Speed of Light and the Causality Problem

In summary: No, I don't mean the basic postulates of relativity, like the principle of relativity itself, or the constancy of the speed of light. I mean extra assumptions that are not logically necessary. This may be e.g. the assumption that the space is homogeneous and isotropic (like in the original Einstein's paper) or the assumption that the space is flat, or some others. I am not a specialist in general relativity, but I am pretty sure that there are a lot of assumptions like that. So, I'd say that the relativity principle with some additional assumptions implies Lorentz transformations.
  • #36
cesiumfrog said:
Let's call your concept an "observation". o:)

There are some facts that two observations must agree on if the transformation that relates them is a spatial rotation, but will disagree on if the transformation is a space and/or time translation. An example of such a fact might be "this observation is of an exploding apple" (it's conceptually cleaner to consider macroscopic facts). In this sense SR would include space-time rotations (i.e. boosts) together with other (purely spatial) rotations, whereas you would place boosts with translations, right?

Let's take this "exploding apple" as an example. Let us denote "observation" O which is at rest with respect to the apple and is made exactly at the time when the apple explodes. Now using inertial transformations we can obtain a few other "observations". For example, we can translate O 1 meter to the North. This new "observation" O-translated will also see the apple exploding, exactly as O. The only difference is that the point of explosion will have different coordinates with respect to the O-translated.

Another example: we can rotate "observation" O around its axis and obtain O-rotated. Obviously, from the point of view of O-rotated there is exactly the same explosion. Simply it is seen from a different direction. These examples show that space translations and rotations of "observations" do not have any significant effect on what "observations" are seeing. The effect is purely geometrical: translated and rotated observers view the same thing from different distances and angles. I will call these transformations "dynamical".

Now, let us consider time translations of "observations". Suppose that we displaced our "observation" O 1 year back in time. We are simply asking what happened to the apple 1 year ago. Apparently, the result of such a transformation is far from being "geometrical" or trivial. One year ago the apple might not even exist. If we displaced O 1 year forward in time, the O-time-displaced "observation" would not see anything but rotten debris from the explosion. This is very different from what O sees. This means that time translations are "dynamical". Their results depend very much on interaction acting in the observed system (e.g., on the type of explosive inside the apple).

The next question is about boosts. What will an O-boosted "observation" see? Will it see exactly the same exploding apple as O? Surely, the apple seen by the O-boosted will have a non-zero velocity. It will be also affected by a relativistic length contraction. But can we be sure that there will be no other effects? For example, the explosion seen by the O-boosted may change its properties. Or, perhaps, O-boosted may not see any explosion at all (if the velocity of the boost is high enough).

If I understand you correctly, you firmly believe that boost transformations must be purely kinematical (i.e., change of velocity, length contraction,...) and independent on interactions that control apple's dynamics. You believe that somewhat similar to space translations and rotations, the effect of boosts is a simple change of space-time coordinates of events without any effect on the inner composition of the system (i.e., exploded versus un-exploded). You seem to be so convinced about this, that you are ready to use the kinematical interaction-independent character of boosts as the third postulate of relativity. If this postulate plays such an important role, you should be pretty sure that it is correct. What is the basis for your belief?

Eugene.
 
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  • #37
meopemuk said:
[..spatially]translated will also see the apple exploding, exactly as O. The only difference is that the point of explosion will have different coordinates [..But a one year ]time-displaced "observation" would not see anything but rotten debris from the explosion. This is very different

No, if you spatially translate your observation by one (light-)year, you will certainly not see the explosion.

It's a bit biased to construct some complicated network of clocks and rulers to (after waiting for distant messages to arrive) describe the explosion as being included in the spatially-translated observation, if you then refuse for that same network to also be used to describe the explosion as included in the time-translated observation.

meopemuk said:
What is the basis for your belief?

I have absolutely no basis for my belief that there exists an objective reality (e.g., that you exist outside of my imagination), but all of my existing experience supports the presumption that changing my velocity does not change what else is real. Admittedly there are a few postulates of modern physics which do contradict my everyday experience, but I also have verification of many consequences of those postulates.
 
  • #38
cesiumfrog said:
No, if you spatially translate your observation by one (light-)year, you will certainly not see the explosion.

I agree about that, but I regard this fact as mere technical annoyance rather than something fundamental.

cesiumfrog said:
but all of my existing experience supports the presumption that changing my velocity does not change what else is real.

I also agree that nobody has seen directly any dynamical effect of boosts. So, apparently these effects are rather weak (at accessible velocities). However, there is an argument, which, in my opinion, demonstrates conclusively that these effects should be real.

