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Temperature on the surface of the sun calculated with the Stefan-Boltzmann-rule In a German Wikipedia page, the following calculation for the temperature on the surface of the Sun is made:
$\sigma=5.67*10^{-8}\frac{W}{m^2K^4}$ (Stefan-Boltzmann constant)
$S = 1367\frac{W}{m^2}$ (solar constant)
$D = 1.496*10^{11} m$ (Earth-Sun average distance)
$R = 6.963*10^8 m$ (radius of the Sun)
$T = (\frac{P}{\sigma A})^\frac{1}{4} = (\frac{S4 \pi D^2}{\sigma 4\pi R^2})^\frac{1}{4}=(\frac{SD^2}{\sigma R^2})^\frac{1}{4} = 5775.8\ K$
(Wikipedia gives 5777K because the radius was rounded to $6.96*10^8m$)
This calculation is perfectly clear.
But in Gerthsen Kneser Vogel there is an exercise where Sherlock Holmes estimated the
temperature of the sun only knowing the root of the fraction of D and R.
Lets say, he estimated this fraction to 225, so the square root is about
15, how does he come to 6000 K ? The value $(\frac{S}{\sigma})^\frac{1}{4}$
has about the value 400. It cannot be the approximate average temperature
on earth, which is about 300K. What do I miss ?
| A rough estimate of a body's temperature in the solar system is
$$T=\frac{280K}{\sqrt{D_{AU}}}$$
if we calculate the AU fraction from the Sun's "edge" to its center, R over D = $4.65x10^-3$, and substitute this into the formula, the Sun's temperature would be about 4100K.
Not very close to your 5776 K, but utilizes the square root of the R D fraction.
The formula reflects effective temperatures. However peak, so called sub-solar temperatures, are $\sqrt{2}$ times effective temperatures, which would yield about 5800K.
Clever Sherlock!
| {
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Oscillations in forces other than the Weak As I understand it neutrino oscillations arise due to the neutrino mass eigenstates being distinct from the neutrino flavour eigenstates.
Flavour eigenstates are the states in which neutrinos interact via the weak force, and so are the eigenstates in which they are created and detected, and consist of a superposition of the the mass eigenstates. The mass eigenstates correspond to how the neutrino propagates through spacetime. The finite neutrino mass causes the relative phase of each mass eigenstate to change as the neutrino propagates, changing the probability that it will be detected in a given flavour eigenstate (hence an electron neutrino can oscillate into a muon neutrino and so on).
My question is, can other forces exhibit similar oscillations? For example, in principle is it possible for an electron to oscillate it's charge state as it travels, to be detected as a positron some distance later? Or are charge eigenstates not the similarly related to mass eigenstates?
| Think that there is some confusion. I assume that you associate particle oscillations with the weak interaction because you consider the specific example of the neutrinos (that indeed can only weakly interact)
See also the wikipedia article about neutral particle mixing:
You need a particle $A$, that is different from its antiparticle $B$, but with the same decay-products $F$. You then get the possibility for oscillations:
$$A → F → B → F → A → ...$$
More concretely: it is also experimentally established that neutral kaons oscillates.
Kaons are produced by the strong interaction in eigenstates
$$\overline{K}^0, K^0$$
but they decay by the weak interaction as CP-eigenstates
$$K_1,K_2.$$
Since quark flavour is not conserved in weak interactions transitions between neutral mesons and their particles are possible.
What is observed is that:
An initially pure beam of $K_0$ will turn into its antiparticle while propagating, which will turn back into the original particle, and so on. (Wikipedia)
| {
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Is it more efficient to stack two Peltier modules or to set them side by side? Is it more efficient to stack two Peltier modules or to set them side by side?
And why?
I have a small box that I want to cool down about 20 K below ambient -- cold, but not below freezing.
(I want to keep my camera cool, so I'm putting in this cool box.
The camera looks through a flat glass window on one side of the box).
The heatsink I have on hand is about twice as wide as the widest Peltier module I originally planned on using.
So there's room to put 2 Peltier modules side-by-side under the heatsink.
Or I could center a stack of 2 Peltier modules under the heatsink.
Which arrangement is more efficient?
I have to cut a bigger hole in the insulation for the side-by-side arrangement, so the unwanted heat that "back-flows" through the side-by-side arrangement is worse.
On the other hand, other effects are worse for the stacked arrangement.
(Is https://electronics.stackexchange.com/ a better place to post questions about Peltier coolers?)
| You'll want a much bigger heatsink!!
(and maybe just one TEC)
If it's being cooled only by convection then maybe a heat sink area* that is 10 times that of the TEC. (maybe bigger)
The classic mistake with a TEC is to make the heat sink too small. With too small a heatsink the hot side of the TEC gets hotter, more thermal leakage through the TEC, it has to work harder to keep the same temperature.. and the whole thing goes into thermal runaway.
*one should really talk about the volume of the heatsink.
| {
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Quick question on sketching wavefunction in well
Usually for an infinite well, the sketch for n=3 level is this:
Now I think if one side of the potential barrier is higher, the particle will be more likely to spend time on the left side than the right side, so the wavefunction should have higher amplitudes on the left (skewed to the left):
| Just so this doesn't slip past:
Now I think if one side of the potential barrier is higher, the particle will be more likely to spend time on the left side than the right side, so the wavefunction should have higher amplitudes on the left (skewed to the left):
This is incorrect. Between A and B the well is deeper, so the particle goes faster. Between B and C the well is shallower and the particle goes slower, so it takes a longer time to cross this region. If you take a snapshot at a random time, it will be more likely to be between B and C, and the wavefunction amplitude there is higher:
As Gert mentioned, for a finite well there is also a slight penetration (tunnelling) of the walls, giving an exponential tail there instead of strict zeros at A and C.
| {
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Speed distribution in 1 dimension In 3D, the maxwell velocity distribution is:
$$f = \left(\frac{\alpha}{\pi} \right)^{\frac{3}{2}} e^{-\alpha v^2} d^3 \vec v$$
To get the speed distribution in 3D, we simply expand $d^3\vec v = 4\pi v^2 dv$
Thus in 3D, the maxwell speed distribution is:
$$w = 4\pi^2 \left(\frac{\alpha}{\pi} \right)^{\frac{3}{2}}v^2 \space e^{-\alpha v^2} d v $$
In 2D, the maxwell velocity distribution is:
$$f = \left(\frac{\alpha}{\pi} \right) e^{-\alpha v^2} d^2 \vec v$$
To get the speed distribution in 2D, we simply expand $d^2\vec v = 2\pi v\space dv$
Thus in 2D, the maxwell speed distribution is:
$$f = 2\pi\left(\frac{\alpha}{\pi} \right) v \space e^{-\alpha v^2} d \vec v$$
In 1D, the maxwell velocity distribution is:
$$f = \left(\frac{\alpha}{\pi} \right)^{\frac{1}{2}} e^{-\alpha v^2} d \vec v$$
Following the same line of thought, how do I get the 1D speed distribution?
| You should simply multiply by two, to get
$$f =2 \left(\frac{\alpha}{\pi} \right)^{\frac{1}{2}} e^{-\alpha v^2} \text d v.$$
This is because your integral over speed will now be from zero to infinity, and you need to 'fold over' the integral from minus infinity to zero. Otherwise, there are no further geometrical factors - the integral $\int\text d\vec v=\int_{-\infty}^\infty \text dv$ is alreadyin the form you need it, and you just need to figure out what to do with negative velocities.
| {
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Where does the term "boost" come from for rotation-free transformations? I had never seen rotation free transformations called "boosts" (I think I have it right) before reading some questions here. I'm too old perhaps. I have not found the etymology after some searching, though it sounds like something V.I. Arnold would think up, or jargon from inertial navigation. Anyone know where/how it started or was popularized? (If it is in MTW or Ohanian (old edition) or Weinberg, I promise I'll facepalm)
| Not necessarily the original source of the term, but the earliest use I can find, occurs in Brandeis University Summer Institute in Theoretical Physics, 1964 Vol. 1: Lectures on General Relativity, where on p. 208:
If we use the term boost for a Lorentz transformation from one frame to another parallel to it but with a uniform velocity relative to it, then we can analyze the motion according to the type of transformation as follows: ...
No particular reason as far as etymology is given, though one can guess some plausible analogies. Notably, one can find easily find some uses within a few years after the above on Google Books, including one briefly gives some synonyms for it,
All of these examples of so-called "pure Lorentz transformations" or "accelerations" or "boosts" (perhaps boost is preferable, as it does not invite misunderstandings) i.e., transformations of the type...
as well as a reason to prefer "boost" (besides simply being less of a mouthful, I suppose).
| {
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Dielectric boundary I am trying to determine why electric field may be confined to a certain region if there is a large difference in the permitivity for example if electric field flows through water and then reaches a water air boundary.
I have also been reading about EM waves, is it possible to model electric field as a wave because if so then the transmission T and reflection R coefficients given below in terms of n which is $\propto \epsilon_{r}^{1/2} $ and so if $\epsilon_{1}>>\epsilon_{2}$ then $n_{1}>>n_{2}$ ;
$\displaystyle R$ $\textstyle =$ $\displaystyle \left(\frac{n_1-n_2}{n_1+n_2}\right)^2, \rightarrow 1$
$\displaystyle T$ $\textstyle =$ $\displaystyle \frac{n_2}{n_1}\left(\frac{2 n_1}{n_1+n_2}\right)^2 \rightarrow 0.$
and so it is clear that the wave is reflected at the boundary, is this approach valid ?
http://farside.ph.utexas.edu/teaching/em/lectures/node103.html
| One way to understand what append at the boundary of two dielectrics is to use the Fresnel formula when you know about the indices of your media.
Then, you have to solve the wave equation (d'Alembert equation) with the boundaries condition given by thoses Fresnel coefficient.
The confinement is due to the boundaries conditions betwen the two dielectrics. To explain it, you don't need the microscopic approach give by your link.
| {
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Is there any operator behind probability, in quantum mechanics? In Quantum mechanics, the probability of finding a particle at position $x$ is given by $|\psi(x)|^2$, where $\psi$ is the wave function. Wonder what is the operator which gives this probability? Is probability the result of any operator acting on $\psi$?
| No. These cancerous probabilities come in because of the probabilistic interpretation, which also brings in all kinds of famous paradoxes that infect QM. By itself, the formalism of the theory requires just the solution of a second order differential equation to calculate $\psi(x,t)$ - a process which is completely deterministic, just like solving something coming out of Newton's second law. (I'm talking about the Schr\"odinger equation solution here, Heisenberg's matrix mechanics gives identical answers.)
The formalism of the theory and this interpretation are absolutely independent issues. One can have another interpretation (e.g. Many-Worlds, besides others that you can find mentioned in the link) tacked on to the same formalism, which would give us a different way of making sense of these answers. But by itself, there is nothing in the formalism of QM that requires or necessitates probabilities.
So, your answer is - no. Probability is not the result of any operator acting on $\psi$.
| {
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Why are the 'color-neutral' gluons confined? What makes the two 'color-neutral' gluons
$(r\bar r−b\bar b)/\sqrt2$ and
$(r\bar r+b\bar b −2g\bar g )/\sqrt6$ different from the pure $r\bar r +b\bar b +g\bar g $ ?
Why don't they result in long range (photon-like) interactions?
| There is no fundamental difference between the gluons $(r\bar{r}-b\bar{b})/\sqrt{2}$ and $(r\bar{b} + b\bar{r})/\sqrt{2}$. The first one is represented by the matrix
$$
Z = \frac{1}{\sqrt{2}}\left(\begin{array}{rrr} 1&0&0 \\ 0& -1 & 0 \\ 0 & 0 & 0\end{array}\right)$$
and the second by the matrix
$$
X = \frac{1}{\sqrt{2}}\left(\begin{array}{rrr} 0&1&0 \\ 1&0&0 \\ 0 & 0 & 0\end{array}\right).$$
However, these two matrices are related by the change of basis
$$
H = \left(\begin{array}{rrr}1/\sqrt{2}&1/\sqrt{2}&0 \\ 1/\sqrt{2}& -1/\sqrt{2} & 0 \\ 0 & 0 & 1\end{array}\right).
$$
It is easy to check this by multiplying matrices and seeing that $HZH^\dagger = X$.
Thus, if you call $(r\bar{r}-b\bar{b})/\sqrt{2}$ "color-neutral" and $(r\bar{b} + b\bar{r})/\sqrt{2}$ "non-color-neutral", it is clear that "color-neutral" is not a property that is invariant under change of basis, and thus is not a meaningful property in quantum chromodynamics.
Actually, neither of these gluons is color-neutral.
| {
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Does the temperature of water determine how much heat will be removed from air used to evaporate it? This is a question about evaporative cooling as used in residential evaporative cooling appliances. This type of cooling uses the heat in the ambient outside air to evaporate water and remove the heat from the air, then push the cooled air inside. The equation to predict the temperature of the resulting air after it's given up its heat to evaporate the water is as follows:
$$T_{output} = T_{dry} - (T_{dry} - T_{wet}) * \epsilon$$
where $T_{output}$ is the output air temperature, $T_{dry}$ is the air temperature of the dry bulb, $T_{wet}$ is the air temperature of the wet bulb, and $\epsilon$ is the cooling efficiency.
For example, on a very dry summer day (dry bulb 95 degrees, wet bulb 60 degrees) my evaporative cooler with 90% efficient media is capable of cooling the air to 63.5 degrees.
However, this equation does not seem to take into account the temperature of the water itself. Does it matter? Intuitively, it would seem to make sense to me that hotter water would be easier to evaporate, since it's closer to its boiling point. Or maybe colder water is better because it will absorb more heat from the air? Or maybe it's a wash because the same amount of heat is required, but with hotter water, more is needed because it will evaporate faster? Help me understand this.
| After evaporating the water is at the same temperature as the ambient air. The heat removed from the air is whatever it takes from the starting condition of the water to get to water vapor at that temperature. In particular, cooling the water will require the air to heat it, so each degree C the water is colder will require about an additional 1 cal/g=4.184 Joules/g from the air.
| {
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Calculating probability current for scattering problem I'm trying to calculate the probability current for a scattering problem. The potential is $V = V_0 > 0$ in $x>0$, with $E>V_0$
So I have in the region $x \le 0$:
$$\psi = \exp(ikx) + R \exp(-ikx)$$
And in $x>0$
$$\psi = T \exp(i \kappa x)$$
I am trying to calculate the probability current, $j = \frac{-ih}{2m} (\bar{\psi}\psi' - \bar{\psi'}\psi)$, in each region and show that it is equal.
However when I calculate the probability current in $x<0$, I get:
$$\bar{\psi} = \exp(-ikx) + \bar{R}\exp(ikx)$$
$$\psi' = ik\exp(ikx) -ikR\exp(-ikx)$$
$$\bar{\psi'} = -ik\exp(-ikx) + ik\bar{R}\exp(ikx)$$
$$\psi = \exp(ikx) + R\exp(-ikx)$$
So:
$$\bar{\psi} \psi' = ik -ikR\exp(-2ikx) + \bar{R}ik\exp(2ikx) - ikR\bar{R}$$
$$\bar{\psi}' \psi = -ik -ikR\exp(-2ikx) +ik\bar{R}exp(2ikx) +ik R\bar{R}$$
Hence,
$$j = \frac{hk}{2m} (1 + R \exp(-2ikx) - \bar{R} \exp(2ikx) - R\bar{R})$$
And in the region $x>0$:
$$j = \frac{\kappa h}{2m}(T\bar{T})$$.
I can show that (by imposing continuity conditions at the boundary):
$$k(1-|R|^2) = \kappa |T|^2$$
So I would expect the first probability current to just be: $\frac{kh}{2m}(1-|R|^2)$.
Any help with this issue is much appreciated! I'm pretty sure I'm making some stupid mistake somewhere, but it's very frustrating as I cannot find it!
