# Proof Minkowski metric is invariant under Lorentz transformation

1. Aug 1, 2013

### 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?

2. Aug 1, 2013

3. Aug 1, 2013

### WannabeNewton

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

4. Aug 1, 2013

### 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.)

Last edited: Aug 1, 2013
5. Aug 1, 2013

### SamRoss

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