What is the largest distance for which the influence of the electric field of a single electron was measured?

I suppose that Rydberg atoms are the best way to find by experiments the largest distance of influence of the electric field of a single electron in a electric dipole. Furthermore how the electrical positive charged nucleus influence this distance? Are there experiments with helium molecules or atoms with only one Rydberg electron?

Update: According to the below fig. from Wikipedia, how there can be a finite ionisation level if the electric field of an electron is infinite? It has to be an empiric value?

quantum field theory - The Spectral Function in Many-Body Physics and its Relation to Quasiparticles

recently, I stumbled accross a concept which might be very helpful understanding quasiparticles and effective theories (and might shed light on an the question How to calculate the properties of Photon-Quasiparticles): the spectral function $$A\left(\mathbf{k},\omega \right) \equiv -2\Im G\left(\mathbf{k},\omega \right)$$

as given e.g. in Quasiparticle spectral function in doped graphene (on arXiv).

It is widely used in many-body physics of interacting systems and contains the information equivalent to the Greens function $G$. For free particles, $A$ has a $\delta$-peaked form and gets broader in the case of interactions.

The physical interesting thing is, as I read, quasiparticles of interacting systems can be found if $A$ is also somehow peaked in this case. I don't understand this relationship, hence my question:

What is the relation of the spectral function's peak to the existence of quasiparticles in interacting systems?

Thank you in advance
Sincerely

Robert

Answer

Dear Robert, the answer to your question is trivial and your statement holds pretty much by definition.

You know, the Green's functions contain terms such as $$G(\omega) = \frac{K}{\omega-\omega_0+i\epsilon}$$ where $\epsilon$ is an infinitesimal real positive number. The imaginary part of it is $$-2\Im(G) = 2\pi \delta(\omega-\omega_0)$$ So it's the Dirac delta-function located at the same point $\omega$ which determines the frequency or energy of the particle species. At $\omega_0$, that's where the spectrum is localized in my case. If there are many possible objects, the $G$ and its imaginary part will be sums of many terms.

This delta-function was for a particle of a well-defined mass (or frequency - I omitted the momenta). If the particle is unstable, or otherwise quasi-, the sharp delta-function peak will become a smoother bump, but there's still a bump.

Because you didn't describe what you mean by "peak" more accurately, I can't do it, either. It's a qualitative question and I gave you a qualitative answer.

Cheers LM

quantum mechanics - Spin and Angular Momentum

What is the difference between spin and angular momentum, are they linked i.e. spin is a form of angular momentum. And is spin what I think it is (the way the electron/particle actually spins)

classical mechanics - Randomness v. complexity

There are a few other topics I found that explore this idea from a different perspective:

My question is on similar lines but from the perspective of complexity and within the framework of classical mechanics. It would seem to me that what we typically describe as "random", for example in probability theory, refers to the unpredictability within a complex system. A coin flip has many different initial conditions: the size and shape of the coin, the way the hand moves when it flips it, the material of the surface the coin lands on, the movement of molecules in the air etc. It also has many many interacting parts, many of which exist on the atomic scale, and whose relationship with other parts change dynamically. Probability is a mathematical model that allows us to tell what possible average outputs the system can produce given all the internal variables that are unknown to us. Is this a fair characterization?

However, if we were, in theory, to know all the initial conditions and had a way of describing all of the interactions of all variables in this dynamic system once it has begun, could we not predict the outcome (assuming theoretical computational power)?

My main question: Is there any kind of dynamic system in classical mechanics in which the outcome would not be predictable and, if so, why not?

Answer

Other answers have pointed to the fact that the question in classical mechanics has only limited applicability in real life (CuriousOne's comment in particular points out that this is at best an academic question), but let's discuss it out of curiosity nevertheless.

Most systems in classical mechanics are deterministic. Here is a simple heuristic: The equations of motion are governed by Newton's laws. If the forces do not depend on the time derivatives, this implies that the equations of motions can be seen as ordinary differential equations of second order. Those can be made into a system of first order differential equations with twice as many variables (the other set of variables is the velocities). Now the theorem of Picard-Lindelöff tells us that given initial conditions, i.e. positions and velocities, the trajectory is usually (meaning when the terms of the theorem are met) unique. This uniqueness is exactly what is meant by "determinism".

For most systems, the conditions of Picard-Lindelöff will hold almost everywhere because of general assumptions about the continuity of forces, however it is possible to construct examples where it fails and where explicitly non-unique solutions can be constructed. One such example is known as Norton's dome.

Qmechanic's beautiful answer on another StackExchange question illuminates why the conditions of Picard-Lindelöff are not met and why this system is indeed not deterministic.

The idea is the following: Given the potential $h(r)=−\frac{2}{3g}r^{3/2}$ and gravity, at the top of this dome, there are at least two solutions of the equation as demonstrated in this answer.

However, there is even another class of solutions (and I'll paraphrase Norton himself from now on):

$$h(t)=\begin{cases}(1/144) (t-T)^4 \quad t>T \\ 0 \quad t \leq T \end{cases}$$

In this solution, the ball at the top of the dome suddenly starts moving and this "suddenly" $T$ can be chosen as you wish. This of course violates determinism.

Why does it work? One heuristic is also given: Newton's equations are time reversible and the solution is the time reversed version where you start from the bottom of the dome and then move upwards and the particle has just enough momentum to reach the top, but not more. Since you can find many trajectories through the top of the dome with zero velocity at the top, this would indicate that we might have a uniqueness problem there.

But beware: It doesn't prove anything, because the same is true if you just consider the top of a hill shaped like a ball half - yet here we don't have a problem with determinism, as one can prove from the initial value problem. The reason is that all trajectories through the top of the hill with zero velocity at the top only reach the top at infinite times. This is NOT true for the Norton dome. Here, the ball reaches the top at finite times!

quantum field theory - The system is topological; so what?

Lately I've been studying topological quantum field theory, mainly from mathematically oriented sources and, while I understand why a mathematician would care about TQFT's, I'm not yet quite sure why a physicist would.

In particular, I'd like to know how TQFT's are used in particle physics. Say I begin with a standard QFT consisting of some gauge and matter fields. Somehow, I'm able to argue that in the infra-red, the theory flows to a Chern-Simons theory with gauge group $G$ and level $k$. Now what? What do I learn about the system once I realise its low-energy behaviour is described by a TQFT? What measurables can I study? What "real-life" predictions can I extract from the system?

Note: here I am only interested in (relativistic) particle physics. I know that TQFT's are also used in condensed matter and quantum gravity, but that's not what I want to understand here.

Answer

Any gapped field theory flows in the infrared to a TQFT which describes the set of operators which are neither screened nor confined. See this paper for a very clear point of view on this: https://arxiv.org/abs/1307.4793 .

In gapped gauge theories like QCD, this is equivalent to specifying the dyon content of the theory ( https://arxiv.org/abs/1305.0318 ). Different possible dyon charge lattices correspond to different topological phases for the center 1-form symmetry ( https://arxiv.org/abs/1308.2926 ).

This can cause some unexpected things to happen. For instance, as one tunes the theta angle, at special values there are topological phase transitions between confining vacua with distinct dyon charge lattices ( https://arxiv.org/abs/1703.00501 ) and this accounts for a singularity in the axion potential, which has consequences for the strong CP problem ( https://arxiv.org/abs/1709.00731 ).

Here is another nice paper about line operators in the Standard Model, which discusses the phenomenological issues quite clearly, in an attempt to first answer what is the center of the SM gauge group, eg. whether it's simply-connected or not ( https://arxiv.org/abs/1705.01853 ). One has to answer this question before asking the ones above about the topological phase of the associated 1-form symmetries.

To put it all from a more abstract perspective, there are no known unitary 3+1D TQFTs which are not some kind of twisted gauge theory. We know some non-abelian ones, but there are no-go results for realizing them in familiar systems of SU(n) gauge fields interacting with fermions ( https://arxiv.org/abs/1106.0004 ). Thus, all of the questions of topological order in the Standard Model are equivalent to questions about the strong CP problem and about the center of the gauge group.

But still we haven't seen a magnetic monopole...

homework and exercises - Make a semi transparent mirror with copper

The question:

How would you make a semi transparent mirror (50% reflection, 50% transmission) with glass with a layer of copper. For light $\lambda$ = 500nm Try to be as realistic as possible

What I've done so far.

The reflection coefficient of copper for 500nm light is 50%. So that's great. But I'm not sure whether that's enough. (It is an exam question, so I guess not). I'm guessing it'll have to do something with skin depth, but I'm not sure.

Any suggestions on how to continue ?

Answer

To a very good approximation the transmission of a metal film falls exponentially with thickness i.e.:

$$ T = e^{-\alpha t}$$

where $\alpha$ is the absorption coefficient given on the web site Alexander mentioned, http://refractiveindex.info/?group=METALS&material=Copper, and at 500nm wavelength this gives $\alpha = 6.4297\times 10^5/cm$. So you just have to solve for $T = 0.5$.

If you want to do the calculation properly it turns into a bit of a nightmare. By one of those strange co-incidences I did exactly this calculation as part of my PhD, and even more amazingly I have my thesis to hand. The reference I used for the calculation was O. S. Heavens, Optical Properties of Thin Solid Films, Butterworths Scientific Publications, London 1955. It's on Google Books here, but annoyingly hasn't been scanned so you can't see the contents.

