quantum field theory - Local guage symmety implies causality

I read in a QFT book that local gauge symmetry implies causality. Could someone please explain that statement and why it's true?

Thank you.

cosmology - How Does Hubble's Expansion Affect Two Rope-Tied Galaxies?

Suppose we have two galaxies that are sufficiently far apart so that the distance between them increases due to Hubble's expansion. If I were to connect these two galaxies with a rope, would there be tension in the rope? Would the tension increase with time? Is the origin of the tension some sort of drag between the expanding space and matter?

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.

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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)?

   
   

nuclear physics - Explanation of energy levels in molecules, atoms, nuclei and their relationship

Why are the energy levels of molecules, the atoms that form them and the nuclei inside the atoms considered separately? Or phrased in a different way- what is it that makes their energy levels so different? What is the relationship between the energy levels of molecules, atoms and nuclei?

I'd interested to learn different interpretations on this question.

Answer

In physics we distinguish between the physics of "atoms and molecules" and nuclei. Atoms and molecules are described by the same theory, thus I will ignore those molecules here completely and only consider the difference between nuclei and atoms.

I suppose you recognize that an atom is a bound system, so is a nucleus a bound system. Maybe you have seen how the energy levels are computed for an atomic system and now you wonder what is the difference to the energy levels of a nucleus.

The simplest form of a natural atom is the Hydrogen atom. It consists of a nucleus, a proton, and an electron. The energy levels of the Hydrogen atom are described by the theory of quantum mechanics. The Coulomb force between the nucleus and the electron defines the $-1/r$ potential which we use in the quantum mechanical approach to solve for the energy levels of the atomic system. This is a very long story, which I illustrate in this short paragraph, and there exist a lot corrections (fine structure, hyperfine structure, lamb shift, darwin term, ...) to the energy level of the simple Hydrogen atom. But their net effect on the energy levels is a minor correction in our problem here, thus it is not what distinguishes the atomic energy levels from those of a nuclei.

When Goeppert and Jensen firstly described the nuclear shell model, they considered a different potential to solve for the energy levels of the nucleus. If we compare the atom and the nucleus, we'll see that there is a fundamental difference in the potential, that we need to describe the nucleus.

1. The nucleus consists of two different types of particles, neutron (uncharged) and proton (charged)


2. In contrast to the atomic potential, which has a center, that's why we also named the Coulomb force a central force, the potential of the nucleus is generated by the neighboring particles in the nucleus itself.



3. On short distances the interaction force between nucleons (protons, neutrons) is stronger than the Coulomb force

For an analytical description of the bound states of the nucleus, people use the Harmonic Oscillator potential or the potential well. This approach leads to reasonable results, which help us understand and predict the energy levels. Yet, I do not know of a standard potential which describes all nuclei correctly. It is an active field of research.

I will also give you an example, how we can estimate the energy scale in the nucleus and the atom just by considering their dimensions. The Heisenberg Uncertainty principle states, that $\Delta x \Delta p \ge \frac{\hbar}{2}$ . For the electron of an atom, we assume that its location $x$ has the uncertainty on the same order of the diameter $d_{atom}$ of the atom , so $\Delta x_{atom} = d_{atom}$. With this we construct the minimal impulse of the electron in the atom:

$$\Delta p_{atom} \ge \frac{\hbar}{2\Delta x_{atom}}$$

Although you might say, this is only the uncertainty of the impulse, yet it is a measure for the order of magnitude the impulse will be.

For a nucleon (proton or neutron) in a nucleus we can make the similar approach with Heisenberg Uncertainty principle:

$$\Delta p_{nuc} \ge \frac{\hbar}{2\Delta x_{nuc}}$$

Since the diameter of a nucleus is orders of magnitudes smaller, the uncertainty of impulse of the nucleon is also by that order larger and so will be the energy levels. Here are the numbers: $d_{nuc} \simeq 1.75\text{fm}$ and $d_{atom} \simeq 0.1\text{nm}$ or $ 100\, 000\text{fm}$.

To give you a short summary:

The nature of the potential, which leads to the bound states in an atom and a nucleus is different and leads to fundamental differences in the energy levels for both. Since the nucleus is more compact, the energy levels will also be greater.

homework and exercises - Derivation of Boltzmann equation in Kolb and Turner's The Early Universe

In the homogeneous and isotropic FRW Universe, the collisionless Boltzmann equation is given by $$ E\frac{\partial f}{ \partial t}-\frac{\dot{a}}{a}|\textbf{p}|^2\frac{\partial f}{\partial E}=0\tag{1}$$ where the phase space distribution $f$ is a function of the particle's energy $E$ and time $t$. See page 116 of The Early Universe by E. Kolb and M. Turner. The number density of the particle species at any time is given by $$n(t)=\frac{g}{(2\pi)^3}~\int d^3\textbf{p} f(E,t).\tag{2}$$ In Kolb and Turner's book, it is mentioned that by using (2) and doing an integration by parts, Eq.(1) can be reduced to $$\frac{dn}{dt}+3\frac{\dot{a}}{a}n=0.\tag{3}$$

How do we derive Eq.(3)?

My attempt

Taking derivatives of (2) w.r.t $t$ and $E$, we find, $$\frac{dn}{dt}=\frac{g}{(2\pi)^3}\int d^3\textbf{p}~\frac{\partial f}{\partial t},\tag{4}$$ $$\frac{dn}{dE}=\frac{g}{(2\pi)^3}\int d^3\textbf{p}~\frac{\partial f}{\partial E}\tag{5}.$$ Eq.(4) trivially gives the first term of Eq.(3) upon integration by $d\Pi=\frac{g}{(2\pi)^3}\frac{d^3\textbf{p}}{2E}$. But the second term becomes $$-\frac{\dot{a}}{a}\int \frac{|\textbf{p}|^2}{2E}\frac{\partial f}{\partial E}\frac{g~d^3\textbf{p}}{(2\pi)^3}.$$

Any help?

Answer

Rewrite $(1)$ as $$\frac{\partial f}{ \partial t}-\frac{\dot{a}}{a}\frac{|\textbf{p}|^2}{E}\frac{\partial f}{\partial E}=0\tag{1}$$ Now integrate with respect to $d^3\textbf{p}$: $$\frac{dn}{dt}-\frac{\dot{a}}{a}\frac{g}{(2\pi)^3} \int d^3\textbf{p}\frac{|\textbf{p}|^2}{E}\frac{\partial f}{\partial E}=0$$ We must express the second integral in terms of $n$. The integrand is rotationally symmetric so $$\int d^3\textbf{p}\frac{|\textbf{p}|^2}{E}\frac{\partial f}{\partial E}=4\pi\int d\mathrm{p} \frac{p^4}{E}\frac{\partial f}{\partial E}=4\pi\int d\mathrm{p}\, p^3\frac{\partial f}{\partial p}\\ $$ where in the last equality we used the chain rule with the dispersion relation $E = \sqrt{p^2+m^2}$. Now integrating by parts: $$4\pi\int d\mathrm{p}\, p^3\frac{\partial f}{\partial p} = -3 \times 4\pi\int d\mathrm{p}\, p^2 f = -3 \int d^3\textbf{p} f = -3n \frac{(2\pi)^3}{g}$$ Putting this back in you get $(2)$.

general relativity - What is the Komar mass of the de-Sitter spacetime?

The Komar mass of some spacetime is defined as an integral (volume or surface, depending on its formulation): https://en.wikipedia.org/wiki/Komar_mass

The de-Sitter metric in static coordinates is ($\Lambda > 0$ here): \begin{equation}\tag{1} ds^2 = (1 - \frac{\Lambda}{3} \, r^2) \, dt^2 - \frac{1}{1 - \frac{\Lambda}{3} \, r^2} \, dr^2 - r^2 \, d\Omega^2. \end{equation} This metric has a timelike Killing vector for $r < \ell$, where $\ell = \sqrt{3/\Lambda}$: \begin{equation}\tag{2} \xi^{\mu} = (1, 0, 0, 0). \end{equation} Thus $\xi^{\mu} \, \xi_{\mu} = g_{00}$. It is spacelike for $r > \ell$.

AFAIK, the Komar mass is defined only for asymptotically flat spacetimes, which the de-Sitter spacetime isn't. So I believe that the Komar mass cannot be defined for the de-Sitter spacetime. Is this true?

If the Komar mass was defined for the de-Sitter spacetime, it should already be well known today, but I can't find it anywhere. So I guess that the answer to my question is affirmative, but I need a clear confirmation of it.

If an observer (say at $r = 0$) has only access to the region inside the horizon located at $r = \ell$, could we define a Komar mass (or an energy of some sort) for that observer?

Answer

One could define "Komar-like" quantities for spacetimes which admit Killing vector fields, in general. But, whether they make sense is another question.

