Galileo and Lorentz transformation

[Al68]

Let me try to make it more clear. Suppose we have an inertial laboratory A, which observes object a. Suppose also that we have another laboratory B which observes object b. The experimental setups A+a and B+b are exactly the same. The only difference is that they are moving with respect to each other. Then the principle of relativity tells us that all results of measurements in A+a and in B+b are the same.

Sure, but that's not analogous to transforming an event from one frame to another.

The principle of relativity does not tell us anything about what observer A will find by making measurements on the object b, or what will be measurement results in B+a.

Yes, it does. It says that object b (and a) will obey the same laws of physics in A as it does in B.

A single object observed by different frames obeys the same laws of physics in both. That's the whole subject of this thread, and the context of the first postulate in SR.

 

[meopemuk]

A single object observed by different frames obeys the same laws of physics in both. That's the whole subject of this thread, and the context of the first postulate in SR.

It is impossible to argue with that. Of course, all objects in all reference frames obey the same laws of physics. The interesting question is "what these laws of physics are?"

For example, how can we find properties of the system 'a' observed from the reference frame B? If the only information given to us are the results of measurements of 'a' performed in the reference frame A. In other words, we need transformation laws, which connect observables in A with observables in B. The principle of relativity alone is not sufficient to obtain these transformation laws. Even the knowledge that inertial transformations connecting frames A, B, C,... form the Poincare group is not sufficient to determine the transformation laws for observables. This problem can be solved if we know the representation of the Poincare group in the Hilbert space of the system 'a'. In classical mechanics, the same task is fulfiled by constructing the appropriate representation of the Poincare group by canonical transformations in the phase space of the system 'a'.

Eugene.

 

[Al68]

It is impossible to argue with that. Of course, all objects in all reference frames obey the same laws of physics. The interesting question is "what these laws of physics are?"

For example, how can we find properties of the system 'a' observed from the reference frame B? If the only information given to us are the results of measurements of 'a' performed in the reference frame A. In other words, we need transformation laws, which connect observables in A with observables in B. The principle of relativity alone is not sufficient to obtain these transformation laws.

I agree with all of that, but I don't see the point.

In the context of an explosion, if the explosion is a result of the laws of physics, then it occurs in all reference frames according to the first postulate. Of course the laws of physics don't tell us how to label the coordinates. Neither does the first postulate.

In fact I can arbitrarily define a coordinate system any way I choose, with arbitrary transformation laws, without violating the laws of physics or the first postulate. But whether or not the explosion occurs or not is completely independent of my choice.

 

[Fredrik]

Eugene, I agree that there's nothing logically inconsistent about your interpretation of the principle of relativity, but don't you see how different it is from standard SR? Just about every calculation in SR is based on the axiom that physical events are (coordinate independent) points in Minkowski space.

Consider e.g. the method we use to prove that two laboratories that don't have the same velocity will measure different lengths of an object that's co-moving with one of the laboratories. The co-moving laboratory measures the proper distance between the endpoints at two events that are assigned the same time coordinate by the coordinate system that's associated with its motion. The other laboratory also measures a proper distance, but between two different events. To understand what the result will be, the first thing we have to do is to figure out which two events that is. We can e.g. draw a spacetime diagram that shows all the relevant events mapped to $\mathbb R^2$ by the coordinate system associated with the motion of the co-moving laboratory. The world lines are vertical in this diagram, and we can prove that events that the other laboratory considers simultaneous are on a line with slope v, and from that we can figure out which two events the other laboratory will consider, and what the result will be.

When we use your way of looking at things, that object may not even exist to the second laboratory. That may not be logically inconsistent, or even in contradiction with every possible interpretation of the principle of relativity, but it certainly isn't special relativity.

 

[Al68]

When we use your way of looking at things, that object may not even exist to the second laboratory. That may not be logically inconsistent, or even in contradiction with every possible interpretation of the principle of relativity, but it certainly isn't special relativity.

It would be pretty difficult to observe an object violate the laws of physics if the object just isn't observed at all. Surely that's not what meopemuk is talking about?

 

[Dale]

I would like to reiterate my suggestion that this is inappropriate for this forum. We have given meopemuk more than enough time to say something remotely related to standard SR, which he has not. The present discussion does not belong here. Particularly the very non-mainstream interpretation of the first postulate.

Of course, for known unstable particles and realistic observer speeds the "boost induced decay probability" is extremely small and cannot be presently observed.

