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I've a problem understanding Absolute Gravitational Potential Energy? My current concept about potential energy is that 'If work is done on a body when it is at a point (taken as zero P.E), it covers some distance. Afterward, when it is allowed to move freely it does the same work. So, I get my work back as if I stored it in that body. But if the body doesn't come back to my assigned value of zero or in other words starts to move away from it then potential energy would be negative'. Refer to the following fig.:
According to my understanding, if the object is taken from point A to B then it had positive P.E.
But if it started to move towards O than it's P.E would be negative.
Correct me if I am wrong.
Now here's the Question:
Absolute Gravitational P.E is defined as 'Work done by Gravitational force for moving a body from a certain position to infinity'
Ok, agreed. So it's obvious that if Earth's center (O) is taken as zero P.E then Absolute P.E would be negative.
Now my book says:
Here U is absolute P.E. r is the distance from the center of Earth.
Consider the following figure:
It can be clearly seen that Workdone'y' is greater than Workdone'x', which is contrary to my textbook.
Or in other words, if Workdone is -50 and -5, which one would be greater?
| In case $x$ the system starts with Gravitational Potential Energy (GPE) of $-5$ and ends with GPE of $0$. So the GPE has been increased by $+5$ because $-5+5=0$. The amount of work done is $+5$.
In case $y$ the system starts with GPE of $-50$ and ends with GPE of $0$. So the GPE has been increased by $+50$ because $-50+50=0$. The work done is $+50$.
More work has been done in case $y$ than in case $x$ because $+50 > +5$.
| {
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Centre of mass and Rotational kinetic energy
A massless rod of length $l$ is pivoted at the upper end and two equal point masses of mass $m$ are attached to it, one at the centre of rod and one at its lower end. Then how much horizontal velocity must be provided to the lower end so that the rod just becomes horizontal?(consider only gravitational force is acting).
My question is -
How to solve above question by applying mechanical energy conservation law on Centre of mass of these 2 masses rather than applying it for individual masses and equating. If we can't do it by centre of mass approach, then why so?
I asked my teacher about it and he said that you can't find out velocity for COM just like that because it's performing rotational motion.
Why is it so? What's the reason behind it?
I have taken the reference plane which passes through pivoted end.
| The CM can be used to calculate the change in the gravitational potential energy of the system in this example, but it does not always give the correct result. It works here because the gravitational field is assumed to be uniform. If the gravitational field were not uniform, using the CM would not give the correct result for the change in GPE. You would have to consider the change for each of the constituent masses separately.
The CM cannot be used to calculate the change in kinetic energy in this example, but it does work in some situations. For example, when the motion is in 1D even though the object is 3D. You can use the motion of the CM to calculate changes in kinetic energy of an object when there is no rotation of the object.
| {
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Why don't spectacles form these weird images? It's an established fact that:
*
*Convex lenses produce inverted images of objects beyond the focus, on the other side of the lens.
*Any object placed at a finite distance from a concave lens appears to be somewhere between the focus and the optical centre when viewed from the other side.
Both these facts come from the lens formula, $$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$$
But I wonder:
*
*why a person wearing convex lenses doesn't see inverted images of objects beyond the focus, and
*why a person wearing concave lenses doesn't feel that the furthest objects are at a distance of $f$ from their eyes.
Hope my question is clear.
An example
Consider a person wearing spectacles with concave lenses of power $-1$ $\mathbf{D}$. The focal length would be $-1$ $\mathrm{m}$. If an object is at an object-distance $u=-3$ $\mathrm{m}$, simple calculations show that the image would be at an image-distance $v=-0.75$ $\mathrm{m}$ away.
But obviously the object doesn't appear to be so close to the person wearing the spectacles.
More confusingly still, the image would be magnified by a factor of $0.25$, and hence would appear to be rather small.
But again, as anyone wearing concave lenses would tell you, that isn't what they see!
Why is this so?
| Objects through spectacles appear both at a different distance and magnified. It turns out these effects cancel out, and the object subtends the same solid angle in your vision as without the spectacles. Your eye cannot directly perceive distance- your brain estimates it in several ways.
One way is by comparing the angular size in your vision: objects thought to be the same physical size with larger angular size appear closer. So since the spectacles do not change the angular size of objects, they appear to be at the same distance.
Using your numbers, the image is $\frac{1}{4}$ as big, but also at $\frac{1}{4}$ the distance away. Some simple geometry shows that the angular size has not changed.
Another way your brain can determine the distance of objects is by the direction your eyes are pointing when both are focused on a single point. Spectacles don't change that direction either (at least not much), and so don't mess with depth perception.
| {
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Relation between particle numbers and collisions between gas molecules inside a closed container I'm not a physicist, nor studying physics, so this may be a dumb or a very hard question, I don’t know. I'm not sure if I used correct tags, feel free to correct them if you feel like it is necesary. Forgive me about my English aswell, it’s not my mother language.
I'd like to know if there is a known relation between the average collisions per particle with other particles (not with walls) and the total amount of particles in a given volume and temperature. I guess its somehow a stochastic mechanic though, but how high is the typical deviation?. Will proximity to walls affect the number of collisions or wall collisions will compensate particle collisions?
My real interest is not to apply this to real particles, but with agents behaving like gas particles in a closed container, so every particle will be moving at the same speed (which if I’m not wrong solely depends on temperature in nature), and perfect elasticity can be simulated as well. I'm not taking gravity into consideration either. The "perfect simple answer" here is a formula relating volume, number of collisions per particle, temperature and amount of molecules (density?).
I’m always happy to learn, so if after a “simple answer” you want to refer me to additional documentation I’ll be happy to look further.
Thank you all.
| For an ideal gas, PV = nRT.
Pressure is proportional to the number of collisions each gas molecule sees-- not just the collisions with walls, but the collisions with other gas molecules as well. Hence how you can have different air pressures at different altitudes even though there are no "walls."
| {
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Why do phones land face down? Layman here.
I'm not sure if this is the case or not, but my anecdotal evidence is that mobile phones, especially large screen phones, tend to fall face down when you drop them; much to the owner's dismay, this leads to cracked screens.
I'm sure there is a scientific explanation for this, so I'd like to know: Why do mobile phones tend to fall and land face first (if so)?
I have a feeling it's related to the way your toast always falls butter side down, or how the shuttlecock always turns toward the same direction, but I'd like to know the explanation.
| I think one of the commentators summed is up nicely, that you are more likely to forget the times when it lands face up.
It's a psychological phenomenon that Christopher Chabris and Daniel Simons touch on in their book: "The Invisible Gorilla".
In short, it comes down to the fact that a phone falling face-down tends to be much more traumatic for the owner of said phone, and the mental trauma caused tends to leave a bigger impact on your memory. Whereas a phone landing face-up is quickly forgotten. Thinking back, one is therefore likely to only recall the phone falling face-down.
| {
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Electrically charged comb reacts to water Correct me if this is wrong, if a negatively charged comb reacts to water(ex. tapwater), the water will bend towards the comb because the water molecule has a slight positively charged "side" and because of that the free electrons on the comb and the water molecule will attract to each other.
But what if the comb was positively charged? Will the water still bend towards the comb because the water molecule also has a negatively charged "side"? Will the water bend away from the comb? (Is this because the water molecule is more positively charged than negatively charged?)
| Either a positive or a negative charge will attract a stream of water.
This happens because the charged object causes the water to become polarized. Say you have a negative comb - the positive side of the water molecules are attracted to it and the negative sides are repelled. This makes the positive sided a bit closer to the comb with the result that the attractive positive-negative force is a bit bigger than the repulsive negative-negative force. The net force is attractive.
With a positive charged object the negative sides of the water molecules would be pulled closer and the attractive force would once again be larger than the (positive-positive) repulsive force.
| {
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Wave function of many particle I read that the wave function of a system of many particles is formed from the product of the wavefunctions of the individual particles. What is the logic behind it?
| If the particles are not interacting then the Schrodinger equation for the 2-particle system will be
\begin{align}
H&=H_1+H_2\nonumber \\
H\Psi(x_1,x_2)&=
\left(-\frac{\hbar^2}{2m_1}\frac{\partial^2}{\partial x_1^2}+V_1(x_1)-
\frac{\hbar^2}{2m_2}\frac{\partial^2}{\partial x_2^2}+V_2(x_2)\right)\Psi(x_1,x_2)
=E\Psi(x_1,x_2)
\end{align}
Using the usual separation ansatz $\Psi(x_1,x_2)=\psi(x_1)\phi(x_2)$ one then obtains a pair
of independent Schrodinger equations
\begin{align}
H_1\psi(x_1)&=E_1\psi(x_1)\, ,\\
H_2\phi(x_2)&=E_2\phi(x_2)
\end{align}
with $E_1+E_2=E$. This by assumption gives a product form. Note that the solutions are multiplicative but the eigenvalues (i.e. the energies) are additive.
Of course if there is an interaction term $V_{12}(x_1,x_2)$ then separability is lost, although one can use the factored form as a basis set if $V_{12}$ can be treated as a perturbation.
If the particles are indistinguishable, then the solutions $\psi(x_1)\phi(x_2)$ must be additional symmetrized:
$$
\Psi_{\pm}(x_1,x_2)\sim \psi(x_1)\phi(x_2)\pm \psi(x_2)\phi(x_1)
$$
with the $+$ and $-$ signs applicable to bosons and fermions, respectively.
| {
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Poisson's equation in regions Suppose I have two regions in space. Region 1 and region 2.
In Region 1 I have a bunch of charges, and in region 2 I have no charges.
Is it true that Laplace's equation is satisfied in region 2?
| Yes, indeed. The electrostatic potential $V$ satisfies
$$\nabla^{2}V\left(\vec{r}\right)=-\frac{\rho\left(\vec{r}\right)}{\varepsilon_{0}}$$
so in a region $U\subset\mathbb{R}^{3}$ in which $\rho=0$, one also has
$$\nabla^{2}V\left(\vec{r}\right)=0$$
Note, however, that it doesn't mean you can always do something with it. For example, most theorems on Laplace's equation will require $U$ to be a connected open subset of $\mathbb{R}^{3}$ (with the topology induced by the Euclidean metric). So taking for instance $U$ to be a bunch of discrete points with no charges won't help you so much.
| {
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Why do you have to include the Jacobian for every coordinate system, but the Cartesian? In physics and engineering it is common to convert between different coordinate systems - spherical, polar, Cartesian, e.t.c. - depending on the problem. Physically, they are all clearly equivalent and it shouldn't matter which one we use.
Nevertheless, when solving problems involving volume or surface integrals, we always have to add the Jacobian matrix to the expression if we are any other coordinate system but Cartesian, just as if we were converting from the Cartesian system.
Take the expression:
$$
\iiint_V \nabla \cdot \textbf{u}\ dV
$$
It seems intuitive that we should be able to go from this expression and express $V$, $\textbf{u}$, $\nabla$, and $dV$ in whatever coordinate system we want at this point, but plugging the spherical coordinates straight into the expression yields the wrong answer - one has to add the Jacobian just as if the first step was expressing everything in Cartesians and converting to polars. Why do Cartesians get such special treatment? My expectation would be because the eigenvectors are constant throughout space, but I would appreciate a more thorough explanation.
| The original question: "Why do you have to include the Jacobian for every coordinate system, but the Cartesian?"
Perhaps you should elaborate on why │J│=1 for Cartesian coordinates, and the rest of your answer will hold perfectly.
Why does some one not write $1 x = 2$ and instead of $x = 2$? This is the same question.
Most people define extent of a rectangular object in $\Bbb{R}^n$ in a Cartesian sense as the product of the $\Bbb{R}^1$ extents. In that case of the transformation of an Cartesian original n-dimensional space to Cartesian $\Bbb{R}^n$ is the identity matrix, which is the matrix equivalent of $1$ for coordinate transformations. But, why bother with putting a 1 in your formulas?
$$
\int _{\varphi _1}^{\varphi _2}\int _{\theta _1}^{\theta _2}\int _{\tau _1}^{\tau _2}f((\left(
\begin{array}{ccc}
\tau & \theta & \varphi \\
\end{array}
\right)\cdot\left(
\begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{array}
\right))^\mathsf{T})d\tau d\theta d\varphi
$$
where $f$ is a real valued function of a row 3-dimensional vector. But, like not writing the $1$ in the algebra equation, why not just use $f(\left(
\begin{array}{ccc}
\tau & \theta & \varphi \\
\end{array}
\right))$? All the matrix multiplication is doing is relabeling $\left(\begin{array}{ccc}
\tau & \theta & \varphi \\
\end{array}
\right)$ as $\left(\begin{array}{ccc} x & y & z \\ \end{array}\right)$.
| {
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Direction of velocity vector in 3D space According to a well-known textbook (Halliday & Resnick), the direction of a velocity vector, $\vec v$, at any instant is the direction of the tangent to a particle's path at that instant, as is illustrated below in 2D.
According to the same textbook, the same holds for 3D. However, the tangent to a curve in 3D is not a line, but a plane! A vector could be in a plane and still take on any direction betwen $0^{\circ}$ and $360^{\circ}$ within that plane.
How do we define and determine the direction of $\vec v$ given a particle's path in 3D? (If possible, include illustrations in your answers).
| Since asked in Physics.SE I'll project the Physics point of view.
This can be answered intuitively. Imagine yourself flying aimlessly in air. At certain instant you are heading to somewhere. Where will your velocity vector point to?
Find a plane containing smallest part of path curve at that instant. The velocity vector must lie in the same plane.
Note: For exactly straight paths it'll be limiting case. Therefore make those impractical straight paths slightly curved ;), then do the above procedure.
| {
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Why is the correlation function a power law at the critical point? I’m taking my first exam in statistical field theory and critical phenomena. I’ve reached a point in which we use the fact that the pair correlation function decays as a power law at the critical point:
$$\left<\psi(x)\psi(0)\right> \sim\frac{1}{x^{D-2+\nu}}$$
to renormalize it to reach the $\epsilon$-expansion in 4 dimensions, which I’m comfortable with.
The thing is, the whole procedure is based on this assumption and I couldn’t find a way to prove it form the topic we previously discussed, which are Landau $\psi^4$ expansion, Hartree fock approximation and normalization or blocking variables.
Can anyone give me a hint on who to proceed? I'm really missing the thing which glues the two things together.
| I cannot derive the behavior from a microscopic theory, but I can motivate it somewhat.
The general relation (away from the critical point) includes a factor of $\exp(-\frac{r}{\xi})$, where $r$ is the distance and $\xi$ is the correlation length, which can depend on, for example, temperature. At the critical point this correlation length diverges, so that there are correlations at all length scales, and only the first factor remains.
As for the reason - as I said, I don't know of a derivation. It has been empirically verified for a number of systems. You may also want to read about universality and critical exponents. This type of power law behavior appears in remarkably many places in nature! The scale invariance of e.g. the Ising model at its critical temperature is a manifestation of this power law, and there is a conformal field theory that describes the Ising model there.
Related questions
What does the behavior of the pair correlation function look like in the vicinity of the critical point?
Universality classes
What happens for the spins around the phase transition
Ising model for dummies
| {
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Why we see things on distances we shouldn't see? One person took experiment from Montenegro and with distance 201 km he saw clearly montains that are 1521 meters high. He was on 189 meters above the sea level. Official calculator says you cannot see less than 1811 meters from that distance. My question is: Why he saw the mountain? Is there any reference that can help me understand that. I'm a mathematician and I calculated that the radius of the Earth should be 7260 km for that to be possible. I need to understand which physical laws are related to this problem and please give me a reference.
| Per @John Forkosh, the refraction phenomena is called "looming":
https://en.wikipedia.org/wiki/Looming_and_similar_refraction_phenomena
| {
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Why is work done equal to $-pdV$ only applicable for a reversible process? In thermodynamics, when we're interested at gases, I know that the work done can be written to be $-pdV$ for a reversible process ($p$ is the pressure of the system, and $V$ is the volume of the system).
This is because $$dW=Fdx=-pAdx=-pd(Ax)=-pdV$$
However, why is it not true also that the work done is $-pdV$ for non-reversible processes as well?
|
why is it not true also that the work done is for non-reversible processes as well?
The general expression for infinitesimal work done by contact pressure forces on a system inside a closed boundary surface is actually
$$
-p_{ext}dV
$$
where $p_{ext}$ is the external pressure. This follows from mechanics, where net work done on a system of mass particles by external forces (after particles of the system undergo displacements $d\mathbf{r}_a$) is
$$
\sum_a \mathbf{F}^{ext}_{-a} \cdot d\mathbf{r}_a
$$
where $\mathbf{F}^{ext}_{-a}$ is force due to all particles that are not part of the system acting on the particle of the system $a$. This can be adapted to the macroscopic description with walls and pistons, the forces are replaced by pressure, and the displacements by change of volume.
One things remains: the pressure is that of external agents, the change of volume is that of system.
If the process is reversible at each stage, the system can be ascribed internal pressure $p$ that has the same value as the external pressure $p_{ext}$. Then it is possible to write the work as
$$
-pdV.
$$
However, in case the process is not so reversible, there may be stages of it where the system does not have single pressure $p$ (imagine the gas is moving inside the enclosure in a turbulent fashion; there is no single pressure, but it depends on the position). Then we cannot express the work as a function of $p$. We can still use the general formula with $p_{ext}$ though.
| {
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Turning a bike, why does it lean? I realize this seems like a pretty simple issue of drawing a free body diagram, but I just can't seem to figure it out.
