Einstein and "spooky actions"

[Mr Spinkles]

The only forces which exist in Relativistic classic mechanics are elctromagnetism and gravity, both of which are limited to c or less. The only possible contention is what shadows look like, but like I said, please do the math for that one.

Let me get this straight: you won't take my word for it, you won't take a physics textbook's word for it, and yet you lack the ability to calculate it yourself? Do you always guess your opinions and stick to them unless someone can educate you out of them? I beg your pardon, PolyHedral, but that is breathtaking arrogance.

Your objection about nothing traveling faster than light: Pass your hand over a lamp of width W. For the sake of argument, say your hand is moving at velocity C, and thus it passes over the lamp in time T = W/C. A screen some distance away, with width L > W, is illuminated by the lamp. A shadow passes over the screen due to your hand passing over the lamp. How fast does the shadow move? The shadow's velocity across the screen is V = L/T = L/W x C, which is greater than C. So for sufficiently fast hands and large screens, shadows can indeed travel faster than the speed of light. This doesn't violate relativity because no information/energy/mass etc. is transmitted faster than C by the shadow. There are other kinds of superluminal motion (e.g. phase velocity) and they don't violate relativity, either.

Your objection about Bob being unable to observe anything outside his past light cone: Bob is initially located at time t = 0 and position x = 0, that is, the vertex of his past and future light cones. He therefore observes, at t = 0, anything passing through that vertex (t=0,x=0). For example, a particle at rest at (t = -10, x = 0) is a vertical line which passes through the vertex (t=0,x=0). Therefore, Bob observes such a particle. The shadow discussed above may start to the left, outside Bob's light cone, but because it travels faster than C is propagates to the right and arrives at (t=0,x=0), so Bob can observe it.

Now consider a particle which is emitted outside Bob's past light cone, let's say, at (t=0,x= -10). Suppose, hypothetically, that this particle travels instantaneously to the right, towards Bob. Then its trajectory is a horizontal line. This would be a form of "nonlocality" in the sense that at one instant in time, t = 0, the particle is present at many points in space, x = -10,-9,...0...+9,+10... Clearly this line passes through the vertex (t=0,x=0), and therefore Bob could indeed observe such a hypothetical particle, if it existed. We do not observe such particles of course, and we believe they don't exist.

In QM we are not (as far as we know) talking about a nonlocal particle but a nonlocal influence or "collapse of the wavefunction". But the same reasoning applies in terms of your objection about whether Bob could observe such a hypothetical influence, or not. Suppose Alice's measurement occurs to Bob's left outside his past light cone, at (t=0,x = -10). According to QM her measurement has an instantaneous or nonlocal influence on Bob's particle, which would be represented as before by a horizontal line passing through (t=0,x=0). Ergo, it could affect Bob and he could observe the effect, too.

However, it is correct to worry that such an influence would threaten causality, the same way that a faster-than-light particle, or a faster-than-light force (such as gravity and electromagnetism, which you mentioned) could violate causality. As I've told you several times now: it turns out, when you consider it carefully, that only a causal influence (like those mentioned above) would violate causality if it traveled faster-than-light. The influence stipulated by QM of Alice's measurement is not a strictly causal one, and therefore, it does not lead to violations of causality.

If you are still unconvinced I encourage you to study the following standard textbooks, some of them even have problems you can work out directly related to what we have discussed:

Tipler, Modern Physics
Griffiths, Introduction to Quantum Mechanics
Griffiths, Introduction to Electrodynamics
Jackson, Classical Electrodynamics
Shankar, Principles of Quantum Mechanics

Once you understand what relativity and QM say, you will be in better position to dispute what they say.

 

[Mr Spinkles]

The only thing I did is to change the initial scenario only very slightly to show that your arguments don't hold.

You switched your argument entirely, from saying that QM is local, to saying that particles aren't "really" entangled in singlet states as described by QM. You seem to be saying that each particle already has a definite spin (up or down) even before measurement. That is the realist interpretation, i.e., the interpretation that QM is an incomplete theory with unaccounted for hidden variables. This is a complete reversal on your part. What's ironic is it doesn't rescue you from nonlocality, if you insist on realism you have to reject locality, due to Bell's theorem, as you even acknowledged earlier (but have forgotten).

