.
Letting $\vm_i=(\cos\theta_i,\sin\theta_i)$ we find
$$H_{i,i+1}=-\frac{J}{2}(3\cos(\theta_i+\theta_{i+1}) + \cos(\theta_i-\theta_{i+1})).$$
The potential between nearest neighbors is minimized when $\theta_i = \theta_{i+1}=0$ or $\pi$, in which case we find $H_{i,i+1}=-2J$.
Thus, the energy of configuration (1) is roughly
$$H = -2Jn$$
where $n+1$ is the number of magnets.
(Here we ignore longer range effects.)
On the other hand, if $\theta_i = -\theta_{i+1} = \pm\pi/2$ we have $H_{i,i+1}=-J$.
Thus, the energy of configuration (2) is roughly
$$H = -Jn.$$
Notice that the configuration
$$\begin{equation*}
\uparrow\rightarrow\rightarrow\cdots\rightarrow\tag{4}
\end{equation*}$$
has potential
$$H = -2J(n-1)$$
since $H_{1,2}=0$.
For large $n$ this is vastly less than the energy of configuration (2).
Some results for particular values of $n$
For two magnets the configuration will be as claimed.
$\hskip2.8in$
For three magnets the energy can be minimized algebraically.
Already we see that the magnets like to be aligned.
We find
$$\begin{eqnarray*}
\theta_1 &=& \pi/2 = 90^{\circ} \\
\theta_2 &=& -\cos^{-1}\sqrt{2/3} \approx -35^\circ \\
\theta_3 &=& \cos^{-1}2\sqrt{2}/3 \approx 19^\circ.
\end{eqnarray*}$$
$\hskip2.6in$
For $50$ magnets the energy can be minimized numerically.
We find, to a tenth of a degree,
$$\begin{eqnarray*}
\theta_1 &=& 90^{\circ} \\
\theta_2 &=& -30.0^\circ \\
\theta_3 &=& 8.2^\circ \\
\theta_4 &=& -2.2^\circ \\
\theta_5 &=& 0.6^\circ \\
\theta_6 &=& -0.2^\circ \\
\theta_7 &=& 0.0^\circ \\
&\vdots& \\
\theta_{50} &=& 0.0^\circ.
\end{eqnarray*}$$
$\hskip1.5in$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/64275",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Maxwell's Equations-Relativity How did Maxwell develop the magnetic field without relativity? Was it purely experimental? I don't see how else he would have developed any understanding for the magnetic field.
| A long time before Maxwell wrote down a unified theory, Oersted discovered a connection between electricity and magnetism. In the development of electromagnetism, there were many bits and pieces of partial knowledge that were discovered and formulated by many different scientists. The popular ones (after whom we've named partial "laws") that immediately come to mind are Faraday (who came up with the concept of a field), Ampere, Gauss, Coulomb, etc. In fact, Faraday put together electric and magentic fields to make motors and generators. If I recall the history correctly, Maxwell understood the connections and put together all these partial results into a coherent mathematical formulation.
As it happened, it was noticed that this theory was not invariant under Galiean transformations and the challenge was to figure out which of the two concepts to be modified. To make a long story short, there were many proposals and notable the Michelson-Morley experiment, but the one that we now know to be correct was Einstein's theory of Special Relativity.
Some references for the history of E&M
*
*http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-013-electromagnetics-and-applications-spring-2009/lecture-notes/MIT6_013S09_res_maxwell.pdf
*http://digitalcommons.sacredheart.edu/cgi/viewcontent.cgi?article=1002&context=wac_prize
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/64351",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Why is there no UV catastrophe (divergence) in turbulence? I have just read that as the Reynolds number is increased, the separation of macroscopic and microscopic scales increases and that this also means that there is no UV catastrophy (or equivalently UV devergence?) in turbulence.
I do not understand what this means, for example does the increased separation of the macro and microscales just mean that the spectrum is broadening? And what exactly would an UV catastrophe mean in the context of turbulence and how can I see (technically and mathematically) that it does in fact not exist?
I have only a rough intuition what it could mean when considering the turbulence problem from a QFT approach, namely that it from this point of view the infinit Reynolds number limit should be renormalizable (?)...
Any comments that would help me to understand what the sentence in the first paragraph exactly means would be appreciated.
| The fluid equivalent to the Ultraviolet Catastrophe is the transition to turbulence. Just as Maxwell's Equations break down at high frequencies and are replaced by a stochastic model (Quantum Mechanics), the Navier-Stokes Equations break down at high Reynolds Number and are replaced by a stochastic model (Kolmogorov Cascade). Reference: Reid, J. (2019) The Fluid Catastrophe, Cambridge Scholars Publishing, Newcastle upon Tyne.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/64499",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Does inertia increase with speed? I have heard that when the speed of the object increase, the mass of the object also increase. (Why does an object with higher speed gain more (relativistic) mass?)
So inertia which is related to mass, increase with speed?
So, if I accelerate on a bus, my mass will increase and my inertia will increase for a while on the bus, until the bus stops?
| For both interpretations, the answer is 'yes' since force still acts in an opposite force on anything which has mass. As you accelerate, your velocity increases and therefore mass will increase. The increase in mass will bring about an opposite force. The greater the mass, the greater the inertia.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/64661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
How does an earthen pot keep water cool? I understand that evaporative cooling takes place thanks to small pores contained in the pot and that allow some water to go through and evaporate. However I couldn't understand clearly whether water inside the pot stays at its original temperature or would it cool further?
If it will become cooler then how?
| Latent heat of evaporation cools the pot in the same way it cools your skin. If the pot is then at a lower temperature than the water in it, heat energy will be transfered to the pot from the water in it by conduction.
Using plant xylem water for evaporative cooling, the desert cicada Diceroprocta
apache can maintain a body temperature as much as 5°C below ambient
(ra=42°C)
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/64716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 1
} |
Comparing Static Frictions In this figure, which of the static frictional forces will be more?
My aim isn't to solve this particular problem but to learn how is static friction distributed . Since each of the rough-surfaces are perfectly capable of providing the $-1N$ horizontal frictional force but why don't they ? This is kind of ambiguity that who will provide a bigger share in total static friction. And as the surface have different $\mu$, so we can't even invoke symmetry.
| In the cases where you have static friction, the forces will always be defined by the looking at the system and applying the constraints(in other words $F_s\le \mu N$ will only give an upper bound). On the other hand when you are dealing with kinetic friction, it can be easily derived from the famous $F_k=\mu N$.
As an example, let's solve this problem(As they say, a good picture is worth a million words).
The fastest way to find the amount of the static friction forces will perhaps be this:
*
*Look at the horizontal forces acting on $m_1$. There is only one such force, $F_{s_1}$. But we know that the object is stationary($\Rightarrow a=0$), ergo, $F_{s_1}=0$.
*Now that $F_{s_1}$ is zero, let's look at the horizontal forces acting on $m_2$. We have:
$$F_{s_2}-F=0 \\ \Rightarrow F_{s_2}=F$$
Or in this case the friction in the bottom of the lower box is $1_N$ and the static friction between boxes is zero.
$\square$
One should perhaps check whether $F_s\le \mu N$ or not. In this case, we can easily verify that:
$$N_1=m_1 g \\ N_2=N_1+m_2 g=(m_1+m_2)g$$
$$\Rightarrow N_1\approx 98_N \\ N_2 \approx 196_N$$
$$0_N\le 0.75 \times 98_N \color{green}{\large{\checkmark}}\\
1_N\le 1\times 196_N \color{green}{\large{\checkmark}} $$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/64780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Optics alignment of a confocal scanning microscope I am facing a challenge in my project regarding optical alignment.
See the figure:
The challenge is with the vertical optical system alignment. I considered placing a mirror and check back if the image and source coincide. But since the light is too low on power (less bright).
How can I align it so well that everything is absolutely vertical and good?
| The basic concept of optical alignment is to perfectly align the beam step-by-step on all optical elements. Enshure your beam has the right height and angle before alignment. Your aim to align the beam on microscope optical axis. Using point sources like (e.g. Flourescent beads, microspheres, Quantum dots, quantum wells, NV centers in diamond) allows an easy alignment to a point spread function on your CCD sensor.
On-axis with an iris aperture
A quick method uses an iris aperture to make a parallel beam to the optical axis. Insert Iris $A$ between M3 and M2. The open iris aperture should allow the beam to transmit in center. Place a mirror perpendicular to opical axis at the translation stage with remove microscope objective and ocular. The beam is reflected on the plane mirror and possibly hits the back side of iris aperture.
Iterate angle and position alignment
Using M3 to center the beam on back of iris apeture, while oving aperture up and down. Possible position errors shoule be aligned with mirror M1 an M2. These have the largest impact on position. Start iterating with angle alignment. This coarse alignment should allow to insert microscope elements and fine align the beam to microscope objective.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/64865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Ducted or open fluid flow, which is best for aerodynamics and lift I'm designing a copter and trying to decide if the propellers should be ducted or open axial flow. I've read some theory on ducted and open air flow but I can't find any where that compares the two. I would prefer to use ducted over open for safety (ie put a safety guard round the propellers so they can't be run into with fingers etc). So are there any major advantages or considerations I should take into account, with ducted vs open axial flow before ploughing ignorantly ahead with one or the other type of flow?
| Actually, ducted fan propeller are much less efficient than open blade propeller soif efficiency is your priority, open propeller is your option
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/64932",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Can a current carrying loop or wire produces no magnetic field? A current carrying wire produces magnetic field around it. We can find the direction by Fleming's Right hand rule. We know change in electric filed produces magnetic field and change on magnetic field produces electric field. It is mutual relationship.
My question is that is there any condition such that current carrying loop or wire produces no magnetic field
| A current carrying loop arranged in the form of a twisted pair of cables
with same diameter, will produce a nearly zero magnetic field.
The use of a coaxial cable will produce a still lower magnetic field.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/64993",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Volume element $\mathrm{d}^4k =\mathrm{d}k^0 \,|\mathbf{k}|^2\,\mathrm{d}|\mathbf{k}| \,\mathrm{d}(\cos\theta) \,\mathrm{d}\phi$ in Minkowski space? Suppose we have an integral
$$\int \mathrm{d}^4k \,\ f(k)$$
we want to evaluate and that we're in Minkowski space with some metric $(+,-,-,-)$.
Is it true that: $$\mathrm{d}^4k = \mathrm{d}k^0\ \mathrm{d}^3\mathbf{k} = \mathrm{d}k^0 \,\,\,\,|\mathbf{k}|^2\,\mathrm{d}|\mathbf{k}| \,\,\mathrm{d}(\cos\theta) \,\mathrm{d}\phi$$
just like in ordinary space?
If not, what are the differences between this and an Euclidean integral in (say) 4 dimensions?
| Yes it is.
The volume form on any (pseudo-)Riemannian manifold $(M,g)$ of dimension $n$, where $g$ is the metric, is given in local coordinates $(x^1, \dots, x^n)$
$$
\sqrt{|\det (g_{\mu\nu})|}dx^1\wedge \cdots \wedge dx^n
$$
where $\det(g_{\mu\nu})$ is the determinant of the metric in these coordinates. In cartesian coordinates, the determinant of the Euclidean metric is $+1$ why the determinant of the Minkowski metric is $-1$. However, the absolute value in the square root factor of the volume form kills the sign difference, so the volume forms are the same.
NOTE. The notational convention in physics is
$$
d^n x = dx^1\wedge \cdots \wedge dx^n
$$
See, for example, Carroll's Spacetime and Geometry eq. 2.95. So, the answer to your question is really "yes" due to notational convention, but then this begs the question "Why would one use $d^nx$ as the volume form for both Minkowski space and Euclidean space?" whose answer is given abave.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/65037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Momentum, Impulse and Newton's Second Law of Motion Newton formulated his Second Law as such:
$$\sum{\vec{F}} = \frac{\delta \vec{p}}{\delta t}$$
and of course, $\vec{p} = m \vec{v}$.
Why is it that if the net force $\sum \vec{F}$ is constant (which implies that the rate of change of momentum is constant), then
$$\frac{\delta \vec{p}}{\delta t} = \Delta \vec{p}$$
In other words, why does the rate of change of momentum equal to the total change in momentum?
If there were previous answers on this, please inform me! Thanks :)
| I think I have figured out the answer, hopefully.
Firstly, let us begin by stating the First Law:
$$\sum \vec{F} = \frac{\delta \vec{p}}{\delta t}$$
When the net force is constant, we it means that there is no change of momentum, in other words,
$$ \frac{\delta \vec{p}}{\delta t} = 0$$
In this case, we know that the function of momentum is a constant, i.e, a straight line function.
This would derive the impulse-momentum theorem:
$$\Delta \vec{p} = \vec{J}$$
Would this be valid?
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/65089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Acceleration due to gravity? I was looking into orbitals and found something I haven't been able to understand.
http://www.math.ubc.ca/~cass/courses/m309-01a/hunter/satelliteOrbits.html
There is a part on the page which states the following:
$$\vec a = \frac{G(M+m)\vec{r}}{r^3}\approx GM\vec{r} / r^3$$ since $m << M$
I'm not sure why this is the case and why we do not use $\vec a = \vec F/m$ here. The cubed radius is really throwing me off. Can anyone further explain this or point me to a reference with a proof? Thanks
| We have
$${\bf F} = \mu {\bf a}$$
where $\mu = M m/(M+m)$ is the reduced mass.
Thus,
$$\frac{G M m}{r^3}{\bf r} = \frac{M m}{M+m}{\bf a}$$
or
$$\begin{eqnarray*}
{\bf a} &=& \frac{G(M+m)}{r^3}{\bf r} \\
&=& \frac{G M}{r^3}{\bf r}\left(1+\frac{m}{M}\right) \\
&\simeq& \frac{G M}{r^3}{\bf r}.
\end{eqnarray*}$$
Notice that $m/M \simeq 10^{-19}$.
The error accrued by neglecting this term is very small indeed.
As mentioned by @Lagerbaer, the ratio you find troubling involves the vector ${\bf r}$ divided by $r^3$.
This can be written as
$$\frac{{\bf r}}{r^3} = \frac{\hat r}{r^2}$$
where $\hat r$ is the unit vector in the ${\bf r}$ direction.
The magnitude of the gravitational force will be the familiar one since
$|{\bf r}/r^3| = 1/r^2$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/65181",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How does tension apply torque on a pulley? How is tension in a string able to apply torque on a pulley? How does string itself able to apply a force on pulley? What is happening inside the pulley?
The pulley has a mass $m$ and is a disc.
| Tasos' answer above is the simplest answer to the question. With this approach you do not have to address the details of the interaction between the string and the pulley. This is a great example of how picking the appropriate system (here the pulley with the string) leads to a simpler evaluation. This assumes no slip between the string and the surface of the pulley.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/65297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
How does relativity explain gravity, without assuming gravity I have seen the "objects pull down on space-time" explanations, but they assume a "pull down" force themselves. Could anyone explain the space-time explanation without assuming gravity in the first place?
| Why don't you want to assume gravity? Gravity it is an experimental fact, a starting point for doing physics. General Relativy is a geometrical theory of gravity, built on the basis of Special Relativity and always having in mind that it should recover the non-relativistic Newtonian theory of the gravitational field.
The "pull down" is a deviation of the flat, Minkowski spacetime, governed by the Einstein Field equations. And in my opinion a not-so-good analogy because it is rather difficult to imagine how to pull down time.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/65363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Relativistic Lagrangian transformations I need to study the relativistic lagrangian of a free particle.
