Understanding the Lorentz Transformation: Unpacking the Mathematical Steps

AI Thread Summary
The discussion focuses on understanding the mathematical derivation of the Lorentz Transformation, specifically how the term (1 - v²/c²) is derived. The user references equations from Einstein's Relativity and attempts to simplify them by factoring and substituting known values. A swimming analogy is introduced to illustrate the relationship between distances and velocities, ultimately connecting it to the transformation. The conversation emphasizes the mathematical steps rather than the physical implications of relativity. The user expresses a desire for clarity in the mathematical process involved in these transformations.
dperez3894
Messages
8
Reaction score
0
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?
 
Mathematics news on Phys.org
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}).
 
Last edited:
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 t<sub>1</sub>= d/(v+c) rto do that. Swimming back up stream his speed, still relative to the shore, is c- v so it takes him t<sub>2</sub> = 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(c<sup>2</sup>- v<sup>2</sup>).

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>t<sup>2</sup>= v<sup>2</sup>+ d'<sup>2</sup>. Solving for t, t<sup>2<sup>= d'<sup>2</sup>/(c<sup>2</sup>- v<sup>2</sup>) and t= \frac{d&#039;}{\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&#039;}{\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&#039;}{\sqrt{c^2- v^2}}

That gives
d= \frac{d&#039; \sqrt{c^2- v^2}}{c}
= d&#039; \sqrt{1- (v/c)^2}.

Do you see how that is connected to the Michaelson-Morley experiment?
 
Last edited by a moderator:
Muzza said:
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&#039; = 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&#039; = 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&#039; = 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.
 
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Back
Top