# Deriving energy-momentum tensor in Schwartz QFT

I don't understand how to derive the chain of equalities right after equation (3.31):

We are told $$\frac{\delta \phi}{\delta \xi^\nu} = \partial_\nu \phi\tag{3.30}$$ and $$\frac{\delta \mathcal{L}}{\delta \xi^\nu} = \partial_\nu \mathcal{L}\tag{3.31},$$ and I'm confused about the line immediately following it:

"Since this is a total derivative, $$\delta S = \int d^4 x \delta \mathcal{L} = \xi^\nu \int d^4 x \partial_\nu \mathcal{L} = 0 \ldots$$"

1. We have $$\frac{\delta \mathcal{L}}{\delta \xi^\nu} = \partial_\nu \mathcal{L} \rightarrow \delta \mathcal{L} = \partial_\mu \mathcal{L} \delta \xi^\nu$$, so why don't we have $$\int d^4 x \delta \mathcal{L} = \delta \xi^\nu \int d^4 x \delta_\mu \mathcal{L}$$ instead of $$\int d^4 x \delta \mathcal{L} = \xi^\nu \int d^4 x \delta_\mu \mathcal{L}$$?

2. Why is $$\partial_\nu \mathcal{L}$$ a total derivative? Isn't it a partial derivative $$(\partial_t, \partial_x, \partial_y, \partial_z)$$?

Later in the page, we go from $$\partial_\nu L = \partial_\mu \left(\sum_n \frac{\partial L}{\partial(\partial_\mu \phi_n)} \partial_\nu \phi_n \right)\tag{3.33}$$ to $$\partial_\mu \left(\sum_n \frac{\partial L}{\partial(\partial_\mu \phi_n)} \partial_\nu \phi_n - g_{\mu \nu} L\right) = 0,\tag{3.34}$$ which seems to suggest $$\partial_\nu L = \partial_\mu g_{\mu \nu} L$$.

1. Why is $$\partial_\nu L = \partial_\mu g_{\mu \nu} L$$ true? I realize it looks like a contraction, but if someone could write out the mathematical details that would be super helpful.

1. We have $$\frac{\delta \mathcal{L}}{\delta \xi^\nu} = \partial_\nu \mathcal{L} \rightarrow \delta \mathcal{L} = \partial_\mu \mathcal{L} > \delta \xi^\nu$$, so why don't we have $$\int d^4 x \delta \mathcal{L} = > \delta \xi^\nu \int d^4 x \delta_\mu \mathcal{L}$$ instead of $$\int d^4 > x \delta \mathcal{L} = \xi^\nu \int d^4 x \delta_\mu \mathcal{L}$$?

Once you "break" the differentiation fraction, and you turn $$\delta \mathcal{L}$$ and $$\delta \xi$$ into finite differences ($$\neq$$ infinitesimal), then they are just variables. Just rename $$\delta\xi^\nu \rightarrow \xi^\nu$$. It's like renaming $$\delta x\rightarrow x$$ in a Taylor expansion.

1. Why is $$\partial_\nu \mathcal{L}$$ a total derivative? Isn't it a partial derivative $$(\partial_t, \partial_x, \partial_y, \partial_z)$$?

$$\partial_\nu \mathcal{L}$$ on its own is just a component of the four-dimensional derivative, as you are pointing out.

But the differential $$\delta \mathcal{L}$$ is written as a contraction $$\xi^\nu \partial_\nu \mathcal{L}$$ which involves all partial derivatives and is hence a total derivative. This is the same as the usual $$\mathrm{d}f = \nabla f \cdot \mathrm{d}\mathbf{r} = \partial_x f \,\mathrm{d}x + \dots$$

1. Why is $$\partial_\nu L = \partial_\mu g_{\mu \nu} L$$ true? I realize it looks like a contraction, but if someone could write out the mathematical details that would be super helpful.

You need to change $$\partial_\nu$$ into $$\partial_\mu$$.

You can contract with the metric once to get $$g^{\lambda\nu}\partial_\nu = \partial^\lambda$$ to get a contravariant derivative, and then contract again to get it covariant: $$g_{\mu \lambda}\partial^\lambda = \partial_\mu$$.

Combined: $$\partial_\mu = g_{\mu \lambda}g^{\lambda\nu }\partial_\nu = \delta^{\nu}_ {\mu}\partial_\nu.$$

I realise now, though, that I am not quite sure how to turn the last line into $$g_{\mu \nu}$$... it might be a typo though (?) because the other term has a contracted $$\mu$$ index, so this one cannot have a free $$\mu$$...

Equation 3.33 or its equivalent 3-34 is simply the Euler Lagrange equation for your system The expression 3-34 is the quadridivergence of the moment energy tensor, it is null indicating the energy moment conservation law