classical mechanics - Derivation of Hamilton-Jacobi theory using canonical transformations

The derivation of the Hamilton-Jacobi equation using canonical transformations is typically done involving a type-2 generating function.

Is it possible to use a another type of generating function, namely type 1, 3 or 4?

quantum electrodynamics - How are classical optics phenomena explained in QED (color)?

How is the following classical optics phenomenon explained in quantum electrodynamics?

According to Schroedinger's model of the atom, only particular colors are emitted depending on the type of atom and the state of its electrons. Then, how it is possible that the same type of matter changes color in relation to temperature? For example, water is transparent with a light blue tint, but snow is white (which means, pretty much, that photons have no particular color in the visible spectrum).

Split by request: See the other part of this question here.

Answer

It's important to distinguish different phases of the material. While it's true that water and snow consists of the same building blocks of $H_2 O$, those individual blocks are actually not as important as the resulting material. You can build a storage room and pyramids just out of bricks (at least in principle).

Gases

Let's start with the simplest case. Simplest in the sense that there is not much going on, just molecules flying around. Let us just assume ideal gas where molecules don't interact with each other. Then to explain everything it suffices to look at just one molecule.

The color of the molecule of course arises because of its absorption spectrum. This in turn depends on the molecule's energy levels. For atoms these are pretty nice discrete levels. For more complicated molecules you also have rotational and vibrational degrees of freedom to take into account and besides discrete levels you'll also see continuous strips that consist of very fine energy levels corresponding to that. For illustration, see the wikipedia article on $H_2$ hydrogen.

In any case, if you are somehow able to obtain that energy spectrum you can then investigate the macroscopic properties by the means of the usual Boltzmann statistics

$$\hat W = Z^{-1} \exp(-{\beta \hat H}) = Z^{-1} \sum_n\exp(- \beta E_n) {\hat P}_n $$

with $Z = {\rm Tr} \exp(-{\beta \hat H})$ being the partition function, $\beta$ the inverse temperature and $P_n$ projector on $n$-th energy level.

Using this you can see that as you increase temperature the energy level distribution will change to occupy higher levels and this in turn will change the probability of individual absorption processes between certain levels.

Now, the only place where you need to talk about QED is the connection between absorption spectrum and energy levels. Recall that in quantum mechanics energy levels are stable. They don't change, so there would be neither emission nor absorption. To resolve this apparent paradox we have to recall that we forgot to quantize the electromagnetic field. If you take this into account then atom's excited energy levels are no longer stable because of the fluctuations of electromagnetic field. Or in particle terms: because the number of particles is not conserved and it's easy to create photon out of nothing. Of course, energy is conserved so only photon's corresponding to the energy difference are possible. And the same can be said for absorption.

Liquids and solids

For now I won't talk about these phases because the answer is both becoming long and because optical properties of solids would constitute a book of its own. As for liquids, I have a feeling that the same analysis as for gases should hold except it's of course no longer sufficient to talk about individual molecules because of non-negligible interactions. But I guess starting from ideal gas's statistics and just switching temperature should give a reasonable first approximation.

---

Remark: Looking back on the answer, I didn't really explain your question completely. But I guess it's because it's quite hard to encompass everything that is going on. Maybe another split and specialization (e.g. to color of solids) of the question would be in order? :-)

optics - Real images and their formation

We're told that a real image is formed when light rays actually converge to a point. That's all good. But what happens if a screen isn't there to take the image on? Is it still there?

Answer

Whether the screen is there or not the image is there but the problem is focussing your eye on a region of air where the image is formed without a screen.

Try the following set up:

Illuminate a 35 mm slide with a light bulb and adjust the lens so that a sharp real inverted image is formed on a screen with the eye to the left of the screen.

Now replace the screen with a sheet of tissue paper (something which allows light through and will at the same time produce a visible image) and form a sharp image on the tissue paper.

Now observe the image from the other side of the tissue paper (to the right of it as in the diagram).
Your eye will have to be at least 25 cm from the screen if you have normal eyesight.

Keep focussing on the image and slowly move the tissue paper slowly to one side so that some of the image is on the tissue paper and some in "mid air".

With a little practice you should be able see the image of the 35 mm slide without the tissue paper being there at all.

The tissue paper was used to enable you to focus on the correct area of space to view the sharp image.

Update as a result of some comments

Set up with a $3.5\,\rm cm$ focal length hand magnifier as the converging lens.
The object is a pin (white) illuminated by a torch which is switched off for the photograph to be taken without contrast problems.
The other pin (red) will be used to located the image of the white pin.

A white screen was placed next to the image pin to show the real inverted image formed by the lens.

Viewing point now from the top of the first picture ie on the other side of the lens from the position of the object pin.

Image is distorted due to a variety of cheap lens defects.

The position of the real image can be conformed by moving ones eye up and down and seeing that the tips of the image and the tip of the image pin do not move relative to one another - a position of no parallax.

---

In the end if you know what you are looking for and approximately where to look just looking through the lens on the side remote from an object will enable you to see the real image in mid air.
This is more difficult if the image is highly magnified.

---

Further update

Note that in the third photograph the image is in focus so the camera "knew" where the image was.
The image pin did help with location but by telling the camera to focus at a certain distance away and no image location pin I would still have been able to get a sharp image on the photograph.

---

Consider the diagram below which shows the formation of an image of an object $ABC$ on the retina of your eye.

A sharp image is formed if all the light which leaves point $B$ on the object arrives at the same point on the retina $B'$.
So all the rays in the cone of light with apex $B'$ shown in the diagram arrive at the same point on the retina $B'$.
The same being true of all points on the object eg $A$ and $C$ which will arrive ar $A'$ and $C'$.
The light from object $ABC$ originates either from the object itself or as a result of light which has been reflected off it.
So you "see" object "ABC" with your eye.

Now how is that different from the arrangement below?

What was the object in the first diagram is now an image of the object $A''B''C''$ formed by the converging lens.
That intermediate image $ABC$ forms an image $A'B'C'$ on the retina which is no different to that in the first diagram.
You "see" intermediate image $ABC$.

Because there are no reference points around it (just air) it is difficult to decide exactly where that image is and you do indeed see it as though it is "in the lens".
By using an image pin (or your finger) you can easily show by the method of no parallax that you are actually looking at an image in mid air.

---

When you use an optical instrument you are looking at an image in the air but you have the advantage of being allowed to move the eyepiece to form an image of that image in the air as well as possibly having some cross hairs which are in the image plane.

general relativity - Would quantum fluctuations cause problems for scalar-field inflation?

Wheeler once said that spacetime would be highly curved at very small scales because of the uncertainty principle for energy-momentum. In which case the spacetime becomes very bumpy and not smooth anymore, which Wheeler called spacetime foam. It seems that such a picture doesn't bother us because in most cases we are dealing with physics of larger scales and the spacetime becomes smooth again at an averaged level over the large scale.

But when we extend the picture to cosmology, problems appear even at a semiclassical level. Now let's consider the $\phi^4$ scalar theory with $$V(\phi)=\frac{\lambda}{4}\phi^4.$$ For the vacuum, because of the uncertainty principle, $\phi$ cannot stay at 0 every where and all the time. If that the field would have definite configuration and definite velocity (field momentum) which violates the uncertainty principle. At the Planck scale $l_p$, the energy should have an uncertainty of $M_p$. Thus for every Plack volume, $\phi$ may take values between $-M_p/(\lambda)^{1/4}$ and $M_p/(\lambda)^{1/4}$ so that $V(\phi)=\frac{\lambda}{4}\phi^{4}\sim M_p^4$.

However in such case, this small patch of space will be driven to inflation. $$a(t)=a_0\exp(Ht),$$ where $$H=\left(\frac{8\pi}{3}V(\phi)/M^{2}_p\right)^{1/2}.$$ Note that the analysis here applies not only to the early universe but also the current universe. But such inflation at small scales will surely cause problems such as the inhomogeneity in our universe which is of course not the case of our observed universe.

So what's the problem with the analysis given above?

Answer

The picture in this post is basically wrong. The uncertainty is used for the length scale of the whole system. For example, if we consider a particle moving in a box of volume $L^3$, we would say that the particle has momentum uncertainty ~$\hbar/L$ and therefore energy uncertainty $\hbar^2/(2m L^2)$. Hence for a large box, the energy uncertainty is nearly zero. But we can not say in this case that, we can look at a smaller region, and in that region, there is a large energy uncertainty which may lead to a large total energy uncertainty if we add them up. That is, we should distinguish the scale of the whole system and the distance from point to point which can be arbitrary small. Likewise, the uncertainty principle does not force us away from the spacetime concept between two arbitrary close points. But it tell us that, a quantum system with a very small scale is meaningless since a black hole will be formed from the large energy fluctuations.

I hope this answer is not misleading.

electromagnetic radiation - Explanation of consistency between Maxwells equations and the existence of photons

I have limited knowledge of QM but I know some about electromagnetism.I have read some about the photon description and I am confused.

How can EM waves be viewed as a continuous, changing field but at the same time be considered as consisting of a finite number of discrete particles?

For example, when considering EM radiation power, the inverse square law applies based upon surface area expansion as you move from the radiating source. How does this work with the photon description?

Answer

The mathematical consistency can be shown, but it needs quantum field theory (QFT) to follow the mathematics. You can read how classical fields and particle emerge from QFT here.