It's
$\ L= - m c^2 \sqrt[2]{1- \frac{|u|^2}{c^2}} $ I need to study the translation, boost and rotation symmetry. I say it doesn't depend of the position, so it has translation symmetry and the momentum will conserve. It's rotation invariant because depends only of the modulus of the speed $|u|$(What is the conserved quantity derived by this symmetry?) What can I say about the boost transformation, I can use the boost of the speeed $u$, but I can't see any conclusion about that.
| Only the action S is a relativist invariant : invariant under translations, rotations, and boosts. The Lagrangian itself is not invariant under boosts.
The action $S$ is : $S = -mc \int ds = - mc \int \sqrt {ds^2} = - mc \int \sqrt {c^2 dt^2 - \vec {dx}^2} = - mc^2 \int \sqrt {1 - (\frac{\vec {dx}}{c \, dt})^2} \,\,\,dt$
$= \int L \, dt$
where L is the Lagrangian : $L= - mc^2 \sqrt {1 - (\frac{\vec {dx}}{c \, dt})^2}$
$ds^2$ is clearly a relativist invariant, and so the action S is too.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/65414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Why does sound move faster in solids? I know that the molecules are closer together in solids, and I know thicker springs also respond carry waves faster than thinner springs, but for some reasons I can't understand why.
The molecules will have a larger distance to move before colliding with another molecule, but in a thicker medium wouldn't that time just be spent relaying the message between multiple atoms? Why is the relaying between a lot of tight knit atoms faster than one molecule moving a farther distance and colliding with another?
| I assume "faster in solids" means faster than in gases.
The speed of a mechanical wave is in general proportional to $\sqrt{k/m}$, where $k$ is some measure of the restoring force (e.g., the tension in a string, or a Young's modulus), and $m$ is some measure of inertia (e.g., the mass per unit length of a string, or the density of the medium).
Compared to a gas, a solid has a density that is greater by about a factor of $10^3$. However, gases are very compressible, while solids and liquids are extremely incompressible. The incompressibility factor overwhelms the inertial factor.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/65580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Can a single molecule have a temperature? A show on the weather channel said that as a water molecule ascends in the atmosphere it cools. Does it make sense to talk about the temperature of a single molecule?
| As stated in the other answers, in the specific scenario depicted in your question, it makes no sense to talk about the temperature of a molecule. But I can't help to widen the picture because there is a very interesting case here: what about the internal degrees of freedom of the molecule? So far everybody has only considered the motion of the whole molecule, because that was the context you implicitly gave, fair enough.
But what about the vibration of the molecular bonds? The rotations of two neighbour parts of the molecule about a bond? This is not as trivial! For small molecules, the number of associated degrees of freedom are too few to warrant talking about temperature. But it is not so for a large protein. There are so many of them that a statistical approach is possible: one can define an entropy, in the usual $\sum p\log p$ way, where $p$ is the probability for a given configuration. Then, as usual, if we have an entropy we can minimise it with the constraint of a given total energy, and a temperature comes out of that.
After writing this, I realised that @Rococo and @JohnRennie had written an answer on a similar line long before mine!
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/65690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "32",
"answer_count": 10,
"answer_id": 9
} |
Why are the magnetic moment and the angular moment related? Why are the magnetic moment and the angular moment related? I've always read everywhere that they are related but found nowhere a satisfactory explanation of the cause
| You can understand, it through Einstein-de Hass effect or other way Barnett effect;
In classical mechanics, when an object of mass $m$ moves circularly, it gives rise to angular momentum. Similarly, when a charge particle moves circularly it gives rise to magnetic moment. So charge particles are not massless, they do have some mass, their orbital motion results into this angular momentum.
As per above effect means that: You have a ferromagnetic substance hanged by a thin fibre. Now, you magnetise the ferromagnetic rod, it starts rotating other way, because of conservation of angular momentum arising from the moving mass charges inside the rod.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/65822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Find E and B from vector potential I have a vector potential given by:
$\mathbf{A}(x,t) = \mathbf{e}_{y}\frac{1}{2} e^{-(x-ct)^{2}/{4a^{2}}}$
Now, the question is "Determine the E and B under the condition that the scalar potential vanishes $V = 0$."
But I'm not quite sure what it means when $V=0$ ?
As far as I can see, the B-field is given by:
$\mathbf{B}=\nabla \times \mathbf{A}$
And then I have that:
$\mathbf{E} = -\nabla V$
So is it just this straightforward ?
That I find the B-field from A, and since $V = 0$, the E-field is zero, or am I doing it wrong ?
Thanks in advance :)
| The expression for the electric field includes derivatives of the vector potential with respect to time.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/65989",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
A Musical Pathway Using a small number of sound emitters, could you create a room where certain nodes emitted particular tones, but no meaningful sound was heard anywhere else.
So, for example, by walking down a certain path, you could hear the tones for "Mary Had a Little Lamb." Is there a generalized algorithm to make particular paths for particular tone sets?
| Wavefield Synthesis can do this but not with a small number of emitters, uses a massive array of speakers create a field effect, and phase alteration can reposition sounds within that field.
http://en.wikipedia.org/wiki/Wave_field_synthesis
Downside is it's not perfect, it takes a big array of speakers placed very accurately, works best on higher frequencies. I've experienced the setup in the Technical University of Berlin, you should contact them, but it's 2400+ speakers embedded in the walls of the room so not exactly portable. I heard a few demonstrations, some using speech and some using musical tones and timbres. It worked about 70% of the time, but it's quite fragile.
http://www.four-audio.de/en/references/wave-field-synthesis.html
I've heard of people doing portable wavefield with about 40 speakers (still expensive given the quality of speaker required) but no reports on its reliability.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/66129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 7,
"answer_id": 1
} |
Harmonic oscillator - wavefunctions I understand now how I can derive the lowest energy state $W_0 = \tfrac{1}{2}\hbar \omega$ of the quantum harmonic oscillator (HO) using the ladder operators. What is the easiest way to now derive possible wavefunctions - the ones with Hermite polynomials?
I need some guidance first and then I will come up with a bit more detailed questions.
| I think the easiest way to do this is to avoid solving differential equations to the greatest extent possible. There is, in fact, a way to use ladder operators and only requires you to solve one, fairly easy differential equation;
First, we note that the ladder operator technique can be used to derive the entire spectrum of one-dimensional harmonic oscillator.
$$
E_n = (n+\tfrac{1}{2})\hbar\omega
$$
The technique can also be used to show that the corresponding, properly normalized eigenvectors satisfy the following properties
$$
a^\dagger|n\rangle = \sqrt{n+1}|n+1\rangle, \qquad a|0\rangle = 0
$$
The left-hand property shows that, once one has one of the eigenstates, every other eigenstate corresponding to a higher eigenvalue can be obtained by applying the raising operator. In particular, if one knows the position basis representation of the ground state, then one can obtain the position basis representation of every other eigenstate by applying the position basis representation of the raising operator.
The right-hand property shows that the ground state is annihilated by the lowering operator. Writing this condition in the position basis, one obtains a simple differential equation for the ground state wavefunction, and then, per the left-hand property, one generates all other wavefunctions.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/66215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Momentum of particle in a box Take a unit box, the energy eigenfunctions are $\sin(n\pi x)$ (ignoring normalization constant) inside the box and 0 outside. I have read that there is no momentum operator for a particle in a box, since $\frac{\hbar}{i}\frac{d}{dx}\sin(n\pi x)=\frac{\hbar}{i}n\pi\cos(n\pi x)$ and this isn't 0 at the end points. Nonetheless, we can write $\sin(n\pi x)=\frac{e^{in\pi x}-e^{-in\pi x}}{2i}$, which seems to imply that there are two possible values of momentum: $n\pi$ and $-n\pi$, each with 50% probability.. Is this wrong? If you measured one of these momenta and the wavefunction collapsed to one of the eigenstates then it wouldn't solve the boundary conditions. So, what values of momentum could you obtain if you measured the momentum of a particle in a box?
Edit: I know that you can't measure the momentum of a particle exactly, but normally after a measurement of momentum, or such a continuous observable, the wavefunction collapses to a continuous superposition of momentum eigenstates corresponding to the precision of your measurement. But in this case since the wavefunction seems to just be a superposition of two momentum eigenstates, the wavefunction must have to collapse to one of them exactly, or so it seems.
| There are two different issues. One of them is the sign of the momentum; the other one is whether the momentum is spread (it's not because of the unnatural boundary conditions).
Concerning the first point, the standing wave (sine) is a real function and every real wave function has the same probability to carry momentum $+p$ and $-p$. So indeed, both of them are equally likely.
But even if you write the sine as a difference of two complex exponentials, it's still true that these exponentials aren't equal to the wave function everywhere – just inside the box – so it's still untrue that the momentum is sharply confined to two values $p$ and $-p$.
To get the probabilities of different momenta, you need to Fourier transform the standing wave – a few waves of the sine. One has
$$\int_0^1 dx\,\sin(n\pi x)\exp(ipx) = \frac{n\pi[-1 +e^{ip}(-1)^n]}{p^2-n^2\pi^2} $$
Square the absolute value to get the probability density that the momentum is $p$. The momentum $p$ should have the natural prefactors $\hbar/L$ etc. and the overall wave function should get another normalization factor for the overall probability to equal one. This changes nothing about the shape of the probability distribution: almost all values of $p$ have a nonzero probability.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/66429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 0
} |
Discovery of $E=hf$? How was the equation $E=hf$ discovered?
Was the proportional expression between energy and frequency of light $E\propto f$ discovered only by experiment? Or is there some logical(theoretical) senses affected?
| The relationship $E = h f$ was proposed by Max Plank in 1899 or 1900 as a way to "fix" a major problem in the existing understanding of the how light was emitted by hot bodies (the so called "ultraviolet catastrophe"). The conventional story holds that Plank did not consider this as fundamental.1
Later Albert Einstein took the idea as a way to explain the photo-electric effect in 1905, bringing the principle that light energy actually came in discrete chucks to the forefront. This work was among that cited by the committee in awarding Einstien's Nobel prize.
The discovery of Compton Scattering in 1923 gave the "photon" a firm place in modern physics.
Quantum field theories eventual came to explain the photon as an excitation of the electromagnetic field.
1 I can't say if that is true or not, but it is the way the Lore goes.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/66556",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Which causal structures are absent from any "nice" patch of Minkowski space? Which "causal separation structures" (or "interval structures") can not be found among the events in "any nice patch ($P$) of Minkowski space"?,
where "causal separation structure" ($s$) should be understood as a function from $n (n - 1) / 2$ distinct pairs (formed from $n$ elements/events of some set $E$) to the set of three possible assignments of "causal separation" (namely either "timelike", or "lightlike", or else "spacelike").
Of particular interest: what's the smallest applicable number $n$? --
for which a corresponding "causal separation structure" can be expressed which can not be found among the events in patch $P$; i.e. such that $E \, {\nsubseteq} P$.
Finally: please indicate under which conditions (on which sort of patches, necessarily different from "any nice patch of Minkowski space") the proposed structure could be found instead; or otherwise, whether it is "impossible" in general.
| It seems you don't us to differentiate between past time-like (null) and future time-like (null). Perhaps you have in mind more general spacetimes, but I'm having a hard time imagining a situation where one could distinguish time-like and space-like separated points but not past time-like and future time-like. If one was allowed to differentiate future and past then one would have a simple example with three points.
Without past and future, the minimal impossible configuration can be made with four points. Since any three points which are all lightlike to each other other determine a null ray one cannot have four points ABCD such that:
A, B and C are lightlike separated from each other, B, C and D are all lightlike separated from each other but A and D are spacelike or timelike separated.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/66636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Interaction of an electromagnetic wave with a two level system in the domain of quantum field theory Suppose I shine an electromagnetic wave on a two-level system. I need to describe how the system evolves in context of quantum field theory i.e. using a quantized EM field in the problem. The first step would be to write down the interaction Hamiltonian. What would it be?
| A single free-space field mode of the quantized electric field in the dipole approximation can be written as:
$$
\vec{\hat{E}}(t) = i\left(\frac{\hbar \omega}{2\epsilon_0 V}\right)^{1/2}\vec{e}(\hat{a}e^{-i\omega t} - \hat{a}^{\dagger}e^{i\omega t}),
$$
with $\omega$ the frequency of the field, V the volume in which the field 'lives' and $a$ and $a^{\dagger}$ the usual lower and raising operators.
(See for instance Chapter 2 Gerry & Knight - Introductory Quantum Optics).
The Hamiltonian for the atom interacting with the quantized field would be:
$H = H_{atom} + H_{field} + H_{int}$, where
$$
H_{int} = -\hat{d}\cdot\hat{E} = -i\left(\frac{\hbar \omega}{2\epsilon_0 V}\right)^{1/2}(\hat{d}\cdot{e})(\hat{a}-\hat{a}^{\dagger}).$$
The operator $d$ is the dipole moment operator.
(Note that the E field in the last equation has been taken in the Schrödinger picture!)
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/66707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Human power on treadmill On an elliptical treadmill a regular person can easily burn 1000 calories in one hour (treadmill reports calories burnt). This translates into:
$$(1\times 10^3\mathrm{cal/hr}\times 4.2\times10^3\mathrm{J/cal})/3.6\times 10^3\mathrm{s/hr} \approx 1.2 \; \mathrm{kW} \approx 1.5 \; \mathrm{hp}$$
On the other hand, Wikipedia says "A trained cyclist can produce about 400 watts of mechanical power for an hour or more..."
Is the problem that the treadmill gives wrong numbers? Or it is true that running using legs and arms - on elliptical machine or cross-country skiing which seems to be similar - a human can produce a lot more mechanical power than cycling? I thought the maximum power is set by the cardiovascular system, so it would be the same, running or cycling.
| This almost a duplicate of How efficient is the human body?. Humans are about 20% efficient at turning food into mechanical energy, so your cyclist generating 400W is metabolising food at about 2kW or about half a Calorie per second or 1800 Calories per hour.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/66780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Can a hybrid vehicle ever be more efficient than a hydrocarbon-only vehicle built with the same parts? Based on the laws of thermodynamics, shouldn't it be theoretically impossible for a non plug-in hybrid vehicle to ever be more fuel-efficient than a vehicle that connects the same engine directly to the wheels without converting it into electricity first?
The only energy input on a non plug-in hybrid is the gas tank, and therefore any energy leaving the vehicle in the form of torque must have originated in that tank. When the vehicle is driven, electrical energy leaves the battery and is converted into kinetic energy by the motors.
Once the battery is depleted, the gasoline engine begins turning, converting chemical energy into kinetic energy, which turns the wheels directly as well as turning an alternator which converts the kinetic energy into electrical energy, charging the battery.
But since all this energy is, no matter what, coming from that gas tank, it really doesn't matter what form it's in when it's turned towards moving the vehicle; it still started out as the same amount of chemical energy!
Furthermore, no conversion of energy can ever be one hundred percent efficient: some is inevitably lost to heat and other forms of unusable energy. So chemical -> kinetic -> electrical -> kinetic will always be less efficient than chemical -> kinetic.
