The Divergence of the Klein-Gordon Energy-Momentum Tensor

[GooberGunter]

Homework Statement

Prove that the Energy-Momentum Tensor for the Klein-Gordon Equation does not diverge
$$\partial_\mu T^{\mu\nu}=0$$

Relevant Equations

$$T^{μν}=∂^μϕ∂^νϕ−η^{μν}L$$
$$L=\frac{1}{2}\partial^2 \phi - \frac{1}{2}m^2 \phi^2$$

I've tried this problem so, so, so so so many times. Given the equations above, the proof starts easily enough:

$$\partial_\mu T^{\mu\nu}=\partial_\mu (∂^μ ϕ∂^ν ϕ)-\eta^{\mu\nu}\partial_\mu[\frac{1}{2}∂^2ϕ−\frac{1}{2}m^2ϕ^2]$$
apply product rule to all terms
$$=\partial^\nu \phi \cdot \partial_\mu \partial^\mu \phi + \partial^\mu \phi \cdot \partial_\mu \partial^\nu \phi - \eta^{\mu\nu}[\frac{1}{2}(\partial_\sigma \phi \cdot \partial_\mu \partial^\sigma \phi + \partial^\sigma \phi \cdot \partial_\mu \partial_\sigma \phi)-\frac{1}{2}m^2\partial_\mu(\phi^2)]$$
And the equation of motion starts to appear:
$$=\partial^\nu \phi \cdot \partial_\mu \partial^\mu \phi + \eta^{\mu\nu}(m^2 \phi \partial_\mu \phi) +\partial^\mu \phi \cdot \partial_\mu \partial^\nu \phi - \eta^{\mu\nu}\frac{1}{2}(\partial_\sigma \phi \cdot \partial_\mu \partial^\sigma \phi + \partial^\sigma \phi \cdot \partial_\mu \partial_\sigma \phi)$$
Finally we can eliminate 2 of the 5 terms
$$=\partial^\nu \phi \cdot (\partial_\mu \partial^\mu \phi + m^2 \phi) +\partial^\mu \phi \cdot \partial_\mu \partial^\nu \phi - \eta^{\mu\nu}\frac{1}{2}(\partial_\sigma \phi \cdot \partial_\mu \partial^\sigma \phi+ \partial^\sigma \phi \cdot \partial_\mu \partial_\sigma \phi)$$
$$=\partial^\nu \phi[0] +\partial^\mu \phi \cdot \partial_\mu \partial^\nu \phi - \eta^{\mu\nu}\frac{1}{2}(\partial_\sigma \phi \cdot \partial_\mu \partial^\sigma \phi + \partial^\sigma \phi \cdot \partial_\mu \partial_\sigma \phi)$$

This is the last step I arrive at
$$=\partial^\mu \phi \cdot \partial_\mu \partial^\nu \phi - \eta^{\mu\nu}\frac{1}{2}(\partial_\sigma \phi \cdot \partial_\mu \partial^\sigma \phi + \partial^\sigma \phi \cdot \partial_\mu \partial_\sigma \phi)$$
​

Every solution I've found online has an error or skips this crucial step.

This solutions manual from a UMD course skips a step between (31) and (32) and too readily removes the 1/2 coefficient in (31). I'm sure its a typo.
This solution from another post just raises the index of the first derivative and lowers the sigma index in the second derivative, but the closest I can replicate is:

$$=\partial^\mu \phi \cdot \partial_\mu \partial^\nu \phi - \frac{1}{2}(\eta^{\mu\nu} \delta^{\sigma}_\nu\partial_\sigma \phi \cdot \partial_\mu \partial^\sigma \phi + \partial^\sigma \phi \cdot \partial_\mu \eta^{\mu\nu} \delta^{\sigma}_\nu \partial_\sigma \phi)$$
$$=\partial^\mu \phi \cdot \partial_\mu \partial^\nu \phi - \frac{1}{2}(\partial^\nu \phi \cdot \partial_\mu \partial^\nu\phi + \partial^\nu \phi \cdot \partial_\mu \partial^\nu \phi)$$

Depending on what you decide to contract, you end up with 2 equations that don't resolve to 0.
$$\partial^\mu \phi \cdot \partial_\mu \partial^\nu \phi -\partial^\nu \phi \cdot \partial_\mu \partial^\nu\phi$$
$$\partial^\mu \phi \cdot \partial_\mu \partial^\nu \phi - \partial^\mu \phi \cdot \partial_\mu \partial^\mu \phi$$

My understanding of tensor notation is still a little shaky, I only studied a chapter from a mathematical methods textbook. Where did I go wrong? Any help is greatly appreciated!

 

[TSny]

This is the last step I arrive at

$$=\partial^\mu \phi \cdot \partial_\mu \partial^\nu \phi - \eta^{\mu\nu}\frac{1}{2}(\partial_\sigma \phi \cdot \partial_\mu \partial^\sigma \phi + \partial^\sigma \phi \cdot \partial_\mu \partial_\sigma \phi)$$​

I think this is good so far.

Show that the first term may be written $$\partial^\mu \phi \cdot \partial_\mu \partial^\nu \phi =\partial_\mu \phi \cdot \partial^\mu \partial^\nu \phi =\partial_\sigma \phi \cdot \partial^\sigma \partial^\nu \phi$$

Show that the second term may be written $$ \eta^{\mu\nu}\frac{1}{2}(\partial_\sigma \phi \cdot \partial_\mu \partial^\sigma \phi + \partial^\sigma \phi \cdot \partial_\mu \partial_\sigma \phi) = \frac{1}{2}(\partial_\sigma \phi \cdot \partial^\nu \partial^\sigma \phi + \partial^\sigma \phi \cdot \partial^\nu \partial_\sigma \phi)$$ Then proceed onward.