cosmology - If space is "expanding" in itself - why then is there redshift?

The "kid's" way of understanding the expanding universe is that: "space" is totally "ordinary", and all the galaxies are expanding through it (like an explosion). Of course, that's wrong.

The usual better explanation is that "space itself is expanding." (Of course, on scales below clusters, gravity pulls "smaller" structures together.)

An even more up-to-date explanation is that the conceptual "metric of space" is "expanding" (here's a typical pedagogic example) which can perhaps be summarized as the "scale is changing".

So ... distant objects are redshifted.

But why? Everything's just expanding -- the very metric of spacetime is expanding.

Indeed, it would seem to me that you would only see redshift (or if you prefer, time dilation of far-away things) strictly in the case of "everyday" motion within the metric of space; the very idea of the actual "metric of space changing!" would seem to be that, those of us internal to that metric of space would have no clue that any such expansion is happening: the scale is just changing for everything.

What's the best way to understand this?

Imagine simply a meter cube in a video game with a few things in it. There is no exterior, it is the universe. I expand the entire thing...

{note...of course, obviously, the 'outside' (shadows etc.) added by the 3D presentation software to clarify the PNG here, have utterly no meaning and do not exist in any way}

... to all the beings inside, I believe absolutely nothing has changed, there'd be no redshift between the objects there.

What's the deal?

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Note too this somewhat similar (related?) question, which came up with the recent 2016 gravitational wave discovery:

How is it that distortions in space can be measured as distances?

Answer

What are the observational/experimental facts:

1)Atoms have definite spectra, with a fixed pattern, a fingerprint of the atom

2) The further away ( measured by luminocity) galaxies all around ours the more shifted the fingerpring pattern towards the red part of the spectrum.

3) This happens uniformly all around.

The model that fits these facts is General Relativity, which predicted the behavior

In the hierarchy of forces , the gravitational force is the weakest. This assures that atoms, matter in general up to the size of galaxies keep their structure, the raisin bread analogy. Gravity is strong enough to keep even clusters of galaxies unaffected and given some assumptions on the energy density and solution of the general relativity equations gravity can fight the expansion and lead to the big crunch,.

Photons are elementary particles that have to obey locally energy and momentum conservation. The expansion of the universe changes their momentum and thus the atomic spectra arrive shifted towards the infrared.

the very idea of the actual "metric of space changing!" would seem to be that, those of us internal to that metric of space would have no clue that any such expansion is happening: the scale is just changing for everything.

It is the fact that matter is bound by forces that are not affected by the expansion that allows us to measure the expansion. Otherwise you are correct, our atoms would also be expanding and we would see no shift in the atomic thumbprints.

Can post-selecting on the screen in the Delayed Choice Quantum Eraser experiment be used to predict the quantum-eraser measurement results?

I'm curious about QM and spent the last 3 days thinking about Delayed Choice Quantum Eraser (DCQE) experiment, but I couldn't solve this issue:

1. 
   

   Assume we do the DCQE experiment so that for the whole experiment (let's say, 1000 particles) finishes before the first particle reaches the detectors 1-4 (using the notation from this diagram from Wikipedia).

   
   


2. 
   
   

   Assume also we're using a switch to control if the idler particle goes to which-way-detector or to the eraser.

   
   


3. 
   

   This schematic shows what we should see in D0, even though we don't know yet R01-04

   
   


4. 
   

   Now, knowing D0, use the switch to send the 100 particles that reached the most-left side of D0 to the eraser detectors (D1 and D2). It seems much more likely that they reach D2 than D1, because it's on peak on D2 (and on rest on D1) even though it's going through a half-silvered mirror. This looks like a contradiction to me.

   
   

So, am I missing something or QM is just that weird?

Answer

You're over-interpreting these sketches - they are only sketches, and their specific details can't really be used to make any real predictions.

Here is a more accurate version of those sketches, with a proper underpinning on a solid model of the experiment's behaviour:

^{Mathematica source via Import["http://halirutan.github.io/Mathematica-SE-Tools/decode.m"]["http://i.stack.imgur.com/P6HYG.png"]}

As you can see "the leftmost part of $D_0$" is equally compatible with the patterns $R_{03}$ and $R_{04}$, as detected on the quantum-eraser detectors 1 and 2.

