Deriving the lorentz transforms

[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

 

[Andrew Mason]

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

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

 

[Thrice]

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

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.

 

[Thrice]

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

 

[Thrice]

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

 

[Thrice]

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

 

[nrqed]

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

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.

 

[Thrice]

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