In this image you can get an intuition of the complexity of the build up of classical E and B fields:

It is just displaying the spins for the photons, which are either +1 or -1, nevertheless the circular polarization is built up by the photons, the connection being the left and right polarization.

This is clear if one understands that it is the complex wavefunctions of the photons which are superposed into a $Ψ$ . The real measurements of E an B come from the $Ψ^*Ψ$ probability distribution, and thus it is not a simple build up.

The wavefunctions of the individual photons are solutions of a quantized Maxwell's equations, and they carry the E and B fields in the complex function:

that is why the emergent classical wave is consistently built up.

thermodynamics - Does the scientific community consider the Loschmidt paradox resolved? If so what is the resolution?

Does the scientific community consider the Loschmidt paradox resolved? If so what is the resolution?

I have never seen dissipation explained, although what I have seen a lot is descriptions of dissipation (i.e. more detailed pathways/mechanisms for specific systems). Typically one introduces axioms of dissipation for example:

entropy $S(t_1) \geq S(t_0) \Leftrightarrow t_1 \geq t_0$ (most often in words)

These axioms (based on overwhelming evidence/observations) are sadly often considered proofs. I have no problem with useful axioms (and I most certainly believe they are true), but I wonder if it can be proven in terms of other (deeper and already present) axioms. I.e. is the axiom really independent? or is it a corollary from deeper axioms from say logic (but not necessarily that deep).

(my opinion is that a proof would need as axioms some suitable definition of time (based on connection between microscopic and macroscopic degrees of freedom))

quantum mechanics - Interference of two non-entangled photons, calculation using tensor product of Hilbert spaces

I'm trying to calculate the interference of two non-entangled photons, like in a double-slit experiment with two photon sources, one behind each slit (follow-up on this question).

The individual particles have the wave function $\psi_i\in\mathcal{H}_i$ where $i\in\{1,2\}$. The tensor product of their wave function is

$$ \Psi(1,2) = c_1\Psi_1 + c_2\Psi_2 = c_1(\psi_1\otimes 1_2) + c_2(1_1\otimes\psi_2) $$

This is what I tried:

$$ \begin{align} |\Psi(1,2)|^2 & = \langle c_1(\psi_1\otimes\mathbf{1}) + c_2(\mathbf{1}\otimes\psi_2) | c_1(\psi_1\otimes\mathbf{1}) + c_2(\mathbf{1}\otimes\psi_2) \rangle \\ & = \sum_i|c_i|^2|\psi_i|^2 + \overline{c_1}c_2\langle\psi_1\otimes\mathbf{1} | \mathbf{1}\otimes\psi_2\rangle + \overline{c_2}c_1\langle\mathbf{1}\otimes\psi_2|\psi_1\otimes\mathbf{1}\rangle \\ & = \sum_i|c_i|^2|\psi_i|^2 + 2\,\mathfrak{Re}\left(\overline{c_1}c_2\langle\psi_1\otimes\mathbf{1} | \mathbf{1}\otimes\psi_2\rangle\right) \\ & = \sum_i|c_i|^2|\psi_i|^2 + 2\,\mathfrak{Re}\left(\overline{c_1}c_2\langle\psi_1|\mathbf{1}\rangle\langle\mathbf{1}|\psi_2\rangle\right) \end{align} $$

Is that correct so far? How to continue to show the interference?

Answer

Photons are quantum mechanical entities. This means that their interactions can be described by Feynman diagrams, which are reinterpreted as creation and annihilation operators operating on ground states with propagators in between.

It has little meaning to talk of two individual photons "interfeering" without writing down the Feynman diagrams. Photon photon interactions are four vertex at lowest order. This means a factor of ~10^-8, as the electromagnetic coupling constant is of order 10^-2.

A Feynman diagram (box diagram) for photon–photon scattering, one photon scatters from the transient vacuum charge fluctuations of the other

The four propagators for energies below gamma energies are highly off mass shell and thus diminish the probabilities even further.

So it is very improbable that two photons will have a measurable interference.

speed of light - Special Relativity, 2nd Postulate -- Why?

As a lowly physics undergrad who has been chewing on this 2nd postulate of special relativity for a year or more, I simply can't wrap my head around reasons why it is true or how Einstein might have been convinced enough to propose this postulate.

Consider Alfred who is riding in a car travelling at 88 m/s with his headlights on and Bernard who is on the side of the road hitch hiking. Why does the light propagating from Alfred's car move at $c$ relative to both Alfred and Bernard, and not at $c$ + 88 m/s relative to Bernard?

The nifty results of special relativity all kind of hinge on this idea, and asking my professors in class hasnt really yielded an answer much more than "because we have never observed a case otherwise".

Answer

Actually given that the first postulate says that all physical laws are the same in all inertial frames, you could replace the second postulate by the postulate: "Maxwell's equations are the physical laws for electromagnetism".

From Maxwell's laws you can derive that the speed of light in vacuum has a specific, constant value, in SI units $c=1/\sqrt{\epsilon_0\mu_0}$. Now there are three possibilities:

1. 
   

   Maxwell's laws are valid only in a specific inertial frame (or rather, in a specific set of inertial frames at rest relative to each other).

   
   

   That's the essence of the aether hypothesis. It would violate the first postulate. Also, experiments failed to measure that preferred frame.

   
   



2. 
   

   Maxwell's laws are not the correct description of electromagnetism (that is, they are valid in no inertial frame).

   
   

   That option would, of course, have been compatible with the first postulate, but not very likely, given the huge experimental support for Maxwell's equations.

   
   


3. 
   

   Maxwell's laws are valid in all inertial frames.

   
   

   If that is the case, then all of the consequences of Maxwell's equations have to be valid in all inertial frames. One of the consequences of Maxwell's equations is the value of the speed of light in vacuum.

   
   

However, it turns out that the only thing from Maxwell's equations you actually need in order to derive special relativity is the constant speed of light. Therefore it makes sense to postulate that directly; that way even if it turned out that Maxwell's theory had to be revised, you don't need to revise relativity as long as the revised theory still predicts a constant speed of light.

Why is the speed of light defined as 299792458 m/s?

Why is the speed of light defined as $299792458$ $m/s$? Why did they choose that number and no other number?

Or phrased differently:

Why is a metre $1/299792458$ of the distance light travels in a sec in a vacuum?

Answer

The speed of light is 299 792 458 m/s because people used to define one meter as 1/40,000,000 of the Earth's meridian - so that the circumference of the Earth was 40,000 kilometers.

Also, they used to define one second as 1/86,400 of a solar day so that the day may be divided to 24 hours each containing 60 minutes per 60 seconds.

In our Universe, it happens to be the case that light is moving by speed so that in 1 second defined above, it moves approximately by 299,792,458 meters defined above. In other words, during one solar day, light changes its position by $$ \delta x = 86,400 \times 299,792,458 / 40,000,000\,\,{\rm circumferences\,\,of\,\,the\,\,Earth}.$$ The number above is approximately 647,552. Try it. Instruct light to orbit along the surface of the Earth and you will find out that in between two noons, it will complete 647,552 orbits. Why it is exactly this number? Well, it was because how the Earth was created and evolved.

If it didn't hit a big rock called Megapluto about 4,701,234,567.31415926 years ago, it would have been a few percent larger and it would be rotating with a frequency smaller by 1.734546346 percent, so 647,552 would be replaced by 648,243.25246 - but because we hit Megapluto, the ratio eventually became what I said.

(There were about a million of similarly important big events that I skip, too.)

The Earth's size and speed of spinning were OK for a while but they're not really regular or accurate so people ultimately switched to wavelengths and durations of some electromagnetic waves emitted by atoms. Spectroscopy remains the most accurate way in which we can measure time and distances. They chose the new meter and the new second as a multiple of the wavelength or periodicity of the photons emitted by various atoms - so that the new meter and the new second agreed with the old ones - those defined from the circumference of the Earth and from the solar day - within the available accuracy.

For some time, people would use two independent electromagnetic waves to define 1 meter and 1 second. In those units, they could have measured the speed of light and find out that it was 299,792,458 plus minus 1.2 meters per second or so. (The accuracy was not that great for years, but the error of 1.2 meters was the final accuracy achieved in the early 1980s.)

Because the speed of light is so fundamental - adult physicists use units in which $c=1$, anyway - physicists decided in the 1980s to redefine the units so that both 1 meter and 1 second use the same type of electromagnetic wave to be defined. 1 meter was defined as 1/299,792,458 of light seconds which, once again, agreed with the previous definition based on two different electromagnetic waves within the accuracy.

The advantage is that the speed of light is known to be accurately, by definition, today. Up to the inconvenient numerical factor of 299,792,458 - which is otherwise convenient to communicate with ordinary and not so ordinary people, especially those who have been trained to use meters and seconds based on the solar day and meridian - it is as robust as the $c=1$ units.

rotational dynamics - Why does the video recorded by a falling, rotating camera seem to stabilize?

Here is a video of a GoPro falling from a plane: http://www.youtube.com/watch?v=QrxPuk0JefA

Any idea what is happening when the image "stabilizes" around 0:35? I think it is because the camera's tangential velocity approaches the speed of the GoPro's rolling shutter.

Follow-up question:

• Why is the image skewed with the left side skewed "up" and the right side skewed "down?"

A challenge:

• Find the axis of rotation (with supporting diagram and explanation of how it was computed).

space travel - How are interplanetary trajectories found?

How are interplanetary trajectories that involve gravity assist maneuvers found?