In relativistic quantum theory we must construct an unitary representation of the Poincare group [itex]U_g [/itex] in the Hilbert space of observed system. This representation tells us how to connect operators of observables in two different reference frames

[tex] F' = U_g F U_g^{-1} [/tex]...(1)

In particular, when g is time translation

[tex] U_g = \exp(\frac{i}{\hbar} Ht) [/tex]

where

[tex] H = H_0 +V [/tex] .....(2)

is the interacting Hamiltonian. Then eq. (1) tells us how observable [itex]F [/itex] evolves in time. The fact that the Hamiltonian contains interaction [itex] V [/itex] explains non-trivial interacting dynamics, such as reactions, decays of apples, etc. Without this interaction term the dynamics of particles would be trivial and boring (all particles move with uniform velocities along straight lines independent on the presence of other particles)

When g is boost with rapidity [itex] \vec{\theta}[/itex], then

[tex] U_g = \exp(\frac{i}{\hbar} \mathbf{K} \vec{\theta}) [/tex]

and eq. (1) tells us how observable F is seen from different moving frames of reference. In particular, eq. (1) should describe for us Lorentz transformations (for time-position, energy-momentum, etc.). It is not difficult to demonstrate that if the operator of boost is non-interacting

[tex] \mathbf{K} = \mathbf{K}_0 [/tex]... (3)

then eq. (1) would lead to usual Lorentz transformation formulas. These formulas would be linear and universal (i.e., independent on the type of observed physical system and interactions acting there).

However, if the boost operator contains interaction term

[tex] \mathbf{K} = \mathbf{K}_0 + \mathbf{W}[/tex]...(4)

then boost transformations of observables computed by formula (1) become non-trivial and interaction-dependent.

The punchline is this: It is known from the theory of Poincare group representations (You can find this discussion, for example, in Weinberg's "The quantum theory of fields", vol. 1)that if the Hamiltonian contains interaction dependence (2), then the boost operator cannot remain interaction-independent as in (3). It must have interaction terms, like in (4). So, necessarily, boost transformations of observables must depend on interactions, i.e., they must be "dynamical".

Eugene.
 
  • #39
meopemuk said:
I regard this fact [that real observations are specific to a time and place, not only a time] as mere technical annoyance rather than something fundamental.

So far I don't follow your full argument, but since you treat time and space on such a different footing right from the outset (despite that no actual observation seems to exhibit this difference), I'm not sure I even find it surprising or interesting that the difference carries through to boosts.

If you're actually proposing something that is different from mainstream, can you describe a simple (hypothetical) experiment to demonstrate the point?
 
  • #40
cesiumfrog said:
... you treat time and space on such a different footing right from the outset (despite that no actual observation seems to exhibit this difference)...

Space (distance) is measured with rulers. Time is measured with clocks. What can be more different than that?

cesiumfrog said:
If you're actually proposing something that is different from mainstream, can you describe a simple (hypothetical) experiment to demonstrate the point?

One possible experiment is described in the references that I gave earlier. Their idea is that the decay law of fast moving particles should deviate from the usual expression given by Einstein's time dilation formula. For known unstable particles these deviations are several orders of magnitude smaller than the precision of modern instruments. So, special relativity remains a very good approximation for all practical purposes.

Eugene.
 
  • #41
Does that mean you think there is an absolute frame of reference?
 
  • #42
cesiumfrog said:
Does that mean you think there is an absolute frame of reference?

No, it doesn't. I accept both Einstein's postulates (the equivalence of all inertial reference frames and the independence of the speed of light on the velocity of the source and/or observer). So, there is no absolute frame. However, I do not accept the third tacit postulate of special relativity that we discussed above (two events coinciding in one reference frame must coincide in all other frames).

Eugene.
 
  • #43
I think what you've shown using QM is that the result of a measurement can depend on the relative velocities of the state preparing system and the measurement apparatus.

It seems like a bit of a jump to deny classical macroscopic reality to the extent of claiming different inertial reference frames will give different answers to such a question as "do these two billiard balls ever collide".
 
  • #44
cesiumfrog said:
It seems like a bit of a jump to deny classical macroscopic reality to the extent of claiming different inertial reference frames will give different answers to such a question as "do these two billiard balls ever collide".

I am not sure I am willing to go as far as claiming that two balls colliding in one frame never collide in another moving frame. However, it seems perfectly reasonable to me that if the balls are charged (i.e., interacting) then the space-time coordinates of their collision may not transform by Lorentz formulas between different moving frames.

Actually, this fact (the impossibility of Lorentz transformations of worldlines of interacting particles) is a well-know fact. There is a theorem that says that if worldlines (or trajectories) in a system of particles transform exactly by Lorentz formulas, then the particles must be non-interacting:

D. G. Currie, T. F. Jordan, and E. C. G. Sudarshan, "Relativistic Invariance and Hamiltonian Theories of Interacting Particles" Rev. Mod. Phys. 35, 350 - 375 (1963)

Eugene.
 
  • #45
This argument is becoming absurd. If "observation" O of "events" A and B with spacetime coordinates a and b determines that they are coincident then, by definition a - b = 0 so a = b. Then if "observation" O' is given by a' = f(a) and b' = f(b) then a' = f(a) = f(b) = b' so a' - b' = 0.