In meopemuk's own words, it is at this time a speculative theory with differences from SR that are so small as to live within the errors of all current theories. Once there is some reputable experimental evidence supporting this theory then it would be appropriate to discuss it in another portion of PF, but even then it will not belong in the relativity sub-forums.

 

[meopemuk]

Eugene, I agree that there's nothing logically inconsistent about your interpretation of the principle of relativity, but don't you see how different it is from standard SR? Just about every calculation in SR is based on the axiom that physical events are (coordinate independent) points in Minkowski space.

Consider e.g. the method we use to prove that two laboratories that don't have the same velocity will measure different lengths of an object that's co-moving with one of the laboratories. The co-moving laboratory measures the proper distance between the endpoints at two events that are assigned the same time coordinate by the coordinate system that's associated with its motion. The other laboratory also measures a proper distance, but between two different events. To understand what the result will be, the first thing we have to do is to figure out which two events that is. We can e.g. draw a spacetime diagram that shows all the relevant events mapped to $\mathbb R^2$ by the coordinate system associated with the motion of the co-moving laboratory. The world lines are vertical in this diagram, and we can prove that events that the other laboratory considers simultaneous are on a line with slope v, and from that we can figure out which two events the other laboratory will consider, and what the result will be.

Fredrik,

I am fully aware that my approach is different from standard SR. It suggests a different solution for the length contraction problem, which does not involve construction of space-time diagrams.

Let me choose two particles on the opposite ends of the stick and denote their position operators (or their position dynamical variables in the classical case) in the frame at rest by $$x_1$$ and $$x_2$$. I use 1-dimensional case for simplicity. Then the length of the stick in the rest frame is $$L = |x_1-x_2|$$. Then I switch to the moving frame description. I denote the boost operator by $$K_x$$, and obtain the length of the stick in the moving frame by usual quantum-mechanical formula

$$L' = e^{-iK_x \theta} L e^{iK_x \theta}$$......(1)

If interactions between atoms in the stick are weak, then this result will not be different from the usual SR length contraction formula

$$L' = L/\cosh \theta$$..........(2)

However, for very strong interactions, results (1) and (2) will be different.

When we use your way of looking at things, that object may not even exist to the second laboratory. That may not be logically inconsistent, or even in contradiction with every possible interpretation of the principle of relativity, but it certainly isn't special relativity.

Yes, this is not the standard special relativity. For example, the operator $$K_x$$ in (1) may contain interaction terms that lead to decays of particles 1 and 2. Then, in the moving frame even the particle content of the stick can be altered. So, strictly speaking, the notion of the "length of the stick" will be altered as well.

I would like to reiterate my suggestion that this is inappropriate for this forum. We have given meopemuk more than enough time to say something remotely related to standard SR, which he has not. The present discussion does not belong here. Particularly the very non-mainstream interpretation of the first postulate.

Even if you don't buy my arguments, I think in our discussion we touch some basic and interesting issues regarding the logical structure of special relativity. A critical discussion of SR postulates could be benefitial in learning this theory by everyone, IMHO.

On the other hand, I understand that I over-used your hospitality on this forum. So, if you decided to lock this thread, I will not be offended.

Eugene.

 

[meopemuk]

In the context of an explosion, if the explosion is a result of the laws of physics, then it occurs in all reference frames according to the first postulate.

The first postulate tells us that (in the notation I've used above) if an explosion is seen in the setup A+a, then an explosion must be seen also in the setup B+b. However, the principle of relativity does not tell us what we should see in the experimental setups A+b and B+a. It might be true that in these combinations Observer+object explosions are not observed. I admit that such a possibility is odd, but it does not violate the principle of relativity at all.

Eugene.

 

[Fredrik]

However, for very strong interactions, results (1) and (2) will be different.

I don't see how the strength of the interaction enters into the calculation. These are the things I think I do see:

The Hilbert space of two non-interacting systems is the tensor product of the two component systems. Each of those admits a representation of the Poincaré group (or its covering group), and we can use those to construct a representation on the tensor product space. For example, the Hamiltonian is defined as $H=H_1\otimes H_2$, and the other generators are defined the same way. An interaction between these systems is defined as a modification of the Hamiltonian H → H+V that entangles the two systems. (What I mean by this is that the time evolution operator applied to an unentangled state gives us an entangled state). When we do this, we also have to modify the boost generator, K → K+W, to ensure that the commutation relations are still satisfied.

You have defined a length operator L, and you're using the modified boost generator to transform it. I'm OK with that. What I don't see is how to verify the statement I quoted from your post.