Examples:

• the MESSENGER spacecraft made flybys of Earth, Venus, and Mercury before getting into orbit


• the Juno spacecraft will flyby Earth (returning after launch) before continuing to Jupiter



• many more...

Are these found using a brute force search of virtually all possible paths or is there a more direct method for finding these helpful alignments?

If I decided to leave for Neptune today (using gravity assist), how would I find a good route and are such tools available on the desktop?

Answer

If I decided to leave for Neptune today (using gravity assist), how would I find a good route and are such tools available on the desktop?

I don't know about the 'find[ing] a good route' stuff, but my understanding is that the main tool used for these calculations is the SPICE Toolkit, and if I understand correctly, it's no longer ITAR controlled, so anyone should be allowed to download it.

I've also never used it myself, so I have no idea what sort of user-interfaces there are to it; we just use it for computing the location of spacecraft via an API.

(and this reminds me -- in the movie 'Starship Troopers', if the computer's able to confirm that the new route is the 'optimal path', why did they rely on humans putting in the path in the first place?)

statistical mechanics - In thermodynamic systems why must the free energy of the system be minimized?

Is this somehow a consequence of the second law of thermodynamics?

cosmology - Seeing cosmic activity now, really means it happens millions/billions of years ago?

A Recent report about a cosmic burst 3.8 billion light years away. It is written as though it is happening now. However, my question is, if the event is 3.8 billion light years away, doesn't that mean we are continuously looking at history, or is it possible to detect activity in "realtime" despite the distance?

Answer

You are absolutely correct. The event happened 3.8 billion years ago, but it gets tedious and confusing to write in the past tense. For example, which one of the following statements do you find easier to understand?

The high-energy radiation continued to brighten and fade for at least a week after the burst 3.8 billion years ago

or

More than a week later, high-energy radiation continues to brighten and fade from its location

As an example, see this article about writing in astronomy, under Miscellaneous Sticky Points:

Remember that statements about astronomical objects should always be in present tense ("the galaxy has a strong color gradient") unless you are specifically talking about a past event or an object that no longer exists ("the supernova progenitor star was a type O supergiant").

Hope that helps.

What is the general statistical definition of temperature?

Temperature in an isolated system is defined as: $$\frac{1}{T} = -\frac{\partial{S(E,V,N)}}{\partial{E}} $$ But I wonder how one can generalize this to a random system. Or for instance to a point in a system. Because in these books about statistical physics they talk often about "temperature gradients in a system", but for these to exist, temperature has first to be defined in every point (although I can't find a general definition).

I hope someone can help me out.

Answer

Length scales are not accounted for properly in your question. When you have a system at local equilibrium where a temperature gradient can be defined then each "point" in this description contains say $10^{10}$ molecules and can be seen as a thermostatistical system at equilibrium. We call that "local" equilibrium because intensive quantities such as temperature and chemical potential might not be uniform throughout the whole system i.e. they may vary from one "point" to another. There are evolution equations of these mesoscopic quantities that deal with such local equilibrium situations. The simplest are the Fourier (for temperature) and the Fick (for particle density) equations but they can be derived from more general equations with a collision kernel such as e.g. the Boltzmann equation.

How is the conservation of momentum satisfied in long-range attraction such as electromagnetism and gravity?

I'm not a physicist, but my understanding is that electromagnetism (including attraction between opposite charges) is mediated by the photon, and gravity is probably (hypothetized to be?) mediated by the graviton.

I'm curious how that works from the point of view of the conservation of momentum. My naive imagination is, if a particle leaves A to the direction of B, doesn't that mean that A would have to change its momentum to the other direction (away from B)? And when B absorb this particle coming from A, shouldn't B now change its momentum to the same direction (away from A)?

How come is it that in the case of gravity and electromagnetism, A and B move towards each other as a result of this interaction?

Answer

If you consider things classically (for the moment forgetting about virtual particles as mediators of the force) things get more clear.

For instantaneous forces (which do not exist in nature), momentum conservation comes from the fact, that the forces in nature fulfil Newtons axiom actio = reactio, meaning, that for two particles, that interact we have the equations of motion:

$$m_x \ddot x = F(x, y)$$ $$m_y \ddot y = -F(x, y)$$

For the time derivative of the total momentum we get:

$$\partial_t P = \partial_t (p_x + p_y) = \partial_t (m_x \dot x + m_y \dot y) = m_x \ddot x + m_y \ddot y = F(x, y) - F(x, y) = 0$$.

That is the total momentum is conserved.

If we consider that the fields causing the forces propagate (and thus the forces are not instantaneous) we have to consider the momentum of the fields and can write local equations for momentum conservation.

Now: Do not take the virtual particle thing too serious. They are in many ways just mathematical artefacts of how we compute things in quantum field theory (so called perturbation theory). Most importantly, do not confuse them with some macroscopic particle. Rather they are "packets" of waves. Furthermore each elementary process conserves momentum (techspeak: the momentum is conserved at all vertices of a Feynman diagram)! As they are a computational device, the virtual particles do not follow the usual rules of propagating particles, but even if a virtual particle starts from A with a moment away from the particle B, it still can reach B and there interact and give B the momentum carried away from A (thus conserving the total momentum).

nuclear physics - Fission fragments interaction

In a nuclear fission reaction, do the fission fragments interact with each other or with other product neutrons and gammas? if the answer is yes then how could we determine the probability of this interactions and what are the conditions for this interactions?

homework and exercises - Riemann tensor in 2d and 3d

Ok so I seem to be missing something here.

I know that the number of independent coefficients of the Riemann tensor is $\frac{1}{12} n^2 (n^2-1)$, which means in 2d it's 1 (i.e. Riemann tensor given by Ricci Scalar) and in 3d it's 6 (i.e. Riemann Tensor given by Ricci tensor).

But why does that constrain the Riemann tensor to only be a function of the metric? Why not a tensorial combination of derivatives of the metric?

What I mean is why is the Riemann tensor in 2D of the form \begin{align} R_{abcd} = \frac{R}{2}(g_{ac}g_{bd} - g_{a d}g_{b c}) \end{align}

and in 3D, \begin{align} R_{abcd} = f(R_{ac})g_{bd} - f(R_{ad})g_{bc} + f(R_{bd})g_{a c} - f(R_{bc})g_{ad} \end{align} where $f(R_{ab}) = R_{ab} - \frac{1}{4}R g_{ab}$?

Wikipedia says something about the Bianchi identities but I can't work it out. A hint I got (for the 2d case at least) was to consider the RHS (the terms in parenthesis) and show that it satisfies all the required properties of the Riemann tensor (sraightforward) and proceed from there - but, I have not been able to come up with any argument as to why there must be a unique tensor satisfying those properties.

Of course I could brute force it by computing $R_{abcd}$ from the Christoffel symbols etc., but surely there must be a more elegant method to prove the statements above.

Help, anyone? I haven't been able to find any proofs online - maybe my Googling skills suck.

Answer

The simplicity of geometry in lower dimensions is because the Riemann curvature tensor could be expressed in terms of simpler tensor object: scalar curvature and metric (in 2d) or Ricci tensor and metric (in 3d). That fact, of course, does not alter the possibility to write Riemann tensor (as well as Ricci tensor and scalar curvature) as a combination of metric derivatives. But each term in such a combination is not a tensor - only the whole object.

Now, let us recapture, why in lower dimensions we are able to reduce the Riemann tensor to a combination of lower rank tensor objects.

For 3d case, definition of Ricci tensor:

$$ R_{ab} = R_{abcd}g^{ac}$$

contains 6 independent components, exactly the number of independent components in Riemann tensor. So, this equation could be reversed, thus expressing $R_{abcd}$ in terms of $R_{ab}$ and $g_{ab}$.

In 2d case we could similarly start with definition of Ricci scalar:

$$ R = R_{ab} g^{ab} ,$$

and reverse it expressing $R_{ab}$ through $g_{ab}$ and $R$. The next step would be to express Riemann tensor with $g_{ab}$ and $R_{ab}$ (and thus through scalar $R$ only).

Time evolution in quantum mechanics

We know that an operator A in quantum mechanics has time evolution given by Heisenberg equation:

$$ \frac{i}{\hbar}[H,A]+\frac{\partial A}{\partial t}=\frac{d A}{d t} $$

Can we derive from this that $$ A(t)=e^{\frac{i}{\hbar}Ht}A(0)e^{-\frac{i}{\hbar}Ht} \qquad ? $$

L.M.: I added $i/\hbar$ in front of $[H,A]$.

Answer

We have to consider the operator, that doesn't explicitly depends on time.

$\frac{\partial A}{\partial t} = 0$

Let's apply commutator formula recursively:

$\frac{d^2 A}{d t^2} = \left(\frac{i}{\hbar}\right)^2[H,[H,A]]$

$\frac{d^3 A}{d t^3} = \left(\frac{i}{\hbar}\right)^3[H,[H,[H,A]]]$

e.t.c.

Then we combine those derivatives in a series for $A(t)$

$A(t)=A(0)+\frac{d A}{d t}t+\frac{1}{2!}\frac{d^2 A}{d t^2}t^2+\frac{1}{3!}\frac{d^3 A}{d t^3}t^3+...$

$A(t)=A(0)+\frac{i}{\hbar}[H,A]t+\frac{1}{2!}\left(\frac{i}{\hbar}\right)^2[H,[H,A]]t^2+\frac{1}{3!}\left(\frac{i}{\hbar}\right)^3[H,[H,[H,A]]]t^3+...$

And then you use this formula to arrive at the result:

$e^XYe^{-X} = Y+\frac{1}{1!}[X,Y]+\frac{1}{2!}[X,[X,Y]]+\frac{1}{3!}[X,[X,[X,Y]]]+...$

determinism - Are living organisms deterministic?