Thanks
| You say
$$j = \frac{hk}{2m} (1 + R \exp(-2ikx) - \bar{R} \exp(2ikx) - R\bar{R})$$
But instead as
$$j=\frac{-i\hbar}{2m}(\bar\Psi\Psi'-\bar\Psi'\Psi)=\frac{-ih}{4\pi m}(\bar\Psi\Psi'-\bar\Psi'\Psi)$$
In $x<0$,
$$\bar{\psi} \psi' = ik -ikR\exp(-2ikx) + \bar{R}ik\exp(2ikx) - ikR\bar{R}$$
$$\bar{\psi}' \psi = -ik -ikR\exp(-2ikx) +ik\bar{R}exp(2ikx) +ik R\bar{R}$$
So
$$j = \frac{-i\hbar }{2m} (\bar{\psi}\psi' - \bar{\psi'}\psi)=\frac{-i\hbar}{2m} (2ik-2ikR\bar R)=\frac{-i\hbar}{2m} .2ik(1-1R\bar R)=\frac{hk}{2\pi m}(1-|R|^2)$$.
And in $x>0$,
$$\psi=T\exp(ikx)$$
$$\bar{\psi} \psi' = ik\bar TT$$
$$\bar{\psi}' \psi = -ik\bar TT$$
$$j=\frac{-i\hbar }{2m} (\bar{\psi}\psi' - \bar{\psi'}\psi)=\frac{-i\hbar }{2m} (2ik\bar TT)=\frac{-i\hbar }{2m} 2ik(|T|^2)=\frac{hk}{2\pi m}(|T|^2)$$
Your mistake was simply subtraction of two
| {
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Interpretation of the WIMP annihilation cross section graph I have some trouble in the interpretation of the WIMP cross-section annihilation versus their mass.
I understand that the lines represent a upper bound on the cross section from the observation. But what do the contours from, e.g., DAMA mean?
| The DAMA collaboration have claimed to have actually detected dark matter, but this is in conflict with the results of other searches. But here one is assuming some model that describes the way dark matter interacts with nuclei and the properties of the dark matter halo. (e.g. spin independent WIMP-nucleus interactions). Then what is going on here is that the DAMA results are consistent with the WIMP mass being in some range and the cross section being in some range, so there will be an area in the figure within which the parameters must be in. However, that area will be in the excluded range of the other searches, as the results from those are inconsistent with the DAMA results.
Now, you can still believe that the DAMA results are correct, because one always has to make assumptions when interpreteting the results of the dark matter searches. E.g. a low mass WIMP is still possible in some models. You can assume that dark matter particles are not WIMPS etc. etc.
| {
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Moment of inertia of a cylinder When I tried to calculate the moment of inertia ($I_C$) of a cylinder (mass M, height H, radius R) around the rotating axis going symmetrically through its middle, I came up with a different result than expected ($\frac{1}{2}MR^2$), but I do not spot my mistake, since my calculation makes perfect sense to me:
$$
I_C := \int_V{ρr^2dV} = \int_0^H{\int_0^{2π}{\int_0^R{ρr^2 dr dφ dh}}} = ρ \int_0^H{\int_0^{2π}{\frac{R^3}{3} dφ dh}} = ρ \cdot 2πH \frac{R^3}{3} = Vρ\frac{R^2}{3} = \frac{1}{3}MR^2
$$
Can anyone spot what's wrong?
| Lets start from the general definition for the moment of inertia
$$I=\int_{0}^{M}r^{2}dm$$
The mass element is $dm=\rho dV$ with $dV=L2\pi rdr$ ($L$ being the length of the cylinder). Substituting you'll get
$$I=2\pi\rho L\int_{0}^{R}r^{3}dr$$
.Taking into account that $\rho=\frac{M}{\pi R^{2}L}$ can you spot you're mistake?
| {
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What is intrinsic gravitational entropy? What is intrinsic gravitational entropy? Does it have to do with dark matter or coarse graining in the universe? Is it unique to general relativity, or there are predictions from quantum mechanics as well? Please explain in detail.
| The statement is by Hawking. Assuming I understand him correctly he means that there is an entropy associated with gravitational fields that is quite separate from any any entropy associated with matter or radiation. In other words this entropy is intrinsic to the gravitational field and will always be present regardless of the arrangement of matter and energy.
The obvious example of this is the entropy associated with a black hole. This entropy depends only on the area of the event horizon and not on whatever matter fell into the black hole.
| {
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When a planet has a high gravity, is it impossible to build and launch a successful chemical rocket to space? Just recently a a large rocky planet has been discovered. "Astronomers have discovered a new type of rocky planet beyond the solar system that weighs more than 17 times as much as Earth while being just over twice the size"
http://www.reuters.com/article/2014/06/02/us-astronomy-exoplanet-idUSKBN0ED29V20140602
Then I was thinking, if aliens lived on this planet, could they get off the planet with a normal chemical rocket, or would the rocket have to be so massive that it could not be built? or so heavy that the amount of fuel required would push the weight too far so that it could never reach space?
I wonder is there are sad planets where aliens never made it into space due to the large amount of gravity.
| It would be possible, though much more complicated than on earth.
Chemical rockets on earth deliver ~3% of it's launch weight to low earth orbit.
On this superheavy planet some 0.1-0.01% of launch weight might be delivered to low orbit.
While human exploration would be extremely hard in such conditions, it would be possible to explore the universe with tiny automated probes.
| {
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Physical Interpretation of four velocity in GR I'm confused about the physical interpretation of the four-velocity $U^\mu=\frac{dx^\mu}{d\tau}$ in General Relativity. I know that it is a tangent vector to a particle's "worldline", but what does this mean more physically?
For example, I am comfortable with what $U^\mu$ means in Special Relativity. In your inertial frame, you cover a distance $\Delta x^\mu$ and your clock says time $\Delta \tau$ has passed, and by taking the limit as $\Delta \tau \to 0$ this defines your $U^\mu$.
But I'm unsure about what $U^\mu$ means in curved space, or even in an accelerated reference frame. In either case the frame is no longer an inertial frame, which makes it confusing to interpret $\tau$, because it's no longer the "proper time in a frame", there is no one frame we are working in.
| The 4-velocity is defined to be normalized, i.e.:
$g_{\mu\nu}u^{\mu}u^{\nu}=-1 \; ,$
so if you choose comoving observers, for which $u_i=0$, then $u_{0} = \frac{1}{\sqrt{|g_{00}|}}$.
From the above formula you can read the difference between the time felt by the comoving observer and the proper time, indeed $\frac{dx^0}{d\tau} = \frac{1}{\sqrt{g_{00}}} \rightarrow d\tau^2 =g_{00}dx^0$.
this should clarify which kind of frame you should consider in your sentence: "..proper time in a frame", there is no one frame we are working in."
| {
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Moon's pull causes tides on far side of Earth: why? I have always wondered and once I even got it, but then completely forgot. I understand that gravity causes high and low tides in oceans, but why does it occur on the other side of Earth?
| First we must understand a little what is meant by "tide." A tide is the difference of gravitational force an object feels across its volume from another object. In the Earth's case the side closest to the moon feels a stronger force pulling it towards the moon than the center of the Earth does, while the side opposite the moon feels a force weaker than the center of the Earth feels. The picture below (taken from this site, which is a great reference as well, especially for explaining some misconceptions about the second lunar tidal bulge) shows this. The center of the earth feels a force toward the moon as calculated by Newton's Law of Gravitation:
$$F=G \frac{m_1 m_2}{r^2}$$
while the other areas of the Earth's surface feels a slightly different force from the moon than the center of the Earth does, as demonstrated by the arrows. The side closest to the moon feels an additional force by virtue of being closer to the moon, as demonstrated by the arrows pointing towards the moon, while the side furthest away feels a less strong force, represented by the arrows pointing away from the moon (here represented as a generic satellite).
The side closest to the moon has a tidal bulge because of the additional gravitational force pulling the sea level higher than the average level, while the side opposite the moon also has a tidal bulge by virtue of the lessened force of gravity it feels being further away from the moon. So, both bulges are caused by the moon; one side feels a greater attraction, while the other side feels a smaller attraction.
| {
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Placing two similarly charged particles in space Now, I will make a hypothetical situation. Assume that we place two similarly charged particles (lets take electrons) in space. Imagine that there is no other force acting on the particles except the repulsive force and the gravitational force of the particles. In other words, only these two electrons are present in the universe. So they are free from any outside interference. Now by nature, these electrons will start moving away from each other due to the repulsive force. Since there is nothing to stop them (gravitational force will only slow them down and not stop them as it is of a lesser magnitude than the repulsive force) they will keep moving and never stop. Over here we exclude expansion of space also for no complications. Now since the particles will keep moving as there is a constant repulsive force acting on them, they will do infinite work because $Work = Force * displacement$ and the displacement over here will keep increasing. Please tell me what is the problem in my thought experiment because it violates conservation of energy.
| as they move farther apart,, the magnitude of potential energy of the system reduces(which is 0 at infinity and infinity at 0 seperation in terms of magnitude),,, and the kinetic energy increases(which in this case tends to a specific value at infinity given by (2q1q2/4.pi.e)rm)^(1/2) where r is initial seperation
),, so the sum is conserved,, which means energy is conserved,,, hence , no it does not violate the principle.
| {
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Is the scalar curvature of the Schwarzschild solution 0? The Schwarzschild solution is meant to be a solution of the vacuum Einstein equations. That is
$$R_{\mu\nu}=0.$$
So, the Ricci tensor must be null for $r>0$.
Now, if the scalar curvature is nothing but the Ricci tensor contracted, and the Ricci tensor is null, the cuvature should be zero.
Nonetheless, I have been told that the curvature of the Schwarzschild solution (in the usual coordinates) is
$$\frac{12r_s^2}{r^6},$$
which is obviously non zero.
What am I making wrong?
| You're correct that $R=0$. $R_{abcd} R^{abcd} = \frac{12 r_s^2}{r^6}$ is the Kretschmann scalar for the Schwarzschild metric, an invariant used to find the true singularities of a spacetime. In this case, only the singularity at $r=0$ is a spacetime singularity, not a coordinate-system one.
| {
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The Helium mass fraction from the Big Bang Nucleosynthesis In Perkin's book Particle Astrophysics (page 144): I do not understand how one comes to the following expression (the second equality with $r$) for the Helium mass fraction due to the Big Bang Nucleosynthesis:
$$Y= \frac{4N_\text{He}}{4N_\text{He}+N_\text{H}}= \frac{2r}{1+r} $$
where $r=N_\text{n}/N_\text{p}$.
The first equality follows from the fact that He is (approximately) 4 times heavier than H:
$$Y = \frac{m_\text{He}}{m_\text{He}+m_\text{H}}= \frac{4N_\text{He}}{4N_\text{He}+N_\text{H}}.$$
However I can't derive the second equality relating $Y$ to $r$:
$N_\text{He}= 2N_\text{p} + 2N_\text{n}$ and $N_\text{H}= N_\text{p} + N_\text{n}$
$$Y= \frac{4N_\text{He}}{4N_\text{He}+N_\text{H}}= \frac{8(N_\text{n}+N_\text{p})}{9(N_\text{n}+N_\text{p})}\quad???$$
| The problem is you have the wrong relations between $\{N_\mathrm{H}, N_\mathrm{He}\}$ and $\{N_\mathrm{p}, N_\mathrm{n}\}$. Every hydrogen contains 1 proton, and every helium contains 2, so $N_\mathrm{p} = N_\mathrm{H} + 2 N_\mathrm{He}$. The neutrons are only contributed to by helium in the accounting: $N_\mathrm{n} = 2 N_\mathrm{He}$. Inverting these relations yields
\begin{align}
N_\mathrm{H} & = N_\mathrm{p} - N_\mathrm{n} \\
N_\mathrm{He} & = \frac{1}{2} N_\mathrm{n}.
\end{align}
It looks like you got the direction wrong, in the sense that there should be fewer helium nuclei than protons or neutrons (the 2's are on the wrong side). Also, "hydrogen" means ${}^1\mathrm{H}$ not ${}^2\mathrm{D}$ unless otherwise stated.
| {
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the energy relations in oceanography description I'm reading the following paper and do not entirely understand a point that the author is trying to make. In page 99 (2 of article) the author refers to the following equation:
$$ Q^{*} = \beta S^{*}e^{-\beta z} + 2B^{*} \delta \left(z\right) $$
and in the following paragraph he states that
"The first term on the right represents the convergence of the penetrating component S^{*} of solar radiation"
My immediate question is does "first term" refer to the $2B^{*}\delta\left(z\right)$ or does it refer to "2B"? i.e. does rhs mean rhs of the equal sign or what?
In addition, can anyone explain, in layman's terms, what this equation is referring to?
| The first term on the right is $\beta S^{*}e^{-\beta z}$
This means that the amount of light penertrating into the ocean decreases exponentially with depth $z$.
The second term on the right is $2B^{*} \delta \left(z\right)$
This takes into account heat transfer processes taking place only at the surface, such as evaportion.
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How can $J_1^2, J_2^2, J_{1z}, J_{2z}$ commute mutually? I'm reading through J. J. Sakurai's Modern Quantum Mechanics book and currently looking at the "Angular-momentum addition" part.
Here, it says you have two options and that one option is to construct simultaneous eigenket $\vert j_1j_2;m_1m_2\rangle$ of $J_1^2, J_2^2, J_{1z}, J_{2z}$ since the four operators commute with each other. I understand that $J_1^2$ and $J_{1z}$ commute, but I'm not sure how $J_1^2$ and $J_{2z}$ can commute intuitively.
"commute" means that one can measure both at once right? But total angular momentum of spin 1 and angular momentum of spin 2 are independent.
Where am I wrong here?
| Technically when two operators $A$ and $B$ commute it means that $AB = BA$, but from a physical standpoint yes it means that both observables can be measured simultaneously, and in that respect you kinda answered your own question. They're independent so they commute. For an analogy think of the spin of the electron in a hydrogen atom and its angular momentum about the CM, or think of the spins of two separate electrons. You can simultaneously measure these two quantities, why should they have anything to do with each other? Using the definition of commutativity it's a simple exercise to verify this: just operate both $J_1^2$ and $J_{2z}$ on the kets you've written and see what you get.
| {
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Heat equation: boundary conditions? Say we have a bar centered at $x=0$, that is heated. We have the 1D Heat equation which we can solve to find a parabolic temperature profile :
$$\kappa \frac{d^2 T}{d x^2}=-\frac{Q}{L S}$$
with
$$T(L/2)=T(-L/2)=T_{\infty}$$
$T_{\infty}$ is the ambient temperature at the end of the bar. This boundary condition alone is enough to solve the equation. However, now say we want to add a very tiny rod at the center of the bar that is connected to the environment to act as a path for heat flow. $R$ is the thermal resistance of the rod. $S$ is the cross sectional area of the bar. $\kappa$ is thermal conductivity of the system. $Q$ is the heat dissipated by joule heating. The corresponding boundary condition is:
$$\frac{T(0)-T_{\infty}}{R_{\rm thermal \ rod}}=\kappa S \left(\frac{d T}{dx}_{x=0^+}-\frac{d T}{d x}_{x=0^-} \right)$$
I have fiddled around with this math for a while but cannot figure out how to implement it.
In particular I am following this paper on the T-Bridge Method for thermal conductivity.
| I think the problem is that while in the first case your differential equation applies to all your domain of interest and you can just use it, in the second situation the DE doesn't apply at x=0. This means you need to solve the DE at the domains $0<x<L/2$ and $-L/2<x<0$ separately, where it does still apply.
When applying it to those domains you should remember that the temperature $T$ is still continuous throughout all the domain because you don't want infinite $\frac{d T}{d x}$; that is, infinite heat fluxes. If you do that, after you solve the DE with $T(0)$ as an undeterminate boundary value, you should find that the right hand side of your third equation is a function of $T(0)$ and thus you can find it in terms of the problem constants $T_{\infty}$, $S$, $R_{\text{thermal rod}}$, etc.
| {
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Why does my house seem to warm faster in summer than it cools in winter? In summers when we switch off the air conditioner, the room seems to instantly get hot again.
But in winter, when we switch off the heater the room seems to remain hot for some time. Why this difference?
| This question can be answered with a nice visual comparison.
