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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

Peter4075
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3 Answers3

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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

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

Vijay Murthy
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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 09 '12 at 20:18
  • Yes lengths are measured in units of $ct$ – Vijay Murthy Mar 09 '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 09 '12 at 21:20
2

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.

pasaba por aqui
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0

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.

John Rennie
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