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.

One thought on “Dominant energy condition

  1. Pingback: Vacuum stress-energy and the cosmological constant | Physics pages

Leave a Reply

Your email address will not be published. Required fields are marked *