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Is it more efficient to drive fast uphill? I know that for a rocket escaping the atmosphere, it's not efficient to travel slowly because even staying stationary consumes a lot of fuel. Does the same apply to a vehicle traveling uphill? In other words, is it more efficient for a car/bike/runner to accelerate and go over the hill quickly because energy is exerted during the climb even to "keep the vehicle stationary" and not just to travel the height and distance?
| Some preliminaries: there is no time limit, no specific initial conditions and there is no wind. The road is straight and empty. There is no speed limit and safety is of no concern. In short, this has to be cheap.
Make sure you arrive at the speed $v > \sqrt{2gh}$. You need an excess to compensate for rolling resistance and air resistance. At the foot of the hill, go out of gear and switch off the engine. If you did this with enough fingerspitzengefühl, you car will reach the top at speed very near zero. Wait until it rolls down the other side. Switch the engine back on, raise the rpm's to the same value and go into the same gear as at the start of the climb. The whole trip has cost you nothing more than some dissipation by air and rolling resistance.
| {
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Fuel in Nuclear physics Why $^{233}\mathrm{U}$ is an efficient nuclear fuel and $^{239}\mathrm{Pu}$ cannot be used in thermal reactors??
I couldn't find a good answer for this question in the introductory course I am taking. Could someone explain this? I know that both need to be breeded!
| You start by looking as to whether the absorption of a thermal neutron will provide enough energy (binding energy of last neutron) to exceed the activation energy (critical energy) for fission.
Both isotopes satisfy that requirement which is shown in the table below which has been taken from this question where the terms are explained.
Next one should look to see is the capture cross section for thermal neutrons to produce a fission is different.
For Pu-239 it is 750 barns and for U-233 it is 531 barns so there is nothing in favour of U-233 so far although if one looks further there is another competing cross section which is for capture of a neutron and then the emission of em radiation as explained below.
After fission both isotopes produce, on average, more than two neutrons per fission which is sufficient to sustain a controlled chain reaction.
The values are approximately 2.9 per fission for Pu-233 and 2.5 per fission for U-233.
Now look at what happens after a neutron is absorbed and here is possibly the answer to your question?
Once a neutron is absorbed the resulting nucleus can either undergo a fission or emit em radiation.
The emission of em radiation from a nucleus is bad news as far as a fission occurring is concerned because now the nucleus has less energy for a fission to be initiated and the probability of there being a fission is massively reduced.
After the capture of a thermal neutron by Pu-239 to form Pu-240 73% undergo a fission and 27% emit a gamma ray (called radiative capture of a neutron) whereas for U-233 which forms U-234 the values are 94% for fission and 6% for the emission of em radiation.
So you will see that less of U-233 is "wasted" as compared with Pu-239.
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How detectable is the error in special relativity experiments because of general relativistic effects at the surface of the Earth? I believe that experiments done on special relativity in a laboratory at the surface of the Earth usually do not consider the effects of general relativity (since the gravitational field at Earth's surface is weak). Of course, errors shall emerge from the approximations, but is this error detectable at all? How much do the measurements deviate from the theoretical predictions because of general relativistic effects and does laboratory equipment have enough precision for these kinds of deviations to be detected?
| Nice question.
One way of stating the equivalence principle (e.p.) is that locally, spacetime is flat, so that special relativity is valid. Therefore any experiment in a small enough laboratory, if the apparatus is in free fall, is predicted by GR to give the same results as in SR. For example, SR predicts that if we release a brass ball and an iron ball at rest relative to one another, they will stay at rest relative to one another. This has been tested in Eotvos-style experiments, and the null results confirm SR and the e.p., but they are not specifically tests of general relativity.
If the experiment is local but the apparatus is not in free fall, but is anchored to the earth, then the e.p. predicts that the results are the same as in a rocket ship in outer space that is accelerating at g. An example of this type of experiment is the Pound-Rebka experiment. So this experiment as well is only really a test of SR+e.p., not specifically GR.
To get a real test of GR at the surface of the earth, the experiment needs to explore a region of spacetime at the surface of the earth that is large enough so that different parts of it do not all have the same acceleration. Tests that fall in this category include the Hafele-Keating experiment and the detection of gravitational waves by LIGO/Virgo.
| {
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Why does plasmon have higher erngy than phonon? In my mind plasmon is movement of electrons and phonon is movement of atoms in an lattice. movement of atoms should have a large energy because atom is larger.
| Fundamentally the plasmon is at higher energy because electrons weigh much less that atoms. In fact phonons are simply plasmons of the atoms.
One can see this from the classical formula for the plasma frequency.
$$\omega = \sqrt{\frac{4\pi n q^2}{m}}$$
Assuming you have an equal concentration of free electrons and atoms, $n$ will be the same for the atom and electron plasma frequencies. However, the nuclei have masses that are usually $10^4$ to $10^5$ times larger than the electron, which translates to a frequencies that are over 100 times smaller for phonons compared to electrons. This is consistent with experiments seeing phonons with milli-electronvolt energies and plasmons with electronvolt energies.
| {
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What's the Lagrangian for a freely rotating rod? This is one of those problems that I thought would be easy, then spent forever on it and realized that I know nothing:
A rigid rod of uniform density has mass $M$ and length $L$ and is free to rotate about its center without a fixed axis. Consider its center to be fixed at the origin of an inertial Cartesian coordinate system, and note that the position of the rod can be specified with the spherical coordinate angles $\theta$ and $\phi$. (Also assume zero gravity.)
*
*What is the Lagrangian of this system, in terms of the coordinates $\theta$ and $\phi$?
First, I calculated the moments of inertia for rotation in the pure $\theta$ and pure $\phi$ directions. For $\theta$, this is the usual $\frac{1}{12}ML^2$, but $\phi$ it becomes $\frac{1}{12}ML^2\sin^2\theta$ assuming the rod is held at fixed $\theta$. Then I used $K_{rot}=\frac{1}{2}I\omega^2$ for each type of rotation and added the two:
$$\begin{align} L &= K-U \\ &= K_\theta+K_\phi-0 \\&= \frac{1}{24} M L^2\dot\theta^2+ \frac{1}{24} M L^2\sin^2(\theta) \,\,\dot\phi^2 \end{align}$$
My main concern is that adding the two energies doesn't seem justified, so my next question is:
*If this is correct, why is adding these two energies justified?
To emphasize the concern, consider this: pure $\theta$-rotation corresponds to $\vec\omega$ in one direction, while pure $\phi$-rotation corresponds to $\vec\omega$ in another direction. These two vectors span only a plane, whereas in general $\vec\omega$ could point anywhere in 3D space, so it isn't at all clear that these angular coordinates simply "add" in a straightforward way.
Trying to find the answer more rigorously, I calculated the inertia tensor (assuming at time $t=0$ the rod is lying in the $z$-$x$ plane) and tried to use $K_{rot}=\frac{1}{2}\vec\omega \cdot (\tilde{I} \vec\omega)$. This leads to my final and most deceptively difficult question:
*What is $\vec\omega$ in terms of $\theta$, $\dot\theta$, $\phi$, and $\dot\phi$?
Also if there is a cleaner or more straighforward approach for solving this problem, I'd love to know about it!
| We can just go back to basics, and integrate the contributions to the kinetic energy along the rod. Let the linear density be $\lambda = M/L$. The square of the velocity in spherical coordinates is
$$v^2 = \dot{r}^2 + r^2(\dot{\theta}^2 + \sin^2 \theta\, \dot{\phi}^2);$$
at all points on the rod we have $\dot{r}=0$ and that the angular part in parentheses is constant. The energy is then
$$E = \frac12 \int_{L/2}^{L/2} \lambda r^2 (\dot{\theta}^2 + \sin^2 \theta\, \dot{\phi}^2).$$
We can just take everything except $r^2$ out of the integral, and we get
$$E = \frac{1}{24} ML^2 (\dot{\theta}^2 + \sin^2 \theta\, \dot{\phi}^2)$$
so your initial guess was right. We can add the two energies because they correspond to orthogonal directions, just like in cartesian coordinates we would say $E = \frac12 m (v_x^2 + v_y^2 + v_z^2)$. You can also see that the thing in parentheses in just $\omega^2$, but hopefully this simpler derivation convinces you that everything is legitimate.
| {
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Difference in calculation of Ginzburg Criterion in Gaussian and Mean-Field theory? In the past I asked this question asking about a subtlety in the Ginzburg criterion for mean field theory and the Gaussian approximation.
I am now having a hard time determining in general what is different between the calculations of the Ginzburg criterion for these two approximations.
Most books only seem to focus on the mean field theory calculation and barley mention the Gaussian.
My question is therefore what is the difference between the derivation of the Ginzburg criterion for the MFA and the Gaussian approximation and why.
(for simplity I am considering a general $\phi^n$ theory)
| I believe this to be explained in (Dimo, 1993; sections 7.5.2 and 7.11)$1$.
Saddle Point Approximation
In the saddle point approximation we have a thermodynamic potential given by$^2$:
$$\Phi_G=-V \frac{r_0^2}{16 u} -\ln \int \mathcal{D} \delta \phi e^{-\mathcal{H}_G}$$
In the mean field approximation we assume that the second term does not affect the thermodynamics of the system and thus require it to be small compared to the first term. The Gaussian approximation allows us to evaluate the second term.
Gaussian Approximation
The criterion set by Gaussian approximation is determined by the integral $I_1(r)$ were:
$$I_1(r)=I_1(0)+\Delta I_1(r)$$
and
$$\Delta I_1(r)=-r \int_k \frac{1}{ck^2(r+c k^2)}$$
this second term comes from some sort of 4-point interaction$^3$. We then need this second term to not contribute much to the value of $r$. This gives the Ginzburg criterion for the Gaussian approximation.
Footnotes
$^1$ Unfortunately I can't get access to the full book so what I say in this answer comes from the Google preview - which was fairly limited.
$^2$ I am using the (approximately) notation of the above source. Since I feel it is self evident and can't actually verify it I will not define symbols.
$^3$ As far as I can tell.
| {
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Can you sail up/down a moving river on a windless day? It's not possible to sail into the wind directly, but only at an angle. It can be shown (it may help for below) by considering the pressure forces on a flat sail and flat keel that sailing upwind is only possible because of the angle between the sail and keel.
Suppose a sailboat is on a long, wide, straight river with a current of $0.5\,\text{m/s}$ on a windless day (there's no wind with respect to the river bank).
*
*Can the sailboat travel faster downstream than $0.5\,\text{m/s}$ by raising its sails and sailing?
*Is there any way the sailboat can sail upstream on the slow-moving river on this windless day?
The source of this question is a bathroom physics problem posted above university physics department toilets near grad offices. Problems are submitted by department members. To the best of my knowledge, none of these were ever homework questions. This question is minimally modified from a problem that was provided by Matt Kleban.
| Yes. If the wind is moving fast enough, and moving sideways across the river. The wind will hit the bottom of the sail if it is at a slight angle, and therefore push it up stream. The speed of wind required depends on a lot of factors having to do with the boat, but it would be possible with high enough winds perpendicular to the riverbank. Similarly it could go faster downstream if you rotated the sail 180 degrees from where it travels upstream.
| {
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Should we take the elevator instead of the stairs to save energy? In our university there's a posted sign that encourages students to take the stairs instead of the elevator to save the university electricity during the hot NYC summers.
When you climb the stairs you generate something like 8x your mechanical energy in metabolism which is dissipated as heat and must be cooled by the air conditioning in our cooled stairwells.
What are plausible assumptions about the energy efficiency limits of air conditioning and mechanical lifting of an elevator? What (fundamental) limits are there that prevent 100% efficiency in the elevator case? Are there other important factors for this analysis?
Is it plausible that the university is wrong? Should we take the elevator instead of the stairs to save the university energy? Do the humans of an energy-efficient future look like Wall-E?
The source of this question is a bathroom physics problem posted above university physics department toilets near grad offices. Problems are submitted by department members. To the best of my knowledge, none of these were ever homework questions. This question is adapted by my toilet memory from a problem that was provided by Andrei Gruzinov.
| I wondered the same thing and so did some empirical testing with my elevator.
TLDR;
The elevator uses a huge amount of energy to move just the elevator.
For 5 floor ascent using my single piston hydraulic lift elevator on the day I tested...
Elevator: ~550KWs
Stairs: 92KWs / (22 calories)
If the elevator is going up anyway, then you should hop on since your additional weight has almost no effect on the total power used.
If you are considering taking the elevator alone, you should take the stairs.
Also keep in mind that most the energy used by the elevator is turned into heat and that heat is generated inside the envelope of the building, so must be rejected by the AC system.
More explanation and data here...
https://wp.josh.com/2013/05/29/elevator-power-usage-should-i-take-the-stairs/
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How can gamma rays affect a human body if they can move through atoms without colliding with them? I found this answer on this SE about why mirrors don't reflect gamma rays, and the answer says that it is because gamma rays are so, so small that they "see mostly empty space between the atoms of the solid.".
If that is true, so gamma rays can move through atoms without colliding with them, why can gamma rays affect people enough that they are said to be dangerous? If they are so small, that would mean it's very unlikely for a gamma ray to accidentally touch an atom, so it would be a very rare situation, which the body should be able to fix without problems due to it's unlikeliness, right?
| It's like this bro, if you look down a perfectly repeating crystal lattice at the top, you will see little holes depending on the packing structure, through which the gamma rays will just pass through the but if you align the nuclei, and have a long enough Crystal, and you turn it a very small degree the holes will close and one nucleus Center will be very close behind but just very small angle from being perfectly behind the one just before it. So if you have a really really long crystal, theoretically, gamma rays will be hitting the nuclei in that crystal lattice more frequently than if you just had it straight on, or something like this?
| {
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Can we measure Earth's gravity without a pendulum? I'm wondering how can we measure Earth's gravity without pendulum? of course we can do it by using Newton's laws of physics but the air friction might change the actual value also how physists measure distant planets gravity that are found by transit's (by detecting dip's in light intensity)? I know g=GM/(r^2)
| The conventional method for measuring the Earth's acceleration of gravity is that of measuring the trajectory of a free-falling body. The instrument that implements this principle is called the absolute free-fall gravimeter.
In this type of gravimeter, the free-falling body is usually a retroreflector (corner-cube) constituting an arm of an interferometer which measures the retroreflector trajectory. The retroreflector falls in a vacuum chamber, but to further reduce the residual friction, it is closed in a so-called drag-free chamber which follows the fall through a servo system, so that the retroreflector is surrounded by still air. The chamber and the servo system allow also for a smooth landing.
The acceleration of gravity is determined by fitting the trajectory to the theoretical equation of motion, which takes also into account the gravity gradient along the fall. The time reference is usually given by an atomic clock.
This kind of gravimeter is commercially available (see e.g. the FG5-X gravimeter).
| {
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Do magnets wear out? Can a magnet ever wear out or lose strength?
If you break a magnet it (seemingly) gets weaker, but what about from normal use?
Or even very heavy use, like placing 2 magnets facing each other, so that they detract from each other, does that strain cause it to wear quicker?
(Note, I'm not looking for a merely yes or no answer; If yes, what will cause it to wear out quicker or slower. If no, why?)
| Yes, a magnet, as time passes, will lose part of his strength. There are two main reasons:
*
*Thermal energy: it causes the disorientation of the atomic magnetic momenta.
*If you have a bar magnet free in space it’s easy to see (using Ampère’s law) that there is inside it a magnetic field $H$ opposite to the magnetisation of the magnet. In order to avoid this phenomenon you should anchor it (that is to say “linking north with south pole“) with a ferromagnet.
This two phenomena will cause atomic magnetic momenta to disorient, and, in so doing, the magnetic strength of the magnet will decrease.
The demagnetisation happens even if you apply a sufficiently strong magnetic field opposite to the one generated by the magnet.
EDIT:
The proof of the existence of a field $H$ inside the magnet is now reported: let’s take a bar magnet as shown in figure
The Ampère’s law tells us that
$$
\int_\gamma H ds =0\; .
$$
Now let’s call $\gamma_1$ the piece of curve inside the magnet and $\gamma_2$ the piece outside with length respectively $L_1$ and $L_2$. The Ampère’s law becomes
$$
\int_{\gamma_1}Hds +\int_{\gamma_2} Hds =0\; .
$$
Let be $H_1$ the mean $H$ field inside the magnet and $H_2$ the one outside. The integral turns into
$$
H_1L_1+H_2L_2=0
$$
From here we have
$$
H_1=-\frac{L_2}{L_1}H_2=-\frac{L_2}{L_1}\frac{B}{\mu_0}\;
$$
And here we have what was to be demonstrated.
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Entropy of the ideal gas - How to derive the $S(U,V,N)?$ I'd like to derive the $S(U,V,N)$ function. In the lecture, we were using the $S=k_b \log(\Omega)$, some combinatorics, approximation, $6N$ dimensional spheres, etc. But I'd like to avoid that if it's possible. So I think I should do it from the first and second law of thermodynamics:
$$\mathrm{d}U=T\mathrm{d}S-p\mathrm{d}V+\mu\mathrm{d}N$$
$$\mathrm{d}S=\frac{1}{T}\mathrm{d}U+\frac{p}{T}\mathrm{d}V-\frac{\mu}{T}\mathrm{d}N$$
And with $pV=Nk_bT$ and $U=\frac{f}{2}Nk_b T$, I could get that
$$\mathrm{d}S=\frac{f}{2}k_b \frac{N}{U}\mathrm{d}U+k_b\frac{N}{V}\mathrm{d}V-\frac{\mu}{T}\mathrm{d}N$$
But I can't get rid of the $\frac{\mu}{T}$. How could I do it?
| From the definition of chemical potential:
\begin{equation}\mu = \left(\frac{\partial U}{\partial N}\right) = cRT \end{equation}
where c = 3/2 for a monatomic ideal gas, c = 5/2 for a diatomic ideal gas.