I disagree that your opinion prevails among physicists, perhaps among the old folks who have learned their QM back some 30 years ago.

Griffiths wrote in 2005 the orthodox interpretation prevails among physicists. No matter what your interpretation, there is no significant disagreement that it is not wrong to say QM is a nonlocal theory, and it is not wrong to write down the singlet state the way I (and Susskind!) did. Those were simply errors on your part, not a matter of interpretation.

But the interesting thing is that eventually they will all be dead. For the young generation and the next ones coming up, that old language of the dinosaur age will disappear, and the sooner the better.

You got your Masters 15 years before I was born. I am the young generation.

 

[zaybu]

In your analysis it doesn't. You just assume a priori that there will be no output towards A, which is begging the question. (Specifically, you seem to say the second beam splitter will not actually split the beam travelling right, but instead always let it pass straight through.)

I'm not assuming anything. That's what the equation gives you. The solution is found by looking at the symmetry of the situation. The top left-hand is exactly like the what's at the bottom right corner. One beam ( the input I), breaks into two beams. At the bottom right-hand corner, the same two beams, flowing exactly like the top left-hand corner, moving into B, must give the same outflow (the output O). Otherwise, the symmetry would be broken, and then you would really have a mysterious case.

You're not making any sense. Does the output towards B match the phase of the input, or not?

I'm not sure what you are hinting at. The data is about the number of particles the detector is counting, it's not measuring differences in phases. Do you have another experiment in mind? This one is about counting photons.

 

[zaybu]

You switched your argument entirely, from saying that QM is local, to saying that particles aren't "really" entangled in singlet states as described by QM.

I didn't switch anything. I changed the scenario slightly to show you that your arguments would no longer make sense. That was the purpose for that.

You seem to be saying that each particle already has a definite spin (up or down) even before measurement. That is the realist interpretation,

No one said that. That's why Alice must perform their experiment, and ONLY after they made their observation can they compare and see that the law of angular momentum is conserved. In your case, you were calculating the total spin BEFORE they even had measured. So of course you ended up with violations. And then you had the audacity to say I believe in violations in the conservation. That's how twisted your arguments are.

QM has its placed in that experiment, but not the way you envision it. That the spin is quantized is from QM, OTOH, conservation of angular momentum is classical, but it still hold in the quantum regime.

i.e., the interpretation that QM is an incomplete theory with unaccounted for hidden variables.

That's you're interpretation. I disagree. QM is complete. That, I have said it, and will say it again: pursuing such enquiry that QM is incomplete is a waste of time.

What's ironic is it doesn't rescue you from nonlocality, if you insist on realism you have to reject locality, due to Bell's theorem, as you even acknowledged earlier (but have forgotten).

Bell's theorem has other issues, one of which is that it is applied to a classical system. So if an experiment violates this theorem, the only valid conclusion is that classical physics can never describe a quantum system. But people like you jump to a different conclusion, which I have argued, is unjustified.

Griffiths wrote in 2005 the orthodox interpretation prevails among physicists. No matter what your interpretation, there is no significant disagreement that it is not wrong to say QM is a nonlocal theory,

It's a position I am quite well aware. But what I'm saying, that this position needs to show evidence. Violations of Bell's isn't one of them.

and it is not wrong to write down the singlet state the way I (and Susskind!) did. Those were simply errors on your part, not a matter of interpretation.

Using a singlet is not wrong in itself. It's how one uses it that can be oh soooo wrong.

You got your Masters 15 years before I was born. I am the young generation.

I suppose all the study I've done after doesn't count. If my wife would hear you, she would club you to death. She had to sacrifice many things for my returning to study.

 

[Reptillian]

The only forces which exist in Relativistic classic mechanics are elctromagnetism and gravity, both of which are limited to c or less. The only possible contention is what shadows look like, but like I said, please do the math for that one.

I've been working on the math for my Dyson Sphere example, right now I'm calculating size of the sphere and stellar angular velocity necessary to give a shadow speed of greater than the speed of light on the sphere in non-relativistic Newtonian physics, then I can compare to a relativistic scenario.