If a bike leans, then it must have had a torque that made it lean. I considered the centripetal force, in this case friction, as a possible source for this torque, but that would create torque in the opposite direction. The only remaining torque causing force in my FBD is weight. But weight was there before the leaning, and would only cause a torque after leaning.
If anything, because of centripetal acceleration, observing from an inertial frame of reference, the bike should lean in the opposite direction because of the introduction of the centripetal force. So the question is, what force causes the bike to start leaning?
| Two things that need to be considered. Often it is not the cyclists weight that causes the lean in. Subconsciously, as people learn to ride a bike, they do something called counter-steering.
So for example coming in to turn right, a cyclist will quickly turn the handlebars left to lean the bike to the right due to centripetal acceleration. Once the lean has been achieved, the handlebars are then turned to the right to turn and balance the bike in a turn. Then to upright the bike, the handlebars are turned slightly harder in until it levels (due to increasing the centripetal force outwards relative to the lean), then the bike is straightened, and the handlebars follow.
People don't believe that they do this, it's quite funny.
| {
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How is it possible that particles maintain continual motion? The kinetic theory of matter (particle theory) states that: "all matter consists of many, very small particles which are constantly moving or in a continual state of motion".
Given the 1st law of thermodynamics, how is this possible? That is, how are particles of a substance in the solid phase able to continually vibrate. And how are particles of a substance in the gas phase able to maintain random motion?
| All particles are always moving because they have thermal energy which makes them vibrate and randomly move around. We could say that when particles get warmer then they vibrate with a higher frequency and move a lot faster. When they get cooler, then the particles move and vibrate slower. There is the theory that if a particle is at $-273\ \mathrm{°C}$ or $0\ \mathrm{K}$, then it would be at a stationary position and would not vibrate at all. But there is no evidence of such case yet.
By the way, that thermal energy that all particles contain for different amounts is believed to come from the big bang.
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Mass and Newton's Second Law While trying to understand the second law of Newton from "An Introduction to Mechanics" by Kleppner and Kolenkow, I came across the following lines that I don't understand:
"It is natural to assume that for three-dimensional motion, force, like acceleration, behaves like a vector. Although this turns out to be the case, it is not obviously true. For instance, if mass were different in different directions, acceleration would not be parallel to force and force and acceleration could not be related by a simple vector equation. Although the concept of mass having different values in different directions might sound absurd, it is not impossible. In fact, physicists have carried out very sensitive tests on this hypothesis, without finding any variation. So, we can treat mass as a scalar, i.e. a simple number, and write $\vec{F} = m\vec{a}$."
The lines above lead me to question:
*
*Why is it not" obviously true" that force behaves like a vector?
*Why is it not impossible for mass values to be different in different directions?
| The accelerating system may be more complex that a single body. Consider an example of a ball in a pipe. If you apply a force to a ball along the pipe, then you would move only the ball inside the pipe. So your effective mass for the equation is the mass of the ball. However, if you apply a force to the ball at 90 degrees, then you would move both the ball and the pipe. So your effective mass is a sum of the mass of the ball and the mass of the pipe. Finally, if you apply a force at any other angle, your effective mass would be the mass of the ball plus the appropriate fraction of the mass of the pipe. Thus in a complex system the effective mass can be different in different directions.
| {
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What is the "surface term"? In Peskin's quantum field theory book, There is a sentence in page 17:
...
More generally, we can allow the action to change by a surface term, since the presence of such a term would not affect our derivation of the Euler-Lagrange equations of motion ...
...
\begin{equation}
\mathcal{L}(x)\to\mathcal{L}(x)+\partial_\mu\mathcal{J}^\mu(x).\tag{2.10}
\end{equation}
What is the "surface term"? Is it just a partial derivative term as $\partial_\mu\mathcal{J}^\mu(x)$?
| *
*We use the divergence theorem to connect boundary terms to divergences, as already mentioned in kryomaxim's answer.
*It should be stressed that the relevant divergence term $d_{\mu}{\cal J}^{\mu}$ is a total spacetime derivative, not a partial/explicit spacetime derivative $\partial_{\mu}{\cal J}^{\mu}$, even if many authors use confusing notation, cf. e.g. my Phys.SE answer here.
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$SU(2)$ to $U(1)$ symmetry breaking Why does electroweak theory have $SU(2) \times U(1)$ to $U(1)$ symmetry breaking. Is it not possible to have simply $SU(2)$ to $U(1)$ symmetry breaking?
| There is the Pati-Salam model that has an electroweak field $SU(2)_L\times SU(2)_R$. With the strong nuclear interactions it is $SU(4)\times SU(2)_L\times SU(2)_R$. There are some problems with the idea of an $SU(4)$ strong nuclear interaction theory. The recent results by the LLNL lattice group might breath some life into this. The $SU(4)$ has $4$ colors with $7$ additional gluons that change color with an additional color neutral gluon. Maybe the additional stuff is associated with dark matter. The symmetry breaking of this theory involves the decomposition of the $SU(2)_R~\rightarrow~U(1)$ that recovers the standard electroweak field theory. The $\sigma_{\pm}$ elements from this broken $SU(2)_R$, which in the $SU(2)_L$ go on to define the $W^{\pm}$, might have some additional physics. These might go into the generation of additional photons or combine to form a sort of $Z_\gamma$. Whether this is the case or bit will ultimately come from experimental data.
| {
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"timestamp": "2023-03-29T00:00:00",
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$dU=dQ$ and $dU=TdS$, but $dQ$ not always equal to $TdS$? Why? $$ dU = dQ+dW $$ $$ dU=TdS-pdV $$ The equations above are always true for a thermodynamic state of a certain system. Now let's say that we have a situation where $dW=0$, this tells us that $$ dU=dQ $$ $$ dU=TdS $$But still I can't write $ dQ=TdS $, since this only works for a reversible change of my system. So if I don't have a reversible system I can work with $ dU=dQ $ and $ dU=TdS $, but I can't work with $ dQ=TdS $.
I get this, but I have been trying to figure out why this is, and I just can't seem to get it.
| Even if dW = 0 (e.g., constant volume), you can't write dQ=TdS for an irreversible process because, for irreversible heating or cooling of a body, T is not constant spatially within the body (i.e., T varies with spatial position). For transient irreversible heating, the temperatures near the boundary are hotter than in the interior, and for transient irreversible cooling, the reverse is true. So what value of T are you supposed to use? If you use the value of T at the boundary at which the heat transfer Q is occurring (i.e., $T = T_B$), you find that $dQ<T_BdS$. More precisely, $$\int{\frac{dQ}{T_B}}<\Delta S$$This is just the statement of the Clausius Inequality.
| {
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"timestamp": "2023-03-29T00:00:00",
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Shouldn't pressure be dependent on volume? Pressure is defined as $\frac{F}{A}$ And while this makes sense when you are talking about pressure at the surface of a fluid, deep within a fluid it doesn't make sense, at least not to me.
It especially doesn't make sense to me if you are in a body of water. You will have forces coming from all directions. The force of gravity will pull you down and this gives the pressure at the top. The buoyant force provides a pressure from below. And water moves at different rates in different places so depending on where you are, you might or might not have equal pressures in each direction.
So shouldn't the total pressure be equivalent to the force of gravity at x distance from the center of mass and thus you would have 0 pressure if the water was perfectly still and you were in a position where gravity and buoyancy are balanced and more pressure in a given direction if there is more force pushing you in that direction than what is opposing it? But even then, it is still 3-dimensional so shouldn't volume be a factor in determining pressure instead of area being a factor?
| If you are in a body of fluid of density $ \rho$ at a depth $h$ below the surface,
The net pressure acting would be
$P= \rho gh$
Remember that pressure is defined to be force acting per unit area of surface i.e., Pressure is always considered to be acting normal to the surface( This is why even though it has a direction, it's considered scalar or more precisely tensor).
So, if were to calculate the force acting on the top surface of a body at a depth $h$ in a fluid of density $\rho$ with top surface area $A$,
$F=(\rho gh)A$
So, your consideration in 3-dimensional sense would be to determine the force acting and not the pressure.
| {
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What law or formula discusses the relationship between pressure and dew point? A lot of times, dew point is focused primarily upon temperature and relative humidity. However, that same point of saturation is affected by the pressure, but I can't find a formula, or law even, that discusses this.
It's possible that my premise of the relationship between dew point and pressure may be inaccurate, but if so, how can we then explain that a refrigerant may be a vapor at 210°F in a 280psi container, have a dew point of 125°F, but also be a vapor at 75°F in a 70psi container.
Whether the substance is water, freon, or any liquid, there should be relationships between pressure and dew points (saturation points, condensation points, etc - they're all meaning the same thing). We can clearly see with freon there's some relationship between pressure and temperatures where the rate of condensation is greater than that of the evaporation, but I can't seem to find any laws/formulas for this.
| Even though I am not an expert on this, I found some information that can help. Check out this website
https://www.vaisala.com/sites/default/files/documents/Dew-point-compressed-air-Application-note-B210991EN-B-LOW-v1.pdf
and this website
https://en.wikipedia.org/wiki/File:Dewpoint.jpg
| {
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Is an electric field a transverse wave not in a vacuum? According to "Introduction to Plasma Physics and Controlled Fusion 3rd Edition", in discussing electromagnetic waves in a plasma, $\vec{E}$ is said to be a transverse wave. Is $\vec{E}$ a transverse wave even in a plasma? I only know the proof that $\vec{E}$ is a transverse wave "in a vacuum". By one of Maxwell equations and by assuming a plane wave solution, we easily get
$\frac{\rho}{\epsilon _0} = \vec{\nabla} \cdot \vec{E} = \vec{k} \cdot \vec{E}.$
$\vec{E}$ is not a transverse wave, is it?
(In the book, an equation $\vec{k} \cdot \vec{E} = 0$ is used. So I have to know whether this equation is truly true.)
|
$\vec{E}$ is not a transverse wave, is it?
There is nothing wrong with transverse ($\mathbf{k} \cdot \mathbf{E} = 0$, e.g., O-mode) or longitudinal ($\mathbf{k} \times \mathbf{E} = 0$, e.g., ion acoustic waves) waves in a plasma.
Is $\vec{E}$ a transverse wave even in a plasma?
It can be, but again it is not required to be. There are lots of waves with longitudinal oscillations of $\mathbf{E}$ that exist in a plasma and lots of work done measuring the electric fields associated with these modes (e.g., the following spacecraft measure electric and magnetic fields Wind, THEMIS, MMS, etc.).
| {
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Physics interpretation of the expectation value of an electrons spin components Suppose I have an electron that is in the spin state
$$\chi =A\begin{bmatrix}3i \\4\end{bmatrix} $$
If I calculate the expectation values of its spin components $S_x$ $S_y$ $S_z$, I get
$$\langle S_x \rangle= 0$$
$$\langle S_y \rangle = -\hbar \dfrac{12}{25}$$
$$\langle S_z \rangle= -\hbar \dfrac{7}{50}$$
I can do the math part easily, however I am having a bit of difficulty trying to apply a physical interpretation to my answers.
My attempt at a physical interpretation
Due to the uncertainty principle, we can not having a well defined $S_x$ $S_y$ $S_z$ component simultaneously. Hence, we can only calculate the expectation value of its components. For $S_x$, the probability that the spin will lie in the $S_x$ component will be zero. For $S_y$ and $S_z$ we the spin has either $-\hbar \dfrac{12}{25}$ or $
-\hbar \dfrac{7}{50}$.
| Average values do not tell you about probabilities (which must be non-negative numbers). In particular, if your particle is describe by a state $\vert\xi\rangle=\frac{1}{\sqrt{2}}\left(\begin{array}{c}1\\1 \end{array}\right)$ the average $\langle S_z\rangle=0$ but the probability of having the system in the spin-up state is $1/2$, as is the probability of having the system in the spin-down state.
Average values are weighted sums of probabilities: in the case of $\vert\xi\rangle$, the average value is
$$
\langle S_z\rangle = P(\uparrow)\frac{\hbar}{2}+P(\downarrow)\left(-\frac{\hbar}{2}\right)
=\frac{1}{2}\frac{\hbar}{2}+ \frac{1}{2}\left(-\frac{\hbar}{2}\right)=0
$$
where $P(\uparrow)$ is the probability of having your system in the spin-up state, and $\hbar/2$ is the value of $S_z$ when the system is in the spin-up state.
Clearly here, the fact that the average value is $0$ does not in any way mean the probability of the spin being along $\hat z$ is $0$, since the system is half the time along $+\hat z$ and half the time along $-\hat z$. In fact, the only two possible values of spin are $\pm\frac{\hbar }{2}$, which excludes $\langle S_z\rangle=0$ for this example. More generally, the only possible outcomes of $S_y$ or $S_x$ are $\pm \frac{\hbar}{2}$, and never numbers such as $-\frac{12\hbar}{25}$.
In general, the probability of a system described by $\vert \chi\rangle$ to be found in a state having definite spin along - say - $\boldsymbol{\hat y}$ is given by
$$
P(\uparrow)_{\hat y}= \vert\langle\chi\vert \uparrow\rangle_{\hat y}\vert^2\, ,\qquad P(\downarrow)_{\hat y}=\vert\langle\chi\vert \downarrow\rangle_{\hat y}\vert^2
$$
where $\vert\uparrow\rangle_{\hat y}$ is the eigenstate of $\sigma_y$ with eigenvalue $+\hbar/2$. In your case, we have
$$
P(\uparrow)_{\hat y}=\frac{1}{50}\, ,\qquad
P(\downarrow)_{\hat y}=\frac{49}{50}
$$
so
$$
\langle S_y\rangle = \frac{\hbar}{2}\frac{1}{50}-\frac{\hbar}{2}\frac{49}{50}=-\frac{12\hbar}{50}\, .
$$
What the average values tell you is the "direction" of the spin. The unit vector $\boldsymbol{\hat n}=(\sigma_x,\sigma_y,\sigma_z)=(0,-24/25,-7/25)$ points in a definite direction, and if you were to choose your quantization axis along this direction, the system would be in a spin-up state, i.e it would be an eigenstate of
$$
\boldsymbol{\hat n}\cdot \vec S=
\frac{1}{25}\left(
\begin{array}{cc}
-7 & 24 i \\
-24 i & 7 \\
\end{array}
\right)\, . \tag{1}
$$
You can indeed verify that your vector $\chi$ is an eigenvector of (1) with eigenvalue $\frac{\hbar}{2}$
| {
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How can entropy increase in a reversible process? According to Wikipedia, a reversible process is "a process whose direction can be "reversed" by inducing infinitesimal changes to some property of the system via its surroundings, with no increase in entropy."
However, for isothermal processes, any reversible heat added to the system at constant temperature increases entropy. So, a reversible process of heating is causing a change in entropy. This seems to contradict the definition above.
Where have I gone wrong?
| When one says that entropy does not increase for a reversible process, the entropy they are talking about is the total entropy of an isolated system. If we consider the universe as our isolated system total entropy is equal to the entropy of system + the entropy of the surroundings.
So, for a reversible process where heat is added into the system, the entropy of the system does increase however the entropy of the surroundings in any reversible process = -(entropy of the system)and ultimately, the total entropy is zero.
| {
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Do higher frequency/energy levels in the EM spectrum mean higher temperatures? I am trying to find concrete evidence that for example, light in the optical spectrum would be hotter than infrared light because it has a higher frequency, and that is directly proportional to energy. Is energy directly proportional to temperature?
If we are to split up the optical spectrum into its components, blue light has a higher frequency than red light, and blue light is hotter than red light. Does this work for the whole spectrum?
| Light of a specific wavelength does not have a well-defined "temperature" - however, it does have energy.
The Planck law tells us that the spectrum of a black body of a certain temperature covers a range of wavelengths, and the Wien displacement law tells us what the peak of that distribution is as a function of temperature.
The actual radiation from a warm body will be the black body radiation modified by the emissivity of the surface - so in principle, a hot surface that has very little emissivity at short wavelengths might appear "redder" than a cooler surface that has low emissivity in the longer wavelengths - but in practice that would be quite hard to achieve.
Cooler objects have essential their entire emission in the (near)-infrared band, but the same rules still apply: the cooler the object, the longer the emitted wavelengths. The microwave background of the universe corresponds to a temperature of about 2.7 K - and it follows the same laws of physics.
| {
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Why X-ray emission is able to exit the volume without essential photo-absorption My question is written in the title
I am currently studying of radiation of plasma
And I can't get it why it works like this.
"Why X-ray can penetrate the volume without photo-absorption."
| There is always some absorption. But maybe the point is that the x-ray emission lines of an element are lower in energy than the corresponding x-ray absorption threshold energies.
For example, the Cu K$\alpha$ and K$\beta$ x-ray emission lines are below the Cu K absorption edge. At energies just above the Cu $1s$ binding energy, the attenuation in copper is about five times stronger.
| {
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Loss in energy during gravitational redshift We know from General Relativity that light needs to expend some energy to overcome the gravitational pull of massive objects. This is done by decreasing it's frequency and so it's colour shifts towards red. But what I don't understand is that where does this energy go? According to law of conservation of energy, the total energy in the universe must be constant. But this is not in the case of gravitational red shift. We can't add it to the gravitational potential energy because as Einstein said gravity is not a force but simply a curvature in space time.
| We know from General Relativity that light needs to expend some energy to move into the upwards direction. This is done by decreasing its frequency and so its colour shifts towards red. According to law of conservation of energy, the total energy in a closed system must be constant. This is true in the case of gravitational red shift too. The energy goes from the light to somewhere. If the light gets reflected back, energy starts going from the aforementioned 'somewhere' to the light.
