Galileo and Lorentz transformation

[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.

 

[Dale]

meopemuk, do you agree with the following statement?

Let S and S' be two coordinate systems that are related to each other through some invertible transformation relation T(A)=A' where A' and A are the coordinates of some arbitrary event in S' and S respectively. Then, if the bomb explodes at event A in S then it explodes at event A' in S'.

 

[meopemuk]

meopemuk, do you agree with the following statement?

Let S and S' be two coordinate systems that are related to each other through some invertible transformation relation T(A)=A' where A' and A are the coordinates of some arbitrary event in S' and S respectively. Then, if the bomb explodes at event A in S then it explodes at event A' in S'.

It is obvious that simple re-labeling coordinates of events cannot make the events change or disappear. So, different "coordinate systems" always agree on the physical nature of events.

However, my point is that inertial transformations of observers are not always reducible to simple re-labeling of coordinates. It is true that space translations and rotations amount to simple coordinate changes. Observers related by space translations and/or rotations see the same explosion at different coordinate points. However, time translations and boosts are different. Two observers related by a time translation may disagree about the explosion. The same with boosts: two observers moving with respect to each other may disagree about the explosion. So, boosts cannot be represented exactly as pseudo-rotations of the Minkowski space-time coordinates. Boost transformations have also non-trivial dynamical components.

Eugene.

 

[Dale]

It is obvious that simple re-labeling coordinates of events cannot make the events change or disappear. So, different "coordinate systems" always agree on the physical nature of events.

Good. This is what I mistakenly thought you were saying, and I am glad to know that your position is not as extreme as I was understanding.

Two observers related by a time translation may disagree about the explosion.

Before reacting to this, let me clarify:

If S and S' are related by the transformations
t' = t + 5
x' = x
y' = y
z' = z

And if the explosion occurs at A = (t, x, y, z) = (1,2,3,4)

Are you suggesting that it is possible that the explosion does not occur at A' = (t', x', y', z') = (6,2,3,4)?

If this is not your claim then are you merely pointing out the obvious fact that at t' = 1 the explosion has not yet occurred? Or are you emphasizing the fact that that for some instantaneous observation made at O' = (6,0,0,0) the light from the explosion at A' has not yet reached O' and so the instantaneous observer does not visually see the explosion?

If none of these are your intent, please explain in detail what you mean.

 

[A.T.]

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.

Says who? The principle of relativity is based on empirical experience. Is there any empirical data to justify the above restriction of this principle?

Observers related by space translations and/or rotations see the same explosion at different coordinate points. However, time translations and boosts are different.

First you want to have them all on equal footing, and now you say they are different?

Two observers related by a time translation may disagree about the explosion.

Only if you mean your "instantaneous observers" who observe a single time coordinate only. Consequently for a space translation you would have to consider observers who observe a single space coordinate. They would disagree on many things as well. I find both concepts rather useless so far.

 

[yuiop]

...

However, my point is that inertial transformations of observers are not always reducible to simple re-labeling of coordinates. It is true that space translations and rotations amount to simple coordinate changes. Observers related by space translations and/or rotations see the same explosion at different coordinate points. However, time translations and boosts are different. Two observers related by a time translation may disagree about the explosion. The same with boosts: two observers moving with respect to each other may disagree about the explosion. So, boosts cannot be represented exactly as pseudo-rotations of the Minkowski space-time coordinates. Boost transformations have also non-trivial dynamical components.

Eugene.

You seem to be implying that because observer(s) translated in time will disagree about about whether an event occurred or not, that it follows that two observers that are separated by a boost will also disagree.

First of all there are some obvious distinctions. Two observers that are at rest with the same physical location and separated only by a time translation are in fact one and the same observer. Tom Yesterday and Tom Today can not see each other, while Tom Stationary and Tom Moving can. Only one way communication exists between Tom Today and Tom Yesterday. Tom Yesterday can leave messages for Tom Today but Tom Today can not leave messages for Tom Yesterday. Tom Stationary and Tom Moving on the other hand can communicate both ways with each other. So the temporally separated pair of observers (T.Yesterday and T.Today) are not in any way comparable to the dynamically separated pair of observers (T.Stationary and T.Moving).

