# Fourier transform of the energy momentum tensor

I am trying to calculate the Fourier transform of the energy momentum tensor of a scalar field, in particular the first term:

$$$$T_{\mu \nu}(x) = \frac{1}{2} \partial_{\mu} \phi(x) \partial_{\nu} \phi(x) + etc$$$$ and first of all this is symmetric in $$\mu$$ and $$\nu$$. Now taking the fourier transform of the field:

$$$$\phi(x)=\int d^4k \exp (-ikx) \phi (k)$$$$ we have

$$$$T_{\mu \nu}(x) = \int d^4k \exp (-ikx) \phi (k) \int d^4q \exp (-iqx) \phi (q) k_{\mu} q_{\nu}$$$$ and already I don't see if this is symmetric in the indices, do I have to make it symmetric?

Then how do I obtain the energy momentum tensor in the momentum space? Because I want to write the previous equation in a form like

$$$$T_{\mu \nu} (x)=\int d^4k \exp (-ikx) T_{\mu \nu}(k)$$$$ to recognize $$T_{\mu \nu}(k)$$ but I have an "extra" integral.

• Perhaps think $p_{\mu} q_{\nu} \equiv 1/2 \, (p_{\mu} q_{\nu} + p_{\nu} q_{\mu})$ following QM postulates and consider that the original $T$ does depend on two operators $\hat{p}$ represented in your $\partial_{\mu}$ and the same for $q.$ Commented Feb 25, 2021 at 23:19

1. Since $$T_{\mu \nu}(x)$$ is symmetric in $$\mu$$ and $$\nu$$, the RHS is also symmetric by default. If it is not manifestly symmetric (like in your example), you can always symmetrize it over the corresponding indices. So in your integral, change $$k_\mu q_\nu \to \frac{k_\mu q_\nu+k_\nu q_\mu}{2}$$.
2. In order to find $$T_{\mu \nu} (p)$$ (where $$p=$$ momentum) from $$T_{\mu \nu} (x)$$, you need to perform the inverse Fourier transform
$$T_{\mu \nu} (p) = \frac{1}{(2 \pi)^4}\int d^4 x \ e^{ipx} T_{\mu \nu} (x).$$
Plug $$T_{\mu \nu} (x)$$ from your formula into the above equation and perform the integral over $$x$$ to get a momentum-conserving delta function. Can you see this?