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## Existence of a symmetry for translational variation.

Posted by peeterjoot on September 12, 2009

[Click here for a PDF of this sequence of posts with nicer formatting]

Considering an example Lagrangian we found that there was a symmetry provided we could commute the variational derivative with the gradient

\begin{aligned}\frac{\delta }{\delta \phi} \mathbf{a} \cdot \boldsymbol{\nabla} \mathcal{L}&=\mathbf{a} \cdot \boldsymbol{\nabla} \frac{\delta \mathcal{L}}{\delta \phi} \end{aligned}

What this really means is not clear in general and a better answer to the existence question for incremental translation can be had by considering the transformation of the action directly around the stationary fields.

Without really any loss of generality we can consider an action with a four dimensional spacetime volume element, and apply the incremental translation operator to this

\begin{aligned}\int &d^4 x a \cdot \nabla \mathcal{L}( A^\beta + \bar{A}^\beta, \partial_\alpha A^\beta + \partial_\alpha \bar{A}^\beta) \\ &=\int d^4 x a \cdot \nabla \mathcal{L}( \bar{A}^\beta, \partial_\alpha \bar{A}^\beta)+\int d^4 x a \cdot \nabla \left(\frac{\partial {\mathcal{L}}}{\partial {A^\beta}} \bar{A^\beta}+\frac{\partial {\mathcal{L}}}{\partial {(\partial_\alpha A^\beta)}} \partial_\alpha \bar{A^\beta}\right)+ \cdots \end{aligned}

For the first term we have $a \cdot \nabla \int d^4 x \mathcal{L}( \bar{A}^\beta, \partial_\alpha \bar{A}^\beta)$, but this integral is our stationary action. The remainder, to first order in the field variables, can then be expanded and integrated by parts

\begin{aligned}\int &d^4 x a^\mu \partial_\mu \left(\frac{\partial {\mathcal{L}}}{\partial {A^\beta}} \bar{A^\beta}+\frac{\partial {\mathcal{L}}}{\partial {(\partial_\alpha A^\beta)}} \partial_\alpha \bar{A^\beta}\right) \\ &=\int d^4 x a^\mu \left(\left( \partial_\mu \frac{\partial {\mathcal{L}}}{\partial {A^\beta}} \right) \bar{A^\beta}+\frac{\partial {\mathcal{L}}}{\partial {A^\beta}} \left( \partial_\mu \bar{A^\beta} \right)+\left( \partial_\mu \frac{\partial {\mathcal{L}}}{\partial {(\partial_\alpha A^\beta)}} \right) \partial_\alpha \bar{A^\beta}+\frac{\partial {\mathcal{L}}}{\partial {(\partial_\alpha A^\beta)}} \left( \partial_\mu \partial_\alpha \bar{A^\beta} \right)\right) \\ &=\int d^4 x \left(\left( a^\mu \partial_\mu \frac{\partial {\mathcal{L}}}{\partial {A^\beta}} \right) \bar{A^\beta}-\left( \partial_\mu a^\mu \frac{\partial {\mathcal{L}}}{\partial {A^\beta}} \right)\bar{A^\beta} +\left( \partial_\mu \frac{\partial {\mathcal{L}}}{\partial {(\partial_\alpha A^\beta)}} \right) \partial_\alpha \bar{A^\beta}-\left( \partial_\mu a^\mu \frac{\partial {\mathcal{L}}}{\partial {(\partial_\alpha A^\beta)}} \right) \partial_\alpha \bar{A^\beta} \right) \\ \end{aligned}

Since $a^\mu$ are constants, this is zero, so there can be no contribution to the field equations by the addition of the translation increment to the Lagrangian.