# Stress-energy tensor from Noether’s theorem

Michael E. Peskin & Daniel V. Schroeder, An Introduction to Quantum Field Theory, (Perseus Books, 1995) – Chapter 2.

Noether’s theorem allows us to find conserved currents from transformations which leave the Lagrangian invariant, up to a divergence. That is, if

$\displaystyle \delta\mathcal{L}=\partial_{\mu}\left(\frac{\delta\mathcal{L}}{\delta\partial_{\mu}\phi}\delta\phi-\mathcal{J}^{\mu}\right)=0 \ \ \ \ \ (1)$

the conserved current is

$\displaystyle j^{\mu}\equiv\frac{\delta\mathcal{L}}{\delta\partial_{\mu}\phi}\delta\phi-\mathcal{J}^{\mu} \ \ \ \ \ (2)$

Suppose we translate the system in spacetime, so that

$\displaystyle x^{\mu}\rightarrow x^{\mu}+a^{\mu} \ \ \ \ \ (3)$

where ${a^{\mu}}$ is a constant 4-vector. Then for infinitesimal displacements, we can write the variation in the field as the first two terms in a Taylor series:

$\displaystyle \phi\left(x^{\mu}+a^{\mu}\right)=\phi\left(x^{\mu}\right)+a^{\nu}\partial_{\nu}\phi\left(x^{\mu}\right) \ \ \ \ \ (4)$

So ${\delta\phi_{a}}$ in 1 is

$\displaystyle \delta\phi_{a}=a^{\nu}\partial_{\nu}\phi\left(x^{\mu}\right) \ \ \ \ \ (5)$

The Lagrangian, being a scalar, can also be expanded in a Taylor series:

 $\displaystyle \mathcal{L}$ $\displaystyle \rightarrow$ $\displaystyle \mathcal{L}+\delta\mathcal{L}\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \mathcal{L}+a^{\nu}\partial_{\nu}\mathcal{L} \ \ \ \ \ (7)$

We can write the last term as a divergence by introducting the Kronecker delta:

 $\displaystyle \mathcal{L}+a^{\mu}\partial_{\mu}\mathcal{L}$ $\displaystyle =$ $\displaystyle \mathcal{L}+a^{\nu}\partial_{\mu}\left(\delta_{\;\nu}^{\mu}\mathcal{L}\right)\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \mathcal{L}+a^{\nu}\partial_{\mu}\mathcal{J}_{\nu}^{\mu} \ \ \ \ \ (9)$

If we now require 1 to be true, that is, we require the Lagrangian to be invariant under translation in spacetime, we get the condition

$\displaystyle a^{\nu}\partial_{\mu}\left(\frac{\delta\mathcal{L}}{\delta\partial_{\mu}\phi}\partial_{\nu}\phi-\delta_{\;\nu}^{\mu}\mathcal{L}\right)=0 \ \ \ \ \ (10)$

The translations ${a^{\nu}}$ are four independent parameters, so this equation actually gives us four separate conserved currents. We could, for example, choose ${a^{0}}$ to be non-zero and the other three ${a^{j}}$ to be zero, or one of the other three ${a^{j}}$ to be non-zero with the remaining three zero, and so on. Because of these arbitrary choices, the divergence term itself must be zero to satisfy the equation in all cases. That is, we can define

 $\displaystyle T_{\;\nu}^{\mu}$ $\displaystyle \equiv$ $\displaystyle \frac{\delta\mathcal{L}}{\delta\partial_{\mu}\phi}\partial_{\nu}\phi-\delta_{\;\nu}^{\mu}\mathcal{L}\ \ \ \ \ (11)$ $\displaystyle \partial_{\mu}T_{\;\nu}^{\mu}$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (12)$

${T_{\;\nu}^{\mu}}$ is the stress-energy tensor which we’ve already met in the context of general relativity.

For the Klein-Gordon Lagrangian (for a real field, for the present, and using Peskin’s definition with a factor of ${\frac{1}{2}}$):

$\displaystyle \mathcal{L}=\frac{1}{2}\partial_{\alpha}\phi\partial^{\alpha}\phi-\frac{1}{2}m^{2}\phi^{2} \ \ \ \ \ (13)$

${\mu=\nu=0}$, we get

 $\displaystyle T_{\;0}^{0}$ $\displaystyle =$ $\displaystyle \frac{\delta\mathcal{L}}{\delta\partial_{0}\phi}\partial_{0}\phi-\mathcal{L}\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \partial^{0}\phi\partial_{0}\phi-\frac{1}{2}\left(\partial_{\alpha}\phi\partial^{\alpha}\phi-m^{2}\phi^{2}\right)\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\dot{\phi}^{2}+\frac{1}{2}\left(\nabla\phi\right)^{2}+\frac{m^{2}}{2}\phi^{2} \ \ \ \ \ (16)$

Using the conjugate momentum

$\displaystyle \pi\equiv\frac{\partial\mathcal{L}}{\partial\dot{\phi}}=\dot{\phi} \ \ \ \ \ (17)$

this component of the stress-energy tensor turns out to be the Hamiltonian density:

