Length contraction and time dilaton

[bernhard.rothenstein]

selleri coordinates

It seems to me that Selleri coordinates would not avoid synchronized clocks, they would just use a non-standard convention of synchronization.

Not only Selleri but Abreu and Guerra (see arxiv and European Journal Physics) as well, use Einstein synchronization in I and nonstandard clock synchronization in I.

 

[DrGreg]

It seems to me that Selleri coordinates would not avoid synchronized clocks, they would just use a non-standard convention of synchronization.

I'm not sure whether you aimed that comment at me or Bernhard.

The only reason I mentioned Selleri coordinates in the first place was to provide a counterexample to the proposition that you could derive the standard formula for Lorentz contraction without a clock synchronisation convention (explicit or implied). If you don't use Einstein-synced coordinates, Lorentz contraction need not be true; e.g. between pairs of Selleri coord systems it's possible to have length dilation instead of contraction. Therefore any proof of Lorentz contraction must make use of a synchronisation convention somewhere, either explicitly or by quoting some other sync-dependent result.

 

[Dale]

Therefore any proof of Lorentz contraction must make use of a synchronisation convention somewhere, either explicitly or by quoting some other sync-dependent result.

Then the simple answer to the OP:

Please let me know if length contraction could be derived without using synchronized clocks.

Is "no".

 

[DrGreg]

Thanks for your help.
Please follow the following approach to Selleri.
Consider the inertial reference frames I and I' in the standard arrangement, I' moving with constant speed V relative to I in the positive direction of the overlapped OX(O'X') axes. At a point M(x,0) we find the clocks C1(x,0) and C2(x,0) at rest in I both synchronized with clock C0(0,0) located at the origin O(0,0) of I. Clock C1(x,0) is sinchronized using an isotropic light signal that propagates with speed C the second with an anisotropic light signal propagationg in the positive directioon of the OX axis with speed c, both emitted from O when C0 reads t=0. When the corresponding light signal arrives at C1 it reads t(E)=x/C whereas clock C2 reads t(a)=x/c, t(E) and t(a) being related by
t(E)=t(a)+(x/C)(1-C/c). (1)
Performing the Lorentz transformations to I' the result is
x'=gx[1-V/C(1-C/c)-gVt(a) (2)
t'(E)=gt(a)+g(x/C)(1-C/c-V/c) (3)
Imposing the condition of absolute simultaneity (dt(a)=0 implies dt'(E)=0 we impose the condition
1-C/c-V/C=0 (4)
and so
c=C/(1-V/C) (5)
with which (2) becomes
x'=x/g-gVt(a) (6)
(3) becoming
t'(E)=gt(a) (7)
recovering the Selleri transformation equations presented by the author as
x'=g(t-x/C) (8)
t'=gt (9)
the weak point consisting in the fact that they do not mention the clocks which display the involved times.
Do you find flows in the lines above which are only an attempt to put your help in my own way of thinking.
Thanks in advance.
t'=gt

Sorry for the delay in replying. I am not able to log into this forum every day.

For the benefit of other readers I should say that this seems to have no connection with the original question of this thread.

Let me try and reword this, as it's currently rather confusing.

Consider the inertial reference frames I and I' in the standard arrangement, I' moving with constant speed V relative to I in the positive direction of the overlapped OX & O'X' axes. At a point M(x,0,0) we find the clocks C₁ and C₂, and at the origin of I we find clock C₀, all three at rest in I.

Clock C₁ is synchronised to C₀ in such a way as to make the speed of light isotropic when measured using C₀ and C₁,

t_E = x/c​

being the time measured by C₁ when a photon is received at M that was emitted from the origin when C₀ time was zero.

Clock C₂ is synchronised to C₀ in such a way as to make the speed of light from O to M anisotropically equal to c_f when measured using C₀ and C₂,

t_a = x/c_f​

being the time measured by C₂ when a photon is received at M that was emitted from the origin when C₀ time was zero.

t_E and t_a are therefore related by

t_E = t_a + \frac{x}{c} \left( 1 - \frac{c}{c_f} \right). (1)​

At this point I get lost: I can't see how you got (2) and (3). There is a missing intermediate step that you will have to supply.

 

[DrGreg]

Thanks to a private message from Bernhard I can continue from where my previous post stopped.

(I measure V in the direction implied by post #27, opposite to the private message. Just a change of sign.)​

Now consider Einstein-synced coordinates (t'_E, x') in the I' frame. These are related to the Einstein-synced coordinates (t_E, x) in the I frame by the Lorentz transform

x' = \gamma (x - V t_E) = \gamma \left(1 - \frac{V}{c}(1 - c/c_f) \right) x - \gamma V t_a (2)
t'_E = \gamma (t_E - V x / c^2) = \gamma t_a + \gamma \left( 1 - \frac{c}{c_f} - \frac {V}{c} \right) \frac {x} {c} (3)​

Now choose c_f to be the value which makes equation (3) independent of x, i.e.

1 - \frac{c}{c_f} - \frac{V}{c} = 0 (4)​

which solves as

c_f = \frac{c}{1 - V/c} (5)​

Substitution in (2) and (3) gives:

x' = \frac{x}{\gamma} - \gamma V t_a (6)
t'_E = \gamma t_a (7)​

That all seems correct to me.

If anyone other than Bernhard is following this, what we have done is define an alternative coordinate system (t_a, x) in the I frame which is related to the I'-frame coordinates (t'_E, x') via equations (6) and (7) instead of the usual Lorentz Transform. Equation (7) shows that both coordinate systems agree in their definitions of simultaneity -- if two events share the same t'_E coord, they must also share the same t_a coord.

Putting t_a = 0, equation (6) shows that a rod of length x' that is stationary in the I' frame is longer in (t_a, x) coordinates, which is the counterexample to the original implied proposition of post #1.