Let's take a pan of water, and put it on the stove at full power. You periodically throw in an ice cube to keep it at room temperature. Obviously, the moment you stop doing this, your pan is going to heat up very quickly. Now, take that same pan and put it in a slightly colder place. Intuitively, you can see that it won't cool down as quickly as it heated up with a big stove.
The same goes for your house in summer. The sun is a massive radiator; your house is hit with about 1kW per square meter on a sunny day. It's only because a large portion is reflected that your airco unit is even slightly capable of cooling your house. Check your airco; let's say it has a cooling power of approx 3kW: it needs to take away 3kW of energy continuously, or else your room will heat up as if there were a heater of 3kW.
Now, how cold does it have to be outside to to need a 3kW heater? Typical R-values are around 6 °K m²/W, which need you need 6 degrees Kelvin (or Celcius) difference between the inside and the outside to lose one watt of power on a square meter of wall. Let's say it's -20°C outside (that's -4 Fahrenheit), and 20°C inside; i.e., 40°C difference. You'll need a whopping 450m² of wall area to need a 3kW heater. Admittedly, this neglects quite some things left and right, but it gets the point across.
TL;DR: your house is not a glowing hot fireball like the Sun, and will not lose heat quite as fast to its surroundings, as it gains heat by solar radiation.
| {
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Why $SU(3)$ and not $U(3)$? Is there a good reason not to pick $U(3)$ as the colour group? Is there any experiment or intrinsic reason that would ruled out $U(3)$ as colour group instead?
| It would be double counting, since total phase rotations of the quark wave function are already part of the model and the photon that makes them into a gauge symmetry already exists.
The total gauge group is SU(3) × SU(2) × U(1), so the question "where has the U(1) gone" has as its answer that it already was included. In a gauge theory you can only use the symmetries that you start with and you can use them only once.
| {
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Energy of an EM Wave and its temperature and amplitude I'm trying to understand why classical physics fails to explain black body radiation.
I'm confused.
According to Boltzmann, energy calculation for em wave is based on temperature.
According to Maxwell, energy calculation for em wave is based on amplitude.
Are those different kinds of energies? How can we determine the energy of an em wave just taking temperature as a parameter, but not amplitude or frequency??
| Maxwell is talking about a single wave. Boltzmann is talking about an ensemble of many many many many many such waves. Boltzmann finds that the average energy of an ensemble of waves depends on the temperature of whatever the waves contact, assuming that we've let enough time go by that the measurable properties of the system no longer change with time (equilibrium, temperature the same everywhere).
Same goes for any other property you want to consider, for example, the distribution of energy with wavelength (black body curve).
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Why can the entropy of an isolated system increase? From the second law of thermodynamics:
The second law of thermodynamics states that the entropy of an
isolated system never decreases, because isolated systems always
evolve toward thermodynamic equilibrium, a state with maximum entropy.
Now I understand why the entropy can't decrease, but I fail to understand why the entropy tends to increase as the system reach the thermodynamic equilibrium. Since an isolated system can't exchange work and heat with the external environment, and the entropy of a system is the difference of heat divided for the temperature, since the total heat of a system will always be the same for it doesn't receive heat from the external environment, it's natural for me to think that difference of entropy for an isolated system is always zero. Could someone explain me why I am wrong?
PS: There are many questions with a similar title, but they're not asking the same thing.
| Take a room and an ice cube as an example. Let's say that the room is the isolated system. The ice will melt and the total entropy inside the room will increase.
This may seem like a special case, but it's not. All what I'm really saying is that the room as whole is not at equilibrium meaning that the system is exchanging heat, etc. inside itself increasing entropy. That means that the subsystems of the whole system are increasing their entropy by exchanging heat with each other and since entropy is extensive the system as whole is increasing entropy. The cube and the room will exchange, at any infinitesimal moment, heat $Q$, so the cube will gain entropy $\frac{Q}{T_1}$, where $T_1$ is the temperature of the cube because it gained heat $Q$, and the room will lose entropy $\frac{Q}{T_2}$, where $T_2$ is the temperature of the room because it lost heat $Q$. Since $\frac{1}{T_1}>\frac{1}{T_2}$ the total change in entropy will be positive. This exchange will continue until the temperatures are equal meaning that we have reached equilibrium. If the system is at equilibrium it already has maximum entropy.
| {
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Blackbody and standing waves I'm reading articles about black body radiation and why classical mechanics fails to explain it. My question is:
Why do EM waves have to be standing wave in a cavity?
|
I'm reading articles about black body radiation and why classical mechanics fails to explain it.
More correctly, people could not explain it with classical physics (pre 20-th century physics). Nobody seriously thought that classical mechanics was sufficient even then - radiation is described by electromagnetic theory, not mechanics.
Why do EM waves have to be standing wave in a cavity?
They do not have to be. Standing waves are only special kind of waves that can occur in perfectly reflecting cavity. This is because the component of electric field in the walls along these walls is zero for perfect conductor.
We talk about standing waves because 1) we consider perfectly reflecting cavity 2) because all possible states of the EM radiation (even those that are not standing waves) can be expressed as sum of such standing waves.
| {
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Has string theory been able to produce masses of elementary particles? Masses of elementary particles in standard model are strange numbers. Is it possible to obtain these masses in string theory (presumably by using very few number of input parameters)?
| If you know which string theory vacuum you have (which compactification, which fluxes etc) and you know how the symmetries are broken, then you could in principle compute the masses.
Alas, nobody knows which is the right vacuum and how the symmetries are broken. Thus, the answer is in principle yes, in practice no.
The question about the number of parameters is difficult because of the way it looks right now you have to pick the vacuum and the mechanism for breaking various symmetries, each of which introduces parameters or at least other choices for which there are no selection criteria. It is conceivable that these choices can be derived at some point, but at least right now it is clearly not possible, which means that you have a presumably huge number of parameters to fix.
| {
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Efficiency of a thermal reactor I was told that the efficiency of any thermal reactor increases if it is operated at higher temperature- in this case, nuclear reactor was referred to. But I cannot seem to understand why...
| Presumably this is in the context of using reactors to generate electricity. Any power generator works by transferring heat from a hot source to a cold sink, and using the heat flow to do work. This work then generates electricity.
The archetypal heat engine is the Carnot engine, and its efficiency at converting heat to work is given by:
$$ \eta = 1 - \frac{T_c}{T_h} $$
where $T_h$ is the temperature of the hot heat source and $T_c$ is the temperature of the cold heat sink (both temperatures need to be in Kelvin). So the efficiency is increased by increasing $T_h$ or decreasing $T_c$. In practive the cold sink is the environment, and there's little we can do to change that. So the only practical way to make a heat engine more efficient is to increase $T_h$ i.e. run it hotter.
The Carnot engine is an ideal case that doesn't exist in the real world, but the behaviour of real electricity generators is basically similar and the same principle applies. That's why you increase the effiency of your power station if you run the reactor hotter.
| {
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Why is the constant velocity model used in a projectile motion derivation? I was re-studying university physics last week, I'm now in the chapter about kinematics in 2 dimensions and specifically the one treating projectile motion. In page 86 of his book (Serway - Physics for scientists and engineers) he derives the equation of the range of the projectile motion to be: $$R=\frac{{v_i}^2\sin2\theta_i}{g}$$
But I don't know why he used one of his assumptions
$\color{red}{\bf Question1:}$ Why $v_{xi}=x_{x\rlap\bigcirc B}$? Where $\rlap\bigcirc {\,\sf B}$ is the time when the projectile stops.
$\color{darkorange}{\bf Question2:}$ Why did he use the particle under constant velocity model to derive that formula, whereas here we deal with a projectile under constant acceleration?
Any responses are welcome, I'm disappointed a lot about those matters!
| I don't understand question 1: where does he equate a speed to a position?
As far as question 2 is concerned, it is basically what DavePhD said, but maybe I can extend it a bit more saying something about the conservation of linear momentum:
Along the x-direction, there is no external force (because gravity points downwards only, assuming a flat surface) so the linear momentum of the projectile is conserved.
Since $p_x = mv_x$, $v_x$ is constant.
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Maximum helicopter height Helicopters or dual/quadcopters stop rising after reaching some height from the ground. What causes that? And what one should do if he want to prevent this, if he want the helicopter to keep rising up?
| In the most simplified view, a helicopter is basically a really powerful fan, which is something that pushes air (and therefore feels a force in the opposite direction). When you get high enough, the air starts thinnning, so there is less to push, so that force becomes too low to hold it up. Imagine the limiting case of outer space: nothing would happen.
There's nothing you can really do to combat this, except run your motors faster and faster as the air gets thinner. Eventually you have to rely on things like rockets that don't need an atmosphere to gain momentum.
| {
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Why is the speed of light arbitrarily the limit? I know Einstein was great and all. Why is it that exactly at the speed of light is where infinite energy is required to accelerate any object with mass? Is it simply because the math of relativity checks out and explains most of everything? Are there any physicists who disagree with Einstein's theory?
| Let's assume, for argument's sake, that the Galilean transformation holds rather than the Lorentz transformation.
Then your questions would become
Why is infinite speed arbitrarily the limit? Why is it that exactly
at infinite speed is where infinite energy is required to accelerate
any object with mass?
I suspect that you wouldn't, in fact, think of asking such questions since they almost answer themselves.
Moreover, infinite speed would be an invariant speed - a speed that is measured to be the same in all reference frames - since (loosely speaking) $\infty + v = \infty$.
If we ask the question "what if there is a finite invariant speed", the mathematical answer is the Lorentz transformation where $c$ is the finite invariant speed.
Indeed, if we let $c \rightarrow \infty$ in the Lorentz transformation, we recover the Galilean transformation.
From this perspective, the result that accelerating to the invariant speed requires infinite energy and is thus impossible, doesn't seem so odd.
To summarize, without postulating that the speed of light is invariant, one can derive the form of the Lorentz transformations from just the principle of relativity. In this form, there is an undetermined finite invariant speed.
That light propagates at the invariant speed is then simply an empirical fact rather than a postulate.
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Bragg diffraction and lattice planes Crystalline substances show, for certain sharply defined wavelength and incident directions, very sharp peaks of scattered X-ray radiation.
From the illustration below we see that we get constructive interference when the path-length difference is a multiple of the wavelength $\lambda$.
In real crystalline materials we have a large amount of closely packed lattice planes. This large amount accounts for the sharp peaks for certain $\theta$. I do not understand how this follows from the Bragg reflection formula $$ n\lambda = 2d \sin \theta , $$ since $d$ is not constant anymore. I understand the model for two lattice planes as in the illustration.
Is it true that $d$ can only take on values of the seperation of lattice planes, so $d$ is defined to be the seperation of points in the reciprocal lattice, or in others words, is $d$ constrained to be the absolute values of some reciprocal lattice vector?
How does the Bragg condition account for very sharp peaks when we let $d$ run through all such absolute values?
| $d$ is constrained to be integer multiples of any ($hkl$) planar separation. The use of Miller Indices really helps realise how many planes there can be to reflect off, especially when it comes to more complex crystal structures. An alternate representation of Braggs law is $n \lambda = 2d_{hkl}\sin\theta$ to explicitly show which crystal plane is satisfying the Bragg condition. You can also define it in terms of the reciprocal space, where the Bragg condition will be satisfied if the change in the wavevector from scattering is equal to a reciprocal lattice vector associated with the crystal structure of the material, $\Delta \mathbf{k} = \mathbf{G}$.
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Difference in decay for muon and anti muon In a couple weeks, I will conduct a lab experiment where I measure the lifetime of the muons from the secondary cosmic radiation. For that, we have two detectors above each other, one will give a start signal, the other will give a stop signal, assuming the muon came to rest in the second detector and decayed.
There is a preparation question that says:
Muons as well as anti muons arrive at the surface of earth. Which nuclear process is possible for muons but not for anti muons? How does this qualitatively affect the measured lifetime?
As far as I know, the decay of the muon goes like this:
$$ \mu^- \to \nu_\mu + \bar \nu_e + \mathrm e^-$$
And for the anti muon, it goes like this:
$$ \mu^+ \to \bar \nu_\mu + \nu_e + \mathrm e^+$$
The Wikipedia page says:
The mean lifetime of the (positive) muon is 2.1969811±0.0000022 µs.[1] The equality of the muon and antimuon lifetimes has been established to better than one part in 104.
What is that difference that they are asking about in the lab course manual?
| The flux of negative muons at the surface of the earth can be smaller, than the flux of positive muons, because of muon capture by proton (which typically follows after the formation of an hydrogen-like atom with the $\mu^-$ in place of the electron). Such a reduction of the flux, if not taken into account properly, could be misinterpreted as shorter life time compared to the one of $\mu^+$.
See e.g. This link http://www2.fisica.unlp.edu.ar/~veiga/experiments.html
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Why doesn't time change in the non-relativistic limit of Lorentz transformations? A simple boost in the $x$ direction is given by:
$$ \Lambda = \begin{pmatrix}
\cosh(\rho) & \sinh(\rho) & 0 & 0 \\
\sinh(\rho) & \cosh(\rho) & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
\end{pmatrix} $$
Which get linearized to the following transformation:
$$
x^0 \mapsto x^0
, \quad
x^1 \mapsto x^1 + \frac vc x^0
$$
How come the zeroth component is not linearized to $x^0 \mapsto x^0 + \frac vc x^1$? Is that because there is another factor $c$ in the time components? Since $x^0 = ct$, that would mean the time is transformed like
$$ t \mapsto t + \frac v{c^2} x,$$
and $c^{-2}$ is just so small that is ignored?
Or is it just to fit the Galilei transformation?
| The Lorentz transformations : \begin{pmatrix}
\cosh(\rho) & \sinh(\rho) \\
\sinh(\rho) & \cosh(\rho) \\
\end{pmatrix}
form a group.
The galilean transformations :
\begin{pmatrix}
1 & 0 \\
v/c & 1 \\
\end{pmatrix}
form a group
But the transformations :
\begin{pmatrix}
1 & v/c \\
v/c & 1 \\
\end{pmatrix}
do not form a group.
If you restrict your transformations to a group structure, which is the simplest hypothesis, you cannot keep the last example of transformations.
| {
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Where does the steam pressure go? Lets say you build a steam boiler and then you connect several turbines to the boiler in series. What limits the number of turbines that you can connect to the boiler in this way? In my mind, it seems that the steam still has to escape, so no matter how many turbines are in the way, the steam will still turn all of them in its attempt to escape the boiler.
| Yes, the same mass flow of water coming out of the boiler must ultimately pass thru all the turbines. However, the part you appear to be missing is that with multiple turbines in series, the pressure drop accross each one is reduced. There is no free lunch here.
The total pressure drop from boiler to ambient (or a condenser in a recirculation system) is ultimately the same. Generally it is more efficient to harness this power in a single turbine, although that turbine might have multiple stages which you could think of as multiple turbines in series. Each stage reduces the pressure and temperature of the steam, thereby taking power from the flow.
It could even be worse, depending on whether you have a closed system or are just venting the output of the last turbine into the atmosphere or the creek behind the plant. In the latter case, the steam could condense and turn to liquid somewhere along the line of turbines. That would make rather a mess with turbines intended to run on gas.
| {
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Sign of gravitational force I'm reading Lanczos's The variational principles of mechanics, and on pp. 80-81 there is an example involving a system made up of $n$ rigid bars, freely jointed at their end points, and the two free ends of the chain being suspended.
The coordinates are chosen so that the $x$ axis is horizontal, and the $y$ axis is pointed vertically downwards. If the rectangular coordinates of the end points of the bars are denoted by $(x_k, y_k)$ and the length of the bars is denoted by $l_k$, then the expression for the potential energy will be of the form
$$ \frac{g}{2} \sum_{k=0}^{n-1} (y_k + y_{k+1}) l_k .$$
My problem with this is: the way I've understood things so far, the potential should have a negative sign because going "down", that is, going in the direction of the force of gravity, should decrease the value of the potential function. But in this example, the opposite appears to be true: going down increases the value of the potential. What am I getting wrong here?
| I) Yes, it appears that the sentence
[...] the $y$-axis vertically downwards [...]
in Ref. 1 p. 81 should have been
[...] the $y$-axis vertically upwards [...]
