# Conservation of improved energy momentum tensor of a real massless scalar field

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.

• Is this in Minkowski spacetime or in a curved spacetime? – Qmechanic Nov 22 '18 at 17:12
• @Qmechanic Minkowski spacetime. – CMB Nov 24 '18 at 9:52

$$\partial_\mu (\partial^\mu \partial^\nu - g^{\mu\nu} \Box)\phi^2 =(\Box \partial^\nu - \partial^\nu \Box)\phi^2=0$$
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)
$$\partial_\mu T^{\mu\nu} = \partial_\mu (\partial^\mu \partial^\nu \phi^2 - g^{\mu\nu}\Box \phi^2) \\ = \partial_\mu (2\partial^\mu (\phi \partial^\nu \phi) -2g^{\mu\nu} \partial_\alpha(\phi \partial^\alpha \phi))\\ = 2\partial^\mu (\partial^\mu \phi \partial^\nu \phi + \phi \partial^\mu \partial ^\nu \phi -g^{\mu\nu} (\partial_\alpha \phi \partial^\alpha \phi + \phi \Box \phi))\\ = 2(\partial_\mu (\partial^\mu \phi \partial^\nu \phi) +\partial_\mu (\phi \partial^\mu \partial^\nu \phi) - \partial^\nu (\partial_\alpha \phi \partial^\alpha \phi))\\ = 2(\Box\phi\partial^\nu \phi +\partial^\mu\phi \partial_\mu \partial^\nu \phi +\partial_\mu \phi \partial^\mu\partial^\nu \phi +\phi \partial^\nu \Box \phi - \partial^\nu \partial_\alpha \phi \partial^\alpha\phi -\partial_\alpha\phi \partial^\nu \partial^\alpha \phi)\\ = 0.$$
• 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. – G. Smith Nov 22 '18 at 17:16