Einstein and "spooky actions"

[PolyHedral]

But we only need to calculate the position and time of the shadow at two points: at the bottom, and top of the screen. If the average speed between these points is greater than c, then the shadow must have traveled faster than c, if not during the entire journey, then at least during part of the journey. In other words, the exact motion of the shadow in between the initial and final positions does not affect our conclusions.

That's because you're only interested in the position of the light source. I'm interested in how the FTL shadow looks like frm other positions, i.e. is it actually possible to transmit FTL events into someone else's lightcone?

 

[Mr Spinkles]

That's because you're only interested in the position of the light source. I'm interested in how the FTL shadow looks like frm other positions, i.e. is it actually possible to transmit FTL events into someone else's lightcone?

For starters, everyone in reference frame S (at rest w.r.t. the light source and screen) agrees based on their observations that the shadow traveled faster than light. This is true whether the observer is positioned at the light source, or the screen, or anywhere else in frame S. Recall what I said to Reptillian in post #389:

The "observer" can be anyone at rest w.r.t. reference frame S. We have to go back to the definition of a reference frame S, which involves an array of observers and clocks at rest w.r.t. each other, at every location in S.

It may help you to imagine that you, personally, are located at the hand initial position. Your colleagues are located at the other 3 positions (hand final position; shadow initial position; shadow final position). You and your colleagues have all paced out the distances and synchronized your watches beforehand. If a hand (or shadow) arrives at someone's location, they record the time and immediately send a light signal to the other 3 colleagues. Therefore, you personally will observe the hand at its initial position, and sometime later, you will receive light signals from your colleagues reporting their observations. From those signals, you personally can reconstruct all the events.

So for example, you receive a light signal from your colleague reporting the observation of the shadow initial position at time 2d/c. But you know (because you are a competent observer) that meant the shadow actually arrived at the screen at time d/c. In relativity, we aren't talking about what people in frame S "see", but rather, what actually happened by the agreement of all competent observers in frame S. So we don't worry about what you vs. your colleagues personally witnessed. Rather, we consider what everyone in frame S agrees actually happened: the shadow actually arrived at time d/c at the screen, even though you personally witnessed only the light signal from your colleague at time 2d/c to report this event.

 

[zaybu]

Exactly Spherical polar coordinates rock! lol

Sprinkles solution is wrong.

Can you tell me how to upload pictures from my computer? The solution is very simple.

w = distance hand travels
d= distance from light to hand
L = distance from light to shadow
S = distance shadow travels
V(shadow) = (w/d) x V(hand) = (S/L) x V(hand)
Note: w<d, S <L
Therefore, V(shadow) is never greater than the speed of light.

 

[zaybu]

Since there hasn't been any refutation of my arguments, and post #378, (see http://www.religiousforums.com/forum/3334045-post378.html ) says it all,

I will declare myself winner of this debate.

 

[Mr Spinkles]

zaybu, it would help avoid confusion if you would at least keep the definitions of variables consistent. For example, don't call the distance from light to shadow "L" because that was already defined as the length of the screen. Call it something different.

 

[Mr Spinkles]

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.

So, just to be clear (you tend to contradict yourself so I have to ask), before Alice measures the particles which have already been prepared in the singlet state, is the total angular momentum known to be zero?

After Alice measures but before Bob measures, is the total angular momentum still zero? What if Alice measures a spin of +hbar/2, and Bob refrains from measuring ... how can total angular momentum remain zero, unless Bob's particle in some sense "really" has -hbar/2 after Alice measures her particle?

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 &#8595;&#8593;, 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.

Okay well now you know that from the beginning, we are assuming the particles start in the singlet state. That is where any hypothetical "spooky action" occurs, so that is what we must discuss.

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.

[/QUOTE]They aren't premises they are conclusions. The only premise is that the theory of quantum mechanics is correct and complete. You have repeatedly not applied that theory correctly (e.g. you thought the way I wrote the singlet state was "wrong", which it wasn't).

 

[zaybu]

So, just to be clear (you tend to contradict yourself so I have to ask), before Alice measures the particles which have already been prepared in the singlet state, is the total angular momentum known to be zero?
After Alice measures but before Bob measures, is the total angular momentum still zero? What if Alice measures a spin of +hbar/2, and Bob refrains from measuring ... how can total angular momentum remain zero, unless Bob's particle in some sense "really" has -hbar/2 after Alice measures her particle?