II) Let us also mention that Ref. 1 p. 29 eq. (17.9) introduces a function $U$ to be minus the potential energy, however, this $U$ seems unrelated to above.
References:
*
*C. Lanczos, The variational principles of mechanics, 1949.
| {
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Why $e$ in the formula for air density? I am reading a book that says that the density of air is approximately $D = 1.25 e^{(-0.0001h)}$, where h is the height in meters. Why is Euler's number $e$ used here? Was a differential equation used in deriving this formula?
| Euler's constant appears naturally in phenomena where the spatial gradient of a quantity (or rate of change with time) is proportional to the quantity itself: $$\frac{\mathrm{d}X}{\mathrm{d}x} = X/x_0$$ ($x_0$ determines the strength of the proportionality, and keeps units straight.)
The solution of this differential equation is $$X=X_o e^{x/x_0}$$ $X_0$ sets the "vertical" scale: it's the value of $X$ at $x=0$. Air density turns out to behave this way. If you want more detail, you should amend the question, or start a new question!
| {
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Why does angular momentum shorten the Schwarzschild Radius of a black hole? Angular momentum causes the event horizon of a black hole to recede. At maximum angular momentum, $J=GM^2/c$, the Schwarzschild radius is half of what it would be if the black hole wasn't spinning.
Can someone explain why angular momentum reduces the Schwarzschild radius?
| Maybe a qualitative answer motivated from thermodynamics: If you let your black hole rotate, you reduce the number of symmetries of your system, this will decrease your entropy $S$ which is proportional to the surface area. The surface area however is for sure monotonic increasing with your Schwarzschild radius, therefore, if your break symmetries, $r$ will decrease.
| {
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Integrating the generator of the infinitesimal special conformal transformation (c.f Di Francesco, Conformal Field Theory chapters 2 and 4).
The expression for the full generator, $G_a$, of a transformation is $$iG_a \Phi = \frac{\delta x^{\mu}}{\delta \omega_{a}} \partial_{\mu} \Phi - \frac{\delta F}{\delta \omega_a}$$
For an infinitesimal special conformal transformation (SCT), the coordinates transform like $$x'^{\mu} = x^{\mu} + 2(x \cdot b)x^{\mu} - b^{\mu}x^2$$
If we now suppose the field transforms trivially under a SCT across the entire space, then $\delta F/\delta \omega_a = 0$.
Geometrically, a SCT comprises of a inversion, translation and then a further inversion.
An inversion of a point in space just looks like a translation of the point. So the constant vector $b^{\mu}$ parametrises the SCT. Then $$\frac{\delta x^{\mu}}{\delta b^{\nu}} = \frac{\delta x^{\mu}}{\delta (x^{\rho}b_{\rho})} \frac{\delta (x^{\gamma}b_{\gamma})}{\delta b^{\nu}} = 2 x^{\mu}x_{\nu} - x^2 \delta_{\nu}^{\mu}.$$
Now moving on to my question: Di Francesco makes a point of not showing how the finite transformation of the SCT comes from but just states it. $$x'^{\mu} = \frac{x^{\mu} - b^{\mu}x^2}{1-2x\cdot b + b^2 x^2}$$ I was wondering if somebody could point me to a link or explain the derivation. Is the reason for its non appearance due to complication or by being tedious?
I am also wondering how, from either of the infinitesimal or finite forms, we may express the SCT as $$\frac{x'^{\mu}}{x'^2} = \frac{x^{\mu}}{x^2} - b^{\mu},$$ which is to say the SCT is an inversion $(1/x^2)$ a translation $-b^{\mu}$ and then a further inversion $(1/x'^2)$ which then gives $x'^{\mu}$, i.e the transformed coordinate.
|
However, I don't really see how to exponentiate the infinitesimal form
to actually get to the finite form
In addition to the detailed answer of @LubošMotl, you may notice that conformal transformations transform light cones in light cones.
This means the following. Starting from a light cone :
$$(x'-a')^2 = x'^2-2x'.a' + a'^2=0\tag{1}$$
you must find that it is also a light cone for $x$, that is, it exist $a$ such as :
$$(x-a)^2 = x^2-2x.a + a^2=0\tag{2}$$
To do that, replace simply $x'$ by its value in function of $x$, by :
$$x'^{\mu} = \frac{x^{\mu} - b^{\mu}x^2}{1-2x\cdot b + b^2 x^2} \tag{3}$$
and, with some algebra, you will find that it is possible to exhibit $a(a',b)$ ($a$ as a function of $a'$ and $b$) , which satisties ($2$).:
$$a = \dfrac{a' + a'^2b}{1-2a'.b+a'^2b^2}\tag{4}$$
This means that the transformation $(3)$ is a conformal transformation whose infinitesimal expression is clearly:
$$x'^{\mu} = x^{\mu} + 2(x \cdot b)x^{\mu} - b^{\mu}x^2 \tag{5}$$
(because only quadratic terms in x are allowed in infinitesimal conformal transformations)
Any other expression of the global special conformal transformation $(3)$, having infinitesimal expression $(5)$, will not transform light cones in light cones.
| {
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Is it ok to have two events $A$ and $B$ so that for one person $A$ occurs before $B$ but for another $B$ preceds $A$ Imagine two laser beams A and B are released at the same moment to bounce between two mirrors, A was moving and B was at rest, doing the calculations I found that for a person at rest B would reach the upper mirror before A because in his frame of reference A travels less distance. but for another person in the same reference frame of A, A would reach the upper mirror first.
Is that OK in relativity!
| Yes, that's ok! You've stumbled upon one of the basic strange phenomena of relativity.
| {
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How stellar aberration is measured? A simple calculation shows that stellar aberration due orbital motion of earth is roughly 20 arcseconds. My questions are:
*
*Practically how this small value is measured?
*Does this value is in the range of accuracy of a 11 inch reflective telescope with a camera?
*And how this measurement originally was done by Bradley in the 18th century?
| The Wikipedia article on angular resolution
https://en.wikipedia.org/wiki/Angular_resolution
is a source of many useful facts relevant for the question. For example, it was empirically established already by the English 19th century astronomer W.R. Dawes that the angular resolution $\theta$ in arcseconds is about
$$ \theta = \frac{4.56}{D} $$
where $D$ is the diameter of the lens' aperture in inches. A similar result may be calculated from the wave optics, too. If you substitute $D=11$ inches, you see that the angular resolution is better than one arcsecond.
Approximately one arcsecond is the limit one gets from simple telescopes in the atmosphere due to the atmospheric effects etc. It's not "safely smaller" than 20 arcseconds but it is still 20 times smaller.
Very good eyes' angular resolution is actually estimated as 20 arcseconds so the people with the sharpest eyes are marginally able to see the stellar aberration with naked eyes. The average healthy eyes' angular resolution is about 3 times poorer, 60 arcseconds.
At any rate, there was no problem to achieve the desired resolution with the 18th century (and even older) telescopes. In fact, any telescope that improves the eyes' resolution just a little bit is enough. State-of-the-art large terrestrial telescopes with adaptive optics
https://en.wikipedia.org/wiki/European_Extremely_Large_Telescope
are able or planned to be able to go to 0.001 arcseconds so there has still been a lot of progress since the 18th century.
| {
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Help explain how direction change relates to acceleration I was doing some simple harmonic motion problems and I came across this picture describing the position, velocity and acceleration of a linear oscillator. At the moment in time when v is 0 the linear oscillator should not be moving, only changing directions. I'm having a hard time understanding why the acceleration is the greatest at that time (according to these graphs), since there is no velocity change. Is it because acceleration is only the difference in velocity at two different points in time and not one? How exactly does the change in direction affect acceleration?
edit: I found another question that answered my question. haha.
| The expression "change in direction" implies some sort of discontinuity in motion, where in the referenced graphs, there is none. One could easily choose a different frame of reference such that the oscillating object appears never to change direction, only periodically speed up and slow down. The fact that the velocity value at the point in question happens to undergo a sign change is nothing more than an artifact of perspective. When you look at the graph, you see that the point of maximum acceleration occurs at the point where the slope of the velocity curve is steepest. It's not a coincidence... acceleration is by definition, rate of change of velocity, which is expressed in graphical form by slope.
| {
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Can an LC oscillator be used to generate visible light? The LC oscillator is most commonly used to generate radio waves for practical use and the frequency $\omega$ of the LC oscillator equals that of the electromagnetic wave so produced. So, can they in principle be used to emit visible light?
The frequency of visible light is on the order of a few hundred terahertz, and the frequency of an LC oscillator is
$$\omega = \frac{1}{\sqrt{LC}}$$
I admit, the product $LC$ does become very small (on the order of $10^{-30}$) when the numbers are plugged in, but making an inductor and a capacitor with small values isn't difficult, is it?
| An electric oscillator for light exists indeed. They are called LASER diodes. Due to the works of a number of Nobel prizewinners (e.g. Einstein predicted the working in 1911), you can buy then in a store and simple powers then with a DC power source. It's a sort of oscillator, not with electric current, but with light.
| {
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Invariance of a tensor under coordinate transformation I know, that a tensor is a mathematically entity that is represented using a basis and tensor products, in the form of a matrix, and changing a representation doesn't change a tensor, is kind of obvious.
So does the invariance of a tensor under coordinate transformation mean what I stated above or does it mean that under a set of particular transformation the representation of a particular tensor also doesn't change.
Quoted from Wikipedia:
A vector is invariant under any change of basis, so if coordinates transform according to a transformation matrix $L$, the bases transform according to the matrix inverse $L^{−1}$, and conversely if the coordinates transform according to inverse $L^{−1}$, the bases transform according to the matrix $ L$.
Can someone please shed some light on this?
| I guess there is two different notions of invariance of tensors. First notion is that if you look at a tensor as a mapping then the first notion of invariance is what you mentioned above. The other notion of invarance is that you do transformation but the " component" of metric does not change. For instance, if we do Lorentz transformation then the Minkowski metric is invariant, meaning that the component will be +1 , -1 , -1 ,-1 .
I might be wrong!!
| {
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How much does the sound definition vary during an LP (Vinyl)? This question came to me when I realized how the linear speed varies while listening to a Vinyl LP.
The linear speed variation has to be compensated with a variation in the resolution of the grooves, that is, since the linear speed decreases, the groove resolution also has to decrease in some measure. What is this measure of reduction, or else, how much does the linear speed reduce? And how does that influence the sound definition?
| Your question is a little misstated, since it's not so much the "groove resolution" as the high linear frequency limit to which the groove can be cut. The stylus (and piezo or magnetic components which convert motion to electrical signal) doesn't really care about linear speed in the along-track direction, just the transverse speed. So long as the groove walls can be cut such that, e.g., 15kHz undulations are faithfully reproduced in the vinyl, the sound will be fine.
There are other problems, including "hiss" and physical damage to the outer grooves where the stylus is moving longitudinally much faster than in the inner grooves, and (depending on just how fancy your turntable is) cross-track skate force and rotational misalignment of the stylus' axes to the groove direction.
| {
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Renormalizability of standard model I'm wonder what precisely is meant by the renormalizability of the standard model. I can imagine two possibilities:
*
*The renormalizability of all of the interaction described by the Lagrangian before spontaneous symmetry breaking (SSB) by the nonzero vacuum expectation value (VEV) of the Higgs field.
*The renormalizability of the Lagrangian obtain from the initial one after SSB, expressed in terms of suitable new fields (which has direct physical interpretation contrary to the fields appearing in initial Lagrangian).
It seems that in case (2) we obtain an effective (nonrenormalizable) theory only and this precisely was the reason to introduce the mechanism of generating mass by nonzero VEV of Higgs field. The original Lagrangian (case (1)) contains only power counting renormalizable vertices so if there are no anomalies then SM befor SSB is renormalizable. However, in physical prediction (actual computations being performed), as far as I know, Lagrangian after SSB is used. Does is require infinite number of counterterms (is it effective theory)?
| The Standard Model Lagrangian before and after spontaneous symmetry breaking (SSB) is renormalizable. To see that recall that the rule is (though it may not be immediately obvious as to why this rule holds) that a theory is renormalizable if all the terms in the Lagrangian are of dimension 4 or less. This is true by design for the Standard Model in which all other terms are omitted.
To see that this property of the SM is unaffected by SSB consider the part of the Lagrangian associated with the Higgs,
\begin{equation}
{\cal L} = \mu ^2 \left| \phi \right| ^2 - \lambda \left| \phi \right| ^4 - \phi \psi _i \psi _j
\end{equation}
where, $ \psi $ are the set of SM fields which have Yukawas (I'm being a bit sloppy here about all keeping terms that are actually SU(2) invariant). SSB implies shifting the Higgs to its vacuum expectation value which is at some value $ v $:
\begin{equation}
\left( \begin{array}{c}
\phi _1 + i \phi _2 \\
\phi _3 + i \phi _4
\end{array} \right)
\rightarrow \left( \begin{array}{c}
\phi _1 + i \phi _2 + v \\
\phi _3 + i \phi _4
\end{array} \right)
\end{equation}
This doesn't change the dimension of the Higgs field, since $ v $ is still of mass dimension $1$ and so each term containing the Higgs won't change dimensions after SSB. Every term will still be at most of mass dimension $4$. Therefore, whether the theory is renormalizable will hold equally well before or after SSB.
| {
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Uniformity in a solenoid I know the magnetic field strength increases as the number of turns in the solenoid increases.
However, I've learnt the field inside the solenoid is usually nearly uniform.
So, does the number of turns in the solenoid effect the uniformity of the field inside the solenoid? Does the field gets closer to uniform as the number of turns increases?
| If the windings in a solenoid are not closely spaced, there will be some inhomogeneity in the field - so more turns per unit length helps. Usually for "ideal" calculations one assumes a continuous sheet of current.
The second thing is the length. For a finite length magnet the field quickly drops off as you move away from the isocenter. The longer the magnet, the less the curvature of the field in the center.
This is why MRI machines have such a long bore - although the region of uniform field is usually only 50 cm or so, the bore (main magnet) is well over a meter long.
| {
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Why must a singularity form inside a black hole? What is the exact reason that normal matter can not exist within an event horizon?
I can understand how a super-dense object like a neutron star could accrete mass until its physical radius is less than its Schwarzschild radius and an event horizon forms around it.
But why can't the neutron star remain inside the event horizon as it was before the event horizon formed? Aside from the escape velocity at the surface reaching $c$, what actually changed?
Why must all matter within an event horizon undergo complete physical collapse to a point?
| We do not know that "normal matter can not exist within an event horizon".
For all we know, aliens may be sitting around and drinking coffee watching us.*
We have theories on what could happen. different theories have different views.
Nothing has been tested, scientifically proven as yet.
Some calculations (the status thereof) on some theories (Hawking' et al) indicate that an event horizon may not even exist at all.
it may seem lame, but the closer you get to a black hole, the longer it takes to reach it (that is towards infinity). Math has not yet managed to get over that graph yet, thus suppositions/theories, not yet facts. it still is a true frontier.
*hey, take my comment as a a tong in cheek. lighten up. great subject. :)
| {
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Cooling induction current generation We know that conducting materials can be heated by electromagnetic induction. Is it possible to generate current using a cooling process?
| Short answer is: If you heat up the whole material, then no.
If you heat up the material in on end (an iron bar e.g.), then yes. The thermoelectric effect will cause a small current to flow between the two ends of different temperature.
It looks like you are mistakenly assuming that electromagnetic induction causes heating, which should then be possible to reverse. Well, the heating is more a side effect. It is caused by the resistance in the material. Whenever current flows - regardless of the cause - resistance in the material will cause heating. Electrons making up the current e.g. will bumb into the material atoms and transfer their energy to this material. Which is heat on the macro scale. This heat loss is an irreversible process - you cannot "put the heat back" to regain a current.