$\displaystyle T^{00}=T_{\;0}^{0}=\frac{1}{2}\pi^{2}+\frac{1}{2}\left(\nabla\phi\right)^{2}+\frac{m^{2}}{2}\phi^{2}=\mathcal{H} \ \ \ \ \ (18)$

The other three components in the first row of ${T_{\;\nu}^{\mu}}$ are

 $\displaystyle T_{\;j}^{0}$ $\displaystyle =$ $\displaystyle \frac{\delta\mathcal{L}}{\delta\partial_{0}\phi}\partial_{j}\phi\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \dot{\phi}\partial_{j}\phi\ \ \ \ \ (20)$ $\displaystyle T^{0j}$ $\displaystyle =$ $\displaystyle \dot{\phi}\partial^{j}\phi=-\dot{\phi}\partial_{j}\phi=-\pi\partial_{j}\phi \ \ \ \ \ (21)$

which is the physical momentum density.

It’s worth noting that all of this is true for classical fields; we haven’t yet applied it to quantum fields.

# Stress-energy tensor in the weak field, low velocity limit

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 22, Box 22.5.

In the weak field limit, the Ricci tensor becomes

$\displaystyle R_{jm}=\frac{1}{2}\left(\partial_{j}H_{m}+\partial_{m}H_{j}-\eta^{nl}\partial_{n}\partial_{l}h_{jm}\right) \ \ \ \ \ (1)$

where

$\displaystyle H_{m}\equiv\eta^{nl}\left(\partial_{n}h_{lm}-\frac{1}{2}\partial_{m}h_{nl}\right) \ \ \ \ \ (2)$

and the ${h_{ij}}$ is the perturbation on the flat metric, so that

$\displaystyle g_{ij}=\eta_{ij}+h_{ij} \ \ \ \ \ (3)$

Because we can introduce a coordinate transformation for the four coordinates in the form

$\displaystyle \left(x'\right)^{i}=f^{i}\left(x^{j}\right) \ \ \ \ \ (4)$

there are four degrees of freedom that we can play with in specifying the form of ${H_{i}}$. It turns out (we may get around to a proof in some future post) that it is always possible to find a coordinate system in which all ${H_{i}=0}$. If we use such a coordinate system then the Ricci tensor is

$\displaystyle R_{jm}=-\frac{1}{2}\eta^{nl}\partial_{n}\partial_{l}h_{jm} \ \ \ \ \ (5)$

so the Einstein equation becomes

$\displaystyle -\frac{1}{2}\eta^{nl}\partial_{n}\partial_{l}h_{jm}=8\pi G\left(T_{jm}-\frac{1}{2}\eta_{jm}T\right) \ \ \ \ \ (6)$

Introducing the d’Alembertian operator

$\displaystyle \square^{2}\equiv\eta^{nl}\partial_{n}\partial_{l}=-\frac{d^{2}}{dt^{2}}+\nabla^{2} \ \ \ \ \ (7)$

we can write the Einstein equation as

$\displaystyle -\frac{1}{2}\square^{2}h_{jm}=8\pi G\left(T_{jm}-\frac{1}{2}\eta_{jm}T\right) \ \ \ \ \ (8)$

As it stands, this equation is a set of uncoupled differential equations for the ${h_{jm}}$ so in principle they can be solved. However, we can invoke another approximation by assuming that the system is in a steady state so that all time derivatives are zero. This doesn’t necessarily mean that the masses are all stationary, since we might have a star rotating at a constant angular velocity. In such cases, the stress-energy tensor ${T_{jm}}$ is constant in time so we would expect ${h_{jm}}$ to be independent of time as well. In that case, we get

$\displaystyle \nabla^{2}h_{jm}=-16\pi G\left(T_{jm}-\frac{1}{2}\eta_{jm}T\right) \ \ \ \ \ (9)$

This equation should look familiar from electrodynamics, where it is formally equivalent to Poisson’s equation for the electrostatic potential in terms of the charge distribution. In that case we had

$\displaystyle \nabla^{2}V=-\nabla\cdot\mathbf{E}=-\frac{\rho}{\epsilon_{0}} \ \ \ \ \ (10)$

where ${\rho}$ here is the charge density. The solution of this equation is

$\displaystyle V\left(\mathbf{r}\right)=\frac{1}{4\pi\epsilon_{0}}\int\frac{1}{\left|\mathbf{r}-\mathbf{r}'\right|}\rho\left(\mathbf{r}'\right)d^{3}\mathbf{r}' \ \ \ \ \ (11)$

By replacing ${\rho/\epsilon_{0}}$ with ${16\pi G\left(T_{jm}-\frac{1}{2}\eta_{jm}T\right)}$ we can find ${h_{jm}}$ as

$\displaystyle h_{jm}=2G\int\frac{2T_{jm}-\eta_{jm}T}{\left|\mathbf{r}-\mathbf{r}'\right|}d^{3}\mathbf{r}' \ \ \ \ \ (12)$

Thus if we know the stress-energy tensor as a function of position, we can work out the perturbations on the flat metric ${h_{jm}}$.