I'll copy the equation for the optical transmission from my thesis, but I imagine one look will make you run for cover. I compared the full calculation to the simple exponential formula and agreement was basically perfect except at very small film thicknesses (below about 5nm) but in any case the metal films break up into islands at these thicknesses so the equation doesn't really apply.

$$ T= n_s \frac{((1 + g_1)^2 + h_1^2)((1 + g_2)^2 + h_2^2)}{e^{2\alpha} + (g_1^2 + h_1^2)(g_2^2 + h_2^2)e^{-2\alpha} + Ccos(2\gamma) + Dsin(2\gamma)}$$

where:

$$ C = 2(g_1g_2 - h_1h_2)$$ $$ D = 2(g_1h_2 + g_2h_1)$$ $$ g_1 = \frac{1 - n^2 - k^2}{(1+n)^2 + k^2} $$ $$ g_2 = \frac{n^2 - n_s^2 + k^2}{(n+n_s)^2 + k^2} $$ $$ h_1 = \frac{2k}{(1+n)^2 + k^2} $$ $$ h_2 = \frac{-2n_sk}{(n+n_s)^2 + k^2} $$ $$ a = \frac{2\pi k d}{\lambda} $$ $$ \gamma = \frac{2\pi n d}{\lambda} $$

where $k$ and $n$ are the extinction co-efficient and refractive index of the metal film and $n_s$ is the refractive index of the glass substrate. The film thickness is $d$ and the light wavelength is $\lambda$. If you do a sample calculation for some test film thickness you'll probably find most of the terms are approximately zero or unity, which is why it approximates to an exponential equation in the film thickness.

Bear in mind that I've hand copied this from my thesis, so there may be transcription errors lurking as traps for the unwary.

particle physics - Why is the Standard model Higgs not a candidate of dark matter (in particular, a WIMP)?

Please consider me as a naive self-learner in this field.

The Standard Model (SM) Higgs boson is electrically neutral and has a mass of around $125$ GeV (which lies in the WIMP window i.e., between 10 GeV to 1TeV). Why is then it is not a candidate of dark matter?

1. 
   

   Surely it can decay into SM fermions and gauge bosons. But are we sure that the decay modes are not kinematically forbidden or insignificant so as to explain the present relic abundance?

   
   


2. 
   

   As I said that the Higgs boson mass lies in the WIMP mass window. How is it then different from a WIMP?

   
   

Answer

Higgs would quickly decay to a mix of products, many of them electrically charged. Thus dark matter is neither Higgs nor former Higgs.

electrostatics - Does the induced charge on a conductor stay at the surface?

My textbook says that when a conductor is placed in an electric field, the electrons in it realign so that the net electric field inside the conductor is zero. There isn't a proof for this. It merely states that the electrons drift opposite to the applied electric field and create an opposite electric field inside the conductor canceling the external one.

I imagine Gauss's Law would provide a simple proof. If we try enclose a point inside the conductor in a gaussian surface, and note that the enclosed charge is zero, so the flux through the surface is zero and so is the electric field at the surface.

When I looked back at my reasoning, I realized that I had assumed that the charges induced on the conductor always stay at the surface, which I had no way of proving. How do I prove this formally?

special relativity - What is a Lorentz boost and how to calculate it?

I know very little special relativity. I never leaned it properly, but every time I read someone saying: "if you boost in the x-direction, you get such and such" my mind goes blank! I tried understanding it but always get stuck with articles that assume that the reader knows everything.

So, what is a Lorentz boost, and how to calculate it? And why does the direction matters?

Answer

Lorentz boost is simply a Lorentz transformation which doesn't involve rotation. For example, Lorentz boost in the x direction looks like this:

\begin{equation} \left[ \begin{array}{cccc} \gamma & -\beta \gamma & 0 & 0 \newline -\beta \gamma & \gamma & 0 & 0 \newline 0 & 0 & 1 & 0 \newline 0 & 0 & 0 & 1 \end{array} \right] \end{equation}

where coordinates are written as (t, x, y, z) and

\begin{equation} \beta = \frac{v}{c} \end{equation} \begin{equation} \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \end{equation}

This is a linear transformation which given coordinates of an event in one reference frame allows one to determine the coordinates in a frame of reference moving with respect to the first reference frame at velocity v in the x direction.

The ones on the diagonal mean that the transformation does not change the y and z coordinates (i.e. it only affects time t and distance along the x direction). For comparison, Lorentz boost in the y direction looks like this:

\begin{equation} \left[ \begin{array}{cccc} \gamma & 0 & -\beta \gamma & 0 \newline 0 & 1 & 0 & 0 \newline -\beta \gamma & 0 & \gamma & 0 \newline 0 & 0 & 0 & 1 \end{array} \right] \end{equation}

which means that the transformation does not affect the x and z directions (i.e. it only affects time and the y direction).

In order to calculate Lorentz boost for any direction one starts by determining the following values:

\begin{equation} \gamma = \frac{1}{\sqrt{1 - \frac{v_x^2+v_y^2+v_z^2}{c^2}}} \end{equation} \begin{equation} \beta_x = \frac{v_x}{c}, \beta_y = \frac{v_y}{c}, \beta_z = \frac{v_z}{c} \end{equation}

Then the matrix form of the Lorentz boost for velocity v=(v_x, v_y, v_z) is this:

\begin{equation} \left[ \begin{array}{cccc} L_{tt} & L_{tx} & L_{ty} & L_{tz} \newline L_{xt} & L_{xx} & L_{xy} & L_{xz} \newline L_{yt} & L_{yx} & L_{yy} & L_{yz} \newline L_{zt} & L_{zx} & L_{zy} & L_{zz} \newline \end{array} \right] \end{equation}

where

\begin{equation} L_{tt} = \gamma \end{equation} \begin{equation} L_{ta} = L_{at} = -\beta_a \gamma \end{equation} \begin{equation} L_{ab} = L_{ba} = (\gamma - 1) \frac{\beta_a \beta_b}{\beta_x^2 + \beta_y^2 + \beta_z^2} + \delta_{ab} = (\gamma - 1) \frac{v_a v_b}{v^2} + \delta_{ab} \end{equation}

where a and b are x, y or z and δ_ab is the Kronecker delta.

momentum - Use of Operators in Quantum Mechanics

I understand the form of operators in use for quantum mechanics such as the momentum operator: $$\hat{\text{P}}=-ih\frac{d}{dx}$$ My question is in what ways can I use it and what am I getting back? For example: if I simply apply th momentum operator to the wave function $$-ih\frac{d}{dx}\Psi$$ Will I get an Equation that will provide the momentum for a given position? Or is that a useless mathematical thing I just did?

If I use it to get an "expected value" by $$\langle \Psi | \hat{\text{ P }} | \Psi \rangle =\int_{-\infty}^\infty \Psi^* \hat{\text{ P } }\Psi$$ am I getting a number representing the probable momentum of that area integrated for? Or the percentage of total momentum there? Essentially is it a probability (if so of what kind?) or a value for the momentum?

I'm trying to understand these basic things because it has always remained unclear. I'm using it to find the momentum of and electron INSIDE the potential energy barrier as it is tunneling, i.e. the electrons between a surface and a Scanning Tunneling Microscope.

Answer

The first thing you did is useless--- multiplying by the momentum doesn't do very much. But if you multiply by functions of the momentum, you can do things like project out the part of the state with a certain momentum.

The momentum operator is most important because if you find its eigenvectors and eignevalues, these are the states of definite momentum.

The expected value is the average of many measurements of the momentum--- it is the average value of momentum measurements. It is given by the expression you wrote down, but only when you integrate over all space. You can't restrict the range of integration to find the momentum in a limited region.

quantum field theory - Differential geometry of Lie groups

In Weinberg's Classical Solutions of Quantum Field Theory, he states whilst introducing homotopy that groups, such as $SU(2)$, may be endowed with the structure of a smooth manifold after which they may be interpreted as Lie groups. My questions are:

• If we formulate a quantum field theory on a manifold which is also a Lie group, does that quantum field theory inherit any special or useful properties?


• Does a choice of metric exist for any Lie group?


• Are there alternative interpretations of the significance of Killing vectors if they preserve a metric on a manifold which is also a Lie group?

electromagnetism - Why is visible light used in Optical fibers (instead of other EM waves)?

Why aren't other electromagnetic waves used in optical fibres instead of visible light? Is it because the wavelength of light fits the internal reflection/refractive index of the material used for the fibre? e.g. Is the material that refracts light cheaper compared to material that refracts EM waves with other wavelengths?

Answer

Almost all fibers (and certainly all telecomms fibers) do not use visible light but infrared light. A typical wavelength used is 1550 nm. At this wavelength the combined effects of absorption and (Rayleigh) scattering in the fiber attains a minimum value, thereby allowing long transmission distances.

newtonian mechanics - If it was possible to dig a hole that went from one side of the Earth to the other...

...And you jumped in.

What would happen when you got to the middle of the Earth? Would you gradually slow down, until you got to the middle and once you were in middle would every direction feel like it was up?

nuclear physics - If we assume that protons don't decay, then will all matter ultimately decay into Iron-56 or into nickel-62?

Wikipedia says that all matter should decay into iron-56. But it also says Nickel-62 is the most stable nucleus.

