particle physics - Matter - Antimatter Reactory Practicality

With current technology, would the energy released by a matter-antimatter annihilation be more than the energy needed to created the antimatter in the first place? Would it be worth it? Just curious, even if was on a small scale, Could one produce a machine that uses annihilation to produce a net energy gain?

Answer

If you need to produce the antimatter from (e.g. electric) energy – e.g. antiprotons may be produced from collisions of proton pairs at the LHC (most of the initial, "invested" energy is the kinetic energy of the protons there) – then you're obviously not getting an economic source of energy because you're converting energy to mass and back which would be ideally energy-neutral except that the efficiency can't be 100 percent so you're losing energy, after all.

In particular, the LHC consumes as much energy as the nearby city of Geneva and the energy you may get from the produced antimatter is surely not enough to pay the energy debt – it's lower by many and many orders of magnitude.

The annihilation would of course be a tremendous source of energy if we found some "antimatter for free" and if we could harness it: one would get 1,000 times energy from a kilogram of fuel than from nuclear fission power plants. That's what $E=mc^2$ implies. This is of course a speculation. We know that there's no significant reservoir of antimatter left e.g. in the Milky Way because it would have already annihilated with the ordinary matter which is pretty much everywhere, in low densities that are enough. In fact, the antimatter and matter used to be almost balanced, 1 billion and 1 of matter's mass versus 1 billion pieces of antimatter. Matter won the bloody conflict so some baryons survived which is why we and the stars are here today. But that's a long time ago. We can't find any cheap antimatter around anymore.

But that doesn't mean that the "antimatter fuel" is impossible in principle. It's exactly as possible in principle as the nuclear fuel and it would be 1,000 times more effective. An antimatter bomb designed to end the world war could weigh just grams etc. The situations are completely analogous. But by trying to produce the fuel that we can't find in our environment, you are missing the point. It's a similar "clever idea" as if you wanted to produce uranium for nuclear power plants if you didn't have it to start with. That's pretty clearly economically unreasonable, isn't it? The case of antimatter is completely analogous. The fact that we don't have enough of this "fuel" around doesn't mean that it's not the most concentrated "fuel" that could exist. It is. Getting $E=mc^2$ of energy from a unit of mass – by complete annihilation – is the maximum you can get out of it.

optics - Refraction: Swimming goggles, short-sightedness, and underwater vision

I have thought of this question due to personal experience. I am short-sighted, and over the last three years my short-sightedness has worsened. Taking a lifeguard certificate again now that I did already three years ago, I noticed while diving that I have almost no vision anymore when trying to find and collect some training rings, that I had no problem to visually identify three years ago. In all cases, I am not wearing any goggles while diving.

So, I came to think whether short-sightedness has an effect that adds to the effect of the eyes being covered by water and the cornea having almost no refraction. This process is explained in other questions here, but I claim that my set of closely related questions is different, not only due to the inclusion of short-sightedness.

I then tried out my normal glasses under water and noticed no effect at all, not positive, not negative. I am now confused as to why the glasses have no effect, but the short-sightedness that they should correct does have a worsening effect. In this, I am assuming, without being able to justify it, that the glasses should have the same effect under water as above water, because, as the water touches them from both sides, the light should have a different path inside the length (less bending at entry and exit), but that the light rays exiting the glasses under water should be parallel to the light rays that would exit it in air. But what is instead the mechanism at work here that renders air-glasses useless under water?

Second, I notice that, when wearing swimming goggles, my short-sightedness is alleviated under water. This means that I can see better under water with swimming goggles and without any contact lenses or corrective glasses than I can above water without glasses or lenses. A quick search on the web found other questions of short-sighted people that noticed this effect, but I could not identify how this is working out for normal-sighted or far-sighted people. How does this work?

Then I pondered, as contact lenses are swimming on the eye, but, when underwater, have also contact with water instead of the air they are designed for, do contact lenses have an effect under water (ignoring the risk of them swimming away quickly)?

Finally, what would be the dimensions of a pair of glasses that actually work under water (be it for normal-sighted or short-sighted persons), when we, in turn, ignore their performance in air? How thick and large would they be?

gravity - How come gas molecules don't settle down?

If the earth's gravity exerts a net downward gravitational force on all air molecules, how come the molecules don't eventually lose their momentum and all settle down? How is the atmosphere is still miles thick after billions of years?

Answer

First of all, gravity does continually accelerate the air molecules. I don't see how this could make them lose their momentum.

What is the net effect of gravity on the atmosphere? Simple, gravity prevents the atmosphere from flying off in space, and instead it keeps it comfortably wrapped around our planet!

The reason why the atmosphere is still thick after billions of years is because you have two net effects on the air molecules, gravity, which keeps it as close as possible to the ground, and inertia, who has the opposite net effect. So as long as the molecules do not slow down they "orbit" our planet.

The reason is the same as why is the moon orbiting the Earth after billions of years. There's a balance between the kinetic energy of the moon and the gravitational potential energy - or a balance between gravity and inertia.

The other answers give you a summary explanation of what determines the temperature of the atmosphere and hence its molecules' average velocity. The reality is way more complicated as the temperature of the atmosphere is not constant with height and you have to take into account many more factors like varying pressure, convection and so on. Modelling the Earth's atmosphere accurately is very complicated.

In conclusion the basic mechanisms are outlined above. I hope they answer your question.

antimatter - How detectors in particle colliders can differentiate neutrons from antineutrons?

Their mass is the same. None of them interacts with EM fields. And their decay (around 1000s) is far too slow to see their decay products yet in the detector.

How is it then possible to differentiate them?

Answer

Detectors at particle colliders are layered like onions around the collision vertex.

The CMS detector at CERN

First there are charged particle sensitive detectors where charged particles leave tracks because of ionisation, but mass density is low so strong interactions do not happen often; their momentum can be measured by the curvature in the imposed magnetic field.

Then there are electromagnetic calorimeters, where photons leave their energy and charged particles continue as tracks.

Then come the hadronic calorimeters with a lot of mass so that strongly interacting particles, hadrons, protons neutrons antiprotons antineutrons deposit their energy. Protons will have a continuous path up to the hadronic calorimeter due to their charge. Antiprotons will have negative charge. Neutrons will deposit energy without a previous track trace. Antineutrons will also deposit energy without a track, except due to the annihilation with matter the shower will be more energetic.

At LHC energies the difference in the multiplicity due to the annihilation for antineutrons will not be distinguishable. At low energies , antineutrons have higher multiplicity showers.

Generally in colliders the existence of antineutrons might be guessed at by conservation of charge and baryon numbers, in low multiplicity events.

Temperature of gases

I can't find any law that states this (maybe the combined gas law does and I'm misinterpreting it?), but Feynman said that if you compress a gas, the temperature increases. This makes sense, for example, a diesel engine (or gas engine with insufficient octane or too high a compression ratio). Also, must thinking about a piston "hitting" particles as it is compressed makes sense that energy is imparted.

But he goes on to say that when the gas expands, there is a decrease in temperature. This used to make more sense to me, but the more I think about it, it doesn't at all. Why would the particles lose energy if the container expands?

cosmology - Does red shift evidence necessarily imply that the universe started from a singularity?

We are taught that the universe began as a singularity - an infinitely small and infinitely dense point. At the beginning of time there was a 'Big Bang' or, more accurately, 'Inflation'.

The main evidence for this is the observation of the red shift of all of the galaxies. This shows us that as time increases, the universe becomes bigger. A logical outcome of this is that going back in time, the universe shrinks. This is then extrapolated back to the beginning of time, where the universe was infinitely small.

However, the only thing that I can see that we know for sure is that the universe used to be smaller. This does not necessarily imply that it used to be infinitely small. How do we know, for example, that the universe doesn't oscillate and that we are in a time where the universe is expanding (and accelerating) and will eventually contract again?

What other evidence is there that suggests we started from an singularity?

Answer

There are 3 observations that support the big bang theory, i.e. origin of the universe in a singularity:

1. The redshift of galaxies, as you already mentioned.


2. The cosmic background radiation.


3. The amounts of different nuclei in the universe, notably the preponderance of light elements like hydrogen and helium.

Each of these alone would probably not be sufficient to support the big bang theory. The redshift of galaxies could be explained by some other theory, some have been suggested by Hoyle and Narlikar in the past. Probably the other two phenomena could be explained independently as well, but it is the conjunction that fits so well with the big bang hypothesis.

Does that settle the matter once and for all? Short answer is no. Since these 3 observations have been made and confirmed, more detailed observations have been added to the mix and this has complicated the story for the big bang model. But that would take us into a longer post. The current model which is the most widely accepted is the so-called Lambda-CDM model.

As for the problem of the universe starting in a real singularity, instead of a very dense state, this is still an open problem related to a yet to be invented (or completed) theory of quantum gravity. Our current understanding of singularities in General Relativity is going back to the Penrose-Hawking singularity theorems. They are of the kind "Here be dragons!" in that they delineate the conditions for singularities to form and point where our knowledge ends. More can not be done, because a singularity is basically a failure of the theory.

metals - Radiation emitted from a real melted piece of steel [1800C] VS emissivity

In a lot of questions I read that a good approximation of the radiation emitted from a hot piece of steel is the black body radiation.

Than I search for the value of the emissivity of molted metal and I find that they are (for a molten Pure Iron) around 0.4 or less.

I know that the emissivity is the ratio between: the radiant exitance of that surface and the radiant exitance of a black body at the same temperature as that surface, so I expect to find a value which is very close to 1 while I get values of about 0.4.

So my question is: Why?

I hope to not forget something important.

Answer

I think you are confusing the fact that molten metal emits radiation with a spectrum that is very much like a black body spectrum with the fact that the material has a certain ability to absorb or emit radiation at all at any given wavelength (the emissivity).

When a physicist says something radiates like a black body, they do not mean that it has an emissivity of 1. What they mean is that spectrum of radiation is similar to that of the black body spectrum.

homework and exercises - When to use Conservation of Energy vs Conservation of Momentum

I am confused as to when each should be used to solve problems.I am not looking for the answer to this question, but I will include it here so you have all the information about what I am asking about.

