# Non-standard clock synchronization

1. Feb 20, 2008

### bernhard.rothenstein

Selleri proposes the following transformation equations
x=g(x'-vt')
t=t'/g
g=gamma.
Einstein's clock synchronization requires
t(E)=t(e)+x/c
t(E) reading of a clock located at a distance x from the origin, t(e) reading of a clock located at the origin when the synchronizing light signal is emitted,
Is there a link between the proposed transformation equations and the synchronization procedure?

2. Feb 20, 2008

### 1effect

DrGreg has already answered this question in a very beautiful way here.

3. Feb 21, 2008

### bernhard.rothenstein

nonstandard clock synchronization

Thanks. I appreciated DrGreg's answer. But what I mention are not Tangherlini but Selleri transformasions.

4. Feb 21, 2008

### 1effect

They are one and the same thing.

5. Feb 21, 2008

### DrGreg

Yes (except that the roles of "primed" and "unprimed" have been reversed). "Tangherlini Transform", "Selleri Transform", "Generalised Galilean Transform" (GGT), all refer to the same thing (though different authors attribute different significances to them). They also sometimes appear under the heading of "Lorentz Ether Theory" (LET), although Lorentz never used them.

Just to spell out the post that 1effect referred to:

Given Einstein coords (x, t) for observer I, define Tangherlini coords (x'T, t'T) for observer I'T by

x'T = g(x - vt)
t'T = t/g
g = gamma = (1 - v2/c2)-1/2

Then use the Lorentz transform (and its inverse) for Einstein-observer I'E (x'E, t'E)

x'E = g(x - vt)
t = g(t'E + vx'E/c2)

and you will obtain

x'T = x'E
t'T = t'E + vx'E/c2

Note that c must be interpreted as the speed of light in the I frame, or equivalently the "two-way speed of light" in the I'T frame, because, in that frame the "one-way speed of light" depends on which direction you measure it in.

6. Feb 21, 2008

### 1effect

Thank you, DrGreg

I am not familiar with the "one-way speed" dependency on the direction of measurement. Does this follow from the definition of the metric in the Tangherlini/Selleri theory? Can you elaborate?

7. Feb 21, 2008

### DrGreg

Consider the equation x'E = ct'E. That describes something moving at the speed of light measured in the I'E frame. Now convert to the I'T frame. You get

x'T = ct'E

t'T = t'E + vt'E/c = (1 + v/c)t'E

From which the Tangherlini-speed of light is c / (1 + v/c).

Similarly, using x'E = -ct'E, the speed in the other direction is c / (1 - v/c).

But the there-and-back (2-way) average speed of light is still c.

You could also deduce this from the metric I quoted here, by solving ds = 0.

In fact, whenever the metric contains "off-diagonal" terms which "mix space and time", like dxdt, the speed of light, expressed in those coordinates, cannot be isotropic (=the same in all directions). (Think about solving ds = 0 and the roots of quadratic equations.)

8. Feb 21, 2008

### 1effect

Thank you, this is what I thought :

$$x=g(x'-vt')$$
$$t=t'/g$$

$$0=x^2-(ct)^2=g^2(x'-vt')^2-(ct'/g)^2$$

Solving
$$g^2(x'-vt')^2-(ct'/g)^2=0$$ for $$\frac{x'}{t'}$$ I get the anysotropic speed.

You are right, the presence of the mixed term in x't' results immediately in anysotropic light speed.
Thank you so much for the incredible stuff, you are awesome!

9. Feb 22, 2008

### DrGreg

Sorry, I just realised I never gave an explicit answer to this question.

The answer follows from my formula for the "one-way Selleri-speed" of light c / (1 +/- v/c).

$$t'_{T}(E)=t'_{T}(e)+\left|x\right|\frac{1+\frac{v}{c}}{c}$$

where c is the "two-way speed of light".

10. Feb 26, 2008

### bernhard.rothenstein

Please give me some hints for deriving by myself the equation you propose.

11. Feb 27, 2008

### DrGreg

It follows from several previous posts.

ONE

This post showed that the 1-way Selleri-speed of light is

c+ = c / (1 + v/c)

(where c is the 2-way speed and also the 1-way Einstein-speed).

TWO

From this post I thought you already knew that

t(E) = t(e) + x/c+ .... (1)

However, I've just noticed your descriptions of t(E) and t(e) in that post are confusing and maybe not quite what I am assuming. To avoid all doubt, I am interpreting:

t(e) to be the time, on the observer's own clock, when a light signal is emitted by the observer at the spatial origin.

t(E) to be the time, on an auxiliary clock located at a constant distance x from the observer, when the light signal is received.

The auxiliary clock is synchronised to make the equation (1) true.

For a given choice of c+ (between 0.5 and infinity), the set of all possible auxiliary clocks defines a (t,x) coordinate system.

Conclusion

Put the above two posts together to get my result.

The reason I originally put |x| instead of x is because of a technicality I haven't raised yet.

If you define your synchronisation operationally via light from the observer, then equation (1) would hold only for positive x. For negative x you'd need to replace x by -x. That's assuming that c+ is the 1-way speed of light outward away from the observer.

However, you might instead want c+ to be the 1-way speed of light in the positive x direction. In this case, equation (1) is correct even for negative x, but then the light is really travelling from the auxiliary clock to the observer (still in the positive x direction). In the case of Selleri coords, this second interpretation is the correct one, so I was wrong to use |x| in my previous post.

From a practical point of view, for negative x you can use the alternative equation

t(E) = t(e) - x/c- .... (2)

where c- is the 1-way speed of light in the negative x direction.

c+ and c- are related by 1/c = (1/c+ + 1/c-)/2 (see this post).