To verify the consistency of the derivation:
From the “fundamental equation” of an ideal gas:
\begin{equation} PV = N RT \end{equation}
\begin{equation} U=c N RT\end{equation}
if you rewrite the equations of state as
\begin{equation} \frac{1}{T}=\frac{cN R}{U}=(∂S/∂U)_{V,N}\end{equation}
\begin{equation} PT=\frac{NR}{V}=(∂S/∂V)_{U,N} \end{equation}
By integrating both, the first equation gives:
\begin{equation} S(U, V, N) =c N R \ln(U) +f(V, N)\end{equation}
and the second
\begin{equation} S(U, V, N) =N R \ln(V) +g(U, N) \end{equation}
Which are consistent only if we can write S in the form:
\begin{equation}S(U, V, N) =cN R \ln(U) +N R \ln(V) +f(N) \end{equation}
where $f(N)$ is a function of N alone namely:
\begin{equation}f(n) = c RN \end{equation}
which accounts also for the fact that $S$ must be extensive in $U, V$, and $N$.
the arguments of the logarithms must be dimensionless (otherwise, the logarithm doesn’t make sense)
\begin{equation}S(U, V, N) =cN R \ln \frac{U}{Nk_1} +N R \ln \frac{V}{Nk_k2} +f(N) \end{equation}
Here $k_1$ and $k_2$, are constants with the dimensions of energy per mole, volume per mole
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Notation question in calculus of variations -- QFT these two integrals below are equal, but I am not understanding where the $x'$ variable comes from.
\begin{align}
I_0&=e^{ i\int d^4x \left\{ \frac{1}{2}\left[ \left( \partial\varphi(x) \right)^2-\varphi(x)^2 \right] -\big[\varphi(x)\big]^4 +J(x)\varphi(x)\right\}}\\
&=e^{ -i\int d^4x'\big[ \frac{\delta}{\delta J(x')} \big]^4 } e^{ i\int d^4x \left\{ \frac{1}{2}\left[ \left( \partial\varphi(x) \right)^2-\varphi(x)^2 \right] +J(x)\varphi(x)\right\}}
\end{align}
The gist of what I have been doing is to write
\begin{align}
I_0&=e^{ i\int d^4x \left\{ \frac{1}{2}\left[ \left( \partial\varphi(x) \right)^2-\varphi(x)^2 \right] +J(x)\varphi(x)\right\}}e^{ -i\int d^4x \big[\varphi(x)\big]^4} \\
&=e^{ i\int d^4x \left\{ \frac{1}{2}\left[ \left( \partial\varphi(x) \right)^2-\varphi(x)^2 \right] +J(x)\varphi(x)\right\}} \left[ 1+\left( -i\int d^4x \big[ \varphi(x) \big]^4 \right) +...\right]\\
&=\left[ 1+\left( -i\int d^4x \left[\dfrac{d}{dJ} \right]^4\right)+... \right]e^{ i\int d^4x \left\{ \frac{1}{2}\left[ \left( \partial\varphi(x) \right)^2-\varphi(x)^2 \right] +J(x)\varphi(x)\right\}}
\end{align}
The point is to pull the $\varphi^4$ term out of the integral by writing $\varphi$ as $d/dJ$ except in this case I have to use the variation $\delta$ instead and I don't see why. I used the $d/dJ$ trick in simpler examples without an integral in the exponent, and I am not seeing the connection to the variational notation $\delta$ which appears in the second of the first two equations above. Obviously, I can recombine the prefactor sum in my final equation into $\text{exp}$, but I do not see the point of the $x'$ variable. I hope it is clear that if I had $\delta/\delta J(x')$ in my last equation instead of $d/dJ$ then I would get the form of the second equation which is the correct form. Why can't I just write it as $d/dJ$ like I did before? Please give me a tip, thanks.
| The issue is really what it means to to compute one functional derivative $\delta/\delta f$. Once we get that part, we can raise it to the nth power and get the result.
How to compute one variation?
\begin{align}
\dfrac{\delta}{\delta J(x')} K&=\dfrac{\delta}{\delta J(x')} e^{ i\int d^4x \left\{ \frac{1}{2}\left[ \left( \partial\varphi(x) \right)^2-\varphi(x)^2 \right] +J(x)\varphi(x)\right\}} \\
&=\dfrac{\delta}{\delta J(x')} e^{ i\int d^4x J(x)\varphi(x)}e^{ i\int d^4x \left\{ \frac{1}{2}\left[ \left( \partial\varphi(x) \right)^2-\varphi(x)^2 \right] \right\}} \\
&=A\dfrac{\delta}{\delta J(x')} e^{ i\int d^4x J(x)\varphi(x)} \\
\end{align}
Let $F=e^{ i\int d^4x J(x)\varphi(x)} $. The next step is what follows. Instead of using $\Delta$, you need to use $\varepsilon$ times a Dirac delta. This is part of the definition of the functional derivative I guess, If anyone want to say a little more about that would be nice, but I think it's just part of the definition. However, wikipedia calls the Dirac delta a "test function" so maybe it is an ansatz of some type. Anyway, the answer follows:
\begin{align}
\dfrac{\delta}{\delta J(x')}F&=\lim\limits_{\varepsilon\to0}\dfrac{F[J(x)+\varepsilon\delta(x-x')]-F[J(x)]}{\varepsilon}\\
&=\lim\limits_{\varepsilon\to0}\dfrac{ e^{ i\int d^4x [J(x)+\varepsilon\delta(x-x')]\varphi(x)}-e^{ i\int d^4x J(x)\varphi(x)} }{ \varepsilon }\\
&=\lim\limits_{\varepsilon\to0}\dfrac{ e^{ i\int d^4x [J(x)+\varepsilon\delta(x-x')]\varphi(x)}-e^{ i\int d^4x J(x)\varphi(x)} }{ \varepsilon }\\
&=F\,\lim\limits_{\varepsilon\to0}\dfrac{ e^{ i\int d^4x \varepsilon\delta(x-x')\varphi(x)}-1 }{ \varepsilon }\\
&=F\,\lim\limits_{\varepsilon\to0}\dfrac{ e^{ i\varepsilon\varphi(x')}-1 }{ \varepsilon }
\end{align}
use l'Hopital's rule
\begin{align}
\lim\limits_{\varepsilon\to0}\dfrac{ e^{ i\varepsilon\varphi(x')}-1 }{ \varepsilon }\stackrel{*}{=}\lim\limits_{\varepsilon\to0}\dfrac{i\varphi(x') e^{ i\varepsilon\varphi(x')}}{ 1 }=i\varphi(x')
\end{align}
therefore
\begin{align}
\left[\dfrac{\delta}{\delta J}\right]^4F=\left[\varphi(y)\right]^4 F
\end{align}
then
\begin{align}
AF+\left(-i\int d^4x \left[\dfrac{d}{dJ}\right]^4 \right)AF+...&=AF+\left(-i\int d^4x' \left[\varphi(x')\right]^4\right)AF+...\\
&=e^{ -i\int d^4x' \big[\varphi(x')\big]^4}AF \\
&=e^{ -i\int d^4x \big[\varphi(x)\big]^4} AF
\end{align}
Plugging in F and A, we get the expected result.
| {
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Is an atomic clock's measurement principle independent of what a light clock provides Special Relativity has the light clock which basically counts how often a photon goes forth and back along a fixed distance.
Our most accurate clock, defining the second (at least as of today), is an atomic clock.
These seem to be two completely independent ways to measure time and define a second. Are they really independent or are there hard mathematical rules that transform the frequencies relevant in the atomic clock into the distance covered by the speed of light?
If they are independent, how can we be sure they match up?
| Starting with Einstein's 1905 paper on special relativity, the implicit operational definition of time has been that time is what a clock measures, and the implication of this definition is that all clocks measure the same time. Before Einstein, people had been pursuing systems of thought in which such things were explained by physical stresses in specific systems. Under such systems, there was no clear reason to think that if clock A and clock B had the same world-line, they would measure the same tim.
| {
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Why does an axially loaded column break along a 45 degrees plane? In this video, at 7:57, the speaker mentions that columns under compression roughly break along a plane that is at 45 degrees to the axis of loading. If this is true, can someone explain why this happens in terms of inter-molecular forces between the column molecules.
| Experimental measurements have shown that the mode of failure of many materials is along a plane of maximum shear stress. For the rod loading that you have described, the maximum shear stress occurs on a plane oriented at an angle of 45 degrees to the load direction.
| {
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Statistical Mechanics: Boltzmann partition function I’m a high school student and was recently studying basic statistical mechanics (for use in physical chemistry).
In the derivation of the Boltzmann derivation or the partition function we arrive at a factor $e^{-\beta T}$.
So I searched for the connection of $\beta$ to temperature and got one proof for that on the wikipedia site. But it seems to make to make the assumptions that
*
*The energy of a system in equilibrium depends only on the temperature.
*At equilibrium the number of microstates is maximized.
*It uses the formula $S=k\ln(W)$.
So I just wanted to know that which of the assumptions is an experimental fact and whether we can arrive at the equation $S=k\ln(W)$ without already using $\beta=1/(k_BT)$ otherwise the proof would become circular.
To improvise, we can use the property of Lagrange multipliers to obain that
$\beta=\frac{d(ln(W))}{dT}$ but I still don’t see how can we reach to the value of $\beta$ from here.
P.S. I am just a beginner so please use simple language if possible.
| From (2) and $\Omega = \Omega_A \times \Omega_B$ one can see that the fractional change of the number of microstates with energy $\frac{1}{\Omega}\frac{d\Omega}{dE}$ must be equal for two systems in thermal equilibrium. This is the thermodynamic beta $\beta.$ At room temperature it is about 4 % per meV.
Now consider a tiny system that is in thermal contact with a large system. The $\beta$ of that large system will not change much when the tiny system absorbs some energy. But its number of microstates will become smaller. The probability will be lower because of that.
For simplicity, assume one microstate per energy for the small system, maybe just one harmonic oscillator. Rewrite the expression for $\beta = \frac{1}{\Omega}\frac{d\Omega}{dE}$ to a differential equation for the large system $\frac{d\Omega}{dE} = - \beta \Omega$. (The minus sign appears because the energy $dE$ is transferred to the small system.)
Finally solve to obtain the Boltzmann factor: $P(E) \propto \exp[-\beta E]$.
The expression is not valid when particles are indistinguishable and Bose- or Fermi-statistics should be used. The assumption that this derivation relies on is the distinguishability of the small harmonic oscillator.
| {
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Can we (in principle) obtain molecular bound systems by modelling fundamental particles and their interactions? Is it possible, at least in principle, to start with the Schrodinger/Dirac/Klein-Gordon equations to model elementary particles and their interactions and to obtain in the end molecular bound systems? In other words, is it possible (in principle) to deduce the laws of chemistry starting from the laws of elementary particles?
| In principle, yes, this is done in theory of atoms and molecular bonds, also known as quantum chemistry. One can begin with Schroedinger's equation and derive much of the observed properties of atoms and molecules - their energy, spectra, bond lengths and stable arrangements of atoms in space (shape of molecules) which reproduce some chemical ideas of molecules (say, the typical angles between C-H bonds - for example, 120 degrees in methane, which is a tetrahedron).
In practice, the calculations get too demanding to do exactly for small molecules, so there are approximate methods, which have some success, but not always. Chemistry is still not merely 'applied theory of Schroedinger's equation' and probably won't be for a long time.
| {
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Density matrix of a partially polarized beam of electrons The following problem is from Huang's Statistical Mechanics (2nd edition):
8.1 Find the density matrix for a partially polarized incident beam of electrons in a scattering experiment, in which a fraction $f$ of the electrons are polarized along the direction of the beam and a fraction $1 - f$ is polarized opposite to the direction of the beam.
Earlier in the chapter, he gives
$$\rho_{m n} \equiv (\Phi_n, \rho \Phi_m) \equiv \delta_{m n} |b_n|^2 \tag{8.10}$$
where $\Phi_n$ and $b_n$ have the same meaning as in
$$\Psi = \sum_n b_n \Phi_n {,} \tag{8.7}$$
the wave function of a system (each $\Phi_n$ is a wave function for $N$ particles contained in a volume $V$ and is an eigenfunction of the Hamiltonian of the system; the $\Phi_n$ are chosen such that together they form a complete orthonormal set). He also defines the density operator, but I'll omit that here.
I am not looking for a solution, just some guidance or a hint. I am having trouble coming up with an expression for the $\Phi_n$'s. For an individual electron, we can write
$$\left[ A \begin{pmatrix} 1 \\ 0 \end{pmatrix} + B \begin{pmatrix} 0 \\ 1 \end{pmatrix} \right] e^{i \mathbf{k} \cdot \mathbf{r}}$$
Can we just use this same wave function for this problem? If so, why? I'm thinking our coefficients $A$ and $B$ should satisfy $|A|^2 + |B|^2 = f$ for the electrons polarized in the direction of the beam and $|C|^2 + |D|^2 = 1 - f$ for the electrons polarized opposite to the direction of the beam.
Another thought: $|b_1|^2 = f$ and $|b_2|^2 = 1 - f$ in $(8.7)$, so that I would expect the density matrix to be a $2 \times 2$ matrix. But I really am unsure.
This is not homework, I am trying to work through problems in QSM in order to read some papers this summer. Any help would be greatly appreciated.
| Generally speaking:
*
*If you prepare a system in the pure state $|\psi\rangle$, then the density matrix of the system in that state will be $\hat \rho = |\psi\rangle\langle \psi|$.
*If you have $N$ preparation procedures which produce the system in states with density matrices $\hat\rho_1,\ldots,\hat \rho_N$ with probabilities $p_1,\ldots,p_N$, where those probabilities must obey $p_j\geq 0$ and $\sum_{j=1}^N p_j = 1,$ then the probabilistic procedure will be described by the density matrix $$\hat \rho = \sum_{j=1}^N p_j \hat \rho_j.$$
In your case you have two preparation procedures with probabilities $p_1=f$ and $p_2=1-f$, so your task is to produce appropriate descriptions of the $\hat \rho_j$ and to add them up correctly to get $\hat \rho$.
| {
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Is the string-net model Hermitian? In Kitaev and Kong's paper, they define the Hermitian inner product on morphism spaces in Eq. (11).
My question is that:
Given that F symbols satisfy the pentagon identity, does that the string-net Hamiltonian (13) is Hermitian follow from the Hermitian inner product on morphism spaces?
Is any related math theorem about the Hermiticity of Hamiltonian and the Hermitian inner product on morphism spaces?
| We should recall that the Hamiltonian is a construction that satisfies hermiticity. The Hamiltonian is not a derived mathematical object from some theorem.
If you follow Kitaev and Kong's exposition, then I think you are right. The hermiticity of the Hamiltonian results from the hermitian inner product of morphisms defined in eq.(11). If you follow the physical motivation of the "fix-point" Levin & Wen model(s), then hermiticity is enforced because we want the Hamiltonian to be hermitian, which in term defines a hermitian inner product on morphisms (this coincides with the defintion in Kitaev and Kong's paper).
| {
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How does rotation of fan blade produce wind? I observe that the blades are slanted to an angle, but if it produce a lower pressure in one region, is it not possible that the wind from both side will go in?
| Take a ceiling fan. If the blades were not slanted, as the instantaneous momentum of the rotation will be in the plane of the blades, only the thin sides would move air, very inefficiently.
Slanting the blades pushes air in a perpendicular direction to the plane of rotation , the air filling up the vacuum behind comes in the same direction as the one pushed, so an overall draft is established. The direction of the wind depends on the direction of the slant with respect to the direction of rotation.
A a draft is built up , directed where change of air is needed.
| {
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How to average the loudness of a audio file from its amplitude? I would like to average the loudness of different sounds so that their means could be used as a dependent variable in a statistical regression, but I have some troubles to understand how loudness works.
Assume that I have large number of different audio files that are all recorded in the exact same conditions, so that a same original sound would have the exact same frequency and amplitude in each file.
I saw that the loudness of a sound is proportional to the square of its amplitude, but also that $\mathrm{dB} = 10\log_{10}(\mathrm{Amp})$. Therefore I do not know which formula to use, and neither if I should first average the amplitude and then apply the formula, or apply the formula first and then average the dB.
If somebody has some answers or some literature to get me on the right track it would be very helpful.
| I don't think there's a generic physics answer to your question. It depends on what you're trying to accomplish in this analysis. You haven't said whether it's a psychology experiment, or a study of the effect of sound waves on materials, or something else. The reason we often use the dB scale for sounds is that the ear-brain system perceives loudnesses in terms of ratios, and the logarithm converts ratios to differences. "Loudness" is not a physics term. If it's going to have a formal mathematical definition, that would be psychoacoustics, not physics.
so that their means could be used as a dependent variable in a statistical regression
Both a mean and a linear regression are linear things. Whether it makes sense to do these linear operations on a variable depends on whether the variable is naturally structured in a linear way. For example, it makes sense to average celsius temperatures, but it doesn't necessarily make sense to average star ratings on rottentomatoes.
| {
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Interaction of Magnetic field with light Can there be any interaction of a varying magnetic field with light? (Please explain using electromagnetic waves as both are)
(Say we have an alternating current of 60Hz and He-Ne laser (632.8 nm wavelength)
What all different kinds of interaction may happen? In free space or in a dielectric medium (Just an example for the sake of analysis))
Thanks
| In matter, there is Faraday rotation.