I don't think any of this looks like an argument against Minkowski spacetime. We're adding that term to the boost generator to preserve the commutation relations that we get from the assumption that spacetime is Minkowski space. So it looks more like an argument for Minkowski spacetime than against it.

Yes, this is not the standard special relativity. For example, the operator $$K_x$$ in (1) may contain interaction terms that lead to decays of particles 1 and 2. Then, in the moving frame even the particle content of the stick can be altered. So, strictly speaking, the notion of the "length of the stick" will be altered as well.

As long as we're assuming the symmetries of Minkowski spacetime are symmetries of the quantum theory, I'd say that it is standard SR, or rather standard special relativistic quantum mechanics. If the particles can decay in one frame while not doing so in the other, then the whole object could disintegrate before the experiment is over. If this is in fact what we get from simply combining SR and QM, I'd say it's a pretty significant result. But I strongly doubt that we can get anything that extreme from SRQM.

 

[Fredrik]

I would like to reiterate my suggestion that this is inappropriate for this forum.
...
In meopemuk's own words, it is at this time a speculative theory with differences from SR that are so small as to live within the errors of all current theories. Once there is some reputable experimental evidence supporting this theory then it would be appropriate to discuss it in another portion of PF, but even then it will not belong in the relativity sub-forums.

The thread has drifted far from the original topic, and isn't even about classical relativity anymore. I agree that this discussion doesn't belong in the relativity forum. But I think it would be OK in the quantum physics forum. Meopemuk has been describing this as a speculative non-standard theory, but most of it is just the standard way to combine SR and QM. Now he's suggesting that this (very mainstream) theory has some implications that we have so far been unaware of (some of them pretty extreme), and even though I think some of his ideas are based on misunderstandings, I still think this is a meaningful discussion.

If some moderator would like to take action because this stuff doesn't belong in the relativity forum, then I would rather have it cut out and put into a new thread in the quantum physics forum, than to have the thread locked or posts deleted.

 

[meopemuk]

I don't think any of this looks like an argument against Minkowski spacetime. We're adding that term to the boost generator to preserve the commutation relations that we get from the assumption that spacetime is Minkowski space. So it looks more like an argument for Minkowski spacetime than against it.

The best argument against Minkowski spacetime is the Currie-Jordan-Sudarshan theorem that I've mentioned earlier

D. G. Currie, T. F. Jordan, E. C. G. Sudarshan, "Relativistic invariance and Hamiltonian theories of interacting particles", Rev. Mod. Phys., 35 (1963), 350.

This theorem says that in any relativistic theory (where both the Hamiltonian and the boost operator contain interaction terms) world-lines (or trajectories) of particles do not transform by usual Lorentz formulas. So, the Minkowski space-time picture is not applicable.

Eugene.

 

[Al68]

The first postulate tells us that (in the notation I've used above) if an explosion is seen in the setup A+a, then an explosion must be seen also in the setup B+b. However, the principle of relativity does not tell us what we should see in the experimental setups A+b and B+a.

Of course it does. It tells us that b will obey the same laws of physics in A as it does in B.

It might be true that in these combinations Observer+object explosions are not observed. I admit that such a possibility is odd, but it does not violate the principle of relativity at all.

Yes, it does. Even in Newtonian physics, the object b obeys F=ma in both A and B. So does object a.

This is the whole reason for the first postulate in SR. That a single object obeys the same laws of physics in different reference frames.

 

[meopemuk]

The first postulate tells us that (in the notation I've used above) if an explosion is seen in the setup A+a, then an explosion must be seen also in the setup B+b. However, the principle of relativity does not tell us what we should see in the experimental setups A+b and B+a.

Of course it does. It tells us that b will obey the same laws of physics in A as it does in B.

If explosion occurs in the setup A+a, then there is no guarantee that the same explosion is seen in the setup B+a. Let me give you an example.

Suppose that object 'a' is a time bomb. Suppose also that two inertial observers A and B are related to each other by a time translation transformation. For example, observer B makes his observations 1 hour earlier than A. Now, we have agreed that A sees an explosion of the time bomb 'a'. One hour before this observation the bomb was intact. Therefore, no explosion is seen in the setup B+a.

If you agree that my example with time translations is correct, then there is only a little step to change this example by replacing the time translation with a boost. My claim is that if observers A and B are related to each other by a boost, we can get a similar situation: explosion is observed in the setup A+a and not observed in the setup B+a.

Eugene.

 

[Al68]

If explosion occurs in the setup A+a, then there is no guarantee that the same explosion is seen in the setup B+a. Let me give you an example.