The relativistic problems I have are: if the speed of the shadow appears greater than the speed of light, then what happens to the Lorentz contraction of the shadow from the point of view of the stationary observer on the interior of the sphere? If the shadow is moving close to but less than the speed of light, the shadow should be Lorentz contracted. I mean, I can watch the shadow move across the interior surface of the sphere, so if it goes faster than light it's length becomes what...imaginary? The resolution appears to be in the amount of time it takes for the light from the shadow boundaries to reach an observer on the interior surface of the Dyson Sphere. Having given it a little thought, I think you're right in that the shadow won't move faster than light.

I beg your pardon, PolyHedral, but that is breathtaking arrogance.

Pot...kettle...black, so very black...

Your objection about nothing traveling faster than light: Pass your hand over a lamp of width W. For the sake of argument, say your hand is moving at velocity C, and thus it passes over the lamp in time T = W/C. A screen some distance away, with width L > W, is illuminated by the lamp. A shadow passes over the screen due to your hand passing over the lamp. How fast does the shadow move? The shadow's velocity across the screen is V = L/T = L/W x C, which is greater than C. So for sufficiently fast hands and large screens, shadows can indeed travel faster than the speed of light. This doesn't violate relativity because no information/energy/mass etc. is transmitted faster than C by the shadow. There are other kinds of superluminal motion (e.g. phase velocity) and they don't violate relativity, either.

Relativistic addition of velocities should be used....also, what's the width of your hand if you're waving it at the speed of light?

 

[PolyHedral]

Let me get this straight: you won't take my word for it, you won't take a physics textbook's word for it, and yet you lack the ability to calculate it yourself?

Of course I won't take anyone's word for it. I'll take anyone's demonstration of it, though.

(I don't lack the ability to calculate it, just the time.)

Your objection about nothing traveling faster than light: Pass your hand over a lamp of width W. For the sake of argument, say your hand is moving at velocity C, and thus it passes over the lamp in time T = W/C. A screen some distance away, with width L > W, is illuminated by the lamp. A shadow passes over the screen due to your hand passing over the lamp. How fast does the shadow move? The shadow's velocity across the screen is V = L/T = L/W x C, which is greater than C.

It might look greater than c, but since the light itself is limited to c, and the shadow is a signal "in" the light, then there's something wrong with that thinking. Most likely because the shadow is not actually a light beam, but a collection.

(Also, you seem to have suggested my hand is travelling a light speed, which is a bit silly. )

The shadow discussed above may start to the left, outside Bob's light cone, but because it travels faster than C is propagates to the right and arrives at (t=0,x=0), so Bob can observe it.

As Reptillian suggested, I doubt it. Just because it looks as though the shadow moves faster than light from the viewpoint of the light source means nothing to Bob - one would actually have to do the calculations of what Bob sees based on the distances involved.

In QM we are not (as far as we know) talking about a nonlocal particle but a nonlocal influence or "collapse of the wavefunction". But the same reasoning applies in terms of your objection about whether Bob could observe such a hypothetical influence, or not. Suppose Alice's measurement occurs to Bob's left outside his past light cone, at (t=0,x = -10). According to QM her measurement has an instantaneous or nonlocal influence on Bob's particle, which would be represented as before by a horizontal line passing through (t=0,x=0). Ergo, it could affect Bob and he could observe the effect, too.

But then Charlie (who is sitting in a spacelike position/velocity such that Alice's measurement happens after Bob's) would find the effect of Alice's measurement to propogate backwards in time. It doesn't matter that it doesn't carry information - you still have a state change happening for no observable reason. This gets even more complicated if Alice is moving quickly relative to Bob - who's measure of "instantaneous" are we using?

As I've told you several times now: it turns out, when you consider it carefully, that only a causal influence (like those mentioned above) would violate causality if it traveled faster-than-light. The influence stipulated by QM of Alice's measurement is not a strictly causal one, and therefore, it does not lead to violations of causality.

Please point to the part of the equations that make a distinction between "causal" and "non-casaul" influences.

 

[zaybu]

I...also, what's the width of your hand if you're waving it at the speed of light?

Zero, hence no shadow...

