I am confused by a statement in Sean Carroll's "Spacetime and geometry".

On page 174-175, he makes the following statement about the physical interpretation of the WEC in the case of a perfect fluid:

"Because the pressure is isotropic, $T_{\mu\nu}t^\mu t^\nu$ will be non-negative for all timelike vectors $t^\mu$ if both $T_{\mu\nu}U^\mu U^\nu \geq 0$ and $T_{\mu\nu}l^\mu l^\nu \geq 0$ for some nullvector $l^\mu$ ".

$T_{\mu\nu}$ is the energy momentum tensor and $U^\mu$ is the four velocity. For a perfect fluid, the two last inequalities respectively reduce to $\rho \geq 0$ and $\rho + p \geq 0$.

I do understand that both inequalities are necessary conditions for the WEC ( $T_{\mu\nu}t^\mu t^\nu \geq 0$) to hold, but they are not obviously sufficient conditions.

They are necessary conditions as $U^\mu$ is a timelike vector and, by continuity, the inequality involving $l^\mu$ should hold as well. However, they are treated as sufficient conditions, i.e. the WEC is used as a synonym for "$\rho \geq 0$ and $\rho + p \geq 0$".

In a metric that does not differ too much from a Minkowskian metric (weak field limit), we can decompose any timelike vector $t^\mu$ as a sum of a well chosen nullvector $l^\mu$ and a multiple of $U^\mu$ or $$t^\mu = l^\mu+ \lambda U^\mu$$ with $\lambda$ a real number. This directly leads to $$T_{\mu\nu}t^\mu t^\nu = \rho (\lambda -U_\mu l^\mu )^2 +p(U_\mu l^\mu)^2$$. This last equation implies that in the weak field limit, the WEC is equivalent to $\rho \geq 0$ and $p \geq 0$. Can somebody help me understanding where is the mistake I am making and why the conditions are sufficient conditions?


2 Answers 2


You have already noted that they are necessary conditions. That they are sufficient follows precisely because the pressure is isotropic. To see this consider a local Lorentz (orthonormal) frame (a.k.a tetrad). Any null vector can be written as $$ \ell^i = N\left(\delta^i_0 + \delta^i_1\right), $$ since we have specified no spacelike directions (really the 1 is just an arbitrarily chosen spacelike index), while $$ U^i = \delta^i_0. $$ The stress-energy tensor is $T_{ij} = \mathrm{diag}\left(\rho,p,p,p\right)$. Then $$T_{ij}U^iU^j = T_{00} = \rho \geq 0$$ and $$T_{ij}\ell^i\ell^j = N^2\left(T_{00} + T_{11}\right) = N^2\left(\rho + p\right) \geq 0$$ is equivalent to that $\rho + p \geq 0$. Finally, an arbitrary timelike vector can be decomposed, as you say $$ t^i = \ell^i + \lambda U^i, $$ with $\lambda^2 + 2N\lambda > 0$ (equal to 1 if normalized), since otherwise $t^i$ would be null or spacelike. Thus
\begin{align} T_{ij}t^it^j &= N^2(\rho + p) + (\lambda^2 + 2N\lambda)\rho \geq 0 \end{align} by the above results. Meaning that all observers observe the energy density to be non-negative.

  • $\begingroup$ Evaluating the WEC inequality in the local Lorentz frame does indeed simplify things a lot and allows to generalize the statement about the decomposition of the timelike vector. Evaluating however the left hand side of the last line leads me to $$T_{\mu\nu}t^\mu t^\nu =\rho (N+\lambda)^2 + N^2p$$ which is different from your result. I think my mistake was in the interpretation of this last inequality. It has to hold for all timelike vectors, i.e. for any value of $N$ and $\lambda$, what leads me to the conclusion that $\rho \geq 0$ and $\rho + p \geq 0$. $\endgroup$
    – jac
    Commented Oct 2, 2016 at 9:55
  • $\begingroup$ @Jac Ah, yes! Sorry, I am used to decomposition into spacelike and timelike vectors. Good spot! Anyway, the last inequality should be equivalent to the previous results (and so necessarily be true given the previous results). I corrected the answer to account for this. $\endgroup$ Commented Oct 2, 2016 at 10:27
  • $\begingroup$ Indeed, using the fact that $t^\mu$ is timelike allows one to prove that the conditions for $\rho$ and $\rho + p$ are not only necessary but also sufficient conditions. It's all clear to me now. Thanks. $\endgroup$
    – jac
    Commented Oct 2, 2016 at 11:35

For future readers of this thread, here is a description of how to show that the decomposition $t^\mu = l^\mu + \lambda U^\mu$, which is used in the accepted answer, always exists (locally).

Let us denote the norm of the timelike vector $t^\mu$ by

$$T = t^\mu t_\mu < 0.$$

We wish to find a null vector $l^\mu$ and a number $\lambda$ such that $t^\mu = l^\mu + \lambda U^\mu$. We will then have

\begin{split} l^\mu l_\mu &= (t^\mu - \lambda U^\mu)(t_\mu - \lambda U_\mu)\\ &= t^\mu t_\mu - 2 \lambda U^\mu t_\mu + \lambda^2 U^\mu U_\mu. \end{split}

Suppose $t^0 = S$ in the rest frame of the perfect fluid source of the energy-momentum tensor, i.e. the frame in which $U^\mu = (1,0,0,0)^\mu$. Then

$$l^\mu l_\mu = T + 2 \lambda S - \lambda^2.$$

For $l^\mu$ to be null we need $l^\mu l_\mu = 0$, and hence

$$\lambda^2 - 2 \lambda S - T = 0.$$

This equation can be solved for $\lambda$ provided that $S^2 + T \ge 0$. But, using locally inertial coordinates to evaluate $T$, we find that

$$T = t^\mu t_\mu = -(t^0)^2 + \sum_{i=1}^3 (t^i)^2 \ge - S^2,$$

so it is always possible to solve for $\lambda$.


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