Are living organisms physically deterministic at any given time? Since it's all physics and chemistry, it leads me to believe they are.

Answer

In the sense that most people mean, I'd say no because quantum mechanics isn't deterministic (at least the way most people think of "determinism").

---

Ultimately, to be more specific, I think this depends on your interpretation of quantum mechanics, and how reductionist you are.

The best you can do with reconciling quantum mechanics with determinism is taking a particular (admittedly relatively popular) interpretation of quantum mechanics (the many-worlds interpretation) which is "deterministic" but not in the sense you were probably thinking of, or in a sense which is very useful.

Reductionism, which is generally the working mode of thinking for scientists, implies that if quantum mechanics & physics is deterministic, then chemistry and biology are too because they are built only out of pieces of nature that are fully described by quantum mechanics and nothing else. If you want to be controversial, you can suppose that there exists "strongly emergent" phenomena which fundamentally can't be traced to quantum mechanics & physics (e.g. a soul which is not described by physics). This generally isn't very popular in experimental sciences because 1. it's not very useful and 2. these ideas generally don't have a great track record.

More on the quantum mechanics bit:
Quantum mechanics gives describes a system of particles and their evolution in time in terms of probabilities, not "deterministic" events, as classical mechanics would. And there's good reason (see Bell's inequalities and Bell tests) to believe that this probabilistic description is as fundamental as you can get for these things. The purest description of the universe is in terms of probabilities.

Probabilities seem manifestly non-deterministic. However, if you subscribe to the many-worlds theory then this is still in a sense deterministic, as quantum mechanics (or a similar theory) would correctly (and deterministically) describe the evolution of the state of the universe. Locally (i.e. in our part of the state which can't really observe other parts of the state) things look probabilistic, but if you could somehow observe the state of the whole universe, you'd deterministically predict how the whole thing evolves using quantum mechanics. If you don't know what I mean by this, I'd start by looking up the concepts of superposition and entanglement in quantum mechanics. As of yet I don't think there is anything testable about the differences between this and other interpretations of quantum mechanics (e.g. Copenhagen), so many people don't like reading into these things too much.

general relativity - A dictionary of string - standard physics correspondences

Motivated by the (for me very useful) remark ''Standard model generations in string theory are the Euler number of the Calabi Yau, and it is actually reasonably doable to get 4,6,8, or 3 generations'' in https://physics.stackexchange.com/a/22759/7924 , I'd like to ask:

Is there a source giving this sort of dictionary how to relate as much as possible from general relativity and the standard model to string theory, without being bogged down by formalism, speculation, or diluted explanations for laymen?

Or, if there is no such dictionary, I'd like to get as answers contributions to such a concise dictionary.

[Edit April 30, 2012:] I found some more pieces of the wanted dictionary in http://www.physicsforums.com/showthread.php?p=3263502#post3263502 : ''branes and extra dimensions have proved to be implicit in standard quantum field theory, where they emerge from the existence of a continuous degeneracy of ground states. That multidimensional moduli space of ground states is where the extra dimensions come from, in this case! Branes are domain walls separating regions in different ground states, strings are lines of flux connecting these domain walls. Furthermore, in gauge theories with a small number of colors, it looks like the extra dimensions will be a noncommutative geometry, it's only in the "large N" limit of many colors that you get ordinary space.''

electric circuits - Why doesn't current pass through a resistance if there is another path without resistance?

Why doesn't current pass through a resistance if there is another path without resistance? How does it know there is resistance on that path ?

material science - Why can't a piece of paper (of non-zero thickness) be folded more than $N$ times?

Updated:

In order to fold anything in half, it must be $\pi$ times longer than its thickness, and that depending on how something is folded, the amount its length decreases with each fold differs.
– Britney Gallivan, the person who determined that the maximum number of times a paper or other finite thickness materials can be folded = 12.

Mathematics of paper folding explains the mathematical aspect of this.

I would like to know the physical explanation of this. Why is it not possible to fold a paper more than $N$ (=12) times?

Answer

I remember that the question in your title was busted in Mythbusters episode 72. A simple google search also gives many other examples.

As for single- vs alternate-direction folding, I'm guessing that the latter would allow for more folds. It is the thickness vs length along a fold that basically tells you if a fold is possible, since there is always going to be a curvature to the fold. Alternate-direction folding uses both flat directions of the paper, so you run out of length slightly slower. This would be a small effect since you have the linear decrease in length vs the exponential increase in thickness.

---

Thanks to gerry for the key word (given in a comment above). I can now make my above guess more concrete. The limit on the number of folds (for a given length) does follow from the necessary curvature on the fold. The type of image you see for this makes it clear what's going on

For a piece of paper with thickness $t$, the length $L$ needed to make $n$ folds is (OEIS) $$ L/t = \frac{\pi}6 (2^n+4)(2^n-1) \,.$$ This formula was originally derived by (the then Junior high school student) Britney Gallivan in 2001. I find it amazing that it was not known before that time... (and full credit to Britney). For alternate folding of a square piece of paper, the corresponding formula is $$ L/t = \pi 2^{3(n-1)/2} \,.$$

Both formulae give $L=t\,\pi$ as the minimum length required for a single fold. This is because, assuming the paper does not stretch and the inside of the fold is perfectly flat, a single fold uses up the length of a semicircle with outside diameter equal to the thickness of the paper. So if $L < t\,\pi$ then you don't have enough paper to go around the fold.

Let's ignore a lot of the subtleties of the linear folding problem and say that each time you fold the paper you halve its length and double its thickness: $ L_i = \tfrac12 L_{i-1} = 2^{-i}L_0 $ and $ t_i = 2 t_{i-1} = 2^{i} t_0 $, where $L=L_0$ and $t=t_0$ are the original length and thickness respectively. On the final fold (to make it n folds) you need $L_{n-1} \leq \pi t_{n-1}$ which implies $L \leq \frac14\pi\,2^{2n} t$. Qualitatively this reproduce the linear folding result given above. The difference comes from the fact you lose slightly over half of the length on each fold.

These formulae can be inverted and plotted to give the logarithmic graphs

where $L$ is measured in units of $t$. The linear folding is shown in red and the alternate direction folding is given in blue. The boxed area is shown in the inset graphic and details the point where alternate folding permanently gains an extra fold over linear folding.

You can see that there exist certain length ranges where you get more folds with alternate than linear folding. After $L/t = 64\pi \approx 201$ you always get one or more extra folds with alternate compared to linear. You can find similar numbers for two or more extra folds, etc...

---

Looking back on this answer, I really think that I should ditch my theorist tendencies and put some approximate numbers in here. Let's assume that the 8 alternating fold limit for a "normal" piece of paper is correct. Normal office paper is approximately 0.1mm thick. This means that a normal piece of paper must be $$ L \approx \pi\,(0.1\text{mm}) 2^{3\times 7/2} \approx 0.3 \times 2^{10.5}\,\text{mm} \approx .3 \times 1000 \, \text{mm} = 300 \text{mm} \,. $$ Luckily this matches the normal size of office paper, e.g. A4 is 210mm * 297mm.

The last range where you get the same number of folds for linear and alternate folding is $L/t \in (50\pi,64\pi) \approx (157,201)$, where both methods yield 4 folds. For a square piece of paper 0.1mm thick, this corresponds to 15cm and 20cm squares respectively. With less giving only three folds for linear and more giving five folds for alternating. Some simple experiments show that this is approximately correct.

quantum mechanics - Can a particle tunnel from inside a black hole?

Event horizon isn’t special from GTR standpoint, and at least in AdS/CFT correspondence gravity can be “removed” from consideration entirely. So can a particle whose wave function is completely inside a black hole at some point in time “smear” to the outside and end up being observed there just by tunneling? I’m not a physicist, so feel free to make any corrections to the question.

electromagnetism - Zero Ohmic resistance in superconductors is a little bit too enthusiastic?

According to this article:

Superconductors contain tiny tornadoes of supercurrent, called vortex filaments, that create resistance when they move.

Does this mean that our description of zero Ohmic resistance in superconductors is a little bit too enthusiastic?

Answer

I think that since superconductors were originally discovered because they exhibited electrical resistances indistinguishable from zero that zero electrical resistivity may have originally been the defining characteristic of a superconductor. But as more was learned about superconductors and the superconducting state it was realized that superconductors can exist in a mixed or vortex state consisting of normal-state vortices in a superconducting medium, and such a system can exhibit energy dissipation due to the movement of the vortices which results in an electrical resistance that is not strictly zero. So, yes, I guess you could say that the use of the term "superconductor" when these materials were first discovered could have been a bit "too enthusiastic" since it suggested that absolute zero resistance was an essential, defining characteristic of superconductors when it's not.

Not an expert in superconductivity and maybe someone else will chime in, but from my perspective as an experimentalist I would say that an operational definition of superconductivity is a very low electrical resistance combined with the Meissner Effect (i.e., magnetic flux exclusion).

gravity - General Relativity 2-Body Closed Form

Is there a closed form solution in general relativity to the 2-body orbit problem?

Answer

There is no general solution for the two body problem in general relativity.

But!