A 15.0 kg block is attached to a very light horizontal spring of force constant 400.0 N/m and is resting on a frictionless horizontal table. (See (Figure 1) .) Suddenly it is struck by a 3.00 kg stone traveling horizontally at 8.00 m/s to the right, whereupon the stone rebounds at 2.00 m/s horizontally to the left. Find the maximum distance that the block will compress the spring after the collision.

I see that the idea suggested here (at Yahoo Answers) is to use conservation of momentum to solve this problem. But I wanted to solve using conservation of energy in the following way:

$K.E =\frac{mv^2}{2}$

$Spring-P.E=\frac{kx^2}{2}$

$KE_i = KE_f + PE_f$

$KE_i - KE_f = PE_f$

Solving the equations, you get

$x = 0.670m$

The correct answer given is 0.387m.

Answer

When to use which law

Your assumption that conservation of energy (considering only kinetic energy) works while dealing with the collision in the above question is not correct.

Conservation of energy (kinetic energy) doesn't appear to work in all kinds of collision. Some of the initial kinetic energy of the bodies are lost as heat and/or part of it is stored in the form of potential energy of the bodies (deformed body). These kind of collisions are called inelastic collisions.

Hence, direct application of conservation of energy with just kinetic energy terms is not possible. In these cases, the problem cannot be solved with just conservation of momentum. You need some experimental input (usually the coefficient of restitution is given).

However, there are cases where conservation of energy (initial kinetic energy = final kinetic energy) is applicable. Such collisions are called elastic collisions.

Conservation of momentum is always valid and safe whereas conservation of energy requires all forms of energy including heat, sound, light, etc to be considered (which ever stated)

Solution to the given problem

In the above problem, the final velocity of the block is already given. By using the conservation of momentum, you can calculate the final velocity of the block.

Initial Momentum of the stone= $mv$ = $24$ $kg.ms^{-2}$

Applying the law of conservation of momentum,

$24 = m_{stone}v_{final-stone} + m_{block}v_{block-final}$

Substituting the given values in the equation, you get the final velocity of the block to be $2 ms^{-2}$.

Before the collision, the block was at its mean position. After the collision the block will begin to oscillate with the same mean position.

The total energy of the system is equal to the kinetic energy supplied by the moving stone. When the block reaches its extreme position, all the energy will exist in the form of potential energy of the spring. Therefore, by applying the law of conservation of energy,

$\frac{mv^2}{2} = \frac{kx^2}{2}$

You get x to be $\sqrt{\frac{3}{20}}$

astronomy - Spaghettification of humans near black holes

A few months ago I was discussing the spaghettification phenomenom with my wife, just for the fun of it. This was when the mass of the super massive black hole from M87 hit the news. The black hole was meant to have 6.7e+9 solar masses, so I thought it would be fun to figure out at which distance the tidal forces become so great, that the forces between your head and feet start getting... uncomfortable.

I defined the distance from feet to head to be 2 m, the head having a mass of 10 kg and one foot having a mass of 2.5 kg (I didn't research the numbers, I just did some wild guesses).

So, the force between head and foot should be:

$$G*m_B*m_H \over (d+2)^2$$ $$-$$ $$G*m_B*2*m_F \over d^2 $$ where $d$ is the distance from foot to the black hole, $m_B$ is the mass of the black hole, $m_H$ the mass of the head and $2*m_F$ the mass of the feet. Assuming you fall with the feet first.

This is one of the plots I made: (note: the plot only takes the mass of one foot into account.)

I made up this short octave script to plot the forces on this values with the distance in AU as the free variable and thought "ouch, that's gonna hurt pretty soon". I haven't checked which part of the body will rip off first at what forces though...

Anyway, what I wanted to know is: Do my calculations make sense? Is my math correct?

Thanks, Alex.

homework and exercises - Bug in linear thermal expansion, $L_0$ must be $0$

Assume we change the temperature of an object with negligible size in $2^{nd}$ and $3^{rd}$ dimensions from $T_0$ to $T_1$ to $T_2$, with all of them pairwise different. We choose a substance with coefficient $\alpha \ne 0$.

$\alpha L_0(T_2 - T_0) = L_2 - L_0 = (L_2 - L_1) + (L_1 - L_0) = \alpha L_1(T_2 - T_1) + \alpha L_0(T_1 - T_0)$
$\Leftrightarrow L_1T_2 - L_1T_1 + L_0T_1 - L_0T_2 = 0$
$\Leftrightarrow (T_2 - T_1)(L_1 - L_0) = 0$
$\Leftrightarrow \alpha L_0(T_2 - T_1)(T_1 - T_0) = 0$

$\Rightarrow L_0 = 0$

Where is my mistake?

Answer

In principle, you need to integrate the relevant equation for linear expansion (http://en.wikipedia.org/wiki/Thermal_expansion#Linear_expansion ). You, however, use just a linear approximation and make some conclusions based on the term quadratic in $\triangle T$. So you use the linear approximation beyond the limits of its applicability.

quantum mechanics - Factor 2 in Heisenberg Uncertainty Principle: Which formula is correct?

Some websites and textbooks refer to $$\Delta x \Delta p \geq \frac{\hbar}{2}$$ as the correct formula for the uncertainty principle whereas other sources use the formula $$\Delta x \Delta p \geq \hbar.$$ Question: Which one is correct and why?

The latter is used in the textbook "Physics II for Dummies" (German edition) for several examples and the author also derives that formula so I assume that this is not a typing error.

This is the mentioned derivation:

$\sin \theta = \frac{\lambda}{\Delta y}$

assuming $\theta$ is small:

$\tan \theta = \frac{\lambda}{\Delta y}$

de Broglie equation:

$\lambda = \frac{h}{p_x}$

$\Rightarrow \tan \theta \approx \frac{h}{p_x \cdot \Delta y}$

but also:

$\tan \theta = \frac{\Delta p_y}{p_x}$

equalize $\tan \theta$:

$\frac{h}{p_x \cdot \Delta y} \approx \frac{\Delta p_y}{p_x}$

$\Rightarrow \frac{h}{\Delta y} \approx \Delta p_y \Rightarrow \Delta p_y \Delta y \approx h$

$\Rightarrow \Delta p_y \Delta y \geq \frac{h}{2 \pi}$

$\Rightarrow \Delta p \Delta x \geq \frac{h}{2 \pi}$

So: Which one is correct and why?

general relativity - How does an object falling into a plain Schwarschild black hole appear from near the black hole?

I know that when viewed from infinity (or from a very large distance from the black hole event horizon), an object that falls into the black hole will appear to slow down and will become more and more red-shifted as it approaches the event horizon. To the far away observer the object will never be seen to enter past the event horizon since the time dilation at the event horizon approaches $\infty$ as the object approaches it.

Conversely, I know that if you are on an object falling into a black hole, you will simply fall past the event horizon and fall faster and faster and will eventually reach the singularity at the center of the black hole in a finite length of time as measured by the observer on the object.

However, what happens if the observer is in an orbit around the black hole at say, the last stable orbit for a material object at a distance of $3Rs$ ? EDIT: (thanks @Ron) Being in orbit or using a rocket engine to hover near the black hole gives a result similar to the observer at infinity: the infalling object will get more and more redshifted as it approaches the horizon but will never be seen to cross the horizon.

EDIT: So the remaining open question now is: what happens if there are two observers that are both falling into the black hole with one observer slightly ahead of the other observer by a small distance? What does the second observer see when the first observer crosses the event horizon? How does it change when both observers have crossed the event horizon?

EDIT: (Thanks @Ron) I now understand that the second observer will only see the first object cross the horizon exactly when the second observer himself crosses the horizon. (I think of it as the photons are just sitting there at the horizon waiting for the observer to hit them.) My only remaining question is, does the redshift of the infalling object just increase smoothly and continuously with time from the observer's point of view as they both cross the horizon and as they head for the singularity? What redshift will the observer see when the first object hits the singularity itself?

Answer

You don't have to orbit, you can just use a rocket to stay put. All observers that can communicate with infinity for all time agree about the infalling object. It gets frozen and redshifted at the horizon.

EDIT: in response to question

The issue of two objects falling in one after the other is adressed in this question: How does this thought experiment not rule out black holes? . The answers there are all wrong, except mine (this is not an arrogant statement, but a statement of an unfortunate fact).

When you are near a black hole, in order to stay in place, you need to accelerate away from the black hole. If you don't, you fall in. Whenever you accelerate, even in empty Minkowksi space, you see an acceleration event horizon behind you in the direction opposite your acceleration vector. This horizon is a big black wall that follows you around, and you can attribute the various effects you see in the accelerating frame, like the uniform gravitational field and the Unruh radiation, to this black-wall horizon that follows you around.

When you are very near a black hole, staying put, your acceleration horizon coincides with the event horizon, and there is no way to tell them apart locally. This is the equivalence principle, in the form that it takes in the region by the horizon where there is no significant curvature.

The near-horizon Rindler form of the metric allows you to translate any experiment you can do in the frame near a black hole to a flat space with an accelerating observer. So if you measure the local Hawking temperature, it coincides with the Unruh temperature. If you see an object fall and get redshifted, you would see the same thing in empty space, when accelerating.

