Lorentz Transformation math

[dperez3894]

I don't know which math forum to put this in but perhaps someone here could help out.

I want to figure out the mathematics behind the Lorentz Transformation so I can understand it better. I've got a copy of Einstein's Relativity which was reprinted by Three Rivers Press a few years ago.

In the Appendix there's a section on the Lorentz Transformation and I've hit a wall in figuring out the steps between the equations.

On page 133 they have the following equation sets;

(5)
x' = ax - bct
ct' = act - bx

(6)

v = bc/a

On page 134 it gave an example where if a snapshot was taken from K'(t'=0) and t was removed from equations (5) where;

x' = ax - bc
0 = ac - bx

and taking into account expression (6), the following equation was derived;

x' = a(1-(v^2/c^2))x

Where did the (1-(v^2/c^2)) come from?

[Muzza]

I'll attempt a purely mathematical explanation, I don't know **** about relativity ;)

Factor out the a and the x, and use the fact that bc/a = v.

x' = ax - bc = a(x - \frac{bc}{a}) = a(x - v) = ax(1 - \frac{v}{x}).

But 0 = ac - bx was given, so bx = ac, so x = \frac{ac}{b}. Substituting into the previous equation yields

x' = ax(1 - \frac{v}{ac/b}) = ax(1 - \frac{bv}{ac}) = ax(1 - \frac{b}{a} \cdot \frac{v}{c}).

But v = bc/a (given), so \frac{b}{a} = \frac{v}{c}. Thus

x' = ax(1 - \frac{v}{c} \cdot \frac{v}{c}) = ax(1 - \frac{v^2}{c^2}).

[HallsofIvy]

Here's a derivation that has nothing to do with relativity:

A man is swimming in a river. The man can swim at "c" m/s in still water and the river is flowing at "v" m/s.

First the man swims a distance d downstream and then back up again. Of course, down stream his speed, relative to the shore, is v+c so it takes him time t₁= d/(v+c) rto do that. Swimming back up stream his speed, still relative to the shore, is c- v so it takes him t₂ = d/(c-v) to do that:total time downstream and back, d/(c+v)+ d/(c-v)= (d(c-v)+ d(c+v))/((c-v)(c+v))= 2dc(c²- v²).

Now, he swims across the river a distance d' and back again. Drawing a vector diagram, it should be clear that he has to angle upstream slightly so that there is a right triangle with length ct (what he would swim if there were no current), vertical leg vt (how the current pushes him downstream), and horiontal leg (across the river) d'. That is: c<sup>2<sup>t²= v²+ d'². Solving for t, t<sup>2<sup>= d'²/(c²- v²) and t= \frac{d'}{\sqrt{c^2- v^2}}.
That's one way: doing exactly the same thing back, the total time to swim out and back is \frac{2d'}{\sqrt{c^2- v^2}}.

Now suppose he finds that the two times are the same. What is the relationship between d and d'?

We have t= \frac{2dc}{c^2-v^2}= \frac{2d'}{\sqrt{c^2- v^2}}

That gives
d= \frac{d' \sqrt{c^2- v^2}}{c}
= d' \sqrt{1- (v/c)^2}.

Do you see how that is connected to the Michaelson-Morley experiment?

[dperez3894]

I'll attempt a purely mathematical explanation, I don't know **** about relativity ;)

Factor out the a and the x, and use the fact that bc/a = v.

x' = ax - bc = a(x - \frac{bc}{a}) = a(x - v) = ax(1 - \frac{v}{x}).

But 0 = ac - bx was given, so bx = ac, so x = \frac{ac}{b}. Substituting into the previous equation yields

x' = ax(1 - \frac{v}{ac/b}) = ax(1 - \frac{bv}{ac}) = ax(1 - \frac{b}{a} \cdot \frac{v}{c}).

But v = bc/a (given), so \frac{b}{a} = \frac{v}{c}. Thus

x' = ax(1 - \frac{v}{c} \cdot \frac{v}{c}) = ax(1 - \frac{v^2}{c^2}).

Thanks. It's been quite a while since I've factored equations.