This makes your statement "It is important to note that it is impossible to have a relativistic theory in which dynamical effects are associated only with time translations ..." dubious, because time translations ARE unique. As far as I know, no experiment has shown that we can travel backwards in time and yet we are free to move forwards or backwards in the spatial dimensions. The time coordinate in the invariant interval, always has a different sign from the three other spatial coordinates because the time coordinate is not exactly the same as other three coordinates.

 

[meopemuk]

Good. This is what I mistakenly thought you were saying, and I am glad to know that your position is not as extreme as I was understanding.

Before reacting to this, let me clarify:

If S and S' are related by the transformations
t' = t + 5
x' = x
y' = y
z' = z

And if the explosion occurs at A = (t, x, y, z) = (1,2,3,4)

Are you suggesting that it is possible that the explosion does not occur at A' = (t', x', y', z') = (6,2,3,4)?

If this is not your claim then are you merely pointing out the obvious fact that at t' = 1 the explosion has not yet occurred? Or are you emphasizing the fact that that for some instantaneous observation made at O' = (6,0,0,0) the light from the explosion at A' has not yet reached O' and so the instantaneous observer does not visually see the explosion?

If none of these are your intent, please explain in detail what you mean.

In your example there are two inertial observers S and S' that are related by a time translation. Observer S' makes his observations 5 hours later than observer S. So, if observer S saw the explosion, then observer S' sees only the aftermath of the explosion (5 hours later). So, these two observers (based on their measurements) make rather different conclusions about the state of the explosive device. It is not appropriate to say that the views of the two observers can be related by a simple re-labeling of coordinates of events.

Note also that observer S cannot "see into the future", i.e., 5 hours ahead. He can confidently say only about what he actually sees - the explosion. Similarly, observer S' cannot "see into the past". From his perspective, the bomb is seen as a bunch of scattered pieces. So, the opinions of S and S' about the state of the bomb are quite different. It is not possible to use the principle of relativity to reconcile these two opinions. The two opinions can be related to each other if we know the dynamical law (the Hamiltonian) which controls the time evolution of the system - the bomb.

Now, if in the above example you replace "time translation" with "boost" you will obtain a similar situation: two observers S and S' (moving with respect to each other) disagree about the state of the bomb. The two conflicting descriptions can be reconciled if we know the dynamical effect of boosts on the state of the bomb. This can be done if we know the interacting "boost generator".

Eugene.

 

[yuiop]

That was, in fact, Lorentz's explanation of the null result of the Michaelson-Morley experiment when he derived the Lorentz transforms. That theory, however, would require that only physical objects contract with motion, not the space between them while Einstein's theory requires that space itself contract and that all motion, not just electromagnetic, slow down. A version of the Michaelson-Morely experiment, called, I think, the "Kennedy experiment" showed that Einstein's theory was right and Lorentz's was wrong.

I do not think you are giving the full story here when you say Lorentz's (theory) was wrong and a lot depends on what you mean by Lorentz's theory. His early idea that relativistic effects could be explained purely in terms of physical length contraction only, due to motion relative to the ether was wrong, but his later ideas of physical length dilation AND physical time dilation due to motion relative to the ether are entirely consistent with the predictions of Special Relativity and the differences between LET and SR are only philosophical. I think the experiment you refer to is more commonly referred to as the Kennedy-Thorndike experiment.

 

[Dale]

In your example there are two inertial observers S and S' that are related by a time translation. Observer S' makes his observations 5 hours later than observer S. So, if observer S saw the explosion, then observer S' sees only the aftermath of the explosion (5 hours later). So, these two observers (based on their measurements) make rather different conclusions about the state of the explosive device. It is not appropriate to say that the views of the two observers can be related by a simple re-labeling of coordinates of events.

So, if I understand correctly you are only saying that the state of the bomb at t=1 (exploding) is not the same as the state of the bomb at t'=1 (intact). Is this a correct characterization of your claim? Is this all you intend to say? Because if so it seems a rather trivial point.