Suppose you are looking at two billiard balls colliding, and you only know the momentum of one of the ball. You can't decide automatically that momentum has been conserved. You need to measure the momentum of the second ball before you can verify it.

OTHO, if you assume that momentum is conserved, and you know the momentum of one of the ball, you can calculate the momentum of the second ball.

In your case, you assume the spin is conserved, and one of the spin is known, you believe that because you can calculate the second spin, the wave collapses automatically, and presto there must be a spooky action at a distance. Do you have any idea how that is totally ridiculous?

They aren't premises they are conclusions.

They're not. It's your sacred belief, a belief that is totally unfounded, totally unproven, and totally unjustified.

 

[zaybu]

zaybu, it would help avoid confusion if you would at least keep the definitions of variables consistent. For example, don't call the distance from light to shadow "L" because that was already defined as the length of the screen. Call it something different.

Regardless of how you label the distances, the speed of the shadow is NEVER greater than the speed of light. Do you agree?

 

[Mr Spinkles]

Suppose you are looking at two billiard balls colliding, and you only know the momentum of one of the ball. You can't decide automatically that momentum has been conserved. You need to measure the momentum of the second ball before you can verify it.

OTHO, if you assume that momentum is conserved, and you know the momentum of one of the ball, you can calculate the momentum of the second ball.

Agreed. I would add (and I presume you would agree) that in this case, it would be appropriate to go one step further. Not only can we calculate the momentum of the second ball, but we can rest assured the second ball "really" has that momentum even if we haven't measured it. In fact, if that were not true, momentum would not be conserved at that moment.

In your case, you assume the spin is conserved, and one of the spin is known, you believe that because you can calculate the second spin, the wave collapses automatically, and presto there must be a spooky action at a distance.

That's a pejorative way of putting it. To be precise, conservation of total angular momentum can be derived from rotational invariance in three dimensions. Additionally, it can be established by experiment. It needn't be arbitrarily "assumed".

Now, once this principle has been established either by theory or experiment (take your pick), the formalism of quantum mechanics tells us what to do. As you yourself said, it's vectors in Hilbert space and eigenvalues and operators and so forth. The logic concerning measurement of entangled states (such as the singlet) leads to the inescapable consequence of "spooky action" (your term). Either the two particles really are "nonlocal" or a measurement really has a "nonlocal" effect on a distant particle, or QM must be wrong or incomplete. But experiments demonstrate QM is correct (with very little wiggle room for error--Aspect measured that if the "wave function collapse" is not instantaneous/nonlocal, it takes a very short time indeed).

There's no wiggle room on that last point. Well, maybe there is a little wiggle room, e.g. the many-worlds interpretation of QM, or an as-yet unproven "nonreal" hidden variables theory. But those are just as conceptually weird as nonlocality, just for different reasons. At any rate, your confused guessing hasn't yet hit upon those (valid) possibilities. Even if you had, we would then simply have different (but equally valid) interpretations of QM. This would render your original accusation that my description of nonlocality was "totally wrong" misleading and overconfident, at best.

Do you have any idea how that is totally ridiculous?

No, personally I do not. Many things in physics seem totally ridiculous to the uninitiated. I learned long ago, it is best not to judge conclusions based on how they subjectively seem; one can only follow, step-by-step, where logic and evidence lead. Even without "spooky action" many established principles of QM and relativity seem totally ridiculous, but that's the universe we seem to be stuck with.

They're not. It's your sacred belief, a belief that is totally unfounded, totally unproven, and totally unjustified.

It's not my sacred belief. I don't want nonlocality to be true, it's just an inescapable consequence of the theory of quantum mechanics (notwithstanding alternative, but equally "ridiculous" interpretations). If anything, rejecting nonlocality seems to be your sacred belief, given the self-contradictory evolution of your attempts to refute it:

(1) Spinkles wrote down the singlet state wrong
(2) Actually, the way Spinkles wrote it down wasn't wrong, but Bob's particle remains in the singlet state after Alice's measurement
(3) No, wait! Actually, there is no singlet state
(4) On fourth thought, there is a singlet state, but we don't know its total spin before Alice's measurement ...

What will you come up with next?

Again I want you to answer my questions, but be specific this time about before/after Alice's measurement:

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

You agreed (contradicting yourself again) in post #378 that before Alice measures the total spin of the particles is zero. So (1) = zero.

What about (2)? Suppose Alice measures spin +hbar/2 on her particle. What is the total angular momentum? Does Bob's particle have a spin, and if so, what is it? Be sure to show explicitly the contribution of Bob's particle (if any) to the total spin. Remember: this is after Alice measures but before Bob measures his particle.

