# Weak Energy conditions for Perfect fluid derivation

I want to derive the restraints $$\rho + p \geq 0\qquad\text{and}\qquad \rho \geq 0$$ which correspond to applying the weak energy condition ($$T^{\mu \nu} v_\mu v_\nu \geq 0$$ for $$v_\mu$$ timelike) on the stress energy tensor $$T_{\mu \nu}$$ of a perfect fluid given by $$T_{ab} = (\rho + p ) u_a u_b + p g_{ab}.$$

Applying the definition: \begin{align*} 0 & \leq T^{\mu \nu} v_\mu v_\nu \\ & = ( (\rho + p)u^\mu u^\nu + p g^{\mu \nu}) v_\mu v_\nu \\ & = (\rho + p) u^\mu u^\nu v_\mu v_\nu + p g^{\mu \nu} v_\mu v_\nu \\ & = (\rho + p) u^\mu v_\mu u^\nu v_\nu + p v^\nu v_\nu \end{align*} I'm not sure how to evaluate $$u^a v_a$$. I know that $$u_a$$ is the four velocity so we have $$u^a u_a = -1$$ in natural units and also that $$v_a v^a < 0$$ since $$v$$ is timelike but I dont see how I can simplify this expression to get the constraints above.

• Are you interested in the weak energy condition or the strong energy condition? They are different concepts, and one does not imply the other – Filipe Miguel Mar 31 at 9:56

## 2 Answers

The statement of the weak energy condition is that

$$\begin{equation} T^{a b} v_a v_b \geq 0 \end{equation}$$

holds for any timelike vector $$v$$.

The first important step is to prove that the weak energy condition implies the null energy condition, if the previous inequality holds for any timelike vector $$v$$, it will also hold for any null vector $$k$$.

I will leave the mathematical details aside, but the rough idea in proving this statement is using the fact that null vectors constitute the boundary of the light cone defined by the timelike vectors, so we can always take a sequence of timelike vectors that converges to any null vector. (i.e $$\forall$$ null $$k$$ $$\exists$$ a sequence of timelike vectors $$\{v_n\}$$ such that $$\lim \, v_n = k$$).

This sequence leaves the first inequality invariant, proving the weak energy condition. Now we have to add to the first constraint that

$$\begin{equation} T^{ab} k_a k_b \geq 0 \end{equation}$$

holds for any null k.

Henceforth I will use the notation $$a_ab^a \equiv a\cdot b$$ and $$a_a a^a \equiv a^2$$.

Focusing first on the statement of the null energy condition

\begin{align*} 0 & \leq T^{a b} k_a k_b \\ & = ( (\rho + p)u^a u^b + p g^{a b}) k_a k_b \\ & = (\rho + p) (u\cdot k) ^2 + p k^2 \\ & = (\rho + p) (u\cdot k) ^2 \end{align*}

where, in the last step, we used the fact that k is null. Using $$(u\cdot k) ^2 \geq 0$$ we have

$$(\rho + p) \geq 0$$

as required.

Now, looking at the statement of the weak energy condition, we end up with

$$0 \leq (\rho + p) (u\cdot v) ^2 + p v^2$$

for $$v$$ timelike.

Writing $$v = \|v\| n$$, where $$\|v\| = \sqrt{-v^2}$$ and $$n^2 = -1$$, we get that

$$\frac{p}{(u\cdot n)^2} \leq \rho + p$$

Now we need just the final bit of mathematics. Due to the Lorentzian nature of geometry in GR, the Cauchy - Schwartz inequality we all know and love from Riemannian geometry, is changed. Specifically, we know that if $$x$$ and $$y$$ are null or timelike

$$|x\cdot y| \ge \|x\| \|y\|.$$

This means that $$(u\cdot n)^2 \ge 1$$ with equality iff $$n = u$$

Hence:

$$\frac{p}{(u\cdot n)^2} \leq p \leq \rho + p \iff \rho \geq 0$$

Q.E.D

I think the trick is to write the timelike vector $$v^\mu = u^\mu + l^\mu$$ where $$u^\mu$$ is the four-velocity and $$l^\mu$$ is some null vector.

If so, you have that $$T_{\mu \nu} v^\mu v^\nu$$ will be non negative for all timelike vectors $$v^\mu$$ if both $$T_{\mu \nu} u^\mu u^\nu \ge 0$$ and $$T_{\mu \nu} l^\mu l^\nu \ge 0$$. It is just a matter of adding vectors.

In that way you show that $$T_{\mu \nu} u^\mu u^\nu = \rho$$ and $$T_{\mu \nu} l^\mu l^\nu = (\rho + p) (u_\mu l^\mu)^2$$, so that:
$$\rho \ge 0$$ and $$\rho + p \ge 0$$ WEC (weak energy condition)

Note:
The question title refers to the SEC (strong energy condition) which is however other than the WEC and reads $$T_{\mu \nu} v^\mu v^\nu \ge \frac{1}{2} T^\lambda_\lambda v^\sigma v_\sigma$$

• Why do you assume you can always pick a null l such that that is true? – Filipe Miguel Mar 31 at 9:53
• @Filipe Miguel. A generic timelike vector can be written as the addition of the 4-velocity and a generic null vector. To convince you try this in the rest frame. Then the inner product is an invariant whatever the reference frame. I ask you to reconsider the negative score to my answer. – Michele Grosso Apr 1 at 10:39
• You are saying that all timelike vectors v are related to the 4 velocity u by a null vector? That is only true if $u^2 + v^2= 2 u \cdot v$. Or that the sum of u with a generic null yields a timelike vector? That is only true if l is future directed wrt to u, but then you loose the iff requirement, you just prove an implication – Filipe Miguel Apr 1 at 12:02
• @Filipe Miguel. I state about the sum. In the rest frame and with one spatial dimension to simplify the example, you have $u^\mu = (1, 0)$, $l^\mu = (l, l)$ and $v^\mu = (1 + l, l)$. If $l \gt 0$ the vector built in that way is timelike and cover the future oriented and the $+x$ direction. If $l^\mu = (-l, l)$ and $l \lt 0$ you cover the future oriented and the $-x$ direction. To cover the past oriented you consider the four-velocity with a minus sign as in the energy condition formula the inner product of two minus sign is a plus. The inner product is an invariant, so demonstrated. – Michele Grosso Apr 1 at 16:38