The point is that the acceleration you need to avoid falling in is only determined globally, from the condition that you stay in communication with infinity. If you stop accelerating so that you see the particle cross the horizon, the moment you see the particle past the horizon, you've crossed yourself.

electromagnetism - Question about physical degree of freedom in Maxwell Theory: Why Coulomb gauge can fix all redundant degree of freedom

Given $4$-potential $A^\mu(x)=(\phi(x),\mathbf{A}(x))$, the vacuum Maxwell equations: $$\nabla^2\phi+\frac{\partial}{\partial t}(\nabla\cdot \mathbf{A} )=0$$ $$\nabla^2 \mathbf{A} -\frac{\partial^2 \mathbf{A} }{\partial t^2} - \nabla(\nabla\cdot\mathbf{A} + \frac{\partial \phi}{\partial t})=0$$

There is redundant degree of freedom(d.o.f) in $A^\mu(x)$: $$A^\mu(x)\rightarrow A^\mu(x) +\partial^\mu \lambda(x) $$

In Coulomb gauge: $$\nabla\cdot \mathbf{A}(x)=0 \tag{1}$$ Vacuum Maxweel equation becomes: $$\nabla^2\phi =0 \tag{2}$$ $$\nabla^2 \mathbf{A} -\frac{\partial^2 \mathbf{A} }{\partial t^2} = \nabla( \frac{\partial \phi}{\partial t}) \tag{3}$$

Then we can always choose $\phi(x)=0$. So question becomes that physical degree of freedom is $A^\mu(x)= (0,\mathbf{A}(x))$ with one constraint $\nabla\cdot \mathbf{A}(x)=0$. Every textbook then says the physical degree of freedom is $2$. But it seems there are still redundant d.o.f , we can always make $$\mathbf{A}(\mathbf{x},t)\rightarrow \mathbf{A}(\mathbf{x},t)+\nabla \Lambda(\mathbf{x})$$ such that $$\nabla^2 \Lambda(\mathbf{x}) =0\tag{4}$$ But above equation is Laplace equation that has nontrivial solutions, harmonic function. For example, $\Lambda(\mathbf{x}) = xyz $.

My questions

1. 
   

   Using $\phi(x) =0 $ and $\nabla\cdot \mathbf{A}(\mathbf{x},t)=0$, I have substract two redundant d.o.f. , why fixing $\Lambda(\mathbf{x})$ further can not substract more redundant d.o.f. ?

   
   



2. 
   

   Many textbooks will argue that $A^\mu(x)$ should vanish at spacial infinity, so Laplace equation $(4)$ with zero boundary condition in infinity has only trivial solution. But why do we have to require $A^{\mu}$ vanish at spacial infinity? For example, a uniform magnetic field has $\mathbf{A}(x) = \mathbf{B}\times \mathbf{r}/2$ which does not vanish in infinity. If you require that $A^\mu$ vanish at spacial infinity, you even cannot get constant electrical or magnetic field solutions from Vacuum Maxwell equations. And even the elctromagnetic wave solution $e^{i k(t -x)}$ is also nonvanishing at spacial infinity. This question has some relation with Coulomb gauge fixing and “normalizability”

   
   


3. 
   

   Why textbook says "Lorentz gauge is Lorentz invariant but cannot fix all redundant d.o.f. Coulomb gauge can fix all redundant d.o.f but is not Lorentz invariant. "? But it's obvious that only Coulomb gauge $(1)$ also cannot fix all redundant d.o.f. We can see that the gauge fixing $\phi =0 $ is not a consequence of $(2)$. It's forced artificially and $\phi=0$ is independent from Coulomb gauge. For example, vacuum Maxwell equations $(2),(3)$ can have uniform electrical field solution $\phi(x)=-\mathbf{E}\cdot \mathbf{r}$, $\mathbf{A}=0$ satisfying only Coulomb gauge $(1)$ but $\phi(x)\neq 0$. If we requires $\phi=0$ and $\nabla \cdot \mathbf{A}=0$, the solution becomes $\phi=0$ and $\mathbf{A}= \mathbf{E} t$.

   
   

gravity - Does light have mass as it is attracted through gravitational force?

I have heard that no one escape from the intense gravitational field of a black hole (obviously that's why it is black). And gravitational force is due to the mass having the body, no mass no gravity, as $$F=GMm/r^2$$ Putting $0$ in one of the two bodies' masses, gives no gravitational force. But light cannot escape from black hole due to its strong gravity, does that mean that light has mass? please correct me.

Answer

Light does not have mass and Newton's theory of gravity doesn't work here. Any massive object, like the Sun, locally deforms the flat fabric of spacetime into a curved one. Light bends because it has to follow the geodesic (shortest distance between two points in a curved space) as Einstein taught us in his general theory of relativity.

Is there a way to prove that a bound state wavefunction can always be chosen real for an arbitrary potential in Quantum Mechanics?

As we can prove many things that always (at least in introductory quantum mechanical problems) apply using an arbitrary potential (like that $E>V_{\rm min}$ or else the solutions are non-normalizable and superpositions of them can't produce normalizable wavefunctions), is there a way to generally prove for an arbitrary potential that bound states always correspond to real functions?

Answer

I) Yes, the time independent Schrödinger equation (TISE) of the form $$\left(-\frac{\hbar^2}{2m} {\bf \nabla}^2 +V({\bf r}) -E \right) \psi({\bf r}) ~=~0 $$ is $\mathbb{C}$-linear and invariant$^1$ under complex conjugation. So if the wave function $\psi$ is a solution with finite square-norm, then so will $\psi^{\ast}$, $$\frac{\psi+\psi^{\ast}}{2}\quad\text{and}\quad \frac{\psi-\psi^{\ast}}{2i}$$

be. The two latter are real solutions, and at least one of them has non-zero square-norm if $\psi$ has non-zero square-norm. Hence we can always choose a normalizable solution to be real. See also Problem 2.1b in Griffiths, Intro to QM, and this related Phys.SE post.

II) Note that the same argument does not apply to scattering states in the continuous spectrum, since the boundary conditions at infinity of $\psi^{\ast}$ could be off.

Also it does not apply to the time dependent Schrödinger equation (TDSE).

--

$^1$ Here we have implicitly used that the eigenvalue $E\in\mathbb{R}$ is real because the Hamiltonian $H$ is self-adjoint. Note that self-adjointness alone is not enough. E.g. $$H~=~\frac{1}{2m}\left(\frac{\hbar}{i} {\bf \nabla}-q {\bf A}({\bf r}) \right)^2 +V({\bf r})$$ is self-adjoint, but the corresponding TISE is not invariant under complex conjugation.

homework and exercises - Is the tension in both ends the same (on a massed string)?

If two blocks lay on a table and I pull the second block, as shown in the picture below. Is the tension in both ends of the string the same? (The string has a mass)

string - About working of tension force

From the past 3-4 days I have been continuously thinking how tension actually works on atomic level. I have even Googled it many times but I was not satisfied with the answers. Can anyone please help me to understand how does tension actually work when we apply a force on a taut rope.

phase transition - Does any substance melt or evaporate when cooled at constant pressure?

Is there any substance with segments of its phase change diagram lines going in a negative direction?

To explain:

Generally, as phase change diagrams go, with heat increasing, and pressure constant, substances tend to evaporate, sublimate or melt; cooling produces the opposite, fusion (freezing), deposit or condensation into liquid from vapor.

But the diagrams are rarely straightforward - water, for example, has a multitude of forms of ice, where not all forms are obtainable through transitions "from any direction". There are many obscure factors that define when each transition can occur.

Is there any substance, that - without transforming into another substance (say, polymerization) - possesses an area of the phase change diagram (possibly way off "room conditions") where the transition goes in the opposite direction - heating leads to a - not necessarily more dense - but 'more solid' state? Something like thermally hardened glue, but without a chemical transition?

Answer

Yes, here's the phase diagram for Helium-3:

Notice that around 3 MPa, an increase in temperature causes a transition from liquid to solid.

Do individual photons increase entropy as they travel?

If I draw a control volume around a single light wave traveling through empty space, is entropy increasing as it goes? How is the degradation of the quality of energy manifested?

Answer

No, light propagating through space doesn't increase entropy. It's generally considered to be a reversible process, for the simple reason that a suitably curved mirror can be used to restore the initial state. Instead, entropy is produced when light is absorbed or emitted from any surface or volume.

In the ideal case of a black body, the entropy produced by emission is $\frac{4}{3}\frac{J}{T}$, where $T$ is the temperature of the body and $J$ is the net rate of emission, in $\mathrm{Js^{-1}}$. This entropy is associated with the light radiating away from the black body, and its entropy doesn't change until it is absorbed by another body.

Is the Hawking radiation of a charged black hole thermal?

Suppose you have a Schwarzschild black hole of mass $M$ and angular parameter $a = 0$ (no rotation).

Question: is it possible to throw a charge $Q$ at a faster rate than it will be re-radiated? Will the radiation profile be still thermal?

If it is thermal, it would mean that even big, cold black holes would emit a lot of energy in the form of electromagnetic radiation just to be able to get rid of the extra charge? A thermal spectra that starts at $511\: \mathrm{KeV}$ (the energy of the lowest charged particles and has very little emitted power at lower energies) would be a very weird thing to call 'thermal'.

There is a differential expression for the increase in temperature as a small charge (compared to the black hole mass) is added to the black hole that one obtains by taking the formula 11.2.17 in this page (Modern Relativity, 2005, David Waite) and deriving against $e$ and $M$ and taking

$$ \delta T = \left( \frac{\partial T}{\partial e} + \frac{c^2}{G} \frac{\partial T}{\partial M} \right) \delta e$$

so this gives a profile for the variation of black hole temperature with charge.

Question: Is it correct to conclude that you can estimate the overall Hawking radiation of a relatively small black hole ($M \approx 10^{18}~\mathrm{kg}$) by adding electric charge at a faster rate than it will be re-emitted by the thermal spectra?

Or is the spectra totally non-thermal, and the radiation will favour throwing away charged particles, while being electromagnetically cold?

electrostatics - Why don't charges build up on wires like plates of a capacitor?

Suppose you have a parallel plate capacitor and you want to charge it by using a battery. You will connect the plates with the two terminals of the battery. Now my book says that the charges will get accumulated on the plates of the capacitor. The amount of charge will be equal to the capacitance times the voltage of the battery. But what about the wires they are also on the same potential why don't charges accumulate on the wires? I have come across this many a times in capacitor circuits where wires don't get any charge, all the charges accumulate on the capacitors but I just don't understand why.

particle physics - What is the mass density distribution of an electron?