There are a few solutions for specific two body problems. These include the Curzon-Chazy metric (Two particles on a cylindrically symmetric axis)

$ds^2 = e^{-2\psi} dt^2 - e^{2(\psi - \gamma)} (d\rho^2 + dz^2) - e^{2\psi} \rho^2 d\phi^2$

and the Israel-Khan metric ("two black holes held in equilibrium by a strut"). Also of interest and related to the Israel-Khan metric :

"In a 1922 paper, Rudolf Bach and Hermann Weyl [3] discussed the superposition of two exterior Schwarzschild solutions in Weyl coordinates as a characteristic example for an equilibrium configuration consisting of two “sphere-like” bodies at rest. Bach noted that this static solution develops a singularity on the portion of the symmetry axis between the two bodies, which violates the elementary flatness on this interval. "

The Gott spacetime is constituted of two cosmic strings also, if that helps.

From "Exact Solutions of the Einstein Field Equations", by the way :

"In Einstein's theory, a two-body system in static equilibrium is impossible without such singularities - a very satisfactory feature of this non-linear theory."

Edit : I know those are all static two body solutions, and not orbit ones, but here lies the problem : orbit solutions are horrible. With a static 2 body soluion, you get to keep rotational symmetry and time symmetry. Once you go full orbit, you will basically lose all symmetries, and you will also get gravitational waves. That is when things become extremely non-linear, and hence hard to solve.

general relativity - Why is spacetime curved by mass but not charge?

It is written everywhere that gravity is curvature of spacetime caused by the mass of the objects or something to the same effect. This raises a question with me: why isn't spacetime curved due to other forces or aspects of bodies?

Why isn't it that there are curvatures related to the charge of a body or the spin of particles or any other characteristics?

Answer

Charge does curve spacetime. The metric for a charged black hole is different to an uncharged black hole. Charged (non-spinning) black holes are described by the Reissner–Nordström metric. This has some fascinating features, including acting as a portal to other universes, though sadly these are unlikely to be physically relevant. There is some discussion of this in the answers to the question Do objects have energy because of their charge?, though it isn't a duplicate. Anything that appears in the stress-energy tensor will curve spacetime.

Spin also has an effect, though I have to confess I'm out of my comfort zone here. To take spin into account we have to extend GR to Einstein-Cartan theory. However on the large scale the net spin is effectively zero, and we wouldn't expect spin to have any significant effect until we get down to quantum length scales.

electromagnetism - Quantitative relation between two charges moving in parallel in two frames of reference

The relevant question is here. The accepted answer may have explained my question in a descriptive manner. However, I want to see how things are related quantitatively.

Imagine we have two charges $q$ moving parallel to each other. The distance between them is $d$.

• 
  

  In the frame where the charges are stationary. We have: $$m_0 a_0=\frac{q^2}{4\pi\epsilon_0d}$$

  
  


• 
  

  In the laboratory frame, the charges also experience a force caused by the magnetic field which is generated by the other charge: $$B=\mu_0\frac{qv}{2\pi d}$$ The total force is: $$F=\frac{q^2}{4\pi\epsilon_0d}-\mu_0\frac{q^2v^2}{2\pi d}=\frac{m_0}{\sqrt{1-v^2/c^2}}a$$

  
  

There is also the relation of $a_0$ and $a$ that relate these two equations of motion. However, it seems I cannot get the right result.

Any help in figuring out how to relate these two situations would be appreciated.

Answer

Actually it's not that difficult (but a neat problem), there's only one crucial step in the development that I will show you, but let's start from a bit earlier.

Let's first write out the two forces on interest here (in terms of magnitudes, as we know already they're on parallel trajectories):

The coulomb repulsion between the charges: $$F_C = \frac{q^2}{4\pi \epsilon_0 r^2} $$ Now since we have moving charges, (hence a current), each charge will experience a Lorentz force in the magnetic field induced by the other charge, using Biot-Savart's law, we have for the magnetic field $B$ : $$B = \frac{\mu_0 I dL}{4\pi r^2}$$ Which should be interpreted as (by analogy to a electrical circuit) "the magnetic field felt at a distance $r$, induced by a wire length $dL$ carrying current $I$".

Note $\mu_0$ is the permeability of vacuum, $\epsilon_0$ is the corresponding permittivity.

The Lorentz force: $$F_L = qBv = \frac{\mu_0 q^2 v^2}{4\pi r^2}$$ where the $I dL$ term in B field is replaced using: $I=q/dt$ and $dL/dt = v$, which gives finally $$I dL=qv.$$

The two forces together (with $F_C$ the repulsive force here): $$\sum F =\frac{q^2}{4\pi \epsilon_0 r^2} -\frac{\mu_0 q^2 v^2}{4\pi r^2}$$ It is clear that the resulting effect of the two forces depends on the strength of the $B$ field which in turn then depends on the velocity of the two charges, if they are slow, the repulsion dominates. So in order to be able to compare their relative strengths, we need a slight rearrangement for the $\sum F$:

$$F_{tot} =\frac{q^2}{4\pi \epsilon_0 r^2} \left(1-v^2 \epsilon_0 \mu_0\right) $$ where $$v^2 \epsilon_0 \mu_0=\frac{F_L}{F_C}$$ But we know that $\epsilon_0 \mu_0 = c^{-2}$ the inverse squared of speed of light in vacuum, substituted in the total force expression: $$F_{tot}= F_C \left(1-\frac{v^2}{c^2}\right) = \frac{F_C}{\gamma^2} $$ Where $\gamma$ is the relativistic factor $(1-v^2/c^2)^{-1/2}$ or often called Lorentz factor, which relates the measurements performed in different inertial frames moving at $v$ with respect to one another.

Last note: if the charges are not in vacuum, the relative permittivity $\epsilon$ and permeability $\mu$ are included in the expressions.

In quantum field theory, how can Compton scattering change the frequency of light?

Classically, when light scatters off matter, the frequency of the light must stay the same. This follows directly from a continuity argument: if you put in $f$ field oscillations per second, you'd better get $f$ oscillations per second out, because you can just follow each peak through. However, we observe a frequency shift in Compton scattering. In the 1920's, this result was paradoxical, and was considered to have no classical explanation.

In quantum mechanics, the frequency shift is explained by treating light as a particle, the photon. However, in quantum field theory, which also produces the correct result for Compton scattering, light is again treated as a field!

• Why does the continuity argument described above for classical fields fail for quantum fields?


• In quantum field theory, Compton scattering is tree-level, and tree-level behavior is equivalent to classical field theory. Therefore, there should be a classical explanation for Compton scattering, i.e. Compton scattering is not a quantum effect. Is this true, and has this been demonstrated?

Note: I am not asking for a quantum mechanical explanation of the Compton effect. I've already seen this plenty of times. My question is how to reconcile the argument that Compton has no classical explanation (in the first paragraph) with my heuristic argument that Compton does have a classical explanation (the last bullet point).

Answer

The "continuity argument" fails for quantum fields because quantum fields are operator-valued distribution that do not take definite "values".

Compton scattering has a classical equivalent, but not in the way you are thinking. In the non-relativistic regime, we get back Thompson scattering where the frequency of the electromagnetic wave doesn't change. The classical picture is an electromagnetic wave scattering off a point particle.

The relativistic Compton scattering from QFT corresponds to a classical high intensity regime, where the electric field of the infalling wave is strong enough to accelerate the electron such that it essentially Doppler shifts the outgoing, scattered wave, cf. "Limits on the Applicability of Classical Electromagnetic Fields as Inferred from the Radiation Reaction" by K. T. McDonald. This effect is variously known as "radiation-pressure recoil", "radiation reaction", "radiation damping" and other names. Classically, this Compton scattering-like effect vanishes when going to low intensities.

The more general observation to make here is that the "classical limit" of a quantum field theory may not be the classical theory we naively expect. Indeed, the presence of the fermionic electron field in the QFT alone, which is absent from classical electrodynamics, should show that the heuristic "$\hbar\to 0$ argument" that shows that generically tree-level computations for an action $S[\phi_1,\dots, \phi_n]$ correspond to classical field theory computations for the same action does not directly imply that tree-level QED computations correspond to CED computation.

quantum mechanics - Is temperature discrete

Because an object's temperature is inversely proportional to the wavelength of blackbody radiation which it emits, physicists have theorized the existence of Planck temperature at around $1.4×10^{32}$ K.

Does this imply that temperature must be discrete? Meaning, the temperature could never be $1.0×10^{32}$ K, because that would imply a wavelength which is a non-integral multiple of the Planck length.

If it does imply this, does that mean that a system, if approaching the Planck temperature, and given more energy, would suddenly and radically change states from $0.7×10^{32}$ K to $1.4×10^{32}$ K? (perhaps much like a Bose Einstein condensate)

Pardon me if this question is bizarre, as I am not a physicist by profession (except maybe of the armchair variety).

Answer

Temperature cannot be discrete because it is an intensive variable:

An intensive property is a bulk property, meaning that it is a physical property of a system that does not depend on the system size or the amount of material in the system. Examples of intensive properties are the temperature, refractive index, density....

In the simple formulation of the kinetic theory of gases,

$$T=\frac{m\overline{v^2}}{3 k_B},$$

where the temperature is directly connected with the mean of the square of the velocity of the molecules, and there is no discreteness in the formulation.