The only ways I can think of a hybrid being more efficient is through energy recovery; the regenerative breaking system turns kinetic energy back into electrical instead of turning it into heat and wasting it like in normal brakes. But will the energy recovered by this method ever be greater than the energy lost by the extra conversions, as well as the increase inertia from the heavy components?
| Regenerative braking is one thing, another thing is that in practice large powerful engines are generally less efficient than small engines (because a substantial fraction of driving time the engine is not fully loaded, see detailed explanation in Wikipedia on Active Fuel Management), and having an auxiliary electric motor allows using a smaller gasoline engine - that's why hybrid wins in overall efficiency.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/66851",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Fermion annihilation amplitude Is there a physical reason why annihilation of 2 identical fermions with mass m to 2 scalars amplitude for $s=4m^2$ (fermions at rest) is zero? For example we can have 2 scalars annihilating in 2 scalars and the amplitude is non-zero if we have a quadrilinear term in the largangian $\phi_1^2 \phi_2^2$. Also a decay of a scalar to 2 fermions at $s=4m^2$ (m - fermion mass) is zero.
I consider a Yukawa lagrangian $Y\bar{\psi}\psi\phi$, an annihilation in t and u channels at the tree level. I actually am interested in amplitude modulus squared and take an average over all spin states.
| For generally valid reasons, the amplitudes analytically depend on the momenta and on $s$ in particular. There can't be any "jumps", the simplest discontinuities. Because the processes you enumerated with the total energy $E_{cm}\lt 2m$ i.e. $s\lt 4m^2$ are strictly prohibited by energy conservation – $2m$ is the minimum energy two fermions may have, as Trimok said – and because of the continuity, the amplitude has to be zero not only for $s\lt 4m^2$ but also for the borderline case $s=4m^2$.
The argument in the previous paragraph was sloppy and neglected the fact that the prohibition of the processes may come from $M=0$ or from the kinematic factors and that's why the cases of bosons and fermions are different. The amplitudes near the threshold of production – in our case, $s=4m^2$ – are increasing as a higher power law i.e. they are more suppressed for the fermions than for the bosons. This ultimately boils down to the normalization of the scalar-like or spinor-like solutions of the free equations of motion.
To fully understand how it works, just do the careful calculation of your simple two cases and carefully watch the powers of $(E_{fermion}-m)$ that appear in the kinematic factors and the dynamic amplitudes both in the bosonic and fermionic case.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/67085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
On Einstein notation with multiple indices On Einstein notation with multiple indices: For example, consider the expression:
$$a^{ij} b_{ij}.$$
Does the notation signify,
$$a^{00} b_{00} + a^{01} b_{01} + a^{02} b_{02} + ... $$
i.e. you sum over every combination of the indices? Or do you sum over the indices at the same time, i.e. they take on the same values:
$$ a^{00} b_{00} + a^{11} b_{11} +... ?$$
| You would sum over every combination of indices that match. So the i should match, and the j should match. For instance, if each is from 1 to 3, you would get:
$a^{11}b_{11}+a^{12}b_{12}+a^{13}b_{13}+a^{21}b_{21}+a^{22}b_{22}+a^{23}b_{23}+a^{31}b_{31}+a^{31}b_{32}+a^{33}b_{33}$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/67295",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Composition of squeeze operators? I'm wondering if it exists a composition law for the squeezing operation ? I guess so for geometric reason, since they are (generalized, and the phase is annoying of course) hyperbolic rotations of the annihilation $a$ and creation $a^{\dagger}$ operators of some bosonic modes.
I define the squeezing operator as $$ S\left(\zeta\right) = e^{\Sigma} \;\; ; \;\; \Sigma = \frac{\zeta^{\ast} aa -\zeta a^{\dagger}a^{\dagger}}{2}$$
for any complex parameter $\zeta$.
I would like to know the rule for $S\left(\zeta_{1}\right)\cdot S\left(\zeta_{2}\right)$ ?
For instance, we know that
$$D\left(\alpha\right)\cdot D\left(\beta\right)=D\left(\alpha + \beta\right) e^{\left(\alpha \beta^{\ast}-\beta\alpha^{\ast} \right)/2}$$
for the coherent / displacement operator $D\left(\alpha\right)=e^{\Delta}$ with $\Delta = \alpha^{\ast} a - \alpha a^{\dagger}$.
A reference for $S\left(\zeta_{1}\right)\cdot S\left(\zeta_{2}\right)$ would be sufficient.
| You will not have easily a closed general formula, because, first, the commutator of $aa$ and $a^+a^+$, so $4(a^+a + 1/2)$ does not commute with $aa$ and $a^+a^+$, and secondly,worse, because the same thing happens for all the high-order commutators of the Baker-Campbell-Hausdorff relations
(see more precisely Chapter "Zassenhaus formula")
It works with the $D\left(\alpha\right)$ because the commutator of $a$ and $a^+$ is $1$, so of course you have $[a, 1] = [a^+, 1] = 0$, and the infinite series of terms begin a short finite list (with $X \sim~ a, Y ~ \sim a^+$ ):
$$e^X e^Y = e^{X + Y}e^{\frac{1}{2}[X,Y]}$$
Maybe, you could calculate first the value of your operator in the ground state, that is :
$$ <0|S\left(\zeta_{1}\right)\cdot S\left(\zeta_{2}\right)|0>$$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/67371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
} |
Can deterministic world view be denied by anything other than quantum mechanics If we ignored quantum mechanics and looked at the world with a deterministic Newtonian view. Does not that mean that there is no randomness and that if all the information of the state of the universe during the big bang is accessible one can predict the state of the universe at any period of time and predict that I am writing this question right now.
Of course something like that denies the free will but I am asking if there is any thing other than quantum mechanics that denies the deterministic world view.
| From the theory of highly dynamic systems and chaos, perfectly classical and in principle deterministic systems can exhibit a behavior, where minute perturbations of the initial conditions are exponentially enhanced over time, and thus arbitrarily small perturbations can, after a finite time span, lead to a state that bears no resemblance at all to the evolved state of the unperturbed system.
Of course, this system is deterministic if you know all the initial conditions exactly. Not just to high precision, but exactly. But that would generally (since the rational numbers are not a dense set and therefore have zero probability to be occurring exactly in a real-life setting) require an infinite amount of information, which is of course impossible - so your system is really, fundamentally unpredictable, although still in principle deterministic - this is pretty much the definition of "chaos".
According to people more clever than me, quantum mechanics is also deterministic, by the way - see the question that @Greg links to in his comment to your question.
I do not know if that rules out "free will", because I have never seen a consistent and coherent definition of what "free will" is, but it should be clear that it leaves ample room for complexity and unpredictability.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/67517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Why do current-carrying wires heat up? Obviously wires heat up too, but why do they heat up? And for the same reason, why do we get electrical burns?
| The current in the conductor is due to the drifting of electrons inside a conductor in a direction opposite to the flow of electrons
During their drifting they collide with their atoms vibrating about their mean position and lose some of kinetic energy to the vibrating atoms which increases the amplitude of the vibration of the atoms thermal energy of the metal with the corresponding rise in temperature of the conductor
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/67639",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Navier-Stokes equations: conservation of momentum The first Navier-Stokes equation (conservation of mass) says:
$\vec \nabla \cdot \vec v=0$
For a stationary flow, the l.h.s of the second equation is (conservation of momentum):
$\rho \frac{D\vec v}{Dt}=\rho (\underbrace{\frac{\partial \vec v}{\partial t}}_{=0} + (\vec v\cdot \vec \nabla) \vec v)=\rho ( \underbrace{(\vec \nabla \cdot \vec v)}_{=0??} \vec v)\stackrel{??}{=}0$
I find that the l.h.s of the conservation of momentum equation is always equal to zero for a stationary field.
I know this isn't true but where am I wrong in this reasoning ?
| Your mistake here is to assume that the multiplication $\vec v\cdot \vec \nabla$ is commutative. It is not; the dot product here is just a convenient mathematical notation. This part of the Wikipedia article on Navier-Stokes equations explains how to interpret this term.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/67699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Dissolving oxygen into water I was wondering how one would actually calculate how much oxygen would dissolve into water given the necessary initial conditions, and what those initial conditions would need to be. I assume they would be pressure, and initial concentration, but I really don't know where I would go from there. Clearly air and water have different concentrations of gases and liquids, despite having been in contact for thousands of years. And once in water, is oxygen still considered gaseous? I assume it is, but why is it called gaseous-what quality of it deems it a gas despite being surrounded by liquid?
| In water the gas is in aqueous state, because gas molecules are interacting with molecules. At equilibrium the concentration of the gas in water and above water are different, with the ratio being a constant of Henry's Law.
Check out http://en.wikipedia.org/wiki/Henry's_law for a list of constants.
The most anomalous constant is for CO2, which despite being 1/500 as abundant in air as O2, in water it is 50/1. this is due to CO2 actually being able to chemically react with water. Granted, this is a special case but for other gas molecules the dominant forces are polar or van der Waals'.
O2 molecules have two free electron pairs for intermolecular interactions while N2 only have one pair. Electron density is a key factor in vdW forces, so this is the reason O2 and N2 have similar equilibrium concentrations in water despite N2 being 4/1 in atmospheric abundance/partial pressure
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/67764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Why do metal objects in microwaves spark? I heard that electrons accumulate at points on metals, and this clearly explains the arcing phenomenon, but how does a microwave make an electron imbalance on the fork?
| The energy produced in a microwave oven is at a macro-scale wavelength (on the order of centimeters), and leaves the emitter as a coherent wave - one wavelength, phase, and polarization. Despite the "stirring fan" and other measures, objects in the oven will still experience (relatively) orderly macro-scale radio waves and therefore exhibit macro-scale effects. Radio waves (including microwaves) will induce voltage differentials and electrical eddy currents in conductors. So, foil (being a good conductor) in a microwave will end up with charges accumulating here and there (dependent on shape and orientation with respect to the incident waves); voltages can be high enough to break down / ionize the air between points on the foil, and separated by sufficient distance to cause the visible sparking and arcing you see.
On the other hand, visible light is at a micro-scale wavelength (in the range of 500 nanometers). Ordinary light sources (daylight, incandescent, fluorescent, etc.) are essentially streams of photons of random wavelength, phase, and polarization(although fluorescent light occurs at more specific wavelengths, it's otherwise a random emission of photons). As such, there is no macro-scale order as occurs with the microwaves produced in a microwave oven, and there is therefore no possibility of corresponding macro-scale effects.
Even laser light, which is coherent (uniform wavelength, phase, polarization) is still at a micro-scale wavelength. It could generate a plasma (dislocating electrons, etc.), but purely due to heating effects. It won't produce a sparking/arcing effect because the wavelength is too short to produce an effect with such a large scale order (voltage differential over millimeters to centimeters).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/67880",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 3,
"answer_id": 0
} |
Ball jumping from water Few days ago I played with ball(filled with air) in swimming pool. I observed interesting phenomenon.
When I released a ball from 3 meters depth the ball barely jumped above the water surface but when I released it from 50 cm depth it shoot out of the water like nothing.
I observed when released from 3 meter depth the ball goes up in zig-zag trajectory but from 50 cm depth is goes in straight line.
I would be very interested in calculating optimal depth from which the ball would jump the highest above the water surface. And as well I would like to calculate trajectory of the ball under water.
It is obvious that simple drag formula won't help here.
I guess that the zig-zag patterns is happening because there might be something like Karman vortex street behind the ball.
So have anyone idea how to calculate this? Or can you point me to the right literature?
Edit: I forgot one observation I made. It seamed to me that the ball when released from 3m depth was rotating when it hit a surface and that might prevent the jump.
| The answer is quite simple. The pressure exerted by the water column on the ball is more at 3m depth as compared to the pressure exerted on the ball at 50cm depth. As a result, due to the buoyancy of the ball it shoots out of the water with a greater force when it is released from a depth of 50cm.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/68055",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 4,
"answer_id": 2
} |
Fusion vs. Fission I understand why fission generates large amounts of energy when the nucleus is split, but then why does fusion generate such large amounts of energy. If fission releases energy when some mass is lost as energy, then shouldn't the fusion process absorb energy to fuse nuclei together?
I also am curious as to where the energy released from fusion comes from, while fission releases some of the energy of the strong nuclear force.
| The strong force is attractive at short range and wins over the electromagnetic repulsion between protons. Pull the protons apart a little and you will get fission because the electromagnetic force wins. Conversely, push protons together and you will eventually get fusion when the strong force takes over from electromagnetic. For large nuclei the electromagnetic energy required for fusion is larger than the energy returned by the strong force - so you will only get net energy out from fission. For small nuclei it's the other way round - the strong force releases more energy than the electromagnetic force takes to trigger fusion. In this case you will only get net enerrgy out from fusion.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/68138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Mechanics question (a block on top of a block) This question is very similar to this one here.
A block of mass $m_1$ is placed on another block of mass $m_2$ lying on a smooth horizontal surface.
The coefficient of friction (static and kinetic) between $m_1$ and $m_2$ is $\mu$.
Find the acceleration of the blocks if the force applied to $m_1$ is $5N$, given that $m_1 = 2kg$ and $m_2 = 4kg$ and $\mu=0.2$.
I can prove the result in the link, and obtain the critical force as $6N$, so for $5N$, they will have same acceleration, which will be equal to $\frac{5}{6}$.
But when I take the the general case, draw free body diagrams, I get these equation -
$$
m_1 a_1 = F - \mu m_1 g
\\
m_2 a_2 = \mu m_1 g
$$
This gives the answer as $a_1 = \frac{1}{2}$ and $a_2 = 1$
The only place where I think I could have made a mistake would be in case of force of friction.
I am taking the maximum value of friction ($4N$), but since the force applied is greater, I presume this can happen.
Can anyone tell me where I go wrong?
| Because you are not pulling with the critical force, $6N$, then static friction $F_f<\mu m_1g$, where $\mu=0.2$.
The equations you get from Newton's second law are:
$$F-F_f=m_1a$$
$$F_f=m_2a$$
Substitute $m_2a$ into $F_f$ in the first equation:
$$F-m_2a=m_1a$$
$$F=a(m_1+m_2)$$
$$\frac{F}{m_1+m_2}=a$$
$$\frac{5 \mathrm{N} }{2\mathrm{kg}+4\mathrm{kg}}=\frac{5}{6}\frac{\mathrm{m}}{\mathrm{s^2}}$$
The applied force is allowed to be greater than the friction force between the blocks because they are not moving relative to each other. Notice that no matter what the force you apply on the top block, both blocks will have non-zero acceleration, because the horizontal surface is frictionless.
From what we have above, you can calculate the friction force as:
$$F_f=m_2a=(4\mathrm{kg})\left(\frac{5}{6}\frac{\mathrm{m}}{\mathrm{s^2}}\right)=3.3\mathrm{N}<4\mathrm{N}=\mu m_1 g$$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/68366",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
No well-defined frequency for a wave packet? There are similar questions to mine on this site, but not quite what I am asking (I think). The de Broglie relations for energy and momentum
$$ \lambda = \frac{h}{p},
\\
\nu = E/h .$$
equate a specific frequency and wavelength to a particle, yet we know that a wave packet
is a linear combination of an infinite range of frequencies and wavelengths. How is it that we (or nature) choose one frequency and wavelength out of the range? Does this have to do with the collapse of the wave packet when measured? And if so, is the resulting measured frequency a random outcome? Similarly, when an electron jumps from one energy level to another in an atom, it emits a photon of frequency
$$ \
\nu = \Delta E/h .$$
Since the photon is not a pure sinusoidal wave, how can a single frequency be ascribed to the photon?
| Let's say an isolated atom emits a photon. The excited state in the atom has some lifetime $\tau$. Through the energy-time uncertainty relation, that gives the excited state some uncertainty in energy $\delta E\sim h/\tau$ (not the same as $\Delta E$, which is a difference in energy between atomic states). The photon then has the same uncertainty $\delta E$ in its energy, which corresponds to an uncertainty in frequency. The photon isn't in an eigenstate of energy.
