Deriving the Canonical Energy Momentum Tensor

Question

In the Mathematics for Physics of Stone and Goldbart the canonical energy momentum tensor is derived by the action principle as follows.

To the action of the form

$$ S=\int \mathcal{L}(\varphi,\varphi_\mu) \, \mathrm{d}^{d+1}x ,$$

where $\mathcal{L}$ the lagrangigan density and $\varphi_\mu = \frac{\partial \varphi}{\partial x^\mu}$ is, we make the variation of the form

$$ \varphi(x) \rightarrow \varphi(x^\mu + \varepsilon^\mu(x)) = \varphi (x^\mu) + \varepsilon^\mu(x)\partial_\mu\varphi+O(|\varepsilon|^2) ,$$ where $x=(x^0,...,x^d)$ is.

Then the resulting variation is

$$\delta S= \int \left( \frac{\partial\mathcal{L}}{\partial \varphi} \varepsilon^\mu \partial_\mu \varphi +\frac{\partial \mathcal{L}}{\partial \varphi_\nu} \partial_\nu(\varepsilon^\mu \partial_\mu \varphi) \right)\mathrm{d}^{d+1}x$$ $$ = \int \varepsilon^\mu \frac{\partial}{\partial x^\nu} \left( \mathcal{L} \delta^\nu_\mu -\frac{\partial \mathcal{L}}{\partial \varphi_\nu} \partial_\mu \varphi \right)\, \mathrm{d}^{d+1}x. $$

I understand that from going line 1 to 2 some sort of integration by parts is done. However when I try to do that I run into some trouble and don't get the second line. Can someone do it explicityl so that I learn where I made the mistake.

EDIT

My steps are

$$ \delta S=\int_\Omega \frac{\delta S}{\delta \varphi(x)} \mathrm{d}\Omega, $$ where $\Omega$ to be integrated region is.

$$=\int_\Omega \delta\varphi\left( \frac{\partial \mathcal{L}}{\partial \varphi} - \partial_\nu \frac{\partial\mathcal{L}}{\partial \varphi_\nu} \right) \mathrm{d}\Omega$$

$$=\int_\Omega \varepsilon^\mu \frac{\partial \varphi}{\partial x^\mu} \left( \frac{\partial \mathcal{L}}{\partial \varphi} - \partial_\nu \frac{\partial\mathcal{L}}{\partial \varphi_\nu} \right) \mathrm{d}\Omega \qquad \because \delta\varphi= \varepsilon^\mu \partial_\mu \varphi $$

$$=\int_\Omega \varepsilon^\mu \left( \frac{\partial \mathcal{L}}{\partial x^\mu} - \partial_\nu \frac{\partial\mathcal{L}}{\partial \varphi_\nu}\cdot \frac{\partial \varphi}{\partial x^\mu} \right) \mathrm{d}\Omega$$

Now I can take out $\partial_\nu$ however $\partial_\nu$ acts only on $\frac{\partial\mathcal{L}}{\partial \varphi_\nu}$ and not on $\frac{\partial \varphi}{\partial x^\mu}$. I thought $\partial_\nu\frac{\partial \varphi}{\partial x^\mu}$ could be zero so that I can take the partial derivative out of the bracket but I don't see why this should be true. If I were to take it out I arrive at the equation in the second line above.

Answer 1

Note that the chain rule in this case, since $\mathcal{L} = \mathcal{L}(\varphi, \partial_{\mu} \varphi)$ reads

$$\partial_{\nu}\mathcal{L} = \frac{\partial \mathcal{L}}{\partial \varphi} \partial_{\nu} \varphi + \frac{\partial{\mathcal{L}}}{\partial(\partial_{\mu} \varphi)}\partial_{\nu} \partial_{\mu} \varphi.$$

Forgetting the second term is the mistake I think you are making.