The Planck temperature is an estimate maximum attainable temperature with the physics we know, to be attained by matter given the constants we know, and has nothing to do with discreteness as normal temperatures are around $300\text{K}$.

complex numbers - Method of pole shifting (feyman's trick) in Scattering theory vs contour deformation trick

I am studying Scattering theory but I am stuck at this point on evaluating this integral

$G(R)={1\over {4\pi^2 i R }}{\int_0^{\infty} } {q\over{k^2-q^2}}\Biggr(e^{iqR}-e^{-iqR} \Biggl)dq$

Where $ R=|r-r'|$

This integral can be rewritten as

$G(R)={1\over {4{\pi}^2 i R }}{\int_{-\infty}^{\infty} } {q\over{k^2}-{q^2}}{e^{iqR}}dq$

Zettili did this integral by the method of contour integration in his book of 'Quantum Mechanics'.He uses residue theorems and arrived at these results.

$G_+(R)={ -1e^{ikR}\over {4 \pi R}}$ and $G_-(R)={ -1e^{-ikR}\over {4 \pi R}}$

I don't get how he arrived at this result. The test book doesn't provide any detailed explanations about this. But I know to evaluate this integral by pole shifting.

My question is how to evaluate this integral buy just deform the contour in complex plane instead of shifting the poles?

newtonian mechanics - Why are Saturn's rings so thin?

Take a look at this picture (from APOD https://apod.nasa.gov/apod/ap110308.html):

I presume that rocks within rings smash each other. Below the picture there is a note which says that Saturn's rings are about 1 km thick.

Is it an explained phenomenon?

Answer

There seems to be a known explanation. I quote from Composition, Structure, Dynamics, and Evolution of Saturn’s Rings, Larry W. Esposito (Annu. Rev. Earth Planet. Sci. 2010.38:383-410):

[The] rapid collision rate explains why each ring is a nearly flat disk. Starting with a set of particle orbits on eccentric and mutually inclined orbits (e.g., the fragments of a small, shattered moon), collisions between particles dissipate energy but also must conserve the overall angular momentum of the ensemble. Thus, the relative velocity is damped out, and the disk flattens after only a few collisions to a set of nearly coplanar, circular orbits.

I think the key is that particles in a thick ring would not move in parallel planes but would have slanted trajectories, colliding all the time and losing their energy very fast.

condensed matter - Is time reversal symmetry broken in (conventional) superconductors?

How can one see it from BCS wavefunction and BCS Hamiltonian? i.e.

$$H_{BCS}=\sum_{k\sigma}\epsilon_k c_{k\sigma}^\dagger c_{k\sigma}-\Delta^*\sum_k c_{k\uparrow}^\dagger c_{-k\downarrow}^\dagger+h.c.$$

and:

$$\Psi_{BCS}=\Pi_k(u_k+v_k c_{k\uparrow}^\dagger c_{-k\downarrow}^\dagger)|0\rangle$$

If it has this symmetry, what significance does it has?

Answer

The Hamiltonian is time-reversal invariant: $c_{k\uparrow}\rightarrow c_{-k,\downarrow}, c_{k\downarrow}\rightarrow -c_{-k,\uparrow}$. You can check that explicitly. The ground state is also invariant, because Cooper pairs are all spin singlet.

One of the significant implications of time-reversal symmetry for s-wave superconductors is the Anderson's theorem: the pairing (e.g. the critical temperature) is not affected by time-reversal-invariant impurities (i.e. non-magnetic), as long as the impurities are not strong enough to cause localization.

We create systems with different values of Energy?

I understand that Energy is a conserved quantity in a system. A number, that's always the same to the system. However, don't we determine such a number? I mean, we can create systems(Or study them usually) and see the amount of Energy tit can have. Correct? Because to me, Energy is a confusing concept... Why was it even created? Work is a good explanation, but what's the use of Energy? It's not something physical, isn't it our manifestation? We created that concept, to explain work? or what?

What's is the importance of Energy, why do we need such a concept? It's not something real, it's just a mathematical expression taken to extremely in Physics. It's such a confusing subject... And everyone in Physics, act's as if Energy is something "given". Why? It's a number! A measurement of work!

And what about E=mc^2?! How do we define Energy here? A number as well?

earth - Is effective gravity constant over the surface of the ocean?

According to Wikipedia:

The Earth is not spherically symmetric, but is slightly flatter at the poles while bulging at the Equator: an oblate spheroid. There are consequently slight deviations in both the magnitude and direction of gravity across its surface. The net force (or corresponding net acceleration) as measured by a scale and plumb bob is called "effective gravity" or "apparent gravity".

and

In combination, the equatorial bulge and the effects of the surface centrifugal force due to rotation mean that sea-level effective gravity increases from about $9.780\,{m/s^2}$ at the Equator to about $9.832\,{m/s^2}$ at the poles, so an object will weigh about $0.5\%$ more at the poles than at the Equator.

But I don't see how this is physically possible.

If the force on a unit of water at the poles was greater than the force at the equator, shouldn't the ocean fall at the poles and rise at the equator? For the same reason that when I step into the bath the water falls where my foot is pressing down on it and rises elsewhere. Surely the force at the surface of a body of water in equilibrium has to be constant.

So is my physical argument wrong, or is the net force on a body actually the same at the poles and equator?

Answer

You suggest that for a body of water to be in stable equilibrium, i.e. to have minimum energy, it should have the same gravitational field everywhere at its surface, as otherwise we would be able to move the water around to lower the energy. This sounds intuitive, but it's just not true.

As an extreme example, consider a puddle of water on Pluto. The gravitational field is weaker there, so by your argument we should be able to harvest energy to sending water to Pluto. But that's completely wrong: it would actually cost an enormous amount of energy to do this.

The right statement is that the gravitational potential is the same everywhere on the surface of water. That has nothing to do with whether the field is the same.

group theory - How does non-Abelian gauge symmetry imply the quantization of the corresponding charges?

I read an unjustified treatment in a book, saying that in QED charge an not quantized by the gauge symmetry principle (which totally clear for me: Q the generator of $U(1)$ can be anything in $\mathbb{R}$) but for non-Abelian gauge symmetries the "charge" are quantized by virtue of this principle. Could someone give a hint (or reference) of the calculation showing that.

Answer

First of all, note that the real Abelian Lie group $U(1)$ comes in two (multiplicatively written) versions:

1. 
   

   Compact $U(1)~\cong~e^{i\mathbb{R}}~\cong~S^1$, and

   
   


2. 
   
   

   Non-compact $U(1)$ $~\cong~e^{\mathbb{R}}\cong~\mathbb{R}_+\backslash\{0\}$.

   
   

Also note that in the physics literature, we often identifies charge operators with Lie algebra generators for a Cartan subalgebra (CSA) of the gauge Lie algebra.

Moreover, note that the choice of CSA generators is not unique, see also this answer. The ambiguity in the convention choice of charge operators is similar to the ambiguity in the convention choice of spin operators, see also this question. We shall from now on assume that we consistently stick to only one such possible convention.

Given a Lie algebra representation, the eigenvalues of the charge operator are called charges.

Now let us briefly sketch some lore and facts related to OP's question (v2).

1. 
   

   We observe in Nature that Abelian and non-Abelian charges are quantized, as accurately described by electric charge, electroweak hypercharge, electroweak isospin and color charges in the $U(1)\times SU(2)\times SU(3)~$ standard model.

   
   
   


2. 
   

   If there exist dual magnetic monopoles, then quantum theory provides a natural explanation for charge quantization. Namely, by playing with Wilson lines, the singlevaluedness of the wavefunction requires that charges are quantized (i.e., to take only discrete values), and that the gauge group is compact, as first explained by Dirac.

   
   


3. 
   

   It is a standard result in representation theory, that for a finite-dimensional representation of a compact Lie group, that the charges (i.e., the eigenvalues of the CSA generators) take values in a discrete weight lattice.

   
   


4. 
   

   If a gauge group contains both a compact and a non-compact direction, i.e. if its bilinear form$^1$ has indefinite signature, it is impossible to define a non-trivial positive-norm Hilbert subspace of physical, propagating, $A^a_{\mu}$ gauge field states.

   
   

--

$^1$ By a bilinear form is here meant a non-degenerate invariant/associative bilinear form on the Lie algebra. For a semisimple Lie algebra, we can use the Killing form.

gravity - What's the difference between proper acceleration and coordinate acceleration?

I'm not a physics student, but I need to understand these concepts in order to explain what exactly an accelerometer measures and not measures, and why. Please, do not explain to me what an accelerometer does: that isn't the direct purpose of this question. I'm not looking for all mathematical equations, but more for intuitions, but at the same time I'm not looking for oversimplifications.

---

If I understood correctly, proper acceleration is measured in units which are multiples of $g=9.81$. It's not the same thing as coordinate acceleration, which is dependent on the choice of coordinate systems (according to this Wikipedia article).

1. 
   
   

   Why does proper acceleration does not depend on a coordinate system, and why coordinate acceleration does?

   
   

   I understood that proper acceleration is measured from an inertial frame of reference, i.e. a frame of reference which is not accelerating. Is this the reason why proper acceleration does not depend on a coordinate system, where coordinate system here actually means a frame of reference?

   
   

   If this is the case, then I suppose coordinate acceleration is the acceleration measured from any random coordinate system (frame of reference).

   
   
   
   
   1. Would coordinate acceleration and proper acceleration be the same if coordinate acceleration was measured in an inertial frame of reference?
   
   
   
   

Apparently, gravity does not cause proper acceleration since an accelerometer would measure $0g$ if in free-fall, i.e. where the only force acting upon the accelerometer would be gravity, in case we consider gravity a force.

