quantum mechanics - Introducing a phase, what changes?

This question is related to: Mach-Zehnder interferometer and the Fresnel-Arago laws

Let us say we have unpolarised wave taking the form: $$\psi=\psi_0 e^{i(kx-\omega t)+i\phi(t)}$$ Where $\phi$ varies randomly with time. If I split this wave into two and send it through e.g. a double slit, one of the beams will experience a phase change due to an optical path length difference. When we combine these two waves one will take the form: $$\psi=\psi_0 e^{i(kx-\omega t)+i\phi(t)}$$ But what about the other?

Their are 3 possibilities: $$\psi=\psi_0 e^{i(k(x+x_0)-\omega t)+i\phi(t)}$$ $$\psi=\psi_0 e^{i(kx-\omega (t+t_0))+i\phi(t+t_0)}$$ $$\psi=\psi_0 e^{i(k(x+x_0)-\omega (t+t_0))+i\phi(t+t_0)}$$
Where $x_0$ and $t_0$ are constants. Which of these 3 is correct and why?

Answer

The two waves are interfering after having followed different paths, so $x$ must be different between the two. But you are observing them at the same time $t$ which must be the same for the two waves. So answer 1 is the good one.

newtonian mechanics - Does a car use friction to move?

When a car's engine injects fuel into the cylinder chambers, the reaction creates a force that generates rotational momentum to the shaft and over the transmission, it translates that power to the wheels, right?

But, something's bothering me, how does the car actually move, is it dependent on the force of friction? Wheels have a contact with the road in one exact point and when the drive gets to rotate them, a force is exerted on the road in the direction of movement and there is friction that is counteracting it.

Is the key to get a car properly moving to have the drive force equal or less than the force of friction to keep that point locked to the ground and use the rolling of the wheel for translational displacement ie. moving forward? If it's bigger than the force of friction, that would cause the wheels to spin in place?

Am I completely off or am I getting something right? If there's a knowledgeable person on this topic, I'd greatly appreciate some insight, perhaps even a bigger response with some basic vehicle physics.

Does the Hilbert space of the universe have to be infinite dimensional to make sense of quantum mechanics?

Does the Hilbert space of the universe have to be infinite dimensional to make sense of quantum mechanics? Otherwise, decoherence can never become exact. Does interpreting quantum mechanics require exact decoherence and perfect observers of the sort which can only arise from exact superposition sectors in the asymptotic future limit?

quantum mechanics - Determine $p_x$ from $[x,p_x]=ihbar $

With $[x,p_x]=i\hbar $, how to determine the form of the operator $p_x$?

nuclear physics - Why are the dineutron and diproton unbound?

It is known that there are no diproton or dineutron nuclei.

Does this mean that two protons or neutrons are not actually attracted to each other? Even if the attraction was weak, wouldn't it cause bound states anyway?

Related: What do we know about the interactions between the protons and neutrons in a nucleus?

Answer

The nucleon-nucleon interaction has a short range, roughly 1 fm. Therefore if there were to be a bound dineutron, the neutrons would have to be confined within a space roughly this big. The Heisenberg uncertainty principle then dictates a minimum uncertainty in their momentum. This amount of momentum is at the edge of what theoretical calculations suggest the strong nuclear force could successfully fight against. Experiments in 2012 give evidence that the dineutron may be weakly bound, or that it may be a resonance state that is close enough to bound to create the same kind of strong correlations in a detector that you would get from a dineutron. So it appears that the strong nuclear force is not quite strong enough, but this is not even clear experimentally.

If the dineutron isn't bound, the diproton is guaranteed not to be bound. The nuclear interaction is the same as in the dineutron, by isospin symmetry, but in addition there is an electrical repulsion.

waves - Would passing horizontally polarized light through a varying width vertical slit allow you to measure the positional (x) amplitude of light?

I have found closely related questions on StackExchange, but (surprisingly) not this exact question. Seems some answers say individual photons do not have amplitude, only when traveling with other photons, forming a wave. But even then, can't the amplitude be measured? If so, what are the range of amplitudes and how do they vary? Based on the frequencies and energies of the light? Does higher frequency correspond to lower amplitude?

Moderators: This question is not a duplicate to this question: Amplitude of an electromagnetic wave containing a single photon

That prior question is more about the formula for calculating electromagnetic energy amplitude of a single photon. My question is about an experimental way to MEASURE the positional (x) amplitude of the wave passing through a specific apparatus. Two very different things. Thanks.

I have updated my question to a more specific thought experiment:

If you had a vertical slit with a horizontal width that could be varied, and you passed horizontally polarized light through this slit, wouldn't the slit block any photons from passing through it if the width of the slit was smaller than the amplitude of the light waves?

Wouldn't we at least get a predicted or average amplitude, even with uncertainty principle?

Wouldn't the energy/frequency/wavelength of the wave also have some correlation to the amplitude?

Answer

Classically (since rob has done a thorough job on the quantum picture), the amplitude of a light wave is not related to any physical extent. It is not the size of the wave in space, it is the strength of the fields (electric and magnetic).

We often draw wavy lines, but if you look closely the transverse axes will be label differently for, say, waves on a string and electromagnetic waves. You should not take those lines to imply a displacement the way they do in ripple on a pond. They just mean differing values of the field.

Classically, you can not filter different amplitudes with slit width. You simple block more light and create more diffraction as the slit grows narrower.

gravity - Orbital mechanics of Dragon's Egg

In the novel Dragon's Egg, the human crew use one asteroid to swing other asteroids in place to counter the gravity of the neutron star. I understood that it was similar to a gravity sling shot, but I wasn't able to fully get how the crew were able to move the smaller asteroids in place using the big one. Can anyone explain that further?

Answer

I know nothing of this book, but I do know a little about N-body gravitational interactions. When N >= 3, you can do just about anything you want with a little propulsion, but it may take a very long time. This has been proposed by NASA (good approximate for hard SF) as a way of sending probes to the outer solar system: http://en.wikipedia.org/wiki/Interplanetary_Transport_Network

As far as the stability questions, it is known that you can put 7 or 8 equal-mass objects on a stable circular orbit around their center of mass: http://adsabs.harvard.edu/abs/1988A%26A...205..309S

After introducing an external mass and its accompanying tidal forces, I imagine there is still some stability regime, which could be pretty close to the NS, especially if the "asteroids" have WD density, their constellation could be on the order of 1km, while the NS radius is probably closer to 15km.

maxwell equations - Is magnetic reconnection reconcilable with magnetic field lines neither starting nor ending?

According to Maxwell's equations, magnetic fields are divergence-free: $\nabla \cdot \mathbf{B} = 0$. If I understand this correctly, this means that magnetic field lines do not start or end. How can we reconcile this with magnetic reconnection?

Answer

One must be very careful in making the step from $\nabla\cdot\mathbf{B}=0$ to a statement such as "magnetic field lines do not start or end".

Consider the field in the region of an X-point type magnetic null (in two dimensions). Take a 'volume' (i.e. an area) centred on the null point, and look at the field lines through the bounding curve. No matter how small you make the volume, you will see an equal number of field lines of equal strength entering and leaving the volume.

At the point of reconnection (in an idealised case) the field lines 'start' and 'end' at an infinitesimal point. Even in the limit that your volume for the purposes of the calculation tends to zero (which defines the scalar field of divergence), you will still have equal flux 'into' and 'out of' that volume.

Note the sentence in this source, where it is stated that "[f]an field lines and spine field lines are notable exceptions to the general tenet that field lines have no beginning or ending – it seems that certain field lines terminate at null points." There is however, as discussed above, no violation of the condition that the field be divergence-free.

Edit: With the amount of attention this post is getting, I feel I should add a couple of points of clarification.

• In no sense am I saying that any such thing as a 'magnetic monopole' exists at a reconnecting X-point. In the resistive MHD picture, at an infinitesimal spatial point and for an infinitesimal time, magnetic field lines essentially lose their identity when they pass through the reconnection region. It makes no sense to talk about 'tracing' a field line across the X-point as we normally do when we plot maps of field lines. All we can say for sure is that the flux into and the flux out of a sufficiently small (formally infinitesimal) volume around the X-point are equal, satisfying $\nabla\cdot\mathbf{B}=0$.


• The sense in which the field lines 'terminate' at the reconnection is a corollary to this; because we can't identify any particular path which carries us smoothly across the X-point along a particular field line, we're forced to admit a discontinuity. This is why MHD equilibrium solvers for example use certain computational tricks to 'skirt round' the X-point in a given configuration rather than modelling the field all the way to the discontinuity.


• The foregoing discussion is valid only as long as the resistive MHD picture is valid; once we get down to scales comparable to the electron gyro-radius, the whole thing requires a self-consistent kinetic approach.

special relativity - Energy-Momentum Tensor under Lorentz Transformation

In relativity, the symmetric energy-momentum tensor is given by $$ T^{ij}, $$ where $T^{00}$ is the energy density and $\frac{1}{c}T^{10}$ is the momentum density. Thus: $$ \left(\frac{1}{c}T^{00}dV, \frac{1}{c}T^{10}dV\right)^{T}$$ is the 4-momentum. Under a Lorentz transformation, this should transform like 4-vectors where $$ \frac{1}{c}T^{00}dV= \left[\frac{1}{c}T'^{00}dV'+\frac{v}{c^2}T'^{10}dV'\right] \left( 1-\frac{v^2}{c^2}\right)^{-1/2}\\dV=dV'\sqrt{1-\frac{v^2}{c^2}}.$$ After simplifications, we have: $$ T^{00}= \left[T'^{00}+\frac{v}{c}T'^{10} \right] \left( 1-\frac{v^2}{c^2}\right)^{-1}$$ But if we apply the Lorentz transformation to the tensor directly we get $$ T^{00}= \left[T'^{00}+\frac{v}{c}T'^{10}+\frac{v^2}{c^2}T\ ^{11} \right]\left( 1-\frac{v^2}{c^2}\right)^{-1}$$ What accounts for the difference? I think the first is wrong but have no idea why.

newtonian mechanics - Is acceleration an absolute quantity?

I would like to know if acceleration is an absolute quantity, and if so why?

