# Dominant energy condition

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 20; Problem 20.10.

The dominant energy condition (DEC) states that if ${a^{i}}$ is any four-vector that is causal, that is, it satisfies the conditions

 $\displaystyle \mathbf{a}\cdot\mathbf{a}$ $\displaystyle \le$ $\displaystyle 0\ \ \ \ \ (1)$ $\displaystyle a^{t}$ $\displaystyle >$ $\displaystyle 0 \ \ \ \ \ (2)$

then we require the stress-energy tensor ${T^{ij}}$ to satisfy the condition that if

$\displaystyle b^{i}=-T^{ij}g_{jk}a^{k} \ \ \ \ \ (3)$

then ${\mathbf{b}}$ is also a causal four-vector. The causal condition is just a way of saying that a four-vector is either timelike (if ${\mathbf{a}\cdot\mathbf{a}<0}$) or lightlike (if ${\mathbf{a}\cdot\mathbf{a}=0}$). The DEC is a condition on the stress-energy tensor which amounts to saying that taking the scalar product of one of its rows or columns with a causal vector cannot produce a non-causal (spacelike) vector. Physically, this says that nothing can move faster than light. Note that it’s not a property that is automatically true of any stress-energy tensor; rather it is a condition imposed on the tensor to make it physically realistic.

We can use the DEC to show that the momentum density of a perfect fluid is always causal. The tensor in the fluid’s rest frame is

$\displaystyle T^{ij}=\left[\begin{array}{cccc} \rho & 0 & 0 & 0\\ 0 & P & 0 & 0\\ 0 & 0 & P & 0\\ 0 & 0 & 0 & P \end{array}\right] \ \ \ \ \ (4)$

The momentum density is defined as the first row (or column) of the tensor:

$\displaystyle \pi^{i}\equiv T^{ti} \ \ \ \ \ (5)$

In the rest frame,

 $\displaystyle \pi^{i}$ $\displaystyle =$ $\displaystyle \left[\rho,0,0,0\right]\ \ \ \ \ (6)$ $\displaystyle \boldsymbol{\pi}\cdot\boldsymbol{\pi}$ $\displaystyle =$ $\displaystyle -\rho^{2}\le0\ \ \ \ \ (7)$ $\displaystyle \pi^{t}$ $\displaystyle =$ $\displaystyle \rho>0 \ \ \ \ \ (8)$

so ${\boldsymbol{\pi}}$ is causal in this frame. In a local orthonormal frame (LOF) the tensor’s components are

$\displaystyle T_{obs}^{ip}=\eta^{ij}\eta_{km}\left(\mathbf{o}_{j}\right)^{k}\eta^{pq}\eta_{rs}\left(\mathbf{o}_{q}\right)^{r}T^{ms} \ \ \ \ \ (9)$

where ${\mathbf{o}_{i}}$ are the orthonormal basis vectors in the LOF. If we plug in the definition 5 we get

 $\displaystyle \pi_{obs}^{p}$ $\displaystyle =$ $\displaystyle T_{obs}^{tp}\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \eta^{tj}\eta_{km}\left(\mathbf{o}_{j}\right)^{k}\eta^{pq}\eta_{rs}\left(\mathbf{o}_{q}\right)^{r}T^{ms}\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left[-T^{ms}\eta_{km}\left(\mathbf{o}_{t}\right)^{k}\right]\eta^{pq}\eta_{rs}\left(\mathbf{o}_{q}\right)^{r} \ \ \ \ \ (12)$

where we got the last line by using the fact that ${\eta^{ij}}$ is diagonal and ${\eta^{tt}=-1}$. The term in square brackets looks like 3, as long as ${\mathbf{o}_{t}}$ is a causal vector. However, this vector is just the observer’s four-velocity ${\mathbf{u}_{obs}}$ measured in the fluid’s frame, so

 $\displaystyle \mathbf{u}_{obs}\cdot\mathbf{u}_{obs}$ $\displaystyle =$ $\displaystyle -1\ \ \ \ \ (13)$ $\displaystyle \mathbf{u}_{obs}^{t}$ $\displaystyle =$ $\displaystyle \gamma>0 \ \ \ \ \ (14)$

Thus ${\mathbf{o}_{t}}$ is indeed causal, so we can invoke the DEC to say that if we define a vector ${B^{s}}$ by