Contrary to what you said, according to my argument total angular momentum is conserved. (1) = 0, and (2) = +hbar/2 (Alice) - hbar/2 (Bob) = 0. Since Bob's particle is in a spin eigenstate, he can measure it again if he likes without changing its spin, so angular momentum continues to be conserved after Bob's measurement, too.

Contrary to what you said, according to your arguments angular momentum is not conserved. But you can prove me wrong, by showing the calculation.

 

[Mr Spinkles]

Regardless of how you label the distances, the speed of the shadow is NEVER greater than the speed of light. Do you agree?

No, I do not.

 

[zaybu]

Now, once this principle has been established either by theory or experiment (take your pick), the formalism of quantum mechanics tells us what to do

.

In this case, you need QM only for the observation that the spins are either up or down. In classical physics, the spin of an object, be it planet earth or a top, can have any value. But at subatomic scale, it is quantized along a given axis. For that, you need QM to explain this result. But for the correlation of Alice's data and Bob's data, plain old fashion classical physics can explain that.

Your spooky action at a distance is a fabrication that is not needed.

Many things in physics seem totally ridiculous to the uninitiated. I learned long ago, it is best not to judge conclusions based on how they subjectively seem; one can only follow, step-by-step, where logic and evidence lead. Even without "spooky action" many established principles of QM and relativity seem totally ridiculous, but that's the universe we seem to be stuck with.

That physics is weird is perpetuated by people like you. Once you understand the mathemsatical framework of QM and GR, there's little mystery left, except how to quantize gravity or explain the hierarchy problem. But these are not weird, just problems we haven't figured out how to solve.

It's not my sacred belief. I don't want nonlocality to be true, it's just an inescapable consequence of the theory of quantum mechanics (notwithstanding alternative, but equally "ridiculous" interpretations).

It's not. It's based on some twisted argument, which is totally superfluous.

If anything, rejecting nonlocality seems to be your sacred belief, given the self-contradictory evolution of your attempts to refute it:

(1) Spinkles wrote down the singlet state wrong
(2) Actually, the way Spinkles wrote it down wasn't wrong, but Bob's particle remains in the singlet state after Alice's measurement
(3) No, wait! Actually, there is no singlet state
(4) On fourth thought, there is a singlet state, but we don't know its total spin before Alice's measurement ...

What will you come up with next?

Scratch that out, and replace them with, Sprinkles believes that the wavefunction is a real wave.

Again I want you to answer my questions, but be specific this time about before/after Alice's measurement:
<B>

(1) Initial total angular momentum = ... ?
​

</B>

(2) Final total angular momentum = ... ?
​

You agreed (contradicting yourself again) in post #378 that before Alice measures the total spin of the particles is zero. So (1) = zero.

If we assume that Alice knows before doing her measurement that the total spin is zero. What else could it be but zero?

What about (2)?

Assuming Alice knows the conservation of angular momementum, she will know that (2) is also zero.

Suppose Alice measures spin +hbar/2 on her particle. What is the total angular momentum? Does Bob's particle have a spin, and if so, what is it?

This is where you are muddling up the case. On a theoretical basis, the total spin is zero, and all parties know that, but until it is measured experimentally, we can't confirm it.

It's the very reason why I gave you the second scenario in which the knowledge of the total spin was not given to Alice and Bob prior to the experiment. But you still can't figure out where your reasoning is totally out of whack.

Be sure to show explicitly the contribution of Bob's particle (if any) to the total spin. Remember: this is after Alice measures but before Bob measures his particle.

Contrary to what you said, according to my argument total angular momentum is conserved. (1) = 0, and (2) = +hbar/2 (Alice) - hbar/2 (Bob) = 0. Since Bob's particle is in a spin eigenstate, he can measure it again if he likes without changing its spin, so angular momentum continues to be conserved after Bob's measurement, too.

Contrary to what you said, according to your arguments angular momentum is not conserved. But you can prove me wrong, by showing the calculation.

Again, you are mixing up the knowledge from theory, which is known by all parties, with the knowledge from observation, which hasn't been completed yet. And for you to say, "according to your arguments angular momentum is not conserved" is dishonest. It is your conclusion based on faulty reasoning. And so to rescue that law, you have to fabricate a spooky action at a distance. But the law of conservation of angular momentum was never violated in the first place.

 

[PolyHedral]

Sprinkles solution is wrong.

