# Form of Lorentz Transformation Using West-Coast Metric

1. Aug 14, 2012

### LittleSailor

This is a fairly trivial question I think. I'm only asking it here because after some googling I was unable to find its answer. I was at one point led to believe that the form of the Lorentz-transformation matrix is dependent on the convention used for the Minkowski metric. Specifically it was my understanding that

[γ, βγ, 0, 0]
[βγ, γ, 0, 0]
[0, 0, 1, 0]
[0, 0, 0, 1]

was the transformation matrix when the West-coast metric, diag(1, -1, -1, -1), is used. This is the inverse of the more commonly encountered

[γ, -βγ, 0, 0]
[-βγ, γ, 0, 0]
[0, 0, 1, 0]
[0, 0, 0, 1]

which I know is correct at least for the East-coast metric, diag(-1, 1, 1, 1). I was working a problem recently and got a result using the former of these transformation matrices that was clearly incorrect. Does the Lorentz transformation's form actually depend on the convention for the metric, or did I concoct this entire distinction? Perhaps I misunderstood one of my professors.

2. Aug 14, 2012

### PAllen

The form of the Lorentz transform does not depend on whether you use +t metric or -t metric. The two transform matrices you give are simply inverses of each other corresponding to boosts in opposite directions.

3. Aug 14, 2012

### jarod765

As a group the lorentz transformations, $\Lambda^{\mu}_{\nu}$ are defined by:

\Lambda^{\mu}_{\nu} \eta_{\mu \rho} \Lambda^{\rho}_{\sigma} = \eta_{\nu \sigma}

Thus, the metric should not affect the transform because you can multiply both sides of the equation to switch metric convention.

4. Aug 14, 2012

### LittleSailor

Haha, I feel silly now--thanks, guys.