Deriving energy-momentum tensor in Schwartz QFT

Question

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 $\partial_\nu L = \partial_\mu g_{\mu \nu} L$.

3. 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.

Answer 1

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.

2. 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 $\mathrm{d}f = \nabla f \cdot \mathrm{d}\mathbf{r} = \partial_x f \,\mathrm{d}x + \dots$

3. 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 $\mu$...

Answer 2

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