It is known that there is no well-defined notion for the mass in asymptotically de Sitter spacetime since in such spacetimes the Killing vector of time translational symmetries is spacelike at future null infinity, and not timelike. All notions of the mass in GR that we know are defined for the (asymptotically) timelike translational Killing vector fields. However, the mass is well-defined in the anti-de Sitter spacetimes because there exists the asymptotically timelike Killing vector fields. In this case, the usual definition of the Komar mass needs a small correction. See for instance this and this, for more information.

quantum field theory - Weak force: attractive or repulsive?

We are always told that there are the four fundamental forces or interactions of nature: gravitation, electromagnetism, and the weak and strong forces. We know that gravitation is attractive, that electromagnetism can be attractive or repulsive depending on the electric charge of the interacting particles, and that the strong force is attractive between quarks.

But when the weak force is mentioned, the description is always something such as 'responsible for radioactive decay', but there is no mention of whether this force is attractive or repulsive. So my question is: is the weak force/interaction attractive or repulsive?

Answer

Since the electroweak interaction is mediated by spin 1 bosons, it is the case that "like (charge) repels like and opposites attract".

In the electroweak case, the charges in question are weak isospin and weak hypercharge.

For weak isospin, there are two isospin charges (or flavors), up and down, and their associated anti-charges, anti-up and anti-down.

So:

• up repels up (anti-up repels anti-up)


• down repels down (anti-down repels anti-down)


• up attracts down (anti-up attracts anti-down)


• up attracts anti-up (down attracts anti-down)


• up repels anti-down (down repels anti-up)

For weak hypercharge, there is just one type of charge and its associated anti-charge.

So:

• hypercharge repels hypercharge (anti-hypercharge repels anti-hypercharge)


• hypercharge attracts anti-hypercharge

Note that electric charge is a certain mixture of weak isospin and weak hypercharge.

Since (left-handed) particles carry both weak isospin and weak hypercharge, both must be taken into account to determine which particles attract or repel under the electroweak interaction.

pattern - The Security to the Party [Part 4]

A party is being held at a local mansion. The host is very rich and his success is because of one thing — his famous recipe for Spaghetti!

The only guests that may attend are people who correctly reply to the guard at the door.

Here's where you come in. You and a friend are trying to steal this recipe. You sneak by and listen to the passwords.

The first guest arrives. The security says "5", and the guest replies "2".

The second guest arrives. The security says "6", the guest replies "2".

The third guest arrives. The security says "3", the guest replies "2".

Your friend thinks he's got it all figured out so he walks up to the door and the security says "2". He replied "2", and is tarred and feathered and sent away in shame.

Another guest arrives, security says "0" and he says "0" to get in.

You walk up and get a "4" from security. What is your response?

Thanks for the ideas, credits to:

Part 1 was created by warspyking and is found here: The Security to the Party

Part 2 in Mew's version The Security to the Party [Part 2]

Part 3 in Kelvin Barsana's version The Security to the Party [Part 3]

Answer

Answer:

Respond $1$.

Explanation:

Put x into binary. The number of 1s in the result is the answer. Here is a key for the given numbers: $$\begin{align}5&=101 &\rightarrow 2 \\ 6&=110 &\rightarrow 2 \\ 3&=11 &\rightarrow 2 \\ 2&=10 &\rightarrow 1 \\ 0&=0 &\rightarrow 0 \\ 4&=100 &\rightarrow 1\end{align}$$

Theoretically, can a black hole collapse until it gets smaller than its Schwarzschild radius?

So, the concept is this:

When a supergiant star collapses, its nucleus becomes a black hole. So, what if that black hole collapses again?

I don't know if it is possible, but what if it happens?

Answer

No. And maybe, but probably not.

General Relativity tells us that time passes more slowly in an intense gravitational field (from the point of view of someone in a less intense region of the field). This has been verified experimentally on a small scale.

Imagine you're near (but not too near) a black hole, in communication with a probe that's dropping into it. As you observe an analog clock on the probe, you'll see its second hand moving more and more slowly as the probe approaches the event horizon, coming closer and closer to a dead stop.

You'll never actually see the probe reach the event horizon; time on the probe, as seen from an outside vantage point, will come arbitrarily close to the moment it reaches it, but it will never actually get there.

The same thing happens to any matter falling into a black hole. It will quickly reach a point where you can no longer see it, but it will never quite reach the actual event horizon.

Inside the event horizon, time passes infinitely slowly (again, this is all from an outside point of view). So once the event horizon forms during the initial collapse that forms the black hole, nothing further happens inside it. Everything inside the event horizon is frozen in time, and cannot collapse any further.

On the other hand, from the point of view of the probe itself (say, if you foolishly volunteered to ride along), local time continues to pass at its normal rate of 1 second per second. You'll pass through the event horizon, and if the black hole is massive enough for the tidal stress at that point to be manageable, you might not even notice. You might see events, including further collapse, continue to occur after you're inside. But if you look back as you're falling, you'll see time in the outside universe pass more and more quickly, and just as you cross the event horizon you'll see eternity pass in a finite amount of time. So from the point of view of a hypothetical observer who's fallen into a black hole, yes, the body that formed the black hole can continue to collapse -- but any such observer might as well be outside the universe, since there's no way to communicate with them.

All this is based on relativity, ignoring quantum mechanics. Current theory (see Hawking radiation) says that quantum effects cause black holes to evaporate. All information about anything that fell into the black hole is lost (i.e., this isn't a way for you to escape after you've fallen in), but all the mass/energy will eventually come out as random radiation. This happens in a finite amount of time from the point of view of an outside observer -- which means that if you're riding the probe, the black hole will evaporate just as you're crossing the event horizon. (You will not survive the experience.)

Disclaimer: I am not an astrophysicist. I hope that one will come along soon and explain how I've gotten this wrong.

thermal radiation - Problem on Emissivity and absorptivity

I have read the radiation chapter, where I have been introduced with the terms emissivity and absorptivity. emissivity tells about the ability to emit heat energy as thermal radiation compare to a black body. and absorptivity is the amount of heat absobed by body devided by the incident heat energy .but my question is ,are these emissivity and absorptivity constant or depends on temperature? Actually I had a misconception that these quantities are intrinsic property of body If I have really wrong concept and they depends on temperature then tell me and Tell me how they depends on temperature I mean I need The mathematical expression

Answer

In a way, these are intrinsic properties of a material that depend on temperature. The temperature of the object dictates what a black body curve looks like and which wavelengths are emitted in what amounts. The emissivity of a material is not always uniform across all wavelengths, which means at different temperatures, you might get different emission profiles depending on the emissivity of the material at each wavelength along the curve. Similarly, the absorptivity can differ with wavelength. This means the absorptivity and emissivity values for a material might be not just different, but variably different. So the material might absorb well at the wavelength of incident radiation and emit poorly at its current temperature, but as a result, it may then heat up into a blackbody range where it is a good emitter. Furthermore, it also means the amount of energy the material absorbs (when pointed at a blackbody object) could depend on the temperature of the object it is pointed at.

There is no mathematical relation to provide you. Absorptivity and emissivity are intrinsic properties of a material. We test materials and empirically find the wavelength dependence of those values. It varies greatly from material to material, which means any time you want to use a different material, you should look up the $\alpha$ and $\epsilon$ values for the wavelengths you expect to be relevant. Usually, however, it is sufficient to look up wavelength ranges, such as visible light, near IR, UV, etc.

quantum mechanics - Open problem? Square of the wave function $Psi(x)_{x_o} = delta(x-x_0)$ of a particle localized at a point $x_0$?

Does anybody know the status of the problem to define the wave function (non-relativistic Quantum Mechanics) of a particle localized at a definite point?

Landau-Lifshitz says in chapter 1 that this function is $\Psi(x)_{x_o} = \delta(x-x_0)$ and gives an explanation that it produces the correct probability density when it is used to span some other arbitrary wave function $\Psi(x)$. The problem is of course that the wave function given above squares to a non integrable function. As far as I know this problem is unsolved. My question is if anybody knows the status quo of this problem. I am sorry if this question may be duplicated, I could not find it amongst the answered questions.

Answer

Mathematically spoken, since you want your wave functions to be square integrable, your wave functions must be in $L^2$ or some subspace thereof. However, you won't find a function in this space that has a support on a countable set of points, since the Lebesgue integral cannot see countable sets (measure 0), hence there cannot be a function (i.e. no wave function) with support in a single point (incidentally, the delta function is not a "function" in a way for that reason).

This tells us that a wavefunction for a particle that is fully localized cannot be defined in the usual setting of square Lebesgue-integrable functions, which is not too tragic, because we don't really think it makes physical sense anyway.

How much is quantum computation changing the interpretation of quantum theory, and, if at all, how?

At the beginning of quantum computation, David Deutsch made a strong claim that the Many Worlds interpretation of quantum theory was at the foundation of his ability to do what he did. There was a lot of interest in his claim, and Many Worlds had something of a resurgence at the time. Amongst other strong claims, I'm particularly aware that Chris Fuchs has pressed a Bayesian perspective that is rather different from the Many Worlds interpretation.