Suppose that object 'a' is a time bomb. Suppose also that two inertial observers A and B are related to each other by a time translation transformation. For example, observer B makes his observations 1 hour earlier than A. Now, we have agreed that A sees an explosion of the time bomb 'a'. One hour before this observation the bomb was intact. Therefore, no explosion is seen in the setup B+a.

If you agree that my example with time translations is correct, then there is only a little step to change this example by replacing the time translation with a boost. My claim is that if observers A and B are related to each other by a boost, we can get a similar situation: explosion is observed in the setup A+a and not observed in the setup B+a.

Eugene.

Your example has nothing to do with whether or not the explosion happens. It either happens or not.

The only thing that varies is the time and space coordinate that is assigned to it. Which means if the explosion is assigned the time noon in A, and 1 pm in B, then it's true that the explosion already happened at 12:30 in A, but had not happened at 12:30 in B. But the difference is only with the time coordinate of the explosion assigned by each frame.

But that's equivalent to saying that observers on the east and west coast disagree about whether or not the Superbowl kickoff happened, since at 5 pm on the west coast it has happened, but at 5 pm on the east coast it hasn't happened.

Does that mean that whether or not the Superbowl kickoff happens or not depends on which time zone you're in?

What about someone who's TV is broke? Does whether or not the kickoff happened depend on when/whether it is actually observed?

 

[A.T.]

For example, observer B makes his observations 1 hour earlier than A.

You mean he opens his eyes for a second, says "Ahh no explosion yet", and keeps sitting on the bomb that will blow him to pieces 1h later?

Now, we have agreed that A sees an explosion of the time bomb 'a'.

Let also say that A sees B sitting on that bomb while it explodes. It is a possible scenario.

One hour before this observation the bomb was intact. Therefore, no explosion is seen in the setup B+a.

Yeah B will not see much with his eyes closed, but he will still notice the bomb blast under his butt.

Bottom line is: You claiming that events don't happen in some frames, just because the observer 'wasn't making his observation' at that particular time. This might be consistent with your private definition of frames of reference and observers, but is rather a useless concept from practial standpoint. And it is definitely not what Relativity says.

 

[meopemuk]

For example, observer B makes his observations 1 hour earlier than A.

You mean he opens his eyes for a second, says "Ahh no explosion yet", and keeps sitting on the bomb that will blow him to pieces 1h later?

Exactly. I should have mentioned probably that in my definition an observer "opens his eyes for a second", makes a record of what he sees and then closes his eyes. If you want to talk about time development of events seen by a reference frame, you actually need to consider a sequence of above "instantaneous" observers connected to each other by time translation transformations.

So, in my example, when A opens his eyes and records his observations, there *is* an explosion. When B opens his eyes and records his observations, there *is no* explosion. So, A and B observe completely different things when looking at the same system - the time bomb. Of course, if B waits for an hour with his eyes closed, he will feel the explosion. But this is not surprising, because "observer B after 1 hour" is exactly the same as A.

Once we agreed about that (I hope we did), I am suggesting to apply the same logic to "instantaneous" observers A and B'. This time observer A is the same as before (i.e., he sees the explosion), but observer B' moves with respect to A with high velocity (no time translations involved). My claim is that it is possible that B' does not see the explosion. If time translation A->B results in "disappearance" of the explosion, then why can't we see the same "disappearance" as a result of the boost A->B'?

Edit: The reason I am using "instantaneous" observers is that only in this case I can exploit the full power of the Poincare group.

I think this should also address the last comment made by Al68.

Eugene.

 

[Al68]

Exactly. I should have mentioned probably that in my definition an observer "opens his eyes for a second", makes a record of what he sees and then closes his eyes. If you want to talk about time development of events seen by a reference frame, you actually need to consider a sequence of above "instantaneous" observers connected to each other by time translation transformations.

So, in my example, when A opens his eyes and records his observations, there *is* an explosion. When B opens his eyes and records his observations, there *is no* explosion. So, A and B observe completely different things when looking at the same system - the time bomb. Of course, if B waits for an hour with his eyes closed, he will feel the explosion. But this is not surprising, because "observer B after 1 hour" is exactly the same as A.

Once we agreed about that (I hope we did), I am suggesting to apply the same logic to "instantaneous" observers A and B'. This time observer A is the same as before (i.e., he sees the explosion), but observer B' moves with respect to A with high velocity (no time translations involved). My claim is that it is possible that B' does not see the explosion. If time translation A->B results in "disappearance" of the explosion, then why can't we see the same "disappearance" as a result of the boost A->B'?