 

[PolyHedral]

I'm not assuming anything. That's what the equation gives you. The solution is found by looking at the symmetry of the situation. The top left-hand is exactly like the what's at the bottom right corner. One beam ( the input I), breaks into two beams. At the bottom right-hand corner, the same two beams, flowing exactly like the top left-hand corner, moving into B, must give the same outflow (the output O). Otherwise, the symmetry would be broken, and then you would really have a mysterious case.

So the lower beam splitter somehow magically knows that the situation is symmetrical?

I'm not sure what you are hinting at. The data is about the number of particles the detector is counting, it's not measuring differences in phases. Do you have another experiment in mind? This one is about counting photons.

Photons, being oscllating electromagnetic fields, have rotational phase. If you overlap two photons of the same frequency and amplitude 180 degrees out of phase with each other, they disappear completely, as seen in the experiment. Your explanation says the interferometer's output is 180 degrees out with the actual (AFAIK) result of the experiment. (This is distinct from phase of polarization.)

 

[zaybu]

So the lower beam splitter somehow magically knows that the situation is symmetrical?

It's not a question of knowledge for the photon. The photon has no mind and can't know. Period.

It's symmetrical in the sense that both beam splitters are made of the same material, and placed at exactly the same angle with respect to the beam. So whether the splitter is at the top left corner, or at the bottom right corner, you should get exactly the same result. If you didn't, then energy would not be conserved, and that would a serious problem.

Your explanation says the interferometer's output is 180 degrees out with the actual (AFAIK) result of the experiment.

There is no phase being measured in the experiment, and the calculation doesn't include any phase whatsoever. So I don't know where you are getting this idea that there is a phase difference at the output. I don't understand what's bothering you. Or where you are getting this idea.

Look at the diagram again:

Top corner :

→ split into →↓
Bottom corner:
→↓ reunite, and this can only give → as an answer.

Remember that vector states in QM are not exactly the same as ordinary vectors, but they follow the same rule of addition (isomorphic).

 

[idav]

That experiment is boggling. Shows that light will find an escape. How does it know, path of least resistance. Same as the double split. When there two its 50/50, one covered and the light goes 100% through the one slit. Fascinating.

 

[Mr Spinkles]

....also, what's the width of your hand if you're waving it at the speed of light?

Zero, hence no shadow...

Oops, sorry, bad notation on my part. Don't interpret the variable "C" as being the speed of light, just assume it's some arbitrary speed of your hand. Then, as I said, "for sufficiently fast hands and large screens, shadows can indeed travel faster than the speed of light".

 

[Mr Spinkles]

No one said that. That's why Alice must perform their experiment, and ONLY after they made their observation can they compare and see that the law of angular momentum is conserved. In your case, you were calculating the total spin BEFORE they even had measured. So of course you ended up with violations. And then you had the audacity to say I believe in violations in the conservation. That's how twisted your arguments are.

You aren't making any sense. If you don't know the initial angular momentum, how do you know the final angular momentum was conserved? You tell me how you calculate these two quantities in regard to the measurements of the singlet state we are talking about:

(1) Initial total angular momentum = ... ?
(2) Final total angular momentum = ... ?

Conservation implies (1) = (2). How do you calculate them? Be specific.

I would say (and my QM textbooks back me up here) that (1) is the "known" or "given" total spin of two fermions in the singlet state. More precisely, the singlet was already "measured" before Alice and Bob got to them, in order to prepare them in the singlet state. So for example, perhaps someone measured the decay of a neutral pion into an electron and a positron. This measurement established the initial total spin of the system is zero, and the electron and positron are in the singlet state. The total spin must continue to be zero after Alice's measures +hbar/2 on her particle. How can that be the case, unless Bob's particle at that moment now carries -hbar/2?

Using a singlet is not wrong in itself. It's how one uses it that can be oh soooo wrong.

What? But, you just said in post# 320:

However, within my explanation: there are no singlet states ...

Are there singlet states, or not? If there are, the mathematical formalism of quantum mechanics tells us what to do, and it does not agree with what you say.

I suppose all the study I've done after doesn't count. If my wife would hear you, she would club you to death. She had to sacrifice many things for my returning to study.

I wouldn't be surprised if your wife wanted to club someone to death, anyway ...