My perception, however, is that quantum computation/information/communication is now much less in thrall to any particular interpretation, unless we say there is such a thing as the minimalist interpretation. It seems that most quantum computation work is reported in journals in as close to the unadorned Fock space formalism of quantum optics as possible, often cut down to a small, finite number of harmonic oscillators, in one of the many formalisms and ansätze of statistical or condensed matter Physics, or, of course, the discussion revolves around qubits. Nonetheless, quantum information is not called quantum particle information —it's not part of particle physics— and the Bell and other inequalities that form the basis of the general rejection of classical particle physics are central instead of being as peripheral as they often seem in particle physics, so it seems, to me, that there is at least something of a change of perception, of zeitgeist, if you will.

Quantum computation for a long time was a subject of keen interest in the academic quantum foundations conferences that I went to back in the day. Is that interest still there? As Peter Shor points out in a comment, this interest is still there, but Is any of that interest felt in the wider Physics community that has relatively little practical interest in quantum foundations? [Peter Shor Answered that too —in a way that agrees with my prejudices, so it must be right— while I was writing the EDIT below, which tries to address one way in which I think the zeitgeist is being affected.]

EDIT: I suppose a significant number of Physicists will occasionally read PRL articles on quantum information that are highlighted by the editors, and it would be hard to avoid all Scientific American, New Scientist, and Nature articles on the subject. Someone considering grant sources would presumably at least briefly consider whether their work could be applied to QI (I'm bored already with the comp../inf../comm..) and occasionally talk to the QI group at their Institution (almost everybody has one of those by now, right?). Insofar as they do, then it seems inevitable that some QI viewpoints must be increasingly out there. Bell's theorem is surely more a topic of informed discussion than it was 20 years ago. To make the question more specific, and to include my bias in a more explicitly leading way, Do people agree with the implication of my second paragraph that “particles” are less in the air than they were in the heyday of Particle Physics? [If you not unreasonably think that's too specific, feel free to Answer the original and more general Question of the title. The more specific Question still admits at least evidence of the sort given by Peter Shor, and shouldn't be thought of or addressed as only a Yes/No question.]

Hat Tip to Roy Simpson for this Question (his is the credit, mine is the blame).

cipher - Counter-Red Spiders

^{An entry in Fortnightly Topic Challenge #35: Restricted Title 1. Title based on this xkcd.}

This puzzle contains hints for 220 black squares (but that's not the point) but is a standalone puzzle.

The counter-red-spider offensive begins...

We need your help to mount the counter-offensive against the red spiders. The following blueprints just came from mission control.

It looks like they're encrypted somehow, but it doesn't appear that difficult; everything appears to be ready to begin construction. Those rules seem a bit fishy to me, but I'm sure it's nothing you can't handle.

(I just hope those spiders didn't muck with the plans. If they did, things could get a bit... complicated).

Hint:

Hint:

If you're feeling disoriented ask for help from your team.

Hint:

Every man has a job.

---

Hint:

This is not like my previous puzzle.

particle physics - Why is the decay of a neutral rho meson into two neutral pions forbidden?

Why is the decay of a neutral rho meson into two neutral pions forbidden? (Other modes of decay are possible though.) Is it something with conservation of isospin symmetry or something else? Please explain in a bit more detail.

fluid dynamics - finding equation of a water droplet

I have a water droplet which is falling down through its gravity. I've supposed that my droplet is an ellipsoid and I want to find the equation that represents this ellipsoid in XYZ coordinate system. Therefore, I put two perpendicular cameras, one in XY plane and the other in YZ plane to obtain two projection views of this droplet. Then, I wrote corresponding equations that relate these two projected views to the 3D equation of my droplet. Then, I solved these equations, but they returned infinity number of answers (for 3D droplet body). So, I added another equation for droplet volume and solved altogether again. Although, the volume equation remarkably confined the number of feasible answers, it could not unique the answer and returns two possible answers for the droplet as shown in below figure (i.e. both of the proposed droplets have the same projection views on XY and YZ planes AND have the same volume as well)

Fig. 1 Two feasible answers from top view (XZ plane) that we don't have any camera on it. Although these two possible solutions have the same projection views on XY and YZ planes, they are different from top view and this can help us in recognition of the correct answer.

However, only one of these droplets (red or blue) is really my falling droplet (that the cameras have captured its frames). Therefore, I need one more piece of information to be able to distinguish the right one. My question is that which equation OR condition I can use (like the one I did for droplet volume) in order to be able to separate my desired answer among two existing possibilities?

Complementary Descriptions:

I am using two high speed cameras @ 4500 fps and since I am investigating a collision phenomena, it really needs this high amount of frame rate (or even more). A 2 mm spherical solid particle has been shot by a launcher and collides to my 2 mm falling droplet. Right before colliding, the droplet undergoes deformations because of some forces applied to it through the particle, particle launcher flow, etc. I have a third camera, but it is 400 fps and also it should be "synchronized" with others that is another issue as well. What I mean by asking this question is to find a "condition" OR another "related equation" to give me the capability to distinguish between these two possible answers (like what I've added for droplet volume). I have also thought about putting a mirror behind the droplet with a specific angle such that I can capture two different views of the droplet simultaneously with one camera OR putting an appropriate array of some laser emitters (or any other kind of pointer) on top plane to mark the droplet in both frames (at least in one point) so that I can use marked point(s) to identify the answer. However, I am still seeking to choose the most convenient and applicable method. All of your helpful ideas are really appreciated. Below is a schematic of my experimental setup.

Fig. 2 Experimental setup from top view.

Answer

Rather than adding more cameras, just add some mirrors.

The problem you have is that you are trying to do tomography with an under sampled system. This is a VERY broad subject - a bit outside of the scope of your question. But mirrors will work. I would recommend that you place them so the images are all in focus - depending on the depth of focus of your camera you may need to play with the "direct" path as well. But for example four mirrors set up as a pair of periscopes would allow one camera to take two images from +- 22.5 degrees. A second setup at 45 degree from that would give two more views for a total it four. This will reduce the degeneracy.

Here is a diagram of what I had in mind:

You can actually buy a device similar to this - see for example https://www.lhup.edu/~dsimanek/3d/stereo/3dgallery5.htm which includes some examples of the pictures taken with the Loreo stereo adapter - it includes a picture not unlike the beamsplitter I drew. Note though that they have the two sets of mirrors offset slightly, and that they are aimed "straight in front". This ensures that the focal planes for the two sides will coincide, but it also means that you may have a hard time getting very close to the subject - although the stagger may help with that. At any rate - I suggest you play around with the setup (using a small bead the size of your drop) until it looks right with a normal camera, before trying to do this with the high speed camera.

It occurred to me that since the drops are quite small, you might be able to simplify the setup with a pair of prisms - the basic idea being that a prism shifts the angle of the light between input and output. It might be tough to get the image you need (magnification, distance, ...) especially since a simple prism bending light through a large angle may have significant chromatic aberration which will make measuring the dimensions hard. Unless, of course, you use a monochromatic light source (or a filter on your camera). If you have a digital color camera you could see whether looking at just one of the R,G,B components of the image would give you a clearer image...

If you include a flat part in your prism, you might be able to increase the number of views to 6, spaced just 15 degrees apart. And since the "straight through" view is then coming through the prism, I believe the optical path lengths will in fact be the same - because basically the prism is a "very crude lens" and lenses do their focusing thing by having the same path length for all rays from object plane to image plane.

These are just some late night thoughts. If you have a buddy in the optics department he can probably come up with far better optical arrangements that achieve the same thing. This is really a bit like making a stereo camera - it's been done as the above link shows; and there is a company that makes attachments for various cameras - they may be able to give you exactly what you need.

PS - although this is a LONG time ago, I took high speed pictures of water drops during my graduate studies - so this is a trip down memory lane for me (and yes I had only one camera and multiple mirrors... - but it was an Imacon that could do 10 Mfps and my 0.4 mm diameter water jets were going at Mach 6...)

cipher - Staff Party, can I go?

So, there's a really cool professor at the college I'm starting to teach at. He throws pretty awesome faculty parties every month, but you can only get in if you are able to figure out the password. This month, he says the password is the title of his favorite movie. It seems like he understands that I'm new, because he emailed me a hint. Unfortunately, all it is is an image and a note! The note says:

Sorry if it doesn't make much sense. I realized I had the wrong keyboard plugged in when I wrote it.

Here is the image that he sent me:

What does it all mean? How can I get a movie title from a box of cereal?

Edit: As it has already been extracted by @Beastly Gerbil, here is the hidden QR code in higher quality:

Answer

Here's some black and white images of the QR code hidden (unsuccessfully :P) in the package to the left of the eye.