I am wondering if the mass density profile $\rho(\vec{r})$ has been characterized for atomic particles such as quarks and electrons. I am currently taking an intro class in quantum mechanics, and I have run this question by several professors. It is my understanding from the viewpoint of quantum physics a particle's position is given by a probability density function $\Psi(\vec{r},t)$. I also understand that when books quote the "radius" of an electron they are typically referring to some approximate range into which an electron is "likely" to fall, say, one standard deviation from the expectation value of its position or maybe $10^{-15}$ meters.

However it is my impression that, in this viewpoint, wherever the particle "is" or even whether or not the particle "had" any position to begin with (via the Bell Inequalities), it is assumed that if it were (somehow) found, it would be a point mass. This has been verified by my professors and GSIs. I am wondering if its really true.

If the particle was truly a point mass then wherever it is, it would presumably have an infinite mass density. Wouldn't that make electrons and quarks indistinguishable from very tiny black holes? Is there any practical difference between saying that subatomic particles are black holes and that they are point masses? I am aware of such problems as Hawking Radiation although at the scales of the Schwarchild radius of an electron (back of the envelope calculation yields $\tilde{}10^{-57}$ meters), would it really make any more sense to use quantum mechanics as opposed to general relativity?

If anyone knows of an upper bound on the volume over which an electron/quark/gluon/anything else is distributed I would be interested to know. A quick Google Search has yielded nothing but the "classical" electron radius, which is not what I am referring to.

Thanks in advance; look forward to the responses.

Answer

Let me start by saying nothing is known about any possible substructure of the electron. There have been many experiments done to try to determine this, and so far all results are consistent with the electron being a point particle. The best reference I can find is this 1988 paper by Hans Dehmelt (which I unfortunately can't access right now) which sets an upper bound on the radius of $10^{-22}\text{ m}$.

The canonical reference for this sort of thing is the Particle Data Group's list of searches for lepton and quark compositeness. What they actually list in that reference is not exactly a bound on the electron's size in any sense, but rather the bounds on the energy scales at which it might be possible to detect any substructure that may exist within the electron. Currently, the minimum is on the order of $10\text{ TeV}$, which means that for any process occurring up to roughly that energy scale (i.e. everything on Earth except high-energy cosmic rays), an electron is effectively a point. This corresponds to a length scale on the order of $10^{-20}\text{ m}$, so it's not as strong a bound as the Dehmelt result.

Now, most physicists (who care about such things) probably suspect that the electron can't really be a point particle, precisely because of this problem with infinite mass density and the analogous problem with infinite charge density. For example, if we take our current theories at face value and assume that general relativity extends down to microscopic scales, an point-particle electron would actually be a black hole with a radius of $10^{-57}\text{ m}$. However, as the Wikipedia article explains, the electron's charge is larger than the theoretical allowed maximum charge of a black hole of that mass. This would mean that either the electron would be a very exotic naked singularity (which would be theoretically problematic), or general relativity has to break at some point before you get down to that scale. It's commonly believed that the latter is true, which is why so many people are occupied by searching for a quantum theory of gravity.

However, as I've mentioned, we do know that whatever spatial extent the electron may have cannot be larger than $10^{-22}\text{ m}$, and we're still two orders of magnitude away from probing that with the most powerful particle accelerator in the world. So for at least the foreseeable future, the electron will effectively be a point.

homework and exercises - Why does current from a grounded electrical outlet flow through me if I stay on the floor made out of a dielectric material?

Why does current from a grounded electrical outlet flow through me when I touch a "hot" terminal, if stay on the floor made of a dielectric material?

Consider an AC voltage source, such as a wall socket, at $220 \, \mathrm{V}$ and $60 \, \mathrm{Hz}.$ I stay on the wood, tile, or another dielectric material, which the floor commonly consists. The floor is dry. I touch only one terminal (not ground), it hurts me. How?

I know for sure, that the screwdriver-indicator works because of resistor with huge resistance, like $2000 \, \Omega$ + body resistance, we get $$ I ~=~ \frac{220 \, \mathrm{V}}{\sim 3000 \, \Omega} ~=~70 \, \mathrm{mA} \,,$$the current flows, bulb lights, we do not die. Hence the resistance of wood is less than $2 \, \mathrm{k} \Omega \, ?$

Answer

As the mains supply is alternating current the charging and discharging of the capacitance of your body relative to free space (actually your other closer surroundings) causes a small AC current to flow and will illuminate the test indicator.

You are feeling this real current because if flows into you from a single point.

Also even a poor conductor like rubber and wood will still conduct a bit.

What is coherence in quantum mechanics?

What are coherence and quantum entanglement? Does it mean that two particles are the same?

I read this in a book called Physics of the Impossible by Michio Kaku. He says that two particles behave in the same way even if they are separated. He also says that this is helpful in teleportation. How can this be possible? Could somebody please explain?

Answer

Coherent (or pure) state

Consider 2 basic states $\lvert0\rangle$ and $\lvert1\rangle$. (If you never heard about states, treat them as ordinary complex vectors.) Here we suppose that $\lvert0\rangle$ and $\lvert1\rangle$ are orthogonal ($\langle 0\lvert1\rangle =0$).

Now, consider $\lvert c\rangle = \frac{1}{\sqrt{2}} (\lvert0\rangle + e ^{i\phi}\lvert1\rangle)$.

$\lvert c\rangle$ is a (normalized) coherent state, because the phase $\phi$ is constant.

We may look at the density matrix, defined as $\rho = \lvert c\rangle \langle c\lvert$ (that is $\langle i\lvert\rho\lvert j\rangle=\rho_{ij} = c_i^* c_j$, with $c_1 = \frac{1}{\sqrt{2}}, c_2 = \frac{1}{\sqrt{2}}e ^{i\phi}$). We have:

$$\rho =\frac{1}{2} \begin{pmatrix} 1&e ^{i\phi}\\e ^{-i\phi}&1 \end{pmatrix}. \tag{1}$$

This density matrix describes a coherent state. You may verify that $\rho^2 = \rho$, that is $\rho$ is a projector onto the coherent state $\lvert c\rangle$.

Now, suppose the phase $\phi$ is random (that is: the phase difference between $c_1$ and $c_2$ is random), so the mean expectation value of $e ^{i\phi}$ is just zero, and we have a density matrix:

$$\rho' =\frac{1}{2} \begin{pmatrix} 1&0\\0&1 \end{pmatrix}. \tag{2}$$

The density matrix $\rho'$ does not represent no more a coherent state, this is simply a classical statistical probability law. The off-diagonal elements of the matrix $\rho$ have disappeared.

In the 2 cases, we are dealing with only one particle, and the probabilities to find the particle in the $\lvert0\rangle$ state or the $\lvert1\rangle$ state, are the same, and are $\frac{1}{2}$.

$$\langle 0\lvert\rho\lvert0\rangle= \langle 1\lvert\rho\lvert1\rangle= \langle 0\lvert\rho'\lvert0\rangle= \langle 1\lvert\rho'\lvert1\rangle = \frac{1}{2}. \tag{3}$$

Entanglement

Entanglement is about at least $2$ particles, for instance the pure state

$$ \frac{1}{\sqrt{2}}(\lvert0\rangle\lvert0\rangle+\lvert1\rangle\lvert1\rangle )\tag{4}$$is a maximally 2-particles entangled state.

From a given entanged state, you may calculate the correlations for the joint measurement of the 2 particles.

It appears that these entangled quantum correlations are stronger than the classical statistical correlations.

Correlations do not mean that you may exchange instantaneous information; see also this previous answer.

waves - How does an acoustic guitar amplify its sound?

An essential part of a guitar is its hollow body. Without it, the strings wouldn't be very loud; as far as I know, the purpose of the body is to set up some sort of resonance and make the sound louder.

How does this work? How can an isolated system amplify sound? Where is the energy coming from?

Answer

It is not amplification! The purpose of the guitar body is to impedance and mode match between the string and the surrounding air.

When a an object vibrates it pushes on the surrounding air creating pressure waves which we hear as sound. A string vibrating alone without the body of the instrument doesn't make a very loud sound because exchange of energy from the vibrating string to the air pressure waves is inefficient. Why is it inefficient?

The fundamental reason is that the string is a lot stiffer than the surrounding air and has a small cross sectional area. This means that as the string vibrates with a given amount of energy, it doesn't actually displace much air. With the same energy in the motion, a larger, more mechanically compliant object (e.g. the acoustic guitar body) would do a better job at transferring the energy into the air and thus into your ears.

The equation of motion of a vibrating string with vertical displacement $u$ and horizontal position $x$ is

$$\frac{\partial^2 u}{\partial x^2} = \frac{1}{v^2} \frac{\partial^2 u}{\partial t^2}$$

where $v = \mu/T$ is the velocity of sound in the string, $\mu$ is the linear mass density, and $T$ is the tension. If you consider, for example, the fundamental mode, then the solution is of the form

$$u(x,t) = \sin(k x) f(t)\,.$$

Plugging this into the equation of motion gives you

$$ \begin{align} -k^2 f &= \frac{1}{v^2} \ddot{f} \\ 0&= \ddot{f} + \omega_0^2 f \end{align} $$ where $\omega_0^2 = (vk)^2 = (2\pi)^2T/(\mu \lambda^2)$ and $\lambda$ is the wavelength of the mode ($k \equiv 2\pi / \lambda$). This is just the equation of a harmonic oscillator.

Now suppose we add air friction. We define a drag coefficient $\gamma$ by saying that the friction force on a piece of the moving string of length $\delta x$ is

$$F_{\text{friction}} = -\delta x \, \gamma \, \dot{u} \, .$$

Note that $\gamma$ has dimensions of force per velocity per length. Drag coefficients are usually force per velocity; the extra "per length" comes in because we defined $\gamma$ as the friction force per length of string.