So could this mean that in the far future, everything could (through quantum tunneling) fuse and/or decay into nickel-62 rather than iron-56?

Question inspired by an interesting comment made on a post here: http://www.quora.com/Do-atoms-ever-deteriorate-over-time/answer/Alex-K-Chen/comment/574730

Answer

The binding energy curve, again in wikipedia, shows iron as the one with the smallest binding energy per nucleon. Though in the table, the following is stated:

56Fe has the lowest nucleon-specific mass of the four nuclides listed in this table, but this does not imply it is the strongest bound atom per hadron, unless the choice of beginning hadrons is completely free. Iron releases the largest energy if any 56 nucleons are allowed to build a nuclide—changing one to another if necessary, The highest binding energy per hadron, with the hadrons starting as the same number of protons Z and total nucleons A as in the bound nucleus, is 62Ni. Thus, the true absolute value of the total binding energy of a nucleus depends on what we are allowed to construct the nucleus out of. If all nuclei of mass number A were to be allowed to be constructed of A neutrons, then Fe-56 would release the most energy per nucleon, since it has a larger fraction of protons than Ni-62. However, if nucleons are required to be constructed of only the same number of protons and neutrons that they contain, then nickel-62 is the most tightly bound nucleus, per nucleon.

One sees that there is a leeway when constructing models in end of the universe scenaria. There is so much speculation in the time lines. Observation tells us that Ni-62 is not abundant, while Fe is. It seems that in the sequence of supernova explosions iron wins out; according to the quote above, this would mean that it is the number of nucleons that is important and the charges statistically arrange themselves.

Anyway, in a continuously expanding universe with a stable proton it is hard to see how the expanding gases of Helium and Hydrogen can tunnel into anything as they expand so that "all matter" would end up as Fe or Ni atoms .

Not to worry, the proton will decay according to most current models of particle physics anyway.

general relativity - Technically, what is a spacetime singularity?

In popular science books and articles, one often finds that the BigBang is a singularity of spacetime, and it is expected to be solved by a successful theory of Quantum gravity.

Technically what is a spacetime singularity? What diverges and why?

I don't have much familiarity with General Relativity except an elementary knowledge of the Einstein's field equations and Schwarzschild solution. It will be great if someone can explain what is a spacetime singularity all about.

Answer

There is considerable technical difficulty in defining what a singularity is in general relativity.

For an "ordinary" function, we say that it has a singularity in $x_0$ if it is undefined in $x_0$. Of course, one may refine this definition to uninclude places where the function is not defined, but can be continued to that point (you take a regular function that is well behaved in $x_0$, now you remove the point from its domain, then it is singular in $x_0$; but you can put that point back into the domain and extend the function to that point by continuity and you're done).

For example the function $$ f(x)=\frac{1}{x} $$ is singular in $x=0$ .

---

Usually in a classical field theory, you can afford your fields not being defined in certain points. For example, if the field $\phi(x,y,z,t)$ represents some quantity, and it blows up in $(x_0,y_0,z_0,t)$, then we consider it simply not being defined there.

In general relativity however, the field theory is about the metric tensor field $g_{\mu\nu}(x)$, which defines the geometry of spacetime via the line element $ds^2=g_{\mu\nu}(x)dx^\mu dx^\nu$, so if $g_{\mu\nu}$ is misbehaving at a point $x_0$, you cannot just exclude that point from the metric's domain, because that would mean spacetime geometry itself is not defined there. Note however, that the misbehaviour of a metric at a point is not necessarily due to a physical singularity, it may be so that the coordinate system becomes ill-behaved at that point. This is another difficulty in determining what a singularity.

Continuing this thread, one usually constructs so-called curvature scalars and examines their behaviours. This is useful, because unlike tensors, whose components depend on the coordinate system used, the values of scalars do not depend on the coordinates. So you can construct scalars like $$ R,\ R_{\mu\nu}R^{\mu\nu},\ R_{\mu\nu\rho\sigma}R^{\mu\nu\rho\sigma}\ \text{etc.} $$ from the metric and see if these have a pathological behaviour at $x_0$ or not. If at least one of them blows up, or becomes undefined in any other means, then you have a physical singularity at that point.

This would be a good definition of a singularity, but there is a slight problem with it. Let $M$ be the spacetime manifold, and let $x_0$ be a point in which one of the curvature scalars blow up, so there is a singularity there. In this case, why don't I define spacetime to be $$ \tilde{M}=M\setminus\{x_0\}? $$ If I remove the offending point from spacetime, I'll have a singularity-free spacetime. No spacetime there, no singularity there, right?

Obviously, you can understand from intuition that this is total bollox. If we had a problem with $x_0$ before, then now we have a problem with spacetime "suddenly ending" at $x_0$. This problem cannot be handwaved away, because even if you make spacetime itself undefined at the offending point, then now you have a hole in spacetime. We can call that a singularity, but how does one define it?

One way to define these singularities is to define them to be a point where geodesics "disappear". Essentially points where you cannot continue a geodesic further. Even if you technically remove the offending point from the manifold, this definition remains valid.

---

Conclusion: A singularity is a point where spacetime geometry itself becomes ill-defined, because of the pathological behaviour of the metric and/or curvature functions. To avoid using technicalities to handwave these away, these are defined as places where geodesics suddenly terminate.

resource recommendations - Number theory in Physics

As a Graduate Mathematics student, my interest lies in Number theory. I am curious to know if Number theory has any connections or applications to physics. I have never even heard of any applications of Number theory to physics. I have heard Applications of linear algebra and analysis to many branches of physics, but not number theory.

Waiting forward in receiving interesting answers!

special relativity - Hypothetical effect of Solar System/Orion Spur traveling near speed of light

I was hoping that some genius could explain what we would observe in the universe around us in the following hypothetical scenario in which the earth itself were traveling at or near the speed of light. I am confident that somebody can tell me that my hypothetical situation cannot exist, but they are not really the point here, though I don't mind being proven dumb.

Scenario: Say that our solar system, jointly with the Orion Spur in which it resides, is traveling at near the speed of light as it prepares to merge with the adjacent Perseus Arm of our galaxy (some 5,000-10,000 ly away). How does that change how we see the Perseus Arm? What do we see as far as the rest of the Milky Way galaxy? How does our speed change what we see in distant galaxies? Or in other words, does the Milky Way galaxy as a whole seem to move faster or slower due to our speed? Do other galaxies seem to be moving faster or slower than their actual motion?

Hope that makes sense and hope somebody besides me finds it fascinating to consider light speed's impact on our perception.

EDIT1: I guess a component of this question is regarding red shifts and blue shifts, as well as how our observation of star evolution and galactic evolution would be impacted. Such as, would distant galaxies appear to be evolving faster or slower than actual and whether we would think they were moving toward or away from us, just due to our own near c motion?

EDIT2: I have reduced my scenarios to just one, as all responders are consistently telling me that such is better. Hope that the change aids the dialogue. My apology as a newby to this forum.

Physical meaning of the Jacobian in relation to Dirac delta function

Is there a physical meaning to the equation $$\delta(x-a)=\dfrac{\delta(\xi-\alpha)}{|J|} \, ?$$ In non-rectangular coordinate systems where the transformation is non-singular, what is the implication of dividing the Dirac delta function by the Jacobian of the transformation to the coordinate system?

semiconductor physics - Effective mass of a particle

I was reading about concept of effective mass on Wikipedia and came across the line that effective mass of a particle can be negative and even zero. How is this possible?

thermodynamics - Why can I touch aluminum foil in the oven and not get burned?

I cook frequently with aluminum foil as a cover in the oven. When it's time to remove the foil and cook uncovered, I find I can handle it with my bare hands, and it's barely warm.

What are the physics for this? Does it have something to do with the thickness and storing energy?

Answer

You get burned because energy is transferred from the hot object to your hand until they are both at the same temperature. The more energy transferred, the more damage done to you.

Aluminium, like most metals, has a lower heat capacity than water (ie you) so transferring a small amount of energy lowers the temperature of aluminium more than it heats you (about 5x as much). Next the mass of the aluminium foil is very low - there isn't much metal to hold the heat, and finally the foil is probably crinkled so although it is a good conductor of heat you are only touching a very small part of the surface area so the heat flow to you is low.

If you put your hand flat on an aluminium engine block at the same temperature you would get burned.

The same thing applies to the sparks from a grinder or firework "sparkler", the sparks are hot enough to be molten iron - but are so small they contain very little energy.

newtonian mechanics - Explanation that air drag is proportional to speed or square speed?

A falling object with no initial velocity with mass $m$ is influenced by a gravitational force $g$ and the drag (air resistance) which is proportional to the object's speed. By Newton´s laws this can be written as:

1. $mg-kv=ma$ (for low speeds)


2. $mg-kv^2=ma$ (for high speeds).

I assume that $k$ is a positive constant that depends on the geometry of the object and the viscosity. But how can one explain that the air resistance is proportional to the velocity? And to the velocity squared in the second equation?

Answer

One's naive expectation would be that as the object moves through the medium, it collides with molecules at a rate proportional to $v$. The volume swept out in time $t$ is $A v t$, where $A$ is the cross-sectional area, so the mass with which it collides is $\rho A v t$. The impulse in each collision is proportional to $v$, and therefore the drag force should be proportional to $\rho A v^2$, with a constant of proportionality $C_D$ (the drag coefficient) of order unity.

