# Einstein's Clock Synchronization Convention

1. Aug 31, 2005

### Aether

Just a friendly reminder, the first postulate of the special theory of relativity, namely that the speed of light c is the same in all inertial frames, only holds true in view of Einstein's clock synchronization convention. There is no experimental basis whatsoever for preferring this convention over absolute clock synchronization. "Thus the much debated question concerning the empirical equivalence of special relativity and an ether theory taking into account time dilation and length contraction but maintaining absolute simultaneity can be answered affirmatively." -- R. Mansouri & R.U. Sexl, A Test Theory of Special Relativity: I. Simultaneity and Clock Synchronization, General Relativity and Gravitation, Vol. 8, No. 7 (1977), pp. 497-513.

This paper by Mansouri & Sexl is the first of a series of three papers, the other two papers are:

R. Mansouri & R.U. Sexl, A Test Theory of Special Relativity: II. First Order Tests, General Relativity and Gravitation, Vol. 8, No. 7 (1977), pp. 515-524.

R. Mansouri & R.U. Sexl, A Test Theory of Special Relativity: III. Second Order Tests, General Relativity and Gravitation, Vol. 8, No. 10 (1977), pp. 809-814.

This series of papers by Mansouri & Sexl is referenced by most, if not all, of the subsequently published experimental tests of Local Lorentz Invariance (LLI).

Last edited: Aug 31, 2005
2. Aug 31, 2005

### JesseM

The experimental aspect is that there's no experiment you can do that will lead to one observer's definition of synchronization being preferred over another, because all the most fundamental laws of physics favored by experiment have the property of Lorentz-invariance.

3. Aug 31, 2005

### pervect

Staff Emeritus
True

False.

There is an extremely good basis for preferring Einstein's convention. This is the conservation and isotropy of momentum.

The primary reason to synchronize clocks is to be able to measure velocities. When we demand that an object of mass m and velocity v moving north have an equal and opposite momentum to an object of mass m and velocity v moving south, we require Einsteinan clock synchronization.

Empirically, this means that we require an two objects of equal masses moving at the same speed in opposite directions to stop when they collide inelastically.

It is indeed *possible* to use non-Einsteinain clock synchronizations, and under some circumstances it is more-or-less forced on us. In such circumstances, one must not remember that momentum is not isotropic.

Note that Newton's laws assume that momentum is isotropic (an isotropic function of velocity). Therfore Newton's laws (with the definition of momentum as p=mv) cannot be used unless Einstein's clock synchronization is used. Some other definition of momentum other than p=mv must be used if it is to remain a conserved quantity when non-standard clock synchronizations are used.

The ability to use Newton's laws at low velocities was what motivated Einstein to define his method of clock synchronization.

Last edited: Aug 31, 2005
4. Aug 31, 2005

### Hurkyl

Staff Emeritus
I would like to remind you that this is analogous to reminding people that there is no experimental basis for using an orthogonal coordinate system when doing plane geometry either. (As opposed to a system where, say, the angle between the x and y axes is 45°)

Your choice of how to define coordinates is not a physical choice -- the result of any physical experiment will be the same no matter what coordinate system you opt to use.

Einstein's coordinates are used because they're nice -- Einstein's coordinate systems are precisely the rectilinear coordinate systems whose coordinate axes are orthogonal. (and non-null)

5. Aug 31, 2005

### Aether

The general linear transformation that Mansouri & Sexl use has three parameters $$(\epsilon{_x},\ \epsilon{_2y}, and\ \epsilon{_3z})$$ that are determined by synchronization procedures. I expect that momentum is conserved when these are taken into account and that it is isotropic, albeit with three extra synchronization parameters to keep track of.

Mansouri & Sexl needed to add three synchronization parameters to transform the time coordinate while maintaining absolute simultaneity with only four coordinates. Six coordinates, three for time, would seem to be a more natural choice.

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6. Aug 31, 2005

### Hurkyl

Staff Emeritus
Except for the fact we do have empirical evidence for a single time dimension.