In vacuum, hypothetical axions would also cause a rotation of the plane of polarization. Experiments have been done with superconducting magnets and ultra-high vacuum. This gave a zero-result, within experimental boundaries. (University of Rochester and Brookhaven)
| {
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Do the ladder operators $a$ and $a^\dagger$ form a complete algebra basis? It is easy to construct any operator (in continuous variables) using the set of operators $$\{|\ell\rangle\langle m |\},$$ where $l$ and $m$ are integers and the operators are represented in the Fock basis, i.e any operator $\hat M$ can be written as $$\hat M=\sum_{\ell,m}\alpha_{\ell,m}|\ell\rangle\langle m |$$ where $\alpha_{\ell,m}$ are complex coefficients. My question is, can we do the same thing with the set $$\{a^k (a^\dagger)^\ell\}.$$
Actually, this boils down to a single example which would be sufficient. Can we find coefficients $\alpha_{k,\ell}$ such that $$|0\rangle\langle 0|=\sum_{k,\ell}\alpha_{k,\ell}a^k (a^\dagger)^\ell.$$ (here $|0\rangle$ is the vacuum and I take $a^0=I$)
| Theorem: any operator $\mathcal O$ may be expressed as a sum of products of creation and annihilation operators:
$$
\mathcal O=\sum_{n,m\in\mathbb N} (a^\dagger)^n(a)^m c_{nm}\tag{4.2.8}
$$
for some coefficients $c_{nm}\in\mathbb C$.
This theorem can be generalised to field theory, where $a,a^\dagger$ are indexed by continuous parameters. The proof of the generalised theorem can be found on ref.1.
For completeness, we sketch the proof here. We proceed by induction. Given $\mathcal O$, we set
$$
c_{00}:=\langle 0|\mathcal O|0\rangle
$$
We now claim that if we are able to fix $c_{nm}$ for all $(n,m)\leq(\ell,k)$ with $(n,m)\neq (\ell,k)$ so that $(4.2.8)$ holds for all matrix elements with $n$- and $m$-particle states, then we can fix $c_{\ell k}$ so that the same holds true for the matrix elements with $\ell$- and $k$-particle states. This is easy to see, because sandwiching $(4.2.8)$ between $\langle \ell|$ and $|k\rangle$, we get
$$
\langle\ell|\mathcal O|k\rangle=\ell! k!c_{\ell k}+\text{terms involving $c_{nm}$ with $(n,m)\leq(\ell,k)$ and $(n,m)\neq(\ell,k)$}
$$
whence the claim follows. By induction, the theorem is proven.
$$\tag*{$\square$}$$
References.
*
*Weinberg - Quantum theory of fields, Vol.1, §4.2.
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Is there a material emitting light when changing phase? I think a substance which emits light when it gets vaporized could have numerous applications, for example in display technology or in leakage detection.
The only problem is: I'm struggling to find a explanation how this should work.
Light emission from atoms stems from the electrons changing there discrete energy states inside the atom.
On the other hand phase change is connected to inter-atom forces and thermo-dynamics.
(This should apply to molecules as well).
Is there a phenomenon, which somehow connects these two mechanism, so that a phase change is always accompanied with photon emission? Or is there a material / compound / element with this property?
As photon emission is always also energy emission I'd suspect this is more likely when changing from solid to liquid or liquid to gaseous. Optical photons and (more-or-less) standard conditions would be preferable, but I'm mainly interested whether such thing exists or could exist at all.
| There is heat emitted or absorbed during phase changes. This is infrared rather than visible light. This is because phase changes involve intermolecular interactions with lower energies than the atomic interactions associated with electrons jumping between different levels in atoms. So there are photons involved in phase changes but they are just not visible.
Just because these photons are emitted does not mean that they are easily observed (see the comment by @AndersSandberg that links evidence of observation of these phase transition photons nevertheless). In a mixed phase medium the photons emitted when bonds are broken may be almost immediately absorbed leading to kinetic energy increases by the free phase.
| {
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How come I can kick a football further when it's moving towards me? If a ball is moving towards me, I can kick it further than if I were to kick it if it was stationary. But surely if the ball is moving in the opposite direction, it should take more force to kick it the same distance as I am accounting for the initial movement.
Does this have a theoretical explanation, or is it that I can perform better technique if it's moving?
| A short answer to this: I will treat the event of the ball being kicked as an elastic collision between two bodies (the foot (with the body attached) moving with initial velocity $u_1$, and the ball approaching with negative velocity $u_2$.
The problem of elastic collisions of 2 particles is a problem from various physics courses, I won't derive it here, but I'll use the result for $v2$, the velocity of the football after the collision occured, given in Wikipedia:
\begin{align}
v_{2}={\frac {u_{2}(m_{2}-m_{1})+2m_{1}u_{1}}{m_{1}+m_{2}}}
\end{align}
The mass of the foot is bigger than the mass of the football, which makes the quantity $(m_2 - m_1)$ negative. As $u_2$ is negative as well, the velocity $v_2$ will increase for bigger absolute values of $u_1$.
Of course this argument is just an approximation to reality, where the foot is connected to the leg (and the leg to the body and so on). An easy way to model this would be to increase the mass of the foot, taking into account that it is attached to a leg and to a body.
| {
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Does an electron accelerated in a linear accelerator lose any energy? In a circular one, it would lose energy due to bremsstrahlung (synchrotron emission), but if it is accelerated from rest to ~GeV energies, does it lose any energy in this acceleration process?
| According to Larmors' formula in Lorentz invariant form the power loss $P$ of a accelerated particle is given by (cgs-units are used, $e$ is the electron charge, $c$ speed of light, $m_0$ the rest mass of the electron):
$$ P = - \frac{2}{3}\frac{e^2 c}{(m_0 c^2)^2}\left[ \frac{dp^\mu}{d\tau} \frac{dp_\mu}{d\tau}\right] $$
where $\tau$ is the proper time, and $p^{\mu}= (E/c, \vec{p})$ are the 4 components of the 4-momentum (E is the energy of the electron and $\vec{p}$ its 3-momentum). If the 4-momentum change rate is written out in its components we get:
$$ P = \frac{2}{3}\frac{e^2c}{(m_0 c^2)^2}\left[ \left(\frac{d\vec{p}}{d\tau}\right)^2 - \frac{1}{c^2} \left( \frac{d E}{d\tau}\right)^2\right] $$
Using $\frac{dE}{d\tau} = v \frac{dp}{dt}$ and replacing the proper time of the particle by the observer time $t$ by $d\tau = \frac{1}{\gamma}dt $ gives ( $\frac{1}{\gamma} = \sqrt{1-\beta^2}$ with $\beta =\frac{v}{c}$):
$$ P = \frac{2}{3}\frac{e^2c \gamma^2}{(m_0 c^2)^2}\left[ \left(\frac{d\vec{p}}{dt}\right)^2 - \frac{v^2}{c^2} \left( \frac{d p}{dt}\right)^2\right]\,\,\, (*)$$
In case of a linear acceleration we can set $\vec{p} = p\cdot \vec{e}_p$ where $p$ is the length of the momentum vector and $\vec{e}_p$ its time-independent (!) unit direction vector ($\frac{d \vec{e}_p}{dt}=0$). Then we get:
$$ P = \frac{2}{3}\frac{e^2c}{(m_0 c^2)^2}\left[ \gamma^2(1-\beta^2) \left( \frac{d p}{dt}\right)^2\right]$$
The term $ \gamma^2(1-\beta^2) =1$.
The parameter which in linear accelerators measures the acceleration typically is the energy gain per length, i.e. $\left(\frac{dE}{dx}\right)$ which is equal to: $\frac{dE}{dx} = \frac{dp}{dt}$. With this in mind we get:
$$ P = \frac{2}{3}\frac{e^2c}{(m_0 c^2)^2} \left(\frac{dE}{dx}\right)^2 $$
The energy gain per meter in a typical linear accelerator is $15$MeV/m, with this value one gets a radiation power in order of $10^{-17}$W (changing the formula to SI-units). So the answer is: Yes, principally there is energy loss in a linear accelerator, but it is so small, that
it can be neglected right away.
In the opposite case one is only interested in the radiation on a circular path, term $\frac{dp}{dt}$ can be set to zero in formula (*), as the length of the momentum vector does not change. However we can use:
$$ \left(\frac{d\vec{p}}{dt}\right)^2 = \vec{F}^2 = (m\cdot a)^2 = \left(\frac{mv^2}{R}\right)^2 $$
where for $a$ the centripetal acceleration $a= \frac{v^2}{R}$ was used.
If however the centripetal acceleration is plugged in we will finally get the well-known formula for the radiation power in a circular accelerator which is tremendously much larger than in case of the linear accelerator.
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Is potential Difference Really a Measure of Electromotive force? If I separate some amount of positive and negative charge a certain distance I will create some voltage. If I then separate the same amount of positive and negative charge a longer distance I will create an even bigger voltage since more work would've been needed to separate them.
Now, the charges with a larger potential difference will experience less of a force of attraction because they're farther apart, and the charges with a lower potential difference will experience a larger force of attraction since they are closer together.
Question 1: If voltage is a measure of the "push" acting on charge, then how can this be so?
Question 2: Voltage is proportional to current. But in this case I would think that the lower voltage scenario would produce higher current because there's a stronger force acting on the charge. What's going on? How does potential difference produce current?
| Let me offer an Old fashioned answer: Voltage is NOT like push! Voltage is more like pressure in a fluid, where now “pressure gradient” causes push. Electric fluid gets pushed from high pressure to low. So voltage gradient or volts per metre gives electric field or force per coulomb. Hope this helps too.
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Does a vibrating string produce changes in tension in the string? A taut string anchored at both ends increases in tension as the string is displaced to a side. When released the string will vibrate at whatever frequency to which the system is tuned.
If this were a simple standing wave it seems that tension would drop as the string moved to the neutral position and increase again as it deflected on the opposing side. Harmonics in this case would not change the behavior.
Is this the way an actual string behaves, or is there some other oscillation pattern that allows tension to remain relatively constant even as the string vibrates? Does a vibrating string always produce changes in tension?
| String vibration, as well as vibration of other mechanical oscillators (springs, pendulums, etc.) involves periodic back and forth transitions between its kinetic and potential energy.
The potential energy of a string is associated with its tension, the kinetic energy - with its velocity.
So, if the tension of a string stayed the same, its potential energy would not be changing, there would not be any transitions between potential and kinetic energy and, therefore, there would not be any vibrations.
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How can there be single electron, while we observe many simultaneously? Wheeler and Feynman's idea of single electron universe says that all the electrons and positrons are in fact a single particle bouncing forward and backward in time.
I don't get the nature of bouncing phenomenon here. We observe multiple electrons simultaneously, so does this mean that the bouncing occur in infinite speed and no time? Is this possible? How should we understand that "bouncing"?
| It was just a "what if" discussion, as this link shows;
Feynman was struck by Wheeler's insight that antiparticles could be represented by reversed world lines, and credits this to Wheeler, saying in his Nobel speech:
“ I did not take the idea that all the electrons were the same one from [Wheeler] as seriously as I took the observation that positrons could simply be represented as electrons going from the future to the past in a back section of their world lines. That, I stole!
It never was a complete model , as there is an asymmetry between electrons and positrons in numbers, and all those other pesky particles discovered since then. But it proved a useful idea of Wheeler with use in Feynman diagrams.
So understanding that "bouncing" , is understanding how to read Feynman diagrams and decide on the limits in the integrals they represent when a calculation is done.
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Breaking of a commutator involving Dirac spinors and gamma matrices I'm trying to understand a particular step in the solution to problem 27 in THIS solution sheet. By the middle of the page, they start with the simplification of this expression
$$\left[s^{\mu}\left(x\right),s^{\nu}\left(y\right)\right]\overset{(1)}{=}e^{2}\left[\overline{\psi}\left(x\right)\gamma^{\mu}\psi\left(x\right),\overline{\psi}\left(y\right)\gamma^{\nu}\psi\left(y\right)\right]\overset{(2)}{=}\cdots$$
and I don't understand how they break this commutator at step (2). Is their algebra correct? Using $\left[AB,C\right]=A\left[B,C\right]+\left[A,C\right]B$,
and identifying $A=\overline{\psi}\left(x\right)\gamma^{\mu}$ and
$B=\psi\left(x\right)$, shouldn't it be
$$
\overset{2}{=}e^{2}\overline{\psi}\left(x\right)\gamma^{\mu}\left[\psi\left(x\right),\overline{\psi}\left(y\right)\gamma^{\nu}\psi\left(y\right)\right]+e^{2}\left[\overline{\psi}\left(x\right)\gamma^{\mu},\overline{\psi}\left(y\right)\gamma^{\nu}\psi\left(y\right)\right]\psi\left(x\right)
$$
The same kind of misplacement of $\gamma^{\nu}$ seems to happen again at step (3) when they break the first commutator. In here, if they were using
$\left[A,BC\right]=\left\{ A,B\right\} C-B\left\{ A,C\right\} $ with
$B=\overline{\psi}\left(y\right)$, $C=\gamma^{\nu}\psi\left(y\right)$,
I guess one would have
$$
\left[\psi\left(x\right),\overline{\psi}\left(y\right)\gamma^{\nu}\psi\left(y\right)\right]=\left\{ \psi\left(x\right),\overline{\psi}\left(y\right)\right\} \gamma^{\nu}\psi\left(y\right)-\overline{\psi}\left(y\right)\left\{ \psi\left(x\right),\gamma^{\nu}\psi\left(y\right)\right\}
$$
instead of what's there. I must be missing something, but I can't see what! Could you please help?
| Using the formula
\begin{equation}
[AB,CD] = A\{B,C\}D - AC \{B,D\} + \{A,C\}DB - C \{A,D\}B \quad,
\end{equation}
and setting
\begin{equation}
A = \bar{\psi}(x) \quad,\quad
B = \gamma^\mu \psi(x) \quad,\quad
C = \bar{\psi}(y) \quad,\quad
D = \gamma^\nu \psi(y) \quad,
\end{equation}
one gets (spinor indices suppressed):
\begin{equation}\begin{alignedat}{9}
&[\bar{\psi}(x) \gamma^\mu \psi(x), \bar{\psi}(y) \gamma^\nu \psi(y)]
\\&=
\bar{\psi}(x)\{ \gamma^\mu \psi(x),\bar{\psi}(y)\}\gamma^\nu \psi(y) - \bar{\psi}(x)\bar{\psi}(y) \{ \gamma^\mu \psi(x),\gamma^\nu \psi(y)\}
\\&+ \{\bar{\psi}(x),\bar{\psi}(y)\}\gamma^\nu \psi(y) \gamma^\mu \psi(x) - \bar{\psi}(y) \{\bar{\psi}(x),\gamma^\nu \psi(y)\} \gamma^\mu \psi(x)
\\&=
\gamma^\mu\bar{\psi}(x)\{ \psi(x),\bar{\psi}(y)\}\gamma^\nu \psi(y) - \gamma^\mu\bar{\psi}(x)\bar{\psi}(y) \{ \psi(x), \psi(y)\}\gamma^\nu
\\&+ \{\bar{\psi}(x),\bar{\psi}(y)\}\gamma^\nu \psi(y) \gamma^\mu \psi(x) - \bar{\psi}(y) \{\bar{\psi}(x), \psi(y)\} \gamma^\nu\gamma^\mu \psi(x)=0\quad.
\end{alignedat}\end{equation}
Actually, a more general statement is true $-$ see the bottom line on page 5 here.
For the detailed solution see Problem 11 here.
| {
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Weird Reflection Pattern in Reading Glasses While fidgeting with a pair of reading glasses, I noticed a strange reflection pattern (shown in video and photo).
I would appreciate it if anyone that knows more about this could help me figure out why there were eight dots in the reflection instead of four, and why there were different colors when all the light sources were an orange/yellow/warm color.
There were eight dots in each frame, four orange, four green. They made a cube shape, each one of the dots being a corner of the cube. The same image could be seen in each frame, and if I focused my eyes it looked 3d.
The picture below shows the light source that the reflection came from, just four ordinary ceiling lights. I know it couldn't have been anything else because it was night, and to test this I turned off all other lights and it was still there. In fact it was then even brighter. I looked into the glasses again in the morning, the effect was still there.
Thanks,
Ella
| You might want to investigate your glasses further with some soap water, like in the photos below. The reflection of a window at the front surface and its reflection at the back surface are green due to the anti reflection coating of the lens, if it is dry. Both reflections could be identified by wetting the front or the back surface with soap water. The soap water forms a thin film that changes the color of the reflection.
| {
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Why isn't work done when carrying a backpack with constant velocity? Here's an excerpt from my textbook:
In the following situations, state whether work is being done, and why or why not.
A) A person carrying a backpack walking across a floor
B) A person shoveling snow across a driveway at a constant speed
The solutions are:
A) No work is being done on the bag because the direction of motion is perpendicular to the direction of force due to gravity
B) Work is being done on the shovel because the direction of motion and the direction of the applied force is the same
I have no issue with situation B, but situation A bothers me. Although I agree that no work is being done by the forces of gravity and the force applied to hold it up, is it not the case that the person is applying a force forward to the bag as they walk, and therefore work is being done by that force? To make things a bit more uniform (since walking is a jerky motion), presuming that the person wearing the bag is riding a bike at a constant speed, that means that they have to apply a force that balances wind resistance.