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Still, you're not entirely wrong, particularly in the sharper formulation you give in the comments:

Isn't it true that the patterns made by the particles that reach R01-04 follow distinct distributions on D0? If so, it seems reasonable to extrapolate that there are some regions that are inverse peaks for D1/D2

Yes, the patterns made on the $D_0$ screen when post-selecting on $D_1$ and $D_2$ detections are indeed different - and, in fact, they're complementary interference patterns, with the peaks on $R_{01}$ lining up with the troughs on $R_{02}$ and vice versa. (This is how they can add up to an interference-less $D_0$ pattern when there is no post-selection. It is crucial that you understand that both $R_{01}+R_{02}$ and $R_{03}+R_{04}$ add up to $D_0$, and what that means - the 1/2 and 3/4 pairs are just different ways of splitting up the $D_0$ counts, depending on information acquired later.)

This means that you can zero in on one of the peaks of the $R_{01}$ fringes, say, the green box below:

If use some fancy switching mechanism to ensure that you send all the particles that fell on that green box over to the $D_1$/$D_2$ quantum-eraser part of the idler-photon side of the experiment, then indeed, as you say,

it seems much more likely that they reach D1 than D2.

Is this a problem or a contradiction? No. The photons are not going through an arbitrary half-silvered mirror - they're going through a precisely calibrated beam splitter. The beam path that reaches $D_2$ includes a contribution from $M_b$ (red beam) and a contribution from $M_a$ (blue beam), and if those beams are coherent, they can interfere both destructively and constructively. Absent any information about what happened to the signal photon on $D_0$, the idler and the signal are entangled, and there is zero relative coherence between those two beams, and $D_2$ will click half the time. However, by post-selecting on $D_0$'s measurements on the green box, you're effectively fixing the phase between the two beams in such a way that they interfere destructively on the $D_2$ side (and constructively on the $D_1$ side), so no light goes through to $D_2$ (on those post-selected runs).

So, basically, what you've described is a fancy way to run the quantum-eraser apparatus in reverse, where by splitting the $D_0$ screen into sectors you're providing information that can be used in a post-selection scheme to recover the interference pattern that comes out of the BS$_\mathrm{c}$ beam splitter.

If that seems weird, then yes,

QM is just that weird.

electromagnetic radiation - Eikonal approximation for wave optics. Why follow the unit vector parallel to the Pointing vector?

The description of the passage from wave optics to geometrical optics claims that light rays are the integral curves of a certain vector field (the Pointing vector direction, normalized to 1). Here are the details, could you fill the blanks:

The wavelength $\lambda$ is much smaller than all other characteristic lengths.

1. 
   

   The setting is a "nice" medium (with spatially varying refraction index $n(x)$) through which an "almost plane" wave is propagating. The wave (if linearly polarized) turns out to be representable by $\vec{E}{(x,t)} = \vec{E}_0\exp(i(\chi{(x)}-\omega t))$ and $\vec{B}{(x,t)} = \vec{B}_0\exp(i(\chi{(x)}-\omega t))$ with constant $\vec{E}_0$ and $\vec{B}_0$.

   
   


2. 
   

   Maxwell's equations imply $(\vec{\nabla}\chi)^2=\frac{n^2\omega^2}{c^2}$ and a time-averaged Pointing vector $\vec{S}=\frac{c}{n}\vec{s}$, where $\vec{s}$ is the unit vector $\vec{s}=\frac{\vec{\nabla}\chi}{n\omega/c}$.

   
   
   


3. 
   

   The integral curves of the field of unit vectors $\vec{s}$ are the light rays.

   
   


4. 
   

   Working through the equations results for this for a ray trajectory $X(\tau)$ (where $\tau$ is just a parameter): $\frac{d}{d\tau}(n\frac{d\vec{X}}{d\tau})=\vec{\nabla} n$

   
   

How does one prove the jump from 2 to 3. Why follow the unit vectors and not the Pointing vectors themselves? Or even, why should the light rays be tangential to the Pointing vectors at all (besides intuition like "light rays should transport energy")?

Could someone give me the proof of 3. or point me to a reference?

Answer

There's a number of interesting points to this.

