# test

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test

Edited by t686
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• 1 year later...

this is a test.

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$\begin{array} + & 0 & 1 \\ 0 & 0 & 1 \\ 1 & 1 & 1 \\\end{array}$

Edited by studiot
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I can see my table in the list in the activity tab, but I can't see it in the thread itself.

This is even after refreshing the thread in the normal manner for MathML.

$\begin{array}{*{20}{c}} + & 0 & 1 \\ 0 & 0 & 1 \\ 1 & 1 & 1 \\ \end{array}$

$\left( {\begin{array}{*{20}{c}} + & 0 & 1 \\ 0 & 0 & 1 \\ 1 & 1 & 1 \\ \end{array}} \right)$

$\begin{array} + & 0 & 1 \\ 0 & 0 & 1 \\ 1 & 1 & 1 \\ \end{array}$

$\left( {\begin{array} + & 0 & 1 \\ 0 & 0 & 1 \\ 1 & 1 & 1 \\ \end{array}} \right)$

Edited by studiot
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• 3 months later...

$\frac {d^2 x^{\mu}} {d \alpha^2} + \Gamma_{\rho \sigma}^{\mu} \frac {dx^{\rho} } {d \alpha } \frac {dx^{\sigma}} {d \alpha}$

Same again, in larger font:

$\frac {d^2 x^{\mu}} {d \alpha^2} + \Gamma_{\rho \sigma}^{\mu} \frac {dx^{\rho} } {d \alpha } \frac {dx^{\sigma}} {d \alpha}$

Test post:

The following is the usual geodesic equation from General Relativity:

$\frac {d^2 x^{\mu}} {d \tau^2} + \Gamma_{\rho \sigma}^{\mu} \frac {dx^{\rho} } {d \tau } \frac {dx^{\sigma}} {d \tau}$

Where $\tau$ is an affine parameter (for example, proper time)  and the path is given by  $x^\mu = x^\mu (\tau)$

Suppose the same path is parameterised another way.  Let  $x^\mu = x^\mu (\lambda)$ where the parameter $\lambda$  is NOT assumed to be an affine parameter.

Then,  $\frac {d x^\mu} {d \lambda} = \frac {d x^\mu} {d \tau} . \frac {d \tau} {d \lambda}$   by the chain rule and since we can assume $\frac {d \tau} {d \lambda}$ exists without loss of generality.

Edited by Col Not Colin
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Latex test:

$\frac {d^2 x^{\mu}} {d \alpha^2} + \Gamma_{\rho \sigma}^{\mu} \frac {dx^{\rho} } {d \alpha } \frac {dx^{\sigma}} {d \alpha}$

Same again, in larger font:

$\frac {d^2 x^{\mu}} {d \alpha^2} + \Gamma_{\rho \sigma}^{\mu} \frac {dx^{\rho} } {d \alpha } \frac {dx^{\sigma}} {d \alpha}$

Test post:

The following is the usual geodesic equation from General Relativity:

[Eqn 1]          $\frac {d^2 x^{\mu}} {d \tau^2} + \Gamma_{\rho \sigma}^{\mu} \frac {dx^{\rho} } {d \tau } \frac {dx^{\sigma}} {d \tau} = 0$

Where $\tau$ is an affine parameter (for example, proper time)  and the path is given by  $x^\mu = x^\mu (\tau)$

Suppose the same path is parameterised another way.  Let  $x^\mu = x^\mu (\lambda)$ where the parameter $\lambda$  is NOT assumed to be an affine parameter.

Then,

[Eqn 2]     $\frac {d x^\mu} {d \lambda} = \frac {d x^\mu} {d \tau} \frac {d \tau} {d \lambda}$   by the chain rule (we can assume $\frac {d \tau} {d \lambda}$ exists).

Hence,

[Eqn 3]     $\frac {d^2 x^\mu} {d \lambda ^2} = \frac {d^2 x^\mu} {d \tau ^2} {\frac {d \tau} {d \lambda}}^2 + \frac {d x^\mu} {d \tau} \frac {d^2 \tau} {d \lambda ^2}$

Combining Eqn 1 and Eqn 2 we obtain,

$\frac {d^2 x^{\mu}} {d \lambda^2} + \Gamma_{\rho \sigma}^{\mu} \frac {dx^{\rho} } {d \lambda } \frac {dx^{\sigma}} {d \lambda} \space = \space \frac {d x^{\mu}} {d \tau} \frac {d^2 \tau} {d \lambda ^2} + \large [ \normalsize {\frac {d \tau} {d \lambda}} \large ] \normalsize ^2 \large ( \normalsize \frac {d^2 x^{\mu}} {d \tau^2} + \Gamma_{\rho \sigma}^{\mu} \frac {dx^{\rho} } {d \tau } \frac {dx^{\sigma}} {d \tau} \large )$

COMMENT:  having a time-limit to edit posts in the sandbox seems un-kind.  Sorry for making a lot of posts, I ran out of time.  Crumbs... this is hard work.  I'll try an equation editor and see if I can import the finished thing.

Edited by Col Not Colin
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• 1 month later...

$\begin{array}{*{20}{c}} {62310721} \\ {\underline {25644387} } \\ {87955108} \\ \end{array}$

$\begin{array}{l} 62310721 \\ \underline {00000007} \\ 62310728 \\ \end{array}$

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