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Why ONLY Maxwell's equations are the basic equations of electromagnetism? In electromagnetism we say that all the electromagnetic interactions are governed by the 4 golden rules of Maxwell. But I want to know: is this(to assume that there is no requirement of any other rule)only an assumption, a practical observation, or is there a deeper theoretical point behind it? Could there be a deeper theory behind assuming that there is not requirement of rules other than Maxwell's equations?
|
In electromagnetism we say that all the electromagnetic interactions are governed by the 4 golden rules of Maxwell. But I want to know that is this only an assumption
It is not an assumption, it is an elegant way of joining the diverse laws of electrictity and magnetism into one mathematical framework.
or a practical observation
The laws of electricity and magnetism were described mathematically by fitting observations and always being validated, i.e. correct, in their predictions. Maxwell's equations not only incorporated them but also by unifying electricity and magnetism mathematically give predictions that have never been falsified.
So yes, they are a mathematical model fitting observations, a very elegant model.
or there exist any theoretical point behind it?
Physics is about observations and the derivation of mathematical models, theories, that will fit them and will also predict new observations to be measured and evaluate the theory. Physics is not about philosophy or mathematics, it is about describing nature using mathematics as a tool.
If there exists a "theoretical point" it is that theoretical physicists try to unify in one mathematical model all the known observations, i.e. continue on what Maxwell has done in unifying electricity and magnetism, by unifying the weak with the electromagnetic, and proposing a unification with the strong in similar mathematical frameworks. The goal being in unifying also gravity, all four forces in one mathematical model
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Analytical problems with Green's function I have a question about the right definition of the Green's function in physics. Why do we introduce (or not) an infinitesimal, positive number $\eta$ to the following definition:
$$\left[ i\hbar\frac{\partial}{\partial t} - \hat{H}(\mathbf{r}) \pm i\eta\right]G(\mathbf{r},t;\mathbf{r'},t') = \delta(\mathbf{r} - \mathbf{r'})\delta(t-t')$$
| This type of problem can be conveniently treated using the complex
Laplace transform. For a function $f(t)$ with $t\geqslant 0$ it is defined as
\begin{equation*}
\hat{f}(z)=\int_{0}^{\infty }dt\exp [izt]f(t),\;Imz>0.
\end{equation*}
Setting $z=\omega +i\eta $ ($\eta >0$ but arbitrary otherwise) we have, with
$\theta (t)$ the Heaviside step function ($\theta (t)=1$ for $t\geqslant 0$
and $0$ otherwise),
\begin{equation*}
\hat{f}(\omega +i\eta )=\int_{-\infty }^{+\infty }dt\exp [i\omega t]\theta
(t)\exp [-\eta t]f(t).
\end{equation*}
Thus $\hat{f}(\omega +i\eta )$ is the Fourier transform of $\theta (t)\exp
[-\eta t]f(t)$ and hence
\begin{eqnarray*}
\theta (t)\exp [-\eta t]f(t) &=&\frac{1}{2\pi }\int_{-\infty }^{+\infty
}d\omega \exp [-i\omega t]\hat{f}(\omega +i\eta ) \\
f(t) &=&\frac{1}{2\pi }\int_{-\infty }^{+\infty }d\omega \exp [-i(\omega
+i\eta )t]\hat{f}(\omega +i\eta ) \\
&=&\frac{1}{2\pi }\int_{\Gamma }dz\exp [-izt]\hat{f}(z),\;t\geqslant 0,
\end{eqnarray*}
where $\Gamma $ is the line $(-\infty +i\eta ,+\infty +i\eta )$.
Now the problem at hand. Setting $\hslash =1$ we deal with the Schr\"{o}
dinger equation
\begin{equation*}
\partial _{t}\psi (t)=-iH\psi (t).
\end{equation*}
Since, by partial integration (note $Imz>0)$,
\begin{equation*}
\int_{0}^{\infty }dt\exp [izt]\partial _{t}\psi (t)=-\psi (0)-iz\hat{\psi}
(z),
\end{equation*}
we obtain, after some rearranging,
\begin{eqnarray*}
i[z-H]\hat{\psi}(z) &=&\psi (0), \\
\hat{\psi}(z) &=&-i[z-H]^{-1}\psi (0), \\
\psi (t) &=&\frac{1}{2\pi i}\int_{\Gamma }dz\exp [-izt][z-H]^{-1}\psi
(0)=\exp [-iHt]\psi (0),\;t\geqslant 0
\end{eqnarray*}
The object $[z-H]^{-1}$ is known as the resolvent of $H$ and plays a key
role in mathematical investigations. The Green's function is the
corresponding kernel in coordinate representation
\begin{equation*}
\hat{G}(\mathbf{x}_{1},\mathbf{x}_{2},z)=\langle \mathbf{x}_{1}|[z-H]^{-1}|%
\mathbf{x}_{2}\rangle
\end{equation*}
and
\begin{eqnarray*}
G(\mathbf{x}_{1},\mathbf{x}_{2},t_{1},t_{2}) &=&\frac{1}{2\pi i}\int_{\Gamma
}dz\exp [-iz(t_{1}-t_{2})]\hat{G}(\mathbf{x}_{1},\mathbf{x}_{2},z), \\
\psi (\mathbf{x}_{1},t_{1}) &=&\int d\mathbf{x}_{2}dt_{2}G(\mathbf{x}_{1},%
\mathbf{x}_{2},t_{1},t_{2})\psi (\mathbf{x}_{2},t_{2}),\;t_{1}\geqslant
t_{2}.
\end{eqnarray*}
Formally $G(\mathbf{x}_{1},\mathbf{x}_{2},t_{1},t_{2})$ satisfies the
equation in the question but now we have a precise description about the $z$
- integral.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Electrostatics - Inserting a brass plate between two charges The question is: if I were to insert a brass plate between two charges, what will happen to the force between the charges? Would it increase, decrease or stay the same?
Does the brass plate increase the value of permittivity of the medium and therefore the force decreases?
The correct answer is that it will increase. But I do not understand how.
| The brass plate is a conductor, so the potential will be the same on both sides. The thickness of the brass plate therefore subtracts from the effective distance between the two charges, making the electric field strength higher in the remaining open space between the charges. This stronger field will cause more force to be experienced by each charge.
Another effect is that the shape of the field will change. Since the conductive plate has the same potential everywhere on its surface, each charge now sees a plane at half the voltage and half the distance, as apposed to the point charge before.
| {
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Is entropy in quantum mechanics emergent or fundamental? Since a quantum mechanical system, even an isolated system containing one particle, can be described by a density matrix, with entropy for the system given by $\langle S\rangle=-k \rho\ln(\rho)$, is not entropy therefore a property of the system like mass or energy?
| You have provided the von Neumann entropy definition which is derived from its density matrix. I would consider it an intrinsic rather than fundamental property, but this is just semantics.
Some recent work by John Baez has investigated the dynamics of quantum entropy called quantropy.
| {
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Why does wavelength affect diffraction? I have seen many questions of this type but I could nowhere find the answer to "why". I know this is a phenomenon which has been seen and discovered and we know it happens and how it happens. But my question is why would wavelength affect the amount of diffraction? I am looking for a very simple logical explanation rather than a complex mathematical answer. Why will a blue ray bend lesser than a red ray through a slit of the size a little bigger than the wavelength of the blue ray? I need an answer that will answer "why" does diffraction depend on wavelength of light.
Image sources: http://www.olympusmicro.com/primer/java/diffraction/index.html
| The question "why does the wavelength affect diffraction", I think, could be best answered by looking at the two extreme cases. Assuming a narrow opening is illuminated:
*
*If the wavelength is much smaller than the width of a slit, wave effects can be completely ignored, because interference effects won't play a role. Consequently, the light waves will pass through the opening like a ray.
*If the wavelength is much larger than the width of a slit, again, no diffraction pattern will be observed. However, the slit now acts as a point source, i.e. the narrow opening becomes the source of a new wave (Huygen's principle).
Consequently, longer wavelengths are redirected more strongly than shorter wavelengths, and hence diffraction is wavelength dependent. Diffraction is understood as the interference pattern of all waves behind the narrow opening. Considering the two extreme cases above, I would then argue that blue light is less diffracted as red light.
| {
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Natural entanglement system I'm a beginner and amateur interested in quantum physics.
I would like to know if entangled systems of natural states exist or whether such systems require human intervention?
Is it possible? Either no or yes, Why?
| In short, entanglement is perfectly normal. I am sure that entanglement is ubiquitous in, say, atoms with more than one incomplete subshell, as well as in some kind of organic molecules, but I am not a quantum chemistry expert and can’t provide an easy-to-realize example.
Generally, any decay process produce particle states that are entangled in some way but, to be specific and illustrative, many particle decay processes produce spin-entangled particles, namely those where a particle of lesser spin decays to particles of (summarily) greater spin. Consider a spin-0 particle, such as π0, that decays to several (two or three) particles with spin, as the most obvious example.
The problems with understanding “entanglement” (which states are “entangled” and which are not) are based on the problems with understanding “quantum state”, that is a tricky (and not very reliable) concept despite its apparent mathematical simplicity. It is not easy to define what does a “quantum state” mean objectively, to exclude assumptions of an observer completely.
| {
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How could the multiverse theory be disproven? Theorists (physicists) suggest that there is the term/entity, the Multiverse that contains a huge number of universes not necessarily like our own.
I personally find this theory very elegant because its explains the probabilistic outcomes of the experiments with fixed conditions on a quantum level.
It's obviously very hard to prove the validity of the theory of multiverse, but
*
*How can we disprove the existence of multiverse?
| Any particular theory of (astro)physics that predicts a multiverse will be disproven the moment experimental results or observations are found to be in conflict with the theory. In physics all theories are falsifiable, so this is not a problem in principle. However, you can then postulate that there may still exist a multiverse and that this idea cannot be proved wrong. But physics is about building falsifiable theories that can explain the physical world, it is not about defending or attacking the idea of multiverses.
A good example is the Many Worlds Interpretation of quantum mechanics (MWI). Obviously MWI is easily falsifiable, if you detect non-unitary time evolution in a well isolated system well within the predicted decoherence time scale, then you have disproven the MWI. A more interesting question is whether one can disprove collapse theories. As pointed out by David Deutsch, there is an experiment that you could in principle do to falsify the idea that the wavefunction undergoes a non-unitary collapse when performing a measurement.
This involves implementing an observer in a quantum computer and then doing a measurement that can have different outcomes. You then let the quantum computer evolve according to the unitary transform that reverses the act of performing the measurement but such that the observer will keep its mememory of having performed an observation. What then happens is that the quantum state of the measured system will have been restored to is original state while in collapse theories the final state will not be the same as the initial state.
| {
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How can a gas support tensile stresses? In working through a rigorous derivation of the compressible Navier-Stokes equations, I find that the momentum flux in the X-direction should be driven not only by the normal pressure gradient $\frac{\partial p}{\partial x}$ and shear stress terms $\frac{\partial(\tau_{yx})}{\partial x}$ and $\frac{\partial(\tau_{zx})}{\partial x}$, but also by the gradient of the normal stress $\frac{\partial(\tau_{xx})}{\partial x}$. It's intuitively clear to me how adjacent lamina moving at different speeds can transfer momentum across their interface, and so the shear stress terms in the momentum equation are readily intelligible. The normal stress term, on the other hand, is far less intuitive because I cannot see how a freely-deforming fluid can support tensile stresses. Positive normal stresses (i.e. compression) are not that hard to understand, but it's proving exceedingly difficult to fully envisage an element of a fluid "pulling on" an adjacent element in a way even remotely analogous to the behavior of a solid under the same conditions. I am also unclear on the difference between "pressure" and "normal stress" in the fluid. How exactly are these terms different? I am interested primarily with gases not liquids.
| It looks like the question boils down (at least in part) to the following: can a fluid have negative ABSOLUTE pressure? This question has been discussed here several times. My take is: it can (although such state is probably metastable in the best case), because the force between two molecules can be attractive. See, e.g., http://www.youtube.com/watch?v=BickMFHAZR0 , where they discuss how trees taller than 10m can deliver water to their top. I don't know though if a gas, rather than a liquid, can have negative pressure. In cosmology, the so-called Chaplygin gas is considered though.
| {
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Difference between weight of water and pressure of water (I didn't even have a basic formal education in physics. I'm learning through the internet out of my own interest, so if there are any silly mistakes, kindly bear with and guide me through.)
Everywhere, everyone is saying that pressure will same be for a given height. How is it possible? When volume changes, doesn't pressure change in calculations?
[Pressure will be 2.5 bars approximately (including atmospheric pressure) for a height of 15 metres from ground level. I checked on many sites that 2.5 bars equals 2.5 kg/cm².]
Now let's take base area as 2000 cm² and height of 15 m as constant in three scenarios:
EACH SCENARIO IS INDEPENDENT AND TUBES ARE SEPARATE, I.E NOT CONNECTED WITH EACH OTHER
In scenario 1 a straight vertical tube from ground level will have capacity of 3000 litres with a weight of 2.5 kg per cm² [3000 litres / 2000 cm² = 1.5 kg/cm² + 1 kg/cm² atmospheric pressure].
In scenario 2 the tube becomes narrow from the base area, resulting in a volume of 2000 litres.
In scenario 3 the tube becomes wide from the base area, resulting in a volume of 4000 litres
Question:
When they say pressure remains the same at a given height, does it mean that weight will be 2.5 kg/cm² in all 3 scenarios, where volume is 3000, 2000, or 4000 litres (base area 2000 cm², height 15 m constant)? How is it possible? Where am I wrong?
| Consider this diagram showing the three columns you describe all connected to the same body of water:
Your question asks whether the three pressures $P_1$, $P_2$ and $P_3$ will be the same. The answer is obviously yes, because the columns are all connected to the same body of water. For example if $P_1 > P_2$ then water would flow from the base of column 1 to the base of column 2 until the pressures became equal.
OK, but the next question is whether the heights of all the columns are the same. Again the answer is yes because if they weren't we could connect the tops of the columns, let water flow between them and we would have a perpetual motion machine.
So we conclude that the pressure is only related to the height $h$ and does not depend on the shape of the column. Specifically, the relationship between pressure and height is:
$$ P = \rho g h $$
where $\rho$ is the density of the fluid and $g$ is the gravitational acceleration (9.81 m/sec$^2$).
| {
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Why does the Walecka model not include pions? The Walecka or $\sigma$/$\omega$-model is an effective theory describing nucleon-nucleon interaction by an exchange of $\sigma$/$\omega$-mesons. Why does it not include interactions by pions?
| The answer given in Walecka's 1974 paper is mostly correct. The one pion exchange contribution to the Hartree energy vanishes in balanced nuclear matter. The same point was made in the 1972 paper of Miller and Green (Phys Rev C5 241) where the same type model was used for doubly magic (finite) nuclei. If exchange is included (Hartree Fock as opposed to Hartree) then the pion would contribute.
| {
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Magnetic field and Newton's third law If a magnet exerts force on a iron block (opposite and EQUAL), does a iron block also exerts force on magnet (via Newton's third law)? If yes then what magnetic property does it has to produce equal and opposite force on magnet considering that its not a ideal environment? If no then is it not the violation of newton's third law?
| Its a violation of Newton third law..because Newton third law only valid when
1)same kind of force ie when there is force of same nature
So electromagnetic and block forces are not of same nature so Newton third will not hold
| {
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Singularity in Newton's gravitational law If $r=0$ in the well know equation $F= G\dfrac{m_1\cdot m_2}{r^2}$, it will not follow that the force will be infinite?
May someone please clarify it to me?
| True point masses and other singularities can wreak all kinds of havoc in Newtonian physics. A couple of examples:
*
*Particles can attain infinite velocity in finite time: Saari, D., and Zhihong J. (1995), "Off to infinity in finite time." Notices of the AMS 42:5.
*Particles can exhibit non-deterministic behavior. See this question, Norton's dome and its equation, and also see Norton, John D. (2008) "The dome: An unexpectedly simple failure of determinism." Philosophy of Science 75:5, 786-798.
Fortunately, true point masses and singularities such as those exhibited by Norton's Dome don't exist in reality.
| {
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Gradient is covariant or contravariant? I read somewhere people write gradient in covariant form because of their proposes.
I think gradient expanded in covariant basis $i$, $j$, $k$, so by invariance nature of vectors, component of gradient must be in contravariant form. However we know by transformation properties and chain rule we find it is a covariant vector. What is wrong with my reasoning?