I might say that since there is no net force there is no work being done, but then situation B would have no work being done as well, since the shovel/snow are moving at a constant speed.
Where did I misread this?
| After reading your statements, I think you may also look at this from the point of view of the work-kinetic energy theorem ($W_{net}=\Delta KE$).
In the first scenario, the bag is moving as constant speed, then $\Delta KE = 0$ which implies no work done. From the point of view of forces acting on the bag, $\vec{a}=\vec{0} \Rightarrow \vec{F}_{net}=\vec{0} \Rightarrow W_{net}=0$
I don't like the solution given by the book because when determining total work done on an object, you should consider all the forces acting on that object, which includes the force of the person holding the backpack against gravity as well. If you want to determine work done by a single force, that's fine as well, but gravity is not the only force acting on the backpack. Sorry you have a poor textbook.
| {
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Expansion wave in a hermetically sealed room Suppose there is a human in a hermetically sealed room who takes in a small amount of air without letting it out. The room is initially at atmospheric pressure. The suction is similar to the case in which a piston is withdrawn to generate an expansion wave in the surrounding media. If I measure the pressure in a region furthest away from the human, will it still be atmospheric? How will equilibrium in pressure be achieved in this case compared to the case when the room is not hermetically sealed (opened to atmosphere)?
| Let's do a simple analysis. Let quantities with subscript "r" and "l" correspond to the room (excluding the lung) and the lung respectively. Since in a sealed room total mass of air $m_0$ must remain constant, we have: $$m_r+m_l=m_0$$
Assuming that temperature is uniform and steady, ideal gas law gives:
$$p_rV_r=m_rRT\\ p_lV_l=m_lRT\\ \therefore\quad p_rV_r+p_lV_l=(m_r+m_l)RT=m_0RT,~\textrm{a constant}\\$$
If initially the pressure was $p_0$ everywhere (same inside room and lung) then $p_0V_0=m_0RT$. Thus: $$p_rV_r+p_lV_l=p_0V_0$$
Now we assume $V_0=V_r+V_l$; this is a good assumption as far as breathing is concerned. Above equation becomes:
$$(p_r-p_0)V_0+(p_l-p_r)V_l=0$$
If $p_0$ is atmospheric pressure, the question whether room pressure is above or below atmospheric subsequent to inhalation concerns the sign of $(p_r-p_0)$. Its sign is opposite to that of $(p_l-p_r)$. If pressures inside the room and lung were equalized at the end of inhalation, i.e. $p_l=p_r$, then room pressure remains atmospheric, i.e. $p_r=p_o$. Maintaining unequal pressure requires external force on the lung, and a non-zero pressure difference can result if breathing was stopped before pressures were equalized. In this case, higher (lower) lung pressure compared to that of room coexists with below (above) atmospheric pressure in the room.
To obtain the result for a room which isn't sealed but open to atmosphere (or a sealed room which is very big compared to the lung), we take the limit $V_0\to\infty$. Since $(p_l-p_r)V_l$ is a finite quantity, we must have $p_r\to p_0$. Therefore in a room which isn't sealed pressure will practically always be atmospheric.
| {
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How to mathematically prove that point charge and infinitesimal volume charge are same? In electrostatics, while deriving certain elementary equations, I have seen all the books just assuming that point charge and infinitesimal volume charge are same.
How can we rigorously, mathematically and formally prove that point charge and infinitesimal volume charge are indeed the same?
| The Dirac delta function formalizes this equivalence mathematically. It took mathematicians some time to make this object rigorous and led to the development of the theory of distributions.
Addendum
$\def\vr{{\bf r}}$The Dirac delta function is defined such that
(1) $\delta(\vr) = 0$ everywhere except at $\vr={\bf 0}$, where it diverges, and (2)
$\int_V \delta(\vr) dV = 1,$
where $V$ is any volume containing the origin.
Now consider the charge density
$$\rho(\vr) = q\delta(\vr).$$
This is a charge density which is zero everywhere except at the origin, where it diverges, and
$$\int_V \rho(\vr)dV = q,$$
that is, the total charge is $q$.
Thus, this is the charge density of a point charge with charge $q$ located at the origin.
All of this can be formalized rigorously using the theory of distributions.
| {
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Relative size of the 2 tidal bulges I am interested in making a simplified model that represents only the effect that the moon has on Earths tides.
In this model I am going to assume that the earth is completely spherical (with no continents or surface variation) and covered entirely by a single ocean of water.
In simple terms I want to know if the tidal bulges on the earth are different sizes or if they are of equal size?
In many diagrams it appears that the bulges are symmetrical, but if the side of the earth facing the Moon is experiencing a greater gravitational force, and the far side a weaker gravitational force, would that not result in the side facing the Moon having a proportionally larger tidal bulge?
I do appreciate that most diagrams are purely illustrative and that the actual difference in size would be too small to practically depict on a visual diagram.
I have attached a diagram, that while exaggerated, I am wondering if it is more accurate that showing the bulges as equal in size?
| The tidal force on the point on the earth nearest to the moon, is roughly 1/20 greater than than on the point furthest from the moon, which is what I think was being asked, not 1/90 as stated in the other answer.
The moon is roughly 60 earth radii from the earth, so if we call the gravitational force of the moon on the centre of the earth 1/60², the tidal force on the nearest point is 1/59²−1/60², the tidal force on the furthest point is 1/60²−1/61², both of which are roughly 1/30 of the gravitational force.
The difference between the tidal force on the nearest and furthest point is thus (1/59²−1/60²)−(1/60²−1/61²) which is roughly 1/20 of the tidal force. This should show up on earth tides, if they could be measured accurately enough.
You can calculate tidal forces using the centrifugal force, as recommended by others, but it is much easier to use differential gravity as I have done.
| {
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Ionization Energy Confusion In my Barron's SAT Physics book, it states that ionization energy is equal to the absolute value of the ground state energy. This doesn't make sense to me because ionization energy is the minimum amount of energy required to eject an electron from an atom, which I would think would require an ionization energy greater than the absolute value of the ground state energy. If the ionization energy is equal to the absolute value of the ground state energy, then wouldn't that mean the electron is now situated on the outer most energy level with energy value 0 eV (electron still in atom)? I speculate that the reason for why I am wrong is that the 0 eV energy level is actually imaginary, which I have concluded based the 0 eV ring being dashed in my book's diagram of the energy levels as shown:
However, in this Khan Academy video https://www.khanacademy.org/science/physics/quantum-physics/atoms-and-electrons/v/atomic-energy-levels at 9:22 the narrator states that if an electron has more than zero energy then it is ejected, which is what I would think would be true.
| The "absolute" energy of the ground state is simply the energy of the ground state with respect to the vacuum. So the 0eV state you describe is a free election that is at rest in the vacuum, it is not a bound state, since its radius is essentially infinity (in comparison to the ground state radius).
In other words, the ionization energy is the energy required to remove an electron from the ground state to the vacuum state where it has zero kinetic energy.
| {
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Can you isolate a system from conduction and convection, but not radiation? Can you create a 'box' that isolates the inside from conduction and convection, but lets infrared and visible light pass?
| Certainly. This is how solar heat collectors work. You surround the thing you wish to heat with double-paned glass windows. Visible and (some) IR go through the glass by radiation, and the glass blocks conduction and convection losses.
Please note that when the object inside the glass box gets hot enough, it will begin to re-radiate in the infrared, and if the glass used in the construction of the box happens to be transparent to IR then you'll start to lose some of the heat that you have collected. This means that for best performance, the glass itself has to be chosen to minimize this. The tradeoff then is that if the glass blocks IR losses, it also blocks incoming IR from outside.
The IR re-radiation rate scales strongly with the temperature of the hot object (as T^4) so if the temperature rise is moderate and not extreme, then those losses will not prevent this technique from working.
| {
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When will velocity and acceleration vectors be perpendicular?
Suppose a particle is moving in the $xy$ plane with
$$x=at, \quad y=at(1-bt),$$
where $a$ and $b$ are positive constants. When will the velocity vector and acceleration vector be perpendicular?
I know that these vectors are perpendicular in circular motion. Should I then use the circle equation
$$ x^2+y^2 = r^2 $$
then substitute $r$ with centripetal acceleration
$$ a_c = v^2/r $$
and then substitute for $v$? But I don't see how I will get an equation that has something to do with vectors.
| No, circular motion is only one of the cases where the acceleration and velocity are perpendicular, you should definitely not be using that.
You have a position vector, $\vec r(t) = [a~t,~~a~t(1-b~t)]$.
Its first and second derivatives are velocity and acceleration vectors, and you want to take the dot product of those two vectors, to see whether that dot product is zero:$$x'(t) x''(t) + y'(t) y''(t) = 0.$$ The first term is trivially zero from the second derivative of $x$, the second derivative of $y$ is a constant that can be divided out unless $a=0$ or $b=0$.
To restate that more physically, your acceleration is constant, and in the y direction. So you are looking for a velocity which is purely in the x direction.
| {
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Why are photo electrons emitted instantly from metal surface just nanoseconds after the light falls upon it? Why are photo electrons emitted instantly from metal surface just nanoseconds after the light falls upon it? How does the quantum theory of radiation explain it? Why can't classical physics explain this?
| The photoelectric effect, which is what you are describing, is one of the basic experimental effects that forced the invention of Quantum mechanics.( The other reasons were black body radiation and the atomic spectra.)
Classically there should not be this behavior, because classically the frequency of the light should not play a role in the ejection of electrons, only the energy of the classical light beam was expected to affect the ejection of the electrons.
Nanoseconds are not "instantly". It is within the $Δ(t)$ allowed by the quantum mechanical solutions for the specific interaction : photon hitting atoms and ejecting electrons, releasing them from the atomic/molecular binding.
| {
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Why is a fermion field complex? The Lagrangian of a fermion field is
\begin{equation}
\mathcal{L} = \overline{\psi} (i\gamma_{\mu} \partial^{\mu} - m)\psi
\end{equation}
It is said that the fermion field $\psi$ is necessarily complex because of the Dirac structure. I don't quite understand this. Why is the fermion field complex from a physical point of view? A complex field has two components, i.e., the real and imaginary components. Does this imply that all fermions are composite particles? For example, an electron is assumed to be a point particle that does not have structure. How can it have two components if it is structureless?
| Fermions are not necessarily complex. Majorana spinors fulfill a reality condition $\Psi=\Psi^\mathcal{C}$, and in the Majorana basis a Majorana spinor is manifestly real. Clearly, only neutral fermions can be described by Majorana spinors, and neutrinos are candidates for such fermions, even though at present it is not clear whether they are Dirac or Majorana particles.
I think that there are some issues with your question. The answer above is on the title "Why are fermions complex". In your question, you suggest that all fermions need to be described by the Dirac equation, which is not the case. There is also the Weyl equation, and some refer to the field equation of Majorana spinors as Majorana equation. And I cannot follow the statement on the degrees of freedom vs. structure either, photons are described by a vector and do not have a "structure" either, at least according to our present knowledge.
| {
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What effect does gravity have on a spinning object? If gravity was the only force present would gravity, beside pulling, also stop the rotation / spin of a small object its pulling towards it over time?
What would be the effect on an uneven object and what would be the effect on a perfect sphere?
| The gravity acting on a body, spherical or not, is, by definition, equivalent to a single force applied to the center of gravity (and, therefore, mass) of that body, which means that it won't produce any torque relative to the com and, therefore, won't cause any angular acceleration.
Hence, the gravity should have no effect on the rotation of the body.
| {
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Rectilinear image projection in image stitching For the pinhole camera model, the mapping from 3D to 2D coordinates is described by a perspective projection(rectilinear projection).
However for the image stitching application, perspective projection will bring some problems. Its primary disadvantage is that it can greatly exaggerate perspective as the angle of view increases, leading to objects appearing skewed at the edges of the frame.
The input images:
Stitched image:
The images are from the link:
https://www.cambridgeincolour.com/tutorials/image-projections.htm
I can't figure out the reason that the left port almost doesn't change much but the right port undergoes much change for the the third image. The overlapping part of the second and third images doesn't change a lot.
Edit:
I draw an diagram to show capturation relationship between two images.
The scene contains two black lines. The red square indicates the first capturation position and the blue square shows the second capturation position. To match two images, the second image has been warped as the following figure shows. It looks like the same transformation that the hall example undergoes. I don't understand why such transformation is required to match the two images.
| The rectilinear projection keeps all straight lines straight. As a result, the image stretches increasingly stronger toward the far edge. The answer to your question is that the right side of the right image is much closer to the right edge of the stitched image and for this reason is stretched more.
In contrast, some other projections avoid this effect, but anavoidably at the expense of at least some straight lines becoming curved. Examples include the obvious fish-eye projection and the cylindrical projection shown on the same site. Note that, while the image is not nearly as much stretched on the sides, the roof line is curved, unlike in the rectilinear projection in your question:
These are just natural creative trade-offs of geometric optics.
| {
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If object size increases, shouldn't magnification decrease by formula According to the magnification formula, magnification is the image size by object size
According to this if object comes near and its size increases shouldn't magnification decrease?
| Magnification of the focused object can be shown to be:
$m = \frac{f}{s-f}$,
where $f$ is the focal distance and $s$ distance from the object to the lens. So if you keep your subject in focus as it comes closer, magnification will indeed increase, not decrease.
In case you only focus the lens once and then move the object around without changing that focus, there is a bit different formula for a defocused object:
$m_d = \frac{s m}{s_d}$,
where $m$ is the magnification at the focused point, $s$ the distance of that focused spot, and $m_d$ and $s_d$ magnification and distance at the defocused spot. As you can see, the relationship still stays the same: magnification increases as the object gets closer.
| {
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Photomulitplyer Tubes and NaI(TI) crystals in Gamma Camera, Energy and spatial resolution I am trying to understand how the PM tubes and NaI(TI) crystals are used for the energy and spatial resolution.
I understand the principle behind how the crystals and PMT work, but very confused to how they relate to the different resolutions.
In one book I am reading it talk about the intrinsic and extrinsic resolution which the in simply terms, with and without a collimator.
I am quite new to this type of technology and if possible could someone explain it in laymen’s terms, every book I have tried to read so far seem to jump straight in the deep end, and I cant seem to build a visualise of what is actually happening.
| For nuclear medicine imaging, a good scintillator will produce flashes of light that are proportional to the energy deposited in the scintillator. A gamma ray that deposits more energy will produce a brighter flash of light. This is where the energy resolution of the detector comes from.
Spatial resolution is determined by how much the light flash spreads out before being detected by the photomultipliers. In a gamma camera, this is largely determined by the thickness of the scintillator layer. A thinner detector layer will have slightly better resolution than a thicker layer, but lower detection efficiency.
| {
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What's the cause of sideways pressure (and why pressure is scalar)? Since I think the two questions are related that's why I decided to put them together. Why pressure is a scalar ? It's the ratio of two vectors (force and directed area) right ?
Also, suppose you have a cylinder filled up with water, and at a point $P$ at distance $h$ from the surface, I find it intuive to understand that there would be a horizontal pressure because of the weight of the water in a column with height $h$ erected from $P$, but I don't understand why there should be a sideways pressure, with the same value as the downward pressure at $P$ too ? What's the cause for it ?
Why the pressure is same in all directions ?
(Please don't use stress tensors/linear algebra to answer my questions since I don't understand them)
| It's because water molecules are slippery. If they were not (e.g. compare with atoms in a solid) then the pressure would not be same in all directions.
When some higher up molecule gets pushed down onto one below, it is very unlikely for them to happen to be exactly at a point where the top one could balance on the one below. Instead it slides off to the side. So this will continue until the other molecules off to the side push back against the ones coming down and slipping off. The whole thing rapidly results in the forces getting to act in all directions and balancing one another: the water will only stop moving when the forces do balance. If they do not balance then the water will slip again and flow until the forces get balanced again. And that means balanced to the side as well as up and down.
The reason for the exact balance in an ideal fluid is that there is nothing to prevent this slipping of one molecule over another. They have no sheer force, no stickiness. In a real fluid there is a bit of stickiness, called viscosity, and then the horizontal pressure does not always have to be equal to the vertical pressure at any given point.
| {
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If gravity reaches infinite intensity on the all event horizon surface, doesn't this fact exclude the black holes central singularity? Aren't this two singularities an excess or an unnecessary redundancy?
| A comment to the question states it is motivated by my answer to Surface gravity and mass of a black hole. Specifically that answer explains that the gravitational acceleration experienced by a user hovering at a distance $r$ from the black hole is:
$$ a=\frac{GMm}{r^2}\frac{1}{\sqrt{1-\frac{r_s}{r}}} \tag{1} $$
and this does go to infinity as $r \to r_s$. But this does not mean that the gravity is in any sense infinite at the event horizon. All this result does is indicate how dangerous it is to draw conclusions from the experience of any particular observer. As Alfred points out in his answer, the spacetime curvature remains finite at the horizon, and indeed is finite everywhere except at the singularity (where it is undefined).
A convenient measure of the curvature is the Kretschmann scalar:
$$ K = \frac{48G^2M^2}{c^4r^6} \tag{2} $$
And this is finite for all $r \gt 0$. Alternatively another measure of the gravity at the horizon is the surface gravity:
$$ \kappa = \frac{1}{2r_s} \tag{3} $$
which again is finite.
