Translational invariance and the derivation of the energy-momentum tensor

Question

Consider a field $\phi$ which transforms as $\phi\rightarrow\phi+\delta\phi$ and say $X\left(\phi\right)=\delta\phi$ is a symmetry of your Lagrangian, which under that transformation changes by a total derivative $\delta L=\partial_{\mu}F^{\mu}$. Noether's theorem tells us there's a conserved current given by,

$$ j^{\mu}=\frac{\partial L}{\partial\left(\partial_{\mu}\phi\right)}X\left(\phi\right)-F^{\mu}\left(\phi\right). $$

Now, when considering spacetime translations we have

$$ x^{\nu}\rightarrow x^{\nu}-\varepsilon^{\nu} $$

$$ \phi\left(x\right)\rightarrow\phi\left(x\right)+\varepsilon^{\nu}\partial_{\nu}\phi\left(x\right) $$

$$ L\rightarrow L+\varepsilon^{\nu}\partial_{\nu}L\left(x\right) $$

$$ \left(j^{\mu}\right)_{\nu}=\frac{\partial L}{\partial\left(\partial_{\mu}\phi\right)}\partial_{\nu}\phi\left(x\right)-\delta_{\nu}^{\mu}L\equiv T_{\,\nu}^{\mu} $$

I'm following these well-known QFT notes by D. Tong, page 14. The thing I didn't understand in this derivation is how the last term, $\delta_{\nu}^{\mu}L$, comes about. I don't understand how $\partial_{\nu}L\left(x\right)$ becomes a $\delta_{\nu}^{\mu}L$. This is most certainly something very simple but it's driving me crazy. Can you help me clarify this?

Answer 1

It appears because you have four directions for possible translations. So Noether's theorem tells you you will have four currents. If you do the derivation you will realize that before the last equation you wrote, everything will be contracted with $\epsilon^\nu$, but since this translation is arbitrary, you will be allowed to factor it away. Then as you can see from the previous equation: $$L\rightarrow L + \epsilon^\nu \partial_\nu L = L + \epsilon^\nu \delta_\nu^\mu \partial_\mu L$$ so you get that in Tong's notation $F^\mu_\nu(x) = \delta^\mu_\nu L(x)$.