For many real-life examples such as a visible photon emitted by a hydrogen atom, or gamma-rays emitted by beta-decay daughters, $\tau$ is very long compared to $h/\Delta E$, so we have $\delta E \ll \Delta E$. The uncertainty $\delta E$ is also often very small compared to the limitations imposed by, e.g., Doppler shifts or the resolution of the detector.
Yes, when you measure the energy of the photon, you get a random outcome. However, there is a quantum-mechanical correlation between this energy and the energy of the atom, so that energy is exactly conserved (not just statistically, on an average basis).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/68506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Neutrinos: how can they carry information about universe? I know that neutrinos are particles with a very small mass and no electric charge. They infrequently interact with matter and so they can give us information about the "old" universe. But how can they do it?
| Note: I am not an expert on neutrinos, so if I have missed anything, someone please let me know.
For one thing, the knowledge of the mere existence of neutrinos is important for a full understanding of our universe. Also you may have heard that for a long time neutrinos were thought to be massless, but the observation of neutrino flavor oscillation requires them to have mass. This is a bit of a problem as the standard model of particle physics doesn't account for non-zero neutrino masses, which is an indication of some new physics, beyond the standard model.
As you mentioned, neutrinos are very weakly interacting with other matter, making it hard to detect them, which is why neutrino detectors are built on such large scales (see http://icecube.wisc.edu and http://www-sk.icrr.u-tokyo.ac.jp/kam/kamiokande.html for example). But this can be useful in terms of things we may want to find out about our universe. This very weakly interacting behavior enables physicist to see the neutrino spectrum from a star, supernova, gamma ray bursts, etc. virtually unchanged on the way from the object to Earth. Meaning that information about objects very far from Earth (and therefore far back in time) can be found where conventional techniques using light are less successful (for example if there is dust or some other object in between Earth and the object of interest - the light would interact whereas the neutrinos pass through pretty much unchanged).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/68909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
If the multiverse theory is true, can there be a Universe where there are different laws of physics? This is probably a very difficult question. But my question is essentially this, if there are other Universes can different laws of physics exist in those Universes and if so, can't there be a Universe where the laws of physics are so different that the multiverse theory has to come out as false?
| this is of course, entirely philosophical, and cannot be answered from physics considerations. But it is a fair question nonetheless, so i'll take a shot.
The only notion of metaverse that makes sense to me is Tegmark's vision of mathematically consistent universes. This merges with the anthropic principle in order to state that universes have few constraints on their laws, which is that they are Gödel-complete, and that they allow sentient beings to develop that can do interesting questions about existence as we are just doing here. Any mathematical set of evolution rules that can give rise to life, will have solutions where intelligent entities that do those questions.
Unless some fundamental reason make the laws of our universe 'special' in a way that makes it unique regarding it's ability to host intelligent life, most probably, there is a huge family, probably infinite, of set of physical laws that can host intelligent life.
Even in the specific case of string theory universes, there might be a lot of different vacuums that can host life. This is still a daunting technical problem to analyse, but at least is approachable in principle.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/68987",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
About the microscopic form of magnetocrystalline anisotropy Currently people write magnetocrystalline anisotropy as
$H_{an}=-K s_x^2$
from its classical counterpart:
$H_{an}=-K ( \sin \theta)^2$
where $K$ is the anisotropy constant, but for spin 1/2, $s_x^2$ is just the identity matrix, which shows no anisotropy, so how to write the correct form of microscopic Hamiltonian?
I also heard that magnetocrystalline anisotropy comes from spin-orbit interaction, is there some paper depict there relations? like deriving the Heisenberg Hamiltonian $S_i \cdot {S_j}$ from Coulomb interaction by Dirac?
| A couple of points:
*
*For $S=\frac{1}{2}$, the expectation value for the "single-ion" magnetocrystalline anisotropy is indeed zero, and it should be! That is, if we consider a lattice of sites labelled by $i$, then $(\sigma_i^x)^2=\mathbb{I}$, the identity matrix, and there is indeed no anisotropy as you say. There is a general result that applies here called Kramer's theorem, which says that, so long as time reversal symmetry remains unbroken (i.e. no magnetic field), then the $S=+\frac{1}{2}$ and $-\frac{1}{2}$ have to be the same energy. So where does the anisotropy come from? Two places: (a) The spin at a given atomic site may be more than $S=\frac{1}{2}$, in which case the spin operators take a different form, and don't square to the identity. (b) The anisotropy may originate from interactions between different atomic sites, so the Hamiltonian reads $\sum_{i,j} \sigma_i^x \sigma_j^x$, which works just fine.
*Although Moriya's paper is the classic reference for this material, the derivation is actually (slightly) incorrect. There was a series of papers in mid 90's that corrected it, for example: Phys. Rev. B 52, 10239–10267 (1995).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/69058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Why is there a line in the middle of the Planck full sky map? Why there is a straight line (or perhaps a flat surface) in the middle of the Planck full sky map?
(source: esa.int)
and zoom
(source: handshakemag.com)
(it makes it seem that the big bang was a collision between two unknown surfaces)
| What you're seeing is the galactic plane of the Milky Way. There are several processes emitting at a variety of wavelengths, and because the emission is so nearby it comes out quite bright on the all sky map. Planck observes at about $\mathrm{mm}$ wavelengths, so the most prominent emission within the galaxy is thermal emission from dust clouds.
Often maps are "cleaned" by coming up with a model for the galactic emission and subtracting it, leaving approximately only extragalactic emission. Alternately if data in the galactic plane is unobserved, the missing region may be fit according to an appropriate model (using the observed region as a boundary condition) and filled in.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/69127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Gravitational field strength and Horizon in Rindler coordinates I came across the following statements in 't Hooft's black holes notes, but not being able to justify them.
The metric in the Rindler coordinates $x=\tilde{x}, y=\tilde{y}, z= \rho \cosh{\tau}, t= \rho \sinh{\tau}$ is
$$ds^2 = -\rho^2 dt^2 + d\rho^2 + d\tilde{x}^2+d\tilde{y}^2$$
*
*Gravitational Field Strength
The actual gravitational Field strength felt by the (Rindler)observer
is inversely proportional to the distance from the origin.
How can I see this from the metric? How do I mathematically and physically justify this claim?
2.
Horizon
Why does the surface $x=t$, act like a Horizon i.e. anything from outside the shaded area shown in the picture can't enter into the shaded region? I think, this can be seen by showing that all time-like curves starting from a point outside the shaded area don't enter the shaded region. But what are all the time-like curves in the Minkowski space-time? and would this assumption still hold for any arbitrary geometry which is locally like the Rindler metric(e.g. the schwarzchild metric near the horizon?)
| 1)Gravitational Field Strength
You have to consider the analogy with a uniform accelerated move, in special relativity, that is :
$z' = \frac{1}{a} ch (a \tau'), t' = \frac{1}{a} sh (a \tau')$
Here $a$ is the acceleration and $\tau'$ is the proper time.
You get : $dz'^2 - dt'^2 = -d\tau'^2$,as wished.
Now, make the coordinate change $\tau = a \tau'$, we get :
$z' = \frac{1}{a} ch (\tau), t' = \frac{1}{a} sh (\tau)$
$dz'^2 - dt'^2 = - \frac{1}{a^2} d\tau^2$.
But this is the same result that the Rindler metric with $d\rho=d \tilde x =d \tilde y = 0$, so we make the identification between $\rho$: and $\frac{1}{a}$, so the gravity strengh (acceleration) is inversely proprotionnal to the space-like quantity $\rho$
2) Horizon
I am not sure to understand all your questions, but you have to be careful about several points :
The Minkowski metric of the form : $-dt^2 + dz^2 + d \tilde x^2 + d \tilde y^2$, linked to the Rindler metric, by the transformations $z= \rho \cosh{\tau}, t= \rho \sinh{\tau}$, is only correct :
*
*near the horizon
*for a small angular region, that could be considered centered at $\theta = 0$
The horizon itself is not a 3 dimensional-surface in space-time, because the horizon corresponds to $\rho=0$, that is $t=z=0$.
This is because the metrics $g_{00} = 0$ at the horizon, so the horizon has no extension in the time coordinates.
So, considering that the horizon is $z^2 - t^2 = 0$ is not correct, the horizon is $t=z=0$.
A correct global representation is then using Kruskal–Szekeres coordinates, which gives a correct global point of view.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/69198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Doubt over Kelvin-Planck statement of thermodynamics' second law I have a doubt over the Kelvin and Planck's statement of thermodynamics' second law, in particular applied to a cycle. Let's take a Carnot cycle as an example, and let's call the first two transformations (the isotherm and the adiabatic) done. Now, isn't it obvious that the machine has to give up heat to go back to the initial state? Isn't it something that follows from the fact that the cycle has to be continuous?
| Every thermodynamic system satisfies the first law during a given process;
$$
\Delta E = Q-W
$$
Here $\Delta E$ is the change in its internal energy, $Q$ is the heat transferred to the system, and $W$ is the work done by the system. For a system undergoing a cyclic process, namely one for which it starts and ends in the same thermodynamic state, one has $\Delta E = 0$, and the first law then tells us that
$$
Q = W
$$
Now, suppose that the system is taking some heat $Q_H>0$ from a reservoir, and turning it into some work $W$. Let's define
$$
Q_\mathrm{exhuast} = Q-Q_H
$$
then we can write
$$
Q_H + Q_\mathrm{exhaust} = W
$$
The Kelvin statement tells us that we must have $Q_\mathrm{exhaust}\neq 0$ because otherwise, we would have $Q_H = W$; the sole result of the process would have been to transform the heat it took in into work. The symbol $Q_\mathrm{exhaust}$ is what one commonly calls the exhaust heat.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/69378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
is it possible to load a transversal wave with a longitudinal wave like in communication engineering for sending information what we do,
we take a high frequency carrier and modulate it with the message signal
so can we do the same thing like take a high frequency carrier (transversal wave)
but the message is not transversal wave rather the message is longitudinal wave
say a sound wave
so can we modulate or load a transversal wave with a longitudinal wave ??
also if there is any work on this regard .. kindly post the link also??
| That is what FM Radio works on.
So lets say we have an entirely analogue setup - a mic as input (RJ talking for example) and that being broadcast over a 91 MHz Signal real-time = Audio wave over a High frequency carrier. There is no digitization as you are not sending 0s and 1s here.
However Digital audio over FM has its advantages since the variations are only either zeros and ones, the amplitude can be made higher so there are no dropped bits. This enables the decoder on the receiver to Maintain hi Fidelity to the original broadcasted signal.
The analogue signals can pickup interference on the way since the source audio signal has very fine variations thus a slight disturbance can alter the signal.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/69509",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Operator on Function of Momentum (QM) I have exactly 0 clue on how to start this problem, but I would be forever grateful for a hint in the right direction.
Given the operators $\hat x=x$ and $\hat p=-i\hbar \frac{d}{dx}$, prove the following relation:
$$
[\hat x, g(\hat p)]=i\hbar\frac{dg}{d\hat p}.
$$
| Deduce the general form of the commutator
$$[\hat{x},\hat{p}^n] $$
write your function as a power series of $\hat{p}$
$$g(\hat{p})=\sum_{n=0}^{\infty}g_{n}\hat{p}^n $$
apply linearity of the commutator and then you should get your result
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/69577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Static Spherical symmetric solution of Einstein's equations with a perfect fluid I am reading Wald for the interior solutions of a static spherical metric. Assume it to be of the form
$$ds^2 = -f(r)dt^2 + h(r)dr^2 + r^2 ( d{\theta^2} \sin^2{\theta}d{\phi^2})$$
Wald states: For a perfect fluid tensor $T_{ab}= \rho u_a u_b + P ( g_{ab}+ u_{ab})$
In order to be compatible with the static symmetry of space time, the
four velocity of the fluid should point in the direction of the static
killing vector $\xi^a$
i.e. $u^a=-(e_0)^a=-f^{\frac{1}{2}}(dt)^a$
EDIT: It also seems $(e_0)_a=f^{\frac{1}{2}}(dt)_a=f^{-\frac{1}{2}}(\frac{\partial}{\partial t})_a$. Please could someone tell, why this is so?
*
*First, why is the the static killing vector $\frac{\partial}{\partial t}$ equal to $-f^{\frac{1}{2}} dt$?
*Second, why is the velocity, along the killing time vector? What would happen if there is a component perpendicular to it? Does this mean, the fluid doesn't move through space?
|
*
*First, why is the the static killing vector $\frac{\partial}{\partial t}$ equal to $-f^{\frac{1}{2}} dt$?
The vectors $\partial/\partial t$ and $-f^{\frac{1}{2}} dt$ are not equal to each other. They're parallel to each other, and the factor of $-f^{1/2}$ is just so that the four-velocity is properly normalized. If you plug $u$ into the metric, you have to get 1.
*
*Second, why is the velocity, along the killing time vector? What would happen if there is a component perpendicular to it? Does this mean, the fluid doesn't move through space?
The Killing vector supplies a preferred frame of reference, one in which the observables (e.g., curvature scalars) stay constant. A perfect fluid is one for which there exists a frame such that the stress-energy tensor is diagonal; in this frame, there is no spatial flux of energy-momentum. So basically this means that in this frame, the fluid doesn't move through space (in the sense that there's no flux).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/69725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Vacuum polarization in QCD and gluon bubbles In analogy to QED, the following Feynman diagram is a diagram contributing to the vacuum polarization effect, leading to anti-screening, asymptotic freedom and running of the strong coupling constant. It can also be interpreted as a correction to the propagator of the gluon.
(Of course, there's no true analogy to this diagram in QED because photons do not couple to themselves, and here we make use of the existence of a 3-gluon vertex. In QED we would have a fermion-antifermion pair instead of bosons in the loop, similar to a gluon going to fermion-antifermion and back in QCD.)
My question now is about the following diagram which has a 4-gluon vertex:
Does this also contribute to vacuum polarization and the effects it entails? Or is it something conceptually different? (It seems different because we don't have the splitting of one object into two and back.)
(PS: Feel free to add more meaningful tags and correct any statements in the above which are wrong.)
| Yes, this is another diagram that contributes to "vacuum polarization", also called self-energy. If you want to make a consistent calculation of the self-energy in perturbation theory, you need to include the two diagrams that you mention, plus the diagram where the gluon splits into $\bar{q}q$, plus, depending on your gauge choice any ghost diagrams.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/69933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
The General Relativity from String Theory point of view I have a hard time understand the statement that
When you only look at the classical limit or classical physics, string
theory exactly agrees with general relativity
Because from what I know, String Theory assumes a fixed space time background (ie, all the strings and membranes interact in a fix background, and their interaction gives rise to fundamental particles that we observe), but General Relativity assumes that the space time background is influenced by what is in it and the interaction between them.
Given that both have very different assumptions, what do string theorists mean when they say string theory agrees with general relativity in a classical limit? Or more specifically, how does string theory--a fix spacetime background theory-- reconciles with the general relativity on dynamic spacetime background part? I can understand a fix, static spacetime in the context of changing, dynamic spacetime background, but I cannot understand a chanding, dynamic spacetime in the context of a fix, static spacetime background.
|
UPDATE: I have written a more complete answer here: How do Einstein's field equations come out of string theory?