What is the aforementioned 'somewhere'? It is the surrounding matter and the surrounding space-time.
| {
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If an astronaut orbits earth, can both Time Dilation and Gravitational Time Dilation affect it? I am very new at all of this stuff and this one thing bugs me very much...
If an astronaut is orbiting earth, it should be experiencing Time a bit faster than those on earth, correct?
Well then because of the speed he is traveling, he will also experience Time a little slower than those on earth.
Correct me if I’m wrong but those should pretty much cancel out, but that sounds completely wrong to me.
If you could answer this question and fix any of my poor logic/ reasoning, that would be awesome :)
| Yes, as astronaut will be at relatively high speed then it will experience time dilation and also earth gravitational pull is not strong compare to others there will be negligible affect of gravitational time dilation.
| {
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What exactly is The Big Crunch? What exactly is The Big Crunch? I want to know exactly what happens. Does the universe come back in on itself entirely? I did research and it says it could result in a pre-Big Bang particle. The universe is expanding quicker so the chances are smaller of it happening but what chance could it still happen.
| For an body to escape another body's gravitational influence, the 1st body's velocity has to exceed the 2nd body's escape velocity, e.g. to get to a speed where the sum of an object's kinetic energy and its gravitational potential energy is equal to zero. The formula for escape velocity is given as V=the square root of 2GM/r, where G is the gravitational constant, M is the mass of the 2nd body, and r is the distance from the 2nd body's center of mass to the 1st body.
If the speed of the universe's expansion does not exceed the escape velocity of the universe's mass, then the universe will begin to collapse in on itself by its own gravity. The temperature of the cosmic microwave background will begin to rise and large galactic structures will start merging together from superclusters all the way to galaxies. When the objects are close enough together for stars to merge, the temperature of the CMB will have risen enough so that it stars will not be able to expel their internal heat and so will explode into gas clouds which then will dissolve into their respective subatomic particles because of the increasing heat. Eventually, everything will either collapse into a black hole or be swallowed into one with all matter in the universe becoming one coalesced singularity.
| {
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Why does a room not warm up faster when I put the heater's thermostat on a higher value? I would say it should warm up faster because the difference in temperature between the room and heater is higher.
Edit: I am talking about a convection heater.
| In addition to answers, I simulated an example setup to show the differences. In your case, only the referance value of the controller is changing.
I have trouble with imgur, so I also posted images online.
Model:
https://s18.postimg.org/71vo9wq0p/image.png
Plot: https://s18.postimg.org/rm0i8el7d/image.png
Comparison: https://s18.postimg.org/4xbb8tw3t/image.png
| {
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Can stellar nucleosynthesis processes other than deuterium-lithium, i.e., oxygen-burning, take place in a reactor? It is known that gravitational confinement and supernovae facilitate the synthesis of heavy nuclides; and, that magnetic confinement is thought to facilitate the deuterium-lithium fusion in human-constructed systems.
Is heavy nuclide synthesis through an artificial mechanism such as magnetic confinement thought possible or impossible?
| A big problem with fusing heavy elements with magnetic confinement is that the reaction rate will peak at higher and higher temperatures. For example, this article shows the cross section for carbon peaking at a few MeV:
https://www.researchgate.net/figure/Carbon-fusion-cross-section-as-a-function-of-energy-from-various-measurements-data_fig1_301842001
Deuterium-tritium, on the other hand, peaks near 50 keV. The trouble with really high temperatures is that radiation losses from bremsstrahlung become much larger and so it's much more difficult to keep energy confined in the plasma. Stars get away with this because they're huge and dense and even radiation takes forever to escape.
| {
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Why won't the water pour out? A couple days ago I was at a restaurant drinking my cup of water through a straw. I childishly put my finger on top of the straw in the cup of water, and I pulled out the straw. I noticed that the water was not pouring out from underneath the straw. I found this very confusing since I only covered the top of the straw, the second I uncovered the top of the straw the liquid then poured out the straw. Can someone please explain why this was the case?
| There is air in the straw between your finger and the fluid level in the straw. If only a few drops of liquid come out of the straw before it stops flowing, then, by the ideal gas law, the air in the head space will have expanded, and its pressure will have dropped to a little lower value than atmospheric. Basically, you have formed a slight vacuum in the head space. The difference in pressure between the bottom of the liquid column and the top of the column is enough to support the weight of the liquid.
| {
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Adding gauge fixing directly by hand is different from by Lagrange multiplier? Why is adding gauge fixing directly different from doing so by Lagrange multiplier? For simplicity, we don't use field model.
Direct method
Consider a system
$$L(x,\dot x,y,\dot y)=\frac{\dot x^2}{2}+\dot x y+\frac{(x-y)^2}{2} \tag{1} \, .$$
This system has gauge symmetry
$$\delta x=f(t),\ \delta y=f(t)-\dot f(t)\tag{2}$$
for arbitrary $f(t)$. Under this transformation,
$$\delta L = \frac{d}{dt}(x f+\frac{1}{2}f^2)\tag{3} \, .$$
The Euler-Lagrange equations are:
\begin{align}
L1 &: \quad \ddot{x}+\dot y- x+ y =0 \tag{4} \\
L2 &: \quad \dot x - x+y = 0 \, .\tag{5}
\end{align}
We see
$$L1= \frac{d}{dt}L2+L2\tag{6}$$
so $(4)$ is not independent of $(5)$ and we need only to solve $(5)$, i.e.
$$\dot x =x-y \tag{7} , .$$
We see $y(t)$ is a gauge freedom, and only fixing $y(t)$ we can solve $x(t)$.
Suppose we choose the gauge $y=0$. Then we solve $\dot x-x = 0$ with result $$x= c e^t \, .\tag{8}$$ with constant $c$ determined by initial value.
Lagrange multiplier method
Now let's try with the Lagrange multiplier method,
$$L'(x,\dot x , y, \dot y, \lambda)= \frac{\dot x^2}{2}+\dot x y+\frac{(x-y)^2}{2} - \lambda y \, .\tag{9}$$
The Euler-Lagrange equations are
\begin{align}
\ddot{x}+\dot y- x+ y &= 0 \tag{10} \\
\dot x - x+y-\lambda &= 0 \tag{11} \\
y &= 0 \, . \tag{12}
\end{align}
Substituting $(12)$ into $(10,11)$ gives
$$\ddot{x} - x =0 \quad \text{and} \quad \dot x-x=\lambda$$
therefore
$$x = c_1 e^{-t}+c_2 e^t \longrightarrow \lambda = -2 c_1 e^{-t} \, .\tag{13}$$
It's obvious that $(13)$ is different from $(8)$.
Why do these two methods give different results?
Note: $y=0$ is a well-defined gauge condition because for any function $y(t)$ I can always choose gauge transformation $f(t)= c e^t + e^t\int_1^{t} e^{-u}y(u)du $ such that $y = 0$
| Let me summarize some key points in my opinion:
*
*$y$ is not a free variable that can be arbitrarily fixed (like a gauge field) but instead is an auxiliary variable.
*The Lagrangian you start with is actually a total time derivative.
*You change the Lagrangian in the Lagrange multipliers case.
Since there are no dynamic term, $g(\dot{y})$ in Eqn.(1), $y$ is an auxiliary variable: You can replace it by its equation of motion, which is $$y=x-\dot{x}$$ After this, you see that your Lagrangian is $$L(x,\dot{x})=\frac{d}{dt}\left(\frac{1}{2}x^2\right)$$ This is the reason why you see a symmetry $$x\to x+f(t)$$ in the first place: No matter how you change $x$, the difference in Lagrangian will be total time derivative as the initial Lagrangian itself is total time derivative.
In general, you cannot add the result of EOM to the Lagrangian: You would get incorrect answers. However, for auxiliary fields which lack dynamical terms, this is possible. In your example, $y$ is not a gauge freedom but is an auxiliary field.
In your Eqn.(9) you change the Lagrangian as you cannot arbitrarily choose $y$. In fact, if you repeat the same calculation with the Lagrangian
$$L'(x,\dot{x},y,\dot{y},\lambda)=L(x,\dot{x},y,\dot{y})+\lambda (x-y-\dot{x})$$
you get the same consistent results.
| {
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Viscosity and energy balance When solving Navier Stokes equations for viscous fluid over rigid surface, the viscous term in the momentum equation accounts for the momentum transfer between the fluid and surface in the near wall region, i.e. part of the fluid momentum is extracted by the act of viscosity. This is manifested as friction, wall shear stress, drag or flow resistance (you name it).
The dilemma that faces me know is the energy balance in this process. Since the kinetic energy and momentum are related by:
$$K=\frac{1}{2}\frac{P^2}{m} $$
Then momentum transfer must implies energy transfer. Since the surface will remain stationary, the kinetic energy should be transformed into other kind of enrgy. I expect that it would transformed into heat.
But since for incompressible flows we usually don't solve the energy equation I'm wondering how can we account for this energy transfer and transformation? and does neglecting that may affect the solution of momentum equation?
| The viscous dissipation of mechanical energy to internal energy is occurring not only at the walls of the duct, but throughout the duct. The local rate at which this is occurring is proportional to the viscosity times the second invariant of the rate of deformation tensor (typically, the square of the shear rate). This is usually referred to as "viscous heat generation." This viscous heat generation certainly does occur in viscous fluids, and, in high viscosity fluids, can cause substantial temperature increase (particularly locally, near the duct wall, where the shear rate is highest). For a complete discussion of the details on this, see Transport Phenomena by Bird, Stewart, and Lightfoot.
| {
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Probability of finding a particle in a region in a state given for a wave function plus a constant This is the problem:
A particle is restrained to move in 1D between two rigid walls localized in x=0 and x=a. For t=0, it’s described by:
$$\psi(x,0) = \left[\cos^{2}\left(\frac{\pi}{a}x\right)-\cos\left(\frac{\pi}{a}x\right)\right]\sin\left(\frac{\pi}{a}x\right)+B $$
For $t>0$, determine the probability of finding the particle between 0 and $\frac{a}{4}$.
So, using some trigonometry and the orthonormal base $\phi_{n}(x)=\sqrt{\frac{2}{a}}\sin(\frac{n\pi}{a}x)$, I can write the wave function as:
$$\psi(x,0)=\sqrt{\frac{a}{32}}\phi_{1}(x)+\sqrt{\frac{a}{8}}\phi_{2}(x)+\sqrt{\frac{a}{32}}\phi_{3}(x)+B$$
I still can’t use the evolution operator. I must find an expression to $B$, so I can put it in terms of the base.
I use: $B=\sum_{n} C_{n}\phi_{n}(x)$ where, after finding the value of $C_{n}$, and noticing that only odd values of n contributes to the wave function:
$$B\rightarrow -\frac{B}{\pi}\sqrt{8a}\sum_{0}^{\infty}\frac{1}{2n+1}\phi_{2n+1}(x)$$
So now, how could I add a constant to the wave function so it’s normalized? It’s just finding the value $B$ using $\langle\psi|\psi\rangle$? Or there is other way? Because using $\langle\psi|\psi\rangle$ I get a quadratic, and I’m not sure that is the way.
| In a $1D$ box your wave function must vanish at $x=0,a$. In particular
$$\psi\left(x=0,t=0\right)=B=0$$
| {
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Does temperature affect the index of refraction? I would like to know the physical reasoning behind the fact that temperature indeed does change the index of refraction of a certain medium.
(Is there an easy experiment for me to try this at home?)
| Light is an electromagnetic radiation having both electrical and magnetic fields . It’s motion through a matter has its interactions with it which depends on the matter through which it has to pass. The molecules themselves are moving with Brownian motion as in liquid or vibrating about its position , a property related to its energy state , which depends on temperature and molecular interactions with electromagnetic wave depends on electron clouds and it’s fields, it encounters while passing through it.
In case of gases, the void space increases with temperature or in layman’s language the density decreases so the hindrances to the passage of light decreases means refractive index of the medium decreases.
Let’s us have some optically active medium and you are able to control the concentration of optically active reagent in liquid, then not only light bends but it’s plain gets rotated depending on the optical reagent concentrations.
Here reason is electromagnetic interactions with optically active medium which not only slows down light but it bends too depending upon the intensity of such interaction on the path of plain polarised light. Here you observe bending ( change in refraction ) and Rotion of plain( due to optical interactions ) .
| {
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Ball swung in a horizontal circle, can the string ever by exactly horizontal to the ground? You swing a ball on the end of a lightweight string in a horizontal circle at constant speed. Can the string ever by truly horizontal? If not, would it slope above the horizontal or below the horizontal? Why?
|
Let's assume the ball is rotating in a circle (in the horizontal plane) with radius $R$ at angular velocity $\omega$.
In order for the ball to stay on its circular path, a centripetal force $F_c$ has to act on it, given by:
$$F_c=mR\omega^2$$
This force is the resultant of the tension in the string $T$ and the weight $mg$. With some simple trigonometry:
$$F_c=T\cos \theta$$
and:
$$T\sin \theta=mg$$
Reworked we get:
$$F_c=\frac{mg}{\tan\theta}=mR\omega^2$$
So that:
$$\tan\theta=\frac{g}{R\omega^2}$$
So $\theta \to 0$, for:
$$R\to \infty$$
and/or:
$$\omega \to \infty$$
So the angle $\theta$ will always be positive.
Note, interestingly, how the angle is independent from the mass.
| {
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Questions about Berry Phase I'm learning about the Berry Phase from the original paper, and from the TIFR Infosys Lectures The Quantum Hall Effect by David Tong (2016).
I have some questions regarding the original derivation of the phase, and an example.
*
*In the derivation, Berry derives the phase as
$$\gamma_n(C) = i \oint_C \langle n (\mathbf{R})|\nabla_{\mathbf{R}}n(\mathbf{R})\rangle \cdot d\mathbf{R} $$
He then proceeds to say that the normalization of $|n(\mathbf{R})\rangle$ implies that $\langle n (\mathbf{R})|\nabla_{\mathbf{R}}n(\mathbf{R})\rangle$ is imaginary. Why does this implication arise?
*In p. 33 of the notes, the example of a spin-1/2 particle in a magnetic field is used. Here the magnitude of the field remains constant but the direction can be changed. First, the Berry connection is calculated using spherical coordinates as
$$\mathcal{F}_{\theta \phi} = -\sin{\theta} $$
And when the coordinates are reverted to Cartesian, it becomes
$$\mathcal{F}_{ij} = -\epsilon_{ijk} \frac{B^k}{2|\mathbf{B}|^3} $$
This is called a magnetic monopole in parameter space. However when this quantity is integrated over the sphere of possible configurations (i.e. the sphere of radius $|\mathbf{B}|$ in parameter space), the following result is presented:
$$\int \mathcal{F}_{ij}dS^{ij} = 4\pi g$$
where $g$ is the 'charge' of the monopole. How does this result come about?
One idea I had for this one is that if we consider the Berry connection in Cartesian coordinates are considered in analogy to Coulomb's Law, then the field produced by the monopole should obey a kind of Gauss' Law. And if we take the quantities' analogues as follows:
$$\frac{1}{4\pi\epsilon} \rightarrow 1 \text{ and } Q \rightarrow -\frac{1}{2}$$
Then we get the result. However I am not happy/convinced with this argument. Can someone shed some light on this?
| For the result $\int F_{ij}dS^{ij}=4\pi g$, just use Gauss's law same as electric field.
For electric field, we have $E=\frac{1} {4\pi \epsilon_0} \frac{q}{r^2}$ and $\oint EdS=\frac{q}{\epsilon_0}=\frac{q}{4\pi \epsilon_0}*4\pi$, where $\frac{q}{4\pi \epsilon_0}$ is the term in E field except $r^2$.