Answer

In standard Newtonian mechanics, acceleration is indeed considered to be an absolute quantity, in that it is not determined relative to any inertial frame of reference (constant velocity). This fact follows directly from the principle that forces are the same everywhere, independent of observer.

Of course, if you're doing classical mechanics in an accelerating reference frame, then you introduce a fictitious force, and accelerations are not absolute with respect to an "inertial frame" or other accelerating reference frames -- though this is less often considered, perhaps.

Note also that the same statement applies to Einstein's Special Relativity. (I don't really understand enough General Relativity to comment, but I suspect it says no, and instead considers other more fundamental things, such as space-time geodesics.)

experimental physics - Radioactivity and quantum superpositions

In the Schrödinger's cat experiment 'there is a tiny bit of radioactive substance, so small that perhaps in the course of the hour, one of the atoms decays'. The rest of the experiment magnifies this into a macroscopic superposition, but I want to know more about the claim that the radioactive decay produces a superposition.

Firstly, has this been experimentally tested? Something along the lines of accelerating radioactive ions so there is a chance that they will decay while in flight (and so change trajectory), and then combining the decayed and undecayed parts to look for interference.

Secondly, the tiny bit of radioactive substance will still contain large numbers of atoms. Won't this cause problems? If the atoms were in a Bose-Einstein condensate, then I would expect that there could be a superposition of 'one (unspecified) atom decayed' and 'no atoms decayed', but they're not, so a specific atom will decay. Won't that mess things up?

photons - Why do lasers require mirror at the ends?

Laser uses mirrors to reflect photons in order to stimulate atoms to emit photons, but why this is so?. I mean, why does a photon stimulate atoms to produce more photons? If a photon made an atom to produce a photon, is not the first photon absorbed by the atom and therefore there is not total gain in the process?

Answer

Lasers are pumped, so that many electrons already have excited states. In such a situation, when a photon passes through the active material, it's unlikely to be absorbed because there's not enough electrons in non-excited states. Much more likely result is that, by mechanism of stimulated emission, excited electrons can lose their energy, producing a duplicate of the photon which passed by. This happens when passing photon has energy very close to energy of radiative transition from excited state to non-excited.

Mirrors make up the optical cavity, which makes photon travel multiple times across the active material, thus making more and more photons duplicate, resulting in exponential growth of electromagnetic energy, which partially exits the cavity because one (or both) of mirrors are not 100% reflective. And the light which exits is the usable light from laser.

Elements and their plasma state

I'm looking for a list of nearly all elements and the temperatures they can reach when in a state of plasma. It is for a project I am working on. However, there isn't a list on google.

Preferably one that is hot enough to survive a drop from orbit.

quantum mechanics - How do electron configuration microstates map to term symbols?

I am trying to understand energy levels of electron configurations. I visited the NIST web site and discovered that the notation used here are called term symbols.

After reading corresponding wikipedia entry I worked through the carbon example in the "Term symbols for an electron configuration" section. So it appears to me that each of the $\binom{t}{e}$ microstates [(in this case $\binom{6}{2}=15$), where $t$ is number of slots in the outside subshell, and $e$ is the number of electrons] will each have an assigned energy level, or term symbol - since in the end there are 15 1s distributed in three different matrices, but only 5 possible term symbols (1 term symbol for 5 microstates, 3 term symbols for 9 microstates, and 1 term symbol for 1 microstate). However, I don't see how a given microstate maps to a specific term symbol.

For example, $M_L = 0$ and $M_S = 0$ is true for 3 microstates, but how to determine which term symbols correspond to them? Does it even make sense to do this?

UPDATE:

Thanks gigacyan, for the detailed answer. By the wording I am not sure at the end if you mean that $M_L = 0$ and $M_S = 0$ can only have the $^1S$ term. If that is true, then the following must be totally off, but I will take a shot anyway:

So are you saying that ANY given microstate cannot be assigned a term symbol (energy level), or that just certain microstates (such as the $M_L = 0$ and $M_S = 0$ case above) cannot be assigned since there are more than one term symbol possibility?

For example, it seems that when there is only one microstate for a given $M_L$ and $M_S$ combination, it is uniquely determined as long as it falls into a matrix that has only one term symbol - for carbon, say $M_L = 2$ and $M_S = 0$ (row 7 of 15 in the microstate table) then it falls into the 5x1 table, which must be $^1D_2$.

But when there is only one microstate for a given $M_L$ and $M_S$ combination but it falls into a matrix with more than one term symbol - for carbon, say $M_L = 1$ and $M_S = 1$ (row 1 of 15 in the microstate table), it falls into the 3x3 matrix, which means that this microstate must be one of $^3P_2$, $^3P_1$, or $^3P_0$, but it is not known which. Is this correct?

at $T approx 0 , text{K}$, will all energy levels within the electronic band structure be occupied up to a certain level?

I saw this from the script of my teacher that I don't understand what does it mean

If we cool down a crystal to an absolute temperature of T ≈ 0K, all atoms of the crystal will exist at their ground states and all energy levels within the electronic band structure will be occupied up to a certain level

At the absolute zero, all electron is at its lowest energy level so the higher level is empty so why all energy levels within the electronic band structure will be occupied up to a certain level. I don't understand what it means. The higher level is empty so why all energy level (including higher level) is occupied at a certain level?

electrostatics - Is potential difference or potential used in defining capacitance?

In my textbook I came across the capacitance of a certain body (i.e. a sphere, not two different spheres as in a spherical capacitor) and in it the formula,

$$Q = CV$$

where $V$ is the potential of the body with respect to the Earth. Now in a parallel plate capacitor, why do we choose the potential difference and not the potential of a single plate to the Earth?

Answer

V in th either situation is potential difference but in the case of an isolated sphere as written in Halliday/Resnick ( Indian edition)

We can assign a capacitance to a single isolated spherical conductor of radius R by Assuming that the " missing plate " is a conducting sphere of infinite radius

So the potential on single sphere comes out to be potential difference

temperature - Thermal Radiation of human body

I've read that all bodies emit electromagnetic radiation of all wavelengths. Does this mean that humans emit gamma rays too (excluding the radiation from radionuclides present in humans)?

quantum field theory - Derive Schwinger-Dyson equations in Srednicki

In eq. (22.20) on p. 135 in Srednicki he defines the functional integral

$$Z(J) = \int\mathcal{D}\phi\,\exp\Big[\mathrm{i}\big(S+\int\mathrm{d^4}y \,J_a\phi_a\big)\Big], \tag{A}$$

where $S$ and $J_a$ are the action and sources respectively (sum over $a$). What I don't get is that when he in eq. (22.21) considers a small variation $\delta Z$ he seem to get the variation of the action inside an integral (I get it without the integral) as follows:

$$0=\delta Z(J) = \mathrm{i}Z(J) \times \Bigg[\int \mathrm{d^4}x\Big(\,\frac{\delta S}{\delta \phi_a(x)}+J_a(x)\Big)\delta \phi_a(x)\Bigg].\tag{B}$$

My attempt:

$$0=\delta Z(J) = \frac{\delta Z}{\delta\phi_b(x)}\delta\phi_b(x)\\[10mm]=\int \mathcal{D}\phi\,\delta\phi_b(x)\Bigg[\frac{\delta}{\delta\phi_b(x)}\mathrm{e}^{\mathrm{i}(S+\int\mathrm{d^4}y\,J_a(y)\phi_a(y))}\Bigg].\tag{C}$$

The box becomes:

$$\Bigg[\frac{\delta}{\delta\phi_b(x)}\mathrm{e}^{\mathrm{i}(S+\int\mathrm{d^4}y\,J_a(y)\phi_a(y))}\Bigg] = \frac{\delta}{\delta\phi_a(y)}\mathrm{e}^{\mathrm{i}(S+\int\mathrm{d^4}y\,J_a(y)\phi_a(y))}\frac{\delta\phi_a(y)}{\delta\phi_b(x)}\\[10mm]=\delta_{ab}\delta^4(x-y)\mathrm{e}^{\mathrm{i}(S+\int\mathrm{d^4}y\,J_a(y)\phi_a(y))}\times\mathrm{i}\underbrace{\frac{\delta}{\delta\phi_a(y)}\Big(S+\int\mathrm{d^4}y\,J_a(y)\phi_a(y)\Big)}_{\Lambda}. \tag{D}$$

Lambda becomes (?) $$\Lambda = \frac{\delta S}{\delta \phi_a(y)}+\int\mathrm{d^4}y J_a(y). \tag{E}$$

What I'm I doing wrong here?

Answer

Let's consider a single scalar field for simplicity. The following step is a misapplication of the functional derivative: \begin{align} \delta Z(J) = \frac{\delta Z}{\delta\phi(x)}\delta\phi(x) \end{align} By definition, one can only take the functional derivative of a functional $F$ with respect to $\phi$ if $F$ is a functional of $\phi$. The functional $Z$ is not a functional of $\phi$ because $\phi$ is being integrated over in the functional integral.

What's going on here is a change of variables in the functional integral. The measure is assumed to be invariant under this change of variables, so what's left is that the terms inside of the exponential can change. To deal with the term involving $S$, we note that under the change of variables $\phi \to \phi + \delta\phi$, the action changes as follows: \begin{align} S[\phi] \to S[\phi + \delta \phi] = S[\phi] + \delta S[\phi] + O(\delta\phi^2) \tag{A} \end{align} and for suitably well-behaved $S$, the first order change of the right hand side (namely $\delta S$) can be written as the integral of the functional derivative of $S$ with respect to $\phi$. To see this, let's assume, for example, that $S$ is the integral of a local Lagrangian density depending on the field and its derivative; \begin{align} S[\phi] = \int d^4x\, \mathscr L(\phi(x), \partial\phi (x)) \tag{B} \end{align} then, under appropriate boundary conditions, we obtain \begin{align} \delta S[\phi] = \int d^4x\, \left[\frac{\partial\mathscr L}{\partial \phi} -\partial_\mu \frac{\partial\mathscr L}{\partial(\partial_\mu\phi)}\right] \delta\phi(x) \tag{C} \end{align} on the other hand, notice that \begin{align} \int d^4x\frac{\delta S}{\delta\phi(x)}\delta \phi(x) &= \int d^4x \,\delta\phi(x)\int d^4y \left[\frac{\partial\mathscr L}{\partial\phi}\frac{\delta\phi(y)}{\delta\phi(x)}+\frac{\partial\mathscr L}{\partial(\partial_\mu\phi)}\partial_\mu\frac{\delta\phi(y)}{\delta\phi(x)}\right] \\ &= \int d^4x \,\delta\phi(x)\int d^4y \left[\frac{\partial\mathscr L}{\partial\phi}\delta(x-y)+\frac{\partial\mathscr L}{\partial(\partial_\mu\phi)}\partial_\mu\delta(x-y)\right]\\ &= \int d^4x\, \left[\frac{\partial\mathscr L}{\partial \phi} -\partial_\mu \frac{\partial\mathscr L}{\partial(\partial_\mu\phi)}\right] \delta\phi(x) \end{align} so in summary, we find that \begin{align} \delta S[\phi] = \int d^4x\, \left[\frac{\partial\mathscr L}{\partial \phi} -\partial_\mu \frac{\partial\mathscr L}{\partial(\partial_\mu\phi)}\right] \delta\phi(x) \tag{D} \end{align} as noted in Srednicki.