• The passage from 1. to 2. is not trivial. If you do the calculation, you will see that the laplacian $\nabla^2\vec{E}$ from the wave equation gives rise to the term in $\left(\nabla\chi\right)^2$ you mention as well as a term in $\nabla^2\chi$. This second term only goes away in the small $\lambda$ limit and it is the essence of the eikonal approximation. It is not a calculation you should wave away: work it out in full and implement the approximation, noticing that locally $\chi(\vec{x})=\vec{k}\cdot\vec{x}+\text{slow factors}$, where $\vec{k}$ is large. (You will of course need to quantify "slow".)


• (The calculation that $\vec{S}=\frac{c^2}{n^2\omega}\nabla\chi$, on the other hand, is trivial.)


• As KDN mentioned, the integral curves of $\vec{S}$ and its unit vector $\vec{s}$ are the same. This follows from the definition of integral curves: they are curves such that the vector field is tangent to them throughout. This is independent of the length of the vector. (In terms of the curve it corresponds to a reparametrization of the "time": it changes the speed but not the direction of the velocity.) Using a unit vector means that light rays will be parametrised by path length.


• One can simply define light rays to be the integral curves of $\vec{s}$ and be happy about it, though of course that is simply missing the physics. The key fact about the light rays, so defined, is that they are everywhere normal to the surfaces of constant $\chi$, i.e. the surfaces of constant phase, i.e. the wavefronts. Plane waves propagate in straight lines normally to the wavefronts in free space, and so do light rays (so defined). It is the normal to the wavefronts that matters when working out Fresnel equations, and therefore the (so defined) light rays will obey Snell's law. Ultimately, proving 3. is a matter of definition: what are light rays? Write down any defining property and you'll be able to prove the integral curves of $\vec{s}$ obey it.


• It is important to note that in isotropic media $\vec{s}$ is not only the local unit Poynting vector, but it is also the local unit wave vector. (Essentially, this is the same point as above.) Intuitively, light rays ought to follow wave vectors because it is wave vectors that tell light waves where to go. In a birefringent (not isotropic) medium the phase propagation direction (wave vector) and the energy propagation direction (Poynting vector) are not necessarily the same (and the snell law does not apply).


• Proving 4. is an interesting exercise (i.e. do it!) but it is essentially trivial. It relies on the identity $\frac{d\vec{X}}{d\tau}=\vec{s}$, which defines light ray curves $\vec{X}(\tau)$, on judicious use of the total derivative $\frac{d}{d\tau}=(\frac{d\vec{X}}{d\tau}\cdot\nabla)$, and some interesting vector calculus manipulations. (Hint: prove $(\nabla\chi\cdot\nabla)\nabla\chi=\frac12\nabla\left(\nabla\chi\right)^2$.) Presumably you know by now that what you get is called the ray equation, what it means, and how to use it, or you would not have stopped there ;).

This looks like enough to get you going but if you have more questions, do ask.

electricity - Why is current slowed down by resistance?

Let's say I wire the negative pole of a battery to the positive pole. Obviously, the battery will short circuit as the electron pool in the negative side will become attracted to the positive side and cause a huge flow.
However, if I add a lamp to the circuit, the electrons again flow because they are attracted to the positive pole, but for some reason only enough of them to power the lamp. Why don't they keep flowing past the lamp and drain the battery just like a short circuit? Why does the resistance dictate the current?

newtonian mechanics - Newton's third law at the quantum level?

let's look at force at the atomic level to understand the newtons third law of motion. I'll use Helium atoms as an example.

Now imagine we start with one atom HE2 stationary, and throw another atom HE1 at it.It is the velocity of HE1 that affects the motion of HE2 , because the system of these two HE2 atoms is isolated , and there is no unbalanced force on each of these . According to physics both the atoms will experience action reaction forces at the same time .

for sake of simplicity lets consider a imaginary hollow box between the two atoms as an overlapping of their atomic forces. As soon as HE1 moves x distance inside this box , there will be a increase of y amount of velocity in HE2 and a correspoding decrease of y in the velocity of HE1 . But there is still a greater velocity in HE1 as compared to HE2 , so the process of changing of velocities will continue up to the time when both the atoms have the same amount of velocity , because after they attain the same amount of velocity they wont be able to enter tat box again .

So , is what i just explained right ?

i dont think so, because we know from our knowledge of head on elastic collisions of equal masses that when HE1 and HE2 will collide, HE2 will gain the velocity that HE1 had and HE1 will itself become stationary.

so can you please correct me , and provide me with the correct answer to understand the newtons third law of motion ?