$\displaystyle B^{s}\equiv-T^{ms}\eta_{km}\left(\mathbf{o}_{t}\right)^{k} \ \ \ \ \ (15)$

then ${B^{s}}$ must be causal. We then get

 $\displaystyle \pi_{obs}^{p}$ $\displaystyle =$ $\displaystyle B^{s}\eta^{pq}\eta_{rs}\left(\mathbf{o}_{q}\right)^{r}\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \eta^{pq}\left[B^{s}\eta_{rs}\left(\mathbf{o}_{q}\right)^{r}\right]\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \eta^{pq}B_{obs,q}\ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle B_{obs}^{p} \ \ \ \ \ (19)$

With this definition, we can calculate

 $\displaystyle \boldsymbol{\pi}_{obs}\cdot\boldsymbol{\pi}_{obs}$ $\displaystyle =$ $\displaystyle \mathbf{B}_{obs}\cdot\mathbf{B}_{obs} \ \ \ \ \ (20)$

However, we know that ${\mathbf{B}}$ is causal because that’s how we defined it in 15 and since its magnitude is a scalar, it is the same in all coordinate systems, so we must have ${\boldsymbol{\pi}_{obs}\cdot\boldsymbol{\pi}_{obs}=\mathbf{B}_{obs}\cdot\mathbf{B}_{obs}\le0}$. As for showing that ${\pi_{obs}^{t}>0}$, we can observe that

 $\displaystyle \pi_{obs}^{t}$ $\displaystyle =$ $\displaystyle \eta^{tq}\left[B^{s}\eta_{rs}\left(\mathbf{o}_{q}\right)^{r}\right]\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -B^{s}\eta_{rs}\left(\mathbf{o}_{t}\right)^{r} \ \ \ \ \ (22)$

Since ${\mathbf{o}_{t}=\gamma\left[1,v^{x},v^{y},v^{z}\right]}$ we see that ${\pi_{obs}^{t}}$ is the Lorentz transformation of the causal vector ${B^{s}}$ and a Lorentz transformation never changes the spacetime nature of a four-vector (that is, timelike remains timelike, etc) so since ${\mathbf{B}}$ is causal, ${B^{t}>0}$ and therefore ${\pi_{obs}^{t}>0}$ as well. This condition is known as the weak energy condition or WEC.

Another property of the stress-energy tensor that can be derived from the DEC is as follows. In the rest frame of a perfect fluid, 4 holds, so the DEC condition 3 says, for some arbitrary causal vector ${\mathbf{a}}$ we get another causal vector ${\mathbf{b}}$:

 $\displaystyle b^{i}$ $\displaystyle =$ $\displaystyle -T^{ij}g_{jk}a^{k}\ \ \ \ \ (23)$ $\displaystyle b^{t}$ $\displaystyle =$ $\displaystyle \rho a^{t}\ \ \ \ \ (24)$ $\displaystyle b^{m}$ $\displaystyle =$ $\displaystyle -Pa^{m}\mbox{ for \ensuremath{m=x,y,z}} \ \ \ \ \ (25)$

Since ${\mathbf{b}}$ is causal, we must have for all choices of ${\mathbf{a}}$:

$\displaystyle \mathbf{b}\cdot\mathbf{b}=-\rho^{2}\left(a^{t}\right)^{2}+P^{2}\sum_{m=x,y,z}\left(a^{m}\right)^{2}\le0 \ \ \ \ \ (26)$

Because ${\mathbf{a}}$ is causal, we have

$\displaystyle \left(a^{t}\right)^{2}\ge\sum_{m=x,y,z}\left(a^{m}\right)^{2} \ \ \ \ \ (27)$

The constraint on ${\rho}$ and ${P}$ in 26 comes in the case of equality in 27, in which case we have

$\displaystyle -\rho^{2}+P^{2}\le0 \ \ \ \ \ (28)$

and since ${\rho>0}$ this amounts to

$\displaystyle \rho\ge\left|P\right| \ \ \ \ \ (29)$

Finally, we can revisit the case of the stress-energy tensor that gives negative pressure, ${T^{ij}=-\Lambda g^{ij}}$, where ${\Lambda}$ is a positive scalar. In this case, if we apply the DEC to some causal vector ${\mathbf{a}}$ we get

 $\displaystyle b^{i}$ $\displaystyle =$ $\displaystyle -T^{ij}g_{jk}a^{k}\ \ \ \ \ (30)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \Lambda g^{ij}g_{jk}a^{k}\ \ \ \ \ (31)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \Lambda a^{i} \ \ \ \ \ (32)$

Since ${\mathbf{b}}$ is a positive scalar multiplied by a causal vector, it too must be causal, so this stress-energy tensor satisfies the DEC.