# Deriving the lorentz transforms

1. Apr 18, 2006

### Thrice

It's a really easy question, I know, but I must be doing something stupid. Can someone please spell out how to get the right hand side matrix form out of the individual equations?

http://img234.imageshack.us/img234/8497/lorentz25wv.jpg [Broken]

Last edited by a moderator: May 2, 2017
2. Apr 18, 2006

### Andrew Mason

If you multiply the matrix by the 4-vector (t', x', y', z') it should result in the 4-vector (t, x, y, z) as set out on the left side of the arrow. I think the matrix is wrong, though. The numerator of the second term in the top row should be v/c^2 and the numerator of the first term in the second row should be v.

AM

Last edited by a moderator: May 2, 2017
3. Apr 18, 2006

### Thrice

See I thought that as well, but they have the inverse of that matrix in the book too & it matches up with the one on http://en.wikipedia.org/wiki/Lorent...rmation_for_frames_in_standard_configuration". Apparently it's gotten by switching V for (-V). Does it work out if you use the 4 vector (ct', x', y', z') & (ct, x, y, z)?

Edit: Right.. it does.. I knew I was doing something stupid sorry.

Last edited by a moderator: Apr 22, 2017
4. Apr 18, 2006

### Thrice

Wait don't go. I have more foolish questions once I figure out how to post in that .. latex is it?

5. Apr 18, 2006

### Thrice

$$\newcommand{\pd}[3]{ \frac{ \partial^{#3}{#1} }{ \partial {#2}^{#3} } } g_{\it ij} \pd{}{V^k}{} (V^i V^j )= 2 g_{\it kj} V^j$$

Last edited: Apr 18, 2006
6. Apr 18, 2006

### Thrice

K got it. Why does that work? Something to do with the symmetry of the metric....

7. Apr 18, 2006

### nrqed

No need for any special symmetry. It follows from ${\partial V^i \over \partial V^k} = \delta^i_k$ and similarly if i is replaced by j. Then you just need to rename a dummy index in one of the terms and you get the answer provided.

8. Apr 18, 2006

### Thrice

Thanks both of you. I don't know where i'd go when my brain isn't working. :)