Proof Minkowski metric is invariant under Lorentz transformation

[SamRoss]

Ok, this should be an easy one but it's driving me nuts. When we take the Lorentz transformations and apply them to x²-c²t² we get the exact same expression in another frame. I can do this math easily by letting c=1 and have seen others do it by letting c=1 but I have never seen anyone actually do it with the c's in there. It doesn't sound like it should be that hard but I just can't get it to work. Can anyone go through it?

 

[SamRoss]

Ok, I found a website that showed how to do it. For anyone out there who might be having the same problem, rewriting c²t² as (ct)² makes a world line of difference!

http://capone.mtsu.edu/phys3100/Syllabus/Examples/CH_1/EX_1_V/EX_1_V__old_/proof.html

 

[WannabeNewton]

It's always the arithmetic isn't it ? Glad it worked out!

 

[Bill_K]

The neatest way:

x' = γ(x - vt)
t' = γ(t - vx/c²)

Take linear combinations:

(1) x' - ct' = γ(x - ct - v(t - x/c)) = γ(1 + v/c)(x - ct)
(2) x' + ct' = γ(x + ct - v(t + x/c)) = γ(1 - v/c)(x + ct)

Multiplying (1) and (2) together, all the leading factors cancel and we get

x'² - c²t'² = x² - c²t²

(Note that x ± ct are the eigenvectors of the transformation. Plus the factors in front are just the Doppler shifts.)

 

[SamRoss]

Wow, I just realized I had simply forgotten to distribute the c squared to everything originally. Talk about dumb!