My second question is: if gradient has been written in covariant form, what is the contravariant form of gradient?
| Gradient is covariant! Why?
The components of a vector contravariant because they transform in the inverse (i.e. contra) way of the vector basis. It is customary to denote these components with an upper index. So, if your coordinates are called $q$'s, they are denoted $q^i$.
Therefore, the gradient (or a derivative if you prefer) is $$\partial_i = \frac{\partial}{\partial q^i},$$ which transform as the inverse of the component transformation ( 1 / contra-variant = co-variant ).
If you're still not convinced... try this!
*
*Propose a coordinate transformation -This is a transformation rule for the contravariant components- (e.g. from Cartesian to Polar)
*Use the chain rule to transform the derivative,
*Check that the transformation of the derivative is the inverse of the coordinate transformation.
Personal Note: Although the notation my homonym Oscar point out is correct [say $\partial^i$], I prefer to avoid it, because is not a derivative wrt the "real" coordinates. Please do not misunderstand my words... It is Ok to define that operator, but should be treated carefully!
Cheers! ;-)
| {
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Is a "shift in the meaning" of Accuracy and Precision occurring? Accuracy and precision are among the most fundamental concepts in experimental physics, and, I always believed, completely unambiguous.
Recently I found that the Wikipedia article on Accuracy and Precision claims that a "shift in the meaning of these terms" is occurring. My first thought was that this must be a joke, a mistake or Wiki vandalism.
However, the ISO 5725-1 (Accuracy (trueness and precision) of measurement methods and results) standard referenced in the article indeed reads
3.7accuracy
[...]
NOTE 3 Accuracy refers to a combination of trueness and precision.
Notably, this seems to imply that the concept of precision is a subset of the concept of accuracy, which is plainly incompatible with the definitions found almost everywhere else that sharply contrast precision and accuracy.
My questions are:
*
*Is such a "shift in meaning" really occurring in the physics/measurement community at large?
*If yes, what is the cause of this change of terminology?
| (Currently editing answer.. previous version at https://physics.stackexchange.com/revisions/135002/7)
| {
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Is there a difference between "average acceleration" and centripetal acceleration? Question adapted from Examkrackers MCAT prep book:
A particle moves along a half circle (diameter=$10\text{ m}$) at a constant speed of $1\text{ m/s}$. What is the average acceleration of the particle as it moves from one side of the half circle to the other side?
A. $0$
B. $0.2/\pi$
C. $0.4/\pi$
D. $1$
The book says C is correct. Acceleration is change in velocity divided by time. Initial velocity is $1\text{ m/s}$ up; final velocity is $1\text{ m/s}$ down. The change in velocity is therefore $2\text{ m/s}$. The time is found from speed equals distance divided by time. Distance is $2\pi r/2$. Thus
$$a= \frac{2}{(2\pi(5)/2)/1} = \frac{2}{5\pi} = 0.4/\pi$$
I thought all of the answers were wrong because I thought they should've used the centripetal acceleration equation: $a= v^2/r$
SO my question like the title: Is there a difference between "average acceleration" and centripetal acceleration?
I searched for a couple hours and couldn't find this issue directly addressed.
| Here the average acceleration can be understood as follows:
The particle going from A to B along a half circle with speed 1m/s can be here viewed as the particle going from A to C and again to A with an initial velocity 1m/s as shown in the figure:
The particle in the half circle, will move under a centripetal acceleration which is always directed towards the center. On the other hand the average acceleration is directed downwards, as it is the final velocity(downward) minus initial velocity(upward) divided by time(vector difference along a line). That is why considering only the y-component of the velocity of the particle in the half circle path, the half circle path is reduced to a straight path for the average acceleration.
As we see in the second figure, there is an acceleration in the downward direction that decelerates the particle when it goes from A to C and accelerate it when it comes to A again. This constant acceleration in the straight path is the average acceleration for the actual half circle path. That is why the average acceleration is calculated as the difference of initial and final velocity divided by time, difference of up velocity and down velocity at A divided by time.
| {
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Wrong calculation of work done on a spring, how is it wrong? So I would have thought that this would be how you derive the work on a spring: basically the same way you do with gravity and other contexts, use $$W=\vec{F}\cdot \vec{x}.$$ If you displace a spring by $x$, then it exerts a force $-k x$, so $F=-kx$, since the displacement is $x$.
So $$W=-kx^2.\qquad \leftarrow\text{ (however, apparently wrong!)}$$
I've seen the correct derivation of work in a spring (with an extra half) and don't doubt that it's correct, but also don't see where my logic fails in this alternate derivation.
| in w=F.X, x is displacement of center of mass of the body not the displacement of system.
here also force is 'kx' and displacement of c.o.m. is 0.5*x,so work done will be 0.5kx^2.
| {
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Ampere's law and Biot-Savart law gives different terms for magnetic field in middle of a current running in a loop
I would like if someone could clarify this issue for me:
When dealing with a current $I$ running in a loop with radius $R$ and looking for the magnetic field in the middle of the loop.
By using Ampere's law, I know that the current $I$ runs through a loop with the same radius $R$, we get that:
$$\oint_c\vec{B} \cdot d\vec{l} = \mu_0 I_{enc} $$
$$B= \frac{\mu_0 I}{2\pi R}$$
and when using Biot–Savart we get that
$$d\vec{l} \cdot \vec{r} = |d\vec{l}||\vec{r}|\sin(\frac{\pi}{2})$$ obtaining:
$$B = \frac{\mu_0 I}{2R}$$
Which is not the same result as with Ampere's law.
I obviously miss something, maybe I can't use Ampere's law?
Anyway, if someone could help me out here I would really appreciate it.
Thanks.
| The Biot Savart law is
$${\bf B} = \frac{\mu_0}{4\pi} \oint \frac{ I\, d{\bf l} \times {\bf r}}{|{\bf r}|^3}$$
In this case $d{\bf l} \times {\bf r} = dl\,|r|$ directed along the loop axis and integrating around the closed loop leads to a B-field magnitude $ B = \mu_0\, I/2R$ as you suggest.
However, I think there is a problem with your application of Ampere's law. This is that
$$ \oint {\bf B} \cdot d{\bf l} = \mu_0 I\, ,$$
where I is the current enclosed by the closed loop around which you do the line integral on the LHS.
Usually, to apply Ampere's law, you choose a loop to integrate over that has either a constant B-field, and/or with a direction that is either parallel or perpendicular to $d{\bf l}$ (so that the scalar product and/or line integral are much simplified). What loop have you done your integral around? Is the B-field constant along this path? I don't think so...
| {
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What will be the effect of placing a light source very close to a photodiode? What will be the effect of placing a photodiode really close to a laser source and what should be the appropriate distance between a light source and photodiode to get maximum output current?
| Looks like the photodiode's ball lens is arranged to focus a collimated beam onto the diode so ideally you would want to create such a beam with a diameter less than the diameter of the lens to avoid large reflections and poor focus as the rays move off the optical axis. As long as the beam is arranged like that the distance shouldn't matter, but it becomes harder to accomplish as the distance increases since it will not be perfectly collimated.
| {
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Why do some hand dryers blow hot air? I am not sure why some hand dryers are blowing hot air and not just air at room temperature.
To me, hair dryers are just a way to dry one's hands using the same principle as when we shake our hands in the air, or when we blow some air over a hot drink.
Given that the blown air temperature's is not enough (is it ?) to vaporize water, why is hot air used?
| Fact:
By increasing the air's temperature, one also increases the amount of water vapor it can hold before saturating.
Speculation:
Therefore, the osmotic pressure on water molecules in your hair increases and the water vaporizes more quickly than with room temperature air.
| {
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Radiation emission and absorption Any object can emit and absorb radiation and the power of emission can be represented by the Stefan-Boltzmann law:
$$P=A\epsilon\sigma T^4$$
In many texts the net power radiated is the difference between the power emitted and the power absorbed:
$$P_{net}=A\epsilon\sigma (T^4-T_s^4)$$
where $$T_{s}$$ is the temperature of the surroundings.
Why can the surrounding and the object share the same $\epsilon$ ?
If we try to find out the radiation emitted from the surrounding it should be $P_s=A\epsilon_s\sigma T_s^4$, and if $\epsilon_s<\epsilon$, we will get a strange result that energy radiated from the surrounding is less than the radiation absorbed by the body from the surrounding. What am I missing?
| The Stefan-Boltzmann law for net power radiated pertains to the object. That is, we're simply asking, how much radiation leaves this object (this depends on the object's emissivity), and how much radiation is absorbed by this object (this depends on the objects absorptivity). The emissivity and absorptivity in the equation you present thus pertain to the object, not the environment. That equation makes some assumptions. I couldn't find a good explanation for why the coefficients are what they are in the net power formula you posted, so I thought I'd take a step back and derive it.
The power emitted per unit area from the surroundings is
$$P_s=\epsilon_s \sigma T_s^4$$
The object will absorb a fraction of that based on its area and absorptivity:
$$P_a=\alpha \epsilon_s \sigma T_s^4$$
The object will emit:
$$P_e=\epsilon \sigma T^4$$
The net power delivered to the object is
$$P_{net} = P_a - P_e = \epsilon\sigma T^4 - \alpha \epsilon_s \sigma T_s^4$$
If the absorptivity and emissivity are equal, and $\epsilon_s = 1$ (blackbody), we get:
$$P_{net} = P_a - P_e = \epsilon \sigma (T^4-T_s^4)$$
So you'd have to assume that the surroundings perfectly emitting, and that the absorptivity and emissivity are equal. The latter is true under thermodynamic equilibrium or local thermodynamic equilibrium. See the Wikipedia page for Planck's law and in particular the section on Kirchhoff's Law.
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Is the Baum Framptom a valid alternative to Big Bang? In the Baum Frampton model, proposed in 2007, because $\epsilon\lt-1$, after the Big Rip the universe starts again empty. The problem of this model is that inverting the arrow of the time, after a certain number of Big Rips it should have been a single universe from which all the sequence of Big Rips started. Is this model compatible with the astronomical observations of the last years and how is it possible to explain in the contest of this model the anomalies in the background microwave radiation?
| I have no idea if it is compatible with observations of 2013. However, I do remember that in one of Frampton's papers, he stated (back in mid 2000s) that the phantom energy density that he was considering was almost ruled out. It is unlikely that any of these models will be confirmed in our lifetime, and it is also unlikely that they will ever be confirmed, but I think that the cyclic model is too beautiful to give up on just yet.
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Why is the pressure inside a soap bubble higher than outside? Apparently, the air inside a soap bubble is under higher pressure than the surrounding air. This is for instance apparent in the sound bubbles make when they burst. Why is the pressure inside the bubble higher in the first place?
| I drew an image to illustrate the forces at play.
For any curved surface of the bubble, the tension pulls parallel to the surface. These forces mostly cancel out, but create a net force inward. This compresses the gas inside the bubble, until the pressure inside is large enough to counteract both the outside pressure, as well as this additional force from the surface tension.
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Pool in a submarine A common theme in aquatic science fiction is the submarine pool/access to the ocean. That terrible TV show Seaquest had it, The Deep & Deep Blue Sea (Samuel L Jackson is standing in front of it when the shark chomps him). My question is how this could possibly work? From what little knowledge I have, I'd say the cabin where the pool resides would have to be pressurized to the water at that depth. The implications are that you'd have to pass through an airlock to get to the room, and that it would only work to a certain depth.
Is this correct, or it too far to the "fiction" side of the science fiction axes?
| Is there a particular way that you think this scheme will fail?
Rather than have an airlock with that particular portion of ship, you can simply pressurize the entire vessel. There are practical reasons why you would not want to do this at great depths (related to how much gas you use and toxicity), but the problems are not related to how the access works.
Given sufficient gas, it will work to any depth. You simply need the air in the vessel to be at the same pressure as the water at the point of access is.
This access is called a moon pool. The wikipedia page has some examples of its use underwater. Moonpool habitat examples
| {
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Which ball touches the ground first? This is a very well known problem, but I can't find an answer in the specific case I'm looking for.
Let's consider two balls :
*
*Ball 1 weighs 10 kg
*Ball 2 weighs 1 kg
*Balls have identical volumes (so Ball 1 is much more dense)
*Balls have identical shapes (perfect spheres)
Let's drop them from a rather important height, on earth, WITH air. (That's the important thing, because all the proofs that I browse take place in a vaccum).
I am arguing with a colleague. He thinks that ball 1 will fall faster in air, and that the two balls will fall at the same speed in a vacuum. I think that the identical shapes and volumes make air friction identical too and that the vaccum has no importance here. Could someone tell who's right and provide a small proof?
| Other answers & comments cover the difference in acceleration due to drag, which will be the largest effect, but don't forget that if you are in an atmosphere there will also be buoyancy to consider.
The buoyancy provides an additional upward force on the balls that is equal to the weight of the displaced air. As it is the same force on each ball, the acceleration resulting from this force will differ based on the mass of the ball.
This is most easily illustrated by considering one as a lead ball and one as a helium balloon - obviously the helium balloon doesn't fall, because it is lighter than the air it displaced. The upward buoyancy force is greater than the downward gravitational force.
In a heavier fluid, like water, this effect is even more pronounced.
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Length contraction and simultaneous length measurements I am just working through an argument from Halliday Resnick to derive the Lorentz contraction (see quote below).
Some paragraphs before this, the authors note that:
If the rod is moving, however, you must note the positions of the end
points simultaneously (in your reference frame) or your measurement
cannot be called a length.
A paragraph later they invoke the following argument:
Length contraction is a direct consequence of time dilation. Consider
once more our two observers. This time, both Sally, seated on a train
moving through a station, and Sam, again on the station platform, want
to measure the length of the platform. Sam, using a tape measure,
finds the length to be $L_0$, a proper length because the platform is
at rest with respect to him. Sam also notes that Sally, on the train,
moves through this length in a time $\Delta t = L_0/v$ where $v$ is
the speed of the train; that is, $$ L_0 = v \Delta t \quad \text{(Sam)} $$ This time interval $\Delta t$ is not a proper time
interval because the two events that define it (Sally passes the back
of the platform and Sally passes the front of the platform) occur at
two different places, and therefore Sam must use two synchronized
clocks to measure the time interval $\Delta t$.
For Sally, however, the platform is moving past her. She finds that
the two events measured by Sam occur at the same place in her
reference frame. She can time them with a single stationary clock, and
so the interval $\Delta t_0$ that she measures is a proper time
interval. To her, the length $L$ of the platform is given by $$ L = v
\Delta t_0 \quad \text{(Sally)}. $$
Then they conclude by dividing the two equations above:
$$ \frac{L}{L_0} = \frac{v\Delta t_0}{v \Delta t} = \frac{1}{\gamma}$$
or
$$ L = \frac{L_0}{\gamma} $$
which is the length contraction equation.
However I don't see in what sense the length was measured simultanous in the derivation above, how is the detailed connection between the statement that the length measurement has to be simulanous and the quoted derivation?
| It's interesting. It is implicit in Sams relation that what Sally calls length is indeed the coordinate difference taken simultaneously in her frame. Sams relation for length actually corresponds to simultaneity in Sally's frame.
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Calculate water flow rate through orifice I'm not very good with fluid physics, and need some help. Imagine the following setup with water contained in-front of a wall with an opening on the bottom:
How do I calculate the water flow $Q$?. I have made some re-search and found I need to (partially) calculate the pressure across the opening (orifice). But I don't know the pressure on the back side of the orifice. Can this be solved in any way?
Note: I'm not saying "please give me the solution, I'm lazy". I want to figure it out myself. But since, in this case, I only found formulas involving calculating pressure drop, I canno't use them to solve the problem. Therefore I'm turning my face to you, to see if there's another way to solve this problem.
Update: The "tank" holding the water is actually a big lake, and the opening is how much the water gate have opened. I need to very precisely calculate how much water flows through the opening.
| I have used the Darcy Formula together with the following formulas for a quick numeric solution (only a few iterations needed)
*
*$$h_f = \frac{\Delta P}{\rho g}$$
*$$ f = {\rm Darcy}(Re)$$
*$$ h_f = f\,\frac{L}{D}\,\left( \frac{v^2}{2 g} \right) $$
*Solve above for $v$
*$$ Re = \frac{\rho D\,v}{\mu} $$
*Go to step 2 until $f$ converges to a value.