Notes. I used integration by parts and the following functional derivative identity in the computations above: \begin{align} \frac{\delta\phi(x)}{\delta\phi(y)} = \delta(x-y) \end{align} which can be proven from the definition of the functional derivative.

newtonian mechanics - How did Newton figure out the law of gravity?

How did Newton figure out the law of gravity? I know there are two other questions answering the same question but I did not understand anything from the answers

Is there a third way of knowing how he did it without using complex formula and derivation.

I am dumb but curious :-)

Answer

To answer the question in your title, he used his newly found fluxions (calculus) to prove that Kepler's laws of planetary motion imply a radial, inverse square law.

Feynman's Lost Lecture is a mixture both of Feynman's attempts to give the simplest possible explanation of how one goes about this derivation and his insights into the history of how Newton did it grounded on Feynman's reading of Newton's Principia. Feynman's treatment is about as simple as you can get, but, as Feynman warns, it's still "God damned hard, there's no question of that." But I'd still recommend this work to you.

In summary:

1. 
   

   Kepler's equal area law implies that the force on a celestial body must always be directed towards the Sun;

   
   


2. 
   
   

   Feynman "reconstructs" Newton's proof that Kepler's first law - that the body's path is an ellipse with Sun at one focus - can be derived from the equal area law and the law that the period $T$ is proportional to the semimajor axis length $L$ to the power of $3/2$, whence

   
   


3. 
   

   One can derive the inverse square law from Kepler's laws as well.

   
   

In my opinion, though, at least as impressive as Newton's analysis of the problem was his unifying realization that the same force that accounted for falling apples might account for the motion of celestial bodies. Although its thorny at first, the maths is the easy part. Insights like this unifying one are in comparison dazzling. Apparently the story of the apple is not altogether apocryphal: Newton seemed to have followed a thought experiment along the lines of "what if this force reached to the top of the highest apple trees, so tall that they might reach the Moon". I explain this in more detail in my answer here. One of Newton's first biographers, William Stukeley, recorded a conversation with Newton in "Memoirs of Sir Isaac Newton's Life". Newton told the apple story to Stukeley, who relayed it as such:

"After dinner, the weather being warm, we went into the garden and drank thea, under the shade of some apple trees...he told me, he was just in the same situation, as when formerly, the notion of gravitation came into his mind. It was occasion'd by the fall of an apple, as he sat in contemplative mood. Why should that apple always descend perpendicularly to the ground, thought he to himself..."

and, in Henry Pemberton's 1728 "A View of Sir Isaac Newton's Philosophy" we find the following passage:

"....that as this power [force of gravity] is not found diminished at the remotest distance from the center of the Earth, to which we can rise, neither at the tops of the loftiest buildings nor ... the highest mountains ... it appeared to him [Newton] reasonable to conclude that this power must extend much farther than was usually thought; why not as high as the moon, said he to himself?"

newtonian mechanics - Consider a horizontal surface with or without friction. Ideally, will a wheel rolling without slipping roll forever in both cases?

Suppose a wheel is rolling smoothly on a horizontal plane i.e., it is rolling without slipping.

Now let's take the two cases of the horizontal plane:

1. 
   

   It has friction

   
   
   


2. 
   

   It is frictionless

   
   

In the first case, as of my knowledge friction doesn't come into play when there is no slipping. So, will the wheel roll forever? If it comes to rest is the answer friction? If not what brings it to rest?

In the second case, there is no friction. So, here also will the wheel roll forever? And if it comes to rest what brings it to rest?

Answer

When the wheel is rolling without slipping, rolling resistance brings the wheel to rest. Here friction is not the right answer let me explain you how.

This friction is not the answer why it comes to rest because friction comes into play to oppose any slippage. The concept of friction most books provide are deficient. The term "rolling friction" is also a misnomer.

The correct explanation for it is rolling resistance for which the word "rolling friction" is often used creating confusions.

Let me explain you rolling resistance and how consideration of friction gives wrong conclusions (I have provided one link explaining rolling motion and correct concept of friction at the end of this answer):

Consider a wheel rolling smoothly. what is the direction of friction force? We might think it must be opposite to the direction of motion thats why it will stop after some time. But, this friction force is providing a torque also making its angular velocity to increase. So, we might think we took the direction of friction force wrongly. So, we take the direction of friction force to be in the same direction of motion which again gives wrong result. Here is the paradox!

Here comes the role of normal reaction. A perfect rigid body doesn't exist. Rolling dynamics in the real world of non-rigid elastic materials is a complex interplay of contact forces due to deformation, and consideration of friction alone can lead to contradictory and unrealistic conclusions. The contact between the wheel and the surface on which it is moving is a surface. This surface is formed due to deformed shape of either the wheel or the surface on which it is moving or both. Because the deformations are not symmetric since the wheel is moving forward, on that surface the reaction forces on the forward portion of the surface is more than that on the backward portion, giving a torque to the wheel in counter-clockwise direction.

Wheel rolling to the right, with surface deformation. The deformation is greatly exaggerated. Normal force components across the deformed region are not uniform in size. They are greater on the forward side, producing a counterclockwise net torque.

This results in slowing down the speed and eventually comes to rest. There are other two cases when the wheel deforms but the surface doesn't and the wheel doesn't deform but the surface does. Please read the content in the link provided.

In the 2nd case if friction is not present the wheel will simply slide forward on the surface. This will no longer be rolling.

Please read this for more details: http://lockhaven.edu/~dsimanek/scenario/rolling.htm

electrostatics - Density of charge induced on a hollow sphere due to eccentric charge inside

Suppose we have a lone hollow metal sphere with net charge equal to $0$. If we were to put a point charge $Q$ inside of the sphere and move it, let's say, away from the sphere center at some distance $d$, determine the distribution of induced charge on the sphere. With no further explanation, the answer in my textbook states that the induced charge will be evenly distributed across inner surface of hollow sphere, no matter where the point charge is, as long as it is inside and not touching the sphere.

Can someone explain in detail or provide information on why this is true?

Answer

The explanation for this is partly Gauss' law and partly the nature of metal.

Metals have free electrons that move to try to compensate for electric fields. The free electrons near the surface on the inside of the sphere will have an uneven distribution if the point charge is off-centre. These electrons will counteract the effect of the positive charge inside the sphere so that there will be no electric field inside the metal shell. Let us say that the charge inside is positive and $+q$ - then there will be extra electrons on the inside of the sphere to counterbalance it - their total charge will be $-q$. Because the metal is neutral overall there will be a lack of electrons on the outer surface and a net positive charge of $+q$ on the outer surface because of the extra electrons. which are on the inside of the sphere.

Now because there is no electric field inside the metal the charge on the outside of the metal sphere will spread itself for the lowest energy configuration by spacing the charges as far apart as possible, which will give an even distribution.

Why can't electrostatic field lines form closed loops?

My physics textbook says "Electrostatic field lines do not form closed loops. This is a consequence of the conservative nature of electric field." But I can't quite understand this. Can anyone elaborate?

Answer

A force is said to be conservative if its work along a trajectory to go from a point $A$ to a point $B$ is equal to the difference $U(A)-U(B)$ where $U$ is a function called potential energy. This implies that if $A=B$ then there is no change in potential energy. This fact is independent of the increase or not of the kinetic energy.

If a conservative force were to form loops, it could provide a non zero net work (because the direction of the force could always be the same as that of the looping trajectory) to go from A and then back to A, while at the same time its conservative character would ensure that this work should be zero; which is a contradiction.

Hence, "conservative force" and "forming loops" are two incompatible properties that cannot be satisfied at the same time.

special relativity - Representation under which Pauli matrices transform

In Peskin and Schroeder's Quantum Field Theory, there is an identity of Pauli matrices which is connected to the Fierz identity, (equation 3.77) $$(\sigma^{\mu})_{\alpha\beta}(\sigma_\mu)_{\gamma\delta}=2\epsilon_{\alpha\gamma}\epsilon_{\beta\delta}.\tag{3.77}$$ The author explains that

One can understand the identity by noting that the indices $\alpha,\gamma$ transform in the Lorentz representation of $\Psi_{L}$, while $\beta,\delta$ transform in the separate representation of $\Psi_{R}$, and the whole quantity must be a Lorentz invariant.

How can one see $\alpha,\gamma$ and $\beta,\delta$ transform in different representation?

visible light - Transparency of materials

Is transparency of material has something to do with inter- or intra-molecular bonding? E.g. both graphite and diamond are carbon, but graphite is opaque and diamond transparent.

newtonian mechanics - What force does the dynamometer show in this case?

So I came across this picture on facebook

My logic says it should show 200N, because you are pulling it with 100N on each side. On the other hand (as one of the facebook comments also stated, also saying is he is wrong he wasted 20k$ on physics degree), if you replace the right side with a wall you get this situation:

In this case I guess everyone agrees that the dynamometer would show 100N force. But if you isolate the dynamometer and "cut" the ropes on either picture, you would get the same thing. Because the system needs to be balanced, there would be 100N on each side.

And I study engineering too but I am confused by this. Why do we get the same distribution of forces, but dynamometer shows a different force?

Is refraction sharp or smooth?