7. Aug 31, 2005

### Aether

Why is it that the invariant interval ds absolutely positively has to be a scalar?

8. Aug 31, 2005

### Hurkyl

Staff Emeritus
Because that's what it's defined to be. I suspect this is not the question you meant to ask.

My comment about the empirical evidence for four dimensions has nothing to do with that: the evidence is the fact that, historically, we've been able to describe any point in space-time that we please by specifying 4 coordinates: where, and when.

9. Aug 31, 2005

### pervect

Staff Emeritus
I don't have the papers you cite, and I couldn't find them or arxiv, either. Your description of them isn't really very enlightening, alas.

The point I want to make is that while it is indeed, possible and sometimes even desirable to use non-Einsteinian clock synchronziation, it is *not* possible to do so and to also assume that Newton's laws work with such a synchronization method.

In other words, when one maks the speed of light anisotropic, one also makes the behavior of matter anisotropic, as well. Light may go "faster" in one direction when you play around with clock synchronziations, but so do racecars, and electron beams, and everything else in the world. (The effect of syncronization "twiddling" is most important for objects which move at high velocities, however).

Basically, one is playing "word games" with the definition of velocity. It is a matter of "convention" that one does not measure the velocity of an airplane by looking at the difference of the clocks at which it takes off in the PST timezone, and the clock at which it lands in the CST timezone. One insists that to get the "fair" speed of the airplane, one uses clocks that are synchronized according to a convention, using the same time-zone for both takeoff and landing times.

Abandoning this convention is possible, but it is not possible to use the "speeds" defined in such an unconventional way with Newton's laws.

10. Aug 31, 2005

### Aether

I'm asking that question because $$ds=c_0d \tau$$, and three time coordinates merely implies that ds has direction; not necessarily that there are two additional "dimensions": one dimension of time + absolute simultaneity.

Only by abandoning absolute simulaneity, or else as pervect suggests perhaps by sacrificing the form of Newton's laws.

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11. Aug 31, 2005

### Hurkyl

Staff Emeritus
Yes, but τ is a scalar, and there aren't three time coordinates.

Of course, in an entirely different theory, you could get entirely different answers.

3 spatial coordiantes + 3 temporal coordinates = 6 dimensions.

No, we did a pretty good job of describing any point in space-time with 4 coordinates, even when we do adopt absolute simultaneity. "Where and when" was not an invention of Einstein.

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12. Aug 31, 2005

### Aether

OK, then we break $$c_0$$ into components.

I'm talking about the issue of momentum/Newton claimed by pervect to be a problem with absolute simultaneity.

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13. Aug 31, 2005

### Perspicacious

You are especially correct on this one point: It's impossible to change the laws of physics by merely resetting clocks.

http://arxiv.org/abs/gr-qc/0409105

The greatest importance is in realizing the enormous role arbitrariness and conventionality play in modern physics.

http://www.everythingimportant.org/relativity/

Last edited: Sep 1, 2005
14. Sep 1, 2005

### Aether

Thank you Perspicacious, I will read those articles.

A particle's 4-momentum is $$p^\mu=mu^\mu$$, and we could define its 7-momentum as $$p^a=mu^a$$ where $$ds^2=c_0^2 d\tau^2=(dx^4)^2+(dx^5)^2+(dx^6)^2$$, and the form of Newton's laws are retained. By moving $$ds^2$$ to the right side of the line element, we have $$0=c_0^2dt^2-(dx^1)^2-(dx^2)^2-(dx^3)^2-(dx^4)^2-(dx^5)^2-(dx^6)^2$$ instead of $$ds^2=c_0^2dt^2-(dx^1)^2-(dx^2)^2-(dx^3)^2$$.

Why isn't this preferable to abandoning absolute simultaneity? Shouldn't abandoning absolute simultaneity be used only as a weapon of last resort?