I think there's a misunderstanding here between what an accelerometer does and what actually proper acceleration is. I believe that the accelerometer would detect $0g$ because it subtracts the acceleration due to gravity from its calculations...

Furthermore, an accelerometer would detect $1g$ (upwards) on the surface of the Earth, apparently, because there's the surface of the Earth pushing the accelerometer upwards.

2. 
   

   Why exactly does an accelerometer measure $1g$ on the surface of the Earth?

   
   


3. 
   

   Coordinate acceleration, I've seen, seems also to be defined as simply the rate of change of velocity. Why isn't proper acceleration also defined as a rate of change of velocity?

   
   


4. 
   

   What's the fundamental difference between coordinate acceleration and proper acceleration (maybe without referring to an accelerometer to avoid a circular definition, which would only cause confusion again)?

   
   
   

quantum mechanics - Some applications of the Einstein-Podolsky-Rosen (EPR) paradox?

If you were to explain the Einstein-Podolsky-Rosen paradox to high school students (age 16, with no particular strength in math), what kind of intuitive example would you provide to make things understandable?

quantum mechanics - Direct Sum of Hilbert spaces

I am a physicist who is not that well-versed in mathematical rigour (a shame, I know! But I'm working on it.) In Wald's book on QFT in Curved spacetimes, I found the following definitions of the direct sum of Hilbert spaces. He says -

Next, we define the direct sum of Hilbert spaces. Let $\{ {\cal H}_\alpha \}$ be an arbitrary collection of Hilbert spaces, indexed by $\alpha$ (We will be interested only with the case where there are at most a countable number of Hilbert spaces, but no such restriction need be made for this construction). The elements of the Cartesian product $\times_\alpha {\cal H}_\alpha$ consist of the collection of vectors $\{\Psi_\alpha\}$ for each $\Psi_\alpha \in {\cal H}_\alpha$. Consider, now, the subset, $V \subset \times_\alpha {\cal H}_\alpha$, composed of elements for which all but finitely many of the $\Psi_\alpha$ vanish. Then $V$ has the natural structure of an inner product space. We define the direct sum Hilbert space $\bigoplus\limits_\alpha {\cal H}_\alpha$ to be the Hilbert space completion of $V$. It follows that in the case of a countable infinite collection of Hilbert spaces $\{{\cal H}_i\}$ each $\Psi \in \bigoplus\limits_i {\cal H}_i$ consists of arbitrary sequences $\{\Psi_i\}$ such that each $\Psi_i \in {\cal H}_i$ and $\sum\limits_i \left\| \Psi_i \right\|_i^2 < \infty$.

Here $\left\| ~~\right\|_i$ is the norm defined in ${\cal H}_i$. Also, Hilbert space completion of an inner product vector space $V$ is a space ${\cal H}$ such that $V \subset {\cal H}$ and ${\cal H}$ is complete in the associated norm. It is constructed from $V$ by taking equivalence classes of Cauchy sequences in $V$.

Now the questions -

1. Why does $V$ have the structure of an inner product space?

2. How does he conclude that $\sum\limits_i \left\| \Psi_i \right\|_i^2 < \infty$?

3. How does this definition of the direct sum match up with the usual things we see when looking at tensors in general relativity or in representations of Lie algebras, etc.?

PS - I also have a similar problem with Wald's definition of a Tensor Product of Hilbert spaces. I have decided to put that into a separate question. If you could answer this one, please consider checking out that one too. It can be found here. Thanks!

Answer

An element of the direct sum $H_1\oplus H_2\oplus ...$ is a sequence $$\Psi = \{\Psi_1, \Psi_2,...\}$$ consisting of an element from $H_1$, and element from $H_2$...etc (countable). This has to have the special property that the sum $$\left\| \Psi_1\right\|^2+\left\| \Psi_2\right\|^2+... $$converges. This convergence is part of the definition of direct sum.

To show that the direct sum is a Hilbert space, I need well defined addition operation. If I have a pair of such things: $$ \Psi = \{\Psi_1, \Psi_2,...\}$$ $$ \Phi = \{\Phi_1, \Phi_2,...\}$$ I can define their sum to be the sequence $$ \Phi+\Psi = \{\Phi_1+\Psi_1, \Phi_2+\Psi_2,...\}$$We need to check that $$\left\| \Psi_1+\Phi_1 \right\|^2+\left\| \Psi_2+\Phi_2 \right\|^2+...$$converges. To do this, we use the fact that the individual terms satisfy $$\left\| \Psi_i+\Phi_i \right\|^2 $$ $$=\left\| \Psi_i\right\|^2+\left\| \Phi_i\right\|^2+(\Psi_i,\Phi_i)+(\Phi_i,\Psi_i) $$ $$ \leq \left\| \Psi_i\right\|^2+\left\| \Phi_i\right\|^2+2\left\| \Psi_i\right\|\left\| \Phi_i\right\| $$ $$ \leq 2\left\| \Psi_i\right\|^2+2\left\| \Phi_i\right\|^2$$So convergence follows from the convergence properties of the individual spaces being summed. So this shows how we can add elements of the direct sum. Scalar products follow in the same way.

To define an inner product, we just add the inner products component wise, i.e $$(\Psi, \Phi) = (\Psi_1,\Phi_1)+(\Psi_2,\Phi_2)+... \ \ \ (1)$$To show the RHS converges, note $$|(\Psi_i, \Phi_i)| \leq \left\| \Psi_i \right\| \left\| \Phi_i \right\| $$But $$2\left\| \Psi_i \right\| \left\| \Phi_i \right\| \leq \left\| \Psi_i \right\|^2+ \left\| \Phi_i \right\|^2$$so the RHS of (1) converges absolutely and we are done - we have a well defined inner product space.

electromagnetism - Isn't the inductor equation negative?

The inductor "resists" change in current. So say you measure the voltage across the inductor from point A to point B - the current is flowing in from A towards B. Now say the current is increasing. The inductor will try to oppose the change by creating a current the opposite direction - from point B to A. To do this it will create a voltage, where point A has a lower voltage than point B in order to "encourage" electrons to flow the opposite way. If this is true though, the voltage measured from point A to B will be negative, so shouldn't the equation be:

$$V_{a \to b} = -L\frac{di}{dt}$$

free body diagram - Does a non-buoyant (denser than water) object (such as a lead diving weight) weigh any less when submerged in water?

I've just been discussing this question with my eldest son, who is an extremely intelligent man, as well as being an engineer, a sailor and a scuba-diver, and he believes that an object heavier than water, such as his anchor, or his diving weights actually becomes less heavy when submerged in water. I described a scenario, whereby a solid steel, or lead weight was suspended from a spring balance and slowly lowered into a body of water, the weight reading on the spring balance would not change. He insisted that it would... I think it would not, because an object that is denser than water would not posess any buoyancy once submerged, therefore the density and weight would remain unchanged. Any comments anyone??? There's a £5 bet riding on this :)

quantum mechanics - Matrix representation of a triplet state

The $SU(2)$ triplet state is typically given in the fundamental representation as a column vector, e.g. \begin{equation} \vec{\Delta} = \left( \begin{array}{c} \delta^{++} \\ \delta^+ \\ \delta^0 \end{array} \right) \end{equation} where I use this notation to be consistent with the references below.

However, I'm reading through papers (in the context of the type II see-saw mechanism, but I don't think it has much relevance) and most references are using what they call the matrix representation of the $SU(2)$ triplet. My instinct was to assume that they were projecting the fundamental representation onto the Pauli matrices:

\begin{equation} \Delta \rightarrow \vec{\Delta } \cdot \vec{ \sigma } = \delta ^{++} \sigma _1 + \delta ^+ \sigma _2 + \delta ^0 \sigma _3 \end{equation}

However, this doesn't quite give me the correct result, which is given by, \begin{equation} \Delta = \left( \begin{array}{cc} \delta ^+ / \sqrt{2} & \delta ^{ + + } \\ \delta ^0 & - \delta ^+ / \sqrt{2} \end{array} \right) \end{equation}

Is there some other way to get this "matrix representation"?

Two sample references where they say they are using the matrix representation can be found here (pg. 124, equation 68) and here (pg. 4 equation 1).

Answer

Typically, the triplet refers to the spin one representation of $SU(2)_L$ of the standard model. Let $I_a$ refer to the generators of $SU(2)_L$. Then the triplet can be organized into a column vector with $I_3$ being diagonal and equal to Diag$(1,0,-1)$. Suppose the triplet has hypercharge $Y$, i.e., the charge associated with $U(1)_Y$. Electroweak symmetry breaking has $SU(2)_L\times U(1)_Y$ broken to $U(1)_{EM}$. The electric charge is given by the formula $Q=I_3+\tfrac12 Y$. If $Q=2$ for the triplet, then we see that $Q=(2,1,0)$ which is written as $(++,+,0)$ and added as a superscript to indicate the charge of a particle in such a triplet. This should explain the notation for $\vec{\Delta}$ that you have given.