My QR Detector decodes this as:

01000011+01111001+00100000+01100011+01101111+00100000+01100001+00100000+01100101+01100001+01110000+01110100+00100000+01111001+01100011+01101101+00101110+00100000+01110101+01110010+01110000+00100000+01111001+01100100+00101110+00100000+01010000+00101110+01111000+00101110+01101110+01101110+01100011+01110010+01100010+01110110+00100000+01000001+01101110+01111001+01100100+01110010+01100111+01101001+01100100+00100000+01111001+01100100+00101110+00100000+01000101+00101110+01100001+01111001+01100100+00100000+01001111+01111001+01100001+01110000+00100000+01100100+01100001+01101111+00100000+01111000+00101110+00101110+01100010+00100000+01100101+00101110+01101111+01111001+01110000+01110010+01100110+00101110+01100101+01110111+00100000+01000011+01101101+01101100+00101110+01110000+01100011+01100001+01101110+00100000+01111001+01110000+01110010+01110010+01101100+01101111+00100000+01100100+01100001+01101011+00101110+00100000+01100101+01110000+01100011+01101011+00101110+01100010+00100000+01111001+01100100+0

This appears to be binary and decodes as:

Cy co a eapt ycm. urp yd. P.x.nncrbv Anydrgid yd. E.ayd Oyap dao x..b e.oyprf.ew Cml.pcan yprrlo dak. epck.b yd

The note from the professor says

"Sorry if it doesn't make much sense. I realized I had the wrong keyboard plugged in when I wrote it".

As @Sleafer points out on the Dvorak Keyboard this is:

It is a dark time for the Rebellion. Although the Death Star has been destroyed, Imperial troops have driven th

So the movie is:

Star Wars V: The Empire Strikes Back

Because the phrase is part of the opening crawl of the movie: (thanks to @gtwebb)

"It is a dark time for the Rebellion. Although the Death Star has been destroyed, Imperial troops have driven the Rebel forces from their hidden base and pursued them across the galaxy.

Evading the dreaded Imperial Starfleet, a group of freedom fighters led by Luke Skywalker has established a new secret base on the remote ice world of Hoth.

The evil lord Darth Vader, obsessed with finding young Skywalker, has dispatched thousands of remote probes into the far reaches of space…."

quantum mechanics - Radio waves within an atom

What effect does the quantum world have on radio waves? For example, if I could shrink myself down and stand on the nucleus (or even smaller sub atomic particles making up the nucleus) with a device which could measure radio waves of the surrounding world (ie with all the signals modern day humanity produces), what readings would I pick up? Would they be the same as normal or would the quantum aspects somehow effect the signals and if so, in what way would they be effected?

newtonian mechanics - Instantaneous cause

The position of an object has a rate of change called velocity. The velocity of an object has a rate of change called acceleration.

Force applied is proportional to the acceleration of the body (according to Newton). However it is known that a force can't be applied instantly. So acceleration has a rate of change too (called?).

Will the cause of the "force acceleration" also have a rate of change? Is this infinite, meaning that there will be always something accelerating something else, or is there any cause in nature that is truly instant?

--

Edit. To make myself more clear. Is there acceleration of the jerk? Is there acceleration of the acceleration of the jerk, and so on...

thermodynamics - Historic derivation of Wien's law

Every book I've read, including a lot of websites, Wikipedia, etc, say that Wien derived this:

$$\rho_\nu(T)=\rho(\nu,T)=\nu^3f\left(\frac{\nu}{T}\right)$$

Being $\rho_v(T)$ the spectral enegy density of a black body for a given temperature and electromagnetic wave frequency. And everywhere it's mentioned that he proved this using thermodynamical arguments in a paper from 1893. I haven't been able to find that paper or that thermodynamical argument, which is what I'm interested in. I've been looking for a few days already.

Does anybody know how he did this?

wavefunction - In Bohmian mechanics, do electrons move inside an atom?

Look at http://www.bohmian-mechanics.net/whatisbm_pictures_hydrogen.html. It is mentioned that in the rest states of a bound electron, the position of the electron is stationary, since the wavefunction's phase is not changing. In the Bohmian picture where the electron DOES have an exact position, it should be supposed to stay at a certain point, since it has no Bohmian trajectory.

However, we KNOW that the electron has a kinetic energy of ~13.6eV, which make it have a velocity of ~2E6 m/s. So isn't this a contradiction in the Bohmian picture of the atom?

homework and exercises - Condition that the Lagrangian energy function $hequivsum_ifrac{partial L}{partialdot q_i}dot q_i-L$ would be same as the mechanical energy $E$

I'm studying Classical Mechanics by Goldstein. I solved a problem but I have a question.

Pro 2.18
A point mass is constrained to move on a massless hoop of radius a fixed in a vertical plane that rotates about its vertical symmetry axis with constant angular speed ω. Obtain the Lagrange equations of motion assuming the only external forces arise from gravity. What are the constants of motion? Show that if ω is greater than a critical value ω0, there can be a solution in which the particle remains stationary on the hoop at a point other than at the bottom, but that if ω < ω0, the only stationary point for the particle is at the bottom of the hoop. What is the value of ω0?

So I proceeded like this solution here.

So here, when we choose only one generalized coordinate $\theta$(polar angle), energy function $h$ is not same as the energy. But in the text (chapter about Lagrangian) it says that if potential $V=V(q)$, $h=E$. For this problem $V=mga \cos \theta$, (or negative, according to how define $\theta$ or axis) so it satisfies the condition that potential only depends on generalized coordinate, not on generalized velocity. So $h$ should be $E$, but apparently not. What is wrong here?

I know that if I set the azimuthal angle as an independent variable, such a contradiction doesn't appear. But I cannot see why I should do that ( the problem says that azimuthal angle is not independent variable, and derivation of $h=E$ says nothing about that.)

Surely something must be wrong with my reasoning, because if the Lagrangian of a system is $L=\frac{1}{2}my'^2+mgy$, we can insert the constant horizontal kinetic energy $\frac{1}{2}mx'^2$ (x is not generalized coordinate here), but that would destroy $h=E$. Can someone explain this to me?

quantum mechanics - Why is time-evolution unitary - The Heisenberg-picture Version

There are various versions of this question already on this site, which attempt to justify / make plausible that the time evolution of quantum mechanical observables is unitary. Most of these questions already employ the assumptions of states being elements of a complex Hilbert space, linearity in the time-evolution, hermitian operator's eigenvectors spanning this Hilbert space and that the squared 2-norm of the projection of a state to the eigenspace of an operator gives the probabillity to measure the eigenvalue of the operator (See for example this or this question).

The answer to the question usually is that time-evolution should preserve the norm of a given state, because we interpret this norm as the overall-probabillity. An Answer also suggests that even if we drop the probabillity interpretation with the squared 2-norm, this paper shows that non-unitary comes with nasty properties like superliminal signalling or distinguishabillity of non-orthogonal states.

However - every of these questions (and even the paper given) answers from the perspective of the Schroedinger-picture only:

I'd like to know if similar reasonings can be given if we are in the Heisenberg-picture from the start (and don't even know about the existence of the Schroedinger-picture). Has something like that ever been formulated? Or is this the Schroedinger-picture more general in that sense?

What I came up with: If I assume (for whatever reason) that $$\hat{A}(\delta t) = \hat{U}^{-1} A(0) \hat{U}$$ and $\hat{A}(0)$ was a normal operator, then $\hat{A}(\delta t)$ should be normal as well (because I at least want $\hat{A}(0)$ and $\hat{A}(\delta t)$ to span the complete hilbert-space with it's eigenvectors. This something that I consider a plausible requirement for operators that correspond to observables. If I choose $\hat{U}$ to be unitary, then $\hat{A}(\delta t)$ is a normal operator - I however don't know how to show that this is the only way for $\hat{A}(\delta t)$ to be normal.

Edit: Since all arguments that have been stated in answers dealing with the Schroedinger-picture only talk about closed (contrary to open) quantum systems, I don't see the need to talk about cases in open quantum systems (or subsystems), where time evolution isn't necessarily unitary.

newtonian mechanics - Can you change the combined center of gravity (COG) of the rider and bike if you move?

In a closed system, I know the rider wouldn't be able to change the combined center of gravity of the system by moving his weight around. If he pushes the bike to lean to the right, he would go to the left by the amount that would keep the combined cog the same.

But a bike with tires and a road under it is not a closed system, the tires wouldn't let the lower part of the bike left or right when tryingo to initiate lean. Does this mean it is possible to push with your bodies inertia to the bike, and the combined cog will be changed? I thing if you do so above the cog of the bike, you could, but I am not sure.

Edit: My question is not about if you can move the center of gravity of the bike at all. My question is about if can you change the combine cog of the system by moving around on the bike?

English is not my mother tongue, sorry for the confusion

thermodynamics - Why is entropy of system same for reversible and irreversible processes?