The effective gravitational terms of the spacetime action, which can be derived from the Polykov action (gravitons are bosons) are --
$$S_{G}=\lambda\int\left(R+\ell_s^2R_{\mu\nu\rho\sigma}R^{\mu\nu\rho\sigma}\right)\mbox{ d}^D x$$
Where we neglected terms of order $\ell_s^4$ and greater. To first-order in $\ell_s$, the string length --
$$S_{EH}=\lambda\int R\mbox{ d}^D x$$
Which is the $n$-dimensional Einstein-Hilbert Action.
The vacuum EFE may also be derived directly from setting the beta functional, which measures the breaking of conformal invariance, to zero:
$$\beta^G_{\mu\nu} = \ell_s^2 R_{\mu\nu}+\ell_s^4R_{\mu\nu}R_{\mu\nu\rho\sigma}R^{\mu\nu\rho\sigma}+... = 0$$
For weak gravity --
$$R_{\mu\nu}=0$$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/70060",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 1
} |
How is energy extracted from fusion? I understand that combining deuterium and tritium will form helium and a neutron. There are three methods to do this (1) tokamak (2) lasers and (3) cold fusion. I would like to know after helium is formed. How is that energy extracted from tokamak and stored?
| If you search the ITER site, ITER being the international prototype fusion reactor which will demonstrated the possibility of getting megawat useful energy from fusion, one sees that their main aim is to demonstrate this feasibility:
The main carrier of energy out of the plasma is the neutron, and methods to efficiently use this energy have not been developed yet, but wait for the commercial prototype.
The helium nucleus carries an electric charge which will respond to the magnetic fields of the tokamak and remain confined within the plasma. However, some 80 percent of the energy produced is carried away from the plasma by the neutron which has no electrical charge and is therefore unaffected by magnetic fields. The neutrons will be absorbed by the surrounding walls of the tokamak, transferring their energy to the walls as heat.
In ITER, this heat will be dispersed through cooling towers. In the subsequent fusion plant prototype DEMO and in future industrial fusion installations, the heat will be used to produce steam and—by way of turbines and alternators—electricity.
They have developed methods for cooling the system and dissipating the energy to the environment.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/70209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
Full time-derivative of a function and Schrodinger equation From Hamiltonian formalism there is well known equation,
$$
\frac{d F}{dt} = \frac{\partial F}{\partial t} + \{F, H\}_{PB},
$$
where $ \{H, F\}_{PB}$ is the Poisson bracket.
After using Hamiltonian formalism in quantum mechanics it transforms into
$$
\frac{d \hat {F}}{dt} = \frac{\partial \hat {F}}{\partial t} + \frac{i}{\hbar }[\hat {H}, \hat {F}]
$$
in the Heisenberg picture, where $[\hat {H}, \hat {F}]$ is the commutator.
How can I get Schroedinger equation from this expression?
| In the Heisenberg representation, the "equivalent" of the Schrodinger equation is :
$$\hat H(t) = \frac{\hat P^2(t)}{2m} + V(\hat X(t))$$
with $[\hat P(t), \hat X(t)] = i\hbar$
If you are looking at eigenstates and eigenvalues of the hamiltonian, you will look for a constant Hamiltonian.
For instance, for the harmonic oscillator, you will have :
$$Constant ~ operator ~ \hat H(t) = \frac{\hat P^2(t)}{2m} + \frac{m \omega^2}{2} \hat X^2(t)$$
For the harmonic oscillator, solutions are :
$\hat X(t) = \sqrt{\frac{\hbar}{m \omega}} (a e^{i \omega t} + a^+e^{-i \omega t})$, with $Constant ~ operator ~ \hat H(t) = (a^+a + 1/2) \hbar \omega$
Here $a$ and $a^+$ are constant operators such as $[a,a^+] = 1$, explicitely the only non-null members of the operator $a$ are $a_{n+1,n} = \sqrt{n}$
Explicitely, $\hat H(t)$ is a diagonal operator, with $H_{nn} = (n+1/2) \hbar \omega$
In this representation, the "wavefunction" is simply a vector in the vector basis upon which these operators act.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/70466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Movie airplanes and suction Having watched a recent action movie (with zombies in it) I wondered whether the suction from a hole in the airplane's hull would really be able to rip out luggage, persons and even seating benches.
To my understanding, "suction" is nothing but the lack of pressure, i.e. the suction in the airplanes cabin can only be proportional to the pressure in there. Bernoulli principle would quickly void the plane of breathable air, justifying the oxygen masks that drop from the cealing. In this short period things would get pushed towards the hole by the air leaving the cabin.
Afterwards though, once the pressure in the cabin has dropped significantly, suction would cease, as there is no more air that can exert force onto objects.
The question now is: Is the above a correct description of the circumstances? And if yes, how large could the forces be? Would they be strong enough to rip people out of their seats or even the whole seat out of the airplane, or is this another cinematic exaggeration?
| I guess this depends on the size of the hole and the altitude of the plane. You are right that the suction effect will last only as long as there exists a pressure differential between the cabin and the outside. The hole's size determines the rate of equilibration. For example, in the James Bond movie Goldfinger, a firearm is triggered in an airplane. Goldfinger flies out the window making some funny noises, too. This is certain to be highly exaggerated and is addressed in German physics professor Metin Tolan's book. In reality, even with a whole window gone, there would be no danger to the passengers from suction. (Lack of oxygen is compensated for by masks, freezing is not a problem as long as the pilot enters descent after the incident.)
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/70562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 1
} |
Biggest experimental validations of postulates of Quantum Mechanics What are some experimental results that validate postulates of Quantum mechanics completely beyond any doubt ? Since there are alternate theories being used by various physicists to describe the same phenomena, I wonder there are some results that are with complete exclusion, described by Quantum Mechanical postulates ?
| There's no such thing as verifying a theory "to complete exclusion." One of the key principles of the scientific method is that theories can't be proven by experiment, only supported (or disproven as the case may be). The scientific reality is that quantum mechanics, like any other theory of its caliber, has been supported by countless experiments, so we assume, for all intents and purposes, that it's true. That's not the same thing as proving it's true.
The consequence of this as it pertains to your question is that there's no experiment that can disprove 100% of all the theories competing with quantum mechanics, since you can't prove quantum mechanics beyond a shadow of a doubt (the same way you can't 100% prove gravity, relativity, etc.). However, competing theories can be falsified on a case by case basis with experiments that show they make inaccurate predictions. So if you name specific theories that compete with quantum mechanics I can find experiments that disproved them.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/70623",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Divergence theorem in complex coordinates This question is related to Stokes' theorem in complex coordinates (CFT)
but, I still don't understand :(
Namely how to prove the divergence theorem in complex coordinate in Eq (2.1.9) in Polchinski's string theory
$$\int_R d^2 z (\partial_z v^z + \partial_{\bar{z}} v^{\bar{z}})= i \oint_{\partial R} (v^z d \bar{z} - v^{\bar{z}} dz ) (1) $$
I may try
$$ \int_R dx dy ( \partial_x F_y - \partial_y F_x) = \oint_{\partial R} (F_x dx + F_y dy)(2) $$, but what kind of substitution I should use to get Eq. (1)?
| Let $\sigma^1$ and $\sigma^2$ be real coordinates on $\mathbb R^2$. Using the results on page 33, we find that
\begin{align}
\partial_zv^z
&= \frac{1}{2}(\partial_1 -i\partial_2)(v^1 + iv^2)
= \frac{1}{2}(\partial_1v^1 + i\partial_1v^2 - i\partial_2v^1 + \partial_2v^2) \\
\partial_{\bar z}v^{\bar z}
&= \frac{1}{2}(\partial_1 +i\partial_2)(v^1 - iv^2)
= \frac{1}{2}(\partial_1v^1 - i\partial_1v^2 + i\partial_2v^1 + \partial_2v^2)
\end{align}
and therefore using $d^2z = dz\,d\bar z = 2 d\sigma^1d\sigma^2$
\begin{align}
\int_R d^2z\,(\partial_zv^z + \partial_{\bar z}v^{\bar z})
&= 2\int_R d\sigma^1\,d\sigma^2\,(\partial_1v^1 + \partial_2v^2)
\end{align}
similarly, for the right hand side we have
\begin{align}
v^zd\bar z
&= (v^1 + iv^2)(d\sigma^1 - id\sigma^2) = v^1d\sigma^1 - iv^1d\sigma^2 + iv^2d\sigma^1 +v^2d\sigma^2 \\
v^{\bar z}dz
&= (v^1 - iv^2)(d\sigma^1 + id\sigma^2) = v^1d\sigma^1 + iv^1d\sigma^2 + -iv^2d\sigma^1 +v^2d\sigma^2
\end{align}
so that
\begin{align}
i\oint_{\partial R} v^z d\bar z - v^{\bar z} d z
&= i\oint_{\partial R} 2i(v^2d\sigma^1 - v^1 d\sigma^2)
= 2\oint_{\partial R} (v^1 d\sigma^2 -v^2d\sigma^1)
\end{align}
The identity in Polchinski is obtained by setting the left and right hand sides equal to one another which, in this case, gives
\begin{align}
\int_R d\sigma^1\,d\sigma^2\,(\partial_1v^1 + \partial_2v^2)
&=\oint_{\partial R} (v^1 d\sigma^2 -v^2d\sigma^1)
\end{align}
which is precisely Stokes' theorem for a region in $\mathbb R^2$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/71168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
How does adding electrons break the angular momentum degeneracy? In the hydrogen atom, the energy does not depend on l. This degeneracy is sometimes called "accidental" (because it does not correspond to some symmetry?). However, there is l dependence in the energy multi-electron atoms. Since the original degeneracy was "accidental", is it no longer correct to talk about what breaks this degeneracy? Physically, I understand this is a result of the lower l-value states being concentrated closer to the nucleus, but is there a way to see that this would be the case in advance, without finding the eigenvalues of the Hamiltonian?
| You could say that the degeneracy for the H-atom is accidential, but this unique property that the energy does not depend on l is because of the fact that you have a pure Coulomb-potential. For this pure Coulomb potential there exists an operator:
$\textbf{R} = \frac{1}{2\mu}(\textbf{p}\times\textbf{l}-\textbf{l}\times\textbf{p})-\frac{e^2}{r}\textbf{r}$
which is derived from the Lenz-Rungevector of classical mechanics and has the properties:
$[\mathcal{H},\textbf{R}] = 0 = [\textbf{l},\textbf{R}]$
So actually it isn't an accidential symmetry!
The correct way to predict what symmetries are present is the machinery of group-theory (or if you want to go to practical calculations representation theory), this provides a systematical way to reduce your system to it's most fundamental symmetry-groups (irreducible representations).
Atomic physics always works the same scheme:
*
*Look for the operators $\mathbf{A}_i$ which belong to the symmetry group of the Hamiltonian (so $[\mathcal{H},\mathbf{A}_i] = 0$) so that they all commute pair-wise, so $[\mathbf{A}_i,\mathbf{A}_j] = 0 \,\forall i,j$.
*Determine the corresponding quantumnumbers and representations of the states, with the right theorems of group theory you can now determine what dependancies your system has.
To go to more numerical results you'll need some approximations ans perturbation theory:
*Determine the assumptions in which you are working (when having a magnetic field for example)
*Apply perturbation-theory (most of the time degenerate perturbation theory)
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/71231",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Hydrogen cloud at the universe's beginning? What prevented all of the hydrogen at the universe's start from coalescing into one gigantic star?
| Pulsar and Chris White have given nice explanations of why the dynamics of the early universe would not lead to the formation of one big starlike object. There is also another, much more generic argument against such a process, which boils down to the existence of cosmological horizons. At any given time in the evolution of the universe, there have been parts of the universe far enough apart that they could never have had any causal relationship -- no signal could have propagated between them, even at the speed of light, even if the signal was emitted immediately after the big bang. If matter in region A and matter in region B are separated in this way, then it's not possible for them to have collapsed into the same object, simply because relativity prevents the matter at A and B from even coming together that fast.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/71423",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Why does the event horizon of a black hole not look like a bright sphere? All infalling matter-energy appears to an external observer as frozen in time at the event horizon. Why then is this horizon not extremely bright due to radiation that is able to escape radially? So should all black holes not appear as a bright sphere?
| As Ben says, black holes that have accretion disks shine very brightly indeed due to heating of the matter in the disk as it falls inwards. However not all black holes have accretion disks. For example the black hole at the centre of our galaxy does not, and I suspect you're asking why it is no longer shining brightly.
If you take a glowing object. e.g. a metal sphere heated to white heat, and drop it into a black hole you'll see its light fade as it approaches the event horizon and in the final stages of the approach to the horizon it would fade to black. This happens because the light emitted by the sphere loses energy as it climbs out of the gravitational well. This effect is called gravitational redshift. This question, Gravitational Redshift around a Schwarzschild Black Hole, has already asked how to calculate the redshift for an object falling into a black hole, but the answer is a bit technical. To summarise, the redshift is given by:
$$ z = \frac{f-f_o}{f_o} = \frac{1}{\sqrt{1 - 2GM/c^2r}} - 1 $$
where $r$ is the distance from the centre of the black hole, $f$ is the original frequency and $f_o$ is the observed frequency. The event horizon is at $r = c^2/2GM$, and at this distance the red shift goes to infinity i.e. the frequency of the light you see falls to zero.
So matter falling into a black hole may well be heated by friction as it falls through the accretion disk, however this is only temporary and as it reaches the event horizon you will see it red shift to infra-red, then radio and eventually to nothing.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/71497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Is it possible for a photon to be at rest? I know it doesn't really make sense if looking at the photon from the wave point of view, but is there any law of physics which prohibits a photon from stopping completely? Thanks.
| Those laws of physics would be Maxwell's Equations. I won't go into too much detail but from those equations you can get a wave equation for light. The speed of the wave is determined by two fundamental constants, $\epsilon_{0}$ and $\mu_{0}$. If the speed of light was variable in anyway then those constants would also be variable. No experiment has ever shown these constants to be different, therefore the speed of light must of constant. Since we know that light is composed of photons then photons must move at light speed.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/71639",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Why does Hawking radiation cause black holes to die? If a particle is being expelled from a black hole and an antiparticle is being driven into it, shouldn't the opposite occur as well and in the same frequencies?
I mean, black holes should emit antiparticle radiation as well and gather energy from it and the black hole energy shouldn't change, right?
What am I missing?
| The following answer is not "rigourous", but it may gives a simple explanation.
Suppose you have a quantum fluctuation just near the horizon, but outside.
This quantum fluctuation create 2 particles, one with a negative energy -E, the other with a positive energy +E.
If the 2 particles stay outside the black hole, they have to anihilate themselves in a time time $t \le \frac{\hbar}{E}$
Now, one of the 2 particles may fall inside the black hole, and there are 2 possibilities ; the escaping particle may have a positive or a negative energy.
The key point is that there is an asymmetry between these 2 cases.
For a particle to be real, its energy has to be positive, but relatively to the time coordinate. With an evolution variable $\tau$, this can be write $\frac{dt}{d\tau}>0$
When the horizon is being crossed (by the infalling particle), we may consider, that there is a change in the nature of the time and radial space coordinate. The time-like coordinate becomes a space-like coordinate, and the radial space coordinate becomes a time-like coordinate.
More precisely, if, outside the black hole, the coordinate are(in units $c=1$) : $z=r+it$, then the "coordinates" inside the black hole are $z \rightarrow z'\sim -iz$
So, $$z'=r'+it'\sim-i(r+it)=(t-ir)$$
So, $t'\sim-r$ and $x'\sim r$
For an escaping particle, the energy must be positive relatively to $t$, so $ E=\frac{dt}{d\tau} >0$, but for the infalling particle the "energy" must be positive relatively to $t'$, that is = $\frac{dt'}{d\tau}>0$, which is "equivalent" to $-\frac{dr}{d\tau}>0$.