In reality, this is only true for a certain range of Reynolds numbers, and even in the range of Reynolds numbers for which it's true, the independent-collision picture above is not what really happens. At low Reynolds numbers you get laminar flow and $C_D\propto 1/v$, while at higher Reynolds numbers there's turbulence, and you get $C_D$ roughly constant.

newtonian mechanics - When does centripetal force cause constant circular motion?

As far as I know, if an object has a speed $s$ and a force is applied to it which generates an acceleration of $s^2/r$, then the object will start moving in a circle of radius $r$. Also, as far as I understand, this circular motion is constant (the radius will remain constant at all times).

However, I've seen physics problems that go something along the lines of a person stands on a merry-go-round, going at a certain tangential speed, find the friction needed for the person to not slip off, and the correct answer would be

$$\mu N m = m \frac{s^2}{r} \implies \mu = \frac{s^2}{N r}$$

Given what I've said in the first paragraph, I don't understand why the person would slip off. If the acceleration is equal to $s^2/r$, why is friction needed?

Answer

To remain on the merry-go-round, the person must be accelerated towards the center. How is the force applied to the person to provide that acceleration? The static friction provides a way for the floor of the merry-go-round to force the person along the circular path. If the floor were to suddenly become frictionless and the person was not otherwise attached, the person would continue moving along a line tangent to the merry-go-round.

general relativity - How do you tell if a metric is curved?

I was reading up on the Kerr metric (from Sean Carroll's book) and something that he said confused me.

To start with, the Kerr metric is pretty messy, but importantly, it contains two constants - $M$ and $a$. $M$ is identified as some mass, and $a$ is identified as angular momentum per unit mass. He says that this metric reduces to flat space in the limit $M \rightarrow 0$, and is given by $$ds^2 = -dt^2 + \frac{r^2 + a^2 \cos^2\theta}{r^2 + a^2}dr^2 + \left(r^2 + a^2 \cos^2\theta \right)d\theta^2 + \left(r^2 + a^2 \right)\sin^2\theta d\phi^2 $$

and $r$, $\theta$ and $\phi$ are regular spherical polar co-ordinates.

But I don't understand why this space is obviously flat. The Schwarzschild metric also contains terms involving $dt^2$, $dr^2$, $d\theta^2$ and $d\phi^2$ but is curved. I always thought that a metric with off diagonal elements implied a curved space, but clearly I was very wrong.

Question: How do you tell if a metric is curved or not, from it's components?

Answer

You tell if a space (or spacetime) is curved or not by calculating its curvature tensor. Or more unambiguously one of the curvature scalars (e.g. Ricci, or Kretschmann) since these don't depend on the coordinate system, but all of the information in the scalars is also contained in the Riemann tensor.

It is not necessarily obvious whether a given metric is curved or flat. You can take a perfectly flat spacetime and express it in some bizarre coordinate system, in which the metric has nonconstant off-diagonal terms. It's a simple exercise to take flat space and use the tensor transformation laws for the metric, with some arbitrary weird coordinate transformation that you just made up. You'll see what I mean.

homework and exercises - Find center of mass and moment applied on beam structure

I have a simple physics problem to solve but it is giving me a slightly difficult time to figure out. The problem:

I have a beam structure with same cross section. It consists of three beam. First in vertical with length 1meter. Second beam joined on top with a angle of 20degrees to the right with lenth 1.1meter and on top of the second beam, a third beam parallel to the ground with 1meter length.

I need to find the center of gravity and how much moment I have on the base of the structure so I would add a mass that would be sufficient to avoid the structure to fall over.

any ideas how?

Answer

I need to find the center of gravity

Center of mass (CoM) is found by adding up all particles $i$:

$$\vec r_{cm}=\frac{\sum m_i \vec r_i}{\sum m_i}$$

$\vec r_{cm}$ is the coordinate vector of the CoM.

Since you have straight beams (assumingly cylinder shaped), start out with symmetry considerations for each of them. The CoM for each beam must be at the exact center (assuming constant density throughout). Put these three CoMs into the formula above to find the combined CoM of the whole structure.

A side note: If the gravitational acceleration $\vec g$ is constant throughout, then the CoM is equal to the center of gravity.

and how much moment i have on the base of the structure so I would add a mass that would be sufficient to avoid the structure to fall over.

Formula for torque about an axis (the tip of the base point):

$$\vec \tau=\vec r_{cm} \times \vec F$$

In your 2D-case simplified to:

$$\tau=r_{cm,\perp} F=r_{cm} F \sin \theta$$

where $r_{cm} \sin \theta$ must be $r_{cm}$'s horizontal component (since gravity $\vec F$ is vertical).

Find this torque of gravity on the structure, and then place an object on the base in whatever manner the problem allows so it gives the opposite torque equal in magnetude.

When photons are emitted, do they accelerate to reach the speed of light?

Photons are considered mass-less particle with a specific velocity but according to the electromagnetic theory, a photon is considered to have both energy and momentum. So what happen when they are emitted, do they accelerate to reach the speed of light or is it somehow instantaneous?

general relativity - Is it foolish to distinguish between covariant and contravariant vectors?

A vector space is a set whose elements satisfy certain axioms. Now there are physical entities that satisfy these properties, which may not be arrows. A co-ordinate transformation is linear map from a vector to itself with a change of basis. Now the transformation is an abstract concept, it is just a mapping. To calculate it we need basis and matrices and how a transformation ends up looking depends only on the basis we choose, a transformation can look like a diagonal matrix if an eigenbasis is used and so on. It has nothing to do with the vectors it is mapping, only the dimension of the vector spaces is important.

So it is foolish to distinguish vectors on the way how their components change under a co-ordinate transformation, since it depends on the basis you used. So there is actually no difference between a contravariant and covariant vector, there is a difference between a contravariant and covariant basis as is shown in arXiv:1002.3217. An inner product is between elements of the same vector space and not between two vector spaces, it is not how it is defined.

Is this approach correct?

Along with this approach mentioned, we can view covectors as members of the dual space of the contra-vector space. What advantage does this approach over the former mentioned in my post?

Addendum: So now there are contra variant vectors and their duals called covariant vectors. But the duals are defined only once the contravectors are set up because they are the maps from the space of contra vectors to $R$ and thus, it won't make sense of to talk of covectors alone. Then what does it mean that the gradient is a covector ? Now saying because it transforms in a certain way makes no sense.

quantum field theory - How to prove $(gamma^mu)^dagger=gamma^0gamma^mugamma^0$?

Studying the basics of spin-$\frac{1}{2}$ QFT, I encountered the gamma matrices. One important property is $(\gamma^5)^\dagger=\gamma^5$, the hermicity of $\gamma^5$. After some searching, I stumbled upon this interesting Phys.SE answer to an earlier question on this forum. Specifically, I am interested in the formula \begin{equation} (\gamma^\mu)^\dagger=\gamma^0\gamma^\mu\gamma^0 \end{equation} which is mentioned but not proven. After consulting a faculty member of my university, I pieced together that the proof must rely somehow on the fact that the $(\gamma^\mu)^\dagger$ also obey the Clifford algebra: $$\{(\gamma^\mu)^\dagger,(\gamma^\nu)^\dagger\}=-2\eta^{\mu\nu}$$ $$\{\gamma^\mu,\gamma^\nu\}=-2\eta^{\mu\nu}$$ (for clarity, I am using $- + + +$ signature for the Minkowski metric). This should imply that there is some similarity transformation relating the two, but I am not well-versed in group theory. I guess that it should somehow turn out that the matrix that acts to transform the two representations of this algebra into each other is $\gamma^0$, which is equal to its inverse $\gamma^0=(\gamma^0)^{-1}$, as can be seen immediately from taking $\mu=\nu=0$ in the Clifford algebra. Then, the similarity transform is in the right form:

$$ (\gamma^\mu)^\dagger=S\gamma^\mu S^{-1}=\gamma^0\gamma^\mu\gamma^0 $$

I have the feeling I've got most of the necessary ingredients. However, I can't seem to be able to make this argument explicit and clear (due to my lack of proper knowledge of group theory). Could someone help me out? It would be much appreciated.

EDIT: I am looking for an answer that does not rely on using a particular representation of the gamma matrices.

soft question - What is a good introductory text to astronomy

What is a good and easy to read introductory text for an adult with limited basic scientific knowledge to astronomy for someone without a telescope and lives in a big city and why do you think that text is good for a beginner?

Answer

A new edition of Discovering the Universe was just published this spring, so it's a good time to buy. My PhD advisor is the managing author of the series now. He is really crazy-focused on not glossing over sticky details that make most textbook authors get sloppy and write descriptions that almost but not quite correct. I respect that. It's for his Astronomy 101 class, which services the lab-science requirements for liberal arts majors and had no prereqs whatsoever. It ranges over just about everything you'd be interested in. He also keeps really careful tabs on how many moons and exoplanets have been discovered and updates all the tables in the back with the very latest in research results. (Although those fields are progressing so rapidly that any printed page goes stale fast.) Lots of really majestic telescopic images. No or not much content on doing visual observations yourself.