 

[idav]

Oops, sorry, bad notation on my part. Don't interpret the variable "C" as being the speed of light, just assume it's some arbitrary speed of your hand. Then, as I said, "for sufficiently fast hands and large screens, shadows can indeed travel faster than the speed of light".

So would the light around said shadow go faster than light?

 

[Mr Spinkles]

Of course I won't take anyone's word for it. I'll take anyone's demonstration of it, though.

(I don't lack the ability to calculate it, just the time.)

I commend you for wanting to see a demonstration before you believe something. But if you won't take the experts' word for it, you shouldn't take your own word for it, either. You can't have it both ways. Like you, I have limited time. Surely you must realize that relativity and quantum mechanics is not something that can be demonstrated, or taught, in a fully convincing way in a post on an internet forum.

So, if you are curious about this subject, you should either (a) give the experts a little credit and take their word for it because you have limited time, or (b) crack open the textbooks I suggested and take the time to dive in and study the demonstration. These are your options.

Please point to the part of the equations that make a distinction between "causal" and "non-casaul" influences.

Okay. Write down the spacetime interval and write down the temporal order of events in different frames for spacelike intervals. Now show yourself (or look it up in a book) that inescapable logical paradoxes would arise if two events with a spacelike interval were causally connected, but such paradoxes would not arise otherwise.

 

[Mr Spinkles]

So would the light around said shadow go faster than light?

No. Light would travel at the speed of light. To get an idea, do the experiment yourself, pass your hand over a lamp and notice your shadow travels faster than your hand--it covers a larger distance in the same time.

Rigorously:

Given:
Light bulb width = W
Screen width = L
Hand velocity = V (no need to assume it's close to the speed of light, as I implied originally)
Speed of light = c

Hand initial position: y = 0, t = 0
Hand final position: y = W, t = W/V

As a check, notice the hand travels at speed V = (W-0)/(W/V-0) = W/W*V = V. As was given.

For convenience call the position of the hand w.r.t. the lamp "y" and the position of the shadow w.r.t the screen uppercase "Y". Additionally, imagine the light rays hitting the top and bottom of the screen forming an isosceles triangle, with sides of length "d" with the screen of length L as the base. Then:

Shadow initial position on screen: Y = 0, t = d/c
Shadow final position: Y = L, t = (W/V) + (d/c)

Notice I have accounted for the fact that the light travels at speed c: your hand interrupts the light beam at time 0, but it takes additional time (d/c) for the shadow to show up on the screen. Your hand also interrupted the light beam at time (W/V), but it took that interruption, too, an additional time (d/c) to reach the screen.

The speed of the shadow is then

V' = L/(W/V + d/c - d/c) = (L/W)*V.

and so for sufficiently large L and small W, V' can be faster than c.

 

[Mr Spinkles]

Relativistic addition of velocities should be used...

See my previous post, above. Also, look up the calculation of phase velocity of a light wave. It is vp = c/n(k), where n(k) is the wavelength-dependent index of refraction of the material. Occasionally n(k) can be less than unity in which case vp is greater than c. In other words, if you follow a particular peak on a wave, it can move faster than c. But that particular peak does not transmit energy/information/force etc. faster than c. So special relativity allows this. Similar story for the group velocity. This is called pulse reshaping and it's a known superluminal phenomenon.

 

[idav]

No. Light would travel at the speed of light. To get an idea, do the experiment yourself, pass your hand over a lamp and notice your shadow travels faster than your hand--it covers a larger distance in the same time.

Rigorously:

Given:
Light bulb width = W
Screen width = L
Hand velocity = V (no need to assume it's close to the speed of light, as I implied originally)
Speed of light = c

Hand initial position: y = 0, t = 0
Hand final position: y = W, t = W/V

As a check, notice the hand travels at speed V = (W-0)/(W/V-0) = W/W*V = V. As was given.

For convenience call the position of the hand w.r.t. the lamp "y" and the position of the shadow w.r.t the screen uppercase "Y". Additionally, imagine the light rays hitting the top and bottom of the screen forming an isosceles triangle, with sides of length "d" with the screen of length L as the base. Then:

Shadow initial position on screen: Y = 0, t = d/c
Shadow final position: Y = L, t = (W/V) + (d/c)

Notice I have accounted for the fact that the light travels at speed c: your hand interrupts the light beam at time 0, but it takes additional time (d/c) for the shadow to show up on the screen. Your hand also interrupted the light beam at time (W/V), but it took that interruption, too, an additional time (d/c) to reach the screen.