The question is then why equation (1) gives an infinite resut while equations (2) and (3) give finite results. Speaking loosely, this is due to the time dilation experienced by an observer hovering close to the event horizon. If we label the time measured by the hovering observer by $\tau$, while the time we measure far from the black hole is $t$, then the time dilation is given by:
$$ \frac{\tau}{t} = \sqrt{1 - \frac{r_s}{r}} \tag{4} $$
As $r \to r_s$ we find the time experienced by the hovering observer $\tau \to 0$. This is the source of the (misleading) claim that time stops at the event horizon. If you'll allow me a rather hand waving argument, acceleration has units of:
$$ \textrm{acceleration} = \frac{\textrm{distance}}{\textrm{time}^2} $$
In the hovering observer's frame time $\tau$ is reduced due to the time dilation so the $1/\tau^2$ term is increased, and indeed goes to infinity at the event horizon. This is the reason the acceleration measured by the shell observer goes to infinity as the observer approaches the horizon.
The surface gravity that I mentioned in equation (3) is in effect the infinite acceleration measured by the shell observer but corrected for the time dilation, so it is a more physically realistic measurement of the gravity at the horizon. You may note that for arbitrarily large black holes this surface gravity can be arbitrarily small.
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Why the pressure at siphon 1 atm at the bottom of the pipe?
I assume the area of horizontal cross section of beaker at the top to be way larger than the area of horizontal cross section of pipe
I think at the same depth pressure of a specific liquid which are connected to each other, as in manometer, should be same. But, in siphon that does not seem to be the case, which is the reason for my question.
Why at the bottom of the pipe of siphon pressure is 1 atm? I know that it is said that pressure is provided by atmosphere. But, why do not the pressure increase because depth increases?
| As @Chet Miller shows, friction effects explain how the pressure at both free surfaces is the atmospheric pressure. One more point. If the elevation at the top of the siphon is too high, the pressure drops to the saturation pressure of the liquid at its temperature, the liquid flashes to vapor and the siphon ceases to function.
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What are the types of infrared scattering? There are various types of scattering of electromagnetic wave. what type of scattering involves infrared electromagnetic wave?
| The question is rather broad, so I will provide a broad answer.
It really depends on the energy level structure of the scatter (atom? molecule? free electron?). Classically, scattering happens because the incoming EM wave induces and drives an oscillating dipole moment in the scattering object, and that oscillating dipole moment radiates. The strength of the induced dipole moment depends on the polarizability of your scatterer at the incident frequency, and of course that depends on the frequency detuning between the incident frequency and the energy spacings between different states.
So what kind of energy level structure is infrared EM relevant for?
Typically, atoms have transitions in the visible to near-infrared frequency.
For free electrons, there is no internal structure. But the incident EM field can oscillate the free electrons and make them radiate. At low frequency (low energy), there is Thomson scattering, and the scattering cross-section is independent of wavelength, so you can create Thomson scattering with your infrared light source. Note that Thomson scattering is the low-energy limit of Compton scattering, which happens at the relativistic scale (and requires X-ray). Finally, the vibrational energy level spacings in molecules is most likely in the infrared region, so you expect to see prominent scattering with infrared light with molecules.
There are also two prominent scattering processes: Rayleigh (elastic, final state same as initial state) and Raman (inelastic, final state different from initial state). There is also Bragg scattering, which refers to the change in the final external state (external meaning motional), whereas Raman usually refers to change in the internal state. Raman scattering is an invaluable tool for studying molecules.
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Why $\rm Pt$-$\rm Ir$ Alloy or Tungsten is used for the tip in scanning tunneling microscopy? Just want to know the properties which qualify these materials to be used as the same.
| "Role of Tip Material in Scanning Tunneling Microscopy", C. Julian Chen, https://www.cambridge.org/core/journals/mrs-online-proceedings-library-archive/article/role-of-tip-material-in-scanning-tunneling-microscopy/B32182D2E29CDC6656348326D7AF243F :"In this paper, we show that atomic resolution in scanning tunneling microscopy (STM) originates from $p_z$ or $d_{z_2}$ states on the tip. Consequently, only a limited selection of tip materials can provide atomic resolution: $d$-band metals, for example, Pt, Ir, Pd, Rh, W, Mo; semiconductors that tend to form $p$-like dangling bonds, for example, Si."
See, in particular, Fig. 1 in this article.
EDIT (5/30/2019): Some additional relevant information from http://www2.cpfs.mpg.de/~ernst/ernst_da.pdf: "One commonly used tip material is platinum iridium (Pt-Ir). This alloy is particularly suitable for the use under ambient conditions because Pt is relatively inert to oxidation. A fraction of Ir is convenient to make the tip harder. Another advantage is the uncomplicated preparation. With some experience, cutting a piece of Pt-Ir wire with a punch or a pair of scissors (see figure 3.1) gives sufficiently sharp tips to achieve atomic resolution. The success rate is rather low, though." Some weak points of the material are also discussed there.
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$SU(5)$ representation and higher anti-symmetric traces In Zee QFT book v2 p.411 eq.16-17, he shows the SU(5) gauge theory anomaly cancellation by the following:
The 1st line in fundamental of SU(5)
$$
tr(T^3)=3(+2)^3+2(-3)^3=30,
$$
is easy to follow, which sums over 3 U(1) charge 2 fermions and 2 U(1) charge -3 fermions, with the cubic polynomial for the anomaly.
The 2nd line in anti-symmetric 10 of SU(5), my understanding is the following
$$
tr(T^3)=3(2+2)^3+6(2-3)^3+(-3-3)^3=-30,
$$
, which sums over 3 U(1) charge 2+2=4 fermions, 6 U(1) charge 2-3=-1 fermions,
and 1 U(1) charge -3-3=-6 fermion with the cubic polynomial for the anomaly.
I kind of give an answer for the above. But I wonder whether there is another easier or succinct way to interpret this
$$
tr(T^3)|_{5^*}
$$
and
$$
tr(T^3)|_{10}
$$
and also the $T$ generator for them in the precise representation theory manner?
| Yes, there is a representation-theoretic way to do this in one step. Zee is taking a slightly different approach, showing the cancellation of the $U(1)^3$ anomaly where $U(1)$ is the hypercharge subgroup of the SM gauge group embedded in $SU(5)$. But actually it's easier to do the whole calculation in one go, looking at the whole gauge group $SU(5)$.
We want to avoid an $SU(5)^3$ gauge anomaly, which means we need to consider triangle diagrams with three $SU(5)$ currents, i.e. evaluate the correlator $\langle j^\mu_a j^\nu_b j^\rho_c \rangle$. For fermions in the loop transforming in a representation $R$, the result is proportional to
$$\text{tr}(T^a_R T^b_R T^c_R) = \frac{i}{2} T_R f^{abc} + \frac14 d^{abc}_R, \quad d_R^{abc} = 2 \text{tr}(T^a_R \{T^b_R, T^b_C\})$$
where I replaced $T^a_R T^b_R$ with the average of their commutator and anticommutator, and $T_R$ is the Dynkin index. One can show the first term does not contribute to the anomaly. As for the second term, there is a unique totally symmetric rank $3$ tensor associated with the algebra up to scaling, so
$$d^{abc}_R = A(R) d^{abc}_F$$
where $F$ stands for the fundamental representation, and $A(R)$ is called the anomaly coefficient.
The total anomaly is proportional to the sum of all the anomaly coefficients. By definition,
$$A(5) = A(F) = 1.$$
From this fact we can construct other anomaly coefficients by the results
$$A(R) = - A(\bar{R}), \quad A(R_1 \oplus R_2) = A(R_1) + A(R_2), \quad A(R_1 \otimes R_2) = A(R_1) d(R_2) + d(R_1) A(R_2)$$
where $d(R)$ is the dimension of the representation $R$. By the first rule,
$$A(\bar{5}) = -1.$$
As for $A(10)$, you need to invoke an additional rule, which is that if something is taking more than two minutes you can look it up, because some mathematician has already computed it for you. (Now you can look them up too, since you know the name!) I looked up $A(10) = 1$, so
$$A(\bar{5}) + A(10) = -1 + 1 = 0$$
as desired. For more details, see Srednicki, Schwartz, or a representation theory book like Ramond.
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Newbie question: Atom identity. How can you talk about two electrons if electrons are identical? How can you talk about two electrons if they are identical (indistinguible)? Does it make sense to let an electron to have an identity by itself?
If they are on diferent places the place they are is a diference (they are not identical).
The act of labeling, naming them make them distinct. If I say let A be an electron and B another one, then they are distingible by name...(they are not identical)
Edit: if electrons are indistinguible by principle, then you cannot name them properly. A tick after you name one of them it could be any other electron in the universe or not exist anymore. You could say let this be electron A, but a tick after, A means nothing. The most you can say is at some point in time there were A and that "electronism" would remain, but this "electronism" would emerge in diferent form as long as "electronism" holds.
I think that the indistinguibility property makes the interpretation of a particle who has an identity and live forever hard to follow. One need to create strange ideas like "it follows all the paths" or "there are several paralel universes"...
| We can do an experiment that distinguishes a box with two electrons in it from a box with just one electron in it. Just measure the electric field near the box and the field near the two electron box will be larger.
We cannot do an experiment that distinguishes between a box with electron A on the left and electron B on the right from a box with the electron A and B swapped. This is what indistinguishability means.
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Stars filling up the night sky Due to the huge number of stars in the universe, will there ever be a time that the night sky is filled up completely with stars such that the night sky is as bright as it is in the daytime?
| No.
The fact that the night sky is not currently filled up with stars is due to the finite speed of light, so that the light of very distant stars hasn't reached Earth yet. However, the universe is expanding, and that expansion is accelerating. This expansion is not bound by the cosmic speed limit of the speed of light and can, in fact, expand faster than light. Thus as the universe continues to accelerate its expansion, at some point the expansion of the regions of space occupied by very distant galaxies will, relative to us, be faster than light and thus the light from those galaxies will never reach us.
Your question is very much along the lines of Olbers' Paradox, and I encourage you to research that further (and specifically the paradox's resolutions) if you want to know more.
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Why arent black holes bright due to the bent light from nearby stars? If black holes change the path of incoming light in a gradual way (from very slight changes to 180 degrees), only depending on how close the light is passing by, the black hole external layer (where a perpendicular light beams are curved but are not doomed to fall into the black hole) would in fact be acting as a light reflecting body since the incoming light from any nearby star (with light beams arriving at all possible distances to the center of black hole) would be reflected in all possible output directions.
In that case, why arent black holes bright due to the contribution of the reflected light from all nearby stars?
| Black holes make gravitational 'lenses'. Lenses don't just bend light toward an observer, they also bend light away. There's situations where a background star can be seen in duplicate, but also where a background of black void can be seen
in duplicate. There is no net light-enhancement.
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What does electrical potential at a point mean? From my understanding, potential difference (or voltage) between point A and point B is the difference in electrical potential at the two points. The potential difference is also, the work done per unit charge in moving charges from point A to point B.
But apparently, the potential at one point (e.g point A) is also measured in voltage? How is it possible that at point A, there is work done per unit charge in moving charges from point A to point A?
| In the definition of electric potential at a point in space, the reference point is usually assumed to be at infinity. So, the potential at point A is defined as the work that needs to be performed to move a positive unit charge from infinity to point A.
| {
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How is work transferred to the system recognised? For example, a potato initially at room temperature $25 \sideset{^{\circ}}{}{\mathrm{C}}$ is baked in an oven that is maintained at $200\sideset{^{\circ}}{}{\mathrm{C}}.$
I made potato as the system and the outer surface of the skin as the system boundary. While the oven and the air inside it is the surroundings.
There is a temperature difference between the skin and the air in the oven which is the driving force of heat transfer (temperature difference).
What about work? Is there transfer through work done?
Isn’t the oven working to produce the heat in the oven which is then transferred to the potato? But work is pressure multiplied by the change in volume. However there’s no change in volume of the potato. So does this mean no work is done?
In summary, how do I identify whether work is done to the system or not?
| While it may be (I haven’t researched it) that the potato “expands” a bit during the cooking process, I would think the amount of work to do this would be extremely small (as JMac pointed out) compared to the heat transfer to the potato that increases its internal energy.
Regarding your question “isn’t the oven working to produce the heat in the oven which is then transferred to the potato? But work is Pressure multiply by the change in volume. However there’s no change in volume of the potato. So does this mean no work is done?”
No, it does not mean no work is done. Not all work is pressure multiplied by volume change. That type of work is properly called “boundary work” (expanding the potato skin). But there are other forms of work. One is electrical work. When the oven’s heating element heats the air in the oven, it is a work transfer (at a rate of $i^2R$ ) from the electrical source to the heating element and then heat transfer from the heating element to the air.
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Why do distinct temperature layers form in bodies of water? This may not be the correct place to ask a question about limnology (which I just found out is the science of lakes) but it seems like the place most likely to produce a good answer.
In my diving lessons I learned that there is a layer of water in lakes where temperature suddenly drops. Judging from the pictures at Wikipedia: Thermocline the teacher was talking about the metalimnion which also sort of makes sense considering the depthof the lake he was talking about.
It is however still no clear to me why these layers form.
The article seems to suggest that the top layers warm-up due to sunlight while the bottom layer stays cold and there is little mixing because the colder water is denser.
But shouldn't that lead to a somewhat linear temperature gradient instead of layers?
Edit:
Here is a nice picture showing the different layers from loon lake:
| Typically, the temperature of a lake is pretty uniform with depth in winter--the surface cools, driving convection, and even if there are periods with heat input at the surface, wind mixing keeps things homogenous. In the summer, there is significant heat input from the sun. Wind still mixes the surface, but the thermal stratification prevents the wind mixing from penetrating to great depth--there is not enough mechanical energy input to overcome the stability of the light water overlying heavy water. In your example figure, wind has mixed the surface layer down to about 5 meters. This is typical: it's called a "mixed layer".
| {
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Trace of 4 Gell-Mann matrices Does any one know what would be $tr(t^a t^b t^c t^d)$, where $t^a$ etc are Gell-Mann matrices?
This came about when analyzing the color factor for the compton effect for QCD. So, must be pretty common, but I could not find a proper reference.
In general is there any reference for trace of arbitrary number of Gell Mann matrices?
| I take the SU(N) generators in the fundamental representation normalized such that
$$
\text{Tr}\left[T^a T^b\right] = \frac{1}{2}\delta^{ab}
$$
The commutator of two generators define the structure constants $f^{abc}$
$$
\left[T^a,T^b\right] = if^{abc}T^c
$$
The anticommutator of two generators is
$$
\left\{T^a,T^b\right\} = \frac{1}{N}\delta^{ab}1 +d^{abc}T^c
$$
where by $1$ I mean the identity matrix and $d^{abc}$ are the "d-symbol" defined as
$$
d^{abc} = 2\text{Tr}\left[ \left\{T^a,T^b\right\}T^c \right]
$$
Then, there is a useful identity
$$
\text{Tr}\left[T^aT^bT^cT^d\right] = \frac{1}{4N}\delta^{ab}\delta^{cd} + \frac{1}{8}\left(d^{abe}d^{cde} - f^{abe}f^{cde}+if^{abe}d^{cde}+if^{cde}d^{abe}\right)
$$
I suggest you this reference http://scipp.ucsc.edu/~haber/ph218/sunid17.pdf where different trace identitites are collected. For your case, look at Equation 75 in Appendix B, page 9.
Check the normalization of the generators before to use this identity.
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A slipping cylinder comes into contact with friction Suppose a cylinder is slipping rigidly on a frictionless horizontal surface. Then, at $t=0$ it reaches a different ground. The coefficient of friction between the cylinder and this new type of ground is $\mu$. What happens?
- assuming usual high-school friction to be the only horizontal force and a homogeneous cylinder with moment of inertia $I=MR^2/2$.
I have come up with this question myself, it is not homework.
Once the cylinder starts slipping on the new ground, kinetic friction will dissipate energy, slowing it down. It will also produce torque and make it roll.
Eventually, the velocity of the contact point will vanish and the cylinder will start rolling without slipping (RWS). This happens when $v=\omega R$ where $v$ is the velocity of the center of mass and $\omega$ is the angular velocity with respect to the center of mass.
Equations of motion are $FR=I\dot{\omega}$ and $F=M\dot{v}$. With $v(0)=v_0$ and $\omega(0)=0$ this leads to the conclusion - using that friction is $F=\mu mg$ and if I made no mistake - that RWS will start at $t=v_0/(3\mu g)$.
Now here comes the actual question. Once the cylinder is RWS, the friction force must actually be zero (otherwise, it would slow down the cylinder without performing work, which is impossible).
However, the friction force is not zero immediately before RWS (it is $\mu mg$). Hence the question: is the friction force dicontinuous? Or have I made some mistake?
| If you analyze a (non-rotating) block in the same situation, you will see similar behavior.
As the block reaches the surface with friction, the frictional force appears and the block decelerates. When the block's speed becomes zero, the modeled frictional force becomes zero.
| {
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Classical theory fails to explain quantization of motions?
I understand everything written here.
But the last point, I cannot get, at all.
How does it point towards Quantization of the two motions, since the energy change is not sudden, but gradual?