Adding this drag term, re-deriving the equation of motion, and again specializing to a single mode, we wind up with

$$0 = \ddot{f} + \omega_0^2 f + \frac{\gamma}{\mu} \dot{f} \, .$$

Now we have a damped harmonic oscillator. The rate at which this damped oscillator decays tells you how fast (i.e. how efficiently) that oscillator transfers energy into the air. The energy loss rate $\kappa$ for a damped harmonic oscillator is just the coefficient of the $\dot{f}$ term, which for us is $\kappa = \gamma / \mu$.$^{[1]}$

The quality factor $Q$ of the resonator, which is the number of radians of oscillation that happen before the energy decays to $1/e$ of its initial value, is $$Q = \omega_0 / \kappa = \frac{2\pi}{\lambda} \frac{\sqrt{T \mu}}{\gamma} \, .$$

Lower $Q$ means less oscillations before the string's energy has dissipated away as sound. In other words, lower $Q$ means louder instrument. As we can see, $Q$ decreases if either $\mu$ or $T$ decreases. This is in perfect agreement with our intuitive discussion above: lower tension would allow the string to deflect more for a given amount of vibrational energy, thus pushing more air around and more quickly delivering its energy to the air.

Alright, so what's going on when we attach the string to a guitar body? We argued above that the lower tension string has more efficient sound production because it can move farther to push more air. However, you know this isn't the whole story with the guitar body because you plainly see with your eyes that the guitar body surface does not deflect even nearly as much as the string does. Note that the guitar surface has much more area than the string. This means that for a given velocity, the frictional force is much higher, i.e. $\gamma$ is larger than for the string.

So there you have it: the guitar body has lower $T$ and higher $\gamma$ than the string. These both contribute to making the $Q$ lower, which means that the vibrating guitar body more efficiently transfers energy to the surrounding air than does the bare string.

Lowering the $T$ to be more mechanically compliant like the air is "impedance matching". In general, two modes with similar response to external force (or voltage, or whatever), more efficiently transfer energy between themselves. This is precisely the same principle at work when you use index matching fluid in an immersion microscope to prevent diffraction, or an impedance matching network in a microwave circuit to prevent reflections.

Increasing the area to get larger $\gamma$ is "mode matching". It's called mode matching because you're taking a vibrational mode with a small cross section (the string) and transferring the energy to one with a larger cross section (the guitar body), which better matches the waves you're trying to get the energy into (the concert hall). This is the same reason horn instruments flare from a tiny, mouth sides aperture at one side, to a large, "concert hall" sized aperture at the other end.

[1] I may have messed up a factor of 2 here. It doesn't matter for the point of this calculation.

rotational dynamics - Why do we use the whole radius vector in the angular momentum, and not just the unit vector?

Why don't we only use the unit vector of the radius? I was thinking that, all we need to know is the component of the linear momentum, perpendicular to it, so why use the magnitude of the radius vector too?

quantum mechanics - After using annihilation operator on vacuum state, why it is $0$ instead of vacuum?

For bosonic systems, why $a|0\rangle=0$ and not $a|0\rangle=|0\rangle$?

Answer

Let's consider the simplest case of a quantum harmonic oscillator, with creation and annihilation operators $a^{\dagger}$ and $a$ respectively. The ground state of our system is, $\lvert 0 \rangle$ which has energy,

$$E_0 = \frac{1}{2}\hbar \omega$$

Every time a creation operator acts, the state $\lvert n \rangle \to \lvert n+1 \rangle$, modulo some constants. Similarly, the annihilation operators lowers the integer $n$. Therefore, if we apply $a$ to the ground state, we reach $n=-1$, which is not allowed,$^{\dagger}$ otherwise our Hamiltonian would be unbounded from below. So the state must be completely annihilated, i.e. zero.

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Suppose we did accept your proposal,

$$a \lvert 0 \rangle = \lvert 0 \rangle$$

It can be shown that such an assumption leads to a contradiction. We may compute the norm of the ground state,

$$ \left( \lvert 0 \rangle \right)^{\dagger} \left(\lvert 0 \rangle \right) = \left( a\lvert 0 \rangle \right)^{\dagger} \left(a\lvert 0 \rangle \right) = \langle 0 \lvert a^\dagger a \rvert 0 \rangle$$

Now, since by the assumption $a\lvert 0 \rangle = \lvert 0 \rangle$, we can make the swap again,

$$\langle 0 \lvert a^\dagger a \rvert 0 \rangle = \langle 0 \lvert a^\dagger \rvert 0 \rangle = \langle 0 \lvert 1 \rangle$$

which is a contradiction, unless we accept $\vert 0 \rangle = \lvert 1 \rangle$, which is clearly not sensible.

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$\dagger$ One of the reasons $n=-1$ is not allowed is as follows: Recall that for the quantum harmonic oscillator, the standard deviations of momentum and position must obey the uncertainty relation,

$$\sigma_x \sigma_p = \hbar\left( n + \frac{1}{2}\right) \leq \frac{\hbar}{2}$$

The lowest value $n$ may take to obey the inequality is $n=0$; any lower and it is violated.

How do you add angular momentum of three or more particles in quantum mechanics?

I'm trying to find some information on how to add the angular momentum of three or more particles, but all the sources I look at deal with only two. In this case I understand that if the angular momentum numbers of the two particles are $j_1$ and $j_2$, then the possible total angular momentum numbers are $J=(j_1+j_2),(j_1+j_2-1)+...+|j_1-j_2|$. However, I don't see how to combine this to three particles.

For example, if I have three protons in a $ 1d_{5/2} $ nuclear energy level (for example), then the protons all have angular momentum $j=5/2$. However, how do I then find the possible total angular momentum of the state? I appreciate that the particles cannot occupy the same state, and hence must have different $m_j$ values which range from $5/2,3/2...-5/2$, and then the $m_j$ is the sum of these, but then how could this be used to find the possible total angular momentum of the state (not just the total $m_j$).

Answer

For every half-integer $j=n/2,n\in\mathbb{Z}$, there is an irreducible representation of $SU(2)$ $$D^j=\exp(-i\vec\theta\cdot\vec J^{(j)})$$ in which the three generators $J_i^{(j)}$ are $(2j+1)\times(2j+1)$ square hermitian matrices. As you probably know, $D^j$ describes states with angular momentum $-j$ to $j$ in integer steps. Given two particles with angular momenta $j$ and $\ell$, we write the total angular momentum of the system as the tensor product $D^j\otimes D^\ell$. We have the standard result $$D^j\otimes D^\ell=\bigoplus_{k=|j-\ell|}^{j+\ell}D^k$$ where on the right it is (conventionally) understood that we write the rotation matrices in order of decreasing angular momentum. Consider, for instance, a meson. This is a composite particle of two spin-half particles. In its ground state, the meson spin representation is $$D^{1/2}\otimes D^{1/2}=D^1\oplus D^0$$ We can thus predict that there are spin-1 and spin-0 mesons, which has been experimentally verified.

Suppose then we wish to find the angular momentum of a baryon. We need $D^{1/2}\otimes D^{1/2}\otimes D^{1/2}$. Convince yourself of the following: Given three matrices $A,B,C$ we have $$A\otimes(B\oplus C)=A\otimes B\oplus A\otimes C$$ Using this, we have $$D^{1/2}\otimes D^{1/2}\otimes D^{1/2}=D^{1/2}\otimes(D^1\oplus D^0)=D^{1/2}\otimes D^1\oplus D^{1/2}\otimes D^0=D^{3/2}\oplus D^{1/2}\oplus D^{1/2}$$ Again, we find baryons with identical quark content in both $3/2$ and $1/2$ states. (There is a slight technicality with the second $D^{1/2}$ involving the Pauli principle.)

The obvious generalization of this is $$D^m\otimes D^j\otimes D^\ell=D^m\otimes\left(\bigoplus_{k=|j-\ell|}^{j+\ell}D^k\right)=\bigoplus_{k=|j-\ell|}^{j+\ell}D^m\otimes D^k=\bigoplus_{k=|j-\ell|}^{j+\ell}\bigoplus_{n=|m-k|}^{m+k}D^n$$ and so on for more particles. From here you can use the Clebsch-Gordan method to construct the actual state vectors for your system.

general relativity - What is a black hole?

Is there a definition of a black hole in a generic spacetime? In some books, for example Wald's, black holes are defined for asymptotically flat spacetime with strong asymptotic predictability, although the definition makes sense without the second condition. Is there a notion of a black hole in general spacetime, not necessarily asymptotically flat? Or is it the case that there is not a "natural" or agreed upon definition?

Answer

What the definition needs to capture is that a black hole is not (1) a naked singularity, or (2) a big bang (or big crunch) singularity. We also want the definition to be convenient to work with so that, for example, it's possible to prove no-hair theorems.

Since we want to exclude naked singularities, it's natural that we require an event horizon. Event horizons are by their nature observer-dependent things. For example, if we have a naked singularity, we can always hide its nakedness by picking an observer who is far away from it and accelerating continuously away from it. Such an accelerated observer always has an event horizon, even in Minkowski space. This example shows that it makes a difference what observer we pick.

Actually, we can't have a material observer at null infinity, since timelike infinity, not null infinity, is the elephants' graveyard for material observers. However, the choice of null infinity is the appropriate one because a black hole is supposed to be something that light can't escape from.

Of course the actual universe isn't asymptotically flat, but that doesn't matter. In practice, all we care about is that the black hole is surrounded by enough empty space so that the notion of light escaping from it is well defined for all practical purposes.

There are other possible ways of defining a black hole, e.g., http://arxiv.org/abs/gr-qc/0508107 .

Is the displacement of a driving oscillator in phase with the driving force?

In a set up such as the following:

I have read in many places that below resonance the driving force is in phase with the harmonic oscillator. I have also read that the driving oscillator is in phase with the harmonic oscillator, however, I would not expect both to be true.

In a harmonic oscillator, I am under the impression that the acceleration (proportional to force) and the displacement are in antiphase. Therefore one would not expect the driving force and driving oscillator to be in phase.

I was wondering if the force acting on the driving force is not the same as the force it applies on the harmonic oscillator, and that perhaps the displacement and applied force of a driving oscillator are in fact in phase.

Is this correct, or do the two original statements, in fact, contradict each other?

Thanks.

