# Finding the correct units for the energy-momentum tensor?

I'm trying to understand the energy-momentum tensor $T^{\mu\nu}$ but I'm confused about the units. My textbook says the components of $T^{\mu\nu}$ are $\mathrm{Jm^{-3}}$. Four-momentum is is given by$$P^{\mu}=\left(E/c,\mathbf{p}\right)=\left(E/c,p_{x},p_{y},p_{z}\right)$$

The $E/c$ component of $P^{\mu}$ has units $\mathrm{Jsm^{-1}}$. The definition of $T^{\mu\nu}$ is “the rate of flow of the $\mu$ component of four-momentum across a surface of constant $\nu$.” Using this definition, how do you get the rate of flow of the $E/c$ component of four-momentum across a surface of constant time (ie the $T^{00}$ component) to have the correct units of $\mathrm{Jm^{-3}}$? Surely you need to multiply $E/c$ by something with $\mathrm{s^{-1}m^{-2}}$ units, but what exactly?

Thank you

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You say

The definition of $T^{\mu\nu}$ is “the rate of flow of the $\mu$ component of four-momentum across a surface [of unit area] of constant $\nu$.”

which means that the dimensions of $T^{\mu\nu}$ ought to be

$$\frac{Momentum}{Area \, \times \, Time.}$$

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Thanks. Just to be clear. Are you saying the units of area are $m^{2}$ even when the surface is time, ie for all four $T^{\mu0}$ components? Than makes sense if time is measured in $ct\equiv m$ units. If that's right, then that's clearer. –  Peter4075 Mar 9 '12 at 20:18
Yes lengths are measured in units of $ct$ –  Vijay Murthy Mar 9 '12 at 21:11
Thanks, though I find it hard to visualise something (the four components of four-momentum) flowing across a surface of unit time per unit time? –  Peter4075 Mar 9 '12 at 21:20

According to http://en.wikipedia.org/wiki/Stress-energy_tensor#Identifying_the_components_of_the_tensor $T_{00}$ is the density of relativistic mass not energy, so you have to divide by $c^2$.

For the momentum components:

$$\frac{dp^\alpha}{dV} = -T^\alpha _\beta u^\beta$$

giving you units of density.

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