And if anything is wrong with the given image, please tell what it is.
| Because the degrees of freedom look like they're 'frozen' at low T. Statistically, we know there's going to be an average energy of $RT/2$ per mole for each degree of freedom. If you have translation, rotation and vibration that makes for a total of 7 degrees of freedom. By classical mechanics, there should be no lower limit to how much energy goes into them, so it should be $C_v = 7R/2$ from the very beginning. Instead, because of QM, there's an energy gap between ground and first excited state for each of these motions, and that means they don't contribute up to the point when $kT \sim \hbar \omega$ for each of them. That's what causes the 'steps' to appear in the heat capacity, and the very existence of those steps is only possible because of quantum effects. That the steps are smoothed out is merely a statistical effect due to the fact that not all modes will activate instantly across the gas.
| {
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Was the universe already expanding before inflation occured? Was the universe already expanding before inflation occured or did inflation cause the universe to start off expanding?
By “cause it to start off expanding” , I mean the cause of the initial expansion.
|
Was the universe already expanding before inflation occured or did inflation cause the universe to start off expanding?
It's not possible for the universe to be in a non-expanding phase and then enter an inflationary phase.
The idea of cosmological expansion is based on two pieces of evidence, one theoretical and one observational. Observationally, Hubble et al. observed the Hubble law. Theoretically, it's not possible to construct a static cosmology in general relativity without fine-tuning.
For the very early universe, we have little clear observational evidence, but the theoretical argument still holds.
All of this is on much more solid ground than anything having to do with the inflation hypothesis. We actually don't know whether inflation is right or not, although it is a very popular idea among theorists.
It's a popular misconception that inflation caused expansion, is needed in order to explain expansion, or is the same thing as expansion. It's not.
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Why does $H_2$ have $C_V$=$7/2 R$ at high temperatures, while the total number of degrees of freedom is 6? The two hydrogen atoms have 6 degrees of freedom in total. Of them, $3$ contribute to translation, $2 $contribute to rotation and $1$ contribute to the vibration.
I know that the vibrations motion is frozen at low temperature due to quantum mechanical effects.
However, then the $C_V$ at high temperature should be $6/2 R$, while experimentally, it is $7/2 R$ (source: Principles of Physics by Walker, Resnick and Halliday)
Edit: The answers reveal that the missing part of specific heat is due to potential energy of vibration. So I am extending the question for clarification. $CO_2$ has total 9 degrees of freedom, of which 3 are translational, 2 are rotational, 4 are vibrational. So, at high temperature, will the $C_V$ of $CO_2$ be $\frac{R}{2} × [3+2+4+4] $? The two 4s are due to kinetic and potential energy of vibrational motion.
| quoting from wikipedia on heat capacity:
Each rotational and translational degree of freedom will contribute R/2 in the total molar heat capacity of the gas. Each vibrational mode will contribute R to the total molar heat capacity, however. This is because for each vibrational mode, there is a potential and kinetic energy component. Both the potential and kinetic components will contribute R/2 to the total molar heat capacity of the gas.
For more general gas, wikipedia [general gas] also give you example of how to calculate the number of degree of freedom and how to apply properly the equirepartion theorem:
For example, triatomic nitrous oxide N2O will have only 2 degrees of rotational freedom (since it is a linear molecule) and contains n=3 atoms: thus the number of possible vibrational degrees of freedom will be v = (3⋅3) − 3 − 2 = 4. There are four ways or "modes" in which the three atoms can vibrate, corresponding to 1) A mode in which an atom at each end of the molecule moves away from, or towards, the center atom at the same time, 2) a mode in which either end atom moves asynchronously with regard to the other two, and 3) and 4) two modes in which the molecule bends out of line, from the center, in the two possible planar directions that are orthogonal to its axis. Each vibrational degree of freedom confers TWO total degrees of freedom, since vibrational energy mode partitions into 1 kinetic and 1 potential mode. This would give nitrous oxide 3 translational, 2 rotational, and 4 vibrational modes (but these last giving 8 vibrational degrees of freedom), for storing energy. This is a total of f = 3 + 2 + 8 = 13 total energy-storing degrees of freedom, for N2O.
For a bent molecule like water H2O, a similar calculation gives 9 − 3 − 3 = 3 modes of vibration, and 3 (translational) + 3 (rotational) + 6 (vibrational) = 12 degrees of freedom.
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Can electrons jump between levels in same shell with constant $L$? I see many cases where electrons jump from various energy levels of some $L$ value to various levels of other $L$ value. But is there any transition in same $L$ level between its different orientations?
$EDIT $
(How to calculate energy difference?)
| They can, following selection rules. As an example the line of hydrogen $\lambda=21 cm$ corresponds to a transition with $\Delta L=0, \Delta S=1$
| {
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A harmoinic oscillator in the Heisenberg picture Considering the Hamiltonian of a harmonic oscillator
\begin{equation}
H=\frac{p^2}{2m}+\frac{m\omega^2 x^2}{2},
\end{equation}
the time evolution of the Heisenberg picture position and momentum operators is given by
\begin{align}
\dot{x}&=\frac{i}{\hbar}[H,x]=\frac{p}{m}\\
\dot{p}&=\frac{i}{\hbar}[H,p]=-m\omega^2x,
\end{align}
from which we get
\begin{align}
\ddot{x}=\frac{\dot{p}}{m}=-\omega^2 x.
\end{align}
We can solve this differential equation with an initial condition $x(0)=x_0$ to obtain
\begin{align}
x=x_0 e^{-i\omega t}.
\end{align}
On the other hand, $p$ may be calculated by inserting $x=x_0 e^{-i\omega t}$ into $\dot{x}=\frac{p}{m}$ as the form
\begin{align}
p=-im\omega x_0 e^{-i\omega t},
\end{align}
but it does not seem to be quantum mechanical because $p$ is represented in terms of $x_0$, so $p$ can commute with $x$. Where I am wrong?
| Your solution $ x = x_0 e^{-i\omega t}$ is not complete. The equation you are solving is second degree. It should depend two initial values $ x = x_0 e^{-i\omega t}+ x_1 e^{+i\omega t}$
| {
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Have any experiments been performed on the magnetic properties of neodymium at high pressure and temperature? As probably everyone here knows, the magnetic property of neodymium disappears at temperatures above ~590°F (310°C) depending on the mix of alloys. Have there ever been any experiments performed to examine how pressure affects the Curie temperature of neodymium alloys? I'm particularly interested in whether or not some neodymium alloys might retain their magnetic properties above their Curie temperature under conditions found, say, within the Earth's outer or inner core.
| Don't think that pure Nd is ferromagnetic (maybe it's antiferromagnetic?), so your question about the Curie temperature really applies to just certain ferromagnetic alloys of Nd. Not aware of any high-P studies on those ferromagnetic alloys offhand. I was a co-author of a paper that looked at the magnetic properties of the heavy lanthanides (which are ferromagnetic) under pressure, and we found that their Curie temperatures dropped with pressure at a $dT_c/dP$ rate of around -10 to -20 K/GPa (See High-pressure magnetic susceptibility experiments on the heavy lanthanides Gd, Tb, Dy, Ho, Er, and Tm). If Nd alloys have $T_c's$ which drop at similar rates, then I wouldn't expect any magnetism in them at the pressure and temperature of the core-mantle boundary (≈140 GPa and 5000 ˚C).
| {
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Why do spheres roll so far? Say you have a sphere and a cuboid, both of equal mass and material. You push both on a ground of the same material over the same period of time with an equal force. Intuition tells you that the sphere would roll much further than the cuboid. Why is this?
The most common model of friction, $F_{\mathrm{fr}}=\mu_k N$, suggests that friction does not depend on surface area, which I understand. Then, if we fit into the equation the coefficient of friction and normal force for each case (the sphere and the cuboid), we would get that the frictional force is equal! So why does it seem like the frictional force on the sphere is so much less than that of the cuboid that it can travel so much further?
| For any wheel, it undergoes sliding and rolling friction as it moves. Do note that sliding and rolling friction act in opposite directions, so they cancel out. In a circular wheel both of the frictions are roughly equal due to contact at a point, so the wheel experiences little or no friction. However, a square wheel has a relatively large sliding friction due to its large faces, making the friction greater than a circular wheel.
Hence, since circular wheels would experience less friction, there is less work done due to friction, and would possess more kinetic energy for it to move further than say a square-shaped wheel.
| {
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Experimental tests of QFT in curved spacetime In terms of QFT in curved spacetime (qftics), are there any known tests that confirm/void it to be a correct prescription for a first order approximation quantum theory of gravity? Obviously the extension makes some sense but how is it confirmed?
I know there are a lot of tests including Lorentz violation in qft, ultimately a possible search route to test qftics, but if there are more specific cases this would be extremely useful to know about.
| *
*By its very definition, QFT in curved spacetime assumes the background spacetime to be fixed and is a framework to treat quantum fields on the fixed spacetime. Hence the question "are there any known tests that confirm/void it to be a correct prescription for a quantum theory of gravity?" is ill-posed; as the mentioned formalism tells us the first-order corrections due to quantum fields only; as opposed to a full understanding of quantum gravity which would also involve taking into account how the geometry changes due to the dynamics of the fields. The key idea is that QFT in curved spacetime is the $O(1)$ correction in a full theory of quantum gravity.
*With regards to the question whether there have been any experimental tests of QFT in curved spacetime, there have been some proposals. You might want to have a look at this and this.
| {
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The dependence of capacitance on the ratio of radii in a cylindrical capacitor Why does the capacitance of two cylindrical capacitors of same length stay the same if the ratio of the outer radii to the inner radii of one capacitor is same as the other.
The capacitance of a cylindrical capacitor is C = (2πel)/(ln(R2/R1)) where e - epsilon symbol, l - length of the capacitor, R2 and R1 are the outer and inner radius respectively.
According to the equation its pretty clear but I want an explanation that's more intuitive, one that does not need equations for an explanation.
| The capacitance of a capacitor, in general, is proportional to the plate area and inversely proportional to the distance between the plates. So, if we keep the ratio between the area of the plates and the distance between the plates the same, the capacitance should remain the same.
If the ratio of the inner and outer radii of a cylindrical capacitor stay the same, it means that both radii have changed by the same factor, which means that the difference between the radii has changed by the same factor and the areas of the plates (since the area of a cylinder is proportional to its radius) have changed by the same factor.
So, since, the ratio between the area of the plates and the distance between the plates have not changed (because both have changed by the same factor), we can conclude, according to our initial observation, that the capacitance should not change either.
| {
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Max Born's statistical interpretation of the wave function How did Max Born derive the probability of finding a particle between $x$ and $x+dx$ at instant $t$?:
$$ \left |\psi(x,t)\right|^2dx$$
Was this result mathematically derived? Or is it just a postulate, like the Schrödinger equation itself?
| This statement is equivalent to the statement that $|\psi(x,t)|^2$ is the probability density function for the particle, because the definition of the probability density $p(x)$ is such that $p(x)dx$ is the probability of an event occurring between $x$ and $x+dx$.
The statement that $|\psi(x,t)|^2$ is the probability density function is known as the Born rule, and in most mainstream quantum mechanics interpretations, it's considered to be a fundamental postulate. Several lesser-known interpretations claim to be able to derive it or otherwise intuit it using statistical arguments (for example, quantum Bayesianism considers the Born rule to be an extension of the Law of Total Probability, and the hidden-measurements interpretation and pilot-wave theory both claim to be able to derive the Born rule from other axioms), but these use a different set of axioms than mainstream quantum-mechanics.
| {
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Can a photon be called a disturbance in the electromagnetic field? Total noob here.
I know that electromagnetic waves do not need a medium to travel. I analogize a photon as a single pulse/oscillation in the electromagnetic field and its propagation is the motion of that pulse through the electromagnetic field. Kind of like the pulse that travels through a string if you jerk it.
| Yes that is a correct understanding, you could also use the pebble in a pond analogy as light/photon seems to spread out. There are many posts on this forum some very good, some not so good. Keep reading and you will get a good feel for the photon etc.
| {
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When are we required to use the Wess-Zumino term? I was recently reading about non-Abelian bosonization, and I had a question concerning the Wess-Zumino term. In particular, I have been reading this short introduction by Ivan Karmazin, which states that
A non-abelian bosonisation introduced by Witten in 1983 allows to translate any fermi theory into local bose theory while having all of the original symmetries conserved.
To ensure that the field theory is scale invariant, we add the Wess-Zumino action
$\Gamma[g]=\frac{1}{24\pi}\int_B d^3 y \epsilon^{ijk}{\bf Tr}\left(
g^{-1} \partial_i g g^{-1}\partial_j g g^{-1}\partial_k g
\right)$
where $g\in SU(N)$. My question is when, precisely, do we need to include the Wess-Zumino term when we bosonize a theory of 1+1 D massless fermions? Does the Wess-Zumino term mainly needed to describe a nonlinear sigma model, or is it more general?
| For starters, the WZ-term is needed to get EOMs that factorize in right- & left-movers, similar to the dual fermionic theory, cf. eq. (15) in Ref. 1.
References:
*
*E. Witten, Non-Abelian Bosonization in Two Dimensions, Commun. Math. Phys. 92 (1984) 455.
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How to put $c$ back into relativistic equations? Many books set the speed of light $c=1$ for convenience. For example, Weinberg in his textbook "Gravitation and Cosmology" (though $G$ is still left as a constant):
$$\begin{align}
\mathrm{d}\tau^2 &= \mathrm{d}t^2 - R^2(t)~[f(r,\theta,\phi)] \tag{11.9.16} \\
t &= \frac{\psi + \sin\psi}{2\sqrt k} \tag{11.9.25}
\end{align}$$
If I now want to run real numbers in SI units, do I
*
*Assume $t'[\mathrm{sec}] \rightarrow t = t'/c = t'/299792458$ where the new time unit is $3.34\ \mathrm{ns}$
*Multiply each $t$ and $\tau$ with $c$:
$$\begin{align}
c^2 \mathrm{d}\tau^2 &= c^2 \mathrm{d}t^2 - R^2(t)~[f(r,\theta,\phi)] \tag{11.9.16} \\
ct &= \frac{\psi + \sin\psi}{2\sqrt k} \tag{11.9.25}
\end{align}$$
*Something else?
| Putting the factors of $c$ and $G$ back can be a tedious business. Many of us do it by (educated) guesswork, followed by checking that the guesses give sensible results.
The rigorous way to do this is using dimensional analysis, that is checking that when adding quantities they must all have the same dimensions, and that if you have some equation:
$$ t = \text{something} $$
The dimension of the $\text{something}$ must be time.
If you take the equation $11.9.16$ for the metric Weinberg writes this as:
$$ d\tau^2 = dt^2 - R^2(t)\left(\frac{dr^2}{1-kr^2} + r^2d\theta^2 + r^2\sin^2\theta d\phi^2\right) $$
where the scale factor $R(t)$ is dimensionless. The right hand side has time + space so to make this dimensionally consistent we have to either multiply the time term by $c^2$ or divide the distance term by $c^2$. Typically we would do the former to get:
$$ c^2 d\tau^2 = c^2 dt^2 - R^2(t)\left(\frac{dr^2}{1-kr^2} + r^2d\theta^2 + r^2\sin^2\theta d\phi^2\right) $$
A similar argument applies to his equation $11.9.25$:
$$ t = \frac{\psi + \sin\psi}{2\sqrt{k}} $$
In this equation $\psi$ is dimensionless and $k$ has dimensions of $L^{-2}$ so the right hand side has dimensions of $L$. To make this dimensionally consistent either multiply by $1/c$ on the right hand side or $c$ on the left hand side.
| {
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Pulley problem, wrong decomposition of forces The mass of the red block is $M$. The rope is inextensible and its mass isn't relevant. The mass of the sphere is $M\sqrt{2}$. The angle between the rope and the horizontal is $45°$.
I'm looking for the acceleration of the body but I'm doing something wrong. I want to understand my mistake.
Well, let's say that $\vec{F_m}$ is the force of the rope (the weight of the sphere). We can decompose it as $\vec{F_{mx}}$ and $\vec{F_{my}}$.
$\vec{F_{p}}$ is the weight of the block.
We arrive that $|\vec{F_{my}}|=|\vec{F_{p}}|$ and this is 100% true, in fact the test asks me to demonstrate it. So there is no friction. Then the only force remained is $\vec{F_{mx}}$. So we can calculate the acceleration in this way:
$a = |\vec{F_{mx}}|/M = Mgcos(45)\sqrt{2}/M=g$
However this isn't the solution, so there is a force which I'm not considerating. The solution in fact is $4.03 m/s^2$
EDIT: The question isn't a duplicate: we're talking about acceleration.
| Consider the tension present in the rope to be T.
So, for the sphere,
$$M\sqrt2g - T = M\sqrt2a_1$$
Similarly, for the block,
$$\frac{T}{\sqrt2} = Ma_2$$
[Here, $a_1$ and $a_2$ are the acceleration of sphere and block respectively.]
Now, assuming that the rope remains taut, the components of $a_1$ and $a_2$ along the rope wouybe equal.
Thus,
$$a_1=\frac{a_2}{\sqrt2}$$
On solving, you should probably reach to the correct answer.
What mistake you were making is
$$\vec{|F_p|} = \frac{T}{\sqrt2}$$
However, this won’t be the case.