But the last expression means only that the particle is an infalling particle, which was our hypothesis. We could say also, for the infalling particle, that the "outside" energy $-E$ becomes a "inside" momentum.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/71718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 1
} |
Why neutrons in nucleus don't decay? In this question it is explained that neutrons in nucleus don't decay because the next state would not be lower in energy than the previous.
How come neutrons in a nucleus don't decay?
But it doesn't say what causes the neutron to know that the next stage is not lower in energy?
| hdhondt is right. A neutron doesn't need to "know" that another state isn't lower in energy to stay in the state is it at. It stays in the state of lowest energy because it is the state of lowest energy (just like a ball at the bottom of a pit).
Furthering hdhondt's ball example: the ball in the pit moves to the bottom (where the gravitational potential energy is the lowest) and, assuming all its kinetic energy dissipates, will stay there. It stays there not because it explored all its possible energy states (all the locations in the pit) and therefore knows which one has the lowest energy but because (again, assuming the kinetic energy dissipates) gravity will cause it to come to rest at the place of lowest potential. I give this as an intuitive picture we've all had experience with. In a nucleus, there are different forces at work but the picture is basically the same. It's not that the neutron knows its in the lowest energy state, but that eventually the neutron will end up in the lowest energy state, just like the ball in the bottom of the pit.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/71818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Why does positronium decay into 2 photons more often than into 3 photons? I cannot find the answer to the above question.
I know that para-positronium is created with a probability of $25\%$ and decays into 2 photons, while ortho-positronium is created with a probability of $75\%$ and decays into 3 photons.
I also know that ortho-positronium has a way longer life time than para-positronium. This, in my understanding, should not affect the number of decays per time, but just means that the ortho-positronium will decay LATER into three photons. But in the end there should be $75\%$ 3-photon-decays and $25\%$ 2-photon-decays. But in reality 2-photon-decay happens about 300 times more often than 3-photon-decay.
What information am I missing?
Thank you!
| I'm afraid the responses so far are either misleading or do not answer the question.
In fact in a dense medium a positron when it forms the longer lived ortho-positronium can "pick-up" an electron from an adjacent atom and then decay into 2 x 511 keV photons long before the 3-photon ortho-positronium state would have decayed. Look up "pick-up effect".
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/71856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 1
} |
Electromagnetic Radiation of Charged particles This question is motivated by similar one.
If an accelerated point charge $q$ radiates with power $W$ then I assume the same particle with charge $-q$ will radiate with the same rate $W$. Now what if we make a dipole with these two charges and accelerate it with the same acceleration? What will be the radiation power?
| According to this paper, New approach to the classical radiation fields of moving dipoles, the answer is:
$P = \dfrac{18d^2a^4}{35c^7} + \dfrac{2d^2\dot a^2}{15c^5}$
Here, $d$ is the fixed electric dipole moment and the acceleration, velocity, and dipole moment are along the $z$ axis.
From the paper:
This formula may be considered as the dipole analogue of the Larmor
formula of the point charge.
And the Larmor formula for a point charge is:
$P = \dfrac{2q^2a^2}{3c^3}$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/72005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
First principle calculation of boiling point of water How can we theoretically calculate the boiling point of water at given pressure (other subtle parameters as well, if any)? What is the most accurate (minimum discrepancy with experimental value) computation that can analytically predict the
boiling point of water?
Possibly we need to invoke quantum mechanics for this. I anticipate that many answer would say that EXAXCT prediction is computationally infeasible, but please give the outline in algorithmic form regardless of computational cost.
My main goal is to learn how quantum mechanics can be applied to phenomena which can be observed by layman. Other example where quantum mechanics is used to predict physical properties of small molecules from first principle are also welcome.
| The boiling point (i.e. the saturated vapor pressure as a function of temperature) is described by the Clausius–Clapeyron equation which is a consequence of classical thermodynamics. Of course if we start investigating why certain thermodynamic potentials used in that equation have to be such and such then eventually we'll need quantum physics to explain the size of atoms and molecules. But there are more direct commonly observed macroscopic consequences of quantum physics which include photovoltaics, lasers, superconductivity. These seem to be closer to what you are looking for.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/72081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Any physical example of an "explosive" differential equation $ y' = ky^2$? I was told that in physics (and in chemistry as well) there are processes that may be described by a differential equation of the form $$ y' = ky^2. $$ That is, the variation of a variable depends from the number of pairs of the elements.
I understand the mathematical meaning of that equation, but I cannot conceive of any physical process which leads to this. Maybe some nuclear reaction could, but I am at a loss. Any hints?
| As described in Wikipedia, world population was growing in just that fashion (hyperbolic) for some time in the 20th century.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/72315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 6,
"answer_id": 4
} |
What are the advantages in using 2 identical capacitors? What advantage might there be in using two identical capacitors in parallel connected in series with another identical parallel pair, rather than using a single capacitor?
| Capacitors are sometimes connected in series to increase the working voltage range, but it is not a good design technique because variances in capacitors can cause a variance in voltage and thus exceed the voltage rating on one of the capacitors.
And if you can't achieve a value of capacitance with one capacitor, sometimes putting two larger ones in series you can get the desired value. This might also be the reason a for the odd combination of values and topology.
If you gave us the actual circuit and some information about the application we might be able to provide some more insight.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/72423",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Minimum height of mirror required to view image I wanted to know the minimum height of mirror required to be able to view a complete image of a person. I considered the following setup:
$HF$ is the person in question. $H$ denotes the head, $F$ the feet, and $E$, the eyes. For the person to see his complete image, a ray each from $H$ and $F$ has to come and reflect into his
eyes ($E$). Let $HE = 0.16m$ and $HF = 1.84m$. $KG$ is the minimum height if mirror required.
Now, since $HI = IE = \frac{HE}{2} = 0.08m$ and $FC = CE = \frac{EF}{2} = 0.92m$, $KG = 1m$.
But this doesn't make any sense. This calculation doesn't take into account the distance of the person from the mirror. It is clear that the distance matters. If I have a really small mirror, and I go far away from it, I can see my whole body; which is not the case if I'm really close to it.
|
In image ,a person of height H is standing against a plane mirror of length AB.We will need the part AF of mirror so that the person will be able to see his full reflection .
In triangle CED we can say $\frac{CD}{CA}=\frac{DE}{AF}$
Or
$\frac{2\not{x}}{\not{x}}=\frac{H}{minimumlength}$
Minimum length =$\frac{H}{2}$
We can see there is no x here ,so distance of person wil not matter
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/72488",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Overtaking with non-constant acceleration I have tried to solve this problem by adding the sum of the displacements during acceleration, constant velocity and deceleration, but it does not work out.
Question:
A car accelerates from rest to $20~\text{m/s}$ in $12$ seconds ($a =5/3~\text{ms}^{-2}$), it travels at $20~\text{m/s}$ for $40$ seconds, then retardation occurs from $20~\text{m/s}$ to rest in $8$ seconds ($a = -2.5~\text{ms}^{-2}$). As the car accelerates an RC car, moving parallel to the car, is moving at $14~\text{m/s}$. When will overtaking occur and what will the distance be? The RC car passes the car just as it starts to accelerate.
I can do this without a problem if acceleration is a constant. Is there a differential equation I can use to compute this as that is my better area or must I stick with the SUVAT equations?
Again, if I could be pointed in the right direction that way I can learn.
| Since the acceleration changes discontinuously, there is no one way to solve it in one step.
One simplification is to switch to the frame of reference of the RC car. You can do this simply by making the initial velocity of the starting/stopping car $-14$ m/s. The question then reduces to: When is the displacement of the moving car zero again?
if you use SAVTU on the first acceleration, you can find out if $S=0$ in less than 12 seconds. If not, you can find out $S$ after 12 seconds, and then see if you get back to $S=0$ at constant velocity of $6$ m/s in less than 40 seconds, and so on...
Explicit solution added;
First, Setting $s=0$ (car catches RC during first acceleration), $a = \frac{5}{3}$, and $u = -14$, and solving for t, we get t = 16.8 sec
It would take 16.8 sec of constant acceleration for the car to catch the RC, which we don't have!
So, finding s at the end of the first acceleration, if $a=\frac{5}{3}$, $u = -14$, $t=12$, we get $$s=-14 \times 12+\frac{1}{2}\times\frac{5}{3}\times 12^2=-48$$The car is 48 m behind the RC at the end of the firat acceleration.
The car is catching the RC at $6 $ m/s, and will make up the 48 m in 8 seconds of constant velocity, which we do have.
Total time is 20 seconds. The RC will cover 280 m, and so will the accelerating car...
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/72539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
What is the meaning of "CW" in LASER? I am reading a user's manual, and the word appears here.
At first, I think "CW" means "center wave".
But later, I find that the meaning of "CW" is "continuous wave".
It makes me confused.
Generally, Laser has a unique frequency. Properly speaking, it has a too small fluctuation of frequency, or line width, such that we can treat the spectrum as harmonic wave which has a unique frequency.
So, literally, dose "continuous wave" means a wave which is not interrupted? Is it the same as "harmonic wave"?
| The choice between continous or pulsed can come down to the power output of the laser. For the same input power you can either output that power at some constant and therefore the average power is equal to the peak, or send the energy out in pulses. In pulsed mode the peak energy can be much higher while maintaining the same average power.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/72606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Can spheres leaking charge be assumed to be in equilibrium? I am struggling with the following problem (Irodov 3.3):
Two small equally charged spheres, each of mass $m$, are suspended from the same point by silk threads of length $l$. The distance between the spheres $x \ll l$. Find the rate $\frac{dq}{dt}$ with which the charge leaks off each sphere if their approach velocity varies as $v = \frac{a}{\sqrt{x}}$, where $a$ is a constant.
This is embarrassingly simple; we make an approximation for $x \ll l$ and get
$$
\frac{1}{4 \pi \epsilon_0} \frac{q^2}{x^2} - \frac{mgx}{2l} = m \ddot{x}.
$$
We can get $\ddot{x}$ from our relation for $v$, so we can solve for $q$ and then find $\frac{dq}{dt}$.
However, in general, $\frac{dq}{dt}$ will depend on $x$ and hence on $t$. The answer in the back of the book and other solutions around the web have $\frac{dq}{dt}$ a constant.
You can get this by assuming that at each moment the spheres are in equilibrium, so that you have $\ddot{x} = 0$ in the equation of motion above.
Does the problem tacitly imply we should assume equilibrium and hence $\frac{dq}{dt}$ is constant, or am I missing something entirely? I.e. why is the assumption of equilibrium justified? I understand reasoning like "the process happens very gradually, so the acceleration is small compared to other quantities in the problem," but I don't understand how that is justified by the problem itself, where we are simply given that the spheres are small (so we can represent them as points) and $x \ll l$ (which we have used to approximate the gravity term in the equation of motion).
| If we continue with the suggestion you made, of obtaining $\ddot x$ from the equation of motion $v=a/\sqrt{x}$ which was provided, and substituting this into the equation $F=m\ddot x$, then we do indeed find that $\dot q$ is not constant. It is only by ignoring the $m\ddot x$ term - by assuming that $v\approx 0$ - that we can reach the result which Irodov intended.
But there is nothing in the question statement which justifies the assumption that $v \approx 0$. No values are given which would enable us to conclude that $\dot v=-(a/2x\sqrt{x})v$ can be neglected so that there is a quasi-static equilibrium.
The conclusion must be that Irodov made an error. He deliberately imposed an unrealistic but fairly simple equation of motion $v=a/\sqrt{x}$ in order to derive an equally unrealistic but simple result (that $\dot q$ is constant). While doing so he failed to state the assumptions which were necessary to obtain this result.
Even the most respected authors and textbooks are fallible.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/72696",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 3,
"answer_id": 0
} |
Can an Incandescent Bulb be considered as a heater? Due to an experiment, I need a small heater (around 70 to 100 watts). I intend to use an incandescent bulb so it can act as a heater. What I wonder here is will a 70 watts Incandescent Bulb be equal to a 70 watts heater?
| Whatever bulb is used, basically all the power consumed will go into heating the environment, with the notable exception being any photons that escape through a window or some such thing.
Different styles of bulb have different visible light-heat ratios, but even the visible light emitted will simply bounce around the room a few times and deposit all its useful energy into random thermal motions. You know these photons cease existing, at least in the visible range, because when you turn off a light, the room essentially instantly goes dark.
Incandescent bulbs work better at heating than, e.g., certain fluorescent bulbs for a given amount of visible lighting because they consume more power (and thus emit more direct heat) per unit desired light. However, even if you designed a light bulb that emitted 100% of its photons in the visible range, that power (small though it would be) would all go into heating its environment.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/72825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Sliding force less or equal? Why is the force required to slide a magnet off a steel plate A LOT less than the force required to directly pull it off?
The force required to pull the magnet can be: 20lb
While the force required to slide the magnet can be: 1lb more/less.
Why is that?
| When you slide a magnet off a plate, you are doing the same amount of work as when you pull it directly off (slightly more, in fact, because of the friction). However, while it's the same amount of work, it's spread out over a much larger distance, and so requires much less force. When you slide a magnet off a plate, the force decreases gradually, starting as soon as the magnet reaches the edge of the plate. When you pull a magnet off a surface perpendicularly, the force is quite large when the magnet is touching it, but decreases quickly as soon as it is a short distance away.
The basic principle is the same as the one that makes levers, double pulley systems, and switchbacks useful. Since work = force × distance, increasing the distance reduces the force required.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/72900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
The relation between Gauss's law and Coulomb law and why is it important that the electric field decrease proportionally to $\frac{1}{r^{2}}$? My question relates to the third MIT's video lecture about Electricity
and Magnetism, specifically from $21:18-22:00$ : http://youtu.be/XaaP1bWFjDA?t=21m18s
I have watched the development of Gauss's law, but I still don't quite
understand the link between Gauss's law and Coulomb law: How does
Gauss's law change if Coulomb law would of been a different one.
I also don't understand why is it so important for Gauss's law that
the electric field decrease proportionally to $\frac{1}{r^{2}}$ ?
For example, what would of happened if the electric field decrease
proportionally to $\frac{1}{r}$ , or $\frac{1}{r^{3}}$ ?
| The $1/|r|^2$ dependence is fundamentally geometrical in nature, stemming from our 3d world.
Differential equations, even PDEs, are sometimes said to have Green's functions. Consider the differential equation
$$\nabla \cdot K = \alpha$$
for some vector field $K$ and some scalar field $\alpha$. This is structurally identical to Gauss's law. This equation has a Green's function $G$ that satisfies
$$\nabla \cdot G = \delta$$
where $\delta$ is the Dirac delta function. This describes, in essence, a point source $\delta$ generating a field $G$.
The Green's function is given by
$$G(r) = \frac{\hat r}{4\pi |r|^2}$$
Thus, any differential equation of the form $\nabla \cdot K = \alpha$ has the same Green's function in 3d--a point source always generates the same basic field.
In short, it is a physical statement to say $\nabla \cdot E = \rho/\epsilon_0$, but once that has been said, Coulomb's law, which describes a point charge, must inevitably follow due to the mathematical structure of the equations and the geometrical nature of 3d space.
(Of course, you're free to work the other way around.)