Refraction: light changes direction of propagation when entering a material with a different refractive index.

Does the direction of propagation of light change sharply and almost instantaneously (as shown in the diagram) or smoothly?

Answer

I don't understand the down vote on the OP. In order for Snell's Law ($n_1 \sin{\theta_1} = n_2\sin{\theta_2}$) to be valid, certain conditions have to be met. Usually we assume the conditions are met, and draw the simple diagrams such as yours. These represent idealizations of real situations. For example, the ideal interface is discrete, changes abruptly, and is perfectly flat, and the materials are homogenous. At a microscopic level, all real interfaces consist of meetings of molecules, so are not discrete, abrupt, nor flat, and materials are not homogeneous (they are made of atoms and molecules).

However, from a macroscopic perspective, if we don't focus attention on the microscopic, the interfaces sure look discrete, abrupt, and flat, and the material homogenous. So we expect Snell's Law to work macroscopically ... as long as we don't look too close to the surface, or too closely at the molecules. And indeed, it does work. In particular, in the idealization the direction of the ray changes abruptly. Macroscopically, if we don't look too close to the surface, it does.

But your question is, really: what happens if we don't take that macroscopic view? Does the ray curve smoothly rather than abruptly change direction?

I doesn't bend. What actually happens is very complicated. I think there are two ways to look at it. In the first, the original rays spawn countless other rays in the interface region, pointing in all directions. Once we focus attention away from the interface region we find that almost all of those rays have destructively interfered, all except those that go off in the macroscopic refraction direction. Another way to look at it is to say that a ray represents a "pencil" of light having a well-defined wavelength and direction. Then, we recognize that the interface region is too small for us to unambiguously define the wave length. Thus, the concept of "ray" does not exist in the interface region.

In any event, the picture of a smoothly curving ray is not the right picture.

Update

Well, whether or not the picture of a curving ray is correct or not depends on how closely you look, and how deeply you want to probe for an answer.

The comments point out a weakness in my answer. I tried to use the language of rays, as suggested by the diagram. Doing so will always eventually lead to trouble. "Ray" can refer to the propagation vector of a plane wave, or it can refer to a tight pencil of light, usually taken to have zero cross section in our imagination. Both of these pictures are mathematical abstractions that are ultimately non-physical.

I did not make the approximation that materials are actually continuous media. You can make that approximation and show that "rays" are curved, but that is not a correct description of what the electric field looks like in the interface region. In the continuous approximation we average the actual electric field over a volume large compared to the size of an atom, but small compared to a wavelength. One ends up with a (average) dielectric constant that is a continuous function of position. The averaged field is called the macroscopic electric field. But within the volume used to do the averaging, the actual microscopic field is varying wildly. This is why I said that either rays are propagating in all directions, or the concept of "ray" does not exist near the surface.

Let's take a different microscopic view. Light is incident on the interface. We'll model the solid as a distribution of polarizable entities, I'll call them atoms, but they could be molecules. They could be distributed regularly, as a crystal, or randomly as in an amorphous solid. The incoming light excites each dipole. Each excited dipole acts as a source of new radiation. The refracted field is the sum of the radiation of all of these dipoles plus the field of the incident radiation. At an atomic level, the field distribution is very complicated.

One can make further progress by doing the averaging procedure. We get an average field, and the fluctuations are smoothed out. This is about all we can do for the amorphous solid, because we don't know where the dipoles actually are. This results in the continuous approximation, and we predict curved rays. But that does not represent the actual electric field! We've removed much of the character by smoothing out the field.

Things are different for a perfect crystal. Here we do know where the atoms are, and we can can add all the contributions from all of the dipoles. The result, as expected, is a wave which propagates in the refraction direction. In this picture, the fields in the surface region are complicated in the way I mentioned in my original answer, and one can't describe what's going on as curved rays.

Interesting aside: when all of those individual contributions from individual dipoles are added up, we find that actually two waves are generated in the refracted material. The first propagates in the direction of the incident wave, but is out of phase with it, causing the incident wave to vanish by destructive interference. The second wave is the actual refracted wave. This result is known as the Ewald-Oseen extinction theorem.

electromagnetism - Why does a changing magnetic field produce a current?

A changing magnetic field induces a current in a conductor. For example, if we move a bar magnet near a conductor loop, a current gets induced in it.

Faraday's law states that

The E.M.F. $\mathcal{E}$ induced in a conducting loop is equal to the rate at which flux $\phi$ through the loop changes with time.

Along with Lenz's law,

$$\mathcal{E} = -\frac{d\phi}{dt}$$

Why is this so? The velocity of the electrons w.r.t. me, the observer is 0, so according to $\vec{F} = q\vec{v} × \vec{B}$, force should be zero in any direction on the electrons in the loop. Then what causes the current to flow and the E.M.F. to be induced? Is the force due to an electric field(the electric field in my reply to Albert in comments) or should I consider the velocity w.r.t. the source of the magnetic field?

Edit : I'm in a frame of reference which is stationary w.r.t. the loop, so from where does the electric force(due to the electric field in my reply to Albert in comments) come from?

general relativity - Wick Rotation in Curved space

So over time I have learned to do exhaustive searches before asking things here. Wick rotations are cool if you are trying to work in qft and make statements about the thermodynamics of some physical thing you are probing. You want to talk about some thermodynamics and who knows may be statistics, and like most people . . . oh Wick rotation is what you would think. Now we are in curved space, do you just throw away this tactic in the dust bin? And yes, I have read up a few things, but it seems I need a proper pointer to where to find how to attack this thing properly. Just really curious.

Answer

Matt Visser's How to Wick rotate generic curved spacetime is a great reference on this subject, which basically summarizes a lot of folklore on the subject.

Addendum (Summary of Paper). This turns out to be an important problem in quantum gravity and QFT in curved spacetime for the obvious reason ("How do we know the usual tricks still work in curved spacetime?").

Visser re-frames the Wick rotation in a more coordinate independent way (the naive $t\mapsto it$ prescription gives incorrect solutions even for de Sitter spacetime).

The more general Wick rotation analytically continues the metric while leaving the local coordinate charts invariant. When we restrict our attention back to flat spacetime, this approach recovers the usual QFT-textbook prescription for Wick rotations.

What does it look like? Well, suppose $g_{L}(-,-)$ is the metric tensor (using MTW-style notation), and $V$ is a non-vanishing timelike vector field. So in $-+++$ signature, $g_{L}(V,V)<0$. The Wick rotation amounts to swapping out this Lorentzian metric for: $$\tag{1}g_{\epsilon} = g_{L} + i\epsilon\frac{V\otimes V}{g_{L}(V,V)}$$ and using this $g_{\epsilon}(-,-)$ metric everywhere instead.

How do we recover the usual Wick rotation? Well, use flat spacetime, so $g_{L}\mapsto\eta_{L}$ and $V\mapsto(1,0,0,0)$. Then the propagator for the scalar field, for example, becomes $$\tag{2}\Delta_{F}(P) = \frac{-i}{\eta_{\epsilon}(P,P)+m^{2}}$$ where $\eta_{\epsilon}$ is the Wick rotated metric tensor for flat spacetime. Eq (2) precisely what you'd find in any generic textbook on QFT. So, good, this generalized procedure --- i.e., Eq (1) --- recovers the usual results we want.

electrostatics - How much electric charge can a copper sphere, made of N atoms, contain?

If somebody wants to charge a cooper sphere made of N atoms, what is the maximum positive charge the sphere can reach? Same question about a negative charge.

Source

For example if the sphere is small, made of 100 - 1000 atoms, can all atoms remain without the electron on the last layer or even without some electrons from the deeper layers?

Note: The indicated answer is for the case when someone wants to charge the sphere with a negative charge and it contain just a qualitative explanation.

general relativity - Does a charged particle accelerating in a gravitational field radiate?

A charged particle undergoing an acceleration radiates photons.

Let's consider a charge in a freely falling frame of reference. In such a frame, the local gravitational field is necessarily zero, and the particle does not accelerate or experience any force. Thus, this charge is free in such a frame. But, a free charge does not emit any photons. There seems to be a paradox. Does a freely falling charge in a gravitational field radiate?

Answer

The paradox is resolved as follows: the number of photons changes when you switch between non-inertial frames. This is actually a remarkable fact and it holds also for quantum particles, which can be created in pairs of particles and antiparticles, and whose number depends on the frame of reference.

Now, a step back. Forget about gravity for a moment, as it is irrelevant here (we are still in GR, though). Imagine a point charge, which is accelerating with respect to a flat empty space. If you switch to the rest frame of the charge, you observe a constant electric field. When you switch back to the inertial frame, you see the field changing with time at each point and carrying away radiation from the charge.

In the presence of gravity the case is absolutely similar. To conclude, switching between non-inertial frames makes a static electric field variable and corresponds to a radiation flow.

Another relevant point: When moving with charge, no energy is emitted, but when standing in the lab frame, there is a flux observed. However, there is no contradiction here as well, as the energy as a quantity is not defined for noninertial frames.

general relativity - Is light affected by gravity? Why?

I would like to know if light is affected by gravity, also, I would like to know what is the correct definition of gravity:

"A force that attracts bodies with mass" or "a force that attracts bodies with energy, such as light"?

Is light massless after all?

newtonian mechanics - Is friction always constant?

Can the norm of the friction force change with time? Notice I said the norm which is in Newtons, can it change with time? Sorry if the question seems a bit confusing but I've done my best to explain.

Answer

I'll assume that you're talking about friction between two solid surfaces sliding over each other. Other comments and answers have pointed out the friction is velocity dependant in gases and liquids, though I would call this viscous drag rather than friction.

When you first learn about friction you're usually taught that the frictional force is given by:

$$ F = \mu L $$

where $L$ is the load (normal force) and $\mu$ is a constant called the coefficient of friction. Your question then amounts to asking whether $\mu$ is really a constant or whether it depends on sliding velocity. The easy answer is to point you to a Google image search for friction coefficient velocity. This finds lots of experimentally measured graphs of $\mu$ against sliding velocity and you'll see $\mu$ is indeed dependant on velocity and isn't a constant.