The correct equation I think would be
$$\vec{|F_p|} = \frac{T}{\sqrt2} + N$$
Where N is the force exerted by the ground in contact on the block
| {
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Virtual work - Tension in a string I am confused about the way Tension or Thrust is used in equation of virtual work. I mean whether to use $ T\cdot dr$ or $-T\cdot dr$ (work done by tension). For instance in the below case :
A string of length a forms the shorter diagonal of a rhombus of four uniform rods each of length b and weight W which are hinged together. If the uppermost rod be supported in a horizontal position , prove that the tension of the string is
$$ \frac{2W(2b^2 - a^2 )}{b\sqrt{4b^2-a^2}}. $$
My Approach:
Assuming T is tension in string BD.
$$ BD = 2b\cos\theta$$
As $ \theta $ increases, the length of the string decreases or movement of string is in the direction of tension, hence the work done by tension should be postive. Hence the equation of principle work would be $$ T\delta(2b \cos\theta) + 4W \delta(b \sin\theta \cos\theta) = 0 $$ but the solution in book says
$$ -T\delta(2b \cos\theta) + 4W \delta(b \sin\theta \cos\theta) = 0 $$
| Work done by Tension in virtual work equation should be simplified in the following way.
If the length of string is increasing (decreasing) -> 'dr' is positive(negative).
If the Change in length is in the direction (opposite) of force -> Tension is postive (negative).
| {
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The definition of the hamiltonian in lagrangian mechanics So going through the "Analytical Mechanics by Hand and Finch". In section 1.10 of the book, the Hamiltonian $H$ is defined as: $$H = \sum_k{\dot{q_k}\frac{\partial L}{\partial \dot{q_k}} -L}.\tag{1.65}$$
And then author affirms that this quantity is constant and takes the derivative $\frac{dH}{dt}$:
$$\frac{dH}{dt} = \sum_k {\ddot{q_k} \frac{\partial L}{\partial \dot{q_k}} + \dot{q_k}\frac{d}{dt}(\frac{\partial L}{\partial \dot{q_k}}) - \frac{d L}{d t}}.\tag{1.66}$$
Now the book writes: "According to the chain rule for differentiating an implicit function over time": $$ \frac{dL}{dt} = \sum_k {\frac{\partial L}{\partial q_k}\dot{q_k} + \sum{}\frac{\partial L}{\partial \dot{q_k}}{\ddot{q_k}} + \frac{\partial L}{\partial t}}.\tag{1.67}$$
And the summing the second and third gives: $$\frac{dH}{dt} = - \frac{\partial L}{\partial t}.\tag{1.68}$$
Now I don't understand how the third equation is derived and also why is the Hamiltonian $H$ is defined in the way it is in the first equation?
| The first equation is the definition of the Hamiltonian. This definition is an application of the Legendre transformation, which converts a function of one variable to a function of another in a particular way to preserve the information contained in the first, but might allow a more practical or useful calculation scheme. It's beyond the scope of this answer to go into the details. The Wikipedia page on the Legendre transform is not a particularly good introduction to the topic.
The third equation is not derived, it's simply an expression of the derivative. A mathematical application of the chain rule to $L(q,\dot{q},t)$.
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Calculating angular velocity of rolling object with just gravity? From what I have learned, you can calculate what the angular velocity of an object will be based on its potential energy.
Say there is a situation where:
*
*acceleration due to gravity = 10 m/s²
*friction = infinite (object is in pure rolling motion)
*we know objects current state (velocity, angular velocity, etc..)
*slope angle = 13° above horizontal
And I need to find:
*
*objects angular velocity one second from current time
As you can see, I don't think I can use the potential energy conversion formula (maybe I can, but I am not seeing it).
Is there a way that I could find the 'future' angular velocity (in this case 1 second ahead) by just knowing the current 'state' of the object and that gravity will be affecting it?
Edit:
In my physics simulation, I find that given these values:
*
*gravity = g = 10 m/s²
*friction = μ = infinite
*radius of sphere = r = 0.5m
*angle of incline = θ = 13°
The angular acceleration is roughly 3 radians/s².
I just realized that if I could find the angular acceleration, I could then predict the angular velocity. Could someone help me find a formula to get the angular acceleration, with just knowing these values?
| Yes, of course you can apply energy conservation or by simple work done by rotating force is given as
The mechanical work applied during rotation is the torque ($\tau$) times the rotation angle ($\theta$):
$$W = \tau \theta$$
so that the work done is equal to $\frac{1}{2}I\omega^2$.
| {
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Why is the gravitational potential energy (or electrical) the work done to move a point mass $m$ from infinity to that point in a gravitational field? To be specific, I'm asking why does the mass need to be moved "at a constant speed"?
My textbook says it is so that no kinetic energy is involved. But shouldn't kinetic energy be involved in order to move the point mass from infinity to a point in a gravitational field? Since kinetic energy is converted to gravitational potential energy.
| If you consider a generic stationary mass distribution, it generates a gravitational stationary field in all the space around:
$$\mathbf G=\mathbf G(x,y,z)$$
You can demonstrate that always exists a function called potential (not potential energy) for which:
$$grad(V(x,y,z))=\mathbf G(x,y,z)$$ in every point of the space. Also you can demonstrate that:
$$\int_{\beta} \mathbf G \cdot \mathbf {dr} =V(\mathbf b)-V(\mathbf a)$$
Where $\beta$ is a line which starts in $\mathbf a$ and ends in $\mathbf b$. You can obtain also that:
$$\int_{\beta} m\mathbf G \cdot \mathbf {dr} =mV(\mathbf b)-mV(\mathbf a)$$
If you define $$U=Vm$$ the potential energy own by a mass, finaly you have:
$$W_{\beta}=\int_{\beta} \mathbf F \cdot \mathbf {dr} =U(\mathbf b)-U(\mathbf a)=\Delta k_{a-b}$$
However the kinetic energy is not involved in the defintion of the potential energy.
| {
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Calculating the Instantaneous speed When we are calculating speed from the graph for a uniform motion the speed which get as an answer is actually the average speed and since in uniform motion the average speed is same as that of instantaneous speed so its correct,right? Now if this is right if we calculate the variable speed with the help of a tangent isnt that also average speed?
| Average velocity between two points in time is given by the slope of the secant line between the two corresponding points on the position vs. time graph.
Instantaneous velocity is found at a point in time. It is given by the slope of the tangent line at that point on the position vs. time graph.
In the case on constant velocity, the slope of the secant line between any two points and the slope of the tangent line at any point are always the same. If the velocity is not constant then it is not generally true that the slope of the secant line between any two points will be equal to the slope of the tangent line at some other point.
Something to keep in mind, in saying average velocity you have to specify between which time points you are referring to. For instantaneous velocity you have to specify at what point in time. There is not a single average or instantaneous velocity.
| {
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Do compact symplectic manifolds play a role in physics? In classical mechanics, the phase space is the cotangent bundle of the configuration space, and it is a symplectic manifold, but not compact.
Do compact symplectic manifolds have physical meaning? Or just of mathematical interest?
| The phase space for a spin-$J$ system is the the two sphere $S^2$ with symplectic structure $\omega= J \sin \theta d\theta \wedge d\phi$, i.e. $J$ times the area 2-form. The unit vector ${\bf n}$ that specfies points on the sphere correponds to the direction of the spin: ${\bf S}= J{\bf n}$.
In general, any Hamiltonian system whose quantum version has finite-dimensional Hilbert space ${\mathcal H}$ will have a compact phase space whose Liouville volume $\int_M\omega^2/n!$ is proportional to ${\rm dim}({\mathcal H})$. A typical example would be the ${\rm SU}(n)$ internal symmetry dynamics described by the Wong equations in gauge theories.
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Why does a bird not get electrocuted when it is sitting on a high tension electric wire? Why does a bird not get electrocuted when it is sitting on a high tension electric wire?
| This is because the bird is not grounded. That is, there is no potential difference for electricity to flow (eg. It is not touching anything conductive that touches the ground - air doesn't count because its conductivity is very poor and hence negligible). But, as soon as the bird touches the pole, the bird fries because now, the electricity flow can occur from the high voltage/potential in the wire to the pole (metallic conductor), which is connected to the ground or at zero potential.
| {
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What direction does electric current flow in when the voltage drop is negative? We were given the below diagram, and we needed to determine whether the unknown component is supplying energy into the system.
$\hspace{200px}$
I thought the unknown component was supplying energy since charge is flowing to a higher potential, but my teacher said it was the other way around.
What does the $`` -2 \, \mathrm{V} "$ label in the diagram mean? Is it:
*
*the difference from $\left(+\right)$ to $\left(-\right);$ or
*the difference from $\left(-\right)$ to $\left(+\right) ?$
I'm sure there is a convention to this sort of stuff but I'm not aware of it yet.
| in these kind of problems first always try to find which component is acting as source (energy supplier) and which is acting as sink (energy consumer)and then,the rest follows
absorb negative signs of voltage by interchanging polarities of upper and lower terminals respectively and that of the current by changing it's direction
now draw modified diagram such that current $+3A$ will leave $+ve$ terminal of battery making it act like source i.e, it supplies/delivers energy/ to the unknown component which makes unknown component as sink i.e, it dissipates/absorbs energy.
| {
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Explanation for observed short while winding coil around iron core I'm winding a coil, 230 turns #23 enameled magnet wire on a 1/2" diameter solid iron core 6" long ( on a lathe, hand feeding the wire )
When finished, I check continuity ( wire to rod ) with a multi-tester.
Tester shows wire is shorted to the core. OK I must have cut through the enamel.
Wind a new one (super careful this time) testing the continuity every 25 turns, after 50 turns shows shorted again.
| Wrapped paper around a new core, wrapped by hand, no short.
Has to be poor technique, or bad wire insulation.
Mystery solved.
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Stress Energy Tensor in language of differential forms The motivation for this is that quantities like the electric current $J$ in maxwell's equations of motion can be expressed as a differential 3-form, so that the continuity equation can be written just as
$$dJ=0$$
Which is really nice because it can all be done without defining a metric tensor!
Now the stress-energy tensor has a similar continuity equation but is generally represented as a symmetric 2-tensor. so it can obviously not be represented as a 3 form, but can it somehow be represented potentially in the language of differential forms so that a metric tensor does not have to be defined?
| With the help of a Killing vector field $\xi$ one can define the current 3-form
$$J_\xi = \star\ \iota_\xi T$$
of which you can then take the exterior derivative to obtain the conservation law
$$\operatorname d J_\xi = 0.$$
Note that the metric is hidden inside both $T$ and $\xi$.
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Intuition behind group velocity
Why is the group velocity of the wave given as $\frac {\partial w
}{\partial k}$?
I understand what the group velocity means but its always given as a definition and I would like to have some intuition behind it. The $k$ derivative confuses me.
| Let there be 2 waves of unit amplitude with similar angular frequencies $\omega_1$ and $\omega_2$, and angular wavenumbers $k_1$ and $k_2$ respectively. Therefore, adding these 2 waves, we have $$y = \sin(k_1x-\omega_1 t) + \sin(k_2x-\omega_2 t)$$ Using the factor formula, we can express this as: $$y = 2 \sin \left( \frac{(k_1x-\omega_1 t) + (k_2x-\omega_2 t)}{2}\right) \cos \left( \frac{(k_1x-\omega_1 t) - (k_2x-\omega_2 t)}{2}\right)$$ $$= 2 \sin \left( \frac{(k_1 + k_2)x-(\omega_1 + \omega_2)t }{2}\right) \cos \left( \frac{(k_1 - k_2)x-(\omega_1 - \omega_2)t }{2}\right)$$ $$= 2 \cos \left( \frac{(k_1 - k_2)x-(\omega_1 - \omega_2)t }{2}\right) \sin \left( \frac{(k_1 + k_2)x-(\omega_1 + \omega_2)t }{2}\right)$$ $$= y' \sin \left( \frac{(k_1 + k_2)x-(\omega_1 + \omega_2)t }{2}\right)$$
The important thing here is that $\omega_1$ and $\omega_2$, and $k_1$ and $k_2$ are very close to each other. Since the frequency of the sine term is the sum of $\omega_1$ and $\omega_2$, while the frequency of the cosine term is the difference of $\omega_1$ and $\omega_2$. The frequencies of the sine and cosine terms will therefore be very different. Therefore, we see that the resultant wave has a frequency of $(\omega_1 + \omega_2)$, but its amplitude, $y'$, will be perceived to be modulated by the cosine term, forming an envelope, which look like this.
The group velocity is the velocity of the envelopes, not the velocity of the individual wavefronts. We know that $$v = f \lambda = \left(\frac{\omega}{2\pi} \right) \left(\frac{2\pi}{k} \right) = \frac{\omega}{k}$$ Therefore, the velocity of the envelope, given by the cosine term, will be equal to $$\frac{\omega_1 - \omega_2}{k_1 - k_2}$$ Since $\omega_1$ and $\omega_2$, and $k_1$ and $k_2$ are very close to each other, this reduces to $$\frac{d\omega}{dk}$$ as claimed.
An intuitive way to think about it is this. The two waves interfere constructively when the cosine term is equal to 1, and destructively if it is equal to 0. Let two successive points of constructive interference be a wavelength by itself, since you are finding their velocity (which is the velocity of the envelope). Ignoring the sine term, this can be a wave on its own, completely described by the cosine term. Thus its velocity, which is the group velocity, will be exclusively calculated from the cosine term by $\frac{\omega_1 - \omega_2}{k_1 - k_2}$.
| {
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Euler-Lagrange equations using $\vec{E}$ and $\vec{B}$ instead of $A^\mu$ We all know that the lagrangian for the free electromagnetic field is given by $$ \mathscr{L} = -\frac{1}{4}F^{\mu \nu}F_{\mu \nu} $$ where $ F^{\mu \nu} = \partial^\mu A^\nu -\partial^\nu A^\mu $ is the electromagnetic field tensor. But also we know that
Lets consider $c=1$ for simplicity. Then, doing the math, the lagrangian can be written as $$ \mathscr{L} = \frac{1}{2} (|\vec{E}|^2 - |\vec{B}|^2) $$
By applying Euler-Lagrange, i.e.
$$ \partial_\mu \left( \frac{\partial \mathscr{L}}{\partial (\partial_\mu \phi_i)} \right) - \frac{\partial \mathscr{L}}{\partial \phi_i} = 0 $$
where $\phi_i$ is each of the components of each field, I find $$ \vec{E} = 0 $$ and $$ \vec{B} = 0 $$ but not the Maxwell equations... What is going on?
| $\vec{E}(x,t)$ and $\vec{B}(x,t)$ are not totally independent variables.
I'm not familar with field theory, but have simpler example with LC circuit, a zero dimensional field theory:
$$
H=T+V=\frac{L}{2}I^2+\frac{1}{2C}Q^2 \quad \sim \quad \frac{1}{2}(|\vec{B}|^2+|\vec{E}|^2)
$$
$$
L=T-V=\frac{L}{2}\dot{Q}^2-\frac{1}{2C}Q^2 \quad \sim \quad \frac{1}{2}(|\vec{B}|^2-|\vec{E}|^2)
$$
Charge $Q$ is the dynamic variable.
$|\vec{E}|$ is propotional to $Q$, $|\vec{B}|$ is propotional to $\dot{Q}=I$,
You can't take current $I$ as independent from $Q$. This is the constraint of the system.
If you insist to take $I$ and $Q$ as independent variable, then you get a different system: a capacitor and a inductor seperately
| {
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Precision in cosmological constant The observed cosmological constant, what is its precision? How precise do we know is it not a fluke or misinterpretation of cosmological data?
| The best values of the cosmological parameters available, are those published by Planck Collaboration 2018.
In it, the values of the Hubble constant and the dark energy density ratio are:
$$\displaystyle H_0 = 67.66 \pm 0.42 \, (km/s)/Mpc$$
$$\Omega_{\Lambda_0} = 0.6889 \pm 0.056$$
Remembering that:
$$\displaystyle \Lambda = \frac{3 H_0^2}{c^2} \, \Omega_{\Lambda_0}$$
We get for the Cosmological Constant the value:
$$\displaystyle \Lambda = 1.106 \times 10^{-52} \pm 0.021 \times 10^{-52} \, m^{-2}$$
The values of the cosmological parameters of Planck Collaboration 2018 are obtained from the CMB, that is to say, when the Universe was 380,000 years old and was composed exclusively of elementary particles without any star or galaxy yet.
Recently, The Dark Energy Survey (DES), studying galaxies in a sphere of 8,000 million years centered on us, (therefore objects of the universe at least 6,000 million years after the origin of the CMB), has obtained the same values, which makes these figures very robust.
More information at Dark Energy Survey reveals most accurate measurement of dark matter structure in the universe
Regards
| {
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Reason for black and white colour We always say that an object is black to a person's eyes if it absorbs all incident light or doesn't reflect anything, and white if it doesn't absorb and reflects all.
My question is which specific property allows black objects to absorb and stops white objects from absorbing.
| A black object is black because it has many internal degrees of freedom that are in the energy range of the visible spectrum. This leads to light in the visible spectrum being absorbed and the energy used to fill those degrees of freedom. In the visible range, those transitions usually correspond to electronic excitations.
| {
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Do quantum fluctuations cause variations in energy density across space? Suppose that the early universe was filled with energy uniformly. Would quantum fluctuations cause variations in energy density across space resulting in regions with higher and lower energy density around the mean?
I read about quantum fluctuations online and still don’t understand how it works so I may have got it all wrong.
|
Suppose that the early universe was filled with energy uniformly. Would quantum fluctuations cause variations in energy density across space resulting in regions with higher and lower energy density around the mean?
The inflation period and the inflaton field were proposed in order to explain the high uniformity of the cosmic microwave background, to a level of
$10^{-5}$, among other observations. The small inhomogeneities observed below that level, are explained by quantum fluctuations at the time of the inflation period.
