So I am reading Goenner's Spezielle Relativitästheorie and I am currently in chapter §4.9.1 Variation under Inclusion of Coordinates: So basically we have: $$\delta W_\zeta=\int d^4x' \mathcal{L}'(x')-\int d^4x \mathcal{L}(x)=0$$ Where the Variation results from the Symmetry 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 $$\bar{\delta}\mathcal{L}=\frac{\partial \mathcal{L}}{\partial u}\bar{\delta}u+\frac{\partial \mathcal{L}}{\partial u_{,\alpha}}\bar{\delta}u_{,\alpha}$$ Moroever: $$\mathcal{L}'(x')=\mathcal{L}'(x)+\frac{\partial \mathcal{L}'}{\partial x}\delta x=\mathcal{L}'(x)+\frac{\partial \mathcal{L}}{\partial x}\delta x$$ 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: $$\int d^4x\left[\bar\delta u \left(\frac{\partial \mathcal{L}}{\partial u}-\frac{\partial}{\partial x^\mu}\frac{\partial \mathcal{L}}{\partial u_{,\mu}}\right)+\left[\mathcal{L}(x)\delta x^\alpha+\frac{\partial \mathcal{L}}{\partial u_{,\alpha}}\left(\delta u - u_{,\beta}\delta x^\beta\right)\right]_{,\alpha}\right]$$

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

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.

So I am reading Goenner's Spezielle Relativitästheorie and I am currently in chapter §4.9.1 Variation under Inclusion of Coordinates: So basically we have: $$\delta W_\zeta=\int d^4x' \mathcal{L}'(x')-\int d^4x \mathcal{L}(x)=0$$ Where the Variation results from the Symmetry 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 $$\bar{\delta}\mathcal{L}=\frac{\partial \mathcal{L}}{\partial u}\bar{\delta}u+\frac{\partial \mathcal{L}}{\partial u_{,\alpha}}\bar{\delta}u_{,\alpha}$$ Moroever: $$\mathcal{L}'(x')=\mathcal{L}'(x)+\frac{\partial \mathcal{L}'}{\partial x}\delta x=\mathcal{L}'(x)+\frac{\partial \mathcal{L}}{\partial x}\delta x$$ 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: $$\int d^4x\left[\bar\delta u \left(\frac{\partial \mathcal{L}}{\partial u}-\frac{\partial}{\partial x^\mu}\frac{\partial \mathcal{L}}{\partial u_{,\mu}}\right)+\left[\mathcal{L}(x)\delta x^\alpha+\frac{\partial \mathcal{L}}{\partial u_{,\alpha}}\left(\delta u - u_{,\beta}\delta x^\beta\right)\right]_{,\alpha}\right]$$

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

Thanks

So I am reading Goenner's Spezielle Relativitästheorie and I am currently in chapter §4.9.1 Variation under Inclusion of Coordinates: So basically we have: $$\delta W_\zeta=\int d^4x' \mathcal{L}'(x')-\int d^4x \mathcal{L}(x)=0$$ Where the Variation results from the Symmetry 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 $$\bar{\delta}\mathcal{L}=\frac{\partial \mathcal{L}}{\partial u}\bar{\delta}u+\frac{\partial \mathcal{L}}{\partial u_{,\alpha}}\bar{\delta}u_{,\alpha}$$ Moroever: $$\mathcal{L}'(x')=\mathcal{L}'(x)+\frac{\partial \mathcal{L}'}{\partial x}\delta x=\mathcal{L}'(x)+\frac{\partial \mathcal{L}}{\partial x}\delta x$$ 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: $$\int d^4x\left[\bar\delta u \left(\frac{\partial \mathcal{L}}{\partial u}-\frac{\partial}{\partial x^\mu}\frac{\partial \mathcal{L}}{\partial u_{,\mu}}\right)+\left[\mathcal{L}(x)\delta x^\alpha+\frac{\partial \mathcal{L}}{\partial u_{,\alpha}}\left(\delta u - u_{,\beta}\delta x^\beta\right)\right]_{,\alpha}\right]$$

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