Carroll & Ostlie is a comprehensive introduction to the math and physics of astronomy, but will require some mathematical background knowledge through basic calculus. I mention this only because it's not clear just how limited your "limited basic scientific knowledge" is. There's no visual component to it at all, so you won't be hampered by your big city life. I used this in my second-year undergrad astrophysics course for astrophysics majors.

quantum field theory - The derivation of the Belinfante-Rosenfeld tensor

• 
  

  It seems me that there is a "difference" (at least apparently) in how the Belinfante-Rosenfeld tensor is thought of in section 7.4 of Volume 1 of Weinberg's QFT book and in section 2.5.1 of the conformal field theory book by di Francesco et. al.

  
  

  I would be glad if someone can help reconcile the issue.

  
  

Schematically the main issue as I see is this -

• 
  

  If the action and the lagrangian density is writable as $ I = \int d^4x L$ and $\omega_{\mu \nu}$ be the small parameter of Lorentz transformation then Weinberg is thinking of $\omega_{\mu \nu}$ to be space-time independent and he is varying the action to write it in the form, $\delta I = \int d^4x (\delta L = (A^{\mu \nu})\omega_{\mu \nu}) $ Then some symmetrized form of whatever this $A^{\mu \nu}$ comes out to be is what is being called the Belinfante tensor. ( Its conservation needs the fields to satisfy the equations of motion)

  
  


• 
  

  But following Francesco et.al's set-up I am inclined to think of $\omega_{\mu \nu}$ as being space-time dependent and then the variation of the action will also pick up terms from the Jacobian and the calculation roughly goes as saying, $\delta I = \int (\delta(d^4x)) L + \int d^4x (\delta L)$. Since under rigid Lorentz transformations the volume element is invariant the coefficient of $\omega_{\mu \nu}$ in the first variation will vanish but the second variation will produce a coefficient say $B^{\mu \nu}$. But both the variations will produce a coefficient for the derivative of $\omega_{\mu \nu}$ and let them be $C^{\mu \nu \lambda}$ and $D^{\mu \nu \lambda}$ respectively.
  Now the argument will be that since the original action was to start off invariant under Lorentz transformations, when evaluated about them, the $B^{\mu \nu}$ should be $0$ and on shifting partial derivatives the sum of $C^{\mu \nu \lambda}$ and $D^{\mu \nu \lambda}$ is the conserved current (..and its not clear whether their conservation needs the field to satisfy the equation of motion..)

  
  

So by the first way $A^{\mu \nu}$ will be the conserved current and in the second the conserved current will be, $C^{\mu \nu \lambda} + D^{\mu \nu \lambda}$ (the C tensor will basically look like $-x^\nu \eta^{\lambda \mu}L$)

Is the above argument correct?

If yes then are the two arguments equivalent?

How or is Weinberg's argument taking into account the contribution to the conserved current from the variation of the Jacobian of the transformation?

Answer

The two derivations are actually identical, except for the fact that Weinberg didn't have the general form of the Noether theorem for symmetries acting on the space-time coordinates as well as on the fields (Equation 2.141 in Di Francesco, Mathieu and Sénéchal's book).
As a consquence, Weinberg had to compute the variation of the action with respect to the Lorentz generators from scratch (including the substitution of the equations of motion).

Furthermor, I wanted to remark that the term depending on the variation of the space time coordinates in the general form of the Noether theorem is not due to the noninvariance of the Minkowski space time measure $d^4x$ as this measure is invariant under both translations and Lorentz transformations. The extra term is due to the dependence of the Lagrangian on the space time coordinates through its dependence on the fields.

Now, both authors use the derivation as a means of computation of the Belinfante & Rosenfeld 3-tensor whose divergence is to be added to the canonical energy momentum tensor to obtain the symmetric Belinfante energy momentum tensor. The principle upon which this computation is based is that the orbital part of the canonical conserved current corresponding the Lorentz symmetry must have the form:

$M^{\mu\nu\rho} = x^{\nu} T_B^{\mu\rho} - x^{\rho} T_B^{\mu\nu} $

with $T_B$ both conserved and symmetric (as can be checked by a direct computation), therefore, they arrange the extra-terms they obtained to bring the Lorentz canonical current to this form and as a consequence they obtain the required tensor to be added.

I wanted to add that both authors use the derivatives of the symmetry group parameters in their intermediate computations, but this is not required. The same currents can be obtained for variation with respect to global constant parameters. If the action were locally invariant (with respect to variable parameters), then the currents would have been conserved off-shell. This is the Noether's second theorem.

Finally, I want to refer you to the this article of Gotay and Marsden describing a method of obtaining a symmetric and (gauge invariant) energy-momentum tensor directly based on Noether's theory.

fluid dynamics - Why using average pressure in calculations gives the most accurate results?

In the saga of trying to answer the question that came into my mind while studying the basic concepts of Fluid Mechanics, "Why textbooks use geometric center to calculate hydrostatic pressure when presenting pressure gauges?", after asking it to my professor of Fluid Mechanics, consulting ~20 textbooks, asking it as a part of this question, asking it on Engineering.SE by the recommendation of a Physics.SE user, I didn't gave up and finally found an answer while reading the comments on this YouTube video:

How a Piezometer Works by Donald Elger

Why is it [the pressure measurement with piezometer] taken from the middle of the pipe?

Elger's answer: The pressure variation across a section of a pipe is hydrostatic; thus, the pressure will vary linearly with radius and the pressure at the center of the pipe is the average pressure. If you use this value of pressure in your calculations, this will be give you the most accurate results. Thus, engineers nearly always apply or measure the pressure at the center of the pipe.

---

With this new information, a new question arose: Why average pressure gives the most accurate results if used in calculations?

Answer

I also asked this question on Quora and started sending requests. Someone answered it. I'll post the answer.

On reading the context for this question, i.e the best location for measurement of pressure along a pipe and why it is the center, it helps to revisit the fundamentals of pipe flow. Essentially the center of the pipe has zero shear stress since the velocity profile is typically symmetric and almost no turbulent shear. If you trace the centerline of the pipe, you would see that the total pressure at the inlet is converted to a mix of static pressure and kinematic pressure, with almost no losses. This is not true near the wall, where there are viscous losses in the boundary layer region and there might be significant turbulence or reverse flow. So the center of the pipe is a cleaner place to read the total pressure or static pressure. Of course, the sensor will cause disturbance in the flow which needs to be accounted for.                                                                                                                           Author: Roopesh Mathur

---

I built an example to complement Roopesh's answer and give an example of the "calculations" that Elger's answer mentions.

Consider an experiment in which a Pitot tube is used and there is a flow with velocity profile given by: $$v(h)=V_{max}\cdot\left(1-\frac{\left|h-R\right|}{R}\right)^{1/7} ,\space 0\leq h\leq2R \space\space$$ Note that $v(h)=v(2R-h)$, so the velocity profile is symmetric, with axis of symmetry passing through $h=R$. Our goal is to determine $V_{max}$. Below is an image that illustrates the experiment:

                                                                              _{(Adapted from Fluid Mechanics - Yunus A. Çengel & Cimbala)}

The Pitot tube can measure the stagnation pressure at a point, where $P_{stag}=P+\rho\frac{v^2}{2}$. If a piezometer is used in conjunction with a Pitot tube, it's possible to calculate the fluid velocity at a specific location, using the static pressure $P$ at of this location, measured with the piezometer, and the stagnation pressure at that location, measured with the Pitot tube:

$$v=\sqrt{\frac{2(P_{stag}-P)}{\rho}}$$ Since $v=v(h)$ , by the velocity profile formula, we have:

$$\left. \begin{array}{r} v=v(h)\\ P=P(h)=\gamma\cdot (h+k)\\ P_{stag}=P_{stag}(h)=P(h)+\rho\frac{v^2(h)}{2} \end{array}\right\} v(h)=\sqrt{\frac{2\left[P_{stag}(h)-P(h)\right]}{\rho}} $$ To determine $V_{max}$ it's necessary to obtain the velocity at a specific height – using the Pitot tube, the piezometer and the Pitot velocity formula – and then replace the experimental value found in the velocity profile formula. At first, we can choose any height to do the measurements!

Roopesh's answer tell us the best height to be chosen in order to get the most accurate result: the height of the pipe centerline ($h=R\space$ in my example), because there we have "zero shear stress" and "almost no turbulent shear". Furthermore in this height there are "almost no losses" in the total pressure. Then, we have:

$$v(R)=\sqrt{\frac{2\left[P_{stag}(R)-P(R)\right]}{\rho}}=\sqrt{\frac{2\left[P_{stag}(R)-P_{average}\right]}{\rho}}$$

And this confirms what Elger said:

If you use this value [average pressure] in your calculations, this will be give you the most accurate results.

---

So, in general, the average pressure gives the most accurate results if used in calculations because there are many applications/cases in which the locations with $P=P_{average}$ are the best places for experimental data collection.

black holes - Nonlinear refraction index of vacuum above Schwinger limit

This question is more about trying to feel the waters in our current abilities to compute (or roughly estimate) the refraction index of vacuum, specifically when high numbers of electromagnetic quanta of a given frequency $\omega$ occupy a spherical symmetric incoming wavefront modes $$\frac{e^{i k r}}{r^2}$$

I'm interested in intensities above the Schwinger limit. Do exist analytical or lattice QED estimates?

Why this is interesting?

