# Absolute simultaneity

1. Mar 5, 2008

### bernhard.rothenstein

Consider please the Lorentz transformation for the time coordinates of the same event
t'(E)=g(t(E)-Vx/cc) (1)
where g stands for gamma and t(E) and t'(E) for times displayed by clocks synchronized following Einstein's procedure.
My question is: with what times displayed by how synchronized clocks should I replace t(E) and t'(E) in order to arrive at absolute simultaneity. The synchronization procedure should be faisible and in accordance with that proposed by Einstein. I know so far the "everyday" clock synchronization which leads to such result.

2. Mar 5, 2008

### tiny-tim

not possible

Sorry, but absolute simultaneity is not compatible with relativity. See http://en.wikipedia.org/wiki/Absolute_time_and_space:

3. Mar 5, 2008

### yogi

I am not sure there are experimental results that cannot be explained just as well using the concept of absolute simultaneity - Franco Selleri spent many years trying to find an experiment that would show which best describes nature - he reported he could not resolve the question based upon empirical data. The point frequently adopted is that SR is self consistent, and therefore it must be correct - but you cannot use the conventions of SR to prove SR - it must be done by some experiment that distinguishes between the two philosophies

4. Mar 6, 2008

### DrGreg

I suppose I'd better explain the background to this. It depends exactly what you mean by "absolute simultaneity".

It's perfectly true that there is no absolute simultaneity using the standard Special Relativity coordinate systems that use the Einstein Synchronisation Convention for synchronising the clocks within each frame.

However, there is nothing to stop you using a non-standard family of coordinate systems, that are synchronised in such a way that all of the coordinate systems in the family agree on what is simultaneous. These coordinate systems are pretty weird, compared with standard coordinates. For instance the one-way speed of light is no longer constant in all directions, measured in these coordinates, and many of the equations of physics include a parameter that takes a different value in each frame within the family. So, on the face of it, in these coordinate systems, the postulates of relativity seem to be broken. But this is really an algebraic illusion. The coordinates are just an algebraic change of variables from the standard coordinates, so they still work correctly as long as you use them in the right way. Geometrically, these weird coordinates are a non-rectangular coordinate system.

To give a concrete example: I'll restrict myself to one space dimension for simplicity. Choose an inertial observer to be a reference point R for all other observers. If $$(t_E(I), x_E(I))$$ are the standard, Einstein-synchronised coordinates for any inertial frame I, define alternative coordinates $$(t_S(I), x_S(I))$$ by

$$x_S(I) = x_E(I)$$
$$t_S(I) = t_E(I) + \frac{v x_E(I)}{c^2}$$

where $$v$$ is the velocity of the I-frame relative to the reference point R. (Note: in the special case where I=R, $$v = 0$$, so $$t_S(R) = t_E(R)$$.)

By applying the Lorentz transform between $$(t_E(I), x_E(I))$$ and $$(t_E(R), x_E(R))$$, it is not difficult to show that

$$x_S(I) = \gamma(x_S(R) - v t_S(R))$$
$$t_S(I) = \frac{t_S(R)}{\gamma}$$

Thus, in "s" coordinates, all of the $$t_S(R)$$ coordinates share the same definition of simultaneity. So within the context of this family of coordinate systems, there is such a thing as "absolute simultaneity". But it's an artificial feature of the coordinate system family and has no physical significance. This is because the choice of R is arbitrary -- choose a different R and you'll get a completely different family of coordinate systems, with their own definition of "absolute simultaneity" that differs from the original choice of R.

There's an even weaker form of absolute simultaneity where you synchronise a clock to the visual image of a (relatively stationary) clock at the spatial origin -- if you look at the origin's clock and you see 12 o'clock, you set your own clock to 12 o'clock too. This is the so-called "everyday" synchronisation to which Bernhard refers, and it gives you a coordinate system with absolute simultaneity. But this is even weirder, because if clocks A and B are relatively stationary and A is synchronised to B, B is not synchronised to A by this definition! This is what you might call "one-way absolute simultaneity".

5. Mar 6, 2008

### DrGreg

To me, the phrase "in accordance with that proposed by Einstein" means Einstein synchronisation, so I don't think you really mean that.

The answer to your question is, in general, for "two-way absolute simultaneity",

$$t'=a t$$

for some a independent of t and x. I'll assume here that t is Einstein-synchronised in the reference frame R, and t' is a time coordinate in another frame I. However, if you want your time coordinate to coincide with proper time in the I-frame, a must be given by

$$a = \frac{1}{\gamma}$$

Have a look at this in Wikipedia (not all that well written), and http://relativity.livingreviews.org/Articles/lrr-2005-5/articlesu9.html [Broken] in "Modern Tests of Lorentz Invariance", by David Mattingly, Department of Physics, University of California at Davis (much better written).

Last edited by a moderator: May 3, 2017
6. Mar 6, 2008

### bernhard.rothenstein

Thank you. Is
$$t_S(I) = t_E(I) + \frac{v x_E(I)}{c^2}$$
a guess. Is there a relationship between that equation, aparent position and actual position?

7. Mar 7, 2008

### DrGreg

I should clarify what I said yesterday. To use the notation of post #4, the condition for "2-way" absolute simultaneity is

$$t_S(I) = a(I) t_S(R)$$ ....(1)

where $$a(I)$$ does not depend on x or t, but may vary from frame I to another.

