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Homework Statement:

a) Given the Lagrangian density
$$\mathcal{L} = \frac 1 2 (\partial_{\mu} A_{\nu})( \partial^{\mu} A^{\nu})$$
Show that the EnergyMomentum tensor is
$$T_{\mu \nu} = \partial_{\mu} A_{\rho} \partial_{\nu}A^{\rho} + \frac 1 2 \eta_{\mu \nu} \partial_{\rho} A_{\sigma} \partial^{\rho} A^{\sigma}$$
b) Show that ##\partial_{\nu} T^{\mu \nu} = 0##
Relevant Equations:

The given Lagrangian
$$\mathcal{L} = \frac 1 2 \partial_{\mu} A_{\nu} \partial^{\mu} A^{\nu}$$
The Energymomentum tensor to show
$$T^{\mu \nu} = \partial_{\mu} A_{\rho} \partial_{\nu}A^{\rho} + \frac 1 2 \eta_{\mu \nu} \partial_{\rho} A_{\sigma} \partial^{\rho} A^{\sigma}$$
We also have to show that ##\partial_{\nu} T^{\mu \nu} =0##
REMARK: First of all I have to say that this Lagrangian reminds me of the Lagrangian from which we can derive Maxwell's equations, which is (reference: Tong QFT lecture notes, equation 1.18; I have attached the PDF).
$$\mathcal{L} = \frac 1 2 (\partial_{\mu} A_{\nu} )(\partial^{\mu} A^{\nu}) + \frac 1 2(\partial_{\mu} A^{\mu})^2$$
I guess here we are dealing with a simplified version of it:
$$\mathcal{L} = \frac 1 2 (\partial_{\mu} A_{\nu})( \partial^{\mu} A^{\nu})$$
a) At first glace we see that the given Lagrangian is invariant under Lorentz transformations because the Lorentz indices ##\mu## and ##\nu## are contracted with the metric (which is a Lorentz invariant object); let's show this explicitely:
$$\mathcal{L} = \frac 1 2 (\eta_{\mu \lambda}\partial^{\lambda} A_{\nu})( \eta^{\mu \rho} \partial_{\rho} A^{\nu})$$
Beautiful! OK that's good because we already know that, by Noether's theorem, there has to be a conserved current associated to the transformation we decide to apply.
I have studied that the Energymomentum tensor is defined to be the conserved current obtained from applying the following transformation: translations. Thus the logic step now is to apply translation to the given Lagrangian to get the current and then compute our specific case.
The general formula for the current ##j^{\mu}## when ##\delta \mathcal{L} = 0## is given by
$$j^{\mu} = \frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\phi_a)}\partial \phi_a$$
I wanted to be careful here and I thought of working with the (I think) equivalent Lagrangian
$$\mathcal{L} = \frac 1 2 (\partial_{\mu} \phi)( \partial^{\mu} \phi)$$
I did the following: first I focused on computing ##\frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\phi)}##
$$\frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\phi)} = \frac 1 2 \eta^{\rho \sigma} \frac{\partial (\partial_{\rho} \phi \partial_{\sigma} \phi)}{\partial(\partial_{\mu} \phi)} = \frac 1 2 \eta^{\rho \sigma} (\delta_{\rho}^{\mu} \partial_{\sigma} \phi + \delta_{\sigma}^{\mu} \partial_{\rho} \phi)$$
Where, note, I carefully swapped ##\rho## by ##\mu## in the original Lagrangian so we can avoid any index confusions
OK, so noticing that we can contract the metric like ##\eta^{\rho \sigma} \delta_{\rho}^{\mu}## I get
$$\frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\phi)} = \frac 1 2 (\eta^{\mu \sigma} \partial_{\sigma} \phi + \eta^{\mu \rho} \partial_{\rho} \phi )$$
We're just left multiplying by ##\partial \phi## to get the current
$$j^{\mu} = \frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\phi)}\partial \phi = \frac 1 2 (\eta^{\mu \sigma} \partial_{\sigma} \phi + \eta^{\mu \rho} \partial_{\rho} \phi )\partial \phi$$
Thus the EnergyMomentum tensor is
$$T^{\mu}_{ \ \nu} = j^{\mu}_{ \ \nu} = \frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\phi)}\partial_{\nu} \phi = \frac 1 2 (\eta^{\mu \sigma} \partial_{\sigma} \phi + \eta^{\mu \rho} \partial_{\rho} \phi )\partial_{\nu} \phi$$
Now the idea would be to generalize this result and be able to get the desired tensor. I do not see how to do so...
How can I proceed from here?