Friction is an emergent property. It happens because when you touch the two surfaces together high points (asperities) on the two surfaces come into contact and bond due to the same sort of forces that hold solids together. To slide the surfaces you have to break these bonds, and that takes energy. The energy dissipation per unit distance of sliding gives the force.

The approximate equation comes about because the area of contact of the asperities is approximately proportional to the load. As you press harder you deform the asperities so they flatten out and form a larger contact patch. As a rough guide you'd expect the pressure at the contact point, i.e. the normal force divided by the real area of contact, to be roughly equal to the yield stress of the solid.

The dependance of $\mu$ on velocity is complicated and can go up or down depending on the system being studied. The rate at which asperities touch and bond depends on all sorts of things so it's hard to say anything definitive about it. All we can say is that there's no reason to expect the rate to be independant of velocity so we shouldn't be surprised to find that it isn't, and also that it varies from system to system..

thermodynamics - How do you prove $S=-sum pln p$?

How does one prove the formula for entropy $S=-\sum p\ln p$? Obviously systems on the microscopic level are fully determined by the microscopic equations of motion. So if you want to introduce a law on top of that, you have to prove consistency, i.e. entropy cannot be a postulate. I can imagine that it is derived from probability theory for general system. Do you know such a line?

Once you have such a reasoning, what are the assumptions to it? Can these assumptions be invalid for special systems? Would these system not obey thermodynamics, statistical mechanics and not have any sort of temperature no matter how general?

If thermodynamics/stat.mech. are completely general, how would you apply them the system where one point particle orbits another?

Answer

The theorem is called the noiseless coding theorem, and it is often proven in clunky ways in information theory books. The point of the theorem is to calculate the minimum number of bits per variable you need to encode the values of N identical random variables chosen from $1...K$ whose probabilities of having a value $i$ between $1$ and $K$ is $p_i$. The minimum number of bits you need on average per variable in the large N limit is defined to be the information in the random variable. It is the minimum number of bits of information per variable you need to record in a computer so as to remember the values of the N copies with perfect fidelity.

If the variables are uniformly distributed, the answer is obvious: there are $K^N$ possiblities for N throws, and $2^{CN}$ possiblities for $CN$ bits, so $C=\log_2(k)$ for large N. Any less than CN bits, and you will not be able to encode the values of the random variables, because they are all equally likely. Any more than this, you will have extra room. This is the information in a uniform random variable.

For a general distribution, you can get the answer with a little bit of law of large numbers. If you have many copies of the random variable, the sum of the probabilities is equal to 1,

$$ P(n_1, n_2, ... , n_k) = \prod_{j=1}^N p_{n_j}$$

This probability is dominated for large N by those configurations where the number of values of type i is equal to $Np_i$, since this is the mean number of the type i's. So that the P value on any typical configuration is:

$$ P(n_1,...,n_k) = \prod_{i=1}^k p_i^{Np_i} = e^{N\sum p_i \log(p_i)}$$

So for those possibilities where the probability is not extremely small, the probability is more or less constant and equal to the above value. The total number M(N) of these not-exceedingly unlikely possibilities is what is required to make the sum of probabilities equal to 1.

$$M(N) \propto e^{ - N \sum p_i \log(p_i)}$$

To encode which of the M(N) possiblities is realized in each N picks, you therefore need a number of bits B(N) which is enough to encode all these possibilities:

$$2^{B(N)} \propto e^{ - N \sum p_i \log(p_i)}$$

which means that

$${B(N)\over N} = - \sum p_i \log_2(p_i)$$

And all subleading constants are washed out by the large N limit. This is the information, and the asymptotic equality above is the Shannon noiseless coding theorem. To make it rigorous, all you need are some careful bounds on the large number estimates.

Replica coincidences

There is another interpretation of the Shannon entropy in terms of coincidences which is interesting. Consider the probability that you pick two values of the random variable, and you get the same value twice:

$$P_2 = \sum p_i^2$$

This is clearly an estimate of how many different values there are to select from. If you ask what is the probability that you get the same value k-times in k-throws, it is

$$P_k = \sum p_i p_i^{k-1}$$

If you ask, what is the probability of a coincidence after $k=1+\epsilon$ throws, you get the Shannon entropy. This is like the replica trick, so I think it is good to keep in mind.

Entropy from information

To recover statistical mechanics from the Shannon information, you are given:

• the values of the macroscopic conserved quantities (or their thermodynamic conjugates), energy, momentum, angular momentum, charge, and particle number


• the macroscopic constraints (or their thermodynaic conjugates) volume, positions of macroscopic objects, etc.

Then the statistical distribution of the microscopic configuration is the maximum entropy distribution (as little information known to you as possible) on phase space satisfying the constraint that the quantities match the macroscopic quantities.

electromagnetism - Dirichlet and Neumann Boundary condition: physical example

Can anybody tell me some practical/physical example where we use Dirichlet and Neumann Boundary condition. Is it possible to use both conditions together at the same region?

If we have a cylindrical symmetric array of lenses to focus ions, which regions it comes the Dirichlet BC and Neumann BC? This types of systems can be simulated using SIMION. I know SIMION solved Laplaces equation. But I am confused with the boundary conditions.

Answer

There is a standard book which contains everything about electrostatics, the Laplace/Poisson equation and boundary conditions: Classical Electrodynamics by J. D. Jackson. Get the book from the library of your choice, read all chapters labeled "Electrostatics", and you will find the answers to all your questions (if you are simulating this, you need to know all this stuff anyway).

The nature of the boundary condition depends on the system you are describing. It means something else if you are calculating heat flow than electrostatics - obviously!

Let's say you have a differential equation (e.g. the Poisson equation $\Delta\varphi(\vec r)=-\frac{\varrho(\vec r)}{\varepsilon_0}$ describing electrostatics, and you solve it for the function $\varphi(\vec r)$. At the boundaries of the region (e.g. a cylinder, a cube, etc.) you have to fix some property of $\varphi(\vec r)$.

• Neumann boundary condition: You fix $\frac{\partial\varphi(\vec r)}{\partial\vec n}=\text{const}$ along the boundary, where $\vec n$ is the normal vector to the surface. It's basically the derivative of $\varphi$ if you go straight away from the surface. It can be a different value for every $\vec r$.


• Dirichlet boundary condition: You fix $\varphi(\vec r)=\text{const}$. It can be a different value for every $\vec r$.

You can only fix one of those two, or the sum (this is called Robin boundary condition).

Physical examples for electrostatics:

• Neumann boundary condition: The aforementioned derivative is constant if there is a fixed amount of charge on a surface, i.e. $\frac{\partial\varphi(\vec r)}{\partial\vec n}=\sigma(\vec r)$.


• Dirichlet boundary condition: The electrostatic potential $\varphi(\vec r)$ is fixed if you have a capacitor plate which you connected to a voltage source. E.g. if you have two capacitor plates which are at 0V and 5V, respectively, you would set $\varphi(\vec r)=0$ at the first plate and $\varphi(\vec r)=5$ at the second plate. That way, you can calculate the capacitance.

For heat flow, fixing the field $u$ (=Dirichlet BC) means fixing the temperature. If you have elements in your system that have a fixed temperature, you would use that one.

If you lenses are based on electrostatics, they probably only have Dirichlet boundary conditions, because that's how you describe a capacitor plate. If you have outer boundaries which are not capacitor plates, you should use a Neumann BC = 0 (in this case, it has nothing to do with fixing the charge), because that's the best BC for simulating an "infinite" system.

quantum mechanics - Directional derivatives in the multivariable Taylor expansion of the translation operator

Let $T_\epsilon=e^{i \mathbf{\epsilon} P/ \hbar}$ an operator. Show that $T_\epsilon\Psi(\mathbf r)=\Psi(\mathbf r + \mathbf \epsilon)$.

Where $P=-i\hbar \nabla$.

Here's what I've gotten: $$\begin{align}T_\epsilon\Psi(\mathbf r)&= e^{i \mathbf{\epsilon} P/ \hbar}\Psi(\mathbf r)\\ &=\sum^\infty_{n=0} \frac{(i\epsilon \cdot (-i\hbar \nabla)/\hbar)^n}{n!} \Psi(\mathbf r) \\ &=\sum^\infty_{n=0} \frac{(\mathbf \epsilon \cdot \nabla)^n}{n!}\Psi(\mathbf r) \\ &= \Psi(\mathbf r) + (\epsilon \cdot \nabla) \Psi(\mathbf r) + \frac{(\epsilon \cdot \nabla)^2 \Psi(\mathbf r)}{2} + \cdots\end{align}$$

This looks somewhat like a Taylor expansion of $\Psi(\mathbf r)$, but it's different than I've seen before -- I've never seen it in terms of a directional derivative. Can you confirm if this is the Taylor expansion of $\Psi(\mathbf r + \mathbf \epsilon)$? Or if not, what I should be getting when expanding $e^{i \mathbf{\epsilon} P/ \hbar}\Psi(\mathbf r)$?

Answer

The Taylor series of a function of $d$ variables is as follows:

$$f(\mathbf{x} + \mathbf{y}) = \sum_{n_1=0}^{\infty}\ldots\sum_{n_d=0}^{\infty} \frac{(y_1-x_1)^{n_1} \ldots (y_d - x_d)^{n_d}}{n_1!\ldots n_d!} \left. \left( \partial_1^{n_1}\ldots \partial_d^{n_d} \right) f \right|_{\mathbf{x}}.$$

where $\partial_i^{n_i} f$ means "the $n_i^{\text{th}}$ order partial derivative of $f$ with respect to the $i^{\text{th}}$ coordinate". Consider only the terms where $\sum_{i=1}^{\infty} n_i = 1$. These are the terms for which there is only one derivative being taken, a.k.a. the "linear terms". Keeping the $0^{\text{th}}$ order term and the linear terms gives

$$f(\mathbf{x} + \mathbf{y}) = f(\mathbf{x})+\sum_{i=1}^{d}(y_i - x_i) \left. (\partial_i f \right)|_{\mathbf{x}} . \qquad (*)$$

Define the displacement $\epsilon = \mathbf{y} - \mathbf{x}$. Then

$$\epsilon \cdot \nabla = \sum_{i=1}^d (y_i - x_i) \partial_i$$

by definition of what $\cdot$ means. Therefore, keeping only up to linear terms, $(*)$ becomes

$$f(\mathbf{y} - \mathbf{x}) = f(\mathbf{x}) + \left. (\epsilon \cdot \nabla)\right|_{\mathbf{x}}f . $$

So you see, your formula is correct, just in a particular notation you may not be used to.