Answer

The driveR is applying a force say $F_0 \cos \omega t$ at the end of the "spring" remote from the "mass".
That end is executing simple harmonic motion and so the displacement of the end of the spring must be proportional to the force.

If there is no damping and you do not set your initial conditions carefully you are bound to have transient behaviour superimposed on steady state behaviour.

With damping you eventually on have steady state behaviour where the amplitude of the "mass" is constant.

To figure out and possibly remember what happens as the frequency of the driveR changes consider this.

You are holding the top of a spring and there is a mass attached to the bottom of the spring.
Very, very slowly you move the top of the spring up a certain distance and then do the same with the top of the spring moving down.
What will happen to the mass at the end of the spring?
Now this is where you have to put the mass into a beaker of water if you are actually doing the experiment or assume a reasonable amount of damping so that the transients die away reasonably quickly an do not mask the steady state motion.
The mass at the end of the spring will follow almost exactly the motion of your hand.
The driveR (your hand) and the driveN (the mass at the end of the spring) move in phase with one another.

As you move your hand up and down quicker but with the same amplitude the mass will start lagging behind the motion of your hand and the amplitude of the mass will increase.
It is difficult but not impossible to show this increase in amplitude of the driveN for constant amplitude driveR as the frequency of the driveR increases.

What is easier to show and see is that at higher frequencies of the driveR the amplitude of the driven decreases to a small value and the spring looks as though at times the top and bottom of the spring is being pulled in opposite direction.
The driveR (your hand) and the driveN (mass) are almost in anti-phase.

Here is a video produced by MIT replacing your hand with a wheel the rim of which is coupled by mechanical linkages to the top of a spring which has a mass at the other end.
The advantage of producing a video is that one can wait for the transients to die away before starting to record the motion.
Notice that the mass is damped using air resistance made larger by having a large horizontal circular disc attached to it.
The video shows very clearly that although the amplitude of the driveR (rod on the right) is constant the amplitude of the driveN, which at low frequency is the sane as that of the driveR, changes as the frequency of the driveR increase going though a maximum and then dropping off towards zero.
The phase changes are also clearly to see.

Another demonstration of mechanical resonance which shows the amplitude and phase relationships all at one time is Barton's pendula (sorry about the music) where an osculating pendulum bob of large mass acts as the driveR and pendula of different length with light bobs act as the driveN systems.
You have to wait for the transients to die down.
I chose this video because it also showed the side view which makes it a lot easier to observe both the amplitude of the motions of the pendulum bobs and their relative phase.

thermodynamics - Generalized work to increase both entropy and energy of a system

Suppose we have an adiabatic container with $N$ ideal gas particles, and each particle consists of two identical atoms so that each possesses a vibrational mode. For simplicity, let us assume that the vibrational modes are approximated by harmonic oscillators. In other words, we have $N$ ideal gases and, simultaneously, $N$ harmonic oscillators in the container.

If we exert an external force to compress the container, some amount of work will be transferred into the container, increasing total kinetic energy of ideal gases. In other words, the number of allowable microstates in a momentum phase space of the gases increases. On the other hand, the shirinkage of a coordinate phase space (= reduced volume) cancels out the expansion of the momentum phase space. Thus, the total number of microstates of the container does not change, resulting in no change in entropy of the container, i.e. $$dS =k_B(dln\Omega)= \frac{1}{T}(dE + PdV - \mu dN - fdX) = 0$$ since $dE = -PdV$ and $\mu dN = fdX = 0$ ($f$ and $X$ stand for generalized forces and generalized coordinates respectively). This result agrees with what we already know: $dS=0$ since $dQ_{rev}/T=0$ in an adiabatic compression of an ideal container.

However, suppose that we radiate an electric field to the container. The field will transfer some work to the container, assuming that the field interacts with electron density polarizations of the gases, which results in excitation of the vibrational modes. As a result, the number of allowed modes inside the container will increase and subsequently, the entropy of the container will also increase (the number of microstates rise), i.e. $$dS=k_B(dln\Omega) >0 \tag{1} \label{1} $$. However, since all work done by the electric field is equivalent to the change of the energy of the container, $$dE=fdX$$ and thus, $$dS=\frac{1}{T}(dE + PdV - \mu dN - fdX) =0 .\tag{2} \label{2}$$ Obviously, equations $\eqref{1}$ and $\eqref{2}$ are contradictory. If the generalized work has a form of $-PdV$, I do not have any problem (as illustrated above). The origin of such contradiction may result from the fact that the generalized work done by the electric field increase both the energy and the total number of allowed microstates inside the container.

How can I solve this contradiction? Shouldn't I interpret the energy transferred by the field as a work? If this is true, why? I mean, how can I judge whether an energy transferred to a system is $TdS$ (~ heat) or $fdX$ (~ work)?

special relativity - Klein-Gordon equation and wave velocity

It looks like solutions of the KG eqn travel faster than light, because if $$\omega^2 - k^2 = m^2$$ then $$\mid\ \omega\mid \ > \ \mid k\ \mid$$ and I thought the wave velocity was $\omega / k$.

How do I interpret this?

I heard something about particles following the packet velocity, which isn't too fast, but why can't these fast wave fronts be used for signalling irrespective of what any packets are doing?

Edit:

In the meantime I've been looking at propagators but it all looks too grandiose to me. To my eyes, the KG equation is nothing but a guitar string where each point on the string is pulled back to the quiescent position by a spring. I could set up a simulation of that and ping it with a tall, narrow gaussian. Sure enough, the waves would go faster with the extra springs because it's all stiffer, and that effect would be stronger at long wavelengths where the springs would dominate the curvature, so we'd have a $d\omega/dk$. But I don't think anything would hit infinity (like this propagator thinks) and I'd be very surprised if the far end of the string stayed completely motionless until the packet velocity arrived (or would it?)

So why all this talk of complex numbers and contour integrals that were never part of the original problem? Even if I promoted the displacement to a complex number, the real and imaginary parts would form two independent systems because all the diffs are second order so there's nothing to mix them up. What am I missing?

The possible dupe doesn't answer this. It tells us what wave packets are but not why the phase velocity can't carry a signal.

nuclear physics - Binding energy in fission

In nuclear fission the total rest mass of the products is less than the mass of the original nucleus. But on the other hand, the binding energy of the products is higher. Doesn't the decrease of the rest mass of the products contribute to the increase of their binding energy? So why then there is enormous energy released?

What is the role spin-networks play in Loop Quantum Gravity?

This is a second question (in what will probably become a series) in my attempt to understand LQG a little. In the previous question I asked about the general concepts behind LQG to which space_cadet provided a very nice answer.

Let me summarize that answer so that I can check that I've understood it properly. In LQG, one works with connections instead of a metric because this greatly simplifies the equations (space_cadet makes an analogy to "taking a square root" of K-G equation to obtain Dirac equation). The connections should determine the geometry of a given 3D manifold which is a space-like slice of our 4D space-time. Then, as is usual in quantizing a system, one proceeds to define "wave-functions" on the configuration space of connections and the space of these functionals on connections should form a Hilbert space.

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Note: I guess there is more than one Hilbert space present, depending on precisely what space of connections we work with. This will probably have to do with enforcing the usual Einstein constraints and also diffeomorphism constraints. But I'll postpone these technicalities to a later question.

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So that's one picture we have about LQG. But when people actually talk about LQG, one always hears about spin-networks and area and volume operators. So how does these objects connect with the space_cadet's answer? Let's start slowly

1. What is a spin-network exactly?


2. What are the main mathematical properties?

Just a reference to the literature will suffice because I realize that the questions (especially the second one) might be quite broad. Although, for once, wikipedia article does a decent job in hinting at answers of both 1. and 2. but it leaves me greatly dissatisfied. In particular, I have no idea what happens at the vertices. Wikipedia says that they should carry intertwining operators. Intertwinors always work on two representations so presumably there is an intertwinor for every pair of edges joining at the vertex? Also, by Schur's lemma, intertwinors of inequivalent irreps are zero, so that usually this notion would be pretty trivial. As you can see, I am really confused about this, so I'd like to hear

 3. What is the significance of vertices and intertwinors for spin-networks?

Okay, having the definitions out of the way, spin-networks should presumably also form a basis of the aforementioned Hilbert space (I am not sure which one Hilbert space it is and what conditions this puts on the spin-networks; if possible, let's postpone this discussion until some later time) so there must be some correspondence between connection functionals and spin-networks.

 4. How does this correspondence look precisely? If I have some spin-network, how do I obtain a functional on connections from that?

Answer

So what are spin-networks? Briefly, they are graphs with representations ("spins") of some gauge group (generally SU(2) or SL(2,C) in LQG) living on each edge. At each non-trivial vertex, one has three or more edges meeting up. What is the simplest purpose of the intertwiner? It is to ensure that angular momentum is conserved at each vertex. For the case of four-valent edge we have four spins: $(j_1,j_2,j_3,j_4)$. There is a simple visual picture of the intertwiner in this case.

Picture a tetrahedron enclosing the given vertex, such that each edge pierces precisely one face of the tetrahedron. Now, the natural prescription for what happens when a surface is punctured by a spin is to associate the Casimir of that spin $ \mathbf{J}^2 $ with the puncture. The Casimir for spin $j$ has eigenvalues $ j (j+1) $. You can also see these as energy eigenvalues for the quantum rotor model. These eigenvalues are identified with the area associated with a puncture.

In order for the said edges and vertices to correspond to a consistent geometry it is important that certain constraints be satisfied. For instance, for a triangle we require that the edge lengths satisfy the triangle inequality $ a + b \lt c $ and the angles should add up to $ \angle a + \angle b + \angle c = \kappa \pi$, with $\kappa = 1$ if the triangle is embedded in a flat space and $\kappa \ne 1$ denoting the deviation of the space from zero curvature (positively or negatively curved).

In a similar manner, for a classical tetrahedron, now it is the sums of the areas of the faces which should satisfy "closure" constraints. For a quantum tetrahedron these constraints translate into relations between the operators $j_i$ which endow the faces with area.

