Deriving Lorentz Transformation: Exploring vx/c^2

[Jeno]

Can i know how does the Lorentz transformation equations are derived?
x'=Y(x-vt)
y'=y
z'=z
t'=Y(t-vx/c^2)

Especially the equation for t', what is term vx/c^2 there implies?
thank you for answering me.
-Jeno-

 

[Hootenanny]

For your reference;
http://galileo.phys.virginia.edu/classes/252/lorentztrans.html

 

[pmb_phy]

See also - http://www.geocities.com/physics_world/sr/lorentz_trans.htm

Best wishes

Pete

 

[pess5]

Can i know how does the Lorentz transformation equations are derived?
x'=Y(x-vt)
y'=y
z'=z
t'=Y(t-vx/c^2)

Especially the equation for t', what is term vx/c^2 there implies?
thank you for answering me.
-Jeno-

Well, you ask a lot here. It's a tall order, unless you desire a quick cheat sheet of sorts, with much assumed and/or omitted. Your question requires Sections 1 thru 3 of Einstein's 1905 OEMB paper, On the Electrodynamics of Moving Bodies. Section 3 is where the derivation takes place, whereby Einstein uses his kinematic model (thought experiment) of an emitter & reflector bouncing light back & forth.

The -vx/c^2 term is a temporal offset based upon distance x. Clocks in sync in the moving frame do not appear in sync per the stationary observing frame, however those moving clocks all tick at the same rate. If one clock reads 12am in some instant of your vantage, the other clock does not. It reads 12am at some other moment in your vantage, prior or later, even though the clocks are 12am AT ONCE per themselves. The delta time between like readouts (per you) depends opon the variable separation x, and the recorded velocity.

pess

 

[Hootenanny]

See also - http://www.geocities.com/physics_world/sr/lorentz_trans.htm

Best wishes

Pete

Nice site Pete, extremely useful reference! I think its going into my bookmarks

 

[Zeit]

Hello,

Pete, there is something I would like to understand at the number (7) of your website :

$$v\gamma t' = v\gamma^2 t - \gamma^2 \beta^2 x = v\gamma^2 t - \gamma^2 vx/c^2$$

Why does $\beta^2 x = vx/c^2$ if $\beta^2 = v^2 / c^2$?

Isn't $v\gamma t' = v\gamma^2 t - \gamma^2 v^2 x/c^2 = v\gamma^2 (t-vx/c^2 )$ ?

Thanks

 

[lightarrow]

Hello,

Pete, there is something I would like to understand at the number (7) of your website :

vgamma t' = vgamma^2 t - gamma^2 beta^2 x = vgamma^2 t - gamma^2 vx/c^2

Why does beta^2 x = vx/c^2 if beta^2 = v^2 / c^2?

Isn't vgamma t' = vgamma^2 t - gamma^2 v^2 x/c^2 = vgamma^2 (t-vx/c^2 )?

Thanks

Infact, there is a mistake. It should be as you say. The last term shouldn't be gamma^2 vx/c^2 but, instead:
gamma^2 v^2x/c^2.

 

[pmb_phy]

Hello,

Pete, there is something I would like to understand at the number (7) of your website :

$$v\gamma t' = v\gamma^2 t - \gamma^2 \beta^2 x = v\gamma^2 t - \gamma^2 vx/c^2$$

Why does $\beta^2 x = vx/c^2$ if $\beta^2 = v^2 / c^2$?

Isn't $v\gamma t' = v\gamma^2 t - \gamma^2 v^2 x/c^2 = v\gamma^2 (t-vx/c^2 )$ ?

Thanks

Its probably beccause I screwed up.

When I create these web pages I first do it with pen to paper. Then I transfer to an MS Word document so I can add equations etc. Then I take that and translated it into a web page. There's plenty of places to go wrong in this process. But thank you so much for pointing it out. I'll put this in my file of web pages to correct. Thanks again.

Best wishes

Pete

 

[Zeit]

No problem, I thank you for it, because I had been searching for this kind of webpage when I found this thread

 

[pmb_phy]

No problem, I thank you for it, because I had been searching for this kind of webpage when I found this thread

Thanks. I got to tell you Zeit. Its a very good feeling when you've done an enormous amount of work and someone finds it of use. Thank you for letting me know.

Best wishes

Pete

ps - Feel free to e-mail me or PM me about anything whatsoever. Okay?

 

[Zeit]

Hello,

Thanks Pete for your proposal, I'll write you if I need to

Good bye

 

[richie'ril]

PETE, Please i want to know the similarities between lorenz and galilean transformation

 

[Integral]

Richie,
How long are you willing to wait for a answer? Pete has not been active for nearly 3yrs.
I suggest you start your own thread.

Necroposting. Thread locked.