Edit: crucially, in 2d and 1d, the Green's functions are different. You probably already know these solutions. The 2d Green's function is something familiar from the field of a uniform line charge, and has $1/|r|$ dependence. The 1d Green's function has constant magnitude but changes direction on opposite sides of the "point" charge--this is the basic feature of a uniformly charged sheet.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/73028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Heating coffee by yelling? Is it a myth that yelling to a coffee mug will heat it? I have been hearing my friend saying that screaming will heat coffee or water.
| This probably comes from an advertisement campaign by Physics Central. Their poster is
It is a silly question; no amount of yelling will appreciably heat a coffee cup. We are very inefficient at turning energy into sound. The page from which I got the photo says that a loud shot puts 0.001 W of power into the air. This is a trivial amount compared either to the rate that a cup of coffee at 50C cools purely by radiation (around 1 Watt, and this is not the main way coffee cools; it mainly cools via evaporation), or to a human metabolism (around 100 Watts).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/73190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Projectile Motion Question involving a ball and a ramp inclined at an angle The question is to finde the initial horizontal velocity of the ball at end of the ramp, where it is released.
I know how to do this using gravitational potential energy and kinetic energy ($v=\sqrt{2gh}$), assuming all potential energy is converted into kinetic energy but the question is asking me to find the error in the experiment.
It gives me the values on $y$ as the ramp is lifted up (it is lifted up about 10 cm each time) and the corresponding values of $x^2$. I drew a graph of $x^2$ vs $y$ and found the gradient.
How can I use this to find the experimental initial horizontal velocity?
This sketch shows what the experiment looks like:
| Vertically:
$\ddot{y}=a$
$\frac{d\dot{y}}{dt}=a\therefore \int d\dot{y}=\int adt\therefore \dot{y}=u\sin\theta +at$
$\frac{dy}{dt}=u\sin\theta +at\therefore \int dy=\int \left ( u\sin\theta +at \right )dt\therefore y=u\sin\theta t+\frac{a}{2}t^{2}$
Horizontally:
$\ddot{x}=0$
$\frac{d\dot{x}}{dt}=0\therefore \int d\dot{x}=\int 0dt\therefore \dot{x}=u\cos\theta$
$\frac{dx}{dt}=u\cos\theta\therefore x=u\cos\theta t$
We know the gradient of the graph of $\frac{x^{2}}{y}=m$ ($m$ is the gradient, measured in $cm$)
Sub $x=u\cos\theta t$ and $y=\frac{a}{2}t^{2}$
$\therefore u\cos\theta=\sqrt{\frac{\frac{a}{2}m}{100}}m/s$
Theoretically, $u\cos\theta$ can be found by considering the conversion of gravitational potential energy to kinetic energy, assuming no energy loses
$\therefore mah=\frac{1}{2}m\left (u\cos\theta \right )^{2}\therefore u\cos\theta=\sqrt{2ah}$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/73334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
The Momentum Operator in QM I've seen the 'derivation' as to why momentum is an operator, but I still don't buy it. Momentum has always been just a product $m{\bf v}$. Why should it now be an operator. Why can't we just multiply the wave function by $\hbar{\bf k}$? Why should momentum be a derivative of a wave function?
| The simplest possible solution to the Schrodinger equation is that of a plane wave: $$ \psi = e^{i(\boldsymbol{\mathbf{k}}\cdot \boldsymbol{\mathbf{r}}- \omega t )} $$ In this scenario, the position of the system is indeterminate, but the momentum is known. Next, take the gradient of this wavefunction. $$ \nabla \psi = ik \psi$$ But we know that $p = \hbar k $, from De Broglie. Substituting this into the equation and rearranging yields $$ \frac{\hbar}{i} \nabla \psi = p \psi $$ This is an example of an eigenvalue equation - an operator acts on a function or vector to produce a constant multiple of the original function. So, we see that the eigenvalue is the measured momentum. Therefore, the operator producing it, $ \frac{\hbar}{i} \nabla $, is called the momentum operator.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/73483",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
How does one prove that Energy = Voltage x Charge? We know
$$E = q V$$
where $E$ is the energy (in Joules), $V$ is the potential difference (in Volts), and $q$ is the charge. Why is this equation true and how we prove it?
| electric field strength is
$$E=\frac Fq=\frac Vd$$
with $V$=voltage, $d$=distance between charged plates
\begin{align}
\frac Fq&=\frac Vd \\
Fd&=qV
\end{align}
but $Fd$=energy
$$\therefore {\rm energy}=qV$$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/73769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 2
} |
Will a diamond helm be more protective against bullets? Will a helm made entirely of diamond be more protective against bullets than standard steel or kevlar helmlets?
| Diamond, though hard, is brittle. The crystalline planes within it can shear relatively easily.
"The toughness of natural diamond has been measured as 2.0 MPa m1/2, which is good compared to other gemstones, but poor compared to most engineering materials." - Wikipedia
The fracture toughness of steel is about 50 MPA.m1/2. The crystals in it are formed into tiny grains and the crystalline structure has many disclocations. This gives steel strength.
For resisting bullets, dissipation of energy by elastic deformation is an important factor. I suspect diamond is much less elastic than steel.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/73851",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How do we know photons have spin 1? Electrons have spin 1/2, and as they are charged, they also have an associated magnetic moment, which can be measured by an electron beam splitting up in an inhomogeneous magnetic field or through the interaction of the electrons's magnetic moment with an external magnetic field in spectroscopic measurements.
On the other hand, a photon is neutral - how can one measure its spin if there's no magnetic moment? How do we know it has spin 1?
| Spin of the photon is an ongoing theoretical research. For classical electromagnetic field the total angular momentum is $\vec{J}=\vec{r}\times <\vec{E}\times \vec{B}>$ (all vectors). In field theory this quantity is $\vec{J}=\vec{L}+\vec{S}$, it is gauge invariant and possible to observe. But! Nobody found a correct way to represent it in the sum of two gauge invariant operators for angular orbital momentum and spin (L and S). This concludes the theoretical knowledge - operator of photon's spin is not known.
It is possible to measure the Z-component of the angular momentum - helicity.
If you are thinking of photon's spin as a MAGNETIC quantity then you are definitely not correct (not wrong either). There is no double slit experiment for photons :) Unfortunately.
You can read further here : http://arxiv.org/abs/arXiv:1006.3876
Frankly speaking photon is not as simple as people are trying to pretend.
Cracking it's nature will need more efforts.
Thank you for thinking.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/73942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "53",
"answer_count": 6,
"answer_id": 1
} |
Compound microscope vs astronomical telescope In compound microscope, we take such an objective lens which has small focal length. While in astronomical telescope, we take such an objective lens which has large focal length.
Why don't we use same objective lens in both? The function of both the devices is to enlarge an object at infinity.
| Wikipedia's article on Focal length explains pretty clearly (emphasis mine):
In most photography and all telescopy, where the subject is essentially infinitely far away, longer focal length (lower optical power) leads to higher magnification and a narrower angle of view; conversely, shorter focal length or higher optical power is associated with a wider angle of view. On the other hand, in applications such as microscopy in which magnification is achieved by bringing the object close to the lens, a shorter focal length (higher optical power) leads to higher magnification because the subject can be brought closer to the center of projection.
So it basically depends on the fact that most of our astronomical sources are at distances larger than $\sim10^{18}$ cm away from us, while our microscopic objects are less than $\sim100$ cm away from us.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/74128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Equations of motion of displacement field We have an action: $$S[\boldsymbol{u}] = \frac{1}{2} \int dt \int d^3x \left\{ \mu (\frac{\partial u_{i}}{\partial t})^{2} - \nu (u_{ii})^{2} - \rho(u_{ij})^{2}\right\} $$
Where $u_{ij} = (\partial_{i}u_{j} + \partial_{j} u_{i} )/2$.
Where index $i = 1,2,3$ denotes the $x$ $y$ and $z$ axis and there is summation over repeated indices.
This models a three dimensional lattice of connected springs in the continuum limit. I want to derive the equations of motion of this system and the dispersion relation.
Using the EL equation I get:
$$\mu\partial_{tt}u_{1} - \nu \partial_{11}u_{1} - \rho \partial_{12}u_{1} - \rho \partial_{13}u_{1} - \rho \partial_{21}u_{2} - \rho \partial_{31}u_{3} = 0. $$
And similarly for $u_{2}$ and $u_{3}$. Is this correct and if so, how can I solve these equations to get the equations of motion? Also, how do I find the dispersion relation for this system?
| Partial answer :
Define functions $u_i(p) = u_i(p^0,p^1,p^2,p^3)$ as Fourier transform of the functions $u_i(x)$
Then take the Fourier transform of your three Euler-Lagrange equations.
You get three equations of kind :
$A_{11}(p)u_1(p) + A_{12}(p) u_2(p) + A_{13}(p)u_3(p) = 0$
(The functions $A_{ij}(p)$ are quadratic functions of the $p^i$)
The system of 3 equations has a non trivial solution $u_i(p)$ if and only the determinant det $A$ is zero.
det $A = 0$ gives you the relation of dispersion, because it is a relation between the $p^i$.
From these 3 equations, maybe you are able to find an manageable expression for the $u_i(p)$, maybe trying expression like $u_i(p) = B_{ijk}p^j p^k \Phi(p)$, and introducing this expression in the 3 equations (which in fact are now linearly dependent, so you can only choose 2 of the 3 equations) may lead you to be able to extract the $B_{ijk}$ from quartic equations in the $p^i$ (but this is only a guess).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/74192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
Why is there a factor of $4\pi$ in certain force equations? I mean to ask why there is $4\pi$ present in force equations governing electricity? Though all objects in universe are not spherical and circular, the constant of proportionality in both equations contain $4\pi$. Why?
| Any differential equation of the form $\nabla \cdot A = \alpha$ and $\nabla \wedge A = 0$ in $n$-dimensions has as its Green's function (that is, the solution for a point source, for $\alpha = \delta$, the Dirac delta function) a field $G$ of the form
$$G(r) = \frac{1}{S_{n-1}} \frac{\hat r}{|r|^{n-1}}$$
where $S_{n-1}$ is the surface area of a unit $n$-ball, which we know to be $4\pi$ in 3d. The factor of $4\pi$ that appears in many such force equations is inherently geometrical, and as has been stated, it is confounding to lump it into a constant, especially if one works outside of 3d (see, for example, the electric field of a line charge, which is a 2d problem and thus has $2\pi$ appear in it, as circumference of a unit circle).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/74254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 7,
"answer_id": 0
} |
Path of least resistance vs. short circuit Some sources on the web claim that "electricity follows the path of least resistance" is not true, e.g. this physics SE question. However, in every explanation of "short circuits", the author says that current flows through the short because it's following the path of least resistance. How do I reconcile these two facts?
| It's more like "Electricity tries to follow the path of least resistance". The sentence is a non-rigorous intuitive tool that helps one quickly make sense of current paths. It's not a physical law.
This comes from the behavior of resistors in parallel --this is where current has a "choice" of which direction to take. The potential difference across a resistor is equal to the current into its resistance ($V=IR$, usually called Ohm's law) in appropriate units.
What happens when you have two resistors in parallel is that the p.d. across them must be the same. Which means that $I_1R_1=I_2R_2$. So, more current flows through the resistor with less resistance and vice versa. Current still flows through the greater resistance.
So most current flows through the path of least resistance. In the case of a short circuit, one of the Rs is (nearly) zero, and thus (almost) all the current flows through it.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/74368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Does alternating current (AC) require a complete circuit? This popular question about "whether an AC circuit with one end grounded to Earth and the other end grounded to Mars would work (ignoring resistance/inductance of the wire)" was recently asked on the Electronics SE.
(Picture edited from the one in the above link)
Though I respect the AC/DC experts there, I think (with the exception of the top answer) they are all wrong.
My issue is that they all assume that AC requires a complete circuit in order to function. However, my understanding is that a complete circuit is necessary for DC, but not AC. My intuitive understanding is that AC is similar to two gas-filled rooms with a pump between them - the pump couldn't indefinitely pump gas from one room to another without a complete circuit (DC), but it could pump the gas back and forth indefinitely (AC). In the latter case, not having a complete circuit just offers more resistance to the pump (with smaller rooms causing a larger resistance).
Is my understanding correct - can AC circuits really function without a complete loop?
More importantly, what are the equations that govern this?
If larger isolated conductors really offer less AC-resistance than smaller AC conductors, how is this resistance computed/quantified? Would its "cause" be considered inductance, or something else?
| Indeed, AC can flow without a "complete circuit" - that's what happens in LC circuits all the time. An LC circuit is technically not complete - the capacitor of LC circuit contains an insulator between its plates and so electrons are unable to flow through the capacitor (unless it fails). Still the oscillations in the LC circuit happen because of alternating current flows inside the LC circuit charging and discharging the capacitor through the inductor.
The resistance to AC current increases as the inductor inductance increases (inductance is the measure of how much the inductor can affect the current flowing through it and so the more inductance the more resistance) and as the capacitor capacitance decreases.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/74625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "47",
"answer_count": 5,
"answer_id": 1
} |
Do higher-order mass moments have any physical meaning?
*
*The zeroth moment of mass of an object is simply its total mass.
*The first moment of mass yields an object's center of gravity (after normalization).
*The second moment of mass yields an object's moment of inertia.
Is there an analogous physical interpretation for the third and higher mass moments?
| In statistics the third moment is used to calculate skewness. I would guess this has a physical analogy. Although I haven't thought this through, I'd guess it would be possible to take a disk and deform it asymmetrically so that the centre of mass and moment of inertia remained the same, but the third moment changed. In this case the third moment would be telling you about the asymmetry of the disk.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/74770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 2,
"answer_id": 0
} |
accelerated charged particles and interaction with magnetic field In high school we are taught that magnetic field perpendicular to velocity of an charged particle experience perpendicular force that causes it to move in circular path by relation
$$qvB=\frac{mv^2}{r}$$ but in drawbacks of Bohr's theory it was proposed that accelerated charged electron orbiting along nucleus will immediately loose energy in form of electromagnetic waves and collapse into nucleus.So My question basically is are we taught wrong about this relation that charged particle according to Lorenz's force will perform circular motion as far as required conditions in equations are provided but will it eventually loose energy in form of electromagnetic radiation and halt it's circular orbit ?
| Yes, as the charged particle is accelerated, it will emit radiation. But it never stops at a point because of the uncertainty principle. Refer Energy of electron spinning in a magnetic field
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/74902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Probabilities in statistical mechanics I am reviewing some concepts in statistical mechanics and am becoming confused with how to calculate probabilities when a system has $N$ non-interacting particles.
For instance, let's say we have $N$ electrons with magnetic moment $\vec{\mu} = (g e/2 m)\vec{S}$. If we apply a strong magnetic field parallel to $\vec{S}$, then
$$ E = - \vec{\mu} \cdot \vec{B} = \pm \frac{g e \hbar}{4 m} = E_{\pm}$$
depending on the orientation of the spin of the electron. Therefore, the partition function for one electron is simply
$$ \xi = 2 \cosh \left(\frac{g e \hbar B}{4 m k T}\right) $$
And the probability to find the electron with spin parallel to the magnetic field is simply $e^{-\beta E_{+}}/Z$. So far, so good.
However, what happens when we have $N$ such electrons? Statistical mechanics says that the partition function for the system is now
$$ Z = \frac{\xi^N}{N!} $$
if we assume that the electrons do not interact with each other.
