Let's consider Bell's spaceship experiment. Forward and backward spaceships are simultaneously and instantly accelerated in frame S.

l is distance between the spaceships.As soon as forward spaceship is accelerated it sends light signal for measurement of distance between the ships.As soon as backward spaceship is accelerated it sends light signal for measurement of distance between the ships.

t_f is time of measurement of the distance by forward spaceship 

t_b is time of measurement of the distance by backward spaceship

t_f = l /(gamma(c+v))+l/(gamma(c-v))=2lc/(gamma(c²-v²))=2 l /(c²-v²)^1/2

t_b = l/(gamma(c-v)+l/(gamma(c+v))=2 l/(c²-v²)^!/2

l'_f is distance measured by forward spaceship

l'_b is distance measured by backward spaceship

l'_f=[2 l/(c²-v²)^1/2]*c/2=gamma*l

l'_b=[2 l/(c²-v²)^1/2]*c/2=gamma*l

Firstly distance between the unstoped spaceships in frame S' was  l /gamma

l'-l/gamma is a  change of distance between the spaceships

l' - l/gamma=(gamma²l-l)/gamma=(gamma²- 1)*l/gamma

then time between stops of the spaceships in frame S' is dt'

dt'=(gamma²-1)l/(gamma*v)

Edited August 19, 20187 yr by DimaMazin

What is the actual question or point?

On 20.08.2018 at 9:36 AM, Markus Hanke said:

What is the actual question or point?

Excuse me. I didn't simplify the equation.

dt'=(gamma²-1)l/(gamma*v)=gamma*vl/c²
It corresponds to Einstein's relativity of simultaneity.

On 8/23/2018 at 2:15 PM, DimaMazin said:

Excuse me. I didn't simplify the equation.

dt'=(gamma²-1)l/(gamma*v)=gamma*vl/c²
It corresponds to Einstein's relativity of simultaneity.

I'm probably going to be sorry for saying this, but your question seemed simple enough? I would answer "only if another observer disagreed with your measurement."

It' s the math that seems somewhat over the top since a simple comic drawing would have shown that one line has to be longer than the other once you correctly place both observers in relation to the events. There actually used to be a comic book about relativity that showed the drawing I speak of. I bought it from Google books.

Why complicate with forward going and backward going space ships.

One ship moving left to right on the x axis with another moving left to right on a diagnol. Flash two lights then then ask why didn't they both didn'tsee the lights flash simultaneously?

But perhaps this isn't what you meant, and your question is far beyond my abilities? Though I am sure I recognize the transformation presented it is the need that puzzles me? Actually I don't understand how it corresponds relativity to simultaneity? 

I found the diagram and had to make a correction to how I explained it but I need to check and see if I'm allowed to copy it before I present it.

The copyright says I need to ask permission to copy. 

The book is called an introduction to relativity by Bruce Bassett the diagram is on pages 40 and 50.

Edited September 9, 20187 yr by jajrussel
A poor memory a really poor memory

On 09.09.2018 at 7:40 PM, jajrussel said:

I'm probably going to be sorry for saying this, but your question seemed simple enough? I would answer "only if another observer disagreed with your measurement."

It' s the math that seems somewhat over the top since a simple comic drawing would have shown that one line has to be longer than the other once you correctly place both observers in relation to the events. There actually used to be a comic book about relativity that showed the drawing I speak of. I bought it from Google books.

Why complicate with forward going and backward going space ships.

One ship moving left to right on the x axis with another moving left to right on a diagnol. Flash two lights then then ask why didn't they both didn'tsee the lights flash simultaneously?

But perhaps this isn't what you meant, and your question is far beyond my abilities? Though I am sure I recognize the transformation presented it is the need that puzzles me? Actually I don't understand how it corresponds relativity to simultaneity? 

I found the diagram and had to make a correction to how I explained it but I need to check and see if I'm allowed to copy it before I present it.

The copyright says I need to ask permission to copy. 

The book is called an introduction to relativity by Bruce Bassett the diagram is on pages 40 and 50.

Relativity of simaltaneity is   

 t'=[t-vx/c²]/(1-v²/c²)^1/2

or      t'=gamma*t - gamma*vx/c²

In the my case t=0    because events of the accelerations of the spaceships are simultaneous in frame S

x=  - l

The my question was because I thought that I have opened another relativity of simultaneity( like you think). But it is the same thing. Our level of knowledge of math is less than professional . Therefore we don't understand some things.

 

Edited September 15, 20187 yr by DimaMazin

1 hour ago, DimaMazin said:

Relativity of simaltaneity is   

 t'=[t-vx/c²]/(1-v²/c²)^1/2

or      t'=gamma*t - gamma*vx/c²

Where is this equation from?

2 hours ago, DimaMazin said:

Relativity of simaltaneity is   

 t'=[t-vx/c²]/(1-v²/c²)^1/2

or      t'=gamma*t - gamma*vx/c²

In the my case t=0    because events of the accelerations of the spaceships are simultaneous in frame S

x=  - l

 

Unless x = 0, that's not true. And you have x = -1

7 hours ago, Strange said:

Where is this equation from?

https://en.wikipedia.org/wiki/Relativity_of_simultaneity

6 hours ago, swansont said:

Unless x = 0, that's not true. And you have x = -1

- l isn't -1 . Use glasses. l is distace between the spaceships.

13 minutes ago, DimaMazin said:

https://en.wikipedia.org/wiki/Relativity_of_simultaneity

- l isn't -1 . Use glasses. l is distace between the spaceships.

And it's still not zero. Use brain.

3 minutes ago, swansont said:

And it's still not zero. Use brain.

Zero is point of observer. x₀=0   then x= l or -l

if x=l then v is negative

if x= -l then v is pozitive

1 hour ago, DimaMazin said:

Zero is point of observer. x₀=0   then x= l or -l

if x=l then v is negative

if x= -l then v is pozitive

And in either case, t' is nonzero.

7 minutes ago, swansont said:

And in either case, t' is nonzero.

In the experiment t=0 not t'.

47 minutes ago, DimaMazin said:

In the experiment t=0 not t'.

I guess I don't understand your setup. You are using primed variables in the S frame or unprimed in the moving frame.

Distance beetween the spaceships is the same before and after the accelerations in S and is l . In S' before the events of accelerations the distance = l/gamma. And after the events of accelerations of both spaceships a distance between the spaceships is gamma * l .  If velosity =0 in S then it is v in S'. If vlosity =v in S then it is 0 in S'. 

t is time between events of acelerations of spaceships in S.    t' is time between events of negative acceleration of spaceships in S' .

What  koncretely is unclear?

You have t_f and t_b which are not measured in the S frame.

6 hours ago, swansont said:

You have t_f and t_b which are not measured in the S frame.

t_f = l /(gamma(c+v))+l/(gamma(c-v))=2lc/(gamma(c²-v²))=2 l /(c²-v²)^1/2

t_b = l/(gamma(c-v)+l/(gamma(c+v))=2 l/(c²-v²)^!/2

The measurements are made in S. Clock indication of event without distance is the same in both frames. The clocks indications of receptions and emissions of light  are measured in S , but they  are the same in S'.

In the OP you state that these are times measured by the spaceships, which are not in the S frame.

7 hours ago, swansont said:

In the OP you state that these are times measured by the spaceships, which are not in the S frame.

They do measure distances  and times in S'. But we can define results of the measurements in S, using velosity of the spaceships as v which is 0 in S' . And I made it.