Edit: The reason I am using "instantaneous" observers is that only in this case I can exploit the full power of the Poincare group.

I think this should also address the last comment made by Al68.

Eugene.

Huh? This thread will soon be locked I think, because it's going around in illogical circles, but what exactly are you claiming? If someone doesn't observe the event in question, then why would you call him an observer in any relevant sense? If he didn't observe the event in question, he's not an observer.

 

[meopemuk]

If someone doesn't observe the event in question, then why would you call him an observer in any relevant sense? If he didn't observe the event in question, he's not an observer.

Both pairs of observers (A, B) and (A, B') observe the same physical system 'a'. They can measure all relevant observables in the system. So, all of them are relevant *observers*. There is no anything unusual in the situation in which one observer in the pair sees an explosion while the other observer does not see it. They are different observers, they are entitled to their own points of view. If things look different from their different perspectives, there is nothing illogical about it.

Eugene.

 

[atyy]

Both pairs of observers (A, B) and (A, B') observe the same physical system 'a'. They can measure all relevant observables in the system. So, all of them are relevant *observers*. There is no anything unusual in the situation in which one observer in the pair sees an explosion while the other observer does not see it. They are different observers, they are entitled to their own points of view. If things look different from their different perspectives, there is nothing illogical about it.

Eugene.

Is this like whether or not a magnetic field is observed is frame dependent? ie. an explosion is not a technical term, and until it is defined with an equation, its frame dependence cannot be decided.

 

[meopemuk]

Is this like whether or not a magnetic field is observed is frame dependent? ie. an explosion is not a technical term, and until it is defined with an equation, its frame dependence cannot be decided.

Yes, it is difficult to give a rigorous mathematical definition of "explosion". So, I would prefer a simpler example of unstable particle. Then, if observer A sees the particle 100% undecayed, then observer B (displaced in time with respect to A) sees a non-zero decay probability (this is just the usual time-dependent decay law). Analogously, observer B' (moving with respect to A) may also see a non-zero decay probability.

In quantum mechanics one can give exact mathematical definition of the decay probability. So, all statements can be made precise an unambiguous.

Eugene.

 

[Al68]

Both pairs of observers (A, B) and (A, B') observe the same physical system 'a'. They can measure all relevant observables in the system. So, all of them are relevant *observers*. There is no anything unusual in the situation in which one observer in the pair sees an explosion while the other observer does not see it.

Of course not, but it's very unusual to use the word observer to describe someone who doesn't observe the event in question, unless there is some specific reason he should have observed the event.

Did this second "observer" observe anything specific that would indicate the explosion didn't happen?

Did he expect to observe the explosion during a specific observation period, but failed to do so?

Is there any specific reason that this second "observer" is relevant to the situation?

 

[meopemuk]

Did this second "observer" observe anything specific that would indicate the explosion didn't happen?

Did he expect to observe the explosion during a specific observation period, but failed to do so?

Is there any specific reason that this second "observer" is relevant to the situation?

It might be easier to answer your questions in the case of unstable particle (rather than exploding bomb). Because in this case all quantities have unambiguous mathematical definitions, and precise results can be obtained.

Let's say, "instantaneous" observer A prepares an unstable particle in his own frame. This means that the particle is seen by A as undecayed with 100% probability.

As before, observer B is shifted in time with respect to A. Since the particle has a finite lifetime, observer B finds a non-zero probability of the particle's decay. The probability of the particle being undecayed is less than 100%. It is known, that this probability decreases (almost) exponentially with the time separation between A and B. All this is well-known and hardly controversial.

Next consider observer B', which is moving with respect to A (without any time shift). A rigorous relativistic quantum theory indicates that observer B' will find the particle undecayed with less than 100% probability. According to this observer, the particle has a non-zero chance to decay even at time 0. If we consider other observers connected to B' by time translations (i.e., moving observers at non-zero times), we will find that no one of them sees the particle undecayed with 100% probability. Moving observers always see decay products with some non-zero probability.

Rather unusual conclusions of the last paragraph follow from the fact (which is not well-known, but rigorously proven) that if there is a decay-inducing interaction in the Hamiltonian, then there should be also a decay-inducing interaction in the boost generator. So, if decays are observed as a result of time evolution, then there should be also decays induced by boosts.

Eugene.

 

[Dale]

All if that seems like completely standard relativity of simultaneity stuff to me. Particularly considering that the wavefunction has spatial extent too.