For example, consider P.A.M. Dirac, The Principles of Quantum Mechanics - Fourth Edition, Oxford University Press, 1958. Section 69 The motion of a free electron, p. 262: "...we can concude that a measurement of a component of the velocity of a free electron is certain to lead to the result +or- c...Since electrons are observed in practice to have velocities considerably less than that of light, it would seem that we have here a contradiction with experiment. The contradiction is not real, though, since the theoretical velocity in the above conclusion is the velocity at one instant of time while observed velocities are always average velocities through appreciable time intervals. We shall find upon further examination of the equations of motion that the velocity is not at all constant, but oscillates rapidly about a mean value which agrees with the observed value."

So, to be clear, it is my intention that $$0=c_0^2dt^2-(dx^1)^2-(dx^2)^2-(dx^3)^2-(dx^4)^2-(dx^5)^2-(dx^6)^2$$ should be viewed in this context of rapid oscillation about a mean value which agrees with the observed value.

Or 1 temporal coordinate + 6 spatial coordinates: where three spatial coordinates are instantaneous, and the other three are observed as averages of the first three (same basis vectors) through appreciable time intervals. Isn't that in fact what is observed in nature?

I submit that ds only appears to be a scalar when observation is averaged over any appreciable time interval, but when observed at one instant of time...it is a vector.

Last edited: Sep 1, 2005
15. Sep 1, 2005

### Hurkyl

Staff Emeritus
Aether: you seem to be confusing reality with the mathematical model.

ds is a scalar because that's what it's defined to be: it's part of the mathematical construct that is Minowski space, which Special Relativity asserts acts as a model of reality.

Even if you are exactly correct about the behavior of reality, that doesn't change the fact that the ds of Special Relativity is a scalar.

I have no idea what that would mean.

16. Sep 1, 2005

### Aether

OK, thanks. I'll go back and look at it again with that in mind.

I mean three spatial coordinate differentials, evaluated instantaneously since according to Dirac they change rapidly.

Last edited: Sep 1, 2005
17. Sep 2, 2005

### DrGreg

I think the vast majority of physicists and mathematicians would say that abandoning symmetry should be used only as a weapon of last resort.

In the end it's a matter of your own personal philosophy.

18. Sep 2, 2005

### Aether

Referencing an arbitrarily chosen frame is an application of your own personal philosophy, but referencing a locally preferred frame is not.

Absolute simultaneity merely implies a locally preferred frame of reference, whereas relative simultaneity implies an arbitrary frame of reference. Where is there an abandonment of symmetry?

Suppose that the SI system were to one day adopt a locally preferred frame as a standard reference: SR could still be recovered completely for transformations between any two arbitrary reference frames by first transforming from one arbitrary frame to the SI standard frame, and then transforming from the SI standard frame to the second arbitrary frame.

For example, the money in your pocket references a locally preferred frame (of sorts); without that, we could all be wondering "how many chickens per gallon is gas going to cost me today...assuming that I can even get any??".

Last edited: Sep 2, 2005
19. Sep 2, 2005

### DrGreg

Suppose I measure the position of an object on my desk using xy coordinates aligned to the side of my desk. Then I want to convert the position to XY coordinates aligned to the side of my room.

Which is easier: perform a direct transformation from one to the other, or use an intermediate step of a "locally preferred frame" of eastings/northings?

(In fact, I think you are saying I really ought not to use a rectangular grid but a parallelogram grid -- a Y-Easting grid?)

The symmetry in SR is that anyone is entitled to assume that their own frame is the preferred frame; there's no need to look for any other preferred frame.

20. Sep 2, 2005

### Aether

It is often easier to perform a direct transform.

All I'm saying is that you have a choice. This doesn't matter much until you try to extrapolate your grid to the entire universe.

That's OK as a local approximation, but it doesn't scale to the universe as a whole. For example, the amount of matter-energy in the universe is finite, and it is distributed in a certain pattern. Let's say that it is evenly distributed throughout a 3-sphere of radius R(t). There is a center to that sphere, the center of mass of the universe, which we can take to be at rest, and that is a locally preferred frame.

Last edited: Sep 2, 2005