So similarly, we have here $F=\frac{g}{r^2}$, then $\oint FdS=g*4\pi$
This works because $\int \frac{1}{r^2}dS=\int \frac{1}{r^2}r^2sin\theta d\theta d\phi=\int sin\theta d\theta d\phi=\int d\Omega=4\pi$ is the solid angle.
| {
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How does Fermat's principle of least time follow from Huygen's principle? When I was first introduced to Snell's Law, I was shown a derivation using Fermat's principle of least time. Using this same principle you can show that the angle of incidence must be equal to the angle of reflection. I was later again shown this, but instead of using Fermat's principle, all of this was derived using Huygen's principle. I think I've read somewhere that Fermat's principle can be mathematically derived from Huygen's principle. How can this be done?
| The only way of doing so that I can think of is demonstrate that Huygen's Principle and Fermat's Principle are both equivalent to the eikonal equation. This involves the calculus of variations. I haven't studied optics in a pretty long time, so maybe there's a simpler derivation that I am unaware of or have forgotten. Chapter 10 of these lecture notes provides a step by step derivation of the equivalence of Huygen's Principle, Fermat's Principle, and the eikonal equation.
| {
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Derivation of Total Momentum Operator for Klein-Gordon Field Quantization I am studying the second chapter of Peskin and Schroeder's QFT text. In equation 2.27 and 2.28, the book defines the field operators:
$$
\phi(x) = \int \frac{d^3p}{(2\pi)^3} \frac{1}{\sqrt{2 w_p}} (a_p + a^\dagger_{-p}) \, e^{ipx} \\
\pi(x) = \int \frac{d^3p}{(2\pi)^3} (-i) \sqrt{\frac{w_p}{2}} (a_p - a^\dagger_{-p}) \, e^{ipx}
$$
The momentum operator is then calculated in equation 2.33. However, my own derivation gives a different answer. I am reproducing my steps hoping that someone will be able to find where I went wrong. Starting with the definition of the momentum (conserved charge of spatial translations):
$$
\mathbf{P} = -\int d^3x \, \pi(x) \nabla \phi(x) \\
\mathbf{P} = -\int d^3x \, \Bigg[ \int \frac{d^3p}{(2\pi)^3} (-i) \sqrt{\frac{w_p}{2}} (a_p - a^\dagger_{-p}) \, e^{ipx} \Bigg]\Bigg[\int \frac{d^3p'}{(2\pi)^3} \frac{1}{\sqrt{2 w_{p'}}} (a_{p'} + a^\dagger_{-p'}) \, \nabla e^{ip'x} \Bigg] \\
\mathbf{P} = -\int \int \frac{d^3p}{(2\pi)^3} \frac{d^3p'}{(2\pi)^3} \int d^3x \, (-i^2) \, \mathbf{p}' \, e^{i(p+p')x} \, \Bigg[\sqrt{\frac{w_p}{2}} (a_p - a^\dagger_{-p}) \, \Bigg]\Bigg[\frac{1}{\sqrt{2 w_{p'}}} (a_{p'} + a^\dagger_{-p'}) \Bigg] \\
\mathbf{P} = -\int \int \frac{d^3p}{(2\pi)^3} \frac{d^3p'}{(2\pi)^3} \mathbf{p}' (2\pi)^3 \delta(p+p') \, \sqrt{\frac{w_p}{2}} (a_p - a^\dagger_{-p}) \, \frac{1}{\sqrt{2 w_{p'}}} (a_{p'} + a^\dagger_{-p'}) \\
\mathbf{P} = -\int \int \frac{d^3p}{(2\pi)^3} (-\mathbf{p}) \sqrt{\frac{w_p}{2}} (a_p - a^\dagger_{-p}) \, \frac{1}{\sqrt{2 w_{-p}}} (a_{-p} + a^\dagger_{p}) \\
$$
Since $w_{p} = w_{-p} = |p|^2 + m^2$, we get
$$
\mathbf{P} = -\int \frac{d^3p}{2(2\pi)^3} (-\mathbf{p}) \sqrt{\frac{w_p}{w_{p}}} (a_p - a^\dagger_{-p}) (a_{-p} + a^\dagger_{p}) \\
\mathbf{P} = \int \frac{d^3p}{(2\pi)^3} \frac{\mathbf{p}}{2} (a_p - a^\dagger_{-p}) (a_{-p} + a^\dagger_{p}) \\
\mathbf{P} = \int \frac{d^3p}{(2\pi)^3} \frac{\mathbf{p}}{2} \bigg[ a_p a_{-p} + a_p a^\dagger_{p} -a^\dagger_{-p} a_{-p} - a^\dagger_{-p} a^\dagger_{p} \bigg] \\
\mathbf{P} = \int \frac{d^3p}{(2\pi)^3} \frac{\mathbf{p}}{2} \bigg[ a_p a_{-p} - a^\dagger_{-p} a^\dagger_{p} \bigg] + \int \frac{d^3p}{(2\pi)^3} \frac{\mathbf{p}}{2} \bigg[ a_p a^\dagger_{p} -a^\dagger_{-p} a_{-p}\bigg] \\
$$
The first integral is odd with respect to p, and vanishes. For the second term, we can formally prove that $a^\dagger_{-p} a_{-p} = a^\dagger_{p} a_{p}$, but we can also argue that from noting that this operator pair creates a particle but then destroys it, with any possible constants only depending on the magnitude of \mathbf{p}. This line of reasoning gives us:
$$
\mathbf{P} = \int \frac{d^3p}{(2\pi)^3} \frac{\mathbf{p}}{2} \bigg( a_p a^\dagger_{p} -a^\dagger_{p} a_{p}\bigg) = \int \frac{d^3p}{(2\pi)^3} \frac{\mathbf{p}}{2} \, [ a_p,a^\dagger_{p}] \\
$$
The commutator here is proportional to the delta function, and hence this expression doesn't match what Peskin & Schroeder, and other QFT books have, i.e.,
$$
\mathbf{P} = \int \frac{d^3p}{(2\pi)^3} \mathbf{p} \, a^\dagger_{p} a_p
$$
UPDATE:
I realized later that my assumption that $a^\dagger_{-p} a_{-p} = a^\dagger_{p} a_{p}$ was wrong. When I was trying to prove this using the expansion of the ladder operators in terms of $\phi(x)$ and $\pi(x)$ I was making an algebra error.
| From
$$\mathbf{P} = \int \frac{d^3p}{(2\pi)^3} \frac{\mathbf{p}}{2} \bigg( a_p a^\dagger_{p} -a^\dagger_{-p} a_{-p}\bigg)$$
you actually have $$
\mathbf{P} = \int \frac{d^3p}{(2\pi)^3} \left(\frac{\mathbf{p}}{2}a_p a^\dagger_{p} - \frac{\mathbf{p}}{2}a^\dagger_{-p} a_{-p}\right) = \int \frac{d^3p}{(2\pi)^3} \left(\frac{\mathbf{p}}{2}a_p a^\dagger_{p} - \frac{-\mathbf{p}}{2}a^\dagger_{p} a_{p}\right)
= \int \frac{d^3p}{(2\pi)^3} \left(\frac{\mathbf{p}}{2}a^\dagger_{p} a_p + \frac{\mathbf{p}}{2}a^\dagger_{p} a_{p}\right) + \mbox{renormalization term.}\\
$$
Dropping the infinite renormalization term due to $\delta({\bf 0})$
$$
\mathbf{P}
= \int \frac{d^3p}{(2\pi)^3} \mathbf{p}\: a^\dagger_{p} a_p \:.
$$
| {
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What happens when a dielectric is inserted in a capacitor connected to a battery? The voltage across the capacitor has to stay the same since it is connected to a fixed voltage supply, which means that the potential before insertion and after insertion is equal. That would mean that the electric field within the capacitor is also equal before and after (since E = -dV/dR).
However, when a dielectric is inserted, it reduces the field since the molecules of the dielectric align themselves in such a way that the moment is opposite to the external electric field, which is also supported by:
K = Eexternal/Ereduced
where K is the dielectric constant.
So when it is connected to the battery, why does the electric field remain the same? What happens within the capacitor and dielectric?
| As you correctly observed, the electric field stays the same in the capacitor after insertion of the dielectric because the applied voltage is constant. This is accomplished by the increase in positive and negative areal charge on the plates of the capacitor which is provided by the battery. Before the insertion there is a vacuum between the plates $K=1$ and the areal charge density on the plates is $$Q=\epsilon_0 E=\epsilon_0 V/d$$ After the insertion of a dielectric with $K>1$ the areal charge density on the plates will be $$Q=\epsilon_0 K E=\epsilon_0 K V/d$$ where $V$ is the applied voltage and $d$ is the distance of the plates of the (parallel plate) capacitor. The electric field has to stay the same because the potential difference between the planes stays the same as well as the plate distance $d$. The polarization of the dielectric tries to reduce the electric field in the space between the plates but this is prevented by the increase of charge on the plates delivered by the battery to hold the potential difference $V$ constant.
| {
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Proof of superposition principle for electric potential Superposition principle of electric force is empirical. However superposition principle of electric potential doesn't seem to be following from it. Is there a formal proof of superposition principle for electric potential?
Edit
$V_1+V_2+...+V_n=\int\vec{F_1}.\vec{dr_1}+\int\vec{F_2}.\vec{dr_2}+....+\int\vec{F_n}.\vec{dr_n}$
From here on how shall I proceed? I can't factor out $.\vec{dr}$ as all paths may not necessarily be same.
| The superposition of the electric force is a consequence of the suporposition of electric fields. That is if we have two electric fields $\mathbf E_1$ and $\mathbf E_2$ then we add them to get:
$$ \mathbf E = \mathbf E_1 + \mathbf E_2 $$
This is equivalent to the superposition of the force because the force on a charge $q$ is:
$$\begin{align}
\mathbf F &= q\mathbf E \\
&= q(\mathbf E_1 + \mathbf E_2) \\
&= q\mathbf E_1 + q\mathbf E_2 \\
&= \mathbf F_1 + \mathbf F_2
\end{align}$$
The field and the force are vectors and they add just like all vectors do. Ultimately this is because the electromagnetic field is described by Maxwell's equations and these are linear. This isn't true of all fields. For example general relativity is a nonlinear theory and in GR gravitational fields do not simply add (though in everyday life the deviations from nonlinearity are negligible).
Anyhow, the potential difference between two points $A$ and $B$ is just the integral of the electric field along a line between those two points:
$$ V_{AB} = \int_A^B \mathbf E \cdot d\mathbf r $$
And since $\mathbf E = \mathbf E_1 + \mathbf E_2$ we can write this as:
$$\begin{align}
V_{AB} &= \int_A^B (\mathbf E_1 + \mathbf E_2) \cdot d\mathbf r \\
&= \int_A^B \mathbf E_1 \cdot d\mathbf r + \int_A^B \mathbf E_2 \cdot d\mathbf r \\
&= V_1 + V_2
\end{align}$$
So the superposition of the potentials follows immediately from the superposition of the field.
| {
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Length contraction speed of light The Special Theory of Relativity tells us that a moving object eg spaceship measures shorter in its direction of motion as its velocity increases.
At the speed of light it would have zero length, but infinite mass.
*
*would the spaceship disappear from the perspective of an observer outside the ship at the speed of light, having zero length? where would it go?
*at the speed of light, with zero length, what would a person inside the spaceship perceive?
*if the object has zero length how can it have infinite mass?
Can the above occur in reality or would it lead to paradoxical results?
| I asked a physicist this same question, and the question was answered in a straight forward fashion, which given my novice understanding of physics is what I had been seeking:
The most important thing to remember here is that you can’t actually get to the speed of light. More than merely hypothetical, this is impossible.
1) applying the equations directly, you get a spaceship with zero length but the usual height and width.
2) at every speed, things appear normal on board the spaceship. You never perceive yourself as moving.
3) it’s not just an issue with being very thin; nothing can have infinite mass.
| {
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Nuclear Fusion: Why is spherical magnetic confinement not used instead of tokamaks in nuclear fusion? In nuclear fusion, the goal is to create and sustain (usually with magnetic fields) a high-temperature and high-pressure environment enough to output more energy than put in.
Tokamaks (donut shape) have been the topology of choice for many years. However, it is very difficult to keep the plasma confined within the walls because of its high surface area (especially in the inner rings).
Why hasn't anyone used spherical magnetic confinement instead (to mimic a star's topology due to gravity)? - Apart from General Fusion
E.g. injecting Hydrogen into a magnetically confined spherical space and letting out the fused energy once a critical stage has been reached?
| Not an expert, but I believe the answer lies in the hairy ball theorem.
You see, for a magnetic field to turn charged particles back from a surface, the field must be parallel to the surface, which means that to have a fully confining geometry you must have a smooth, everywhere non-zero, and continuous vector field mapped onto a surface.
But the theorem say that you can't do that on a sphere (or topologically equivalent shape).
| {
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Strange pattern on car windows
A couple of days ago I was in a friend's car, and I noticed this pattern on the windows; I took a picture of the sun through the window to make it clearly visible.
The night before had been quite cold, but I don't think that the temperature went below $0$ °C, even though I am sure that it did some days before.
I can speculate that the phenomenon originated from some condensation/freezing of humidity on the outside of the car window, so I searched the web for pictures of water condensation and frost patterns (and also water staining) on car windows, but couldn't find anything similar.
What could be the origin of this intricate pattern?
| From your question, I can guess that the weather is rainy in your region.
When you drive a car in the rain, the water drops pass your windows at an angle. This, plus wind and other winter stuff causes the path of the drops to twist and jiggle like in this photo
I would also guess that the rain stopped while still driving, so the water could've evaporated in this pattern. The sunlight then makes those residues more pronounced when you took the picture.
Take a look at the following picture from a google search of 'water stains on glass'. To me it looks similar to your photo, just without the effect of moving window (keep in mind that the residues in the water may differ from one place to another due to pollution and etc., so the stains don't have to look the same).
| {
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Which is the physical meaning of the evanescent wave? I was reading that for an incident angle greater than the critical angle, there will be a total internal reflection.
The cosine of the refraction angle is therefore an imaginary number.
If we make a straight forward derivation we get an exponential attenuation factor for the refracted wave. And this wave (named Evanescent wave) only propagates along the boundary.
When we replace the cosine of the refraction angle on the Fresnel's formula, the amplitude of the reflected wave is the same of the incident wave, i.e the real part of Fresnel's reflection coefficient is equal to the unity.
More calculations gives that there is no flow energy across the boundary and I guess this is why the total amplitude is reflected backward, i.e without losing energy.
My questions are here:
What is the physical meaning of the evanescent wave?
If there are no flow energy why is there a propagation factor (along the boundary) for this wave?
If this wave is not propagating how we can understand the existence of this wave?
If there is no flow energy across the boundary but there is energy, why is the amplitude of reflected wave the same of the incident wave? is there no loss of energy?
thanks a lot.
| The evanescent wave is a form of light tunneling. It does propagate perpendicular to it. If a sufficiently thin sheet of air is bordered by glass of both sides, then the extinction at incidence beyond the limiting angle is not complete, and some light will come through on the other side.
It should be noted that an evanescent wave does propagate, along the interface.
| {
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Free electron gas in magnetic field I have a question regarding the calculation of "spin up" number of particles in a free electron gas, when placed in a uniform magnetic field $\textbf B$. In my lecture notes it's said that the spin up number of particles equals: $N_\uparrow=\int g(\epsilon-\textbf B\mu_m)f(\epsilon)\space d\epsilon $, where $g$ is the states density function $g=\frac{dN}{d\epsilon}$ and $f$ is the Fermi-Dirac distribution function $f(\epsilon)=\frac{1}{e^{(\epsilon-\mu)/k T}+1}$.
My question: If the density of states function changes: $g(\epsilon)\rightarrow g(\epsilon-\textbf B \mu_m)$, why doesn't the F-D function change as well, so that $f(\epsilon)\rightarrow f(\epsilon-\textbf B \mu_m)$ ?
This is connected to the calculation of the Pauli paramagnetism.
| In this picture of the Pauli paramagnetism, all energy states with "spin-up" electrons are shifted by an energy $$\Delta \epsilon= B \mu_m$$ Thus the density of states of the "spin-up" electrons as a function of energy $g(\epsilon)$ is shifted to higher energies by $\Delta \epsilon$ while the density of states of the "spin-down" electrons is shifted down by $-\Delta \epsilon$. The chemical potential $\mu$, which is the only system parameter in the FD distribution function of the electron, gas stays the same. Thus explaining the paramagnetism by a reduction in the total number of "spin-up" electrons and an increase in total number of the "spin-down" electrons in the electron gas in the magnetic field.
| {
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How does tension in a rope change when you cut it? Suppose we have a mass $m$ dangling from the ceiling on a vertical rope of length $\ell$ with uniform mass density $\lambda$ per unit length. The weight of the mass $mg$ is balanced exactly by the tension in the rope.
Now, suppose the rope is cut at the top of the ceiling. After a moment, the tension in the rope will be (pretty much) zero throughout and the mass will be in free fall. But, presumably the process is actually continuous, and over some period of time the tension in the rope will decrease from its initial value $T(y)$ (depending on the distance $y$ from the ceiling). How does $T(y)$ evolve over time?
| It would be very very hard to calculate this function.
Suppose that you're cutting the rope with a scissor and during that time, the rope will look like this:
Points $A$ and $B$ will have different mass densities per unit length as $B$ would be more stretched than $A$. This implies that each vertical section of the rope will have a different mass density and obviously, a different tension which would altogether make it more harder to calculate that function.
| {
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Lorentz Symmetry Group as continuous symmetry for limit of discrete spacetime There is a variety of models of quantum field theory, where discrete spacetime is used as technical support, or even suggested as physical reality. As far as I know, all of such models faced serious problems with restoring Lorentz Symmetry in continuous limit.
Is this general property of discrete models they have problems with restore continuous symmetries in general or it is certain property of Lorentz Symmetry Group and alike, in 4 dimensions?
| There is absolutely no issue with restoring continuous $O(4)$ symmetry using lattice regularization. The idea is well understood at this point. The lattice spacing is simply a cut-off, and one looks for regions in the phase diagram of the model where the correlation length is much larger than the lattice spacing, making the cut-off effects disappear. In a model with a continuous phase transition one can tune a physical quantity to be constant as one takes the correlation length larger and larger, thus one is able to make predictions about quantities in physical units in the continuum limit.
The lattice QCD and lattice gauge theory communities have been doing this for decades being some of the only theorist to progress alongside with experimental high energy physics with pre- and post-dictions of the standard model.
visit [hep-lat] to see recent papers in lattice field theory, but lattice literature runs for a long time.
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Path integral kernel dimensions and normalizing factor I am currently reading Quantum Mechanics and Path Integrals by Feynman and Hibbs. Working on problem 3.1 made me wonder why the 1D free particle kernel: $$ K_0(b,a) = \sqrt\frac{m}{2\pi i \hbar(t_a - t_b)} \exp \left(\frac{im(x_b - x_a)^2}{2\hbar (t_b - t_a)} \right)$$
has dimensions $1/\text{length}$ .