The only way for your 3rd postulate to be false is for there to be no such transformation function (implying a preferred reference frame) or for logic to not hold.
 
  • #46
DaleSpam said:
This argument is becoming absurd. If "observation" O of "events" A and B with spacetime coordinates a and b determines that they are coincident then, by definition a - b = 0 so a = b. Then if "observation" O' is given by a' = f(a) and b' = f(b) then a' = f(a) = f(b) = b' so a' - b' = 0.

The only way for your 3rd postulate to be false is for there to be no such transformation function (implying a preferred reference frame) or for logic to not hold.

This is exactly the point I am trying to make. I am suggesting that there is no single universal function [itex] f [/itex] (a.k.a. Lorentz transformation) that can be applied for all events. The transformation of the (space-time coordinates of the) event a may be given by a function [itex] f_a [/itex], and transformation of the (space-time coordinates of the) event b may be given by function [itex] f_b \neq f_a [/itex]. The transformation functions [itex] f_a [/itex] and [itex] f_b [/itex] may depend on the type of event they are acting on, in particular, on whether the event occurs in an interacting system or in a non-interacting system.

It is not so unusual to have different transformation functions [itex] f_b \neq f_a [/itex] for different events. If we consider the transformation of time translation (instead of the boost discussed above), then it would become obvious that there is no universal function [itex] f [/itex] that would tell us how all physical events develop in time. The time evolution of any system depends on interactions in the system.

Of course, there are also examples to the contrary: i.e., the effect of rotations and space translations is independent on the nature of the event and on interactions, so the equality [itex] f_b = f_a [/itex] is perfectly valid. I am trying to say that boosts may have properties more similar to time translations than to rotations.

Eugene.
 
  • #47
meopemuk said:
This is exactly the point I am trying to make.
If you honestly think that you and I are trying to make the same point then that goes pretty far in explaining the absurdity of this discussion.

If a = b then [itex] f_b = f_a [/itex] by substitution. QED. It ain't rocket science, it ain't physics, it ain't even advanced math.

Your 3rd postulate is just silly, if it doesn't hold then the universe cannot be described by math, therefore the universe is fundamentally illogical and there is really no point to doing science since there are no laws and no rules anyway.
 
  • #48
This is exactly the point I am trying to make. I am suggesting that there is no single universal function ... (a.k.a. Lorentz transformation) that can be applied for all events.

I'm unclear in my own mind if the idea is self consistent (I suspect it is probably not, but I'm not positive).

Is there any literature on this idea, i.e. any peer-reviewed papers that discuss it?
 
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  • #49
pervect said:
Is there any literature on this idea, i.e. any peer-reviewed papers that discuss it?

E. V. Stefanovich, "Is Minkowski space-time compatible with quantum mechanics?",
Found. Phys., 32 (2002), 673.

Eugene.
 
  • #50
DaleSpam said:
If "observation" O of "events" A and B with spacetime coordinates a and b determines that they are coincident...

If a = b then [itex] f_b = f_a [/itex] by substitution. QED.

Hi DaleSpam,

I am sorry for using a sloppy notation. I actually meant to consider transformation functions [itex] f_A [/itex] and [itex] f_B [/itex] (rather than [itex] f_a [/itex] and [itex] f_b [/itex]). So, these functions depend on the physical nature of events A and B (rather than on their space-time positions a and b). For example, event A can be a "collision of two neutral billiard balls" and event B can be a "collision of two charged billiard balls". Then, it is possible that boost transformation formulas for these different kinds of events are different [itex] f_A \neq f_B [/itex] . So, if the two events happen to have the same space-time coordinates in the reference frame O ([itex] a=b [/itex]), they do not necessarily have the same space-time coordinates in the moving reference frame O' [itex] f_A(a) \neq f_B(a) [/itex]. I hope I made it clear now.

Eugene.
 
  • #51
Yes, that is clear. It is not as trivial as I had thought at first glance, but it is still too poorly defined to discuss reasonably. Once you figure out what [itex]f_A[/itex] and [itex]f_B[/itex] are and can generate some testable predictions then you will have something worth discussing further.

My guess is that you will find any such transformation incompatible with the relativity postulate.
 
  • #52
DaleSpam said:
My guess is that you will find any such transformation incompatible with the relativity postulate.

As I mentioned earlier, two Einstein's postulates remain exactly valid in this approach. If you decide to check these claims for yourself, you are welcome to look at already quoted (peer reviewed) references:

E. V. Stefanovich, "Quantum effects in relativistic decays", Int. J. Theor. Phys., 35, (1996), 2539.

E. V. Stefanovich, "Is Minkowski space-time compatible with quantum mechanics?",
Found. Phys., 32 (2002), 673.

Eugene.
 
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