Lorentz Transformation: Get Answers to Your Questions

[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 t_a and x_a, and instead of using t' and x' I'd like to use t_b and x_b
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
Twin Paradox asymmetry
Twin Paradox symmetry
Motion in space
Lorentz and Doppler
Universe Frame of Reference.
But before those, I'd like to understand Lorentz first.
Thanks

 

[Vanadium 50]

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.

 

[Stephanus]

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.

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

Consider Einstein's Train example...
You have a train with an observer at the midpoint between the ends. you also have an observer standing along the tracks..

[PLAIN]http://home.earthlink.net/~jparvey/sitebuildercontent/sitebuilderpictures/trainsimul2.gif[/QUOTE]

I think this has to do with Lorentz contraction. Otherwise if wee look the bottom picture, the lights can't reach the observer at the same time from the train reference.
But to do that, the railway somehow must be contracted, otherwise the front of the train will reach the front mark at the SAME TIME with the back of the train.
And yet, the front train is in the middle of the light sphere wrt front train. So does the back train, it's in the centre of the back light sphere.
But before really understanding this picture, I really like to fully understand Lorentz transformation.

 

[A.T.]

I really like to fully understand Lorentz transformation.

This might help:

 

[PeroK]

Or this:

http://www.amazon.com/dp/1891389610/?tag=pfamazon01-20

 

[Stephanus]

This might help:

Excelent, excelent, thanks.

 

[Stephanus]

This might help

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

 

[harrylin]

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

Better use (see post https://www.physicsforums.com/threads/length-contraction.817911/page-4#post-5137038):

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

BTW, the other dimensions are really easy:

y' = y
z' = z

From that follows, as explained in the other thread, for t_a=t_b:
$x'_b - x'_a = \Delta x' = \gamma \Delta x$
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

 

[Stephanus]

Better use (see post https://www.physicsforums.com/threads/length-contraction.817911/page-4#post-5137038):

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

BTW, the other dimensions are really easy:

y' = y
z' = z

From that follows, as explained in the other thread, for t_a=t_b:
$x'_b - x'_a = \Delta x' = \gamma \Delta x$
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

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.

 

[harrylin]

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.

Yes, once more: nothing happens with the lateral dimensions.

 

[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.