OK, so should I just leave my answer like the following?

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- Aug 9th 2012, 01:30 PMAA23Re: Timelike, Spacelike and Null Seperation
OK, so should I just leave my answer like the following?

- Aug 9th 2012, 01:37 PMJohnDMalcolmRe: Timelike, Spacelike and Null Seperation
I'm not sure you understood my last post. I was saying that the order you write the matrices matters for when you do the multiplication. You should have

$\displaystyle g' = \left( \begin{array}{cc} 1 & \cos{\theta} \\ 0 & \sin{\theta} \end{array} \right) \left( \begin{array}{cc} 1 & 0 \\ \cos{\theta} & \sin{\theta} \end{array} \right) $

Then carry out the matrix multiplication to get a single 2x2 matrix for g'. - Aug 9th 2012, 01:42 PMAA23Re: Timelike, Spacelike and Null Seperation
Sorry John I did make an error reading your post. So using what you have said I have arrived at a final answer of

(see attachment) - Aug 9th 2012, 01:47 PMJohnDMalcolmRe: Timelike, Spacelike and Null Seperation
Look at the expression I have in Post #32 and compare it to your expression for g'.

- Aug 9th 2012, 01:52 PMAA23Re: Timelike, Spacelike and Null Seperation
(Giggle) I'm sorry John I totally misread you post again, I think now I have the correct answer.

Thank You - Aug 10th 2012, 02:57 AMJohnDMalcolmRe: Timelike, Spacelike and Null Seperation
$\displaystyle 1 + \cos^{2}{\theta} \neq \sin^{2}{\theta} $