For the sake of typing less, I'm going to explain this in one dimension, but nothing in the actual calculation is restricted to one dimension.

The state vector can be thought of as a linear combination of eigenvectors of the position operator

$$|\Psi\rangle = \int_x \Psi(x) |x\rangle \qquad (1)$$

where $\hat{x}|x\rangle = x|x\rangle$. Bringing in a bra $\langle y |$ from the left and using $\langle y | x \rangle = \delta(x-y)$ in (1) gives

$$\langle y | \Psi \rangle = \Psi(y) . \qquad (2)$$

Combining (1) and (2) gives

$$|\Psi\rangle = \int_x |x\rangle \langle x| \Psi \rangle \qquad (3)$$

which shows that $\int_x |x\rangle \langle x | = \text{Identity}$. Let's really understand what all this means: Eq. (1) just says that you can write a vector as a linear combination of basis vectors. In this case, $\Psi(x)$ are the coefficients in the linear combination. In other words, the wave function is just the coefficients of a linear combination expansion written in the position basis. Eq. (2) makes this explicit by showing that the inner product of a position basis vector $|y\rangle$ with $|\Psi\rangle$ is precisely $\Psi(y)$. Eq. (3) just expresses a neat way of expressing the identity operator in a way which we will find very useful. Note that this works with any basis. For example,

$$\int_p |p\rangle \langle p | = \text{identity} .$$

We can use this to express a position eigenvector in terms of momentum eigenvectors,

$$|x\rangle = \int_p |p\rangle \langle p | x \rangle = \int_p e^{-i x p/\hbar}|p\rangle ,$$ where we've used the fact that $\langle x | p \rangle = e^{i p x/\hbar}$ [1].

Now let's get back to the question at hand. Define $|\Psi'\rangle \equiv e^{i \epsilon \hat{p}/\hbar}|\Psi\rangle$. Let us evaluate $\Psi'(y) \equiv \langle y|\Psi'\rangle$: \begin{align} \Psi'(y) &= \langle y | \Psi' \rangle \\ &= \langle y| e^{i \epsilon \hat{p}/\hbar} |\Psi \rangle \\ &= \langle y| e^{i \epsilon \hat{p}/\hbar} \int_x \Psi(x) |x\rangle \\ &= \int_x \Psi(x) \langle y | e^{i \epsilon \hat{p}/\hbar} |x \rangle . \end{align}

To proceed we need to compute $e^{i \epsilon \hat{p}/\hbar} |x \rangle$. We can do this using our expression for the identity in terms of momentum states,

$$ \begin{align} e^{i \epsilon \hat{p}/\hbar} |x \rangle &= e^{i \epsilon \hat{p}/\hbar} \left( \int_p |p\rangle \langle p | \right) | x \rangle \\ &= \int_p e^{i \epsilon \hat{p} / \hbar} |p\rangle \langle p | x \rangle \\ &= \int_p e^{i \epsilon p / \hbar} |p\rangle \langle p | x \rangle \\ &= \int_p e^{i \epsilon p /\hbar} |p\rangle e^{-i p x/\hbar}\\ &= \int_p e^{-i(x - \epsilon) p / \hbar} |p\rangle \\ &= |x - \epsilon \rangle \end{align}$$

This is actually the thing you should remember about this question:

$$e^{i\epsilon\hat{p}/\hbar}|x\rangle = |x - \epsilon \rangle .$$

This is an excellent equation because it tells you something about how an the $e^{i \epsilon \hat{p} /\hbar}$ operator changes a position eigenvector without having to write it out in a particular basis.

Now that we have that, we can go back to what we were trying to compute,

$$ \begin{align} \Psi'(x) &= \int_x \Psi(x) \langle y | e^{i \epsilon \hat{p}/\hbar} |x \rangle \\ &= \int_x \Psi(x) \langle y | x - \epsilon \rangle \\ &= \int_x \Psi(x) \delta(y - (x - \epsilon)) \\ &= \Psi(y + \epsilon) . \end{align} $$

This is the equation you wanted to prove. If I messed up any minus signs I hope someone will edit :)

[1] I'm not proving that, and if you want to know why that's true it would make a good question on this site.

Mnemonics to remember various properties of materials

I'm trying to figure out how to remember that

• hardness: how resistant it is to deformation


• toughness: how resistant it is to brittle failures



• stress: force on a surface area


• strength: ability to withstand stress without failure


• strain: measurement of deformation of a material

Does anyone know of a mnemonic or easy way? I only know these from googling them, and I'm finding it tricky to remember them.

Answer

I always used to confuse stress and strain: most of my mnemonics involved making words out of initial letters.

When you're stressed, you show the strain.

Stress is what is applied to the material, strain is what it does in response - I always used to get these the wrong way around.

E equals Fl/ea

Young's Modulus = (force × length) / (extension × area)

Good luck with the others: I suggest imagining a hard man (grizzled veteran) who is actually secretly limp wristed and camp (he bends unlike hard materials), a tough guy (showy, probably with twin pistols) who literally goes to pieces in difficult situations (imagine his brittle bones snapping) and a strong man (lifting a dumbbell) who is crushed to a pancake by the weight.

electric circuits - Physical meaning of Impedance

So I have been thinking about the way impedance is defined for electrical systems, and the way it is derived. Even after looking through some websites, I cannot seem to grasp something, which every website I visited seemed to skip on.

In an ideal resistor, if we take the ratio of the voltage to the current at ANY time, the ratio is a constant. This ratio is then the resistance (also the impedance) of the resistor. I have been interpreting this as the opposition the resistor bears to the current flow through it upon application of a voltage. I thought of it like a constant that relates the otherwise-would-be-infinite current to the applied voltage.

Coming to a capacitor, there seems to be some sort of change in the definition of impedance. The impedance for a capacitor is no longer the ratio of voltage to the current, but the ratio of the complex voltage to the complex current. In other words, mathematically it is equivalent to $$\dfrac{1}{i\omega C},$$ and not $$\dfrac{\tan(\omega t)}{\omega C},$$ for a capacitor with capacitance C at a frequency w.

I am finding this difficult to interpret because the ratio of the voltage to current is definitely varying in time. How can the analogy of a resistor's impedance be applied to the impedance of a capacitor when the ratio of the voltage across it to the current through it is changing (it's even 0 or undefined sometimes). Why is the magnitude of the quantity $$\dfrac{1}{i\omega C}$$ now the resistance of the capacitor circuit? What happened to the plain old voltage to current ratio? Have books been skipping this subtlety, or I am just not understanding something very simple?

Clarification:

$$ if \hspace{.5 pc} v(t) = V_m \sin (\omega t), \hspace{0.5 pc} then \dfrac{d V(t)}{dt} = \omega V_m \cos (\omega t) $$ $$ so \hspace{0.5 pc}the\hspace{0.5 pc}ratio\hspace{0.5 pc} \dfrac{v(t)}{i(t)} = \dfrac{V_m \sin (\omega t)}{C\omega V_m \cos (\omega t)} = \dfrac{\tan (\omega t)}{C\omega} $$

$$ In \hspace{0.5 pc}phasor \hspace{0.5 pc}notation, \hspace{0.5 pc}this \hspace{0.5 pc} ratio\hspace{0.5 pc} seems\hspace{0.5 pc} to\hspace{0.5 pc} be\hspace{0.5 pc}different \hspace{0.5 pc}in \hspace{0.5 pc}that: \\ \dfrac{\mathbf{V_c}}{\mathbf{I_c}} = \dfrac{1}{Cj\omega} = -\dfrac{1}{C \omega }j $$ $$ and \hspace{0.5 pc} \Re\{-\dfrac{1}{C \omega }j \} = 0$$ $$but \\ |\dfrac{\mathbf{V_c}}{\mathbf{I_c}}| = \dfrac{1}{C \omega } $$

Why does the magnitude of the ratio of the complex voltage to the complex current now suddenly carry a physical meaning (if my understanding is correct, it is the resistance which can be measured in ohms, just like in the resistor). Also, why is going from the complex domain to real domain not possible for this ratio, because clearly the real part of the complex impedance is 0 (@Alfred Centauri mentions that the impedance is not itself a phasor). I understand that the math works, but the thing that is not making sense to me is this ratio of two complex quantities and the apparent emergence in its physical meaning.

Answer

In general, the voltage across and current through a capacitor or inductor do not have the same form:

$$i_C(t) = C \dfrac{dv_C}{dt} $$

$$v_L(t) = L \dfrac{di_L}{dt} $$

Thus, in general, the ratio of the voltage across to the current through is not a constant.

However, recalling that:

$$\dfrac{d e^{st}}{dt} = s e^{st} $$

where s is a complex constant $s = \sigma + j \omega$, we find that, for these excitations only:

$$i_C(t) = (sC) \cdot v_C(t)$$

$$v_L(t) = (sL) \cdot i_L(t)$$

In words, for complex exponential excitation, the voltage across and current through are proportional.

Now, there are no true complex exponential excitations but since:

$$e^{j \omega t} = \cos(\omega t) + j \sin(\omega t) $$

we can pretend that a circuit with sinusoidal excitation has complex exponential excitation, do the math, and take the real part of the solution at the end and it works.

This is called phasor analysis. The relationship between a sinusoidal voltage its phasor representation is:

$$ v_A(t) = V_m \cos (\omega t + \phi) \rightarrow \mathbf{V_a} = V_m e^{j \phi}$$

This is because:

$$v_A(t) = \Re \{V_me^{j(\omega t + \phi)}\} = \Re\{V_m e^{j \phi}e^{j \omega t}\} = \Re\{\mathbf{V_a} e^{j \omega t}\} $$

Since all of the voltages and currents in a circuit will have the same time dependent part, in phasor analysis, we just "keep track" of the complex constant part which contains the amplitude and phase information.

