# 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

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

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

• 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. Mar 9, 2012 at 20:18
• Yes lengths are measured in units of $ct$ Mar 9, 2012 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? Mar 9, 2012 at 21:20

Notation: I will denote SI units between brackets, not to be confused with a multiplication.

Lets start by a well known case, the stress tensor of a perfect fluid:

$$T^{\alpha \beta} \, = \left(\rho + {p \over c^2}\right)u^{\alpha}u^{\beta} + p g^{\alpha \beta}$$

from it, we could say that stress energy tensor units are from momentum density ($$\rho u^\alpha$$) multiplied by velocity $$u^\beta$$ (momentum flow), or pressure $$p$$ multiplied by metric $$g^{\alpha\beta}$$.

• When spacetime basis is $$(ct,x,y,z)$$, all stress energy tensor components (in contravariant and covariant form) have the same units $$T^{\alpha\beta} \left[ \left( \frac{kg\,m}{s} \frac{1}{m^3} \right) \frac{m}{s} = \frac{kg}{ms^2} \right]$$ equal to pressure units or energy density [ $$\frac{J}{m^3}=Pa=\frac{kg}{ms^2}$$ ]. It corresponds to what is said in the book that is mentioned at the question.

• When spacetime basis is $$(t,x,y,z)$$ we have (assume $$i,j \in \{1,2,3\}$$):

$$T^{00} \left[kg/m^3\right]$$ : mass spatial density.

$$T^{0j}, T^{i0} \left[ \frac{kg}{m^3} \frac{m}{s} = \frac{kg}{m^2s} \right]$$ : momentum per unit of space volume.

$$T^{ij} \left[ \frac{kg}{m^3} \frac{m}{s} \frac{m}{s} = \frac{kg}{ms^2} \right]$$ : energy per unit of space volume.

These units corresponds to wikipedia paragraphs about stress energy tensor.

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.