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Why infrared absorption is a nonlinear technique? I am looking for a good explanation explaining why infrared absorption technique is essentially nonlinear (eg. for carbon monoxide quantification).
When using UV/visible/near-IR absorption technique, Beer-Lambert Law often is valid and then you have quite straight way to convert transmittance into absorbance which linearly respond to concentration (provided BLL limitations are not met).
Above technique relies on electronic transitions, when IR absorption technique relies on vibrational and rotational transitions for molecules which has permanent or transient dipole moments. I would like to figure out why such technique is essentially nonlinear.
| Edit: putting the summary of discussion in comments into my answer.
Beer-Lambert law assumes that every photon has equal probability to be absorbed by every molecule. It is only valid for sufficiently monochromatic light
– that is, the bandwidth of light source should be smaller than the width of the absorption line. When absorption is measured with a laser tuned to a particular rovibrational line of a molecule, the Beer-Lambert law works well.
The method discussed here uses a broadband light source that covers a spectral range containing many rovibrational lines of $CO$, but also regions between those lines, where light is not absorbed at all. There is no simple model that could describe such behavior. In theory, one can calculate the absorption spectrum of a molecule, including temperature- and pressure-dependent line broadening, and then calculate it's convolution with the spectrum of the light source. In practice, using gas samples of known composition one can build a calibration curve Absorption vs gas concentration and use it to convert absorption to gas concentration.
.
My old answer described another case where Beer-Lambert law may be not valid:
The only nonlinear effect that comes to my mind is saturation. Beer-Lambert Law is valid if the population of the ground state is not depleted. However, vibrational bands in mid-IR have very strong absorption, and with intense light source (like laser) one could deplete the ground state - see answer to this question.
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How would an X-ray mirror work? I was wondering if light can be reflected how can someone reflect X-ray of what material does it need to be made of and is its design completely different to that of our original mirrors? Does this mean during long-space voyages in which radiation is an problem why can scientists not develop large panels of X-ray mirrors and Gamma-Mirrors and simply reflect the radiation off rather than worry about that?
| Unfortunately X-ray and gamma mirrors are impossible to build the way you think - mainly because there is much less interaction with the matter comparing to UV - it will go through all materials commonly used for making mirrors. Even for EUV light (wavelength of 13.5nm) building effective mirrors is a royal pain.
As wavelength of X-Rays is very small (down to 0.01nm) you cannot build dielectric mirrors, as this would require depositing dielectric layers which are less than 1 atom thick.
It is only possible to reflect X-Rays at a very shallow angle (see Kirkpatrick-Baez X-Ray lens design), so you cannot use this for shielding of radiation coming from all directions. Gammas are even worse - the only thing you can do is to bring alot of mass.
| {
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Boundary Element Method or Boundary Integral Method Computational Aspects I have to solve a Helmholtz equation inside a simply connected domain. I know that in general the boundary integral can be written as,
$$\phi(x)=\int_V G(x,x') \rho(x')\ d^3x'+\int_S \left[\phi(x')\nabla' G(x,x')-G(x,x')\nabla'\phi(x')\right] \cdot d\hat\sigma'$$
Since my problem is to solve the Helmholtz equation with the Dirichlet boundary condition the boundary integral can be written as ($\rho$ is zero inside the boundary and $\phi(x)=0$ on the boundary),
$$\phi(x)=-\int_SG(x,x') \nabla'\phi(x') \cdot d\hat\sigma'$$
My problem is, how to implement this boundary integral numerically in order to get the eigenvalues and eigenfunctions inside the entire boundary. How can I transform this into a matrix form to obtain the eigenvalues and eigenfunctions?
By the way, the domain is two-dimensional and I am solving this equation in connection with quantum chaos (Quantum chaotic billiard).
| May I understand that, since the domain is 2D, the surface integral is actually the following: $\vec{E}\cdot d\vec{\sigma}' = \vec{E}\cdot\hat{n}dl$, where $\vec{E}$ is an arbitrary 2D vector, $\hat{n}$ denotes the normal of the domain boundary and $dl$ is the line element ? Suppose I understand you correctly and then I propose to switch to the momentum space.
Let $\phi(\vec{x}) = \sum_k e^{i\vec{k}\vec{x}}\phi_{\vec{k}}$. On substitution , we get
$$\phi_\vec{k} = \sum_{\vec{k}'} F(\vec{k},\vec{k'})~\phi_{\vec{k}'},$$ where the matrix $F$ is given by $$F(\vec{k},\vec{k}') = -\frac{i}{N}\int_{body}d^2\vec{x}\int_{boundary} dl' e^{i(\vec{k}\vec{x}-\vec{k}'\vec{x}')}\cdot(\vec{k}'\hat{n}')\cdot G(\vec{x},\vec{x}').$$ $N$ denotes the number of points in k-space. I think this formula allows numerical computation.
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Is there a substance that doesn't reflect OR absorb light from the visible light spectrum? Is there a substance that doesn't reflect or absorb visible light but may reflect light from another spectrum? Is there a theoretical substance that would have these properties?
EDIT:
Sorry I wasn't quite clear with my original question. I've updated it to express more what I was thinking. Would a substance be "invisible" if it didn't reflect or absorb light? Does a real or theoretical substance like that exist? I assume we can see glass and so on because it refracts light and to a certain extent reflects it.
| Lol it's all around you. We call it air. It doesn't absorb or reflect light so it's usually invisible but it does reflect intense light as found in a laser beam.
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Why are magnetic fields so much weaker than electric? In EM radiation, the magnetic field is $ 3*10^8$ times smaller than the electric field, but is it valid to say it's "weaker". These fields have different units, so I don't think you can compare them, but even so it seems like we only interact with the electric field of EM radiation, not the magnetic field. Why is this?
| It's a quirk of units: notice that the conversion between them is dimensionful, and has the value $3\times 10^8\,\mathrm{m/s}$, which is the speed of light.
In the CGS system both fields have the same units, and field-squared is an energy density.
In SI units, the energy density for a configuration of fields is given by
\begin{align}
\frac{dU}{dV} &= \frac12\left( \epsilon_0 E^2 + \frac1{\mu_0} B^2 \right)\\
&= \frac{\epsilon_0}2 \left( E^2 + c^2 B^2 \right)
\end{align}
which tells you that, given the field strength ratio $c$ you have calculated for electromagnetic waves, the energy of an EM wave is shared equally between its electric and magnetic components.
We do tend to usually think about the interaction between EM waves and materials in terms of the electric field component, but that's a bias because it's relatively easy to liberate free electric charges, and magnetic interactions only occur at second order. Spend some time around folks thinking about plasma physics, or wrap your head around the behavior of the sun's magnetic field reversal cycle, for a very different perspective.
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Where does wave frequency come from? I am trying to wrap my head around where do oscillations in electromagnetic waves come from. As an example if I would take a string of guitar and ring it, it would produce a certain sound based on the amount of vibrations per second. That amount of vibrations would be the sum of moves of string per amount of time, e.g there is one oscillation happening many times until string runs out of energy.
When I see the visible light it must be same thing something is vibrating and all the oscillations must the the sum of 'something' of one.
What is producing that one oscillation?
| The thing which is "vibrating" is the electromagnetic field, namely its $\vec E$ and $\vec B$ vectors. The animations here show precisely this.
Of course, it's not that some particles vibrate in this case. The electromagnetic wave can exist without any matter at all — all it needs is the field, which is present everywhere.
But, if we have some charges around, they can be made to vibrate and produce the electromagnetic wave. After the wave is produced, the charges can be removed — the wave is a self-sustaining entity.
At the very beginning of electrodynamics there was a model of luminiferous aether, which assumed that there was some medium in which the electromagnetic waves propagate like mechanical waves. But this model had serious problems, which lead to development of special relativity theory.
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The Alcubierre drive and closed timelike curves Under what conditions would it be possible to create closed timelike curves, assuming an Alcubierre drive could exist? Would it be possible to have the latter without the former?
See here for information on the chronology protection conjecture.
| If somehow, it is only possible to create one alcubierre drive, and it can never turn around, then you won't get closed timelike curves. Otherwise, any construction is going to have them.
The reason is that you can "zoom out" far enough that the distortions to spacetime caused by the drive are no longer present, and the person flying the drive then just looks like a spacelike curve in the spacetime. At that point, you're just travelling along spacelike curves, and all the special-relativistic causality objections apply.
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Kinetic energy and potential energy variation over distance in SHM When you compute the average potential energy of a horizontal spring mass system from the mean position to the positive amplitude A, the value comes out to be $\frac{1}{6}kA^2$. For the average kinetic energy over the same range and direction, it is $\frac{1}{3}kA^2$, which is double the average potential energy. What the physical explanation of the different average values of PE and KE?
P.S. No mathematical explanations please, i.e. area under the graph, etc, only explanations in terms physics of the event are appreciated. The question does not involve time averages of PE and KE. Snapshots of derivations can be uploaded if requested.
| One can't discuss the difference between factors of $1/3$ and $1/6$ without mathematics. The difference between these two numbers – and generally, any fact about any numbers – is all about mathematics.
If all values of $x$ between $0$ and $A$ were equally likely, the average value of $kx^2/2$ would be $kA^2/6$ as you say because the average value of $X^2$ for $X$ uniformly distributed between zero and one is
$$\int_0^1 dX\,X^2 = \left.\frac{X^3}{3} \right|^1_0=1/3$$
However, when the harmonic oscillator (you talk about a spring which is a harmonic oscillator) oscillates, it oscillates harmonically, via sines and cosines, so it spends much more time near the $|x|=A$ extreme points where the speed is low than it spends in the vicinity of $x=0$ where the speed is high.
If you compute the average value (over time) of $kx^2/2$ in this oscillating motion, the result will be proportional to the average value of $\cos^2 t$ over time which is equal to $1/2$. So the average kinetic energy of the oscillating motion will be $kA^2/4$ – a number that doesn't appear in your list of results at all.
Similarly, the maximum value of the kinetic energy is $mv_{\rm max}^2/2 = kA^2/2$ at the maximum achieved exactly when the potential energy $kx^2/2$ has the minimum value (zero). Similarly, the minimum value of the kinetic energy is $0$ exactly when the potential energy $kx^2/2$ is maximized i.e. at $|x|=A$.
The average contribution of the kinetic and potential energy to the total energy is the same for the harmonic oscillator – both averages are $kA^2/4$ – a fact that is guaranteed by the "virial theorem".
No factor of $1/3$ ever appears in the correct average values for the harmonic oscillator (just like it doesn't appear in the right solution to the sleeping beauty problem).
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Does Free Will Theorem imply that quantum mechanics plays crucial role in our brain’s functioning (consciousness)?
*
*Does Free Will Theorem imply that quantum mechanics plays crucial role in our brain’s functioning (consciousness)?
*Is opposite statement of Free Will Theorem right: If elementary particles have a certain amount of free will, then so must we?
Because to me elementary particles does have a bit of free will – quantum mechanics guarantees that nobody can predict what one is going to do, say in double slit experiment.
*So Penrose was right and origins of our consciousness lie in the laws of quantum mechanics?
*Is the only way our free will can come from is that of quantum mechanics?
| I am under the impression that the uncertainty principle is simply an epistemic principle in that our uncertainty is only a function of our inability to make a measurement weak enough to not destroy the prior information we had about the system. This is the reason we have a wave function in the first place. To the free will question, what do you mean when you say free will? Do you mean that the particles have a choice in their evolution? The answer is of course not.
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Airplane on a treadmill I've heard conflicting answers, and would like to see the record set straight:
An jet/propeller airplane is traveling on a giant treadmill at takeoff speed. Will the plane takeoff, or will it remain on the runway, and why?
| An airplane's propulsion does not depend on friction between its wheels and the runway so the relative motion of the runway to the body of the airplane has no effect on the plane's motion$^1$. For example an airplane can take off from ice, where the friction between the wheels and the runway surface is effectively zero.
So the plane would take off normally if you placed it on a giant treadmill running at any speed.
$^1$ there would be some effect because there will be some friction in the wheel axles. However the forces due to friction in the wheel axles will be tiny compared to the thrust generated by the plane's engine.
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How to measure trajectory regularity? I have two animal running trajectories. A regular one with repeated back and forth running between point A and B, like the one on top in the figure. The other one is very irregular, animal paused and turned around a lot in the middle. Is there any algorithm to measure the regularity of a trajectory, like repeated activity on the top? And compare the extend of regularity between the two trajectories? By 'more regular trajectory', I mean animal run with higher mean speed, less turn back and less pauses in the middle, more laps within the same amount of time. The figure above is for demonstration purpose, since animal can not run straight with constant speed anyway.
| If it were me, I would take the velocity as a function of time, and split into constant segments. For real data you will likely have to determine a threshold for what constant means, but in either case, split the trajectory up into "constant" velocity segments. Then compute distance traveled in each of these segments. Looking at the histogram of distances from a path, you should be able to characterize its "regularity". Animals tend to have power law distributed length segments, also known as levy flight, whereas a completely regular path would have only a few distances making up its segments.
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Can a neutrino act as a virtual particle between two electrons to mediate an electron-electron fermonic interaction? Can a neutrino act as a virtual particle between two electrons to mediate electron-electron fermonic interaction analogous to how a photon acts as a virtual particle between two electrons to mediate a bosonic electron-electron interaction?
What would the Lagrangian look like for an electron-electron interaction mediated by neutrinos?
| I'm a bit rusty on my qed, but I'll give this a shot. The simplest case would be described by a diagram similar to:
But the $e^--e^--\nu_e$ vertex doesn't exist (also note that I can't draw the required arrow on the neutrino) - the vertices of the standard model (with the exception of vertices involving the Higgs and neutrino oscillations) are:
With these, the closest interaction to what you describe that I can see is:
There are virtual neutrinos as you specified, but also virtual $W$ bosons.
If you rotate that diagram 90 degrees, there's an $e^--e^+$ scattering mediated by virtual neutrinos and $W$ bosons, but again, not quite what you asked for.
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Diffusion and Drift currents in a PN junction In a forward-biased PN junction, the potential barrier decreases, allowing more majority carriers from one side to diffuse to the other side where they are minority carriers. After they cross the potential barrier, they form a diffusion current, the drift current of minority carriers is insignificant, then they recombine with majority carriers and form a drift current under the effect of the applied electric field.
Why do minority carriers form a diffusion current not a drift current after they cross the potential barrier? It is counter-intuitive that the main current is diffusion when there is an applied electric field.
This is according to all the microelectronics book I'm currently reading. There is one which says this can be proved but without providing anything. Can someone please provide a proof for this.
|
Why do minority carriers form a diffusion current not a drift current after they cross the potential barrier?
Because the electric field is zero outside the depletion region, so a drift current cannot be driven.
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Calculating Hubble's constant at earlier times I want to calculate Hubble's constant at some redshift $z$. I have found the following formula:
$$H^2=H_0^2\left(\Omega_m\left(1+z\right)^3+\Omega_{\Lambda}\right)$$
Now it's obvious that at higher redshift $z>0$ (earlier times) the Hubble's constant $H$ increases, which should not be the case. Why can't I use this formula and which formula should I use then?
|
so, it's actually only numerically solvable? and is there no analytic expression for $H(z)$?
In my answer I'll be using the scale factor $a$ instead of the less wieldy redshift $z$. The two are simply related by $a = 1/(1+z)$.
