# The physical interpretation of the weak energy condition (WEC)

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?

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.

• 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 $\rho + p \geq 0$.
– jac
Commented Oct 2, 2016 at 9:55
• @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. Commented Oct 2, 2016 at 10:27
• 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.
– 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 $$\lambda$$.