Answer

The reason the Newton's law is referred to as the Newton's laws is because the are applicable to newtonian range objects only. The world of quantum mechanics has no relation with the laws of motion at all. The world of quantum mechanics is random and is not predictable, which is a direct violation of the newton's laws.

The point that you make about the force between the Helium atoms is a function of distance between them should be more refined as square of the distance between them as you compare it with a spring.

Let us consider this imaginary assumption of a stationary atom. Which is not possible according to a the laws of quantum mechanics, unless it is in a crystal lattice. But when you throw the another atom at it, The stationary one will feel the repulsion and the moving one also feels the repulsive force no doubt, but the moving one coming to stop and the stationary one moving is not really acceptable, because as soon as the so called stationary atom feels the repulsion it starts the motion in the same direction as that of the moving atom, but moving atom will not come to rest but will surely reduce its motion.

Also the atoms do not move in the newton's world but they have the De'Brogle motion which is of a wave nature and not a straight linear one, so a complex mathematics is required to predict its effect. And the reason I say predict is because it cannot be calculated with absolute precision. So questioning the Newtons third law in the context of a quantum world is not acceptable as it is not meant for it.

photons - Is double slit interference due to EM/de Broglie waves? And how does this relate to quantum mechanical waves?

I'm really confused about the fact that there seems to be two types of waves at play: the EM wave, which I understand to be an actual fluctuation of EM fields in space, and this other type of bulk "wave" that's referred to in explanations of the double slit experiment, whereby the bulk light emanates as a circular, water-like wave from the two slits. The circular path of the bulk wave seems to be what results in the fringe pattern, depending on the phase (the EM phase?) of the two waves at the point where they interfere at the detector. These bulk-like waves turn out to be a consequence of quantum mechanics and even occur with single photons, but at the time the debate seemed to be between particles and these circularly-emanating waves. However, is the only reason they can interfere at all because they are fundamentally EM waves that undergo destructive interference when in opposite phase? There is a difference between the waves in question and EM waves, right?

If the only reason there are dark fringes is because the quantum mechanical distribution causes physical EM waves to interfere, how can electrons generate a fringe pattern as well? I thought the de Broglie waves were complex mathematical abstractions and not representative of physical electron waves. Are de Broglie waves in any way analogous to EM waves? Does the fringe pattern fro electrons occur because of destructive interference of de Broglie waves? Is this fundamentally different from the photon interference?

Answer

An electromagnetic wave is a field, it has a (possibly zero) value at every point in space at every time.

A wavefunction is not a field. It does not have a value at every point in space. For $n$ particles it is a function from $\mathbb R^{3n}$ into the joint spin state of the system. For the simplest case of one particle with spin zero it looks like a function from $\mathbb R^3$ into $\mathbb C$ so you might think it is a complex scalar field. It isn't and that's going to bite you later. And then everything will be confusing and mysterious. If you want to think quantum mechanics is confusing and mysterious then a good way to do that is to think a wavefunction is a complex scakar values field in space.

other type of bulk "wave" that's referred to in explanations of the double slit experiment, whereby the bulk light emanates as a circular, water-like wave from the two slits.

That does not happen. But you also need to be clear, there is a classical double slit experiment with light where the electromagnetic field does spread through space and pass through slits and form interference patterns that have nothing to do with quantum mechanics.

The circular path of the bulk wave seems to be what results in the fringe pattern, depending on the phase (the EM phase?) of the two waves at the point where they interfere at the detector.

Even uncharged particles experience interference when you take quantum effects into account. So it's just the phase of the (possibly zero) spin (which still has a phase even when it's zero spin).

These bulk-like waves turn out to be a consequence of quantum mechanics and even occur with single photons, but at the time the debate seemed to be between particles and these circularly-emanating waves.

If you are going to study history, you have to ask about a specific moment of history and a specific person since each person's understanding changes at specific moments in history.

However, is the only reason they can interfere at all because they are fundamentally EM waves that undergo destructive interference when in opposite phase?

No. Again, uncharged objects can still interfere. And the interference doesn't happen in physical space like $\mathbb R^3$ it happens in configuration space like $\mathbb R^{3n}$.

There is a difference between the waves in question and EM waves, right?