Now for a triangle giving its three edge lengths $(a,b,c)$ completely fixes the angles and there is no more freedom. However, specifying all four areas of a tetrahedron does not fix all the freedom. The tetrahedron can still be bent and distorted in ways that preserve the closure constraints (not so for a triangle!). These are the physical degrees of freedom that an intertwiner possesses - the various shapes that are consistent with a tetrahedron with face areas given by the spins, or more generally a polyhedron for n-valent edges.

Some of the key players in this arena include, among others, Laurent Friedel, Eugenio Bianchi, E. Magliaro, C. Perini, F. Conrady, J. Engle, Rovelli, R. Pereira, K. Krasnov and Etera Livine.

I hope this provides some intuition for these structures. Also, I should add, that at present I am working on a review article on LQG for and by "the bewildered". I reserve the right to use any or all of the contents of my answers to this and other questions on physics.se in said work, with proper acknowledgements to all who contribute with questions and comments. This legalese is necessary so nobody comes after me with a bullsh*t plagiarism charge when my article does appear :P

homework and exercises - How to correctly determine the stopping distance of a coasting bicycle when considering aerodynamic drag?

Given values of $C_{rr}$, $C_{d}{A}$ and $\rho$(Air density) how can I correctly determine the distance and time taken to coast from some $\upsilon$1 to $\upsilon$2?

Answer

Assuming that your rolling resistance is independent of velocity, and that the force of rolling friction is $f_{rr} = -mgC_{rr}$, you can write the equation of motion as

$$F = m \frac{dv}{dt} = - \left(m g C_{rr} +\frac12 \rho v^2 C_d A\right)$$

From Wolfram Alpha we learn that the solution for

$$\frac{dv}{dt} = -\left(a + b v^2\right)$$

is

$$v(t) = -\sqrt{\frac{a}{b}} \tan\left(\sqrt{ab} (c_1 + t)\right)$$

You need to think about this a little bit to understand that the situation where you are decelerating happens then $c_1+t<0$. Getting the values of $a$ and $b$ right:

$$\begin{align} a&= gC_{rr}\\ b&= \frac{\rho C_dA}{2m}\end{align}$$

The result is

$$v(t) = -\sqrt{\frac{2mgC_{rr}}{\rho C_d A}}\tan\left(\sqrt{\frac{\rho C_d A ~g C_{rr}}{2m}}\left(c_1+t\right)\right)$$

You find the integration constant $c_1$ from the initial velocity (put $t=0$; you will find that $c_1$ must be negative), and the evolution of velocity with time follows. Interestingly, there is a finite time to come to a complete stop. That doesn't happen when you have "pure" quadratic drag - it's the rolling friction that dominates at low speeds.

Update

Just to check that things work as expected, I wrote a quick Python program that computes the velocity according to the above expression, incorporating also the effect of slope (note - if the slope is such that the object would accelerate, you get a "math domain error". This is not a hard thing to fix, but it would make the code more complicated to read, so I left that out for now.)

Running the code with three values of slope (where negative slope = downhill) gave the following plot; you can see that the slope of -5° almost exactly cancels the rolling resistance of 0.1 (arcsin(0.1) = 5.7°), leaving just the quadratic drag; if you set the quadratic drag coefficient $C_ad$ to zero, the velocity ends up almost completely unchanged. So yes, this is believable.

And the code (this is not meant to show "good Python", just something I threw together for a quick demo):

# rolling resistance and quadratic drag
import math
import numpy as np

import matplotlib.pyplot as plt

# pick some values for mass etc:
# these have obvious meanings, and SI units
m = 1.
g = 9.81
crr = 0.1
cda = 0.05
rho = 1.2
v0 = 10.


# convert to numbers we use in the formula
b = rho*cda/(2*m)

# a function that allows me to use degrees for slope:
def sind(theta):
    return math.sin(theta*math.pi/180.)

def vt(t,a,b,c1):
# implement the expression I derived

    temp = -np.sqrt(a/b)*np.tan(np.sqrt(a*b)*(c1+t))
# if velocity goes negative, things go awry
    stop = np.where(temp<0)
    if np.prod(np.size(stop))>0:
        # set all elements past the point where v first goes negative to zero
        temp[stop[0][0]:]=0
    return temp

# range of time for simulation:
t = np.linspace(0, 15, 500)


plt.figure()

# calculate for a range of slopes
for slope in np.arange(-5,6,5): 
  a = g*(crr + sind(slope))
  c1 = math.atan(-v0*math.sqrt(b/a))/math.sqrt(a*b)
  plt.plot(t, vt(t,a,b,c1), label='slope = %d'%slope)

plt.xlabel('time (s)')

plt.ylabel('velocity (m/s)')
plt.title('coasting down with rolling and quadratic friction')
plt.legend()
plt.show()

quantum field theory - BRST symmetry, gauge invariance and longitudinal gauge bosons

While quantizing a non-Abelian gauge theory covariantly, we first demand that the BRST charge acting on the physical states of the Hilbert space must be zero. However such physical states still have an unequal number of ghost and antighost particles and as the book is claiming, longitudinal bosons. To get rid of them he then applies the ghost number operator and picks out those states from the physical Hilbert space which are invariant under it.

So here's my question. I've often come across the statement that BRST symmetry is somehow related to gauge invariance. Is this true? If it is, why do you impose the further requirement that the gauge bosons must only be transversely polarized? I mean shouldn't the BRST invariance (which implies gauge invariance which again implies transverse polarizations) be enough to guarantee that?

Answer

1. 
   

   Ref. 1 only shows [via a non-abelian generalization of the abelian eq. (13.31)] that the physical Hilbert space ${\cal H}_{\rm phys}$ [defined to be BRST-closed and have zero ghost number] can not have longitudinal gluons by using the Lorenz gauge condition and the EL equation for the Lautrup-Nakanishi auxiliary field $F^a$.

   
   

   Longitudinal gluons are possible with other gauge-fixing conditions.

   
   
   


2. 
   

   For the connection between gauge and BRST symmetry, see e.g. this Phys.SE post.

   
   

References:

1. A. Das, Lectures on QFT, (2008); chapter 13.

Does the potential energy of a single object exist?

I know that potential energy exists for a system of objects. But why? Is there no potential energy for a single object system? I assume that there is.

Consider the earth-object system. Suppose an external force lifts the object above the earth. Then the work done by it gets stored in the earth-object system as potential energy.Right?

Now, my question is that is the potential energy always zero for a single object system.

MY REASONING

If we consider the object only as the system and lift it above the Earth, we are doing positive work on it and filling it with energy and the gravitational field of the Earth is doing negative work on it and draining it of energy. Thus whenever the object is moving it has kinetic energy which is fine. But when it is not moving,i.e. when its weight is equal to the force applied by us, the net work done is zero and so the object should not have any kinetic energy. Does it have potential energy?

Forces as vectors in Newtonian mechanics

I seem to be confused about the nature of forces as vectors, in the basic Newtonian mechanics framework.

I know what a vector is as a mathematical object, an element of $R^3$. I understand that if a vector is drawn in a usual physical way as an arrow in space, it can be seen as a mathematical vector by translating it to begin at 0 and seeing where the arrow tip ends up. Generally it seems the word "vector" is used in such a way that a vector remains the same vector if it's translated arbitrarily in space (always corresponding to the same mathematical vector).

But now let's say I have a solid object, maybe a metal cube, with some forces acting on it: I push at it with a stick in the center of one facet, it's held by a rope in a different corner, etc. To specify each force that is acting on the cube it doesn't seem enough to specify the vector: I also need to specify the place of application. The cube behaves differently if I push it in the center as opposed in the corner etc.

I'm reading through J.P.Den Hartog's Mechanics that teaches me how to find the resultant force on the cube. I need to sum forces one by one using the parallelogram law, but I should always be careful to slide each force along its line of application, until two forces meet. I could just translate them all to start at the same point and add, but then I won't find the right resultant force, only its direction and magnitude; I will still need to find its line of application (maybe using moments etc.)

So let's say I'm calculating the resultant force "the right way": by sliding arrows along their lines until tails meet, adding, repeating. What am I doing mathematically? (it's not vector addition, that would correspond to just translating them all to 0 and adding) What mathematical objects am I working with? They seem to be specified with 4 free parameters: 3 for direction/magnitude of the vector and 1 more to displace it to the correct line of application; the location at the line of application seems irrelevant according to the laws of statics.

quantum field theory - What is the exact relationship between on-shell amplitudes and off-shell correlators in AdS/CFT?

In this answer to a question, it is mentioned that in the AdS/CFT correspondence, on-shell amplitudes on the AdS side are related to off-shell correlators on the CFT side.

Can somebody explain this to me in some more (technical) details, maybe by an explanatory example?

Answer

First, a reference article, by Witten, http://arxiv.org/abs/hep-th/9802150v2.pdf

I'll try to expose the basic idea, with a flat space-time. Suppose you have a relativistic scalar field theory, on a flat space-time domain, with boundary. The equation of the field is :

$$\square \Phi(x) = 0$$ (fields on-shell)

Now, define the partition function

$$Z = e^{−S(\Phi)}$$, where $$S(\Phi) = \int d^nx \,\partial_i \Phi(x)\,\partial^i \Phi(x)$$ is the action for the field $\Phi$

After this, you make a integration by parts (using the above fied equation) , and Stokes theorem, and you get:

$$S(\Phi) = \int d^nx \, \partial_i \Phi(x)\,\partial^i \Phi(x) = \int d^nx \,\partial_i(\Phi(x)\,\partial^i \Phi(x))$$ $$= \int_{Boundary} d \sigma_i \,(\Phi(x)\,\partial^i \Phi(x)) $$

Now, suppose that the field $\Phi(x)$ has the value $\Phi_0(x)$ on the boundary. Then, you can see that $S$ and $Z$ could be considered as functionals of $\Phi_0$, so we could write $Z(\Phi_0)$:

$$ Z(\Phi_0) = e^{ \,( -\int_{Boundary} d \sigma_i \,(\Phi(x)\,\partial^i \Phi(x)))}$$

Now, the true calculus is not with flat space-time, but with Ads or euclidean Ads,so in your calculus, you must involve the correct metrics, but the idea is the same.