The speed of the shadow is then

V' = L/(W/V + d/c - d/c) = (L/W)*V.

and so for sufficiently large L and small W, V' can be faster than c.

This is brilliant. Cool trick with waves of photons.

 

[zaybu]

You aren't making any sense. If you don't know the initial angular momentum, how do you know the final angular momentum was conserved? You tell me how you calculate these two quantities in regard to the measurements of the singlet state we are talking about:

(1) Initial total angular momentum = ... ?
(2) Final total angular momentum = ... ?

Conservation implies (1) = (2). How do you calculate them? Be specific.

The first scenario was: initially, a source is at rest, decays into two particles... In the second scenario, Alice and Bob are given that information by someone after they have measured and tabulated their spins. So, in both cases, the initial total angular momentum is ZERO.

I would say (and my QM textbooks back me up here) that (1) is the "known" or "given" total spin of two fermions in the singlet state. More precisely, the singlet was already "measured" before Alice and Bob got to them, in order to prepare them in the singlet state.

If you are writing your equation from a God's point of view, of course you could do that, since in the singlet state, you are going to put your ↓↑, and that means you've already factored in the conservation of angular momentum. But if you're Alice or Bob, unless you were told before the experiment of that info, you don't know this. That's why I gave you the second scenario, in which it was clear they did not have that info while doing their experiment. In that case, they cannot write a singlet state. It's after they have made their measurements, and have compared their data, they see that for each case, the total angular momentum is zero.

Our disagrement comes to this:

(1) you believe that the state vector (or using the expression "wavefunction" from the old school) is a real wave.

I don't.

(2) you believe when a measurement is taken, the wavefunction collapses everywhere instantaneously.

I don't.

And so, you reason that there is a spooky action at a distance. I'm saying: your conclusion is based on two faulty premises.

 

[Reptillian]

See my previous post, above. Also, look up the calculation of phase velocity of a light wave. It is vp = c/n(k), where n(k) is the wavelength-dependent index of refraction of the material. Occasionally n(k) can be less than unity in which case vp is greater than c. In other words, if you follow a particular peak on a wave, it can move faster than c. But that particular peak does not transmit energy/information/force etc. faster than c. So special relativity allows this. Similar story for the group velocity. This is called pulse reshaping and it's a known superluminal phenomenon.

See my previous post, above. Also, look up the calculation of phase velocity of a light wave. It is vp = c/n(k), where n(k) is the wavelength-dependent index of refraction of the material. Occasionally n(k) can be less than unity in which case vp is greater than c. In other words, if you follow a particular peak on a wave, it can move faster than c. But that particular peak does not transmit energy/information/force etc. faster than c. So special relativity allows this. Similar story for the group velocity. This is called pulse reshaping and it's a known superluminal phenomenon.

I'm familiar with this phenomenon and with the fact that the group velocity transmits the information.

Here's an interesting link: http://www.philiphofmann.net/book_material/notes/groupphasevelocity.pdf

As for your calculation: the problem I see is that you've assumed that the shadow will travel across the screen in the same amount of time as the time it took for your hand to move across the source. Time dilation is necessary since the shadow will be moving at relativistic speeds.

Here's the setup I have so far:

Already you see my objection...if the speed of the shadow is greater than c, then the shadow size becomes imaginary...since the shadow size is related to the real variables like the size of your hand, the distance to the screen, and the distance to the source then you understand my objection. None of those are imaginary numbers so they can't cancel the i...what does it mean for a shadow to have imaginary size? The resolution I think is time dilation. The shadow on the screen, or on the surface of my Dyson Sphere, lags behind and moves more slowly than you'd expect it to.

 

[zaybu]

if the speed of the shadow is greater than c, then the shadow size becomes imaginary...since the shadow size is related to the real variables like the size of your hand, the distance to the screen, and the distance to the source then you understand my objection.

You missed an important point:

http://www.religiousforums.com/forum/3333643-post367.html