How Do You Solve the Lorentz Transformation Condition Using Rapidity?

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
2 replies · 2K views
ozone
Messages
121
Reaction score
0
Hello,

I was working out of a text for my own knowledge and I ran into a slight snag which has become bothersome. First I was asked to find the condition on a linear transformation for which the dot product [itex]u \cdot u= u_0^2 - u_1^2{}[/itex] is preserved. Easily I found that for [itex]u' = \Lambda u[/itex] our condition is [itex]\Lambda^T \eta \Lambda = \eta[/itex] where [itex]\eta[/itex] is the metric tensor.

However the tricky part came in when I was asked "Solve this condition in terms of ra-
pidity.". It was quite clear that I was asked to derive the lorentz transformation matrix but I wasn't sure the best way to go about doing this from only this condition.

My best attempt was to populate [itex]\Lambda[/itex] with 4 unknown functions [itex]f_1,f_2,..[/itex] of our rapidity [itex]\phi[/itex]. This gave me 3 equations namely

[itex]f_1*f_2 = f_3*f_4[/itex]
[itex]f_1^2 - f_3^2 = 1[/itex]
[itex]f_2^2 - f_4^2 = -1[/itex]

From which it seems clear that [itex]f_1 = cosh(\phi), f_2=sinh(\phi),...[/itex]

However I have two questions/problems with my approach. (1) I do not know how to select a sign for my functions (it seems completely arbitrary what sign cosh or sinh take on, but most texts seem to have a definite convention and (2) my method does not seem completely rigorous and I was hoping to find one that was superior.
 
Last edited:
Physics news on Phys.org
Hi ozone!

(1) Yes there is a sign issue here. There are two possibilities for [itex]\det \Lambda[/itex]: +1 and -1. This means that the Lorentz group splits into two. The transformations with [itex]\det \Lambda[/itex] = +1 are the ones you're familiar with (called the "restricted Lorentz group") and the transformations with [itex]\det \Lambda[/itex] = -1 involve parity flips, time reversals and things like that.

(2) I think your approach is fully rigorous :smile:
 
Alright fair enough, I have heard of the lorentz transformations having a + or - determinate, so this must be the manifestation of that, thank you.