# Jeans' theorem

1. Mar 22, 2008

### noospace

I'm trying to get from the magnetic vector potential

$\vec{A}(\vec{x},t) = \frac{1}{\sqrt{\mathcal{V}}}\sum_{\vec{k},\alpha=1,2}(c_{\vec{k}\alpha}(t) \vec{u}_{\vec{k}\alpha}(\vec{x}) + c.c.)$

where
$c_{\vec{k}\alpha}(t) = c_{\vec{k}\alpha}(0) e^{-i\omega_{\vec{k}\alpha}t}$
$\vec{u}_{\vec{k}\alpha}(t) = \vec{u}_{\vec{k}\alpha}(0) e^{i \vec{k} \cdot \vec{x}}$
c.c. = complex conjugate.

to Jeans' result

$H := \frac{1}{2}\int d^3\vec{x} (E^2 + B^2) = \sum_{k,\alpha}\left(\frac{\omega_{k\alpha}}{c}\right)^2 2c_{\vec{k}\alpha}^\ast c_{\vec{k}\alpha}$.

The first question I have is why we add the complex conjugate term at all. In classical electrodynamics, I'm used to dealing with the complex fields?

Tod derive the result, my approach was to define the complex vector potential

$\widetilde{\vec{A}}(\vec{x},t) = \frac{1}{\sqrt{\mathcal{V}}}\sum_{\vec{k},\alpha=1,2}c_{\vec{k}\alpha}(t) \vec{u}_{\vec{k}\alpha}(\vec{x})$

and use the relation

$H = \frac{1}{2}\int d^3\vec{x} \frac{1}{2} (\Re[\widetilde{\vec{E}}\cdot \widetilde{\vec{E}}^\ast]+\Re[\widetilde{\vec{B}}\cdot \widetilde{\vec{B}}^\ast])$

where $\widetilde{\vec{E}} = - \frac{1}{c}\frac{\partial \widetilde{\vec{A}}}{\partial t}, \;\widetilde{\vec{B}} = \nabla \times \widetilde{\vec{A}}$.

The first term is easy to deal with and gives (after using the normalization $\frac{1}{\mathcal{V}}\int d^3\vec{x} \vec{u}_{\vec{k}\alpha}\cdot\vec{u}_{\vec{k'}\alpha'}^\ast = \delta_{\vec{k}\vec{k}'}\delta_{\alpha\alpha'}$):

$\frac{1}{4}\int d^3 \vec{x}\Re[\widetilde{\vec{E}}\cdot \widetilde{\vec{E}}^\ast] = \frac{1}{4}\sum_{k,\alpha}\left(\frac{\omega_{k\alpha}}{c}\right)^2 c_{\vec{k}\alpha}^\ast c_{\vec{k}\alpha}$

The second term is a bit more problematic. Using the identity

$(\vec{k} \times \vec{u})\cdot(\vec{k}' \times \vec{u}') = \vec{k} \cdot\vec{k}' \vec{u} \cdot\vec{u}' - \vec{k} \cdot{u}' \vec{k}' \cdot\vec{u}$

I get

$\widetilde{\vec{B}} = \nabla \times \widetilde{\vec{A}} = \frac{1}{\sqrt{V}}\sum_{\vec{k},\alpha} i c_{\vec{k}\alpha} \vec{k} \times \vec{u}_{\vec{k}\alpha}$

so

$\frac{1}{2}\int d^3\vec{x} \frac{1}{2}\Re[\widetilde{\vec{B}}\cdot \widetilde{\vec{B}}^\ast]$

$= \frac{1}{4}\sum_{\vec{k},\alpha=1,2}\left(\frac{\omega_{\vec{k}\alpha}}{c}\right)^2 c_{\vec{k}\alpha}^\ast c_{\vec{k}\alpha} - \frac{1}{4}\frac{1}{V}\sum_{\vec{k},\vec{k}',\alpha,\alpha'}\int d^3 \vec{x} c^\ast_{\vec{k}'\alpha'}c_{\vec{k}\alpha} (\vec{k}\cdot \vec{u}^\ast_{\vec{k}'\alpha'})(\vec{k'}\cdot \vec{u}_{\vec{k}\alpha})$

where I have used the normalization condition to evaluate the first term. I'm stuck on how to evaluate this last integral. I know that I have to use the Gauge condition that $\vec{k} \cdot \vec{u}_{\vec{k}\alpha} = 0$ but I'm not sure how to incorporate it.

Any help would be greatly appreciated.

Last edited: Mar 22, 2008
2. Mar 22, 2008

### noospace

I suppose what I want to show is that the term

$\sum_{\vec{k},\vec{k}',\alpha,\alpha'}\int d^3 \vec{x} c^\ast_{\vec{k}'\alpha'}c_{\vec{k}\alpha} (\vec{k}\cdot \vec{u}^\ast_{\vec{k}'\alpha'})(\vec{k'}\cdot \vec{u}_{\vec{k}\alpha})$

vanihses. For then,

$\frac{1}{2}\int d^3 \vec{x}\frac{1}{2}\Re[\widetilde{\vec{E}}\cdot\widetilde{\vec{E}}^\ast + \widetilde{\vec{B}}\cdot\widetilde{\vec{B}}^\ast] = \frac{1}{2}\sum_{\vec{k},\alpha}\left(\frac{\omega_{\vec{k}\alpha}}{c}\right)^2 c^\ast_{\vec{k}\alpha}c_{\vec{k}\alpha}$

which is uncannily similar to what I set out to show. I know $\vec{k} \cdot\vec{u}_{\vec{k}\alpha}$ so I can reduce the sum to a sum over $\sum_{\vec{k} \neq \vec{k}',\alpha,\alpha}$