Thus, the ratio of the phasor voltage and current, a complex constant, is called the impedance:

$$\dfrac{\mathbf{V_c}}{\mathbf{I_c}} = \dfrac{1}{j\omega C} = Z_C$$

$$\dfrac{\mathbf{V_l}}{\mathbf{I_l}} = j \omega L = Z_L$$

$$\dfrac{\mathbf{V_r}}{\mathbf{I_r}} = R = Z_R$$

(Carefully note that though the impedance is the ratio of two phasors, the impedance is not itself a phasor, i.e., it is not associated with a time domain sinusoid).

Now, we can use the standard techniques to solve DC circuits for AC circuits where, by AC circuit, we mean: linear circuits with sinusoidal excitation (all sources must have the same frequency!) and in AC steady state (the sinusoidal amplitudes are constant with time!).

---

So my question is why does the magnitude of the ratio of the complex voltage to the complex current now suddenly carries a physical meaning (if my understanding is correct, it is the resistance which can be measured in ohms, just like in the resistor).

Remember, the complex sources are a convenient fiction; if there were actually physical complex sources to excite the circuit, the phasor representation would by physical.

The physical sources are sinusoidal, not complex but, remarkably, we can mathematically replace the sinusoidal sources with complex sources, solve the circuit in the phasor domain using impedances, and then find the actual, physical sinusoidal solution as the real part of the complex time dependent solution.

Here's an example of the physical content of impedance:

Let the time domain inductor current be:

$i_L(t) = I_m \cos (\omega t + \phi)$

Find the time domain inductor voltage using phasors and impedance. The phasor inductor current is:

$\mathbf{I_l} = I_m e^{j\phi}$

and the impedance of the inductor is:

$Z_L = j \omega L = e^{j\frac{\pi}{2}}\omega L$

Thus, the phasor inductor voltage is:

$\mathbf{V_l} = \mathbf{I_l} Z_L = I_m e^{j\phi}e^{j\frac{\pi}{2}}\omega L = \omega L I_m e^{j(\phi + \frac{\pi}{2})}$

Converting to the time domain:

$v_L(t) = \omega L I_m \cos (\omega t + \phi + \frac{\pi}{2})$

Note that the magnitude of the impedance shows up in the amplitude of the sinusoid and the phase angle of the impedance shows up in the phase of the sinusoid.

quantum field theory - Calculation of Feynman invariant amplitude with internal global symmetry indices: trace overand isospin

This is a complete rewriting of the older post, making more clear the problem.

The issue here is to compute the $$|M|^2 =4a^2(\delta_{ad}\delta_{bc} - \frac{1}{2} \delta_{ab}\delta_{cd}) ^2 (u_{1b}^{\mu}\bar{u}_{1 \mu a} u^{\mu}_{1'b}\bar{u}_{1'\mu a})\\ \times(u_{2'd}^{\nu}\bar{u}_{2'\nu c}u_{2d}^{\nu}\bar{u}_{2\nu c}) $$ invariant Feynman amplitude, and in fact, the search for a possible, if it exists, way for writing this as a trace both on $SU(2)$ and Lorentz indices.

One of terms of the above amplitude is: $$T_1 = \delta_{ad}\delta_{bc}\delta_{ad}\delta_{bc} \omega = 4 (u_{1b}^{\mu}\bar{u}_{1 \mu a} u^{\mu}_{1'b}\bar{u}_{1'\mu a})(u_{2'a}^{\nu}\bar{u}_{2'\nu b}u_{2a}^{\nu}\bar{u}_{2\nu b}),$$where $\omega $ is the series of spinors before contracting the deltas and the 4 coefficient comes from the contraction of deltas.

Question: I'm having a problem seeing how could I simplify this expression. Can I simultaneously reform this term to a trace? The possibilities are a trace over:

1. Just spin indices; but what happens then with the internal indices?


2. Internal isospin trace; but what then happens with spin indices?


3. A trace over both indices?

Thank you.

---

Details

Assume a Lagrangian with an interaction term of the form $L_{int}=\bar \psi \phi \cdot \tau \psi$ , globally symmetric under the action of $SU(2)$ group. Think that the field $\psi$ describes a nucleon as an isospin SU(2) doublet with entries the proton and neutron and the field $\phi$ an isospin triplet (in the adjoint representation) of SU(2) formed by the three scalar pions.

The problem is to calculate the invariant Feynman amplitude. We have, writing all the indices explicitly:

$$M=a(q^2) (\bar{u}_{1'\mu a}\tau^{ab}_k u^{\mu}_{1b})( u_{2'\nu c}\tau^{cd}_k\bar{u}^{\nu}_{2d}) ,$$ where $\tau_k$ are the $SU(2)$ generators; Pauli matrices and q is the interaction momentum of the propagator. The complex conjugate then is: $$M^{*}=a^{*}(q^2) (\bar{u}_{1\mu a}\tau^{ab}_k u^{\mu}_{1' b}) (u_{2\nu c}\tau^{cd}_k\bar{u}^{\nu}_{2' d}) .$$

This way one should recognise the scalar nature over both $\mu, \nu $ Lorentz indices and over $a,b,c,d, k $ isospin indices. A similar calculation can be found in chapter 46 of Srednicki, without the extra internal symmetry and it's indices. There, it's simple then to recognise the trace over spin indices of the $|M|^2 =MM^{*} $.

For the generators there is the following completeness relation: $\tau_{ab}^k \tau_{cd}^k = 2(\delta_{ad}\delta_{bc} - \frac{1}{2} \delta_{ab}\delta_{cd}) .$

So finally, the amplitude is:

$$|M|^2 =4a^2(\delta_{ad}\delta_{bc} - \frac{1}{2} \delta_{ab}\delta_{cd}) ^2 (u_{1b}^{\mu}\bar{u}_{1 \mu a} u^{\mu}_{1'b}\bar{u}_{1'\mu a})\\ \times(u_{2'd}^{\nu}\bar{u}_{2'\nu c}u_{2d}^{\nu}\bar{u}_{2\nu c}) $$

The delta identity gives three terms; the first, let's say $T_1$, is: $$T_1 = \delta_{ad}\delta_{bc}\delta_{ad}\delta_{bc} \omega = 4 (u_{1b}^{\mu}\bar{u}_{1 \mu a} u^{\mu}_{1'b}\bar{u}_{1'\mu a})\\ \times(u_{2'a}^{\nu}\bar{u}_{2'\nu b}u_{2a}^{\nu}\bar{u}_{2\nu b}),$$ where $\omega $ is the series of spinors before contracting the deltas and the 4 coefficient comes from the contraction of deltas.

general relativity - Is spacetime simply connected?

As I've stated in a prior question of mine, I am a mathematician with very little knowledge of Physics, and I ask here things I'm curious about/things that will help me learn.

This falls into the category of things I'm curious about. Have people considered whether spacetime is simply connected? Similarly, one can ask if it contractible, what its Betti numbers are, its Euler characteristic and so forth. What would be the physical significance of it being non-simply-connected?

Answer

I suppose there are many aspects to look at this from, anna v mentioned how Calabi-Yao manifolds in string theory (might?) have lots of holes, I'll approach the question from a purely General Relativity perspective as far as global topology.

Solutions in the Einstein Equations themselves do not reveal anything about global topology except in very specific cases (most notably in 2 (spacial dimensions) + 1 (time dimension) where the theory becomes completely topological). A metric by itself doesn't necessarily place limits on the topology of a manifold.

Beyond this, there is one theorem of general relativity, called the Topological Censorship Hypothesis that essentially states that any topological deviation from simply connected will quickly collapse, resulting in a simply connected surface. This work assumes an asymptotically flat space-time, which is generally the accepted model (as shown by supernova redshift research and things of that nature).

Another aspect of this question is the universe is usually considered homogenous and isotropic in all directions, topological defects would mean this wouldn't be true. Although that really isn't a convincing answer per say...

newtonian mechanics - What has the potential energy: the spring or the body on the spring?

Particles have gravitational potential energy due to its position in the gravitational field. We say the particle has potential energy and not the Earth (the body doing the work). Why is it not the same with a spring doing work on a body?

It is my understanding that we can define a potential energy function for all systems being acted on by a conservative force. Since the spring force is conservative, why can't we define a spring (elastic) potential energy for the body? Why is the potential energy defined only for the spring?

spacetime - In Einstein's General Relativity, do the space-time dimensions curve?

In Einstein's General Relativity, do the space-time dimensions curve according to the positions of stars, planets, and masses?

newtonian mechanics - Norton's dome and its equation

Norton's dome is the curve $$h(r) = -\frac{2}{3g} r ^{3/2}.$$ Where $h$ is the height and $r$ is radial arc distance along the dome. The top of the dome is at $h = 0$.

Via Norton's web.

If we put a point mass on top of the dome and let it slide down from the force of gravity (assume no friction, mass won't slide off dome), then we will get the equation of motion $$\frac{d^2r}{dt^2} ~=~ r^{1/2}$$ (Not just me, lots of sources give this answer).

But this equation of motion doesn't make sense. Because as $r$ becomes large, the tangential force is also becoming large. The tangential force should always be less than or equal to the drive force from gravity. What am I seeing wrong?

homework and exercises - Relativistic charged particle in a constant uniform electric field

I'm doing some special relativity exercises. I have to find $x(t)$ and $v(t)$ of a charged particle left at rest in $t=0$ in an external constant uniform electric field $\vec{E}=E_{0} \hat{i}$, then with that velocity I should find the Liénard–Wiechert radiated power.

I will show you what I did but I feel that it is wrong.

We should solve the equation of motion given by

$$ \tag{1}\frac{dp^{\mu}}{d\tau} = \frac{q}{c} F^{\mu \nu}u_{\nu} $$

The four-velocity is given by

$$ u^{\mu} = (u^{0},u^{1},u^{2},u^{3}) = \gamma (c,v^{1},v^{2},v^{3}) $$

where $v^{\alpha}$ are the components of the three-velocity. The four-momentum is

$$ p^{\mu} = mu^{\mu} $$

This will give us four equtions where two of them will give a constant velocities and the other two are

$$ \tag{2}\frac{d\gamma}{d\tau} = -\frac{qE_{0}}{mc^{2}}\gamma v_{1} $$

$$ \tag{3}\frac{d\gamma}{d\tau} v_{1} + \gamma \frac{dv_{1}}{d\tau} = \frac{qE_{0}}{m} \gamma $$

Replacing $(2)$ in $(3)$ gives

$$ \tag{4}\frac{dv_{1}}{d\tau} = -\frac{qE_{0}}{mc^{2}} (v_{1})^{2} + \frac{qE_{0}}{m} $$

The solution of the ODE $(4)$ gives something like

$$ \tag{5}v_{1}(\tau) = A\tanh{(B\tau)} $$

This component of the three-velocity is in terms of the proper time $\tau$ and the problem ask me to find the velocity in terms of the time $t$. So my attempt was to solve

$$ \tag{6}\frac{dt}{d\tau} = \gamma (\tau) = \frac{1}{\sqrt{1 - \frac{(v_{1}(\tau))^{2}}{c^{2}}}} $$

and then replacing this solution for $\tau$ in $(5)$. But the solution of $(6)$ is this. Which doesn't make any sense to me.