Notice that I am not sure of the equivalence of the following Lagrangians
$$\mathcal{L} = \frac 1 2 (\partial_{\mu} \phi)( \partial^{\mu} \phi)$$
$$\mathcal{L} = \frac 1 2 (\partial_{\mu} A_{\nu})( \partial^{\mu} A^{\nu})$$
I am sure there is a straightforward way to solve a) anyways.
b) To be honest I do not know how to actually prove that ##\partial_{\nu} T^{\mu \nu} = 0##. In the notes I am studying (Tong, page 15 example 1.3.2) he asserts that it does satisfy it but he doesn't prove it explicitly.
Could you please give me a hint on this one?
Thank you.
$$\mathcal{L} = \frac 1 2 (\partial_{\mu} A_{\nu} )(\partial^{\mu} A^{\nu}) + \frac 1 2(\partial_{\mu} A^{\mu})^2$$
I guess here we are dealing with a simplified version of it:
$$\mathcal{L} = \frac 1 2 (\partial_{\mu} A_{\nu})( \partial^{\mu} A^{\nu})$$
a) At first glace we see that the given Lagrangian is invariant under Lorentz transformations because the Lorentz indices ##\mu## and ##\nu## are contracted with the metric (which is a Lorentz invariant object); let's show this explicitely:
$$\mathcal{L} = \frac 1 2 (\eta_{\mu \lambda}\partial^{\lambda} A_{\nu})( \eta^{\mu \rho} \partial_{\rho} A^{\nu})$$
Beautiful! OK that's good because we already know that, by Noether's theorem, there has to be a conserved current associated to the transformation we decide to apply.
I have studied that the Energymomentum tensor is defined to be the conserved current obtained from applying the following transformation: translations. Thus the logic step now is to apply translation to the given Lagrangian to get the current and then compute our specific case.
The general formula for the current ##j^{\mu}## when ##\delta \mathcal{L} = 0## is given by
$$j^{\mu} = \frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\phi_a)}\partial \phi_a$$
I wanted to be careful here and I thought of working with the (I think) equivalent Lagrangian
$$\mathcal{L} = \frac 1 2 (\partial_{\mu} \phi)( \partial^{\mu} \phi)$$
I did the following: first I focused on computing ##\frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\phi)}##
$$\frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\phi)} = \frac 1 2 \eta^{\rho \sigma} \frac{\partial (\partial_{\rho} \phi \partial_{\sigma} \phi)}{\partial(\partial_{\mu} \phi)} = \frac 1 2 \eta^{\rho \sigma} (\delta_{\rho}^{\mu} \partial_{\sigma} \phi + \delta_{\sigma}^{\mu} \partial_{\rho} \phi)$$
Where, note, I carefully swapped ##\rho## by ##\mu## in the original Lagrangian so we can avoid any index confusions
OK, so noticing that we can contract the metric like ##\eta^{\rho \sigma} \delta_{\rho}^{\mu}## I get
$$\frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\phi)} = \frac 1 2 (\eta^{\mu \sigma} \partial_{\sigma} \phi + \eta^{\mu \rho} \partial_{\rho} \phi )$$
We're just left multiplying by ##\partial \phi## to get the current
$$j^{\mu} = \frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\phi)}\partial \phi = \frac 1 2 (\eta^{\mu \sigma} \partial_{\sigma} \phi + \eta^{\mu \rho} \partial_{\rho} \phi )\partial \phi$$
Thus the EnergyMomentum tensor is
$$T^{\mu}_{ \ \nu} = j^{\mu}_{ \ \nu} = \frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\phi)}\partial_{\nu} \phi = \frac 1 2 (\eta^{\mu \sigma} \partial_{\sigma} \phi + \eta^{\mu \rho} \partial_{\rho} \phi )\partial_{\nu} \phi$$
Now the idea would be to generalize this result and be able to get the desired tensor. I do not see how to do so...
How can I proceed from here?
Notice that I am not sure of the equivalence of the following Lagrangians
$$\mathcal{L} = \frac 1 2 (\partial_{\mu} \phi)( \partial^{\mu} \phi)$$
$$\mathcal{L} = \frac 1 2 (\partial_{\mu} A_{\nu})( \partial^{\mu} A^{\nu})$$
I am sure there is a straightforward way to solve a) anyways.
b) To be honest I do not know how to actually prove that ##\partial_{\nu} T^{\mu \nu} = 0##. In the notes I am studying (Tong, page 15 example 1.3.2) he asserts that it does satisfy it but he doesn't prove it explicitly.
Could you please give me a hint on this one?
Thank you.
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