The last step is to say that there is a relation between, the Generating function of correlation functions of CFT operators $O(x)$ living on the boundary, and the partition function $Z$

$$_{CFT} = Z(\Phi_0)$$

The RHS and LHS terms of this equation should be seen as functionals of $\Phi_0$ You can make a development of these terms in powers of $\Phi_0$, and so you got all the correlations functions for the CFT operators $O(x)$

$$ \sim \frac{\partial^n Z}{\partial \Phi_0(X_1)\partial \Phi_0(X_2)...\partial \Phi_0(X_n)}$$

So, ADS side, we are using on-shell partitions functions (because field equations for $\Phi$ are satisfied)

Now, CFT/QFT side, the correlations functions $$ are, by definition, off-shell correlation functions (by Fourier transforms, there is no constraint about momentum). To get scattering amplitudes, we simply need to put the external legs on-shell.

hilbert space - Linearity of quantum mechanics and nonlinearity of macroscopic physics

We live in a world where almost all macroscopic physical phenomena are non-linear, while the description of microscopic phenomena is based on quantum mechanics which is linear by definition. What are the physics points of connection between the two descriptions?

Answer

Linear in the quantum mechanics has nothing to do with its complexity. A two-state spin can be described by a simple 2-by-2 matrix; however, 30 interacting spin, in general, must be described by a 1 billion by 1 billion matrix. It grows exponentially as the number of spins increases, for $10^{23}$ spin, you may need a matrix of size $2^{10^{23}}$. It is not easy to understand and not simple in most sense. If you learn some statistical mechanics, you will know that this number is large enough to have new emergent phenomenon.

quantum mechanics - Allowed Wave Functions of System

Given a single-particle system with Hamiltonian $H$, what constraints can be put on the wave function at a particular point in time $\psi(x)$? Of course $\psi(x)$ must obey boundary conditions given by $H$. However, in situations where $H$ does not yield strict boundary conditions (e.g. the harmonic oscillator) can $\psi(x)$ be any normalised function? My intuition says no for the following reason: QM textbooks say that any valid wave function can be represented as a linear combination of the eigenstates of $H$. In the case of the harmonic oscillator, the eigenstates of $H$ are a countably infinite set. However the set of all normalised functions is uncountably infinite. It seems to me that one cannot change between one complete set of basis functions to another and reduce the "dimensionality" somehow. This makes me think that the eigenstates of $H$ are not complete, that they imply some constraints on $\psi(x)$. This is not an area of mathematics that I understand well. Can somebody help me out?

Answer

Generically, any square-integrable function is an admissible wave function, and the space of square-integrable complex functions indeed has uncountable dimension as a vector space over $\mathbb{C}$.

And it is also true that the eigenstates of the Hamiltonian span the space of states, and that they are countably many. This is the content of the spectral theorem - eigenstates of bounded self-adjoint operators form an orthonormal basis of the Hilbert space (I will ignore the subtlety of free/non-normalizable "eigenstates" here, and assume the Hamiltonian is bounded above and below).

The point is that the notion of "basis" in the context of infinite-dimensional Hilbert space is not the notion of "basis" from finite-dimensional linear algebra, where the "span" of a set is the set of the finite linear combinations (a basis that forms a basis of a vector space in this sense is sometimes called a Hamel basis). When one speaks of a basis in the context of a Hilbert space, one means a Schauder basis instead:

The "Schauder span" of a set is the set of all convergent infinite series made out of the vectors in that set. This relies on the additional topological structure a Hilbert space carries through the norm on it, and on the completeness of that norm. This span includes the usual linear span, but is larger. In particular, a countable set can span a vector space of uncountable dimension in this sense, exactly like all reals are limit points of sequences of rationals, and there are countably many rationals, and uncountably many reals.

particle physics - Working out the charge of a W Boson

When dealing with particle interactions, is there a straightforward way to work out the charge on the W boson?

In particular, the interactions I need to know the charge for are those involving a proton, neutron, an electron/positron and an electron neutrino/antineutrino. Is it easiest just to remember the charge (for AQA AS Physics exam)?

Answer

You have to look at the Feynman diagram to know if you can assign a charge.

• 
  

  Time-like $W$s carry the same charge as the initial and final states.

  
  


• 
  

  Space-like $W$s are ambiguous and you can not assign a unique charge to them, because you can't say which vertex came first in time.

  
  

For interaction that can occur through either time-like or space-like channels you still have the ambiguity.

cosmology - Experimental evidence for parallel universes/multiverses

My idea of physics is that it is a collection of mathematical laws relating observables. And that one can perform alot of mathematical derivations on these laws to produce new laws between observables. My question is how does one translate a mathematical equation into 'there exist other universes like ours'?

How does one derive that there exist other universes, what phenomena do they explain?

Which observables suggest other universes?

quantum mechanics - Interference of Schrödinger's Cat states?

A "Schrödinger's cat state" is a macroscopic superposition state. Quantum states can interfere in simple experiments (such as the Mach-Zehnder/Hong-Ou Mandel/etc). Can Schrödinger's cat states be made to interfere?

I've come up with a thought experiment myself that might show some kind of interference, but I'm not sure if it makes sense.

Let's first just work out an example of a Schrödinger's cat state:

I can take a photon, impinge it on a 50/50 beam splitter - and now the light is in a superposition of reflecting (going "right") or transmitting (going "left"):

$$|photon\rangle = |L\rangle - |R\rangle$$

Now if I add two detectors for L and R, and kill the cat when I detect L. I obtain a superposition state for the cat. But more specifically, before I check the box, I can see that I have created entanglement between the photon's path if the cat is alive or dead:

$$\implies |A\rangle\otimes |L\rangle - |D\rangle \otimes |R\rangle$$

So from here, this is an entangled state that has a defined phase. Can I come up with a way of interfering these states so, similar to a Mach-Zehnder quantum interference experiment, that I can recover just an alive (or dead) cat with 100% probability. Quantum Mechanically, you'd expect that should be the case.

So how about this experiment: After measuring the state of the photon in L or R, in addition to killing the cat when the photon is is in L - we create an output photon associated with the direction the photon was found in, and we send it through the beamsplitter again. You can see it illustrated in this picture:

Now the result I get as an answer here seems to be different in the cat's perspective than in the external perspective. In the cat's perspective - it cannot see the phase of the photon. So if it was $\frac{1}{\sqrt{2}}(|L\rangle - |R \rangle) $ we would only see 50% left and 50% right - and thus we could not see (or prepare) the phase properly for the second part. So we don't see any interesting quantum interference that occurs when the second photon is send out - it's just another 50/50.

But if we consider the measurement to simply entangle possiblities as discussed: $$\implies |A\rangle\otimes |L\rangle - |D\rangle \otimes |R\rangle$$ I don't see why this phase wouldn't remain the same. Unlike the previous case, I would think that this negative sign would be truely assigned to the quantum phase of the photon generation - since we have not "collapsed" anything yet by not observing it. But now when we send it - we should expect the probability amplitudes of the entangled states to interfere, similar to Mach-zender interference. Then we should get either $|A\rangle \otimes |L\rangle$ or $|D\rangle \otimes |R\rangle$. I think this would indicate that the cat can "cheat death" in the perspective of an external observer that does not check the first sequence and only checks the second.

Why are these results inconsistent with eachother? My guess is that one of these two methods are incorrect.

Anyway, overall I am looking to understand a thought experiment in which macroscopic cat state could exhibit some type of quantum interference.

Answer

In your 'cat perspective' argument you say the outcome is 'just another 50/50'. Then in your 'external perspective' argument you say the outcome is 'either $|A⟩⊗|L⟩$ or $|D⟩⊗|R⟩$'. In other words in both arguments you have a cat either alive or dead, with 50% probabilities. This is indeed the correct conclusion, and I don't think there is any inconsistency.

If we present your apparatus a bit more fully, the math goes as follows. Let the three parts of the system be first photon, cat (and associated apparatus), second photon. Then the state after first photon is detected is $$ (|D⟩⊗|L⟩−|A⟩⊗|R⟩)⊗|0⟩ $$ where I have followed your words and diagram in saying the cat is dead (D) for the left photon state (L). The final zero state is the second photon which has not been emitted yet. Then the second photon gets on its way, giving the state $$ |D⟩⊗|L⟩⊗|l⟩−|A⟩⊗|R⟩⊗|r⟩ $$ where $l$ and $r$ are states of second photon. After second photon goes through the beam splitter we have $$ |D⟩⊗|L⟩⊗(|l⟩+|r⟩) −|A⟩⊗|R⟩⊗(|l⟩-|r⟩) $$ which is equal to $$ (|D⟩⊗|L⟩ − |A⟩⊗|R⟩)⊗|l⟩ + (|D⟩⊗|L⟩ +|A⟩⊗|R⟩)⊗|r⟩ $$ (I have dropped normalization factors and used a slightly loose notation for the photon modes, following a reasonably standard practice). In this final state we have the outcome we know to expect: the apparatus which detected the first photon and the cat are in correlated states. The cat is either alive or dead (whether in one-world, or in any one world).

Now perhaps I have misunderstood, and you intended there to be just one photon throughout. In this case you still need something to indicate the state of the apparatus which detected the photon on its journey and killed the cat, or not, accordingly. In the above argument the states labelled $|L⟩$ and $|R⟩$ can serve as states of such an apparatus.

Strictly speaking the above is much too rough to treat a real living breathing cat, or indeed almost anything larger than about a millimetre, because it fails to account for the cosmic background microwave radiation, the gravitational wave environment, the magnetic field noise, the thermal radiation, etc. etc. Any of those would easily be large enough to have very significant effects on the quantum phase of a cat. However if we take it that the 'cat' here is really something smaller such as a single molecule, then the treatment is precise enough to serve a useful purpose.