The Mansouri-Sexl condition that

$$a(I) = \frac{1}{\gamma}$$ ....(2)

applies only in the special case when $$t_S(R) = t_E(R)$$, i.e. when our family of coordinates uses Einstein sync in the reference frame R. This is true of the example I gave in post #4 (the Selleri or Tanglerhini coordinates) but not true for Leubner's "everyday" coords.

The condition for Leubner coords is

$$x_S(I) = x_E(I)$$
$$t_S(I) = t_E(I) - \frac{|x_E(I)|}{c}$$

if you define it literally as described, i.e. with the light signal travelling outward from the spatial origin to the secondary clock. However if you modify this slightly and agree always to synchronise "from left to right" (in the 1-D case), i.e. in the positive x direction, i.e. whoever has the smallest x-coord transmits the signal and the person with the largest x-coord receives the signal, then you can use the modified equation

$$t_S(I) = t_E(I) - \frac{x_E(I)}{c}$$

which defines a "2-way" absolute simultaneity.

If you do the calculation with Lorentz tranforms you should get (if I haven't made a mistake)

$$t_S(I) = \gamma \left(1 + \frac{v}{c} \right) t_S(R)$$

for "left-to-right Leubner sync", confirming my equation (1) above.

If you wanted to come up with a 3-space-dimension version of Leubner-sync which was 2-way absolute, the only way I can think of doing it would be by a 3-stage process. For a given clock at (x,y,z) introduce some intermediate clocks at (x,0,0) and (x,y,0). Perform a 2-way sync along the x-axis from (0,0,0) to (x,0,0); then along the y-axis to (x,y,0); then along the z-axis to (x,y,z). That's a rather awkward way to do it. The equation would be

$$t_S(I) = t_E(I) - \frac{x_E(I) + y_E(I) + z_E(I)}{c}$$

You can of course perform an "outward" sync direct from (0,0,0) to (x,y,z), but that would only be "1-way" absolute simultaneity:

$$t_S(I) = t_E(I) - \frac{\sqrt{x_E(I)^2 + y_E(I)^2 + z_E(I)^2}}{c}$$

I believe this formula for $$t_S(I)$$ is the only one that is compatible with
$$t_S(R) = t_E(R)$$ and (1) and (2). Of course, in 3-dimensions it becomes

$$t_S(I) = t_E(I) + \frac{\textbf{v} \cdot \textbf{x}_E(I)}{c^2}$$.

I just got this from the Mansouri-Sexl paper referred to via the references in my earlier post. I'm not quite sure what you mean by "apparent position" and "actual position" in this context.

8. Mar 8, 2008

### bernhard.rothenstein

photographed space-time coordinates and clock sinchronization

Consider a pointlike source of light S' at rest in I'. Its potion is characterized by the polar coordinates (r,theta). the fact that it emits a light signal at a time t'(e) is associuated with the event E'(r,theta.t'(e). The light signal is received at the origin O' at a time t'(r) .
we have
t'(r)=t'(e)+r'/c . (1)
In Leubner's approach t'(r) represents the reading of a clock C'(0,0) located at the origin O' synchronized a la Einstein with the other clocks of I', t'(e) representing the reading of a clock C'(r',theta') located in front of S' when the light signal is emitted.
Is that analogy correct. Is it of practical importance?

9. Mar 10, 2008

### DrGreg

You seem to have got the light going the wrong way, to the origin instead of from the origin, which is what I thought Leubner was doing. But apart from that, your equation (1) is essentially the same as my "equation for outward sync" in my last post.

The attached diagram may help explain my notation and the equations I've used.

The left-hand diagram shows the "outward" synchronisation procedure with light going from the origin ("e") to a second clock ("r") (at a fixed position r relative to the origin). In s-coordinates, the second clock is synced so that a person at the second clock sees, with their eyes, both clocks synced.

The centre diagram shows the surfaces of simultaneity (thick lines) for observers in two different frames (I=green and R=purple). Where the two sets of lines are parallel we have absolute simultaneity. Note that there's a region (x(R)x(I) < 0) where the lines are not parallel (except for the surfaces through the origin). The situation is even worse if my diagram is considered as a 2D cross-section of 3D cones (or 4D hypercones). So "outward" Leubner sync does not give absolute simultaneity everywhere, only within some regions of spacetime.

One other thing wrong with "outward" Leubner coords is that they are not "inertial", in the sense that inertial motion does not give a constant velocity. An inertial object passing through the origin would show a sudden change of speed in "outward" Leubner coords.

The right-hand diagram is the same as the centre diagram but for "left-to-right" Leubner sync, i.e. when the sync signal always travels in the positive x direction i, instead of outward from the origin. In this case we really do have absolute simultaneity everywhere, and the coordinates are inertial.

For the above reasons, in my opinion the polar $(r, \theta, \phi)$ approach is not worth considering if your objective is to obtain absolute simultaneity everywhere.

Michele Vallisneri, Relativity and Acceleration, Ph.D. thesis. See chapter 4, page 41 onward, with Reichenbach's $\epsilon = 0$.

Rizzi, Ruggiero and Serafini, "Synchronization Gauges and the Principles of Special Relativity", Foundations of Physics 34 (2004) pp. 1835-1887.

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