What about the state of the bomb at t'=6? Are you unwilling to assert that the state of the bomb at t'=6 is the same as the state of the bomb at t=1 (exploding)?

 

[meopemuk]

First you want to have them all on equal footing, and now you say they are different?

This is a great question! Yes, in Wigner-Dirac relativistic theory (either classical or quantum, does not matter) all 10 types of inertial transformations between observers are treated on equal footing. All of them are members of the Poincare group. There is no preference.

In order to use the Poincare group of inertial transformations in (quantum) physics we need to define a unitary representation of the group in the Hilbert space of the observed system. (If you are more comfortable with classical physics, you can make replacements "Hilbert space => phase space" and "unitary representation => representation by canonical transformations". All arguments will remain valid.) Only then we can apply various inertial transformations to state vectors of the system and/or operators of observables. Only then we can say how the physical system is seen by different observers. There is an infinite number of ways how one can build a unitary representation of the Poincare group in the Hilbert space. So, we need to choose a unique way which agrees with observed physics.

It is easy to build a non-interacting representation of the Poincare group in the Hilbert space of any N-particle system. However, this representation is not interesting for obvious reasons. So, let us build another representation, which takes interactions into account. From experience we know that results of time translations depend on interactions between particles. Therefore, the Hermitian representative of the generator of time translation (the Hamiltonian) must have an interacting form: H = H_0 + V. What about 9 other generators? We can confidently say that space translations and rotations do not have any interacting effects. These transformations remain the same as in the non-interacting case. Their Hermitian generators are non-interacting P = P_0, J = J_0.

Dirac was first to notice that in the situation described above it is not possible to assume that the Hermitian representative K of the generator of boosts remains non-interacting. Poincare group properties demand that, unlike P and J, the operator K must contains interaction terms K = K_0 + W. Therefore, boost transformations (similar to time translations) must induce non-trivial dynamical changes in the state of the system.

In this theory all inertial transformations are treated on equal footing. However, their effect on physical states can be rather different.

Only if you mean your "instantaneous observers" who observe a single time coordinate only. Consequently for a space translation you would have to consider observers who observe a single space coordinate. They would disagree on many things as well. I find both concepts rather useless so far.

Yes, I use "instantaneous observers", and it is not difficult to imagine how such observers can be realized in practice. I don't buy the space-time symmetry, so I am not going to conclude that "space-local" observers must exist as well. I don't even understand how such "space-local" observers can exist. They can't see beyond the infinitesimally small space region around them? To me it's just nonsense.

Eugene.

 

[meopemuk]

So, if I understand correctly you are only saying that the state of the bomb at t=1 (exploding) is not the same as the state of the bomb at t'=1 (intact). Is this a correct characterization of your claim? Is this all you intend to say? Because if so it seems a rather trivial point.

Yes, the point I make about the state of the bomb is rather trivial and standard. It does not deserve much discussion.

What about the state of the bomb at t'=6? Are you unwilling to assert that the state of the bomb at t'=6 is the same as the state of the bomb at t=1 (exploding)?

I disagree with your use of time labels t'=1, t'=6, etc. They create an impression that observer S' (or observer S) can see into the past or into the future. I insist on using the notion of "instantaneous" observers. These observers can see only what is before them in just one time instant. So, they assign only one time label to all their measurements. They read this label from the clock that they use.

The use of instantaneous observers is important for

1. treating all inertial transformations (including time translation) on equal footing.
2. Using the full power of the Poincare group
3. Desribing the results of time evolution and boost transformations as action of the Poincare group representation in the Hilbert space (or phase space) of the physical system.

The point is that when you use "long-living observers", then time translations are (sort of) losing their non-trivial dynamical status. From the point of view of "long-living" time-shifted observers S and S' there is no much difference in the bomb behavior. Both of them see the same explosion, simply they see it at different times. So, it appears that time translation is not more complicated than changing the value of the parameter t.

With my choice of "instantaneous observers" it becomes obvious that time translations have a non-trivial dynamical effect. It is also easier to make the point about the similar dynamical effect of boosts.

Eugene.