Every possible difference imaginable. An EM wave is a field in space and has electromagnetic fields (with six components) as its values. A Quabtum wavefunction goes from configuration space and is spin valued.

I thought the de Broglie waves were complex mathematical abstractions and not representative of physical electron waves.

It's a gauge theory so a wavefunction does have too many degrees of freedom. But it's pretty close.

Are de Broglie waves in any way analogous to EM waves?

You can add two together, just like electromagnetic waves.

Does the fringe pattern fro electrons occur because of destructive interference of de Broglie waves?

Yes. But keep in mind the interference fringes happen in regions of configuration space. And what you see when you look at the screen are residuals of that.

Is this fundamentally different from the photon interference?

I don't know what you are saying there.

You mentioned a classical double slit experiment. Are you saying the experiment was fundamentally different?

You could actually pass water waves through some slits and you notice an interference. The interference happens in a similar way for every other kind of interference. But since the wave has more space to move through sometimes when one particle hits the screen another particle has been deflected a different direction, so the wave in configuration space doesn't overlap and so no interference happens. The interference happened in the overlap. This is why interference fringes can be "destroyed" really its like deflecting the beam going through the right slit down and the beam through the left slit up so they don't overlap any more. But if the up and down in the the position of a different particle the residual of just that particle looks lined up when the waves are not overlapping.

Running the particles through one at a time is designed to control for the possibility that the many objects going through are just pushing against each other (like water does). If it's just one particle at a time going through then it has to be something else.

operators - The Physical Meaning behind a Commutator

I've just been introduced to the idea of commutators and I'm aware that it's not a trivial thing if two operators $A$ and $B$ commute, i.e. if two Hermitian operators commute then the eigenvalues of the two operators can be measured with certainty simultaneously.

But what is the physical significance when two operators do not commute such as to give a certain value? For example the position and momentum operator do not commute and give a value of $i\hbar$. What is the significance of the $i\hbar$?

Answer

As you said if two operators commute they share eigenvectors. Physically this means that you can have a definite value for both. For example in the hydrogen atom the Hamiltonian $H$, which is the energy, and $J^2$, the magnitude of angular momentum, commute. A hydrogen atom can be in a state of definite energy and definite angular momentum. However, the position operator $x$ does not commute with $H$, so in a state of definite energy the electron doesn't have a well-defined position.

Then conversely the commutator measures the inability for two quantities to have definite values in the same state. More quantitatively, we have the general Heisenberg uncertainty principle $$\Delta A \Delta B \ge \frac{1}{2} |\langle [A,B] \rangle |$$ that is, the product of the uncertainties in $A$ and $B$ is at least half the (absolute value of the) expectation value of their commutator. By uncertainty we mean the usual standard deviation, $$\Delta A = \sqrt{\langle A^2\rangle - \langle A \rangle^2 }.$$

For the position operator $x$ and the momentum operator $p$, the commutator is just a scalar, $i\hbar$; its expectation value is always $i\hbar$. We thus get the most famous instance of the Heisenberg principle $$\Delta x\Delta p \ge \frac{\hbar}{2}.$$

Now you could ask why should we have $[p,x] = i\hbar$ of all things. Well, in the Hamiltonian formulation of classical mechanics there is an operation called the Poisson bracket, $\{F,G\}$. The Poisson bracket has the same algebraic properties as the commutator (they are both brackets in a Lie algebra) and satisfies $$\{ x_i, p_i \} = \begin{cases}1 & i = j \\ 0 & i \neq j\end{cases}.$$ So suppose that you know that whatever quantum mechanics is, quantum states are vectors and observables are operators, and you want to figure out how those operators should be related. Then it would be tempting to just try $$[x,p] \overset{?}{=} 1.$$ The problem is that $x$ and $p$ should be Hermitian (so that expectation values are always real). Then $[x,p]$ must be anti-Hermitian. But that's not a big problem, you can just multiply by $i$: $$[x,p] \overset{?}{=} i.$$ That's okay algebraically, but $x$ has units of length and $p$ has units of momentum, so we need to put a constant there to get the right units too: $$[x,p] \overset{?}{=} i\hbar.$$ I still put a question mark there because this is really just an educated guess, but experiments show that this is the correct commutation relation to use. (Well, you also have to measure $\hbar$ somehow.)