The time for 2 light pulses in a moving vehicle

[powermind]

DaleSpam,
can you write down the steps how did you derive the times (t_A=v, t_B=-v)?
But pelase be attention that when I am saying the light is switched on, I am not talking about how long the beam takes to travel from point to other! but I mean it is switched on at the specific time only. So, we use the travel's length to get when the light was switched on excatly.

Each frame has own clock. All observers in the frame share the clock of the frame and all may not say same answer depending on the distance between the light and each observer. If O1 saw the light at T1, O2 saw it at T2 and O3 saw it at T3, by the substraction we can get the exactly time that agreed by all observers.

T = T1 - d1/c = T2 - d2/c = T3 - d3/c​

Nugatory,
I agree with you but The A_O1 distance equals to B_O1 distance!

You said:

If those two distances are different ...

You did not tell me what will happen If not? I guess you will conclude that A and B are switched on simultaneously according to O.

All of you start with the expression: A is switched on before B according to O! And this is I would like from you to prove it!

Many thanks,

 

[Nugatory]

I agree with you but The A_O1 distance equals to B_O1 distance!

If the A_M distance is equal to the B_M distance, then the A_01 distance will not be equal to the B_01 distance. It's not possible, and if you think you've set things up in such a way that it is, you'll have made a mistake somewhere (most likely using the position of the receiving eyeball when the light leaves the source, instead of the position of the receiving eyeball when the light reaches it).

 

[powermind]

How is it impossible?! This is very strange!
Besides, there is no relationship between A_M and A_O1. So, saying "If A_M ..., then A_O1 ..." is incorrect.

 

[Dale]

DaleSpam,
can you write down the steps how did you derive the times (t_A=v, t_B=-v)?

Sure. In the M frame we have that A turns on at:
$t'=0$
$x'=-L_0/2$

Using the Lorentz transform (http://en.wikipedia.org/wiki/Lorentz_transformation#Boost_in_the_x-direction) we have:
$t=\gamma(t'+vx'/c^2)$
$t=(0+v(-L_0/2)/c^2)/\sqrt{1-v^2/c^2}$

Using units of length such that $L_0=2\sqrt{1-v^2/c^2}$ and units of time such that $c=1$ and simplifying we have:
$t=(v(-2\sqrt{1-v^2}/2))/\sqrt{1-v^2}=-v$

Similarly in the M frame we have that B turns on at:
$t'=0$
$x'=L_0/2$

As above, using the Lorentz transform we have:
$t=\gamma(t'+vx'/c^2)$
$t=(0+v(L_0/2)/c^2)/\sqrt{1-v^2/c^2}$

As above, using units of length such that $L_0=2\sqrt{1-v^2/c^2}$ and units of time such that $c=1$ and simplifying we have:
$t=(v(2\sqrt{1-v^2}/2))/\sqrt{1-v^2}=v$

 

[powermind]

Thanks DaleSpam.
What does minus sign mean when you say t = -v?

 

[Dale]

The minus sign means that it happens before a nearby O frame synchronized clock reads t=0.

 

[powermind]

Is this logical?!

Suppose M turns the lights on at t’. At this time (t'), O will calculate his time so that t = $\gamma$t’. Definitely the lights should be turned on not before the time (t) according to O.

But we notice that A is switched on before t according O which is not logical!

Can you please explain what the wrong is?

 

[Dale]

Is this logical?!

Yes, it is logical. You are simply forgetting to include the relativity of simutaneity.

Suppose M turns the lights on at t’. At this time (t'), O will calculate his time so that t = $\gamma$t’.

This is only correct for x'=0. The complete formula is $t=\gamma(t'+vx'/c^2)$. For x'=0 that simplifies to what you wrote, but A and B are located at $x'=\pm L_0/2$, not at x'=0, so you cannot use the simplified formula.

Can you please explain what the wrong is?

My recommendation is to avoid the simplified formulas. Always use the full Lorentz transform formulas (http://en.wikipedia.org/wiki/Lorentz_transformation#Boost_in_the_x-direction). They will automatically simplify to the appropriate formulas when they do apply, but you will avoid mistakes like this from using the simplified formulas when they don't apply.

 

[powermind]

DaleSpam,

Q. When did M switch the light on?
A. At t'.

Q. At t' what was the time according to O?
A. t = $\gamma$t'. (x' = 0 or using time dilation equation)

I think you agree right now!

So, the switching event happened at t' according to M or at t according to O.

Is there any problem?

Ok! how was the light A switched on before t?
How can I say "O sees the light A before it is switched on" ??! This is very strange!

You can believe in the relativity of simultaneity but not on that way, can't you?!

 

[Dale]

Q. At t' what was the time according to O?
A. t = $\gamma$t'. (x' = 0 or using time dilation equation)

I think you agree right now!

No, I definitely disagree. As I already explained, that equation doesn't apply here since x'≠0 for A and B. The correct equation is $t=\gamma(t'+vx'/c^2)$.

You cannot use an equation when it doesn't apply, and you cannot neglect the relativity of simultaneity when it suits you. Always use the Lorentz transform. It includes all relativistic effects, and will automatically simplify to the time dilation equation when it is applicable, which it isn't here.

Neglecting the relativity of simultaneity by improperly using the time dilation equation is a key mistake that you have been repeatedly making since the beginning of the thread. I don't know how to be more clear about it. You have been corrected on this point multiple times by multiple people.

 

[Nugatory]

Ok! how was the light A switched on before t?
How can I say "O sees the light A before it is switched on" ??! This is very strange!

You can believe in the relativity of simultaneity but not on that way, can't you?!

For M, the A light source was switched on at t', and the flash of light reached M's eyes at his time $t'+\frac{d'}{c}$ where $d'$ is the distance between where the light source was when the light was switched on and where M's eyes are when the flash of light reached them. It's easy for M to measure $d'$ because the light source and M's eyes are both at rest relative to M.

For O, the A light source was switched on at t, and the flash of light reached O's eyes at his time $t+\frac{d}{c}$ where $d$ is the distance between where the light source was when the light was switched on and where O's eyes are when the flash of light reached them.

Although ${t}\neq{t'}$ and $d\neq{d'}$ both observers see something perfectly reasonable: the light source is switched on, the flash of light travels from the light source to their eyes at speed $c$, and they add the travel time to the emission time to get the arrival time.

 

[powermind]

DaleSpam, sometimes generic talking can not transfer an idea in correct way. Nugatory has clarified what is my mistake. Unluckily I knew that before his post :)

After Nugatory's post we can make a scenario: O will see light A first then M will see A and B at the same time finally O will see B. I hope I am correct

I would like to thank you all, PeterDonis, Nugatory, DaleSpam and all who help me. Thank you for your patience.

Thank you very much.
Will see you back ...

 

[Dale]

Excellent. I am glad that Nugatory's comments were helpful.