More generally this kernel has dimensions $1/(\text{length})^d$ , where $d$ is the number of dimensions of the system. Why is that?
On the other hand it is stated, that the kernel can be interpreted as a probability amplitude. In my understanding this would imply its dimensions to be $\sqrt{1/(\text{length})^d}$ in position space, because the kernels absolute square can be interpreted as a probability density.
Is there a physical interpretation for the normalizing factor $\sqrt{m/2\pi i \hbar(t_a - t_b)} $ other than "fixing" the equivalence to the Schrödinger equation?
To clarify my question, how can a probability amplitude have dimensions $1/(\text{length})^d$ ? Its absolute square would have dimensions $1/(\text{length})^{2d}$ which are not the proper dimensions for a probability density in my understanding.
| This is a consequence of $K(a,b)=\langle x_a,t_a|x_b,t_b\rangle$, which you quote, and the requirement $\displaystyle \int\!\text{d}x\,|x,t\rangle\langle x,t|=1$ (which generalizes to $\displaystyle \int\!\text{d}^dx\,|\vec x,t\rangle\langle\vec x,t|=1$ in $d$ spatial dimensions). In principle, you could label the states by some dimensionless "scale factor", but this is not the path the majority of physicists follows.
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How can one calculate work produced by a flow? I know differential work is $p dV$, since all standard Thermodynamics textbooks cover this. But, almost every case that they cover is about a gas (continuum-mechanics) that has low density, so its pressure does not vary with vertical coordinates. Since thermodyanimics deals with equilibrium states, all exercises are about constant-pressure gases or pressure in a $p = f(V)$ form. I'm talking about a pressure field $p = f(x, y, z)$, that could be obtained from a Navier-Stokes solution, for example. I thought it could be something like $W = \iiint\limits_{V_{2}} \, p dV - \iiint\limits_{V_{1}} \, p dV$
Being $V_{2}$ and $V_{1}$ the final volume frontier and the initial volume frontier, respectively. But I didn't find any references in standard books, and I don't know if $dW = p dV$ remains valid for non-equilibrium process.
| In the derivation of the energy balance equation in continuum mechanics, the rate of doing work at the boundary of the system is calculated by dotting the force per unit area (on the boundary) by the velocity vector (rate of displacement), and integrating over the area of the boundary. The force per unit area on the boundary is obtained using the Cauchy stress relationship, by dotting the stress tensor with a unit outwardly directed normal to the boundary. The (isotropic) pressure (as would be obtained from the solution to the Navier Stokes equation) is one contributor to the stress tensor, which also includes viscous stresses. The rate of doing work at the boundary of the system as a result of the pressure portion of the stress tensor would be the local pressure times the local normal component of velocity, integrated over the boundary. This is the extension of the simpler P(dV/dt) version of the work on the surroundings that we learned in thermodynamics, and reduces to that version if the volume of the gas is changing only as a result of a piston moving in a cylinder. Also, if the divergence theorem is applied to the pressure work done at the boundary of the control volume, we obtain the volume integral given in my comment above.
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Is nuclear power desireable in the long term, given the fact that it's an unnatural heat input to our planet? I've been reflecting on whether we want nuclear at all in the long term (compared to renewables like wind, solar, and hydro). There's a certain amount of heat (energy) entering our planet and leaving it. Greenhouse gases reduce the amount leaving, causing the planet to warm up. Nuclear power increases the input because it's energy that would not be released here without us. But the question to ask is what's the significance of the energy input from nuclear power. Say for example that future society becomes fully powered on fusion reactors, the energy input from these reactors would be roughly $10^{22}$ joules/year (approx 20 times 2013 world energy consumption). We can compare to the total energy input from the sun, which is $10^{25}$ joules/year. From these numbers, the input from nuclear would be about 0.1% the total solar input. Is that enough to cause disturbance to the energy balance of our planet and to worsen any global warming symptoms?
| Why would the nuclear energy of unstable elements not turn into heat without us chain reacting it and using it for electricity first?
The radioactive isotopes also give off heat spontaneously as they decay to more stable elements naturally.
It would be interesting to see an estimate of how much energy is generated by spontaneous radioactive decay by all elements in the whole earth as a system. I would be very surprised if our nuclear output would be anywhere near the total amount of energy of all decaying atoms in earths crust and cores, (but I don't have any source for that.)
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Interval Preserving in Minkowski Space The squared line element in any spacetime is given as $$ds^{2}=g_{ab}dx^{a}dx^{b}.$$ The use of tensors helps us to infer that the line element in some other frame would be $$ds'^{2}=g'_{ab}dx'^{a}dx'^{b}$$ where simply $dx'^{a}=\frac{\partial x'^{a}}{\partial x^{b}} dx'^{b}$.
My question is, in special relativity, there is further a condition on the line element that it should $$c^{2}(s-t)^{2}-(x_{1}-y_{1})^{2}-(x_{2}-y_{2})^{2}-(x_{3}-y_{3})^{2}=c^{2}(s'-t')^{2}-(x'_{1}-y'_{1})^{2}-(x'_{2}-y'_{2})^{2}-(x'_{3}-y'_{3})^{2}$$ which gives us the Lorentz transformations. How can we prove this condition using the postulates of special relativity?
Also where and how do we employ the condition that the frames we are transforming to are inertial?
| In fact THIS IS a postulate, the second postulate of Einsteins's special relativity, velocity of light must be invariant in any inertial frame of reference, so mathematically, for a light wave we have : $cdt^2-dx^2-dy^2-dz^2=0$ and $cdt'^2-dx'^2-dy'^2-dz'^2=0$ in another inertial frame, and so $ds^2=ds'^2$ ; this is the origin of your 'further condition'. General relativity agree with this just in local frames.
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Quantum thermodynamics: better than classical? I am wondering about the use of the so-called quantum thermodynamics and its theoretical advantage: can quantum machines outperform the classical analogues (supposing of course the absolute temperature is NOT zero)?
| To answer your question I must first clarify some ideas!
Engines transform some form of energy, such as thermal, chemical, mechanical, or electrical energy into useful work. Their efficiency, namely, the ratio of the output work to the input energy, is restricted to 1 at most by energy conservation. Engines converting mechanical energy into work may, in principle, approach unit efficiency. By contrast, the efficiency of heat-to-work conversion in a cyclic heat engine that operates between cold and hot thermal baths is independent of the specific design and limited by the universal Carnot bound. This bound follows from the second law of thermodynamics under the reversibility condition. Over the past two decades, extensive studies of thermal machines in the quantum domain have sought to reveal either fundamentally new aspects of thermodynamics or unique quantum advantages compared to their classical counterparts. The study has shown a remarkable similarity with macroscopic thermodynamical results, thus raising the issue of what is quantum in quantum thermodynamics.
What we have seen is that quantum thermodynamic machines do not rely on quantumness or they are truly quantum, exhibit "quantum advantage" but do not contradict the second law of thermodynamics. By "quantum advantage" I want to mean that some results are the thermodynamical equivalence of all engine types in the quantum regime of small action with respect to Planck’s constant. They have the same power, the same heat, and the same efficiency, and they even have the same relaxation rates and relaxation modes.
Yet, despite the great progress achieved, both theoretically and experimentally, the potential of quantum thermal machines for useful quantum technology applications only begins to unfold.
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A star with its radius not much larger than its Schwarzschild radius I was asked about the question today:
suppose that you are observing from afar a spherically symmetric star of mass $M$. Its radius $R$ is $\textbf{not}$ much larger than its Schwarzschild radius. Due to the gravitational bending of light, you can see not only the front side of the star but also the part of its back side? what's it like and how much of the back side can you see?
I don't have any idea about how can one see the back side of the star in this question. If I can see the back side, then there must some light emit from the back side to me. But this confuses me. Any suggestion?
| If space surrounding the observable universe wasn't filled with matter (which would be a universe that very fast resulted in a big crunch, but let's assume this for the sake of the argument), our visible universe would be a black hole with a radius much greater than the Schwarzschild radius. The light inside the radius would be destined to travel forever inside it, but light emerging from matter outside the radius could travel away from the universe (or from the star you mentioned in your question). Whatever the direction of the emerging light (between tangent and perpendicular) on the opposite side of where you're "standing", it can never reach you because it's bent not enough to make such a path to reach your eyes and make you see the backside of the universe (or star).
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Why does it take time to melt ice when the loss of magnetization of a material at its Curie temperature is immediate? Everything is in the title: Why does it take time to melt ice when the loss of magnetization of a material at its Curie temperature is immediate?
Have you an explanation for this difference?
| The melting of ice is a first order phase transition.
First order phase transition involve the release/absorption of a fixed amount of latent heat per unit volume. A finite amount of time is needed for the release/absorption of such heat, and thus for completing the transition.
The loss of magnetization of a ferromagnet at the Curie temperature is a second order phase transition, where there is no latent heat involved.
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Electric potential just outside a spherical shell We know that as we get closer and closer to a point charge, the electric potential approaches infinity.
Since electric potential at the surface of a spherical shell is finite (Gauss law) , so on moving away from the surface it would fall. In other words, it would be finite as well.
But if I think this way: Take an infinitesimal charge element on the surface so that we can take it as a point charge, then potential due to this charge would be similar due to a point charge. So as we approach this element, shouldn't potential go to infinity.
What's wrong with this reasoning? Please help.
| You may be somewhat confusing electrical field and electrical potential. But in both cases, the flaw in your reasoning is the same: the distance from the infinitesimal charge may be infinitisimal small but so is the charge (infinitesimal small). You cannot avoid a proper integral or using Gauss law.
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Doubled velocity problem - alternative calculation Working through Tipler/Mosca I got stucked at a very simple task.
A stone is thrown up vertically (1 dimensional problem) with an initial velocity v and it is reaching a height of h. What happens to the height if you double the initial velocity
The solution is : solve both an equation for v and 2v for h:
$h= \frac{v^2}{2g}$
$h(new)= \frac{(2v)^2}{2g}$
then divide h(new) by h. All variables will cancel out and you will receive a ratio between h(new) and h which is 4. So the answer is: Double initial velocity will cause 4times higher throw. I understand that but I have a question about my first attempt which failed badly:
Why is my attempt running into bad results:
$v=\sqrt{2g\Delta y} \,\,\,\,|*2$ I tried to double the velocity
$2v=2\sqrt{2g\Delta y} \>\>\>|(...)^2$
$4v^2=4*2g\Delta y =8g\Delta y$
$\frac{4v^2}{8g}=\Delta y$ And here I am ending up with useless results.
Can someone explain why my attempt was determinated to fail ?
To not violate homework law: I am trying to understand what was my error in my strategy to solve the problem. The problem is solved and I understand the technique.
| $\Delta y$ denotes maximum height in initial case when the initial velocity is $v$. Now, that you have fixed $\Delta y$ you cannot treat it as a variable for different velocities because $\Delta y$ is $h_{max}$ (which is a variable) for a specific case when initial velocity is $v$ . The new maximum height is not equal to $\Delta y$.
It doesn't matter how you manipulate the velocity-$\Delta y$ equation, you will always end up with the same result because manipulating this equation will bring the same change to both side of the equation.
$\Delta y=\frac{v^2}{2g}=\frac{4v^2}{8g}=H_{max,initial}$
If you want the correct result, you need a general equation with variables, viz., $v$ and $h_{max}$.
$$2gh=V^2$$
From the given data, $h_{max}=\Delta y$ and $V=v$
$\therefore 2g\Delta y=v^2$ $\tag 1$
But on doubling velocity, $h_{max}=H_{new}\neq\Delta y$ and $V=2v$
$\therefore 2gH_{new} ={(2v)}^2$ $\tag 2$
Now you will get your desired result.
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Crater experiment for Newton's second law of motion I am looking for interesting, easy to do experiments to Newton's second law of motion. This is for a high school, so nothing is known about momentum and the law is just $\mathbf{F}=m~\mathbf{a}$.
I found that some people do "crater" experiments dropping objects into flour (see for instance here), however I am still trying to understand how exactly this relates to Newton's law.
As far as I understand all objects dropped from the same height arrive with the same velocity at the flour. The flour provides a force on the object that decelerates it from its velocity down to zero velocity.
If I assume that the force is constant during impact (Does this make sense at all?), I get from Newton's law that the deceleration (absolute value) is smaller for the heavier objects, therefore they leave a deeper impact crater.
Does this sound like a good interpretation of the experiment?
They also mention measuring the width of the crater. How can this be used?
Are there any other (easy to do) experiments related to Newton's second law?
| Put an elastic string across a table and launch rolling marbles at it. Redo the experiment by weaving two strings, and then three strings. Notice that the stopping time gets shorter with more strings, and that more strings means a bigger force. A bit of fine-tuning is necessary to get a large enough difference of stopping distance (time).
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Finding friction coefficient of materials against the hand with modified tilt/pull test I want to find the friction coefficient of various fabrics against human skin. I don't want to cut the material to a smaller size.
Would the following tilt test work?
*
*Get a rigid, square board. Likely the board with be 18 x 18 inches.
*Attach the fabric to the board.
*Flip the board upside down so the fabric is now on the underside of the board.
*Balance the board, material side down, on my hand.
*Conduct tilt test by tilting hand slowly until board begins to slide and record the angle.
Does the fact that the board is much larger than my hand matter to the results of the tilt test?
Alternatively, would the following pull test work?
*
*Get a rigid, square board. Likely the board with be 18 x 18 inches.
*Attach the fabric to the board.
*Flip the board upside down so the fabric is now on the underside of the board.
*Balance the board, material side down, on my hand. Also ensure the board is level.
*Pull on the board with a spring scale and record the results.
Does the fact that the board is much larger than my hand matter to the results of the pull test?
| What I think about this is following. I request others to correct me if I am wrong-
The normal reaction by the hand would be also due to weight of the board. This will increase the frictional force and so you have to keep track of the component of weight of board which is acting on your hand in your first experiment.
One more thing is that you must ensure that the friction between the board and fabric does not affect your experiment. You can ensure this by attaching the fabric properly to the board so that the board does not slip over fabric.
I don't thing size of board will matter from the experiment point of view and the only thing that should matter about the board is its weight.
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How can we explain the linear relation between the temperature scales? In the proof of the relation F=(9/5)C+32 (degrees celsius to fahrenheit) we assume that there is a straight-line (linear) relation between the two temperature scales: F=mC+b. Then we need two "points" to find the equation for that line.
How can we explain to a student the very form (i.e. linear) of the relation among these temperature scales (or between kelvin and degrees celcius, rankine and degrees fahrenheit as well as kelvin and rankine)?
| Since all of these scales could be printed on the same mercury thermometer, and because each scale is made of equal-length units, a linear relationship between scales is the only one possible.
This linear relationship is still true even if there is no mercury in the thermometer.
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How to determine movement made by ultrasonic load speaker? What we have:
Supposed that we have a loudspeaker-based ultrasonic levitator. including a low voltage sine wave generator, an amplifier and a loudspeaker + a metal plate and some clamp-stands +some small polystyrene balls to levitate.
Main idea:
This levitator is designed to work at a frequency (or pitch) of around 25 kHz which is above the audible range of humans.
How can we determine how much will be the movement of small
polystyrene balls?
Details about other variables:
*
*The sine wave generator, generates a sine wave in the frequency range
20-30 kHz and an amplitude in the range 1-5 Volts (peak-to-peak).
*The amplifier: 100 Watts, able to amplify old fashioned analogue signals.
*The load speaker: high power tweeter or compression horn tweeter that works fine up to 30 kHz.
*The object: small polystyrene balls with the range 1-2 mm.
*Distance of plate and speaker: 50 mm.
This is very important that the exact movement of matters determined by mathematical formula.
| Thank you for your comments but we need an equation to find the exact amount of movement; I saw someones told about there will be no movement, I should tell as we experienced in laboratory actually it works and we noticed those variables we brought in the last section are influencing the distance between plate and polyester balls.
If any body else thinks it won't work, I upload the pictures of experience to make sure about it:
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L-band for search radars? I'm making my way through the Linesman-related section's of Gough's Watching the Skies. Linesman was developed to counter the carcinotron jammer.
The main solution was the Type 85, which had 12 frequencies switching each pulse at random.
However, they also deployed a second radar, the Type 84. This was a normal single-frequency pulse radar, but in the L-band rather than the S-band. This was to be used as an early-warning system.
I'm trying to understand why... for one, as a single-frequency system it would be highly vulnerable to jamming, and for another, the longer wavelength would mean lower resolution or much larger antennas.
Gough mentions a single possible reason for this, in passing, that L-band was less susceptible to clutter. I'm not sure why this would be.
Does anyone have any ideas on why the L-band would be better for early warning?
| S-Band is used for weather radar (signal for NWS, clutter for DOD)--that's in long range application. It's used for final approach in air traffic track surveillance. L-band is used for air-route surveillance (ARSR) and defense application (e.g. AN/FPS-117) in the long range (200 nautical miles).
Range resolution is determined by bandwidth, which is typical around 1 MHz, and that can easily be achieved in L or S band.
Regarding frequency hopping, not having it is a weakness (in military applications).
Aside: L-Band is currently allocated for Earth observations, surveillance, and geo-navigation (GPS), so if you want to do Earth science from space, you have to worry about RFI from the other applications.