This is where I get confused. Now, if we want to find the probability that 75% of the electrons have energy $E_+$, then the Boltzmann argument doesn't hold anymore:
$$ \frac{0.75 N }{N} \neq \frac{e^{-\beta E_{+}}}{Z} $$
If the Boltzmann ratio doesn't hold, how can one proceed to calculate the aforementioned probability?
| It seems like you want to consider $N$, distinguishable, non-interacting spins with spin $1/2$, so let's assume we don't have to consider subtleties arising because of identical particles. When there are $N$ distinguishable, the Hilbert space $\mathcal H_N$ of the system has dimension $2^N$, and when they don't interact, the energy eigenbasis can be labeled by sequences
\begin{align}
(+\tfrac{1}{2},+\tfrac{1}{2},+\tfrac{1}{2},-\tfrac{1}{2},-\tfrac{1}{2},\dots, -\tfrac{1}{2},+\tfrac{1}{2},+\tfrac{1}{2},-\tfrac{1}{2})
\end{align}
which specify that particles $1$ through $3$ are spin up, particles $4$ and $5$ are spin down, etc. We denote a general such state as $|\mathbf m\rangle =|m_1, m_2, \dots m_N\rangle$ with $m_i=\pm1/2$. The probability that the system is in a given state $|\mathbf m\rangle$ is
\begin{align}
\mathrm{prob}(\mathbf m) = \frac{e^{-\beta E_\mathbf{m}}}{Z}
\end{align}
where the partition function is
\begin{align}
Z = \sum_{\mathbf m}e^{-\beta E_\mathbf{m}}
\end{align}
and $E_{\mathbf m} = E_{m_1} + E_{m_2} +\cdots+ E_{m_N}$ is the total energy of the state $|\mathbf m\rangle$. Now, to determine the probability that the system is in any state for which $3/4$ of them are up, we simply determine all sequences $\mathbf m$ for which this condition holds, and we sum up the corresponding probabilites. Since $3/4$ of the spins are up if and only if the the numbers $m_i$ add up to $(\frac{3}{4}N\cdot \frac{1}{2} + \frac{1}{4}N\cdot(-\frac{1}{2}) = \frac{1}{4}N$, the probability we're after is
\begin{align}
\mathrm{prob}\left(m_1+\cdots+m_N = \tfrac{1}{4}N\right) = \sum_{m_1+\cdots+m_N = \frac{1}{4}N} \frac{e^{-\beta E_\mathbf{m}}}{Z}
\end{align}
where the right hand side represents a sum over all sequences $\mathbf m$ whose elements sum to $N/4$, and now you just have to figure out how to perform the sum.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/74984",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Calculate exact time of Solar midnight I want to calculate real time of midnight (Solar midnight), but I am unable to find any formula or algorithm for that. I have times of sunrises and sunsets for every day of year. How can I get Solar midnight times from them?
Thank you all.
| The formula for calculating the Local Time equivalent for Local Solar Time is $LST=LT+\frac{TC}{60}$, where $TC$ is the Time Correction factor $TC=4(Longitude-LSTM)+EoT$. $LSTM$ is the Local Solar Time Meridian, calculated by $15º-ΔT_{GMT}$, where $ΔT_{GMT}$ is the difference, in hours, from GMT. (For instance, here on the east coast of the US, this value is -4 in the summer and -5 in the winter.) $EoT$ is the Equation of Time, $EoT=9.87sin(2B)-7.53cos(B)-1.5sin(B)$, where $B=\frac{360}{365}((day\ of\ year)-81)$.
If you rearrange that first equation to calculate $LT$ in terms of $LST$, you'll get $LT=LST-\frac{TC}{60}$. Set $LST$ to 12 to calculate solar noon and 24 to calculate solar midnight.
Equations from here.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/75314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Do bifacial solar panels operate at lower temperature than normal solar panels? Bifacial solar panels have a glass backplane instead of the usual aluminum and are designed to absorb and convert to electricity light incident on either face. Many manufacturers claim that their bifacial panels operate at lower temperature than normal (monofacial) panels because bifacial panels transmit the 20% of the solar spectrum that is too low-energy to be absorbed by silicon. See for example this manufacturer's datasheet. Meanwhile, the only reputable study I could find (apologies for the paywall) was of a lab-scale cell and not a commercial module. That study argues that convection is the dominant heat transfer mechanism and so the differences in optical absorption and emission aren't relevant. However, temperature differences of a few degrees centigrade can have a noticeable impact on solar cell efficiency and useful life. Anybody have a compelling argument or know of other relevant experimental work on this topic?
| For what it's worth, lg310's seem to run a little cooler than gxb300's, at least on my roof.
See also http://dx.doi.org/10.1016/j.egypro.2014.02.148
but read Table 4 carefully. The differences seem to be small.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/75409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Why do chimneys have these spiral "wings"? While walking around I noticed something very peculiar. Many chimneys had spiral "wings", while others didn't. I came up with two possibilities:
*
*The wind circles around the chimney upwards which pushes whatever gases being released even higher into the sky.
*The wind circles around the chimney downwards which prevent the chimney from going left or right and rather "push" it downwards to make it more steady.
I feel that both of those possibilities are silly. So, why do some chimneys have those spiral "wings"? And why other taller chimneys don't?
| The spirals are used to prevent the formation of Kármán vortex sheets downwind of the chimney. They work by diverting the wind upwards on one side of the chimney and downwards on the other, creating a three-dimensional airflow pattern that disrupts the vortex sheet. Without them, the vortex shedding could cause vortex-induced vibration in the chimney, which in strong winds might be enough to damage the (relatively thin-walled and flexible) chimney.
Here's a very nice animation of Kármán vortex shedding from Wikipedia, courtesy of Cesareo de La Rosa Siqueira:
The circle on the left represents the (smooth) chimney, viewed from above, with wind coming from the left; the cyan and purple dots are tracer particles showing how air passing the chimney on either side joins the vortex train generated behind the chimney. The Kármán vorticity is an essentially two-dimensional phenomenon; in three dimensions, each vortex would basically be a tall rotating column of air, with one end on the ground and the other end joining together with the adjacent vortices in a complex turbulent region at the altitude of the top of the chimney.
As the vortices are shed on alternate sides of the chimney, each one imparts a counter-force on the chimney itself. Under suitable conditions, these oscillating forces could drive the chimney itself to vibrate from side to side. The helical projections on the chimney prevent this by disrupting the vortices as they form, or at least causing them to form out of phase at different altitudes.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/75459",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "81",
"answer_count": 1,
"answer_id": 0
} |
Projectile motion of a charged particle in a homogeneous electric field I'm reading an article about projectile motion, but I'm having some trouble with how the author found the equations of motion when a homogeneous electric field is considered.
To allow an immediate comparison as an example of non-gravitational
acceleration the corresponding motion of a point charge in a static
homogeneous electric field is tackled. We choose the charge $q$ and
the electric field $E$ in the direction of the negative $z$ axis such
that $qE=mg$ and use the same dimensionless quantities as before (time
is measured in units of $c/g$ and space in units of $c^2/g$). Also we
use lower case symbols in the inertial reference frame (t,x,z) to
facilitate comparison. With:
$t'^2=1+x'^2+z'^2$
the equation of motion $x''=0$,$z'' = -t'$ gives immediately:
$x'=x_0'$, $x=x_0'\tau$ and $z'=z_0'-t$.
All terms are derived with respect to the proper time $\tau$. I don't understand how $qE=mg$ implies $t'^2=1+x'^2+z'^2$ and where $z''=-t'$ comes from.
| I think:
1) The first equation is just a rearrangement of the standard expression for the norm of the four-velocity $u=\frac{dx}{d\tau}$ in special relativity: $u_\alpha u^\alpha=1$ (with $c$=1).
2) Since $qE/m=g$, the acceleration in the z-axis $\frac{du^3}{d\tau}$ due to the electric field is obtained from the standard electromagnetic force equation in SR as:
$$ \frac{du^3}{d \tau} = \frac{q}{m} {F^3}_\gamma u^\gamma = -\frac{q}{m} E u^0 = -g u^0 $$
which is the second equation (with $g=1$).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/75544",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What is phenomenological equation and phenomenological model? I come across these terms in some papers. My understanding is that it is an equation or model describing a phenomenon. Usually, the equations are given and claimed to be true with only some explanations and justification, but not derived from the first principle. The reason is often a bit obscure to follow, on what base they can do that with high confidence? How can they derive it in the first place? If the only purpose is to generate the expected phenomenon, how can they justify it is really the equation governing the phenomenon?
Is there good definition and explanations of these two terms? I expect few solid physics examples should make it easy to understand.
| You could read the definition in dozens of dictionaries and encyclopedias like this
http://en.wikipedia.org/wiki/Phenomenology_%28science%29
My feeling is that the term is related to the word "Phenomena" that is something that is observed. Thus, the phenomenological models and equations describe rather instrument readings than some fundamental processes behind it. For example, there is a relaxation time $\alpha$ in quantum physics which is often taken as a constant in the exponential factor $e^{-\alpha t}$ since the experimental results show nearly exponential decay of the non-equilibrium particle concentration. That is the phenomenological approach. Alternative approach is to compute this quantity using fundamental physics and taken into account scattering processes standing behind this exponential decay.
Usually, phenomenological method utilizes a kind of scientific intuition. Soon or later, most phenomenological quantities are proved by theoretical computations if they describe the experimental situation with satisfactory accuracy.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/75695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
} |
Minimal Extension of Wave Equation to Include Dispersion Let's say you are modeling some process with the wave equation $\frac{1}{c^{2}}\frac{\partial^{2}\psi}{\partial t^{2}} = \nabla^{2}\psi$. You wish to improve your model by including dispersive effects, but you want your model to be as simple as possible for computational tractability.
What is the minimal appropriate model for phase velocity, say $c \approx c_{0} + c_{2}\omega^{2}$? and how should the wave equation be altered?
| What you want to do is change the wave equation into a Klein-Gordon equation:
$$\frac {1}{c^2} \frac{\partial^2 \psi}{\partial t^2} - \nabla^2 \psi + \alpha^2 \psi = 0,$$
where $\alpha$ is a constant of appropriate dimension and usually (in quantum theory) given by
$$\alpha=\frac {m c}{\hbar}.$$
Inserting an ansatz of the form
$$\psi=e^{i(kx-\omega t)}$$
yields the dispersion relation
$$\omega^2=c^2(k^2+\alpha^2),$$
from which one can deduce an expression for the phase velocity, given by
$$v_{phase}=c\sqrt{1+\frac{\alpha^2}{k^2}}.$$
You might consider reading these lecture notes for more insights.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/75772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
What is the difference between phase difference and path difference? The path difference is the difference between the distances travelled by two waves meeting at a point. Given the path difference, how does one calculate the phase difference?
| Let's assume that, two stones are thrown at two points which are very near, then you will see the following pattern as shown in the figure below:
let's mark the first point of disturbance as $S_1$ and the other as $S_2$, then waves will be emanated as shown above. By having a cross-sectional view, you will see the same waves as shown in the figure below (in the below explanation wavelengths of waves emanated from two different disturbances is assumed to be the same).
The waves emanating from $S_1$ has arrived exactly one cycle earlier than the waves from $S_2$. Thus, we say that, there is a path difference between the two waves of about $\lambda$ (wavelength). If the distance traveled by the waves from two disturbance is same, then path difference will be zero. Once you know the path difference, you can find the phase difference using the formula given below:
$$\Delta{X}=\frac{\lambda\cdot\Delta{\phi}}{2\pi}$$
Here, $\Delta{X}$ is path difference, $\Delta{\phi}$ is phase difference.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/75882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 6,
"answer_id": 2
} |
Why do meteors explode? A report on the Chelyabinsk meteor event earlier this year states
Russian meteor blast injures at least 1,000 people, authorities say
My question is
*
*Why do meteors explode?
*Do all meteors explode?
| Airburst, I think, it is called.
An article on wired.com covers exactly this question
the rock was already going very fast when it entered the Earth’s atmosphere. There is no way the air could get it down to terminal velocity – there just wasn’t enough
distance for a rock this large. But this air resistance is essentially the reason that it explodes
Some of these Ruskeor pieces are colored blue. These are the pieces that interact with the air. So, the air pushes on these front pieces to slow them down, but how does the
rest of the rock slow down? Simple, the blue pieces push on the other pieces. So, in a way, this rock is being crushed. Crushed because the air resistance force pushes on the front, but not the rest of the rock.
The author chose to call the meteor 'Ruskeor' as it was a meteor over Russia.
The article referenced provides the calculations if you are interested
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/76045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 0
} |
Does light loses its energy when it passes through denser medium? I know it does not because it emerges out of denser medium at 300,000 KM per second, but according to $E=mc^2$ and given that speed of light decreases inside denser medium with refractive index greater than 1, does not it suggest that energy of light inside denser medium is less?
| Speed of light is a function. When light traverses through a non-vacuum medium, its speed decreases compare to its speed in vacuum. (That is why, index of refraction, n, as a ratio of speed of light in vacuum per speed of light in non-vacuum, increases).
It means, perhaps, the decreasing of speed of light when light traverses through non-vacuum medium contributes to loss of energy, where energy can be expressed as E = h c/\lambda.
CMIIW. Thank you.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/76240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 4
} |
Controlling Neutrinos for Communication Neutrinos travel straight through earth at the speed of light. Therefore, it seems that they could be great for intercontinental communication. Of course, I assume a lot still needs to be learned about detecting, producing and controlling neutrinos before they can be used for the practical purpose of communication.
My question: In principal, could neutrinos be manipulated similarly to radio waves for the purpose of communication? I mean, modulation, filtering, etc. ?
| We already know a lot about detecting, producing and controlling neutrinos.
Production in a controllable (switchable) way requires a particle accelerator. Enormous energy is not required but high current is. These are complex and expensive facilities. There are not something you can buy off-the-shelf.
Likewise detection requires large devices (tons to megatons of active volume, depending) simply because the neutrino--nucleon cross-section is insanely small at achievable energies. Again these are very complex and expensive devices (not the least because they are individually designed, constructed and tuned).
It also requires a team of experts to keep both the beam and the detector on-line.
The technical challenges are considerable, and the use cases for which the extra cost can be justified are almost certainly non-existent.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/76531",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
What does imaginary number maps to physically? I am taking undergraduate quantum mechanics currently, and the concept of an imaginary number had always troubled me. I always feel that complex numbers are more of a mathematical convenience, but apparently this is not true, it has occurred in way too many of my classes, Circuits, Control Theory and now Quantum Mechanics, and it seems that I always understand the math, but fail to grasp the concept in terms of its physical mapping. Hence my question, what does imaginary number maps to physically?
Any help would be much appreciated
| If you really wanted to, you could formulate the laws of physics without using imaginary numbers - see, for example, Can one do the maths of physics without using $\sqrt{-1}$?
Let's say you need to do a fourier decomposition of a function $f$ in order to find how some responds to being driven at $f$.
If you think back, fourier decomposition is normally introduced with only real numbers, and a function $f$ was decomposed into odd and even parts, which can be represented as infinite series of cosines and sines respectively, and you could certainly do that. However, if we view $f$ as the real part of some complex-valued function, the equations become much simpler.
If this makes you wonder whether the complex-valued function is "more real" than the real part, and what is the physical significance, you could see $f$ as $\frac{1}{2}(z + \bar{z})$ where $z$ is the complex function; then, $z$ and $\bar{z}$ both have no physical significance, but they formally satisfy the equations governing the system, hence their sum must also.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/76595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 7,
"answer_id": 3
} |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.