I read that entropy change of universe is zero in a reversible process but positive in a irreversible process,then doesn't it mean that entropy change of system of both the processes must be different?

I also read that in a reversible cycle, entropy change is zero for example a carnot cycle,and entropy change in an irreversible cycle is positive for example if we consider friction in a carnot cycle. Doesn't this show that entropy change of system in both processes is different?

electromagnetism - Is reversal of magnetic polarity in a planet an instantaneous occurence?

Just what the title states -

Does reversal of magnetic poles in a planet refer to the point in time when reversal is complete?

OR

Does it refer to the entire drawn out process (assuming the poles flip gradually from 0 through 180 degrees?

Answer

As far as I know the term "reversal of magnetic poles" doesn't have a strict definition, so I suppose different commentators might use it in different ways. However I suspect most of us would use it to describe the whole process.

You describe the process as "drawn out" but no-one knows how long it takes because the dynamics of the Earth's core are poorly understood. On a geological timescale the process looks instantaneous, but then geological timescales are pretty long. Models suggest it could be pretty quick, though how realistic the liquid sodium models are is open to debate.

quantum field theory - How Uncertainty Principle, Vacuum fluctuations and Energy Conservation coexist in QFT?

Recently I had a debate about the uncertainty principle in QFT that made me even more confused..

Because we use Fourier transforms in QFT, we should have an analogue to the usual Heisenberg uncertainty principle in QFT between the 4-vector of space-time coordinates and the conjugate momentum, but I found no reference for that, so is my guess is wrong?

We know that there is no universal Hermitian operator for time in QM, even so there is an uncertainty principle for time and energy, well, in QFT time is just a parameter, the same as the spatial coordinates, so is there an uncertainty principle for energy in QFT?

The last question made me confused regarding the energy conservation law in QFT: we use this law in QFT during calculations of propagators only (as I remember), it means we are using it with "bare" particles, while we suppose that these particles don't "interact" with vacuum fluctuations, so does that mean energy conservation law is a statistical law?

This brings to my mind the vacuum expectation value, that we suppose is zero for any observer, but it is zero statistically. At the same time we usually use Noether's theorem to deduce that energy is conserved (locally at least, and not statistically).

I believe I'm missing something here, can you please advise me?

mathematics - A knockout tournament

$12345$ players take part in a knockout tournament. In each round players are paired up; each pair plays a game with the winning player advancing to the next round (no ties). If there are an odd number of players at the start of a round, one is randomly selected to automatically progress. This continues until one player is delared champion.

How many games are played in total?

Answer

In every game one person loses. There is only one winner, so there must be $12344$ losers. Therefore there are $12344$ games.

riddle - Who is Pete's favourite scientist?

Pete is back.

Pete met his old friend Steve at the hospital at the end of March. They both went to the same history class.

Pete: Hey!

Steve: Pete? Pete, is that you?

Pete: Yes. How are you? Haven't seen you in ages

Steve: I am fine. What about you?

The conversation went on for a while. They talked about a lot of things: how they both nailed that one history test without learning anything, how they went to France together for a summer, how they pranked their teacher, and which topic in their class they liked the most. They even both had a favourite scientist. For Steve it was Nikola Tesla.

Who is Pete's favourite scientist?

---

Since I'm not a native speaker it would be nice if you could help me to fix the grammar in this riddle. Please do not change the conversation.

Answer

Is it

Rene Descartes

who was

born in 1596 (the number of words in each line), France, at the end of March

electromagnetism - How can there be net linear momentum in a static electromagnetic field (not propagating)?

I understand from basic conservation of energy and momentum considerations, it is clear in classical electrodynamics that the fields should be able to have energy and momentum. This leads to the usual Poynting vector and energy density relations for electromagnetic fields.

However, I do not know how to interpret situations where there is a net linear momentum in a static electromagnetic field. The fields aren't propagating. It doesn't make sense to me that momentum can be divorced from motion.

As a concrete example to discuss: Consider a massless string of length $L,$ with a spherical shell on each end with a magnetic dipole moment m and positive charge q. The radius of the sphere $R\ll L,$ or alternatively, consider the dipoles to be perfect "point" dipoles. Let the string be along the $y$ axis, with one dipole at the origin and the other at $y=+L.$ If the magnetic dipole at the origin is oriented in the $-z$ direction, and the other dipole in the $+z$ direction, if you calculate the total linear momentum in the fields, the answer is:

$$p_\text{em} = m q \frac{\mu_0}{2 \pi L^2}\, \hat{x}$$

While this is an unstable equilibrium, it is an equilibrium. So classically the state can remain static with no need to evoke other external entities, interactions, etcetera. So there doesn't seem to be any of the usual potential pitfalls to save us here.

Please, can someone explain how a static field can have momentum?

Answer

This is a fairly subtle question! Griffiths recently published a paper on this.

Hidden momentum, field momentum, and electromagnetic impulse:

Electromagnetic fields carry energy, momentum, and angular momentum. The momentum density, $ϵ_{0}(E\times B)$, accounts (among other things) for the pressure of light. But even static fields can carry momentum, and this would appear to contradict a general theorem that the total momentum of a closed system is zero if its center of energy is at rest. In such cases, there must be some other (nonelectromagnetic) momenta that cancel the field momentum. What is the nature of this “hidden momentum” and what happens to it when the electromagnetic fields are turned off?

EDIT:
Free version of the above link.

statistical mechanics - Is there any proof for the 2nd law of thermodynamics?

Are there any analytical proofs for the 2nd law of thermodynamics?

Or is it based entirely on empirical evidence?

Answer

It's simple to "roughly prove" the second law in the context of statistical physics. The evolution $A\to B$ of macrostate $A$, containing $\exp(S_A)$ microstates, to macrostate $B$, containing $\exp(S_B)$ microstates, is easily shown by the formula for the probability "summing over final outcomes, averaging over initial states", to be $\exp(S_B-S_A)$ higher than the probability of the inverse process (with velocities reversed). Because $S_B-S_A$ is supposed to be macroscopic, such as $10^{26}$ for a kilogram of matter, the probability in the wrong direction is the exponential of minus this large difference and is zero for all practical purposes.

The more rigorous versions of this proof are always variations of the 1872 proof of the so-called H-theorem by Ludwig Boltzmann:

H-theorem

This proof may be adjusted to particular or general physical systems, both classical ones and quantum ones. Please ignore the invasive comments on the Wikipedia about Loschmidt's paradoxes and similar stuff which is based on a misunderstanding. The H-theorem is a proof that the thermodynamic arrow of time - the direction of time in which the entropy increases - is inevitably aligned with the logical arrow of time - the direction in which one is allowed to make assumptions (the past) in order to evolve or predict other phenomena (in the future).

Every Universe of our type has to have a globally well-defined logical arrow of time: it has to know that the future is being directly evolving (although probabilistically, but with objectively calculable probabilities) from the past. So any universe has to distinguish the future and the past logically, it has to have a logical arrow of time, which is also imprinted to our asymmetric reasoning about the past and the future. Given these qualitative assumptions that are totally vital for the usage of logic in any setup that works with a time coordinate, the H-theorem shows that a particular quantity can't be decreasing, at least not by macroscopic amounts, for a closed system.

density - Why is ice less dense than water?

The answers to this question explain that ice is less dense than water because it has a "crystal structure", but they dont explain what exactly that is and why this happens, also I saw this answer from another site stating that not all ice is less dense than water.

What is the "crystal structure" that ice has? Why is ice structured that way? Can ice be more dense than water, and if yes, how and when?

Answer

I'm sure you have seen photographs of snowflakes up close. You will notice that there are hundreds of small crystals of ice. This is the crystal structure of ice. You don't see ice cubes with a crystal structure because they freeze too fast. The water doesn't have enough time to move into the crystal lattice when you freeze the water. This web site shows how the molecules line up in the crystals.

Yes, some ice is denser than water. If you put pressure on regular ice, and give it time to rearrange, the molecules will move into a new crystal lattice which results in the ice being more dense than water. In the first ice crystal, there are spaces between some of the molecules which is not there in the second crystal structure.

With extreme pressure, you can have frozen water at 100 °C.

Differences between astronomy, astrophysics and cosmology?

What is the main difference between Astronomy, Astrophysics, and Cosmology?

I have the impression that astronomy is a subject that runs parallel to physics but it is outside the physics field. This is based on the division of departments present in many universities: the department of physics being separated from the department of astronomy. The difference between cosmology and astrophysics is more obscure to me. I have the impression, though, that cosmology is more concerned with the structure of spacetime and universe models while astrophysics is more concerned with stellar life cycles, physical properties of stars, galaxies etc.

Which is the field that had more activity/ scientific breakthroughs recently?

In real life, are those classifications even important? Are there many differences between the education process of a future astronomer, astrophysicist or cosmologist?

Somebody, please, give examples of famous astrophysicists, cosmologists and astronomers at the present time.