Also, wrt this absurd idea that a bomb could blow in one frame but not anothern, it is illogical, but afaik not contrary to the first postulate. If you have two reference frames then you have some coordinate transforms as well. If event a has coordinates A in one frame which are transformed to coordinates A' in another frame then we know mathematically that a occurs at A'.

 

[Fredrik]

The best argument against Minkowski spacetime is the Currie-Jordan-Sudarshan theorem that I've mentioned earlier

D. G. Currie, T. F. Jordan, E. C. G. Sudarshan, "Relativistic invariance and Hamiltonian theories of interacting particles", Rev. Mod. Phys., 35 (1963), 350.

This theorem says that in any relativistic theory (where both the Hamiltonian and the boost operator contain interaction terms) world-lines (or trajectories) of particles do not transform by usual Lorentz formulas. So, the Minkowski space-time picture is not applicable.

Does anyone else think so? The article I linked to in #37 seems to think that the theorem is only relevant for Lagrangians with terms of at least 6th order in c^-1. (Doesn't that mean that all renormalizable field theories are safe?). This book seems to be saying that the theorem is only an problem for Hamiltonian action-at-a-distance theories, and not for field theories.

Let's say, "instantaneous" observer A prepares an unstable particle in his own frame. This means that the particle is seen by A as undecayed with 100% probability.

As before, observer B is shifted in time with respect to A. Since the particle has a finite lifetime, observer B finds a non-zero probability of the particle's decay. The probability of the particle being undecayed is less than 100%. It is known, that this probability decreases (almost) exponentially with the time separation between A and B. All this is well-known and hardly controversial.

Next consider observer B', which is moving with respect to A (without any time shift). A rigorous relativistic quantum theory indicates that observer B' will find the particle undecayed with less than 100% probability. According to this observer, the particle has a non-zero chance to decay even at time 0. If we consider other observers connected to B' by time translations (i.e., moving observers at non-zero times), we will find that no one of them sees the particle undecayed with 100% probability. Moving observers always see decay products with some non-zero probability.

I can't really make sense of this. To prepare an unstable particle as 100% undecayed should mean to produce it in an interaction, but there isn't a well-defined moment when the particle was created. Whatever A intends to measure, he would have to add up amplitudes associated with different events that he can think of as possible events where the particle "might have been created". So there isn't a moment where he can say that the probability is 100%. The closest match for the scenario you're describing that I can think of, is a particle (let's say a muon) that makes a track in a bubble chamber. The formation of the first bubble is a fairly well-defined classical event, and I suppose that at least after the fact, we can say that the particle existed there with probability 1. The thing is, A, B and B' will all agree about which bubble was the first, and if "the particle exists there with probability 1" is a valid conclusion for A, then it's a valid conclusion for B and B' too.

 

[meopemuk]

This book seems to be saying that the theorem is only an problem for Hamiltonian action-at-a-distance theories, and not for field theories.

Yes, usually two "explanations" are suggested of why CJS theorem can be ignored. One of them says that Hamiltonian theories are not good. Another one proposes to reject the idea of particles (and their worldlines) and consider (quantum) fields only. I think that both these "explanations" are not adequate.

First, each relativistic quantum theory must involve a Hilbert space and a representation of the Poincare group in this Hilbert space. Then, inevitably, we have 10 Hermitian generators (H - time translations, $$\mathbf{K}$$ - boosts, $$\mathbf{P}$$ -space translations, $$\mathbf{J}$$ - rotations), which are characteristic for Hamiltonian theories. Quantum field theories also follow the same pattern. See S. Weinberg "The quantum theory of fields", vol. 1.

By the way, quantum field theories have a theorem analogous to the CJS theorem. It is the famous Haag's theorem, which establishes that "interacting" quantum fields cannot have simple Lorentz-like transformation laws.

Second, it is true that traditional QFT does not provide any description of interacting particles (their wave functions, time evolution, etc.). This theory focuses only on properties described by the S-matrix. For such properties, only the movement of particles in the asymptotic regime is relevant. So, QFT is not a full dynamical theory. For example, it is impossible to talk about worldlines of interacting particles even in the classical limit of QFT. These deficiencies can be overcome in the "dressed particle" approach, which is basically a Hamiltonian direct interaction theory. CJS theorem cannot be ignored in the "dressed particle" approach.