I think that I'm misunderstanding something or missing something that will give me a easier solution to this problem. I thought it because in the Liénard–Wiechert radiated power I sould do $dv_{1}/dt$ which is almost impossible to do it without WolframAlpha.

special relativity - Tachyons and Photons

Is there a particle called a "tachyon" that can travel faster than light? If so, would Einstein's relativity be wrong? According to Einstein no particle can travel faster than light.

When does a particle go through the Higgs Field?

This is a short and simple question...

I have been reading my book on particle physics and quantum physics when I had thought of a question that it failed to answer: "Does a particle enter/interact with the Higgs Field when created, or at some other time? And if the answer is neither of these two, does it constantly go through/react with the Higgs Field?"

I have been looking for this everywhere (here, Wikipedia and other sites) and haven't found an answer yet. Does anyone know that answer (or possible answers) to this?

Answer

Does a particle enter/interact with the Higgs Field when created, or at some other time?

After reading your question a couple of times as well as your comments, it occurs to me that you're picturing something like this: a massless particle is created, interacts once with the Higgs field to acquire a permanent classical like mass which it then 'keeps'.

But, this isn't a valid picture at all. A much better picture is to make an analogy with how photons become 'massive' within a superconductor.

Essentially, in this picture, space is an electroweak superconductor with the apparent mass of particles due to continuous interaction with the superconducting 'fluid'. From the Wikipedia article "Higgs Mechanism":

The Higgs mechanism is a type of superconductivity which occurs in the vacuum. It occurs when all of space is filled with a sea of particles which are charged, or, in field language, when a charged field has a nonzero vacuum expectation value. Interaction with the quantum fluid filling the space prevents certain forces from propagating over long distances (as it does in a superconducting medium; e.g., in the Ginzburg–Landau theory).

There's much more to this but hopefully this gets you pointed in the right direction.

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

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

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

Answer

It just happens to be a coincidence.

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

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

nuclear physics - Gaseous fission: Has it even been demonstrated experimentally?

I've been reading quite a bit about gas-core reactors, a theoretical reactor design where the fissioning of Uranium(along with Plutonium & possibly Thorium)occurs in gas phase. The result is that the heat of the reaction converts the gaseous nuclear fuel into plasma which can be contained in a magnetic bottle. The most feasible design for such a reactor is cylindrical metal reactor vessel with a magnetic solenoid where the electromagnets push inward radially; confining the plasma. I would imagine that such a reactor would need an inner lining of neutron reflecting material to deflect neutrons and bounce them back and forth across the chamber. But do to the chemical properties of Uranium Hexafluoride gas it might be more prudent to use a single, supercritical solid fuel rod assembled vertically inside of a vacuum solenoid. The fuel rod would then be bombarded by intense microwaves from directly above to convert it into plasma after the electromagnets are turned on. But the question remains if it is even possible to compress the plasma to a high enough density to initiate fission. Has this experiment ever been tried? If so, what were the results?

Vectors of polarizations from vector boson field solution

Let's have the solution for vector boson Lagrangian in form of 4-vector field: $$ A_{\mu } (x) = \int \sum_{n = 1}^{3} e^{n}_{\mu}(\mathbf p) \left( a_{n}(\mathbf {p})e^{-ipx} + b_{n}^{+} (\mathbf p )e^{ipx}\right)\frac{d^{3}\mathbf {p }}{\sqrt{(2 \pi )^{3}2 \epsilon_{\mathbf p}} }, $$ $$A^{+}_{\mu } (x) = \int \sum_{n = 1}^{3}e^{n}_{\mu}(\mathbf p) \left( a^{+}_{n}(\mathbf {p})e^{ipx} + b_{n} (\mathbf p )e^{-ipx}\right)\frac{d^{3}\mathbf {p }}{\sqrt{(2 \pi )^{3}2 \epsilon_{\mathbf p}} }, $$ where $e^{n}_{\mu}$ are the components of 3 4-vectors, which are called polarization vectors. There are some properties of these vectors: $$ e_{\mu}^{n}e_{\mu}^{l} = -\delta^{nl}, \quad \partial^{\mu}e^{n}_{\mu} = 0, \quad e_{\mu}^{n}e_{\nu}^{n} = -\left(\delta_{\mu \nu} - \frac{\partial_{\mu}\partial_{\nu}}{m^{2}}\right). $$ The first two are obviously, but I have the question about the third. It is equivalent to the transverse projection operator $(\partial^{\perp })^{\mu \nu}$ relative to $\partial_{\mu}$ space. So it realizes Lorentz gauge in form of $$ \partial^{\mu}A^{\perp}_{\mu} = 0, \quad A^{\perp}_{\mu} = \left(\delta_{\mu \nu} - \frac{\partial_{\mu}\partial_{\nu}}{m^{2}}\right)A^{\nu}, $$ which is need for decreasing the number of components $A_{\mu}$ as vector form of representation $\left(\frac{1}{2}, \frac{1}{2} \right)$ of the Lorentz group by one (according to the number of components of spin-1 field).

I don't understand how to interprete this property. Can it be interpreted as matrice of dot product of 4 3-vectors $e_{\mu}$, which makes one of component of $A_{\mu}$ dependent of anothers three?

Answer

In a sum on polarizations, like $\sum_\lambda~e_{\mu}^{\lambda}(k)~e_{\nu}^{\lambda}(k)$, there is a fundamental difference if you are considering all the polarizations, or only the physical polarizations .

If you take all polarizations, the sum is equals to $g_{\mu\nu}$, and it is in fact a normalizations of the $e_{\mu}^{\lambda}(k)$. This sum is non-physical.

$$\sum_{all ~polarizations~~ \lambda}~e_{\mu}^{\lambda}(k)~e_{\nu}^{\lambda}(k)=-g_{\mu\nu} \tag{1}$$

If you consider only the physical polarizations, this sum is physical, and you will get the pole of the propagator, which is a physical quantity too :

$$\sum_{physical ~polarizations~~ \lambda}~e_{\mu}^{\lambda}(k)~e_{\nu}^{\lambda}(k)=-(g_{\mu\nu} - \frac{k^\mu k^\nu}{m^2})\tag{2}$$

The propagator here is :

$$D_{\mu\nu}(k) = \frac {-(g_{\mu\nu} - \frac{k^\mu k^\nu}{m^2})}{k^2-m^2} \tag{3}$$

classical mechanics - Is the principle of least action a boundary value or initial condition problem?

Here is a question that's been bothering me since I was a sophomore in university, and should have probably asked before graduating: In analytic (Lagrangian) mechanics, the derivation of the Euler-Lagrange equations from the principle of least action assumes that the start and end coordinates at the initial and final times are known. As a consequence, any variation on the physical path must vanish at its boundaries. This conveniently cancels out the contributions of the boundary terms after integration by parts, and setting the requirement for minimal action, we obtain the E.L. equations.

This is all nice and dandy, but our intention is finding the location of a particle at a time in the future, which we do not know a priori; after we derive any equations of motion for a system, we solve them by applying initial values instead of boundary conditions.

How are these two approaches consistent?

Answer

I) Initial value problems and boundary value problems are two different classes of questions that we can ask about Nature.

Example: To be concrete:

1. 
   

   an initial value problem could be to ask about the classical trajectory of a particle if the initial position $q_i$ and the initial velocity $v_i$ are given,

   
   


2. 
   
   

   while a boundary value problem could be to ask about the classical trajectory of a particle if the initial position $q_i$ and the final position $q_f$ are given (i.e. Dirichlet boundary conditions).

   
   

II) For boundary value problems, there are no teleology, because we are not deriving a (100 percent certain deterministic) prediction about the final state, but instead we are merely stating that if the final state is such and such, then we can derive such and such.

III) First let us discuss the classical case. Typically the evolution equations (also known as the equations of motion(eom), e.g. Newton's 2nd law) are known, and in particular they do not depend on whether we want to pose an initial value question or a boundary value question.

Let us assume that the eom can be derived from an action principle. (So if we happen to have forgotten the eom, we could always rederive them by doing the following side-calculation: Vary the action with fixed (but arbitrary) boundary values to determine the eom. The specific fixed values at the boundary doesn't matter because we only want to be reminded about the eom; not to determine an actual solution, e.g. a trajectory.)

IV) Next let us consider either an initial value problem or a boundary value problem, that we would like to solve.

Firstly, if we have an initial value problem, we can solve the eom directly with the given initial conditions. (It seems that this is where OP might want to set up a boundary value problem, but that would precisely be the side-calculation mentioned in section III, and it has nothing to do with the initial value problem at hand.)

Secondly, if we have a boundary value problem, there are two possibilities:

1. 
   

   We could solve the eom directly with the given boundary conditions.

   
   


2. 
   

   We could set up a variational problem using the given boundary conditions.

   
   

V) Finally, let us briefly mention the quantum case. If we would try to formulate the path integral

$$\int Dq ~e^{\frac{i}{\hbar}S[q]}$$

as an initial value problem, we would face various problems:

1. 
   

   The concept of a classical path would be ill-defined. This is related to that the concept of the functional derivative $$\frac{\delta S[q]}{\delta q(t)}$$ would be ill-defined, basically because we cannot apply the usual integration-by-part trick when the (final) boundary terms do not vanish.

   
   


2. 
   

   To specify both the initial position $q_i$ and the initial velocity $v_i$ would violate the Heisenberg uncertainty principle.