Answer

Richard Feynman has nice words about science. It is not bad to read chapter 3 of "Feynman Lectures on Physics". The main point of his lecture is that "there is no strict boundary between different fields of science", "nature doesn't concern what we call its parts!"

So, we can't look for a line that divides celestial works into astronomical or astrophysical or cosmological. Although the main keywords about these fields are respectively: "observation of celestial bodies", "machinery of celestial bodies", "evolution of cosmos as a whole". Surely, they use findings of each other.

Recently, cosmology has encountered with big questions, really big! Dark matter and dark energy are the most challenging ones. But it is interesting that if there weren't astronomers and their numerical data, cosmology could not progress.

Usually the students of astronomy focus on classical mechanics and optical methods to be able to observe celestial bodies. Astrophysicists concern general relativity and nuclear physics as two important tools to describe stars or galaxies. But cosmologists are interested in modern theories too, especially string theory. Therefore the courses which they pass should be related to these subjects.

nuclear physics - Binding energy, fission and fusion

My high school physics book doesn't elaborate the idea of binding energy and how it's related to fission and fusion adequately in a way that made me have wrong thoughts about these ideas.

What i understand after doing some research is that: - Binding energy is the energy that has to be given for nucleons to separate them from each other.

• Nucleons in stable nucleus have negative energy considering the energy of a free static nucleon to be the reference energy.

My questions and confusions:

• 
  

  in fission, does the nucleus divide because of the collision between the neutron and the heavy nucleus, or because the nucleus would become unstable after the mass number has increased.

  
  


• 
  
  

  this question is related to the above one ; neutrons are supposed to be the main factor of nucleus stability because it contributes in the strong force. why does adding a new neutron or more to any nucleus without changing the number of protons, make the nucleus unstable?

  
  


• 
  

  My book mentions that when alpha decay happens ,a decrease in mass turns into kinetic energy gained by the products. does it mean the decrease of mass because of the lost neutrons and protons or what?

  
  


• 
  

  the process of losing mass for energy and vice versa in fission and fusions, shouldn't the mass for the neutrons shot out of the mass increase back and that's it? where would energy come from?

  
  

newtonian mechanics - Deriving Newton's Third Law from homogeneity of Space

I am following the first volume of the course of theoretical physics by Landau. So, whatever I say below mainly talks regarding the first 2 chapters of Landau and the approach of deriving Newton's laws from Lagrangian principle supposing Hamilton's principle of extremum action. Please keep this view in mind while reading and answering my queries and kindly neglect the systems to which Action Principle is not applicable:

If we use homogeneity of space in Euler-Lagrange equations, we obtain a remarkable result i.e. the conservation of momentum for a closed system.

Now, this result, using the form of Lagrange for a closed system of particles, transforms into $ \Sigma F = 0 $ . Now, how from this can we conclude that the internal forces that particles exert come in equal and opposite pairs?

Is it because for 2 particles this comes out as $ F_{1} + F_{2} = 0 $ and we take the forces exerted by particles on one other to be independent of other particles (i.e. Superposition Principle) as an experimental fact?

I doubt it as whole of Newtonian Mechanics is derivable from Lagrangian Mechanics and supposed Symmetries. So, according to me, a fact like Newton's Third Law should be derivable from it without using an additional experimental fact.

I have an idea to prove it rigorously. Consider two particles $i$ and $j$. Let the force on $i$ by $j$ be $F_{ij}$ and on $j$ by $i$ be $k_{ij}F_{ij}$. Now the condition becomes $\Sigma (1+k_{ij})F_{ij}=0$ where the terms to be included and rejected in summation understood. As this must be true for any value of $F_{ij}$, we get $k_{ij}=-1$. I don't know if this argument or refinement of such an argument holds or not. I can see many questions arising in this argument and it's not very convincing to me.

I would like to hear from you people as to if it is an experimental result used or not? If not, then is the method given above right or wrong? If wrong, how can we prove it?

Addendum

My method of proof uses the fact of superposition of forces itself, so it is flawed. I have assumed that the coefficients $k_{ij}$ are constants and don't change in the influence of all other particles which is exactly what superposition principle says.

As the superposition of Forces can be derived by superposition of potential energies at a point in space and potential energy being more fundamental in Lagrangian Mechanics, I restate my question as follows:

Is the principle of superposition of potential energies by different sources at a point in space derivable from the inside of Lagrangian Mechanics or is it an experimental fact used in Lagrangian Mechanics?

I, now, doubt this to be derivable as the fundamental assumption about potential energy is only that it is a function of coordinates of particles and this function may or may not respect superposition.

Answer

The derivation in Landau and Lifschitz is making some additional implicit assumptions. They assume that all forces come from pair-interactions, and that the pair forces are rotationally invariant. With these two assumptions, the potential function in the Lagrangian is

$$V(x_1,\ldots,x_n) = \sum_{\langle i,j\rangle} V(|x_i - x_j|)$$

And then it is easy to prove Newton's third law, because the derivative of the distance function is equal and opposite for each pair of particles.

This type of derivation is reasonable from a physical point of view for macroscopic objects, but it is not mathematically ok, because it omits important examples.

No rotational invariance, no third law

Dropping the assumption of rotational invariance, but keeping the assumption of pair-wise interaction, one gets the following counterexample in 2 dimensions, with two particles (A,B) with position vectors $(A_x,A_y)$ $(B_x,B_y)$ respectively:

$$V(A_x,A_y,B_x,B_y) = f(A_x-B_x) + f(A_y - B_y) $$

where $f$ is any function other than $f(x)=x^2$. This pair potential leads to equal and opposite forces, but not collinear ones. Linear momentum and energy are conserved, but angular momentum is not, except when both particles are on the lines $y=\pm x$ relative to each other. The potential is un-physical of course, in the absence of a medium like a lattice that breaks rotational invariance.

Many-body direct interactions, no reflection symmetry, no third law

There is another class of counterexamples which is much more interesting, because they do not break angular momentum or center of mass conservation laws, and so they are physically possible interactions in vacuum, but they do break Newton's third law. This is the chiral three-body interaction.

Consider 3 particles A,B,C in two dimensions whose potential function is equal to the signed area of the triangle formed by the points A,B,C.

$$V(A,B,C) = B_x C_y - A_x C_y -B_x A_y - C_x B_y + C_x A_y + A_x B_y$$

If all 3 particles are collinear, the forces for this 3-body potential are perpendicular to the common line they lie on. The derivative of the area is maximum by moving the points away from the common line. So you obviously cannot write the force as any sum of pairwise interactions along the line of separation, equal and opposite or not. The forces and torques still add up to zero, since this potential is translationally and rotationally invariant.

Many body direct interaction, spatial reflection symmetry, crappy third law

If the force on k particles is reflection invariant, it never gets out of the subspace spanned by their mutual separation. This is because if they lie in a lower dimensional subspace, the system is invariant with respect to reflections perpendicular to that subspace, so the forces must be as well.

This means that you can always cook up equal and opposite forces between the particles that add up to the total force, and pretend that these forces are physically meaningful. This allows you to salvage Newton's third law, in a way. But it gives nonsense forces.

To see that this is nonsense, consider the three-particle triangle area potential from before, but this time take the absolute value. The result is reflection invariant, but contains a discontinuity in the derivative when the particles become collinear. Near collinearity, the forces perpendicular have a finite limit. But in order to write these finite forces as a sum of equal and opposite contributions from the three-particles, you need the forces between the particles to diverge at collinearity.

Three body interactions are natural

There is natural physics that gives such a three-body interaction. You can imagine the three bodies are connected by rigid frictionless struts that are free to expand and contract like collapsible antennas, and a very high-quality massless soap bubble is stretched between the struts. The soap bubble prefers to have less area according to its nonzero surface tension. If the dynamics of the soap bubble and the struts are fast compared to the particles, you can integrate out the soap bubble degrees of freedom and you will get just such a three-body interaction.

Then the reason the bodies snap together near collinearity with a finite transverse force is clear--- the soap bubble wants to collapse to zero area, so it pulls them in. It is then obvious that there is no sense in which they have any diverging pairwise forces, or any pairwise forces at all.

Other cases where you get three body interactions directly is when you have a non-linear field between the three objects, and the field dynamics are fast. Consider a cubically self-interacting massive scalar field (with cubic coupling $\lambda$) sourced by classical stationary delta-function sources of strength g. The leading non-linear contribution to the classical potential is a tree-level, classical, three-body interaction of the form

$$V(x,y,z) \propto g^3 \lambda \int \,\mathrm d^3k_1\mathrm d^3k_2 { e^{i(k_1\cdot (x-z) + k_2\cdot(y-z))} \over (k_1^2 + m^2) (k_2^2 + m^2)((k_1+k_2)^2 + m^2)}$$

which heuristically goes something like ${e^{-mr_{123}}r_{123}\over r_{12}r_{23}r_{13}}$ where the r's are the side lengths of the triangle and $r_{123}$ is the perimeter (this is just a scaling estimate). For nucleons, many body potentials are significant.

