# On the relativity of times

1. Sep 21, 2006

### myoho.renge.kyo

"on the relativity of times"

tB - tA = rAB / (c - v) and t'A - tB = rAB / (c + v)

"The Principle of Relativity", p 42, by A. Einstein.

consequently,

c = [rAB / (tB - tA)] + v and c = [rAB / (t'A - tB) - v

let rAB = x' and (tB - tA) = t' so that

c = (x' / t') + v

since x' = x - v*t

c = [(x - v*t) / t'] + v so that

x = t'*(c - v) + v*t

and let rAB = x and (t'A - tB) = t so that

c = (x / t) - v so that

x = t*(c + v)

consequently,

t'*(c - v) + v*t = t*(c + v)

t'*(c - v) = t*(c + v) - v*t

t'*(c - v) = t*c

t' = t*c / (c - v)

t' = t / (1 - v / c)

t' is not t / sqrt(1 - v^2 / c^2). why? thanks! (6:00 pm thru 7:00 pm, 9/21/2006)

Last edited: Sep 21, 2006
2. Sep 22, 2006

### lightarrow

Unfortunately I don't have that book, and even if I had it, wouldn't it be better if you explained the problem you are considering? What are A, B, tB, tA, v, x ecc.?

Why do you call rAB = x and then, rAB = x' ? You have already written that x'=x-v*t, so they are not the same thing! How can you find a correct solution if you do it? It's impossible!

Maybe you should read better what you write!

Last edited: Sep 22, 2006
3. Sep 22, 2006

### bernhard.rothenstein

a look at Leo Karlov "Paul Kard and Lorentz free special relativity" Phyhs.Educ. 24 165 1989" and at Asher Peres "Relativistic telemetry" Am.J.Phys. 55 516 1987 could be illuminating.

the best things a physicist can offer to another one are information and criticism

4. Oct 5, 2006

### pess5

Generally, when we say x,t and x',t', we imagine 2 sets of spacetime variables, one set for each of 2 observing perspectives, say frame k & frame K.

Your derivation leadin (above here) defines x',t' for the outbound segment of the ray, and x,t for the reflection segment of the same ray. However, it's all from a single observer vantage. Hence there is nothing in your derivation which relates one frame to the other.

So, when you obtained this ...

You are not relating one frame wrt the other. You are instead relating the outbound duration to the return leg duration per a single frame vantage.

The equation in your question ...

is the Fitgerald Contraction, which requires the comparison of 2 frame perspectives against the backdrop of light.

5. Oct 9, 2006

### myoho.renge.kyo

thank you so much. i really appreciate your help. i am having so much dificulty with this. i think i need to understand multivariable calculus and differential equations in order to at least understand p. 44 in The Principle of Relativity. but i would like to understand what you mean.

for example, how can i relate one frame of reference, say K (the stationary system) with K' the moving system using the following:

tB - tA = rAB / (c - v) and t'A - tB = rAB / (c + v)

A. Einstein on p. 42 in The Principle of Relativity, states, "observers moving with [K'] would thus find that [tB - tA is not equal to t'A - tB], while obsevers in [K] would declare that [tB - tA = t'A - tB]." thanks again. in the mean time i am going to read Fitzgerald Contraction.

Last edited: Oct 9, 2006
6. Oct 10, 2006

### pess5

You'd need to relate these ...

outbound path ...

tB-tA = rAB/(c-v) proportional_to TB-TA = rAB/c​

return path ...

t'A-tB = rAB/(c+v) proportional_to T'A-TB = rAB/c​

The proportionality for both legs are identical, and assumed linear.

One observer records t & r, while the other observer records T & r.

Note that ...

One should not assume going in that r=r.

TB-TA = TA'-TB, but tB-tA <> t'A-tB.​

Einstein recognized that all inertial observers experience equal times for the outbound & return light path lengths, given the emitter & refector are at rest with that observer. So the linear constant of proportionality which relates the frames must relate a reflection point which is not at the center of your observed round trip interval, to the center of his round trip interval ... since the emitter & reflector are at rest with him and not you. You see them in motion!

pess5