I can't really make sense of this. To prepare an unstable particle as 100% undecayed should mean to produce it in an interaction, but there isn't a well-defined moment when the particle was created. Whatever A intends to measure, he would have to add up amplitudes associated with different events that he can think of as possible events where the particle "might have been created". So there isn't a moment where he can say that the probability is 100%. The closest match for the scenario you're describing that I can think of, is a particle (let's say a muon) that makes a track in a bubble chamber. The formation of the first bubble is a fairly well-defined classical event, and I suppose that at least after the fact, we can say that the particle existed there with probability 1. The thing is, A, B and B' will all agree about which bubble was the first, and if "the particle exists there with probability 1" is a valid conclusion for A, then it's a valid conclusion for B and B' too.

You've probably misunderstood my definitions of observers A, B, and B'. We've agreed that all these observers are "instantaneous". They "open their eyes" for a short time interval only.

Observer A opens his eyes at the exact time instant when the unstable particle is prepared. So, he cannot see any track in the bubble chamber. He can see at most one bubble at time 0.

Observer B is shifted in time with respect to A. So, he opens his eyes some time t after the unstable particle is prepared. He will definitely see particle tracks. Some of these tracks will be straight (meaning that the particle remains undecayed). Other tracks will be branched-off (meaining that the particle has decayed). The ratio of branched-off tracks to the total number of tracks is the decay probability from the point of view of B.

The situation with observer B' is a bit more complicated. We need to decide how our experimental setup is split between the observed system and the measuring apparatus. We have agreed that our physical system is the unstable particle only. Therefore, the bubble chamber should be considered as measuring apparatus, i.e., a part of the laboratory or observer. Therefore, if the question is "what the moving observer sees?" then to answer this question we need to use a moving bubble chamber (while the device preparing unstable particles for us should remain the same as in the two other examples, i.e., at rest). Needless to say that using bubble chambers moving with high speeds is a very very difficult proposition.

Moreover, your use of a single bubble (at time zero) as an indicator of the undecayed particle is not reliable. Bubbles are created due to the presence of a charged particle. The charge is conserved in decays. Therefore, by looking at the bubble you cannot say whether this bubble was created by the original unstable particle or by its decay products.

Eugene.

 

[atyy]

Second, it is true that traditional QFT does not provide any description of interacting particles (their wave functions, time evolution, etc.). This theory focuses only on properties described by the S-matrix. For such properties, only the movement of particles in the asymptotic regime is relevant. So, QFT is not a full dynamical theory. For example, it is impossible to talk about worldlines of interacting particles even in the classical limit of QFT. These deficiencies can be overcome in the "dressed particle" approach, which is basically a Hamiltonian direct interaction theory. CJS theorem cannot be ignored in the "dressed particle" approach.

In the standard introductions to QFT, they always say why eg. the Dirac equation cannot apply to single particles, because of the sea of negative energy states. Does this not arise in your approach?

 

[meopemuk]

In the standard introductions to QFT, they always say why eg. the Dirac equation cannot apply to single particles, because of the sea of negative energy states. Does this not arise in your approach?

I enjoy this discussion very much, but it is dangerously drifting far beyond the boundaries of this forum.

Regarding "standard introductions to QFT", it is not easy to find a good one. My primary recommendation is S. Weinberg, "The quantum theory of fields" vol. 1. This book has the best explanation of the logic of QFT without using such unnecessary ideas as Dirac equation and the sea of negative energy states.

For the dressed particle approach you can check the references in post #30.

Eugene.

 

[atyy]

I enjoy this discussion very much, but it is dangerously drifting far beyond the boundaries of this forum.

OK, thanks for your time. We'll KIV a discussion some other time in another forum when I've read your work more carefully.

 

[A.T.]

We've agreed that all these observers are "instantaneous". They "open their eyes" for a short time interval only.

Given your definition of an observer, the principle of relativity is not applicable at all.

The principle of relativity states that all inertial observers experience the same laws of physics. But your observes don't experience any laws of physics, because they just observe a static snapshot.

 

[meopemuk]

Given your definition of an observer, the principle of relativity is not applicable at all.

The principle of relativity states that all inertial observers experience the same laws of physics. But your observes don't experience any laws of physics, because they just observe a static snapshot.

The "snapshot" definition of observers covers all laws of physics just fine. For example, if you are interested in the dynamics (time evolution) observed at rest you should consider a sequence of "instantaneous" observers A(t) parameterized by the time parameter t and obtained from the observer A(0) by applying time translations. Then the time evolution of a physical system 'a' prepared in the laboratory A(0) can be obtained by stitching together measurement results of all A(t).

If you want to describe the time evolution in the moving frame, you can first define a moving observer C(0) at time 0 by applying a boost transformation to A(0). Then you apply time translations to C(0) to get a time sequence of moving observers C(t'). The time parameter t' is now measured by the clock attached to this set of moving instantaneous observers.