The forces from the crappy third law are not integrable

If you still insist on a Newton's third law description of three-body interactions like the soap bubble particles, and you give a pairwise force for each pair of particles which adds up to the full many-body interaction, these pairwise forces cannot be thought of as coming from a potential function. They are not integrable.

The example of the soap-bubble force makes it clear--- if A,B,C are nearly collinear with B between A and C, closer to A, you can slide B away from A towards C very very close to collinearity, and bring it back less close to collinear. The A-B force is along the line of separation, and it diverges at collinearity, so the integral of the force along this loop cannot be zero.

The force is still conservative of course, it comes from a three-body potential after all. This means that the two-body A-B force plus the two-body B-C force is integrable. It's just that the A-C two body force is not. So the separation is completely silly.

Absence of multi-body interactions for macroscopic objects in empty space

The interactions of macroscopic objects are through contact forces, which are necessarily pairwise since all other contacts are far away, and electromagnetic and gravitational fields, which are very close to linear at these scales. The electromagnetic and gravitational forces end up being linearly additive between pairs, and the result is a potential of the form Landau and Lifschitz consider--- pairwise interactions which are individually rotationally invariant.

But for close packed atoms in a crystal, there is no reason to ignore 3-body potentials. It is certainly true that in the nucleus three-body and four-body potentials are necessary, but in both cases you are dealing with quantum systems.

So I don't think the third law is particularly fundamental. As a philosophical thing, that nothing can act without being acted upon, it's as valid as any other general principle. But as a mathematical statement of the nature of interactions between particles, it is completely dated. The fundamental things are the conservation of linear momentum, angular momentum, and center of mass, which are independent laws, derived from translation invariance, rotational invariance, and Galilean invariance respectively. The pair-wise forces acting along the separation direction are just an accident.

newtonian mechanics - Can we deviate a bullet from its target? Did anybody try this?

Can we deviate a bullet from its target? Means by applying any strong field at the target? Is there any technique so far like this!

word - Can You Cipher It? #2

Link to the first cryptogram: Can You Decipher It?

Cryptogram #2: kk yn hc im ym nu lr ys rl ua pb at rn lt mk eu ue ui re is zi pw

ii ui bu rr pe dl ss bp rg qk kk zf pw– ew pb oz ev xd ep qb li at ct

Hint #1

SQUARESQUARESQUARESQUARE

Hint #2 (related to cryptogram #1)

What that guys name from the last cipher? I forgot

Hint #3 (related to cryptogram #1):

Where was that from... I seem to be forgetting

Answer

Partial(?)

Hint 1 means

It's a foursquare cipher.

Hint 2 and 3 give us the two keys,

Robert Frost and "Fire and Ice", the author and poem referenced in the last puzzle.

This gives the answer(?) as

moving on your next clue this a four lettered answer xmcrees(?) catch phrase from over x– by the way what time is it

So

McCree's catchphrase is "you know what time it is" and he's from overWATCH, but that's a five letter word?

Oh, but the phrase references

the answer, high NOON.

riddle - Puzzle: Who is the winner of the Great Puzzling Trophy?

It’s the 10th annual award ceremony for the Puzzling StackExchange community! As the host and his attendants are handing out the awards to the best members of the community, they suddenly realize that they have forgotten who the recipient of the Great Puzzling Trophy was!

In a hurry, the host rushes back to the Great Puzzling Room to find his presentation notes. But when he arrived there, they were gone!

In despair, the host slumped onto the floor of the Great Puzzling Room, only to discover a typed note left on the table where the Great Puzzling Trophy originally sat. It contained the following paragraph, which he had seen an attendant leave there before he had gone onto the stage:

“Being even and fair sometimes leads to great difficulties, like when somebody yearns for something they greatly desire but they end up not really being right. The winner of today's best prize has mastered the art in designing a riddle that will leave many questioning whether he is even human. May this individual thus come forward and receive the greatest reward for everything they will succeed in accomplishing over the time they will remain in our community for. As you know, over the course of their time spent well here, never have they ever created one single mediocre puzzle!”

Alongside the note, the host also find a word search torn out of an old newspaper left on the table:

This is it. Can you help the host find the winner of the trophy?

Bonus question: How does the number pattern correspond to one of the important keywords from the word search?

EDIT: I apologise for the redundancy of the spiral. I only noticed some time after posting the third hint. Apologies to those who were mislead by it.

If absolutely needed, these hints are for those who are really struggling to solve this riddle.

Hint #1:

There are three and only three words that have been intentionally put into the word search.

Hint #2:

Still struggling with the word search? Here are the three clues for the words: 1. Twisting and Curling, 2. Send Back Home, 3. Prizewinner.

Hint #3:

This sequence may be slightly helpful: 1, 2, 3, 5, 7, 10, 13, 17, 21, 26, 31, 37, 43, 50, 57, 66, 73, 82, 91, 100.

A bit about me:

This is my first puzzle on this community. I’d absolutely love any feedback at all on anything I could have improved or did really well in. I love this community and the high standard of puzzles that have been posted, and I’m looking forward to posting some more of my own in the future!

Created and posted in a day, this puzzle was inspired by many by the user I have nominated as the winner of the trophy. His puzzles have really inspired me to create one of my own. Once you've solved this riddle, go check him out!

Answer

Partial

The word search gives 3 words as FIRST SPIRAL LETTERS

If the same block is treated as spiral we don't get any meaningful sentence.

Highlighting the letters as per the pattern(1, 2, 3, 5, 7, 10, 13, 17, 21, 26, 31, 37, 43, 50, 57, 66, 73, 82, 91, 100) obtained using spiral gives

Being even and fair sometimes leads to great difficulties, like when somebody yearns for something they greatly desire but they end up not really being right. The winner of today's best prize has mastered the art in designing a riddle that will leave many questioning whether he is even human. May this individual thus come forward and receive the greatest reward for everything they will succeed in accomplishing over the time they will remain in our community for. As you know, over the course of their time spent well here, never have they ever created one single mediocre puzzle!”
Though, taking the sequence of words from the paragraph and the first letters gives BEASTLY GERBIL HAS WON!

Bonus Question

Go upvote F1Krazy's answer for this part.

homework and exercises - Apparent depth is supposed to be independent of viewing angle but this seems inconsistent with critical angle refraction

When rays from the object at some depth are draw so as to be incident at the critical angle (or some angle just lower) the ray is refracted along the surface. Retracing this line would give you an image directly above but on the surface of water. How then can one state that apparent depth doesn't depend on viewing angle?

quantum field theory - Why cannot massless particles carry charges?

How to show that massless particles do not carry charges from QFT's point of view?

Answer

Massless particles can carry and do carry confined charges (gluon in QCD) and they may carry charges under spontaneously broken generators (photon is transforming e.g. under the generators of $SU(2)$ associated with the W-bosons) but they cannot carry charges under unconfined $U(1)$ force like electromagnetism.

The reason may be explained in different ways. One of them is using "running couplings". Renormalization implies that the electromagnetic coupling constant "runs". The dimensionless value of $e$, while naively constant in classical physics, has a subtle dependence on $\log E$ according to quantum mechanics thanks to the shielding by polarized particle-antiparticle pairs.

In the real world, the running begins at low energies comparable to the electron (or positron) mass, the lightest charged particle. At energy scales below the electron mass, there is basically no running anymore. If the electron mass were sent to zero, the coupling would keep on running to arbitrarily low energies. The coupling is weakening at lower energies. It would be zero in classical physics if a massless particle species existed.

It's hard to decide whether such a situation would be inconsistent. It would surely be different than it is in our world where the fine-structure constant has a finite, well-defined low-energy limit, $1/137.036\dots$.

The world with massless charged particles would either be inconsistent; or had a vanishing coefficient in front of Coulomb's law for macroscopic low-energy objects (which would be enough to say that we shouldn't call it "electromagnetism" anymore); or would demand an infinite magnitude of the coupling for any finite energy scales which is probably inconsistent.

One might worry that the arguments based on running are unreliable. There is a more direct, experimental way to approach the question: We know that massless particles don't exist because they would be easily pair-produced at colliders – and pretty much in any process – because the probability of their pair-production is governed by the (known to be finite) fine-structure constant and can't be suppressed, and there can't be a shortage of energy, either (because they're massless).

Concerning the "loopholes" that are important in the theoretical literature, supersymmetric field theories are full of massless charged particles. But whenever they occur, the behavior of the forces at long distances is very different from what we know in the real world. These $U(1)$ forces are typically a part of a non-Abelian symmetry that is either confining like QCD; or is conformal (scale-invariant) when the negative contribution to the running from gauge bosons is cancelled by some other charged particles.