In general the first Friedmann equation can be written as :
$$
H^2 = \left( \frac{\dot{a}}{a} \right)^2 = H_0^2 \left( \sum_i \Omega_i \times a^{-3(1+w_i)} + \Omega_k a^{-2} \right)
$$
Where $w$ is the equation of state parameter of each effective fluid $i$. Now let's take the case of a flat universe ($\Omega_k = 0$) where a single $\Omega_i$ dominates ($\Omega_j \approx 0$ for $j \neq i$). We get that :
\begin{eqnarray*}
\left( \frac{\dot{a}}{a} \right)^2 &\propto& a^{-3(1+w_i)} \\
\dot{a} &\propto& a^{-3/2(1+w)+1}
\end{eqnarray*}
Defining $\gamma$ such that $a \propto t^\gamma$ we get :
\begin{eqnarray*}
t^{\gamma - 1} &\propto& t^{-3/2 (1 + w) \gamma + \gamma} \\
\gamma - 1 &=& -3/2 (1 + w) \gamma + \gamma \\
\gamma &=& \frac{2}{3 (1+w)}
\end{eqnarray*}
This analytic solution is valid for all $w$ except when $w = -1$. There you get an exponential solution instead, and I'll leave it as an exercise for the reader to convince himself of it. As an example, for a flat matter dominated universe ($\Omega_m = 1$ and $\Omega_k = 0$) we get $a \propto t^{2/3}$ because $w_m = 0$.
What happens when we're not in such a simple scenario and there is more than one contributing effective fluid ? In that case there the analytic solutions are more intricate, and a guide to some of them may be found here
Note that you had an analytic form for $H(a)$ right from the start, and if you want one for $H(t)$ then it's pretty straightforward to plug in the solution I just gave you into the Friedmann equation.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is there inductance to a DC circuit? When a DC circuit is carrying current, large amounts or small, is there induced-emf due to the inductance? Or is it only applied to AC circuits?
| If the change is withing the time constant of the circuit, DC will in fact produce on going back emf that opposes the original current flow thus control current. inductive reactance then comes into play initiated by the inductance change thus when the current flow changes a frequency is applied to the DC thus inductive reactance initiated by inductance.
inductance or current change can cause inductive reactance.
Thank you user191954.
Marathonman
| {
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"url": "https://physics.stackexchange.com/questions/129441",
"timestamp": "2023-03-29T00:00:00",
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Why do liquids boil when their vapor pressure equals the ambient pressure? Given that the boiling point of a liquid is the temperature at which the vapor pressure is equal to the ambient (surrounding) pressure, what significance does a liquid's vapor pressure have in the formation of bubbles that happens at and above the boiling point?
The definition of boiling point seems to imply that the pressure inside of the bubbles must be at least as great as the liquid's vapor pressure in order to balance the outside pressure, but is there any particular reason why the pressure inside of the bubbles is related to the vapor pressure?
The vapor pressure seems to be a measurement describing the tendency of the molecules to escape from the surface of the liquid, but I don't see how that relates to bubble formation within the liquid.
This question has bothered me for a while, so any help would be much appreciated.
| When you heat water on the stove top, you see bubbles forming on the bottom of the pot. The bubbles are created where the heat applied (if you move the pot, you see the bubbles forming in a different spot) and is sufficient to convert the liquid into a vapor (less heat would just heat the water). These bubbles form even though the water is below its boiling point (when the bubble detaches and rises, it sometimes disappears, which means the water absorbed the heat from the bubble).
Once the water reaches the boiling point, the water doesn't increase in temperature. It just evaporates at the speed needed to equal the amount of heat added to the water. If the heat is enough, it was boil.
but is there any particular reason why the pressure inside of the bubbles is related to the vapor pressure?
At the bottom of the pot, the pressure would be the vapor pressure plus the depth of the liquid.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/129512",
"timestamp": "2023-03-29T00:00:00",
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Average Velocity of a body moving in a circle with constant speed $v$
A Body is moving with constant speed $v$ along a circle of radius $R$. Find the average velocity of the body from time $t = 0 $ to $t= \frac{R}{3V}$.
My attempt at the question:
Let distance traveled along the circle be $d$.
$$ d= \left(\frac{R}{3v}\right)\cdot v
\Rightarrow d= \frac{R}{3}$$
Let angle subtended by arc be $\theta$. Then
$$ \theta = \frac13 rad. $$
By sine rule I can find out the displacement $S$:
$$S = 2R\sin(1/6) $$
Hence average velocity :
$$\text{average velocity} = \frac{S}{t}
\Rightarrow \text{average velocity}= 6v\sin(1/6) $$
But the correct answer is $\dfrac{3v}{\pi}$.
I need some help here as to how to solve it.
| Just off the top of my head, $T=\frac{2\pi R}{v}$ so the time $t=3R/v$ is a fraction $3/2\pi<1/2$ of the total period. Just less than one half, which means that the angle sub tended is not $\theta=1/3$ actually $\theta = 3rad\sim\pi$ so that most of the one direction of motion along the semi circle will have canceled out and the other component will be responsible for most of the motion.
I think you should set up a vector quantity for the position and derive the velocity by taking derivatives. find the position after a time $t'=3R/v$, find the change in position between the starting point and ending point, find the distance by squaring the change in position then divide by the time.
$$\vec r=Rcos(\omega t)\hat x + Rsin(\omega t)\hat y$$
$$v=\omega R$$
| {
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"timestamp": "2023-03-29T00:00:00",
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What is $c + (-c)$? If object A is moving at velocity $v$ (normalized so that $c=1$) relative to a ground observer emits object B at velocity $w$ relative to A, the velocity of B relative to the ground observer is
$$ v \oplus w = \frac{v+w}{1+vw} $$
As expected, $v \oplus 1 = 1$, as "nothing can go faster than light".
Similarly, $v \oplus -1 = -1$. (same thing in the other direction)
But what if object A is moving at the speed of light and emits object B at the speed of light in the exact opposite direction? In other words, what is the value of $$1 \oplus -1?$$
Putting the values in the formula yields the indeterminate form $\frac{0}{0}$. This invites making sense of things by taking a limit, but $$ \lim_{(v,w)\to (1,-1)} \frac{v+w}{1+vw}$$ is not well-defined, because the limit depends on the path taken.
So what would the ground observer see? Is this even a meaningful question?
Edit: I understand $1 \oplus -1$ doesn't make sense mathematically (thought I made it clear above!), I'm asking what would happen physically. I'm getting the sense that my fears were correct, it's physically a nonsensical situation.
|
I'm getting the sense that my fears were correct, it's physically a
nonsensical situation.
Applying a formula outside of the context in which it was derived will likely produce nonsensical results.
In the derivation of the relativistic velocity addition formula, and using your notation, there is an object B with uniform velocity $w$ in some inertial reference frame (IFR) A.
Further, A has velocity $v$ in the lab frame of reference.
The velocity of B in the lab frame is then given by
$$\frac{v + w}{1 + \frac{vw}{c^2}}$$
Now, when you stipulate that $v = c$, you're attempting to apply this result outside of the context in which it was derived. In particular, in SR, there are no IFRs with relative speed c.
Put another way, an object with speed c in one IFR has speed c in all IFRs; an object with speed c has no rest frame.
Since, by stipulation, A is an IFR in which B has speed $w$, it must be the case that $v < c$.
Less precisely, only one of the velocities in the relativistic velocity addition formula can validly be set to c - both cannot be c since that would presume an IFR with relative speed c exists.
| {
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"url": "https://physics.stackexchange.com/questions/129636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 2
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How to approximate acceleration from a trajectory's coordinates? If I only know $x$- and $y$- coordinates of every point on a trajectory without knowledge of time information, is there any way to approximate Cartesian acceleration angle at each point? Time interval between every two points is very small, ~0.03 second.
| Yes. Provided you are only interested in the direction of the acceleration, and not it's magnitude. And further assuming your time samples are equally spaced, you can take the second derivative of the path and this will be proportional to the acceleration.
A decent method in practice would be to use a second order central finite difference scheme wherein you say that:
$$ a_x(t) = x(t-1) - 2x(t) + x(t+1) $$
and
$$ a_y(t) = y(t-1) - 2y(t) + y(t+1) $$
this will give you decent estimates for the cartesian components of acceleration at every time, caveat to an overall scaling in magnitude that you won't know without knowing the actual timing, but the direction should be alright.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/129701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Potential Energy Concept Imagine a book that we lift it with a force that is exactly equal to the force of gravity so the forces cancel out and the book moves with a constant velocity.
Consider the situation after the book has been lifted, and it has come to rest once again. According to the work and kinetic energy laws
$\Delta W = \Delta K.$
This seems to hold here, since both are zero. That's okay, but where did the increase in the potential energy come from? Is energy not conserved?
| If you lift the book with a force that is exactly equal (but opposite) to the force of gravity acting upon it the book won't go anywhere. After all, that's exactly what a table does when you leave the book on it.
To get to book going you need to lift with a force greater in magnitude than that of gravity, resulting in a net force and an associated acceleration. Thus the book gains velocity and hence kinetic energy, which is converted into potential energy as the book rises.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/129777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proving the equivalency of the potential energy of a system of charges and the work required to assemble a system of charges This is a very cool, and highly beneficial problem in my opinion. I feel as though truly understanding this proof would broaden anyone's conceptual understanding of electric potential.
My textbook asks me to utilize the identity:
$\bigtriangledown(\phi\bigtriangledown\phi) = (\phi\bigtriangledown)^2 + \phi\bigtriangledown^2\phi$
and the divergence theorem to prove that the potential energy of a system of charges $U_E = \cfrac \pi8\int_{entire-field}E^2dv$ and the work that it takes to assemble a charge distribution $U_w = \cfrac 12\int \rho\phi dv$ are "not different" for all charge distributions of finite values.
So far, I have decided to substitute $-\bigtriangledown^2\cfrac{1}{8\pi}$ in for $\rho$ so that I can use the given identity [ I also had to rearrange the ordering to $\phi\bigtriangledown^2\phi = \bigtriangledown(\phi\bigtriangledown\phi) -(\phi\bigtriangledown)^2 $] and ascertain that $U = \cfrac \pi8\int \bigtriangledown(\phi\bigtriangledown\phi)dv - \cfrac \pi8\int \phi\bigtriangledown^2\phi dv, $ but now I am stuck. I know that the next step must have to do with using the divergence theorem to simplify this thing though. Any help would be greatly appreciated.
| Hint:
The divergence theorem tells us that the divergence of a vector field integrated over a region $R$ with boundary $\partial R$ equals the integral of that vector field dotted with the outward-pointing normal along the boundary;
\begin{align}
\int_R dV\, \nabla\cdot \mathbf v=\int_{\partial R} dS\, \mathbf v\cdot\mathbf n.
\end{align}
If we are integrating over all space, then this is like integrating over the inside of a sphere but in the limit that this sphere is infinitely big. So we have
\begin{align}
\int_{\mathbb R^3} dV\, \nabla\cdot\mathbf v = \lim_{r\to\infty}\int_{B_r}dV\,\nabla\cdot \mathbf v = \lim_{r\to\infty} \int_{S_r}dS \,\mathbf v\cdot \mathbf n
\end{align}
where $B_r$ denotes the closed ball of radius $r$ (namely the inside of a sphere of radius $r$ along with its boundary) and $S_r$ denotes the sphere of radius $r$. Now, on the surface of a sphere of radius $r$, the area element and outward-pointing normal are
\begin{align}
dS = r^2\sin\theta d\theta d\phi, \qquad \mathbf n = \hat{\mathbf r},
\end{align}
so we get
\begin{align}
\int_{\mathbb R^3} dV\, \nabla\cdot\mathbf v = \lim_{r\to\infty} \int_{S_r} d\theta\,d\phi\,\sin\theta\,r^2 v_r
\end{align}
If $v_r$ decreases sufficiently rapidly with $r$, then the expression on the right will be zero. In other words, we have found that
Provided the radial component of a vector field falls off sufficiently rapidly with $r$, the integral of the divergence of the vector field over all of space vanishes.
Try to use this fact along with things you know about how the electric field of a finite distribution of charge behaves very far from the distribution.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/129910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Seesaw type-1 and integrating out heavy fields Let's assume seesaw 1 type of generation of left neutrino Majorana mass:
$$
L_{m} = -G_{ij}\begin{pmatrix} \bar{\nu}_{L}& \bar{l}_{L}\end{pmatrix}^{i}i\sigma_{2}\begin{pmatrix}\varphi_{1}^{*} \\ \varphi^{*}_{2} \end{pmatrix}\nu_{R}^{j} - M_{ij}(\nu_{R}^{T})^{i}\hat{C}\nu_{R}^{j} + h.c.
$$
After using the unitary gauge and shifting the vacuum we can get the mass terms
$$
\tag 1 L_{m} = -\frac{1}{2}\begin{pmatrix} \nu_{L} & \nu_{R}^{c}\end{pmatrix}^{T}\hat{C}^{-1}\begin{pmatrix} 0 & \hat{m}_{D}^{*} \\ \hat{m}_{D}^{\dagger} & \hat{M}^{\dagger} \end{pmatrix}\begin{pmatrix} \nu_{L}\\ \nu_{R}^{c}\end{pmatrix}+h.c.,
$$
where $\nu^{c} = \hat{C}\bar{\nu}^{T}, \quad \hat{m}^{ij}_{D} = \eta G^{ij}$,
and interaction terms with the Higgs field $\sigma $:
$$
\tag 2 L_{int} = -\sigma G_{ij} \bar{\nu}_{L}^{i}\nu_{R}^{j} + h.c.
$$
We can "diagonalize" $(1)$ if we assume that $||m_{D}|| << ||M||$:
$$
L_{m} \approx \nu_{L}^{T}\hat{M}_{1}\nu_{L} - \nu_{R}^{T}\hat{M}_{2}\nu_{R} + h.c., \quad \hat{M}_{1}= -\hat{m}_{D}^{T}\hat{M}^{-1}\hat{m}_{D}, \quad \hat{M}_{2} = \hat{M}.
$$
But how to "delete" interaction term $(2)$ (i.e., to integrate out heavy field $\nu_{R}$)?
Or this interaction is the prediction?
| The interaction term disappears because when you integrate out the right handed neutrinos you insert in a linear combination of the other fields in their place. For example I expect that we would have something of the form,
\begin{equation}
G _{ ij} \sigma \bar{\nu} _L ^i \nu _R ^j \rightarrow G _{ij} G _{jk}\sigma ^2 \bar{\nu} _L ^i \nu _L ^k
\end{equation}
This isn't an exact expression since you are doing the matching after diagonalization in which case you are integrating out the interaction basis right handed neutrinos as opposed to the mass basis right handed neutrinos.
To calculate exactly what this term is you can use tree level matching or integrate out the field using the equation of motion.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/130121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Quantum Entanglement and BB84 protocol Can Quantum entanglement can be used to strengthen the security or increase the efficiency of communication via the BB84 protocol? If so, how?
| All different QKD protocols are well covered in most Quantum Information textbooks, e.g. Quantum Information by Jaeger. Trying to give a complete, understandable coverage of the matter is impossible in a post like this, so I'll just give you an overview:
In the E91 scheme, entangled photon pairs are used between Alice & Bob, and unlike the single-photon measurements performed in BB84, here Bell states are used. The first direct consequence of using entangled qubits, is the provision of fully randoms bits to both parties where the bits are correlated with each other(ideally with maximally entangled two-qubit states.)
Unlike BB84, Alice does not provide the prepared qubits, instead a distributor is used, which creates entangled-pairs, and sends each one of them to both Alice and Bob. They perform their measurement using their sets of basis (2-3 different bases). Here also the choice of basis is usual done randomly and all their measurements performed independently. Then they retain only the bits obtained from measurements using the same basis, because knowing they're using entangled-pairs, their results would be perfectly correlated.
Finally bear in mind that the biggest difference here is the fact that Eve will not be able to detect qubit states without inducing errors in the corresponding subspace, whereas in the usual single photon BB84 Eve has the possibility (with a certain probability of course) to measure bits without changing their states, had she luckily chosen the same basis as Alice had used to prepare the qubit's polarization. All in all, this protocol makes Eve's influence more easily "noticeable" due to the nature of the E91 protocol.
To sum up, this was only a crude overview, the protocol has as many nuances as the famous BB84 does, even more, and to able to say whether E91 is more efficient or not is rather difficult without a full understanding of both, or without specific scenarios in mind. Just keep in mind that the main idea is to have fully random shared-key established between Alice and Bob, without any of its bits having been leaked in the process. In this regard E91 may provide a more secure establishment using entangled qubit pairs, and their encryption methods/algorithms remain among the existing classical ones.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/130213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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