Then, the principle of relativity tells you the following. If you prepare a physical system 'c' in the laboratory C(0) in the same fashion as the system 'a' was prepared in A(0), then the results of measurements of 'c' by C(t') at t'=t will be the same as results of measurements of 'a' by A(t). This is exactly what is meant by the expression "dynamical laws of nature are frame invariant".

The benefit of using "instantaneous" observers (instead of "long-living" ones) is that in this case all ten types of Poincare transformations can be treated on equal footing. Time translations can be regarded as changes from one observer to another. In the case of "long-living" observers, the status of time translations is different from space translations, rotations, and boosts. This would not allow us to use the full power of the Poincare group.

Eugene.

 

[A.T.]

The "snapshot" definition of observers covers all laws of physics just fine. For example, if you are interested in the dynamics (time evolution) observed at rest you should consider a sequence of "instantaneous" observers A(t) parameterized by the time parameter t...

Your "sequence of instantaneous observers" A(t) is just the usual "long-living" observer. And for these "long-living" observers it is still true: If the bomb explodes at some A(t_a) it will also explode at some B(t_b).

But your "instantaneous observers" still don't observe any actual physics taking place. And the principle of Relativity applies to them only in the most trivial sense: They all experience the same laws of physics: none.

The benefit of using "instantaneous" observers (instead of "long-living" ones) is that in this case all ten types of Poincare transformations can be treated on equal footing.

Unfortunately we humans are "long-living", therefore we are more interested in the "sequence of instantaneous observers". And the principle of relativity is useful to us only, if it applies to these "sequences of instantaneous observers".

Bottom line is, you use two different definitions of "observer":

- For your Poincare-transformations-on-equal-footing it is the "instantaneous observer"
- For the Principle of Relativity it is the "sequence of instantaneous observers" or "long-living observer"

 

[yuiop]

He doesn't need to stop. He can pass the clocks very closely on timeout. And if A=B they explode and kill him. So according to the guy at rest to the clocks he's dead. But in his own frame he's fine because A>B.

This doesn't happen in SR because it is not a multiple universe theory.

You can call this the third postulate if you want.

On the other hand it can happen in GR. In Schwarzschild coordinates an event where two light rays cross below the event horizon does not happen in KS coordinates where the ingoing ray is in this universe and the outgoing ray is another universe. KS coordinates clearly show that GR involves two universes.

 

[Dale]

Kev, I disagree with that interpretation. Just because some specific coordinate chart does not cover a given region of spacetime does not in any way imply that the uncovered region of spacetime is in a different universe. Particularly in spacetimes where there exist other coordinate charts that do cover the region.

 

[A.T.]

KS coordinates clearly show that GR involves two universes.

Putting aside the validity of this interpretation, we are discussing here two observers who are both able to observe the same bomb. So they and the bomb are all in the same universe in the GR sense.

 

[meopemuk]

Your "sequence of instantaneous observers" A(t) is just the usual "long-living" observer. And for these "long-living" observers it is still true: If the bomb explodes at some A(t_a) it will also explode at some B(t_b).

This is where we disagree. The principle of relativity tells us that if we have two identical bombs 'a' and 'b' prepared in the frames A(0) and B(0), respectively, and if observer A(t) sees an explosion of the bomb 'a', then it is guaranteed that observer B(t) will see exactly the same explosion of the bomb 'b'.

The principle of relativity does not tell us anything certain about what observers A(t) can say about the bomb 'b' and what observers B(t) can say about the bomb 'a'. The principle of relativity does not allow you to connect measurements performed by different observers on the same system. For example, the principle of relativity alone is unable to predict system's dynamics (which in my interpretation is a sequence of observations made by time-translated observers on the same physical system). In order to make such predictions you need more than just principle of relativity - you need a full dynamical description of the system. In particular, you need to know the Hamiltonian and the boost operator of the system.

Eugene.

Edit: Another example: Suppose that the bomb 'a' that observer A(0) has is 1 meter in length. You say that the principle of relativity alone can tell you whether observers B(t) will see this bomb's explosion or not. But can this principle answer a simpler question: what is the length of the bomb 'a' from the point of view of B(t)? From SR we know that the bomb must be seen contracted according to B(t). However, the principle of relativity alone is not sufficient to predict the length contraction or derive the full formalism of SR. You need a few other principles, such as the constancy of the speed of light and the "coincidence condition" that we discussed earlier.