# Relativistic Poisson Equation

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### #69 Dubbelosix

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Posted 14 January 2018 - 03:43 PM

Alternatively the LHS can be written like

$\partial_{\mu} J^{\mu}J_{\mu} + J^{\mu}J_{\mu} \frac{1}{\sqrt{-g}}(\sqrt{-g}\Gamma^{\sigma}_{\mu \sigma})$

Edited by Dubbelosix, 14 January 2018 - 03:43 PM.

### #70 Dubbelosix

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Posted 14 January 2018 - 03:46 PM

The object acting on the square of the currents appears to be formally similar to a simple definition of the Covariant derivative which just consists of the derivative and correction terms in form of Christoffel symbols:

$\nabla_{\mu} = \partial_{\mu} + \Gamma_{\mu}$

### #71 Dubbelosix

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Posted 15 January 2018 - 11:53 AM

Some explaining must be done about where the most logical continuation comes from noticing there is the square of the currents and that usually implies a commutation law. Though, it is more nicer to look at the whole thing with two covariant derivatives

$[\nabla_{\mu}J^{\mu}, \nabla^{\mu} J_{\mu}]$

### #72 Dubbelosix

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Posted 15 January 2018 - 12:09 PM

Now, it's actually not too difficult to understand how to write out the expression given in the last post. The commutation of two covariant derivatives leads to, with indices $i$ and $j$ as

$[\nabla_i, \nabla_j] = (\partial_i + \Gamma_i)(\partial_j + \Gamma_j) - (\partial_j + \Gamma_j)(\partial_i + \Gamma_i)$

$= (\partial_i \partial_j + \Gamma_i \partial_j + \partial_i \Gamma_j + \Gamma_i\Gamma_j) - (\partial_j \partial_i + \partial_j\Gamma_i + \Gamma_j \partial_i + \Gamma_j \Gamma_i)$

$= -\partial_i \Gamma_j + \partial_j \Gamma_i + [\Gamma_i, \Gamma_j]$

Edited by Dubbelosix, 20 January 2018 - 07:01 AM.

### #73 Dubbelosix

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Posted 15 January 2018 - 12:12 PM

Later this can be written in a more nicer tensorial way involving other indices, but this is a nice start.

### #74 Dubbelosix

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Posted 20 January 2018 - 07:06 AM

The interesting part of the derivation (for me) was leading up to the equation

$\partial_{\mu}\phi\partial^{\mu} \phi + \partial^{\mu} \phi \frac{\partial_{\mu} \sqrt{g}}{\sqrt{-g}}\phi = \nabla_{\mu}\phi\nabla^{\mu}\phi = J_{\mu}J^{\mu} + \frac{J_{\mu}J^{\mu} }{\sqrt{-g}}\sqrt{g}$

The Nordstrom energy tensor is

$- 4 \pi T_{matter} = \phi \Box \phi$

Knowing the contracted Einstein tensor in the special case of four dimensions is simply

$\mathbf{G} = \frac{2 - n}{2}R = T$

In which $T$ is identified with the stress energy tensor and $\mathbf{G}$ the Einstein tensor in which $R$ is the scalar curvature.Knowing this then, we might have a form of the Einstein equation which involves the Covariant derivative:

$\nabla_{\mu} \mathbf{G} = \partial_{\mu}T_{matter} + J^{\mu}J_{\mu} \frac{1}{\sqrt{-g}}(\sqrt{-g}\Gamma^{\sigma}_{\mu \sigma})$

Which is obtained by considering a derivative into the current equation at the beginning of this section.

In which substitutions have been made:

$J^{\mu} = \partial^{\mu} \phi$

$J_{\mu} = \partial_{\mu} \phi$

Supposing there is any truth in the last mathematical statements, then the equivalent form of the equation which can been rearranged and then contracted using the killing vector, is often considered a conservation law in itself.

$\nabla_{\mu} (\sqrt{-g}\mathbf{G}^{\mu}_{\nu}) \xi^{\nu} = \frac{2 - n}{2}( \sqrt{-g}R^{\mu}_{;\mu \nu}) \xi^{\nu} = \partial_{\mu}( \sqrt{-g}T^{\mu}_{\nu}) \xi^{\nu} + J^{\mu}J_{\mu}(\sqrt{-g}\Gamma^{\mu}_{ \mu \nu}) \xi^{\nu}$

At least, it is this term $\partial_{\mu} (\sqrt{-g}T^{\mu}_{\nu}) \xi^{\nu} = 0$ which is identified as the conservation law in the textbooks.

Edited by Dubbelosix, 20 January 2018 - 07:08 AM.