Can you tell me how to upload pictures from my computer? The solution is very simple.

w = distance hand travels
d= distance from light to hand
L = distance from light to shadow
S = distance shadow travels
V(shadow) = (w/d) x V(hand) = (S/L) x V(hand)
Note: w<d, S <L
Therefore, V(shadow) is never greater than the speed of light.

Your equations suggest that the shadow can never be faster than the hand, which is wrong. The first form of the equation also suggests that moving the screen further away won't make a difference, which is also wrong.
If my algebra is correct, then the equation you're actually looking for is, given all the velocities are linear,

velocity of shadow = distance to screen from source x velocity of object / distance to object from source

(Except I can't be bothered with corrections for the screen being flat. :run

Obviously, if the distance to the screen is much greater than the distance to the source from the object, then the apparent velocity of the shadow will be much greater than that of light.

 

[zaybu]

Your equations suggest that the shadow can never be faster than the hand, which is wrong.

There's a simple way to check that. Do the experiment. Put a light bulb on, put your hand at a certain distance from it. Move it up or down, and see if the shadow of your hand moves faster that your hand. Let me know about it.

 

[idav]

No spooky action. It would be like sending head and tails of a coin in two directions. When one measures heads no doubt the other will measure tails.

 

[zaybu]

No spooky action. It would be like sending head and tails of a coin in two directions. When one measures heads no doubt the other will measure tails.

That's a false analogy. It implies that Alice and Bob are measuring the same particle, but they're not, they are measuring two different particles. The condition imposed on these two particles is that angular momentum is conserved, but that's not due to any spooky action at a distance but due to rotational invariance.

 

[idav]

That's a false analogy. It implies that Alice and Bob are measuring the same particle, but they're not, they are measuring two different particles. The condition imposed on these two particles is that angular momentum is conserved, but that's not due to any spooky action at a distance but due to rotational invariance.

It was just something simple but your right. I haven't seen anything indicating that measuring one particle will suddenly change the spin of the entangled particle.

 

[zaybu]

It was just something simple but your right. I haven't seen anything indicating that measuring one particle will suddenly change the spin of the entangled particle.

Good.

Spinkles believes that the wavefunction in QM is a real wave, which it isn't. But if that is his starting point, then to rescue that, he needs to invoke a spooky action at a distance. His argument falls apart when it is realized that the wavefunction is not real, but simply one of many mathematical concepts in the QM framework.

 

[Reptillian]

The dot is moving but the light is still going the normal speed. A photon from the laser pointer will hit point A while a completely different photon will hit point B both of which going normal speed.

No no ... ! Hold your horses. You are making things way more complicated than necessary. Trust me.

Break the problem into tiny, manageable pieces. You turn on your laser pointer and aim it at a screen 1 light year away. For 2 years, you continue to see no spot. Then, 2 years later, you see the spot. (**Note: it's also correct to say that 1 year later, the laser dot hits the screen. But we'll stick to your convention of adopting the POV of the guy pointing the laser.)

You turn off your laser pointer. For 2 years, the spot remains. Then after 2 year, the spot disappears.

You move your laser pointer. For 2 years, the spot does not move. Then after 2 years, it moves.

With me so far?

Ok, I understand what you're saying now.

One worry I had was that I would be able to send a message to someone and he'd be able to relay the message to a third guy faster than I'd be able to, but the triangle inequality takes care of that.

A guy on the screen doesn't see the dot as moving faster than light even though we do. Going back to the light year sized spherical setup I mentioned earlier with the source moving pi/2 radians in five seconds. (and now I see why the dot moves across the screen in 5 seconds) It helped me to imagine that I have guys with their own laser pointers positioned on the screen at thetas {0, pi/10 , pi/5 , 3pi/10 , 2pi/5 , pi/2} If each guy shines his pointer at the guy at 0 once the initial dot hits his position, then the guy at 0 will get a signal from the 1st guy about a third of a year after he see's the original dot. Now if each guy sent a signal to every other guy as soon as he received the initial dot, they could use that information to conclude that the source sees the dot move faster than light.

Does that seem reasonable, or am I just rambling?

Can you tell me how to upload pictures from my computer?

Yeah, go to Quick Links then Pictures & Albums to upload. Then you can link to your picture in responses. I can't remember if newer members can upload pictures.

 

[zaybu]

Yeah, go to Quick Links then Pictures & Albums to upload. Then you can link to your picture in responses. I can't remember if newer members can upload pictures.

Thanks, here's the solution:

 

[PolyHedral]

What about circumstances where theta is greater than 1?