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Generalised Coordinates in 3D Rotation If you have N particles on a surface of a rigid body and the rigid body is rotating about some axis, we say there are six generalised coordinates for the system (N particles on the surface) and set up the lagrangian.
The constraints we know are
*
*The distance between any two particles is invariant
*The angles between the line joining any particles is invariant as well
and that's it.
We also know a formula for finding the number of genralised coordinates
i.e difference between number of degrees of freedom (3N) and number of constraints (which is N(N-1)/2).
Clearly using this formula doesn't give 6 generalised coordinates.
Where's the mistake and how to count the number of generalised coordinates?
Note : Even I know the argument as to if you know 3 points on surface, you can determine the position of any other particle on the surface. But my question is about counting the generalised coordinates using the above formula.
| Your mistake is to assume that the number of constraints is $N(N-1)/2$. This number is only true for $N\leq 4$.
Let us label the particles by $1,2,\ldots,N$ and say that $\bar{ij}$ denotes the constraint between particles $i$ and $j$. For one particle there is no constraint. For two particles there is one constraint (see the diagrams bellow). If $N=3$ we have three constraints: $\bar{12},\bar{13},\bar{23}$. Add a fourth particle and we gain three others constraints: $\bar{14}$ and $\bar{24}$ fix the distance from particle $4$ to particles $1$ and $2$ and $\bar{34}$ which forbids the fourth particle to rotate about the axis joining $1$ and $2$. This fixes the distance from $4$ to $3$. This gives a total of six constraints. Finally consider adding one more particle. Now things get different. The new constraints $\bar{15},\bar{25}$ fix the distance to $1$ and $2$. The only freedom the sixth particle still have is to rotate about the line joining $1$ and $2$. Therefore just one more constraint, say $\bar{13}$, is enough to rigidly fix the sixth particle. The total number of constraints in this case is six.
As you can see in this construction, the first particle added zero constraints, the second one added one constraint, the third one added two constraints and the nth ($n\geq 4$) one added three constraints. The total number of constraints for $N\geq 5$ particles is therefore $1+2+3(N-3)=3(N-2)$. Hence there are $3N-3(N-2)=6$ degrees of freedom.
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IR Laser focusing question This is for a laser cutter.
10.4k IR laser outputs a beam at approximately 1/4" diameter, 400 watts. The beam is then focused using a single ZnSe lens with a 2.5" or 5" focal length. The final cutting kerf ends up being about a 0.009" wide gap. We are interested in a tighter focus.
We have been told that it can't be done by a couple of suppliers.
Is it possible to focus a 1/4" IR beam to a spot size tighter that 0.009"?
| The minimum spot size can be estimated by calculating the beam waist using Gaussian beam optics (wikipedia):
$$
\text{smallest spot size}= 2\omega_0=\frac{2\lambda}{\pi NA}
$$
Where $\omega_0$ is defined as the beam radius at the focus point, $\lambda = 10.4$ µm is the wavelength, and NA is the numerical aperture of your optical system. We can calculate NA as follows:
$$
NA=\frac{D}{2f}
$$
Where $D$ = 0.25" is the beam diameter, and the focus length of the lens $f$ = 2.5" or 5". So $NA$ = 0.05 or 0.025 respectively.
This means the smallest spot size $2w_0$ = 132 µm = 0.0052" for the 2.5" lens and 265 µm = 0.0104" for the 5" lens. I would say your supplier is ballpark right, the 0.009" you achieve is close to the physical limit. A few remarks:
*
*you should use the 2.5" lens, this should give you a smaller focus (but typically smaller focal lengths give higher aberrations, which brings us to the next point)
*the exact laser beam profile is quite important. Is your beam gaussian?
Aberrations in the beam distort the shape of the focus point and can make the spot size bigger.
*you did not mention your definition of your beam diameter, this is a science on itself: wikipedia. If you want the last drop out of your optical system, these things get important.
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Probability of finding a particle with a particular value of momentum Given a particle's wave function, what is the general method of finding the probability distribution of momentum (i.e., the probability of finding that particle with a particular value of momentum)?
For example, given
$$
\psi(x) = a e ^ { ikx } + b e ^ { -ikx },
$$
what is the probability of measuring $ p_x = \hbar k $ ?
| First of all, I assume that you know that the probability (density) distribution is given by the squared amplitude of the wavefunction. In your example, you are given the wavefunction in the position basis, so it gives you the position probability density. If you want the momentum probability density, you have to change basis to get $\psi(p)$.
The standard way to change basis is to use a "resolution of the identity". The manipulation is easiest in bra-ket notation. You start out with
\begin{equation}
\psi(x) = \langle x | \psi \rangle = a\, e^{i k x} + b\, e^{-i k x}.
\end{equation}
But you want $\psi(p) = \langle p | \psi \rangle$. You can find it with this manipulation:
\begin{align}
\langle p | \psi \rangle
&= \int_{-\infty}^{\infty} \langle p | x \rangle \langle x| \psi \rangle\, d x \\
&= \int_{-\infty}^{\infty} \frac{1} {\sqrt{2\pi \hbar}} e^{-i p x / \hbar} \langle x| \psi \rangle\, d x \\
&= \int_{-\infty}^{\infty} \frac{1} {\sqrt{2\pi \hbar}} e^{-i p x / \hbar} \left( a\, e^{i k x} + b\, e^{-i k x} \right)\, d x.
\end{align}
Here, the resolution of the identity I used was $\int_{-\infty}^\infty |x\rangle \langle x |\, dx$. For a one-dimensional system, that's just the identity operator, so you should be able to just plonk it into any old expression you want.
I'll leave it as a homework exercise to do the integral and take the squared magnitude of the result. But here are a couple hints. First, I happen to know that your solution $\psi(x)$ is a sum of two momentum eigenstates, so you should almost always get zero probability of measuring any given momentum — except for those two particular momentum eigenvalues. That is, your answer will be zero everywhere except for two values related to $k$ and $-k$. Second, you might want to read up on the Dirac $\delta$ function — specifically how it can be expressed as an integral.
| {
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Work done in moving a charge, from near a conducting plane to infinity A point charge $q$ is located at a distance $l$ from the infinite conducting plane. What amount of work has to be performed in order to slowly remove this charge very far from the plane? (Irodov 3.55)
I understand there is a method of images for solving such problems. I did use it, but I have three reasonable ways to use it, yet only one is giving correct answer.
Method 1: $W_{\text{conservative force}}=-\Delta U_{\text{of that force field}}$
So, since our charge will be at rest finally at infinite distance ("slowly remove"), so by Work-Energy theorem, $W_{\text{electrostatic force}}+W_{\text{external agent}}=0$, hence, $W_{\text{external agent}}=\Delta U_{\text{electrostatic}}=0-\frac{-q^2}{4\pi\epsilon_0\cdot(2l)}=\frac{q^2}{8\pi\epsilon_0l}$
But this is the incorrect answer.
Method 2: $W_{\text{external agent}}= \int\text{Force}\cdot{\text{Displacement}}=\int^\infty_{2l}\frac{q^2}{4\pi\epsilon_0\cdot(x^2)}dx=\frac{q^2}{4\pi\epsilon_0}(\frac 1{2l})=\frac{q^2}{8\pi\epsilon_0l}$
Same as the answer in method 1, and still wrong.
Method 3: $W_{\text{external agent}}= \int\text{Force}\cdot{\text{Displacement}}=\int^\infty_{l}\frac{q^2}{4\pi\epsilon_0\cdot((2x)^2)}dx=\frac{q^2}{16\pi\epsilon_0}(\frac 1{l})=\frac{q^2}{16\pi\epsilon_0l}$
In this attempt, I only changed the variable of integration, and it gave the correct answer.
My question:
Method 1 and 2 are completely logical according to me and they should give a correct answer. Yet, only 3 gives the correct answer. So, what is the logical mistake in method 1 and 2?
| Method of images is being used only to solve the problem, there is not any real point charge behind the plane. So in order to find the work done in dismantling the system if we follow the usual process by which we dismantle the two point charges placed at a distance, if we move one charge with respect to origin while keeping mirror image charge fixed then we have altered the original problem as the infinite plane does not remain conducting plane then. So we have to dismantle in a way (imaginary) such that by moving charge $q$ by $dx$ amount the distance between two charges increase by $2\ dx$. Work done /PE is half of that in case of two real point charges.
| {
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In concerts, why is the audience never heard as off-tune? In concerts, when the audience sing, it sounds almost okay; not out of tune as if they sing all harmonious. However, if a small group of people sing, we can notice the off-tune singing.
Is it simply related to the average value of the sound waves or does it have a special name? I hope I made myself clear with the question.
| Its the relative number of people singing the same false note. In a small group of, e.g. 4 people, one is a large fraction of the total number. For the same effect in a crowd, you would need 25% singing false at the same time in the same way to have a similar effect.
| {
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Direct product of spin representations Consider a system of two 1/2-spins. Under some conditions the Hilbert space can be decomposed into the direct sum of spin-0 and spin-1 representations:
$\frac12\otimes\frac12=0\oplus1$.
When I write this formula on the board, I immediately get an objection that $1/4$ is not equal to 1 ! My question is as follows, how to explain this equation to the audience of physicists. Preferably in one or two sentences, concise, mathematically correct, but without going into much mathematical details.
| Each spin-1/2 particle is associated with a $(2\times\frac{1}{2}+1)$=2-dimensional vector space $\mathbb{V}$ as far as its spin degree of freedom is concerned. A composite system of two spin-1/2 particles is associated with a 4-dimensional vector space which is a direct product $\mathbb{V}_1\otimes \mathbb{V}_2$ of two 2-dimensional vector spaces $\mathbb{V}_1$ and $\mathbb{V}_2$. Under a similarity transformation a $4\times 4$ matrix representing an element of $SU(2)$ that acts on the space $\mathbb{V}_1\otimes \mathbb{V}_2$, can be reduced to a block-diagonal form consisting of block matrices of dimensions $3\times 3$ and $1\times 1$ acting on invariant subspaces of dimensions 3 and 1 respectively.
In technical terms, it means that the 4-dimensional representation is reducible into a 3-dimensional and 1-dimensional irreducible representations, and symbolically written as $2\otimes 2=3\oplus 1$ which respectively corresponds to three triplet states of spin-1 and one singlet state of spin-0 of the composite system.
| {
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Milky Way: where does it end? The milky way galaxy, as is typical for spirals (and others?), has a relatively flat rotation curve except very near the center:
Eventually, at great enough distances, almost the entire mass of the galaxy will be enclosed and the rotation curve must become nearly Keplerian and fall off (dark matter, if it is important at all at these scales, would make the curve fall off even faster). Do we have any data that shows the curve becoming nearly Keplerian at great distances? Do we even have an idea of what that distance would be?
| Over at astronomy they have a post looking at this from the astronomical perspective. There is no clear boundary and it comes down to arbitrary definitions or decisions.
The rotation curve has been measured out to almost 100kpc. See this article for a nice and current review. From that paper comes this figure:
You can see that we have measurements beyond 100kpc, with the curve clearly decreasing once you go beyond 20 or 30kpc. (Don't think much of the two data points going back up at ~200kpc, that is well within the uncertainty and thus not significant)
For comparison, note that the Andromeda Galaxy is over 700kpc away. Since the two galaxies have similar mass, you could also argue that the Milky Way ends half way between us and there.
| {
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In what sense are quasiholes and quasiparticles "excitations" in Fractional Quantum Hall (FQH) systems? In the theory of Fractional Quantum Hall states, one often sees quasi-holes and quasi-electrons (or quasi-particles) being called "excitations" from the ground state initially given by Laughlin (Jastrow-Laughlin style wavefunctions). Usually when one speaks of excitations, it is to a higher energy level. However from the wavefunctions of the quasi-holes and quasi-particles, it seems to me that we are still in the Lowest Landau Level. My question is: in what sense are these "excitations"?
| I will first repeat some semiconductor physics and then move to the fractional quantum hall effect.
In an undoped semiconductor there are many real electrons and real protons that interact. For $T=0$ they condense into a stable neutral state. Let´s call this state the vacuum. We can create create quasi-electron and quasi-holes as fermionic excitations from this vacuum state. These quasi-electrons and quasi-holes are the non-interacting electrons and holes that provide charge transport in a semiconductor.
For the fractional quantum hall effect, there is not just one vacuum state but there are many. Each Laughlin-wavefunction-like state could be considered as the vacuum. So you can switch from one vacuum to another by changing the magnetic field. And in each of these vacua you can create excitations in the form of quasi-particles and quasi-holes. The properties of these quasiparticles will be different for each vacuum. These quasiparticles are typically anyons (somewhere between a fermion and a boson) and have fractional charge.
There are many good introductory texts about the fractional quantum hall effect. For instance this one by Girvin.
| {
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Balancing an orange on a spoon experiment I did a little experiment with balancing an orange on a spoon.
And i realized i don't really understand why i feel a stronger force under both fingers, the further my forefinger is from the orange
(As shown in the first photos with the red arrows)
And why i feel a weaker force under both fingers the closer my forefinger is to the orange.
(As shown in the second photo with the cyan arrows)
| Three forces are acting in this scenario. Two are downwards: the gravitational force on the orange and the force applied by your thumb. The third is upwards: the force applied by your forefinger. Since the scenario is static, the sum of all three forces is zero (counting for example the upward direction as positive and downward as negative).
$$
F_{orange} + F_{forefinger} + F_{thumb} = 0
$$
or equivalently:
$$
F_{thumb} = - (F_{orange} + F_{forefinger})
$$
The torque also needs to be zero, otherwise the spoon will start to rotate even when the sum of the forces is zero. Torque is defined as the distance to some chosen reference point times the force perpendicular to the line that connects to that reference point. The reference point can be chosen freely, so let's pick the orange as the reference point. The torque relative to this point is then:
$$
D_{forefinger}F_{forefinger} + D_{thumb}F_{thumb} = 0
$$
where $D_{forefinger}$ is the forefinger's distance to the orange, and $D_{thumb}$ is the thumb's distance to the orange. Since the orange has zero distance to itself, the $F_{orange}$ term has been omitted.
Solve this equation for $F_{thumb}$:
$$
F_{thumb} = - \frac{D_{forefinger}}{D_{thumb}} F_{forefinger}
$$
Now we can eliminate $F_{thumb}$ and solve for $F_{forefinger}$:
$$
F_{forefinger} = - \frac{D_{thumb}}{D_{thumb}-D_{forefinger}} F_{orange}
$$
Here we can see that the magnitude of $F_{forefinger}$ is equal to $F_{orange}$ when $D_{forefinger}$ is exactly zero, i.e. when the forefinger is placed directly under the orange. It gets larger the closer the forefinger is moved towards the thumb, since that makes the factor $\dfrac{D_{thumb}}{D_{thumb}-D_{forefinger}}$ grow. The magnitude of $F_{thumb}$ also grows since the sum of all three forces must be zero, and $F_{orange}$ is not changing.
| {
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Why in an ideal gas as the volume increases, both the pressure and the temperature drop? I was reading Fundamentals of Physics(Halliday, 10th ed.), chapter 19, when I saw a statement "Figure 19-15a shows our usual insulated cylinder, now containing an ideal gas
and resting on an insulating stand. By removing mass from the piston, we can
allow the gas to expand adiabatically. As the volume increases, both the pressure
and the temperature drop."
But I don't understand why temperature can't stay constant as volume increases and pressure decreases? Why both the pressure and the temperature decrease?
| The Combined Gas Law which is a combination of three other gas laws, states that
"The ratio between the pressure-volume product and the temperature of a system remains constant." (Wikipedia).
It is expressed mathematically as ${PV\over T} = k, \qquad\tag 1$ where $P$, $V$, $T$ and $k$ are the pressure, volume, temperature and a constant (with units of energy divided by temperature) for the particular body of the gas.
For the same body of gas that changes between state 1 and 2:
${P_1V_1\over T_1} = {P_2V_2\over T_2}\tag2$
In your question, the process is adiabatic (no energy or mass transfer), so as volume increases to $V_2$, pressure will drop to $P_2$ in inverse proportion (amount of gas is constant) according to a sub-law of the Combined Gas Law called Boyle's Law, and the temperature drops to $T_2$, which we can be calculated using another sub law (Gay-Lussac's Law). This is an example of adiabatic cooling.
| {
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Integral form for partition sum After reading the derivation of the discrete partition sum, it seems like my book finds the generalization to the continuous case trivial. Just change the summation to an integral and multiply with an differential.
However, I don't understand the validity of this, and I'm wondering where the differential came from. I would understand it if it was a Riemann-sum and we let the step-size go to zero, but I see no step in the expression. Hope someone could help me understand, even though this may be more of a mathematical question than a physics one.
Thanks in advance!
Edit:
This is classical statistical mechanics.
The partition sum is defined as follows:
$$Z = \sum_{i=0} \exp(-\beta E_i)$$
A few pages later, the partition sum for an ideal gas is calculated, and the expression is given as
$$Z = \frac{1}{h^3 N!} \int{ \exp(-\beta E)}dx_1...dp_N$$
Which clearly is much more convenient for calculations in the continuous case. There is still no explanation on how the discrete sum now is an integral, and this is what confuses me.