There is nothing inconsistent about these theories – supersymmetric theories like that are more consistent, not less consistent – but the actual physics in that world would not allow the macroscopic charged static objects we know from the first encounter with the Coulomb law simply because all massive charged particles would be more or less guaranteed to be unstable. Mathematically, these theories are theories with massless charged particles. But physically, the phenomena in such Universes wouldn't follow the history of electromagnetism started by the Coulomb law etc., so from an empirical viewpoint, the corresponding $U(1)$ interaction shouldn't be considered a special case of "our" electromagnetism.

spectroscopy - Machine to identify substances

I was looking at Wikipedia's article on mass spectrometers, and realized that the method described was to heat a sample into its gas phase and separate the ions by mass using a large magnet. While I know this technique works well, and is used extensively, would it not work better/more easily to heat a sample into a its gas phase, shine a very bright light through it, and apply normal spectroscopy? The only downside I can see, is that it would require a large amount of sample to use.

I was thinking of something like this:

Answer

The traditional spectroscopy you describe is still used; different techniques apply better in different circumstances.

For example, there are chemical bonds whose energy is in the ultraviolet. UV spectroscopy is good if you are dealing with such molecules, perhaps things with lots of $\require{mhchem}\ce{C=C}$ bonds. The information you get can tell you about the connectivity of the molecules (which atoms are connected to which others).

Lower energy transitions in the infrared tend to be between different rotational and/or vibrational states of molecules. Think of a molecule bending and flexing -- the energy scales are not as high as actually breaking bonds or liberating electrons. Thus IR spectroscopy is well suited for studying molecules according to their flexibility.

In between is optical spectroscopy. And of course there is no clear delineation in nature as to what information can be gleaned from what whavelengths.

At the more extreme ends of the electromagnetic spectrum there are still further types of spectroscopy.

NMR (aka MRI) uses radio waves to essentially study the electronic environment around hydrogen atoms. An $\ce{H}$ in $\ce{H2}$ will be different from the one attached to $\ce{O}$ in $\ce{CH3OH}$, which will be different from the three attached to the $\ce{C}$ in the same molecule. This is because the proton is affected by the surrounding electron cloud, which is distorted by nearby nuclei. The spin states available to protons are very close in energy and so are probed by low-energy frequencies.

High-energy X-rays are not so good at exciting transitions to other bound states (they tend to quickly ionize materials and break them), but they can still be used to "image" molecules, as was done in figuring out the shape of DNA.

All of these examples used electromagnetic waves to probe a substance, and all of them work very closely with the electrons. If you want to probe the structure of a nucleus with EM waves, you'll need much higher (gamma ray) energies. But there are other ways to probe things. Mass spectrometry is, at its most basic, just pushing on an object with a known force, watching how much it deflects, and inferring its mass. Mass spectrometers can easily tell the difference between isotopes of the same element, which is much more difficult if you only directly query the electrons.

So in summary yes, spectroscopy as you describe is very much used in practice to analyze substances. Each technique has its own niche.

riddle - What is it in my eyes that will result in humanity's demise?

The following stanzas
Are hints to the answer
Each a part of the end
Their piece they defend

I am found out in the ground
I can be bought by the pound
I'm put on the line

So in the end you can dine

I am a basic desire
That many require
I get you places
And contribute in races

Some find me childish
Some surpress me in kindness
But even if brief
After our encounter you'll feel a sense of relief

You have all you need
Now complete this one deed
What is it in my eyes
That will result in humanity's demise?

capacitance - Parallel RLC circuit, how branching currents may each be larger than source current at resonance?

How can the branching currents individually be greater than the source current?

fourier transform - Superposition of waves with different initial phase in Quantum Mechanics

In Quantum Mechanics, if a particle's state is a superposition of many states, then we say that its position is well-defined (by the Heisenberg uncertainty principle, because here we have ill-defined momentum). In the addition of each wave of momentum $p$ using Fourier series, where do we involve the initial phase of each wave that we are adding?
For example, say we add those two waves and get the function that I aforementioned. But if we add up those waves with only one of them having an initial phase of $\pi/12$ (making it $\exp[i(k x + \pi/12)]$ ), won't we get a function different from the one from the first addition? So we still add up the same states of definite momentum but get a different function.

Is there something wrong with my reasoning? Where do we involve the phase of each added wave(I believe it is something in complex analysis which I don't know about)? As someone stated at the comments, this at heart is a Fourier series question, so some mathematics indicating where the phase comes into play in Fourier series would be much appreciated.

NOTE: this question was edited so as to delete some things that I wrote which where wrong, so you might see some answers that try to explain something that might now be missing from the question but was present before.

Answer

There are some correct things and some incorrect things in your question. But let me just give the main points.

First of all, if you add two states together with different relative phases, you get different states. For instance,

$$\frac{1}{\sqrt{2}}|\psi_1\rangle + \frac{1}{\sqrt{2}}|\psi_2\rangle$$

is a different quantum-mechanical state than

$$\frac{1}{\sqrt{2}}|\psi_1\rangle + e^{i\pi/12}\frac{1}{\sqrt{2}}|\psi_2\rangle.$$

There are in fact a (continually) infinite number of different states that you can make out of $|\psi_1\rangle$ and $|\psi_2\rangle$,

$$\cos\theta|\psi_1\rangle + e^{i\phi}\sin\theta|\psi_2\rangle,$$

parameterized by the quantities $\theta$ and $\phi$.

For concreteness, and directly related to your question, if we work in the position representation, you can form the equal-amplitudes superposition of plane-wave states and make a delta-function, which can be thought of as a state of well-defined position (although such states do not actually physically exist). One possibility is

$$\delta(x) = \int\frac{dk}{2\pi}e^{ikx}$$

(In the abstract ket-space of position eigenvectors $|x\rangle$, this state would be the state $|0\rangle$.)

However, if you add a relative phase $e^{-ikx_0}$ to each plane-wave, then this becomes

$$\delta(x-x_0) = \int\frac{dk}{2\pi}e^{ikx}e^{-ikx_0},~~~~~~(1)$$

which in the abstract ket space is $|x_0\rangle$. To see this maybe a little more clearly, we can re-write the previous equations in the abstract ket space completely:

$$|0\rangle = \int{dk}|k\rangle.$$ $$|x_0\rangle = \int{dk}e^{-ikx_0}|k\rangle.~~~~~~(2)$$

In other words, we are adding up the momentum states $|k\rangle$ with different relative phases, so we get different states.

Most simply, if you add two plane waves with different relative phases, you obviously get different functions:

$$e^{ikx}+e^{-ikx}=2\cos{kx}$$

but

$$e^{ikx}+e^{i\pi}e^{-ikx}=2i\sin{kx}$$

Cleaning up some misconceptions

In Quantum Mechanics, if a particle's state is a superposition of many states of definite momentum, then we say that it's position is well-defined

This is not in general true. We say that the "position is well-defined" only if the superposition of momentum states is of a particular form (see equations (1) and (2) above). And it is always necessary to add this caveat: there really is no such thing as a state of well-defined position; it a mathematical convenience that, if used carefully, can be used as an approximation of a narrow wave-packet.

So say we add up two states of definite momentum (sine functions) so we get a function in position space.

This might just be an imprecise statement, but it's important to understand that a state is an abstract quantity that can be represented in either the position or momentum representation. The states of definite momentum are the eigenstates of the momentum operator, and when written in the position representation (i.e. as functions of $x$) take the form of plane waves:

$$|k\rangle = \int dx \langle x|\psi_k\rangle |x\rangle,\\ \langle x|\psi_k\rangle=\psi_k(x)=\frac{1}{\sqrt{2\pi}}e^{ikx}.$$

If you add up two momentum eigenstates, you can represent this function in position space, or you can represent this function in momentum space.

For example, say we add those two waves and get the function that I aforementioned. But if we add up those waves with only one of them having an initial phase of $\pi/12$ (making it $\exp[i(kx+\pi/12)]$), won't we get a function different from the one from the first addition?

Yep. See above.

So we still add up the same states of definite momentum but get a different function.

Yep! Nothing to see here really. I feel like this is the crux of the problem, but I can't tell what it is you're actually confused about.

...so some mathematics indicating where the phase comes into play in Fourier series would be much appreciated.

Under suitable niceness conditions of the functions, suitable boundary conditions etc., a complex-valued function $f(x)$ can be expanded in a Fourier series as

$$f(x) = \sum_{n=-\infty}^{\infty} A_n e^{i2\pi nx},$$

where the $A_n$'s are in general complex, so they can be represented in polar form as $re^{i\phi}$. And the point is, if you change those phases $e^{i\phi}$, you get out a different function $f$. That's it.

Final note

An overall complex phase is irrelevant. That is, $|\psi\rangle$ is the same state as $e^{i\phi}|\psi\rangle$, but as soon as you are forming superpositions, then the relative phases matter, as I've discussed.