Lorentz Transformation

1. Jun 9, 2015

Stephanus

Dear PF Forum,
First, I'd like to thanks this forum for helping this much and so far.
I have a question about Lorentz Transformation. Lots of questions actually

http://en.wikipedia.org/wiki/Lorentz_transformation#Boost_in_the_x-direction
Instead of using t and x, I'd like to use ta and xa, and instead of using t' and x' I'd like to use tb and xb
So here is the equation.
$t_b = \gamma(t_a - \frac{v_ax}{c^2})$
$x_b = \gamma(x - v_at_a)$
$y_b = y_a$, won't be used
$z_b = z_a$, won't be used
okay...

$\gamma \text{ is } \frac{1}{\sqrt{1-\frac{v_a^2}{c^2}}}$
Before I go any further, can I just use 2 dimensions?
1 time and 1 spatial (x), without y and z?
And after this thread, I'd like to go back to my previous threads to understand them
Motion in space
Lorentz and Doppler
Universe Frame of Reference.
But before those, I'd like to understand Lorentz first.
Thanks

Last edited: Jun 9, 2015
2. Jun 9, 2015

Staff Emeritus
You know, by starting six threads on slight variations of the same subject, you maximize the chances for confusion. It's also a good idea to get the basics understood first before firing off half a dozen slightly different complications.

3. Jun 9, 2015

Stephanus

Actually, no.
Those threads actually almost a month ago.
I asked one question, then I realize that there's more that I have to learn.
I started another, still I realize that it's less basic. I want to get to the most basic knowledge.
I start this one because of this in someone thread that I read

Last edited by a moderator: May 7, 2017
4. Jun 9, 2015

A.T.

This might help:

5. Jun 9, 2015

6. Jun 9, 2015

Stephanus

Excelent, excelent, thanks.

7. Jun 10, 2015

Stephanus

Thanks A.T, but could you tell me.
I want to study Lorentz formula
Boost only in x direction.
Can I use just 2 dimensions, before I ask further?
I want to understand it.
Can I just use
$t_b = \gamma(t_a - \frac{vx}{c^2})$
and
this only?
$x_b = \gamma(x_a - vt_a)$
Thanks

Last edited: Jun 10, 2015
8. Jun 10, 2015

harrylin

$x' = \gamma(x - vt)$
$t' = \gamma(t - vx/c^2)$

BTW, the other dimensions are really easy:

y' = y
z' = z

From that follows, as explained in the other thread, for ta=tb:
$x'_b - x'_a = \Delta x' = \gamma \Delta x$
That means that according to how clocks are synchronized in S, lengths in S' appear length contracted.

Note that for relatively small speeds (say <0.001c or < 300 km/s) γ ≈ 1 so that:
x' ≈ x - vt
t'≈ t - vx/c2
That's just to highlight relativity of simultaneity in the Lorentz transformations

9. Jun 10, 2015

Stephanus

Perhaps 1 spatial dimension (plus 1 time dimension) is enough for boost in x direction?
http://en.wikipedia.org/wiki/Lorentz_transformation#Boost_in_the_x-direction
So that the calculation is somewhat simpler.

10. Jun 10, 2015

harrylin

11. Jun 11, 2015

Ibix

As long as you're happy to consider everything moving in one line, x and t is fine.

This is why trains are popular for SR examples.