Now to go to the matrix representation, one uses the fact that the spin-one representation of $SU(2)$ is also its adjoint representation. Since the three entries in $\vec{\Delta}$ are all $I_3$ eigenstates, in order to write the matrix representation using Pauli sigma matrices, you need to write the Pauli matrices in a basis where they are the analog of eigenstates of $I_3$. Such a basis is given by $\tfrac12\sigma_\pm =\tfrac12( \sigma_1\pm i\sigma_2)$ and $\tfrac12\sigma_3$. So the natural guess would be something like $$ \Delta \rightarrow \frac12 (\delta^{++}\ \sigma_+ +\delta^+\ \sigma_3 + \delta^0\ \sigma_-)\ . $$ There will be variations in convention in that $\delta^{++}$ will be paired up with $\sigma_-$ instead of $\sigma_+$ as I have done. I might also be missing some factors of $\sqrt{2}$ but I am sure you get the general idea.

homework and exercises - Tensor of inertia of a hollow cube

I have found the tensor of inertia of a rectangle of sides $a$ and $b$ and mass $m$, around its center, to be $$I_{11}=ma^2/12,$$ $$I_{22}=mb^2/12,$$ $$I_{33}=(ma^2 + mb^2)/12.$$ All other elements of that tensor are equal to zero. I would now like to use this result to determine the tensor of inertia of a hollow cube of side a around its center of mass.

I realise I have to use the parallel axis theorem. I also knoww that the correct equation is I$_{11}$=I$_{22}$=I$_{33}$=ma$^2$/12+ma$^2$/12+ma$^2$/6+4(ma$^2$/12 + m(a/2)$^2$)=5/3*ma$^2$ I simply do not understand why this is correct. Could anyone please explain why this is the correct way to calculate the desired tensor of inertia? Also, why would I be summing all the diagonal elements in my tensor for the rectangle?

Answer

We will work in units where the mass of each face is $1$ and where the length of the side of the cube is $1$.

The contribution to the moment of inertia of each of the top and bottom faces is, using your result for the moment of inertia of a rectangle, $\frac{1}{6}$.

By symmetry, each of the four other faces has the same contribution to the moment of inertia. Let's calculate the contribution of one of them. Let's project the system along the axis about which we are calcuating the moment of inertia. This operation has no effect on the moment of inertia. Our face now becomes a rod. We know the moment of inertia through the center of the rod is $\frac{1}{12}$, but we are calculating the moment of inertia at a distance $\frac{1}{2}$ away from the center of the rod. Using the parallel axis theorem, we get that the moment of inertia is $\frac{1}{12} + (\frac{1}{2})^2 = \frac{4}{12} = \frac{1}{3}$. Then our total moment of inertia is $\frac{1}{6} + \frac{1}{6} + 4*\frac{1}{3}= \frac{5}{3}$. (The first two terms come from the top and bottom face, the last term comes from the four side faces.)

Putting units back in we get the moment of inertia is $I=\frac{5}{3} M a^2$, where $M$ is the mass of a face and $a$ is the side length of the cube (also the side length of a face).

What is meant by phase, phase difference, in (and out of) phase in wave terminology?

What is meant by phase, phase difference, in (and out of) phase in wave terminology? How do you get the relation $$y=A\sin(\omega t + \phi)?$$
Since the graph of sin function is identical to that of a sin wave, I know why the sine function is used in the wave equation, but how do you get $\omega t + \phi$ as the sine angle?

Can the apparent equal size of sun and moon be explained or is this a coincidence?

Is there a possible explanation for the apparent equal size of sun and moon or is this a coincidence?

(An explanation can involve something like tide-lock effects or the anthropic principle.)

Answer

It just happens to be a coincidence.

The current popular theory for how the Moon formed was a glancing impact on the Earth, late in the planet buiding process, by a Mars sized object. This caused the break up of the impactor and debris from both the impactor and the proto-Earth was flung into orbit to later coallesce into the Moon. So the Moon's size just happens to be random.

Plus the Moon was formed closer to the Earth and due to tidal interactions is slowly drifting away. Over time (astronomical time, millions and millions of years) it will appear smaller and smaller in the sky. It will still always be roughly the size of the Sun but total solar eclipses will become rarer and rarer (they will be more and more annular or partial). Likewise in the past, it was larger and total eclipses were both longer and more common.

general relativity - Does any particle ever reach any singularity inside the black hole?

I am not a professional physicist, so I may say something rubbish in here, but this question has always popped in my mind every time I read or hear anyone speak of particles hitting singularities and "weird things happen".

Now to the question at hand, please follow my slow reasoning... As far as I've learned, to reach a black hole singularity, you must first cross an event horizon. The event horizon has this particular property of putting the external universe on an infinite speed to the falling observer. Now due to the Hawking radiation, and knowing that the cosmic background radiation is slowly dimming, sooner or later every black hole in this particular instance of inflation we are living in will evaporate, according to an external observer of said black holes.

This means that every black hole has a finite timespan, as long as this universe survives that long. Now if we go back to the falling observer, we had already established that such an observer would see the outside universe "speed up" infinitely. This means that when the falling observer "hits" the event horizon, he will (or it if we speak about particles, which is clearer in this case), be immediately transported in time towards the final evaporation moment of the black hole. Either this or the particle gets some weird treatment. My point is, such a particle never gets to the singularity, because it has no time to get to it. The moment it crosses the event horizon, the black hole itself evaporates.

Where am I wrong here?

special relativity - Why is not ${(Lambda^T)^mu}_nu = {Lambda_nu}^mu$?

I am following lecture notes on SR. The author writes that the following is equivalent:

$$\Lambda^T\eta\Lambda = \eta \iff \eta_{\mu \nu} {\Lambda^\mu}_\rho{\Lambda^\nu}_\sigma = \eta_{\rho \sigma}. \tag{1}$$ This surprises me, because

$$ {(\Lambda^T)^\mu}_\nu = {\Lambda_\nu}^\mu.\tag{2} $$

And so I expected it to be $$\Lambda^T\eta\Lambda = \eta \iff \eta_{\mu \nu} {\Lambda_\rho}^\mu{\Lambda^\nu}_\sigma = \eta_{\rho \sigma}.\tag{3}$$ Why is this wrong?

Answer

1. 
   

   OP's three equations should read $$\Lambda^T\eta\Lambda ~=~ \eta \quad\iff\quad (\Lambda^T)_{\rho}{}^{\mu}~\eta_{\mu \nu}~ \Lambda^{\nu}{}_{\sigma} ~=~ \eta_{\rho \sigma} ,\tag{1'}$$ $$ (\Lambda^T)_{\nu}{}^{\mu} ~:=~\Lambda^{\mu}{}_{\nu} ,\tag{2'} $$ $$\Lambda^T\eta\Lambda ~=~ \eta \quad\iff\quad \Lambda^{\mu}{}_{\rho}~\eta_{\mu \nu}~ \Lambda^{\nu}{}_{\sigma} ~=~ \eta_{\rho \sigma} .\tag{3'}$$

   
   



2. 
   

   In more detail: Let $V$ be $n$-dimensional $\mathbb{R}$-vector space with a basis $(e_{\mu})_{\mu=1, \ldots, n}$. Let $V^{\ast}$ be the dual vector space with the dual basis $(e^{\ast \nu})_{\nu=1, \ldots, n}$. Let $$\Lambda~=~e_{\mu}~ \Lambda^{\mu}{}_{\nu}\otimes e^{\ast \nu}~ \in~V\otimes V^{\ast}~\cong~{\cal L}(V;V)$$ be a linear map from $V$ to $V$. Let us call the positions of the indices on $\Lambda^{\mu}{}_{\nu}$ for the NW-SE convention, cf. a compass rose. Let $$\Lambda^T~=~e^{\ast \nu}~ (\Lambda^T)_{\nu}{}^{\mu}\otimes e_{\mu}~\in~V^{\ast}\otimes V~\cong~{\cal L}(V^{\ast};V^{\ast})$$ be the transposed linear map from $V^{\ast}$ to $V^{\ast}$. Note that $(\Lambda^T)_{\nu}{}^{\mu}$ is written in the SW-NE convention. Let $$\eta~=~e^{\ast \mu}~\eta_{\mu\nu}\odot e^{\ast \nu}~\in~{\rm Sym}^2V^{\ast}~=~V^{\ast}\odot V^{\ast}$$ be an (indefinite) metric, i.e. an invertible element in the symmetrized tensor product. A (pseudo)orthogonal map $\Lambda$ satisfies by definition $$\Lambda^T\eta\Lambda~=~\eta.$$ See also this related Phys.SE post.

   
   

homework and exercises - Why do electric field lines start and end at 90 degree at the surface of a conductor?

There is one property of electric lines of forces which states that:

Electric field lines start and end at 90 degree at the surface of the conductor.

But why is that so? Is there any proof for this statement or is it observed by any experiment?

Answer

The reason is the same as why the electric field inside a conductor is zero: if it isn't zero, the free electrons undergo a force and move (rearrange) until they don't feel a force any more. If the electrons don't feel a force, the electric field must be zero.

At the surface of a conductor, the free electrons feel a force perpendicular to the surface, but can't go out of the conductor. If there were a component parallel to the surface, the force on the electrons also has a parallel component and the electrons would move until that component is zero.

homework and exercises - How is Capacitance affected by a seperation distance and with or without a battery?

The questions specifically says this :

A parallel-plate capacitor initially has capacitance $C$. The distance between the plates is then doubled, with a $9.0V$ battery connected. The battery is then disconnected, and the plate area is doubled. Finally, a $20V$ battery is connected across the plates.

• What is the new capacity?

I know that when the battery is connected and the separation is doubled, the capacitance is halved. But from then on I am not sure what to do. I tried to look it up online but had no success.

• So my question is basically, what happens when we increase the distance separation of a capacitor without a battery? and then add a large voltage?

Thanks guys!