Edit: Thanks for the suggestion to read the other question. However, it seems to me that the question addresses the limit quantum $\rightarrow$ classical, and that my question addresses the limit discrete $\rightarrow$ continuous. There are indeed some similarities, but without knowledge of quantum statistical mechanics I did not manage to find the answer I wanted.
| Let's take one classical particle confined in a volume $V$. The number of states per energy interval $ \Delta E$ will be: $\Delta i=(4 \pi p^2\Delta p V)/h^3 = 4 \pi (2 m^3 E)^{1/2} \Delta E V)/h^3$ and we see that $ \Delta E_i=\Delta E /\Delta i $ tends to zero when $V$ tends to infinity. For one particle partition function we have: $Z_1=\sum (\ e^ {-\beta E_i} \Delta E_i/\Delta E_i)=[[\sum (4 \pi /h^3)(2 m^3 E_i)^{1/2} \ e^ {-\beta E_i} \Delta E_i]]V$ , the expression in pair brackets is a Riemann sum which can be replaced by the integral when $V$ tends to infinity. Of course, an example of a single particle is taken for simplicity
| {
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Increased angle should lead to decrease height I'm having trouble understanding the answer for the following problem:
*
*The diagram below represents a 10-kilogram object at rest at point A. The object accelerates uniformly from point A to point B in 4 seconds, attaining a maximum speed of 10 meters per second at point B. The object then moves up the incline. The object comes to rest at a vertical height of S (point D) when $\theta$ = $ 30 ^\circ$. If $\theta$ were increased to $ 40 ^\circ$, the object would come to rest at a vertical height at: Less than S, Greater than S, or the Same?*
Given the restraints in the problem, I expected the answer to be less than S, because all conditions, including the velocity, and acceleration of the box stays the same in both problems. However, an increased angle should intuitively lead to a decrease in the vertical position where the box comes to a rest. However, the answer key states otherwise: the height remains the same.
| If one ignores friction (say the surface is very smooth) then this is a classic problem on conservation of energy.
The total energy at the start of the motion is entirely kinetic:
$$
E=\frac{1}{2}mv^2 \tag{1}
$$
whereas, when the box reaches it maximal height, the velocity will be $0$ (otherwise it would continue going up) and at that precise moment all the energy is potential:
$$
E=mgh \tag{2}
$$
By conservation of energy, (1) and (2) are equal so
$$
\frac{1}{2}mv^2=mgh \qquad \Rightarrow \qquad h= \sqrt{\frac{v^2}{2g}}\, .
$$
Besides not depending on how you reach the maximal height $h$, this maximal height is also independent of the mass of the box, i.e. a $5$kg box would reach the same height as a $10$kg box if one ignores friction.
Of course in a more realistic case there would be some friction but, in the case of a roller coaster for instance, the friction is sufficiently small so that the cart always reaches maximum height irrespective of the mass of the cart and the people in it, so neglecting friction is not de facto unrealistic, at least in some situations.
| {
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Rho meson decay to three neutral pions In my assignment, I am asked to show whether $\rho^0\to3\pi^0$ is allowed. From the particle data group, I cannot find the decay mode; hence, I am guessing it is not allowed. Based on the knowledge that $\rho^0\to2\pi^0$ is not allowed, I focus on parity. However, the action contains an angular momentum of 3 bodies. Then I get stuck. Can someone tell me whether this action is allowed? Or at least give me some hint. (I have not learned weak interaction and strong interaction yet.)
| I confess I might be overthinking this, having missed something more obvious (C? -- violated by the weak interactions).
The G-parity of the ρ is +, so it prefers to decay to 2π s, since the π has G parity -. (It requires an antisymmetric state because of angular momentum, as you already see.)
But G-parity relies on isospin, which is broken somewhat in the strong decay, so there is also the 3π mode at the $10^{-4}$ level, where all 3π s are different, instead of your 3π0 which has to be fully symmetric.
You cannot first combine the two π0s to a vector by an L=1 as you notice, by their unavoidable symmetry. So you have to combine them to a symmetric S-wave spinless neutral dipion, and then combine that antisymmetrically with an L=1 to the remaining neutral π0 to net a vector ρ0.
However, this is impossible; ignoring normalizations,
$$
(12+21)3-3(12+21)= 123+213-312-321= (123-321)+(-312+213),
$$
but the terms in the first and second final parentheses are 1-3 antisymmetric, and 3-2 antisymmetric respectively, which is impossible for identical neutral pions.
This argument fails, of course, (phew!), for the extant decay mode π0π+π- which is allowed, although ferociously suppressed.
| {
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Apparent decrease of the force between two electrons Suppose two electrons are moving along parallel paths at constant velocity. Then wouldn't the repulsion of the two electrons be slightly reduced because of the magnetic fields?
If so, how is the apparent decrease in force between the two particles explained in the frame in which both electrons are at rest?
| There is no decrease of the electrostatic repulsion force (and magnetic field) between the two electrons in the co-moving reference frame where the electrons are at rest. The magnetic field appears only for a reference frame that is not moving with the electrons. This can be understood in the framework of Special Relativity which gives a simple transformation of electric and magnetic fields between inertial reference frames. See e.g., D.J. Griffiths, Introduction to Electrodynamics, Prentice-Hall 1999.
| {
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Is there evidence that a = dv/dt and a = F/m are always equivalent? If the rate of change in velocity in a particle (of mass m) caused due to a force F is dv/dt, then
F = m dv/dt
It may be argued that this is how we define force. But my question is:
Can there be any kind of force, which is so strange that no matter how we write the formula for the force, we will find that F = m dv/dt fails in at least some cases?
| The formula you mentioned is for a special case where the mass doesn't change. The general case is, $$\vec{F}=\frac{d \vec{p}}{dt}=\vec{v}\frac{dm}{dt} + m\frac{d \vec{v}}{dt}$$ where $\vec{p}$ is the momentum of the object.
For example, a missile loses mass by burning fuel, so your formula will not suffice to describe its motion.
| {
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Why is the constraint when maximizing entropy on energy rather than momentum? State in classical mechanics is specified by the position and the momentum. Yet, in statistical mechanics, the usual approach is to maximize the entropy of a system subject to an energy constraint. This gives a probability distribution of the energies, and then other quantities such as velocities can be calculated as a function of energy.
Since momentum is conserved, why not maximize the entropy subject to a momentum constraint rather than energy constraint?
| In general, every conservation law is taken into account when maximizing entropy.
For example, consider an ideal gas in a box that's flying at nearly the speed of light, where the gas has some low temperature $T$ in the frame of the box. In our original frame, the maximum entropy configuration consistent with energy conservation has the ideal gas particles moving at speed about $c$ with respect to each other, and hence at some enormous temperature. But this isn't what happens because it violates momentum conservation; instead the gas just remains at temperature $T$.
It's typically easy to ignore linear and angular momentum because we can just transform into a frame where they're zero. In the context of early-universe cosmology, there are other conservation laws, such as the conservation of charge or baryon number, that must be taken into account.
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Software for simulating gas of hard particles (i.e. polyhedra?) I was wondering if there existed a kind of software I could use to simulate a gas of polyhedra, such as tetrahedra. They would interact through entropic interactions only, i.e. excluded volume.
I'm not an expert in simulation so any kind of suggestion is welcome. This is just for recreational purposes.
| HOOMD-blue is another excellent choice. It can leverage high-performance parallel Monte Carlo simulations to look at the collective behaviour of particles with very complex shapes. Here is the documentation for the Monte Carlo package, and here is the paper that describes its usage and implementation.
Note that HOOMD-blue exposes python bindings, meaning that it can be run from a python script, which makes it very easy to manipulate the particles' degrees of freedom for the subsequent analysis.
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Raychaudhuri equation v/s Einstein Field equations Raychaudhuri equation describes the evolution of a geodesic congruence in a spacetime.
While Einstein field equation tells about the curvature of spacetime. This helps in determining the motion of particles on various geodesics.
How are these two description different? Are they telling the same physics in two different ways? If we have Einstein field equations then why do we need Raychaudhuri equation?
|
While Einstein field equation tells about the curvature of spacetime. This helps in determining the motion of particles on various geodesics.
I don't think this is a good characterization of the Einstein field equation. The EFE relates curvature to stress-energy. "Matter tells spacetime how to curve, and spacetime tells matter how to move." The EFE doesn't just define a measure of curvature, it allows us to predict that curvature when we have matter with known properties.
The Raychaudhuri equation is a theorem that can be proved from the EFE, so in that sense we could possibly get along without it. It doesn't imply the EFE, because it's a purely geometrical equation and has no information about matter fields. However, it tells us nontrivial things that we wouldn't have known just by staring at the EFE. For example, it is used in proving the singularity theorems, which were contrary to many relativists' intuitions.
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Issue with the work of a reversible process
You consider a cylinder with a piston inside which can move without friction, and with diathermic walls. The piston divides the cylinder into two rooms: A-B, filled with two ideal gases. There is also a valve that connects B and the outside. Let's consider just the room A as our thermodynimic system.
In the figure above are drawn two consecutive equilibria states of a generic reversible process, each described with a vector of thermodynamics variables. We assume that during the disequilibrium state, the piston moves of $dx$. You can observe that the only way to make the piston move is to use the valve; if the valve stays closed you can only have an isochoric process.
However, now we are intrested in calculating the work that our system (room A) has exchanged during the transition 1->2. In the equilibria states the net force acting on the piston is $0N$ because the pressure in the two rooms is the same, so the piston is firm. Using the kinetic energy theorem between the two equilibra states you obtain:
$$dk=dW=0 $$
While I now that the right expression should be:
$$dW=p dv$$
But I don't understand where it comes from, since it is in contrast with my treatment. And also, which pressure should I consider in this formula? Someone could show me exactly which point of my treatment is wrong?
| In general, the 1st law would say that $dU=\delta Q +\delta L$ but you have postulated a purely mechanical energy exchange, that is one with $dU=\delta L$. This is called an adiabatic process.
If you further assume that the gas is internally frictionless, then the partition will oscillate back and forth just as an ideal spring would do.
If the gas has internal friction, then the oscillation of the partition will be dissipated and the partition will stop at a point where the pressures will be equal, i.e, $\delta p = 0$. Read the subject called "the adiabatic piston".
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Given a mass and position as a function of time, find work
A single force acts on a $7.0 kg$ particle-like object in such a way that the position of the object as a function of time is given by $x = 3.0t − 4.0t^2 + 1.0t^3$, with $x$ in meters and $t$ in seconds. Find the work done on the object by the force from $t = 0s$ to $t = 7.0 s$.
My thinking:
$dW = Fdr$ where $F = force$, $dr =$ change in position
$F = ma$ where $m =$ mass, $a =$ accelleration
$a =$ second derivative of position with respect to time $= 6t-8$
$dW=m(6t-8)(3.0t − 4.0t^2 + 1.0t^3)dt$
integrate from $0$ to $7$ seconds and I'm off by a couple orders of magnitude. Where did I go wrong?
| The work done is equal to the change in Kinetic energy.
$v(0)=3m/s$
$v(7)=94m/s$
That gives: $\text{Work} = \text{KE}_7 - \text{KE}_0 =31kJ$
| {
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Is there an analogue of the LSZ reduction formula in quantum mechanics? In quantum field theory the LSZ reduction formula gives us a method of calculating S-matrix elements. In order to understand better scattering in QFT, I will study scattering in non-relativistic quantum mechanics and that question ocurred to me.
| In QFT, LSZ formula is a tool to obtain S-matrix from correlation function. In QFT correlation function is very easy to calculate(free and interacting theories). When outgoing particles became on-shell then we can relate S-matrix element to correlation function.
I QM all the particles are always on-shell. In non-relativistic limit scattering matrix(amplitude) should be compared to Born-approximation. So if we Fourier transforms the $i \cal{M}$ (non-relativistic limit) back into position space, then we can see the behavior of potential.
In QM, if we can calculate correlation function, then we can easily obtain the S-matrix element. Calculation of correlation function is not easy in QM. But on the other hand calculation of S-matrix element is very easy in Born approximation.
But what we can do is to write an action of which Schrodinger equation is just the equation of motion of that action.
$$S = \int_{xt} \psi^\dagger \left(i{\partial \over \partial t} + {\nabla^2\over 2m}\right)\psi - \psi^\dagger(x) \psi(x) V(x).$$
After that we can do the usual trick using generating functional. After taking the functional derivative with respect to current($J$), we can find correlation function.
$${\displaystyle Z[J]=\int {\mathcal {D}}\phi e^{i(S[\phi ]+\int d^{d}xJ(x)\phi (x))}~,}$$
Ofcourse, propagator are different in QM as compared to QFT. In this way we can find S-matrix element using correlation function.
So at last, In quantum mechanics, LSZ formula is not much use. But we can take the usual LSZ formula and go into the non-relativistic limit.
| {
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Cosmic ray shielding for electronics on Earth There are plenty of topics discussing cosmic ray shielding, but they all put emphasis on spacecraft shielding, where things like radiation and mass are a big problem.
So I decided to ask a question from a purely "particle physics" perspective.
Evidently, cosmic rays and their byproducts are the predominant cause of soft errors, or single bit errors, meaning that they are well able to penetrate the atmosphere, building walls and fairly thick metal equipment racks, with enough energy to flip a bit.
But on the other hand, the prevalence rapidly increases with altitude, meaning that the atmosphere does a decent job of absorbing most of those or significantly diminish their potential. Which leads me to suspect that there is a significant dynamic range to those particles, if the atmosphere is capable of absorbing most but those that manage to get through are still energetic enough to penetrate dense materials and cause damage.
Which brings up the question, what kind of materials provide the best kind of shielding of electronic equipment from that radiation? And how much of it. Considering that outside the context of spacecraft mass is not that much of an issue.
I've read that dense materials are good at scattering them into less energetic particles and hydrogen rich materials are good at absorbing them. I happen to have 4mm thick lead sheets and 20 mm thick PVC panels, so I am wonder whether a lead/pvc/lead/pvc sandwich will be effective at it, or will it take something more heavy duty?
| The cosmic rays that reach us are muons. They are very penetrating: first the atmosphere (equivalent to 10 meters of water). At sea level about 1 muon per cm$^2$ per minute. Still with a highly relativistic kinetic energy of a few GeV.
Then the concrete of the building. They penetrate tens of meters into the ground. So your lead sheet won't make much of a difference.
| {
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Why do charged particles deflect one way but not the other in a magnetic field? I am well aware that a charged particle moving in a magnetic field will experience a force perpendicular to that magnetic field. But why is it that positive and negative particles experience a force in opposite directions?
What exactly determines the direction that a given charge will experience a force? I.e. why does a negative particle experience a force in one direction and not the other?
| The problem with the why in your question is that it will give rise to another why in the explanation to explain why that explanation occurs and so on; giving rise to an infinite number of whys. But to kick things off, I'll try and give an initial explanation:
Charges deflect in a magnetic field dependent upon their charge sign because of:
*
*The principle of Special Relativity
*The electric force
deflects charges dependent upon their sign
We can move to a laboratory moving at the same instantaneous velocity as the charge where it now appears stationary. Any magnetic field here cannot affect the charge because it's not moving; leaving only an electric field, if any, that can affect it. This electric field will deflect that the charge in one of two ways depending upon its sign, which will also be seen in one of two ways in the original laboratory where the charge was moving.
Now we're left with another two questions which I can't give an answer to:
*
*Why is Lorentz symmetry (the
principle of relativity) engrained within physics?
*Why does an electric field deflect a charge dependent upon its sign?
| {
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Weak isospin current I cannot understand the product of a Dirac gamma matrix and a Pauli matrix in this formula of the weak isospin current:
$$J_α^i(x)=\frac12\bar \psi_L(x)\gamma_\alpha\tau^i\psi_L(x),$$ where $γ_α$ is a gamma matrix and $\tau^i$ a Pauli matrix.
(From Manda & Shaw, Quantum Field Theory 2nd, page 393.)
Can anybody help me?
| Short answer: this is just an (outer) tensor product, not a matrix product. I agree that the notation is a bit confusing though.
To see this, try substituting the left chiral duplet of fermions, i.e. for leptons:
$$ \psi_L = \left( \begin{matrix} e_L \\ \nu_e \end{matrix} \right). $$
The matrix $\tau_i$ is sandwiched between the row $\bar{\psi}_L$ and the column $\psi_L$, thus mixing electrons and neutrinos.
The Dirac gamma-matrix $\gamma_{\alpha}$ is acting on spacetime components of the individual fermions. I.e. to give you some concreteness, let's calculate $J_{\alpha}^1$:
$$ J_{\alpha}^1 = \left( \begin{matrix} \bar{e}_L & \bar{\nu}_e \end{matrix} \right) \gamma_{\alpha} \left( \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} \right) \left( \begin{matrix} e_L \\ \nu_e \end{matrix} \right) =\bar{e}_L \gamma_{\alpha} \nu_e + \bar{\nu}_e \gamma_{\alpha} e_L. $$
Here both summands are of form $\bar{\varphi} \gamma_{\mu} \psi$ with $\varphi, \psi$ – Dirac spinors. These transform as 4-vectors (easy to prove) and are, of course, well known from QED where the electromagnetic current was given by $\bar{e} \gamma_{\mu} e$.
One important difference is that here only the left-handed part of the electron field is nonzero.
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Are there any physical quantities whose units are defined using n-th roots, logs, sins, etc. of SI units? All the SI derived units I know are defined as products of SI units raised to an integer power (e.g. the coulomb is measured in $\text{s}\cdot\text{A}$, the pascal in $\text{kg}\cdot\text{m}^{-1}\cdot\text{s}^{-2}$, etc.).