At the time of inflation, there are no observers, only interactions of fields. The inflation model assumes perturbations in the metric (see link above) at that period . The wikipedia article on cosmic inflation states
It explains the origin of the large-scale structure of the cosmos. Quantum fluctuations in the microscopic inflationary region, magnified to cosmic size, become the seeds for the growth of structure in the Universe
it refers to quantum fluctuations as
quantum fluctuation (or vacuum state fluctuation or vacuum fluctuation) is the temporary change in the amount of energy in a point in space, as explained in Werner Heisenberg's uncertainty principle.
The map of cosmic microwave backround is the "measurement" of the fluctuations at the time of inflation. Certainly space time is perturbed during inflation and the cosmological constant allows for energy fluctuations, which are reflected in the CMB map, a snapshot of the universe at the time of photon decoupling (to come in line with the other answers)
| {
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Why isn't average speed defined as the magnitude of average velocity? Speed is usually defined as the magnitude of (instantaneous) velocity. So one could assume that average speed would be defined as the magnitude of average velocity. But instead it is defined as
$$s_{\textrm{average}} = \frac{\textrm{total distance traveled}}{\textrm{total time needed}}$$
which generally speaking is not equal to the magnitude of the corresponding average velocity.
What historical, technical or didactic reasons are there to define average speed this way instead of as the magnitude of average velocity?
| As you wrote these quantities are different and give you different information so why would you want to call the magnitude of the average velocity the average speed?
If you are travelling in a car it is the average speed for the journey which you might want even if the start and finish point were the same.
What use would it be to say that the magnitude of the average velocity was zero?
| {
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How can the solutions to equations of motion be unique if it seems the same state can be arrived at through different histories? Let's assume we have a container, a jar, a can or whatever, which has a hole at its end. If there were water inside, via a differential equation we could calculate the time by which the container is empty.
But here is the thing: through the differential equation, with initial condition, I shall be able to know everything about the container: present past and future.
But let's assume I come and I find the container empty. Then
*
*It could have always been empty
*It could have been emptied in the past before my arrival
So this means I am not able to know, actually, all its story. Past present and future.
So it seems there is an absurdity in claiming that the solution of the differential equation is unique. Where am I wrong?
| An ODE for some physical situation is just some mathematical model of reality. Such models always have limitations and almost always only holds in a probabilistic sense.
Let's take a very simple example and assume the container is filled with atoms of some radioactive material that decays with a half-life $T$. The ODE describing the expected number of atoms in the jar is $N'(t) = -\frac{N(t)}{T}$ with solution $N(t) = N(0)e^{-t/T}$ which is uniquely determined by specifying $N(0)$. If we wait for a long time then eventually $N(t) \ll 1$ for which we don't expect to find any radioactive atoms in the jar. So even if we start of with $N(0) = 1000$ or $N(0) = 2000$ atoms we will in both cases end up with an empty jar for large $t$.
This does not contradict uniqueness of the solution to the respective ODE since the ODE describes the expected number (a probabilistic quantity) not the actual number of atoms. The solution to the ODEs will have a non-zero value for any time $t$ even if the jar is empty (but we can't access this value with our one observation of the system).
| {
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Magnetic field of a solenoid Why, when explaining the magnetic field of a solenoid, are the individual "quasi-rings" split up into two layers with opposite direction of current (see attached pictures; where the x-es and the dots indicate opposite directions). The original picture given clearly states the direction of the current, so why does it "split up into two"?
| that cross and dots indicates the direction of current "cross" means current enters that point and corresponding "dot " means current leaves that point - as shown in figure (in my figure current is in opposite direction to yours but i hope you'll understand)
solenoid is nothing but stack of rings with same central axis carrying current in same direction (here in CW direction looking from right side )
and that two loops in your figure indicates nothing but direction of magnetic field away from the axes of solenoid . remember it looks same as like we get in the ring
in gist i want to say take a current carrying ring as shown below and put them togeather with there central axis coinciding you will get same magnetic field lines like solenoid
| {
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How to simulate possible trajectories of particles after $\beta$ decay? I'm a programmer trying to simulate the movement of the particles involved in $\beta^-$ decay, or at least an approximation of it, for fun, in a 2D universe. I would like to keep the simulation realistic in some ways though, so I would like to obey conservation of energy and conservation of momentum.
So far I have calculated the total energy and momentum of the original neutron (and checked to ensure there is enough energy for $\beta^-$ decay to occur). I have made the proton continue along the original trajectory of the neutron at the same velocity, and created a neutrino traveling at near light speed (trajectory as yet undecided). Then I subtracted the momentum of the proton from the original momentum (per axis), leaving me with the available momentum to distribute between the electron and neutrino (per axis). Then I took the total energy and subtracted from it the total energies of the proton and neutrino and the mass-energy of the electron, leaving me with the kinetic energy of the electron, which I then used to find its velocity (sans trajectory).
This leaves me with a system of equations,
$$ |v_n|^2 = \mathbf{v_{xn}}^2 + \mathbf{v_{yn}}^2 $$
$$ |v_e|^2 = \mathbf{v_{xe}}^2 + \mathbf{v_{ye}}^2 $$
$$ p_x = m_e \mathbf{v_{xe}} + m_n \mathbf{v_{xn}} $$
$$ p_y = m_e \mathbf{v_{ye}} + m_n \mathbf{v_{yn}} $$
where $_e$ and $_n$ stand for electron and neutrino, and $T$ stands for kinetic energy. All unknowns are bolded (4 total).
*
*Are there infinite solutions?
*Given that, how can I pick a solution at random?
| Yes there are an infinite number of solutions.
Indeed, this is how the neutrino was discovered. If a neutron simply decayed to a proton and an electron then the electron KE would always be the same fixed value. However when we measure the electron energy in beta decay we find there is a spread of energies. Wikipedia shows a typical spectrum of electron energies in beta decay:
This can only happen if there is some third body involved, and it was for that reason that Pauli proposed a neutrino was also produced in beta decay.
| {
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Why is the pressure same in a horizontal plane in a static fluid? One of the common explainations involve a cylindrical flask filled with fluid and Newton's second law. But such explainations is specific. What could be possible explanation in a case in which the fluid is filled in some randomly shaped flask?
| Let's take two random points, $x_1$ and $x_2$, lying in the same horizontal plane in a liquid inside a randomly shaped flask.
Now let's draw a very skinny right circular horizontal cylinder, such that points $x_1$ and $x_2$ lie in the centers of its two circular bases , each having area A.
The only horizontal forces acting on the two bases of the cylinder are $p_1A$ and $p_2A$. Since the bases of the cylinder are very small, we assume here that the pressure at each of the bases is uniform.
Naturally, for the cylinder to be in static balance, $p_1A$ has to be equal $p_2A$, and, therefore, $p_1$ has to be equal $p_2$.
If, due to a complex shape of the flask, points $x_1$ and $x_2$ cannot be connected directly, we'll draw a series of horizontal and vertical cylinders going down from both points until a direct horizontal connection can be made. Then, we'll apply the same logic we used earlier for each horizontal cylinder and the same logic you have referred to for each vertical cylinder.
| {
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How to calculate the dipole potential in spherical coordinates I want to calculate the dipole potential in spherical coordinates. I know that the potential can be calculated with
$$ \phi = - \int \mathbf E \cdot\mathrm d\mathbf r,$$
but I don't know the electric field. I would say
$$ \mathbf E = \frac{1}{4 \pi \epsilon_0 r^3} ( 3p\,\hat{\mathbf {r}} \cos(\theta) -\mathbf p) $$
is the electric field in spherical coordinates but I'm not sure because I didn't calculate this formula, in fact I got it from a book and don't understand the way it's calculated.
|
see above is a dipole with electric dipole moment $\mathbf{\vec{p}}$
since you know potential due to dipole it means
$$\phi(r,\theta)=\dfrac { \mathbf{\vec{p}}\cdot\mathbf {\hat r}}{4\pi \epsilon_{0}r^2}=\dfrac{p\cos\theta}{4\pi\epsilon_{0}r^2}$$
To find radial component of field
$$E_{r}=-\dfrac{\partial \phi}{\partial r}=\dfrac{2\ p \cos\theta}{4\pi\epsilon_{0}\ r^3}$$
To find angular component of electric field
$$E_{\theta }=-\frac{1}{r}\cdot\frac{\partial \phi}{\partial \theta}=\dfrac{p\sin\theta}{4\pi\epsilon_{0}\ r^3}$$
therefore,
$$\mathbf{E}_\text{dip}(r,\theta)=\dfrac{p}{4\pi\epsilon_{0}\ r^3}\left(2\cos\theta\ \mathbf {{\hat r}}+\sin \theta\ \mathbf {\hat \theta}\right)$$ (in polar cordinate form)
$$\mathbf{E}_\text{dip}=\dfrac{1}{4\pi\epsilon_{0}\ r^3}\left(3(\mathbf {\vec p} \cdot \mathbf {{\hat r}})\mathbf{\hat r}-\mathbf{\vec p }\right)$$
(in coordinate free form)
$$\mathbf{E}_\text{dip}(r,\theta)=\dfrac{1}{4\pi\epsilon_{0}\ r^3}\left(3(p\cos\theta)\mathbf{\hat r}-\mathbf{\vec p }\right)$$
| {
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Why don't we use infrared light to heat food? Why don't we use infrared (IR) or even the far IR just to heat food in a microwave oven instead of, of course, the conventional 2.45 GHz microwaves? Don't people call IR heat waves?
| We do use (near) infrared radiation to heat food – whenever we toast food or grill (UK)/broil (US) by beaming infrared downwards on to food! The point is that the infrared is strongly absorbed by the food we cook in this way, and doesn't penetrate significantly beyond about a millimetre. So the surface of the food is strongly heated – seared, toasted or scorched! What lies below the surface is cooked much more slowly, mainly by conduction of heat from the surface.
Microwaves are not as strongly absorbed and penetrate much further, so the food is 'cooked from the inside'. The microwaves are mainly absorbed by water molecules that are sent into a vibratory/rotatory motion by the electric field of the microwaves acting on the (polarised) molecules. These are forced oscillations, but not at resonance; the frequency of the microwaves (about 2.4 GHz) is not a natural frequency for the molecule. If it were, the microwaves would be absorbed by the surface layer, and we'd have another grill or toaster!
Edit (prompted by comment below). I don't mean to give the impression that water molecules are the only ones that absorb microwaves. Fats are also strong absorbers.
| {
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$\sqrt{-1}$ coefficient in a function In a simple harmonic oscillator with $\ddot{x} = -\omega^2x$, it can be shown through differentiation that one solution can be given by $\dot{x} = i \omega Ae^{i \omega t}$. What does the factor of $i$ do here? What effect does it have on velocity?
| $$
\ddot x + \omega^2 x =0
$$
Is a second order linear differential equation so there will be two solutions as the basis functions adding which you can get any solution. What you have to note is that the equation allows complex solutions. That doesn't mean you have to take the complex solution. In particular if you start with purely real initial conditions $x(0), v(0) \in \mathbb R$, the evolution will not make the solution develop an imaginary part. You should try to convince yourself of this using either analytical or numerical solutions.
Thus the solution you write can never be a physical solution. So your question is meaningless. You have to construct solutions that are real using the basis $e^{\pm i\omega t}$. Or you could simply work in the basis $\sin(\omega t), \cos(\omega t)$ as long as the coefficients multiplying them are real.
So the bottomline is that the mathematical equation allows non-physical solutions and it is upto the physicist to filter them to reflect experimental observations.
| {
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How are different metrics for spacetime constructed? is a metric something derived? for example where does the extremal Reissner Nordstrom metric come from?
| A metric in general relativity is a solution to a wave equation called the Einstein field equation (EFE). If you're familiar with Maxwell's equations for electromagnetism, the EFE is very similar. It relates the field to its sources. In a vacuum, its solutions are waves.
The Reissner Nordstrom metric is a general solution to the Einstein field equations that satisfies certain other conditions. It's asymptotically flat, i.e., it looks like empty space at large distances. It's static and stationary. It contains an electric field. It's axially symmetric.
I don't know if that's what you were asking, or if you understood all that and wanted to know how people would find closed-form solutions. There is a variety of techniques for finding closed-form solutions, and none of them is a general technique -- all of them are just tricks that work in certain cases. For example, one way of guessing certain solutions in spherical coordinates is to write down a metric of the form $ds^2=Adt^2-Bdr^2-r^2d\Omega^2$, then apply the EFE to get a differential equation for $A$ and $B$ as functions of $r$. This is probably how Reissner and Nordstrom did it, since they were working in the very earliest years of GR. Circa 1970, I think people developed new, coordinate-independent techniques that didn't depend on writing down a metric in a particular coordinate chart. I believe these methods were more algebraic, but I don't know much about them.
| {
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Does anti-blue coating work less effective on high power lens? I purchased a spectacle from an online website with anti-blue coating. The features of this product is mention here.
Once I receive it, I started using the product without testing the anti-blue coating, recently, I visited the optics store and out of curiosity I started to test the anti-blue feature of my specs, only then I realised that my specs is letting the (70-80% intensity) blue light to pass through without obstructing it.
When I called the online store they gave me explanation that I have high power hence it is not blocking it. My eye power is -5.25 and -5.75 without any cylindrical power.
I searched on the internet and couldn't find out any explanation to scientifically prove that their explanation is correct. Because as per my understanding it is a simple coating over the lens which doesn't allow to blue light to pass and so it should not affect my power.
Can anyone please help me understand how this is possible, I mean if the lens power matters or not?
| The blue is short wavelength and the coating might be designed for flat surface.
So the higher curvature will change a bit the angle of incidence and can make the coating less effective.
It is likely a design issue of the coating.
| {
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How can I calculate the impulse caused by a nuclear bomb on structures? I want to calculate the impulse caused by a nuclear bomb on structures. Is there a function that I can use to calculate the amount of impulse acted upon by a nuclear bomb depending on its size? Also, is there a function relating impulse to overpressure? How would I calculate overpressure? I am very new to this topic, and google hasn't been much help!
| I presume you are addressing only blast (pressure damage from the shock wave) and not thermal effects. The blast magnitude depends on the position where the weapon is detonated: high altitude, low altitude, or on the ground. I presume you are considering low altitude air detonation.
There are detailed relationships and graphs for estimating the overpressure. The ones I have are official use only, so unfortunately, I cannot provide them here. I would be highly skeptical of information on the web.
Sorry I cannot provide more details here.
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Klein-Gordon-Equation contains no Spin I have a question about an argument used in Schwabl's "Advanced Quantum Mechanics" concerning the properties of the Klein-Gordan-Equation (see page 120):
Since the eigenenergies of free solutions are $E= \pm \sqrt{p^2c^2+m^2c^4}$the energy states aren't bounded from below. But I don't understand why that then K-G-equation provide a scalar theory that does not contain spin and then could only describe
particles with zero spin.
Intuitively, I guess because that spin can't regard by a scalar equation but I find this "argument" too squishy and would like to know a more plausible argument.
| The components of every field must satisfy the KG equation, regardless of its spin. This makes sense, since it is nothing more than just the Einstein energy momentum relation.
The idea is that the KG equation doesn't require that particles be spinors, so it can't be the full story since we know particles have spin (i.e. they transform under the 2 representation of the Lorentz group).Sure, it would hold for all components of a spinor, but any good theory should force us to conclude that an electron field is spin 1/2.
However, maybe there is a more fundamental equation that encodes the spin information of a particle and yields the KG equation. This is the Dirac equation. The Dirac equation necessitates that (what we now call) spin 1/2 particles are represented by the direct sum of Weyl spinors. Specifically a direct sum of the $2$ and $\bar{2}$ representations of the Lorentz group. This precisely means that, say, the electron field necessitates the existence of an anti electron field, which together are described by a spin 1/2 field and an anti spin 1/2 field.
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How do I know that gauge fields are bosons? QED and the Dirac equation have field operators $\psi$ interact with a gauge field $A^{\mu}$.
We identify $\psi$ as a fermionic field and $A^{\mu}$ as a gauge boson - the photon.
Do we or can we know that one is a fermion and the other is a boson?
Or do we get that information from the commutation relations when we quantise the theory?
| The possible Lorentz invariant wave equations is a relatively limited set of equations. Most wave equations one can write will not be Lorentz invariant under any transformation law you can think of.
We know all the possible finite dim Lorentz invariant field equations. One of the properties of these equations is that each of them have a distinct number associated with them that relates to their angular momentum. We need some representation theory to exactly explain what I mean by this, but trust me that there is a number. This number represents how rotations effect the equations, and it is in fact the "Equation's spin". When quantised, this is the particle's spin.
So once you see a particle's wave equation, you can always tell the spin of the particle. From this point you can look at the gauge field's differential equation and immediately tell what is the gauge particle's spin. You can also ask the reverse question - what are the possible wave equations of a spin 7/2 particle?
The main point is, that the possible Lorentz invariant field equations is such a limited set, that we can tell from the wave equations themselves everything there is to know about the particles they represent. If the particle violates these rules, it cannot be Lorentz Invariant.
The kinetic term of the gauge fields, i.e. $ F_{\mu \nu}F^{\mu \nu}$, is only Lorentz invariant if it has spin 1. You can write different Lagrangians with different kinetic terms, and have gauge fields with different spins.
(The technical name for the "equation's spin" is the total angular momentum of the representation of the Lorentz group of the field)
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/428080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
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