Variation of a Lagrange density Symmetries

Question

So I am reading Goenner's Spezielle Relativitästheorie and I am currently in chapter §4.9.1 Variation under Inclusion of Coordinates p. 129. So basically we have: $$\delta W_\zeta=\int d^4x' \mathcal{L}'(x')-\int d^4x \mathcal{L}(x).$$ Where the Variation results from the infinitesimal transformations through: $$x\rightarrow x'=x+\delta x=x+\zeta.$$ Now you can define a total variation: $$\delta u:=u'(x')-u(x)$$ and a local variation: $$\bar{\delta}u=u'(x)-u(x).$$ It follows that only the local variation commutes with the partial derivative. Now: $$d^4x'=(1+\zeta^\sigma_{,\sigma})d^4x$$ and $$\bar{\delta}\mathcal{L}=\frac{\partial \mathcal{L}}{\partial u}\bar{\delta}u+\frac{\partial \mathcal{L}}{\partial u_{,\alpha}}\bar{\delta}u_{,\alpha}.$$ Moreover: $$\mathcal{L}'(x')=\mathcal{L}'(x)+\frac{\partial \mathcal{L}'}{\partial x}\delta x=\mathcal{L}'(x)+\frac{\partial \mathcal{L}}{\partial x}\delta x.$$ Then the author writes that the variation can be written after using the commutation of the partial derivative with the local variation and partial integration as: $$\delta W_\zeta~=~\int d^4x\left[\bar\delta u \left(\frac{\partial \mathcal{L}}{\partial u}-\frac{\partial}{\partial x^\mu}\frac{\partial \mathcal{L}}{\partial u_{,\mu}}\right)+\left[\mathcal{L}(x)\delta x^\alpha+\frac{\partial \mathcal{L}}{\partial u_{,\alpha}}\left(\delta u - u_{,\beta}\delta x^\beta\right)\right]_{,\alpha}\right].$$

I tried to get there from the first equation, but didn't get anything close to this. Would really appreciate someone's help.

Answer 1

Comments to the question (v3):

1. OP's last formula is the standard expression $$ \delta W~=~ \int \! d^4x~ \left[ {\rm EL} \cdot\bar{\delta}u + d_{\mu} j^{\mu}\right], \tag{A} $$ for the variation of the action $W=\int\! d^4x~{\cal L}$ (= Wirkung in German). Here $$ j^{\mu} ~=~ p^{\mu} \cdot \bar{\delta}u + {\cal L} ~\delta x^{\mu}, \qquad p^{\mu}~:=~\frac{\partial {\cal L}}{\partial u_{,\mu}}, \tag{B} $$ is the bare Noether current, and ${\rm EL}$ is the Euler-Lagrange expression, so Goenner is certainly correct.

2. Hints for the derivation of eq. (A):$$ \tag{C}\delta W~=~ \int \! d^4x~ \left[\delta{\cal L}+ {\cal L}~ d_{\mu}\delta x^{\mu} \right], $$ $$ \delta {\cal L}~=~\bar{\delta} {\cal L} + \delta x^{\mu} ~d_{\mu}{\cal L} \tag{D} ,$$ and $$\bar{\delta} {\cal L} ~=~ {\rm EL} \cdot\bar{\delta}u + d_{\mu}[p^{\mu} \cdot \bar{\delta}u] \tag{E} .$$

Answer 2

Thanks for the hint. This is what I got so far:

$$\delta W=\int d^4x' \mathcal{L}'(x')-\int d^4x \mathcal{L}(x)$$

Now since $d^4x'=d^4x(1+(\delta x^\alpha)_{,\alpha})$ this equation reduces to:

$$\delta W=\int d^4x \mathcal{L}'(x')-\mathcal{L}(x)+(\delta x^\alpha)_{,\alpha}\mathcal{L'}(x')$$

Whic is just: $$\delta W=\int d^4x \delta\mathcal{L}+(\delta x^\alpha)_{,\alpha}(L+\delta L)$$ The second term in the second bracket can be ignored since it yields a second order term (as far as I know) so we get: $$\delta W=\int d^4x \delta\mathcal{L}+(\delta x^\alpha)_{,\alpha}L$$ Now since $\delta L = \frac{\partial\mathcal{L}}{\partial u}\delta u + \frac{\partial\mathcal{L}}{\partial u_{,\alpha}}\delta u_{,\alpha}+\frac{\partial \mathcal{L}}{\partial x^\alpha}\delta x^{\alpha}$ This becomes: $$\delta W=\int d^4x \frac{\partial\mathcal{L}}{\partial u}\delta u + \frac{\partial\mathcal{L}}{\partial u_{,\alpha}}\delta u_{,\alpha}+\frac{\partial \mathcal{L}}{\partial x^\alpha}\delta x^{\alpha}+(\delta x^\alpha)_{,\alpha}L$$ Now the last two terms can be termed together by reversing the product rule and using the reverse product rule on the second term yields: $$\delta W=\int d^4x \left(\frac{\partial\mathcal{L}}{\partial u}-\left(\frac{\partial \mathcal{L}}{\mathcal{u}_{,\alpha}}\right)_{,\alpha}\right)\delta u+(\frac{\partial\mathcal{L}}{\partial u_{,\alpha}}+\delta x^\alpha\mathcal{L})_{,\alpha}$$ This is exactly what Goenner got except for that last term. Could somebody please point out what went wrong? Is it that first order approximation? thanks