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So I'm supposed to find that the improved energy momentum tensor of the scalar field $\phi$ satisfying the evolution equation $\Box \phi = 0$ is conserved. The improved energy momentum tensor is:

$T^{\mu\nu} = \mathcal{T}^{\mu\nu} + b(\partial^\mu \partial^\nu - g^{\mu\nu} \Box)\phi^2$

Where $b$ is a real constant. In other words: $\partial_\mu T^{\mu\nu} = 0$

Knowing that $\partial_\mu \mathcal{T}^{\mu\nu} = 0$ means that necessarily the derivative of the second term must be $0$. Here's how I go about solving this, I'll set $b =1 $ for simplicity

$\partial_\mu (\partial^\mu \partial^\nu - g^{\mu\nu} \Box)\phi^2 =(\Box \partial^\nu - g^{\mu\nu}\partial_\mu \Box)\phi^2 + (\partial^\mu \partial^\nu - g^{\mu\nu} \Box)2\phi\partial_\mu\phi = (\Box\partial^\nu - \partial^\nu \Box)\phi^2 + 2(\partial^\mu \partial^\nu \phi - g^{\mu\nu}\Box\phi)\partial_\mu \phi = 2(\partial^\mu \partial^\nu\phi\partial_\mu\phi)$

I really can't see where I'm wrong. I've tried it in different ways (first expanding the parenthesis term times $\phi^2$) and then deriving but I end up with the same result.

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  • $\begingroup$ Is this in Minkowski spacetime or in a curved spacetime? $\endgroup$ – Qmechanic Nov 22 '18 at 17:12
  • $\begingroup$ @Qmechanic Minkowski spacetime. $\endgroup$ – CMB Nov 24 '18 at 9:52
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$\partial_\mu (\partial^\mu \partial^\nu - g^{\mu\nu} \Box)\phi^2 =(\Box \partial^\nu - \partial^\nu \Box)\phi^2=0$

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Looks like I wasn't applying correctly the chain rule to the terms $\partial_\mu(\partial^\mu \phi^2)$. Now I believe it's alright. (Note that $\Box \phi = 0$ because the evolution equation)

$ \partial_\mu T^{\mu\nu} = \partial_\mu (\partial^\mu \partial^\nu \phi^2 - g^{\mu\nu}\Box \phi^2) \\ = \partial_\mu (2\partial^\mu (\phi \partial^\nu \phi) -2g^{\mu\nu} \partial_\alpha(\phi \partial^\alpha \phi))\\ = 2\partial^\mu (\partial^\mu \phi \partial^\nu \phi + \phi \partial^\mu \partial ^\nu \phi -g^{\mu\nu} (\partial_\alpha \phi \partial^\alpha \phi + \phi \Box \phi))\\ = 2(\partial_\mu (\partial^\mu \phi \partial^\nu \phi) +\partial_\mu (\phi \partial^\mu \partial^\nu \phi) - \partial^\nu (\partial_\alpha \phi \partial^\alpha \phi))\\ = 2(\Box\phi\partial^\nu \phi +\partial^\mu\phi \partial_\mu \partial^\nu \phi +\partial_\mu \phi \partial^\mu\partial^\nu \phi +\phi \partial^\nu \Box \phi - \partial^\nu \partial_\alpha \phi \partial^\alpha\phi -\partial_\alpha\phi \partial^\nu \partial^\alpha \phi)\\ = 0.$

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  • $\begingroup$ When you have two differential operators, $D_1$ and $D_2$, and a function $f$ that they operate on, then $D_1(D_2 f)$ is simply $(D_1 D_2)f$. It isn’t $(D_1 D_2)f + D_2(D_1 f)$ like you were doing in your question. You were applying the chain rule for a product as if a differential operator were one of the multiplicands. The term after the + sign shouldn’t have been there at all. You correctly applied the chain rule in your answer, to $\phi^2$, but this is unneccessary and obscures why the “improved” term is conserved. $\endgroup$ – G. Smith Nov 22 '18 at 17:16
  • $\begingroup$ Yes, absolutely, my bad. Thank you very much! @G.Smith $\endgroup$ – CMB Nov 22 '18 at 19:02

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