Usually i try to make my questions as self-contained as possible, but i believe it might be interesting to others why i'm interested in nonlinear vacuum refraction indices, so here it goes:

Let's review what classical theory says about our ability to create micro black holes with electromagnetic radiation. Our sun produces about $10^{26}$ watts of power, so in principle in the future we could harness all that power. The question trying to be answered here is: is the energy output from Sol enough for a sufficiently technically advanced humanity to create micro black holes?

Let's suppose we focus a spherically symmetric beam of light in a single focal point with the purpose of condense enough energy to create a black hole, well, the Schwarzchild radius of a given flow of energy is

$$ R = \frac{ G E }{c^4}$$

substituing constants,

$$ R = 10^{-45} E $$

Now, since this energy propagates electromagnetically as radiation, it needs to stay long enough inside the critical radius so that the black hole forms. If the (radial) refractive index inside a small enough region is $n$, then the light will stay inside the radius a time $T$

$$ R = \frac{cT}{n}$$

equating both terms we are left with an expression for the power that needs to be delivered in a spherical region in order to create a black hole

$$ \frac{cT}{n} = 10^{-45} E $$

$$ \frac{10^{53}}{n} = \frac{E}{T} = P $$

So, assuming a refractive index of vacuum of 1, it means that we require $10^{53}$ watts focused in a spherical region in order to create a black hole. This is $10^{27}$ more power than what would be available for a humanity that managed to create a Dyson shell around our sun!

But if the refractive index could be managed to be above $10^{30}$ before reaching the focus region, then that future humanity would have a chance to create such micro black holes

Even a less generous increase of the refractive index could be useful, if they could store the energy of the sun for a few years and then zapping it into an extremely brief $10^{-20}$ wide pulse, in a similar fashion as to how the National Ignition Facility achieves 500 TeraWatt pulses (by chirp-pulse compression)

newtonian mechanics - 3-body system centre of mass

I have a three body system of point masses that represent mercury, earth and the sun. I want them to orbit about a common centre of mass, but I think the centre of mass will move. I need the centre of mass to always remain at the origin.

If I move the bodies a certain amount, find the new position of the centre of mass and subtract it from the new positions of the planets, does that mean the centre of mass is still at the origin? Or do I also need to do some calculation using the velocity of the centre of mass?

electromagnetism - Displacement current - how to think of it

What is a good way to think of the displacement current? Maxwell imagined it as being movements in the aether, small changed of electric field producing magnetic field. I don't even understand that definition-assuming there is aether. (On the topic of which, has aether actually been disproved? I read that even with the Michelson-Morley experiment the aether wasn't disproved.)

Answer

Maxwell's equations in a vacuum have induction terms. (1) There is a term saying that a time-varying magnetic field produces an electric field. (2) There is a term saying that a time-varying electric field produces a magnetic field.

Among people who insist on giving hard-to-remember names to all the terms in Maxwell's equations, #2 is called the displacement current. The name is a bad one, because it's not a current, i.e., it has nothing to do with the motion of charged, material particles. The only reason it has the misleading name is that it adds to the current term, and Maxwell, who made up the name, wasn't sure what its ultimate origin was.

The importance of term #2 is mainly that it allows the existence of electromagnetic waves. In an electromagnetic wave, the changing E field induces the B field, and the changing B field induces the E field.

There are elementary reasons that term #2 has to exist. For example, suppose you have a circular, flat Amperian surface $S_1$ and you shoot a single charged particle perpendicularly through its center. In this situation, Maxwell's equations without term #2 predict that the magnetic field at the edge of the surface will be zero, then infinite for an instant, and then zero again after that. But if we construct a similar Amperian surface $S_2$ with the same boundary but an interior surface that is bowed out rather than flat, we get a prediction that the infinite field occurs at a different time. This proves that we can't get away with leaving Maxwell's equations in a form with all the terms except term #2.

The deeper reason for term #2 is that it's required by relativity. Only with term #2 do Maxwell's equations have a form that is the same in all frames of reference.

measurement problem - In the double slit experiment, what happens when there is more than one observer trying to see which slit the same electron goes through?

From all the demonstrations Iv been able to find of Heisenberg's double slit experiment, whenever an observer tries to "see" which slit an electron passes through it collapses the wave function. My question is what would happen when you have two observers or detectors trying to detect the same electron, will they see the same thing? does the wave function only collapse to one possibility no matter who or what the observer?

special relativity - The origin of the value of speed of light in vacuum

Meaning, why is it the exact number that it is? Why not 2x10^8 m/s instead of 3? Does it have something to do with the mass, size or behavior of a photon? To be clear, I'm not asking "how we determined the speed of light". I know there isn't a clear answer, I'm really looking for the prevailing theories.

quantum explanation of doppler effect

How would quantum mechanics explain doppler effect?

And just for curiosity, is there any effect similar to doppler effect occuring at quantum level?

homework and exercises - Vectors, Component Addition, and Significant Figures

I have two vectors $\vec{A}$ and $\vec{B}$ and I need to find the x- and y-components of $\vec{C} = \vec{A} + \vec{B}$. Here's what I have so far:

$$|\vec{A}| = 50.0 \mathrm{m}, \theta = -20.0^\circ$$ $$|\vec{B}| = 70.0 \mathrm{m}, \theta = 50.0^\circ$$

$$C_x = |\vec{A}| \cos (\theta) + |\vec{B}| \sin (\theta)$$ $$C_y = |\vec{A}| \sin (\theta) + |\vec{B}| \sin (\theta)$$

Now, according to my professor, this is the solution for C_x:

$$C_x = 50.0 \cos(-20.0^\circ) + 70.0 \cos(50.0%^\circ)$$ $$C_x = 46.98^\circ + 45.0^\circ$$ $$C_x = 92.0^\circ$$

What I'm wondering is how the rounding works here. I got $46.98$ for $50.0 \cos(-20.0^\circ)$ and $44.99$ for $70.0 \cos(50.0^\circ)$. Why is $44.99$ rounded to $45.0$? If anything, shouldn't it be rounded to $45.00$? What am I missing here?

Answer

I think the real question is actually posed most directly in your comment (so you might want to consider editing some of this into the original question):

I'm more concerned about understanding whats going on and making sure that I know how to do it. I've heard that you shouldn't worry about significant figure rules until you have your final answer. How many decimal places should you round a number like 50.0 cos(-20.0) to? Do you always round to 2 decimals as in my problem?

Yes, you are correct that you should never actually round a number off until you are done with the calculation. However, when you are writing out your intermediate steps, it's common practice to write rounded values, rather than copying every digit your calculator shows you, just to avoid burdening the reader with a lot of extra digits that don't really add anything interesting. Keep in mind that this convention only affects what you write. You still keep the number to full precision in your calculator.

As for choosing the number of digits to write out, you can use the significant figure rules, which go like so:

• 
  

  Addition and subtraction: find the last significant digit of each number, and choose the one with the larger place value. That place should be the last significant digit of your result. Another way to think of this rule is that a digit in the sum (or difference) is not significant unless both digits that were added to produce it are significant. So, using gray shading to designate insignificant digits:

  
  

  $$\begin{align*}3&.146309\\+2&.71\\ =5&.85\color{red}{6309}\end{align*}$$

  
  

  So you would round to the last significant digit in this case, i.e. you would write out $5.86$. But if you use this result again:

  
  

  $$\begin{align*}5&.85\color{red}{6309}\\+4&.93101\\ =10&.78\color{red}{7319}\end{align*}$$

  
  

  This time you would again round to the last significant digit, and write out $10.79$.

  
  


• 
  

  Multiplication and division: your result should have the fewest number of significant digits of either of the numbers you're multiplying or dividing.

  
  

  $$\begin{align*}&253.1\\\div &45\\ = &5.6\color{red}{2444\ldots}\end{align*}$$

  
  
  

  and if you multiply this by the earlier result,

  
  

  $$\begin{align*}&5.6\color{red}{2444\ldots}\\\times &10.78\color{red}{7319}\\ = &60.\color{red}{67268\ldots}\end{align*}$$

  
  

These rules are a simplification of a slightly more complex (but more accurate) system, the error propagation rules, which physicists normally use in research. Unfortunately, the error propagation rules for functions like the sine and cosine can't be simplified quite so easily, so in practice people often just use the multiplication/division rule (write out the fewest number of significant digits) for everything else not mentioned here. Of course, you have to remember that, except for final results, it's really not that important how many digits you write, since you should never be rounding "behind the scenes" in your calculator anyway.

statistical mechanics - What does Liouville's Theorem actually mean?

Basically, the mathematical statement of Liouville's theorem is:

$$\frac{\partial \rho }{\partial t}= -\sum_{i}\left(\frac{\partial \rho}{\partial q_i}\,\dot{q_i}+\frac{\partial\rho}{\partial p_i}\,\dot p_i\right)$$

While I could comprehend the derivation which is nicely done in Reif's Fundamentals of Statistical and Thermal Physics, I could not get what this theorem actually wants to imply.

The Wikipedia article mentions:

It asserts that the phase-space distribution function is constant along the trajectories of the system [...]

What does this mean?

What does the word trajectory mean in the present context?

Is $\rho$ not a function of time?

Can anyone please clarify what that quoted line actually means?