But are there any meaningful physical quantities for which it makes sense to consider different operations, such as n-th roots, logarithms or exponentials of SI units (things like $\text{m}\cdot\sqrt{\text{s}}$, or $\ln{\text{kg}}\cdot e^{2\cdot\text{cd}}$)? Or trigonometric functions, or even more exotic operations?
And if not, is there a specific reason why there aren't/there can't be any?
| It does not normally make sense to put anything but a unitless input into a transcendental function. However, as discussed in this question, it does make sense to take logs of quantities that have units. The result is that you get an additive constant that would depend on the choice of units, and this constant is often of no interest. The classic example would be using the slope of a log-log plot to find the exponent in a power law. However, the units of the input are not passed through to the output. So for example, if you take the log base 10 of 100 kg, you get 2 plus a constant, but the 2 doesn't have units of log-kilograms. The choice of kilograms is present in the additive constant. Therefore there is no such thing as a unit of "log-kilograms."
Roots are used in units all the time. For example, I teach a lab in which we drop balls from unequal heights $h_1$ and $h_2$ and measure the time between the two hits in order to measure $g$ with pretty good precision. The expression for $g$ involves the quantity $\sqrt{h_1}-\sqrt{h_2}$.
| {
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Does a photon have any measurable size as a "particle"? Does a photon have any measurable size as a "particle"? If we determine its exact location and take away all motion in an isolated static environment, how large is it? Is it sub-Planck in size? If there is no motion in this static environment, is the photon also sub-Planck in terms of time?
| In the standard model of particle physics which is continuously validated by experiments, the photon is a zero mass,spin one, point ,elementary particle in the table of particles that build up all that we know. Its energy is $h*ν$ where $ν$ is the frequency of the classical electromagnetic wave that it can build up in superposition with zillions of other photons.
If we determine its exact location and take away all motion in an isolated static environment, how large is it?
Zero mass means it is moving with speed c in all Lorentz frames as all zero mass particles, so this scenario can never happen.
Does a photon have any measurable size as a “particle”?
The size of the footprint of the photon will depend on the measuring instrument. The footprint will be bounded by the Heisenberg uncertainty principle .
Here one observes the footprints of single photons, in a double slit experiment.
The footprint is a dot within the measurement errors, and certainly bounded by the uncertainty principle, since h, the Planck constant, is very small, macroscopic measurements always obey it.
Thus , a photon, as described by current quantum mechanical theories is a point particle, and it demonstrates an effective size according to the boundary conditions of the measurement.
| {
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Trouble in understanding AM modulation Amplitude modulation is in fact, superimposing the low frequency transmission signal into a high frequency carrier signal, right?
So, if the transmission signal can be represented as $c(t)=A_c \sin (\omega_ct )$ and the carrier wave can be represented as $c(t)=A_m \sin (\omega_mt)$, then upon superimposing and simplifying the equation using trigonometric identities we will have
$$
c_m(t)=A_c \sin (\omega_ct )+\frac{\mu A_c}{2}\cos(\omega_c-\omega_m)t-\frac{\mu A_c}{2}\cos(\omega_c+\omega_m)t,
$$
but this equation has the carrier wave and two sinusoidal waves with frequencies $\omega_c-\omega_m$ and $\omega_c+\omega_m$. So where is the signal which is transmitted? What we've got is a corrupted signal with uneven frequencies i.e. $\omega_c-\omega_m$ and $\omega_c+\omega_m$
Correct me if I'm wrong at any point.
| You are correct in your assertion that the AM procedure, when applied to a monochromatic signal, will generate a radio emission that, spectrally, has a spike at the carrier and then two sidebands on either side, separated from the carrier frequency $\omega_c$ by the modulation frequency $\omega_m$.
More specifically:
*
*The strength of the signal is encoded in the depth of the modulation and therefore on the strength of the sidebands when compared to the carrier.
*The frequency of the signal is encoded in the separation of the sidebands with respect to the carrier.
*The phase of the signal is encoded in the phase of the sidebands with respect to the carrier.
This means that the amplitude-modulation procedure has successfully encoded all the relevant information of your signal into a higher-frequency radio beam that can be easily transmitted. It is then the job of the detector (i.e. your consumer radio) to decode that information into audio signals that can be played by a speaker; how the decoder does that is up to the device, but the important thing is that the full information is there to be decoded.
The decoding procedure itself can be done by forgetting about this sideband business and just looking at the signal in the time domain with an envelope detector, or you can explicitly use this carrier-sideband structure with a locally-generated carrier that you then use to extract information from the sidebands in a synchronous detector.
| {
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References on how dimensionality relates to inverse square laws https://en.wikipedia.org/wiki/Spacetime#Privileged_character_of_3.2B1_spacetime
Does Coulomb's Law, with Gauss's Law, imply the existence of only three spatial dimensions?
Why are so many forces explainable using inverse squares when space is three dimensional?
This wikipedia article claims that the existence of inverse square laws (like Newton's Law of Gravitation) has a connection with the fact that our universe has three large scale spatial dimensions.
Can someone point me to some book/paper where this topic is treated in a detailed and formal manner?
| It's pretty simple. In 3 dimensions when you transmit a wave uniformly in all directions the energy at any distance as it propagates is the same. Since the area of a sphere at distance r is $4\pi r^2$ the energy per unit area at any distance is inversely proportional to $r^2$. If the spatial dimensions were n it'd be inversely proportional to $r^{n-1}$
| {
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The symmetry group and representation of spin-$N$ particle I am confused with the symmetry group and the representation of spin-$N$ particles, and will appreciate any help or suggestions of reference.
There are $2N+1$ internal states for a (massive) spin-$N$ particle. These internal states define a $2N+1$-dimensional Hilbert space. It seems to be resonable that, the associate symmetry group is $SU(2N+1)$.
However, it seems also possible that, the $2N+1$ states correspond to the $2N+1$-dimension irreducible representation of $SU(2)$.
Are there some relations between the above two situations, or I just confuse some basic concepts?
| It is not the same. A $2N+1$ dimensional representation of SU(2) would have only three generators: $L_+, L_-$ and $L_z$. In particular, $L_+$ and $L_-$ only connect states with $\Delta m= \pm 1$, i.e. can only connect "neighbouring" states (this is loose terminology). In addition, the size of the matrix elements are quite restricted as the basic commutation relation $[L_+,L_-]=2L_z$ must be preserved by the representation.
On the other hand, generators of $su(2n+1)$ are of the form $C_{ij}$ with $i,j=1,\ldots, 2n+1$ which have non-zero matrix elements between $i$ and $j$, i.e. these can connect "non-neighbouring" states. The size of the matrix element of $C_{ij}$ is $1$ and the matrices must satisfy
$[C_{ij},C_{k\ell}]= \delta_{jk}C_{i\ell}-\delta_{j\ell}C_{kj}$.
In other words, yes it is possible to construct a $2n+1$ dimensional representation of $su(2)$ but this representation is NOT the same as the defining $2n+1$ dimensional representation of $su(2n+1)$.
As a simple example, consider the Gell-Mann matrices for $su(3)$. You can easily enough construct a $3$-dimensional irrep of $su(2)$ of angular momentum $\ell=1$, but the three matrices for $L_+,L_-$ and $L_z$ do not look anything like the 8 Gell-Mann matrices; it is possible to write some but not all the $su(2)$ generators are linear combos of the Gell-Mann matrices, but it is certainly not possible to write the 8 linearly independent Gell-Mann matrices in terms of the 3 linearly independent angular momentum matrices of dimension $3$. If anything, there are two diagonal Gell-Mann matrices but only one diagonal $su(2)$ matrix.
| {
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Symmetries of the theory that are not symmetries of the action nor of the measure? Can anybody think of an example of any theory in which there is a transformation law that does not leave the action nor the path integral measure invariant, but such that the product of both is invariant so that the transformation is a symmetry?
| The theory of a chiral fermion $\psi$ coupled to a Maxwell field in four dimensions has a famous anomaly in the chiral transformations
$$ \psi \rightarrow \psi'=e^{i \gamma_5 \theta} \psi $$
which arises from the non-invariance of the fermion path integral measure
$$
D\psi' D\bar\psi' = D\psi D\bar\psi \, {\rm\exp}\left( i\frac{\theta}{(4\pi)^2}\int F\wedge F\right)\,.
$$
This can be fixed by adding a pseudoscalar (axion) $a(x)$ to the theory, that couples to the Maxwell field as follows
$$
\mathcal{L}_a = (da + A)\wedge * (da + A) + a \frac{1}{(4\pi)^2}F \wedge F
$$
and transforms under the symmetry with a shift
$$
a \rightarrow a' = a - \theta.
$$
The non-invanriace of the classical action cancels the quantum anomaly, giving an example of what you are looking for. This is the toy version of a more general story that goes by the name of "Green-Schwarz anomaly cancellation mechanism".
| {
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Spontaneous Combustion My teacher posed this question in class and it stumped me, primarily based on my lack of fundamentals of thermodynamics:
Peter Griffin was sitting around when all of a sudden he spontaneously combusts (Based on an episode). Is this possible and if not, under what circumstances is this possible?
My reasoning is
$${\rm Peter + O_2 \to H_2O + CO_2 }$$
There are more gas particles so entropy must increase, $\Delta H$ is negative since it is a combustion reaction and so according to
$${\Delta G = \Delta H - T \Delta S}$$
Meaning free energy decreases... Which doesn't seem to make sense because it's highly unlikely that humans spontaneously combust. This got me thinking that the reaction itself is spontaneous. It just does not proceed because it cannot start the reaction due to insufficient activation energy (Particles not travelling fast enough).
My teacher said $\Delta H$ is positive which makes no sense to me... In a combustion reaction energy is released. He said it only occurs at very high temperatures because $T\Delta S$ is large enough to counter the positive $\Delta H$.
Is my teacher right or is my reasoning more correct?
| Good Effort!
We can make an analog case for "spontaneous". Assuming there is a hill, when you put a ball on top of the hill, it will spontaneously roll down the hill. On the contrary, if you put the ball at the foot of the hill, it will not spontaneously roll up the hill. Potential energy plays a role here.
For a chemical reaction, it is the same, if free Gibbs change between an initial state and a final state is negative, the mixture will spontaneously react changing from initial state (Peter + $O_2$) to the final state ($H_2O+CO_2$).
$\Delta H$ is another thing. If it is negative, as you can be sure, the free Gibbs energy is negative. So it is a spontaneous process. If it is positive, which is more useful in practical application, it can be spontaneous if it is less than $T \Delta S$. Otherwise it is not spontaneous.
| {
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When does Rolling Friction come into play? According to Wikipedia, "Rolling resistance, sometimes called rolling friction or rolling drag, is the force resisting the motion when a body (such as a ball, tire, or wheel) rolls on a surface."
But at the same time, it is the static friction that is responsible for the forward movement of the wheels of a vehicle, according to many sources. But by the above definition, should it not be the rolling friction that comes into play?
I had also found rolling friction to be defined as follows:
"Rolling resistance tends to be a catch all term for the energy dissipated in the many moving parts as a vehicle moves."
(For more, please visit Rolling resistance and static friction )
1) So, which definition is the right one?
2) Also, which friction is applicable for the motion of the wheels of a vehicle?
3) The motion of the wheels of a vehicle is considered "rolling without slipping". I know this sounds stupid, but does the condition "rolling and slipping" exist?
| Static friction is what keeps the object rolling (where molecules of the object and the surface essentially act like cogs in a wheel).
Rolling friction refers to the processes of rolling that cause kinetic energy to be turned into heat (which by definition can't be static friction). For example, a rolling basketball becomes slightly deformed (the part touching the ground is smushed in/not perfectly spherical) as it rolls - more so the less it's inflated. That deformation absorbs kinetic energy as heat.
If you have an axle in a car, then you can lump the friction of the axle into rolling friction since it's not a scientifically rigorous definition.
For rolling and slipping, imagine tires on ice. Slipping leads to kinetic (and not just static) friction.
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How to calculate the maximal Lyapunov exponent(s) of a multidimensional system? I've been reading up about Lyapunov exponents for a university group project on chaos theory and I'm a little confused as to how they are calculated for a system of multiple dimensions.
From what I can tell, the maximal Lyapunov exponent $\lambda$ for some 1-d map $f(x_{n})=x_{n+1}$ is:
$\lambda \approx \frac{1}{n}\sum\limits_{i=0}^{n-1}ln|f'(x_{i})|$
Where, if I understand things correctly, $f'(x_{i})$ is the derivative of $f$ at the ith value of $x$.
What my question is, is how does this extend to the multi-dimensional case? Say I had a pair of coupled maps representing a 2d system, $f(x_{n}, y_{n})=x_{n+1}$ and $g(x_{n}, y_{n})=y_{n+1}$, how is the Lyapunov exponent found?
| The extension to several dimensions is natural:
*
*instead of a scalar, the state variable is a vector, and
*the derivative is substituted by the system's Jacobian.
Some earlier papers for numerical methods for calculating the Lyapunov exponents are mentioned in the Wikipedia entry and a more complete entry is found in the Scholarpedia, but textbooks on nonlinear dynamics (e.g., Ott, 1st ed., Eq. 4.40) offer more easier-to-follow how-to's.
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Why is a black body visible? A black body absorbs all energy. It doesn't reflect or transmit energy. It also absorbs all light and doesn't reflect any light. Why, then, can we see it? For instance, burnt platinum is 98-99% black body and yet is visible.
| Different visible wave lengths make sense of different colors in our brain.If there are no visible wavelengths, it makes sense of black. In respect of black body, it doesn’t reflect any visible elecro-magnetic wave as you mention.So it’s black and visible
| {
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What is the shortest distance between electron and positron before they are annihilated? I just want to know how does an electron felt the presence of a positron before they are converted into energy? Also how does the electron tell if it is positron or proton if this makes any difference?
| In QED, the basic annihilation process is described by a $\bar{\psi }{{A}^{\mu }}{{\gamma }_{\mu }}\psi $ interaction term, in which the electron $(\psi )$, positron $(\bar{\psi })$ , and photon $(A)$ field operators act at exactly the same point. However, there are small, higher-order corrections to this term, which arguably make it non-local. (This is fancy technical jargon for John Rennie’s caveat, when interactions become significant.)
This is the process that plays in ${{e}^{-}}{{e}^{+}}\to {{\mu }^{-}}{{\mu }^{+}}$ . Some different happens in ${{e}^{-}}{{e}^{+}}\to \gamma \gamma $, where the incoming electron and positron are annihilated at two different points.
The idea that two particles must have zero separation to annihilate may seem surprising in classical terms, but not in quantum mechanics, where particles of definite momentum have undefined positions.
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$Q$ value of beta plus decay (positron emission) I am unable to understand Q value for positron emission. The general reaction is as follows:
$$p \to n + e^+ + \nu$$
$$ ^A_ZX \to ^A_{Z-1}Y+ e^+ + \nu \tag{1a}$$
This reaction $(1a)$ was giving in my text. First question is that where did one electron go? We began with $Z$ electrons but on the right side it seems only $Z-1$ electrons are present. Maybe this is the cause of confusion.
So I rather took this reaction $(2a)$ for positron emission:
$$^A_ZX \to ^A_{Z-1}Y+ e^+ + e^-+ \nu \tag{2a}$$
Now writing the Q value, we find the mass defect
$$[m_n(^A_ZX)-(m_n(^A_{Z-1}Y)+m_{e^+}+m_{e^-} +m_\nu)]$$
here $m_n(^A_{Z}X)$ are mass of nuclei. Now we can rewrite in terms of atomic mass numbers $m_a(^A_{Z}X)$ as
$$\Delta m=[(m_a(^A_ZX)-Zm_e)-((m_a(^A_{Z-1}Y)-(Z-1)m_e)+m_{e^+} + m_{e^-}+m_\nu)] $$
$$\Delta m= (m_a(^A_ZX)-m_a(^A_{Z-1}Y) - 3m_e -m_\nu) \tag{2b} $$
but as you can see, if we use reaction $(1a)$ then we will get
$$\Delta m= (m_a(^A_ZX)-m_a(^A_{Z-1}Y) - 2m_e -m_\nu) \tag{1b} $$
Am I wrong somewhere? I have highlighted key points(mistakes) in italic.
|
We began with Z electrons but on the right side it seems only Z−1 electrons are present.
During beta plus decay we consider that electrons do not participate in the decay for easier calculations.(In reality they in fact must be included) Only nuclei participate. So their are no electrons on the left side and just a positron on the right which came from one of the protons.
The equation (2a) their are no electrons on the left but you added an electron on the right(assuming only nuclei participate in the decay), that cannot happen. That also violates the law of conservation of charge as a proton should convert into a neutron and positron but after introducing the electron you have decreased the overall charge by one.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/389442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
What does work mean as a concept and not as a mathematical equation? I know that work = Force times displacement or the dot product of the force and the displacement. But what does work mean in the physical world?
Just like we know that in the physical world, density is the compactness of a substance and we can deduce that via the formula of density which is the amount of mass occupying per unit volume, what does work mean then in the physical world?
I thought of the idea that work is a multiple of force by displacement s. But what does that mean in the physical world?
| Work means that the barge owner pays the tugboat owner, in return
for being moved. Without the concept of work, many
exchanges involving force and distance would be
much less comprehensible.
Work also explains to the tugbboat owner why the vessel's fuel
consumption goes up during times that the tow cable is under
strain.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/389629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
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