Answer

$$\frac{\partial \rho }{\partial t}= -\sum_{i}\left(\frac{\partial \rho}{\partial q_i}\,\dot{q_i}+\frac{\partial\rho}{\partial p_i}\,\dot p_i\right)$$

This means that if we have a function of $t, p, q$ namely $\rho(t,\vec p,\vec q)$ and we have a trajectory that is a curve in $(p,q)$ space, namely $q_i(t), p_i(t), i=1\ldots N,$ then:

$$ \frac{\mathrm d}{\mathrm dt} \rho(t, \vec p(t), \vec q(t)) =\frac{\partial \rho }{\partial t}+ \sum_{i}\left(\frac{\partial \rho}{\partial q_i}\,\dot{q_i}+\frac{\partial\rho}{\partial p_i}\,\dot p_i\right) $$

How if $\rho$ is constant along trajectories, then LHS is 0 and the equation you have written follows directory.

So:

• a trajectory is any curve in 2N dimensional space described in $q_i$ and $p_i$ coordinates


• $\rho$ is a function of both time and $\vec q$ and $\vec p$



• whole concept is just an application of a chain rule.

quantum electrodynamics - How are classical optics phenomena explained in QED (Snell's law)?

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

• Reflection and Refraction

Are they simply due to photons being absorbed and re-emitted? How do we get to Snell's law, for example, in that case?

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

Answer

Hwlau is correct about the book but the answer actually isn't that long so I think I can try to mention some basic points.

Path integral

One approach to quantum theory called path integral tells you that you have to sum probability amplitudes (I'll assume that you have at least some idea of what probability amplitude is; QED can't really be explained without this minimal level of knowledge)) over all possible paths that the particle can take.

Now for photons probability amplitude of a given path is $\exp(i K L)$ where $K$ is some constant and $L$ is a length of the path (note that this is very simplified picture but I don't want to get too technical so this is fine for now). The basic point is that you can imagine that amplitude as a unit vector in the complex plane. So when doing a path integral you are adding lots of short arrows (this terminology is of course due to Feynman). In general for any given trajectory I can find many shorter and longer paths so this will give us a nonconstructive interference (you will be adding lots of arrows that point in random directions). But there can exist some special paths which are either longest or shortest (in other words, extremal) and these will give you constructive interference. This is called Fermat's principle.

Fermat's principle

So much for the preparation and now to answer your question. We will proceed in two steps. First we will give classical answer using Fermat's principle and then we will need to address other issues that will arise.

Let's illustrate this first on a problem of light traveling between points $A$ and $B$ in free space. You can find lots of paths between them but if it won't be the shortest one it won't actually contribute to the path integral for the reasons given above. The only one that will is the shortest one so this recovers the fact that light travels in straight lines. The same answer can be recovered for reflection. For refraction you will have to take into account that the constant $K$ mentioned above depends on the index of refraction (at least classically; we will explain how it arises from microscopic principles later). But again you can arrive at Snell's law using just Fermat's principle.

QED

Now to address actual microscopic questions.

First, index of refraction arises because light travels slower in materials.

And what about reflection? Well, we are actually getting to the roots of the QED so it's about time we introduced interactions. Amazingly, there is actually only one interaction: electron absorbs photon. This interaction again gets a probability amplitude and you have to take this into account when computing the path integral. So let's see what we can say about a photon that goes from $A$ then hits a mirror and then goes to $B$.

We already know that the photon travels in straight lines both between $A$ and the mirror and between mirror and $B$. What can happen in between? Well, the complete picture is of course complicated: photon can get absorbed by an electron then it will be re-emitted (note that even if we are talking about the photon here, the emitted photon is actually distinct from the original one; but it doesn't matter that much) then it can travel for some time inside the material get absorbed by another electron, re-emitted again and finally fly back to $B$.

To make the picture simpler we will just consider the case that the material is a 100% real mirror (if it were e.g. glass you would actually get multiple reflections from all of the layers inside the material, most of which would destructively interfere and you'd be left with reflections from front and back surface of the glass; obviously, I would have to make this already long answer twice longer :-)). For mirrors there is only one major contribution and that is that the photon gets scattered (absorbed and re-emitted) directly on the surface layer of electrons of the mirror and then flies back.

Quiz question: and what about process that the photon flies to the mirror and then changes its mind and flies back to $B$ without interacting with any electrons; this is surely a possible trajectory we have to take into account. Is this an important contribution to the path integral or not?

What is the definition of a "UV-complete" theory?

I would like to know (1) what exactly is a UV-complete theory and (2) what is a confirmatory test of that?

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  Is asymptotic freedom enough to conclude that a theory is UV-complete?

  
  

  Does it become conclusive a test if the beta-function is shown to be non-perturbatively negative?

  
  

  If the one-loop beta-function is negative then does supersymmetry (holomorphy) immediately imply that the beta-function is non-perturbatively negative and hence the theory is proven to be UV-complete?

  
  


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  Or is vanishing of the beta-function enough to conclude that theory is UV-complete?

  
  

  Again similarly does supersymmetry (holomorphy) guarantee that if the 1-loop beta function is vanishing then it is non-perturbatively so and hence the theory is UV-complete?

  
  


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  Does superconformal theory necessarily mean UV-complete? We know that there exists pairs of superconformal theories with different gauge groups - related by S-duality - which have the same partition function - what does this say about UV-completeness?

  
  
  


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  Does a theory being UV-complete necessarily mean that it has a Maldacena dual? (...and isn't that the meaning of AdS/CFT that every UV-complete theory has a string dual and vice-versa?..)

  
  

Answer

A UV complete theory is one whose correlation functions or amplitudes may be calculated and yield unambiguously finite results for arbitrarily high energies.

Yes, asymptotic freedom is enough for UV completeness because the UV limit of the theory is free and therefore well-defined. Whenever the coupling constant is small and the beta-function is negative at one-loop level, the higher-loop corrections should be small so that the exact beta-function should be negative, too. Such implications become even easier to make with SUSY.

Yes, exactly scale-invariant (and especially conformal) theories are UV complete if they're consistent at any scale because they predict the same thing at all scales due to the scale invariance. However, scale invariance at the leading order doesn't imply exact scale invariance. So no, one-loop vanishing of the beta-function may be coincidental and – even in a SUSY theory – the full beta-function may still have both signs.

Yes, superconformal theories are a subset of conformal theories so they're UV-complete. The S-duality exchanges descriptions with a different value of the coupling which is a different quantity, and therefore an independent operation/test, from the UV-completeness and scale invariance that relates different dimensionful scales.

It is believed that all conformal theories must have "some" Maldacena dual in the bulk although whether this dual obeys all the usual conditions of a "theory of quantum gravity" or even a "stringy vacuum" is unknown, especially because we can't even say what all these conditions are. The non-AdS/non-CFT correspondence would in principle work for all UV-complete theories but it's much less established and more phenomenological than the proper AdS/CFT correspondence that works with the exactly conformal theories only.

homework and exercises - Lasers : Threshold Pump Power for Laser Oscillation

I was working my way through some basic laser problems , when I cam across this one :

Consider the ruby laser for which we have the following values of the various parameters:

$N =$ $1.6$ x $10^{19}$ $cm^{-3}$ ; $t_{sp}$ = $3$ x $10^{-3}$ $s$ ; $v_p$ = $6.25$ x $10^{14}$ $Hz$

• Find the threshold pump power for laser oscillation , $P_t$ = $Nh\nu$ $/$ $2$$t_{sp}$



• If we assume that the efficiency of the pumping source to be $25 $% and also that only $25$% of the pump light is absorbed on passage through the ruby rod, then the electrical threshold power comes out to be how much ?

I am able to solve the first part but am all thumbs in the second part - can someone help me out ?

^{Disclaimer: This is not a homework question . I am preparing for a physics exam and was solving these questions / examples from the book recommended by my instructor.}

Answer

The efficiency of the pumping source is $x$ means that $x$ amount of electrical power is converted to energy which is useful for pumping the laser medium. The absorption of the pump is $y$ means that $y$ amount of the energy from the pump source is actually pumped into the medium to generate the population inversion necessary for lasing. The total amount of electrical power $P_E$ which makes it into the laser medium is therefore $x\ y\ P_E$.

forces - Mathematical Reasoning for Fluid Pressure as a Scalar

This question from a while ago and answers/comments to this question from earlier today both make heavy mention of the fact that fluid pressure is a scalar. Although this information was surprising to me, it makes sense intuitively, especially when considering Pascal's Principle and liquid depth pressure. However, I struggle when thinking about the mathematical reasons why pressure is a scalar, and the answers to the first question linked either appealed to intuition, seemed contentious, or unsatisfying.

My confusion comes from the fact that pressure is often defined at the high school/undergrad level as $$P\equiv \frac{F}{A}$$ where this not just a relationship, but often presented as the equation definition of pressure. Force is a vector, and I had assumed that area was a scalar. Therefore, I had always figured this was a scalar multiplication problem, and the product would also be a vector. Now that I know that pressure is a scalar, I just can't see how you would get a scalar for pressure out of this relationship. I see two possibilities:

1) For reasons I don't currently understand, area is treated as a vector in continuum physics. As a result, some type of vector algebra must be applied to divide a vector by another vector in order to yield a scalar. If this is the case, I would like to see at least a framework for this math in the answer.

2) If possibility 1 is not the case, then the other possibility in my mind is that $P=\frac{F}{A}$ is not truly the definition of pressure, and there exists some more basic definition for pressure that would explain its scalar nature.

Are either of these explanations correct, or is there a third possibility that I have not considered?

I've tried to make this a substantially different question than either of the other two questions that I've linked to, but if that does not appear to be the case, please let me know.

Answer

I would say pressure is better defined by $$ \vec{F} = P \vec{A}. $$ Yes, we are defining a quantity without having it all alone on the left-hand side. And yes, area is a vector. And as you guessed trying to divide one vector by another leads to trouble, so we won't do it.

Let me explain where this comes from and what it is shorthand for. In continuum mechanics we are often interested in flows of some quantity or another. In particular, we might want to consider flows of momentum, where these flows can go in any direction.

Let's work in Cartesian coordinates for simplicity. Then one can imagine $x$-, $y$-, or $z$-momentum flowing in the $x$-, $y$-, or $z$-direction, where the two coordinates are chosen independently. That is, there can be $x$-momentum moving in the $x$-direction, or it can be moving in the $y$-direction, or whatever. The rate of flow of $i$-momentum ($i$ being $x$, $y$, or $z$) per unit area in the $j$-direction is the $ij$-component of the stress tensor $T$: $T_{ij}$. And the time-rate-of-change of momentum is just another way of saying "force."¹

The way this tensor works is you choose a vector area $\vec{A}$. This is a vector with magnitude given by the same scalar everyone is familiar with, and also with direction given by the direction orthogonal to the surface. The application of $T$ to $\vec{A}$ gives another vector,² which is the force acting on the surface with area $\vec{A}$.

Now stress tensors appear in all sorts of places in physics, but they all behave similarly. In particular, their diagonal components are things like the $x$-force in the $x$-direction, which is your intuition for pressure: it directly pushes on things, as opposed to dragging them like shear. In fact, the off-diagonal elements are the shear forces per unit area.

In an isotropic medium, like a well-behaved fluid, the diagonal elements are in fact all the same: "$x$-pressure" is identical to "$y$-pressure." Call these values $P$. In the absence of shear, the stress tensor has a particularly simple representation: $$ T = \begin{pmatrix} P & 0 & 0 \\ 0 & P & 0 \\ 0 & 0 & P \end{pmatrix}. $$ Then clearly for any area $\vec{A}$ we have $\vec{F} = T \vec{A} = P \vec{A}$.

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¹Many treatments would just omit the whole talk of momentum I gave and express everything up to now as forces (per unit area). But the momentum way of looking at things works particularly well for generalizing the 3-dimensional stress tensor to the 4-dimensional stress-energy tensor used in relativity, and this is my field of study. Just thinking of forces rather than changes in momentum over time might work better in statics problem (there is still momentum flow, but it's hard to visualize since "nothing" is moving).

²If you are not familiar with tensor manipulations, it suffices here to think of the components $T_{ij}$ arranged as a matrix, and to think of the components $A_j$ arranged in a column vector. Then matrix multiplication tells you the components of the column vector $T \vec{A}$ are $F_i = T_{ij} A_j$ (summing over repeated indices).

radiation - Why doesn't diamond glow when hot?

In an answer to this SE question, the respondent explains that heating a perfect diamond will not cause it to glow with thermal blackbody radiation. I don't quite follow his explanation. I think it comes down to: there is no mechanism for diamond to generate light in the visible region of the spectrum.

He mentions that interband transitions are well out of the visual range, so there will be no contribution from that.

He mentions that the Debye temperature for diamond is > 2000 K. I presume that the argument here is that optical phonons will be frozen out, too. (But diamond doesn't have infrared-active phonons, does it?)

So is that why hot diamond doesn't glow?

I suppose that if one considers real (not ideal) crystals, imperfections, impurities, and the existence of surfaces lead to the possibility of emission mechanisms, and thus glow. In fact it might be the case that a finite but otherwise perfect crystal might have an extremely faint glow.

Is this basically the reason that hot diamond does not glow? Further elucidation welcome.

Calculating the mechanical power of a water pump

Say I want to pump water from one container to another. The water levels are 3 meters apart, and I want to pump 10 litres per hour. I figure the mechanical power necessary, assuming no losses, is:

$$ \require{cancel} \dfrac{10\cancel{l}}{\cancel{h}} \dfrac{kg}{\cancel{l}} \dfrac{\cancel{h}}{3600s} = \dfrac{0.0028kg}{s} $$

$$ \dfrac{0.0028kg}{s} \dfrac{3m}{1} \dfrac{9.8m}{s^2} = \dfrac{0.082kg\cdot m^2}{s\cdot s^2} $$

$$ \dfrac{0.082\cancel{kg}\cdot \cancel{m^2}}{\cancel{s}\cdot \cancel{s^2}} \dfrac{\cancel{J} \cdot \cancel{s^2}}{\cancel{kg} \cdot \cancel{m^2}} \dfrac{W\cdot \cancel{s}}{\cancel{J}} = 0.082W $$

But, I know from practical experience that real centrifugal pumps that can work at a 3m head are big and certainly require orders of magnitude more electrical power. What explains the difference?

Intuitively, I figure this must be because the pump must exert some force to balance the force of gravity from pushing water backwards through the pump, siphoning the water back to the lower container, then exert yet more force to accomplish what was desired, pumping to the higher container.

1. How is this force calculated mathematically?


2. Assuming an ideal electric centrifugal pump, can we establish the electrical power required by the pump, given the difference in heights between the containers?


3. Does this apply to all pumps, or just centrifugal pumps?

Can fusion and fission happen at the same time, in the same place?

Can fusion and fission happen at the same time, in the same place? I was talking with a friend, and he thinks that fission and fusion happen at the same time at the sun, is that true? I guess this would cause perpetual motion, then I speculated that they could happen, but in different proportions and I believed that fusion is happening in a higher proportion than fission. What do you think?

I've searched on wikipidia, and I still found no mention to fission happening in the sun.

newtonian mechanics - Purely mechanical description of how gravity causes a gyroscope to precess

I know the vector equation that relates torque to moment of inertia and angular momentum. What I want to know is what physical mechanisms actually occur to keep the gyroscope from falling. Where is the upward force coming from and how, mechanistically, does this get produced?

classical mechanics - Why is the d'Alembert's Principle formulated in terms of virtual displacements rather than real displacements in time?

Why is the d'Alembert's Principle formulated in terms of virtual displacements rather than real displacements in time?

EDIT In other words, which step of the derivation of D'Alembert's principle (or Largrange's equation) will not work if one uses real displacements and why?

quantum mechanics - Is uncertainty principle a technical difficulty in measurement?

Is the uncertainty principle a technical difficulty in measurement or is it an intrinsic concept in quantum mechanics irrelevant of any measurement?

Everyone knows the thought experiment of measuring the position of an electron. One can detect electron's position by hitting it with a photon, due to Compton scattering the collision of the photon with electron will change electron's momentum. This experiment is used to explain uncertainty principle to layman, but it is over simplified, isn't it? It also gives an impression that if there was a better suited experimental method the uncertainty principle becomes irrelevant.

I personally think it is intrinsic as it arises from the non-zero commutator of position and momentum operators irrespective of the measurement process. Am I right?

EDIT: My question is similar to certain extent to this question and this question. The answers there are nice but they focus on explaining basics of quantum mechanics more than they comment on the technical difficulty part. In the answers of question 2 there are statements like "So, it's not a knowledge limit" and "you're sort of correct when you say it's an observational limit" without further comments

To summarize, assume hypothetically we managed to find a way in the future where we can have a look at an electron without disturbing it by measurement or causing its wave function to collapse, would the uncertainty principle still hold in such a case?? Why/Why not?

Answer

Yes, the experiment is oversimplified, because the uncertainty principle is not about "disturbance through measurement". Although that's what Heisenberg said (one of the things he said), it turned out you can't interpret it that way in a very rigorous sense.

Whether there is something like "disturbance through measurement" that gives rise to an uncertainty relation is currently under heated debate in the quantum foundations community (see the work of Ozawa and recently some collaborators on the one hand and the work of Busch, Lahti and Werner on the other, if you want, I can look up some references).

That said, your opinion is correct in the sense that this is exactly how you derive the uncertainty relation. With position and momentum, one could ask the question "but why don't position and momentum commute" and then one can turn to the Fourier transformation and remark that the uncertainty relation is something that holds for any wave (water, electromagnetic, etc.), because the Fourier transform tells us that a small wave packet must consist of a lot of frequencies and a wave with only one frequency is infinite in space, etc. Now, since we have wave functions, we have this phenomenon in quantum mechanics as well. This means that indeed the uncertainty principle in our formalism does not require any measurement, it is an intrinsic property of the wave function in phase space.

EDIT: Even with your extended question, assuming everything else would hold, the uncertainty principle should still be there. It just tells you that the product of the variances of momentum and position are lower bounded, which comes from the wave-function itself. There is no reference to any measurement in the uncertainty principle other than that you need to measure to actually compare anything.

Being more concrete, I would say the following: Given a state of an electron (i.e. a preparation scheme that prepares the exact same physical electron over and over again), you can measure momentum and you will obtain a probability distribution according to the wave function (repeating the experiment multiple times). Then, assuming you have no disturbance in measurement, you measure the position of the exact same state. In that case, this will also have some probability distribution according to the wave function. These two measured probability distributions have variances whose product is lower bounded. That's what the uncertainty principle tells you.

The question of whether or not you could have information without disturbance (at least asymptotically) is still a matter of debate...