Example We can look at this equation for the case of a perfect fluid, where the stress-energy tensor is

$\displaystyle T_{ij}=\left(\rho_{0}+P_{0}\right)u_{i}u_{j}+P_{0}g_{ij} \ \ \ \ \ (13)$

where ${\rho_{0}}$ and ${P_{0}}$ are the fluid’s density and pressure in its rest frame, and ${u_{i}}$ is the fluid’s four-velocity in the observer’s frame. We need to find the numerator of the integrand in 12 for each component. First, we work out the stress-energy scalar ${T}$:

 $\displaystyle T$ $\displaystyle =$ $\displaystyle g^{ij}T_{ij}\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(\rho_{0}+P_{0}\right)g^{ij}u_{i}u_{j}+P_{0}g^{ij}g_{ij}\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\left(\rho_{0}+P_{0}\right)+4P_{0}\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\rho_{0}+3P_{0} \ \ \ \ \ (17)$

since ${g^{ij}u_{i}u_{j}=-1}$ and ${g^{ij}g_{ij}=\delta_{i}^{i}=4}$. Since ${T}$ is a scalar, this result is valid in all coordinate systems. Also, we haven’t yet used any approximations, so this result is valid for all perfect fluids, even ones where the density and pressure are large.

Now let’s assume that ${\rho_{0}}$ and ${P_{0}}$ are small and so we keep only up to first order terms, so any product of ${\rho_{0}}$ or ${P_{0}}$ with ${h^{ij}}$ can be ignored. Thus

 $\displaystyle T_{ij}-\frac{1}{2}g_{ij}T$ $\displaystyle \approx$ $\displaystyle T_{ij}-\frac{1}{2}\eta_{ij}T\ \ \ \ \ (18)$ $\displaystyle 2T_{ij}-g_{ij}T$ $\displaystyle \approx$ $\displaystyle 2T_{ij}-\eta_{ij}T\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle \approx$ $\displaystyle 2\left(\rho_{0}+P_{0}\right)u_{i}u_{j}+2P_{0}\eta_{ij}-\eta_{ij}\left(-\rho_{0}+3P_{0}\right)\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2\left(\rho_{0}+P_{0}\right)u_{i}u_{j}+\eta_{ij}\left(\rho_{0}-P_{0}\right) \ \ \ \ \ (21)$

There are 3 cases to consider. First, ${i=j=t}$ and use ${\eta_{tt}=-1}$. Further, we’ll assume that the spatial velocity components are all small, so ${u_{t}\approx-1}$ and ${u_{i}u_{j}\approx0}$ if ${i}$ and ${j}$ are both spatial indices.

 $\displaystyle 2T_{tt}-\eta_{tt}T$ $\displaystyle =$ $\displaystyle 2\left(\rho_{0}+P_{0}\right)u_{t}u_{t}+\eta_{tt}\left(\rho_{0}-P_{0}\right)\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2\left(\rho_{0}+P_{0}\right)-\left(\rho_{0}-P_{0}\right)\ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \rho_{0}+3P_{0} \ \ \ \ \ (24)$

Now suppose ${i=t}$ and ${j}$ is a spatial index (or vice versa; since everything is symmetric it makes no difference). Then

 $\displaystyle 2T_{tj}-\eta_{tj}T$ $\displaystyle =$ $\displaystyle 2T_{tj}\ \ \ \ \ (25)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2\left(\rho_{0}+P_{0}\right)u_{t}u_{j}\ \ \ \ \ (26)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -2\left(\rho_{0}+P_{0}\right)u_{j} \ \ \ \ \ (27)$

Finally, if both ${i}$ and ${j}$ are spatial indices we get, if ${i\ne j}$

 $\displaystyle 2T_{ij}-g_{ij}T$ $\displaystyle =$ $\displaystyle 2\left(\rho_{0}+P_{0}\right)u_{i}u_{j}+\eta_{ij}\left(\rho_{0}-P_{0}\right)\ \ \ \ \ (28)$ $\displaystyle$ $\displaystyle \approx$ $\displaystyle 0 \ \ \ \ \ (29)$

since the first term involves the second order term ${u_{i}u_{j}}$ and in the second term ${\eta_{ij}=0}$ if ${i\ne j}$.

Now if ${i=j}$ we get

 $\displaystyle 2T_{ii}-g_{ii}T$ $\displaystyle =$ $\displaystyle 2\left(\rho_{0}+P_{0}\right)u_{i}u_{i}+\eta_{ii}\left(\rho_{0}-P_{0}\right)\ \ \ \ \ (30)$ $\displaystyle$ $\displaystyle \approx$ $\displaystyle \rho_{0}-P_{0} \ \ \ \ \ (31)$

again, since the first term has the second order factor ${u_{i}^{2}}$ and in the second term ${\eta_{ii}=+1}$ if ${i}$ is a spatial index.

Therefore, if we know ${\rho_{0}}$ and ${P_{0}}$ as functions of position we can work out the perturbations ${h_{ij}}$ to the metric.

# Conservation of four-momentum implies the geodesic equation

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 22, Box 22.1.

The stress-energy tensor obeys the conservation of four-momentum

$\displaystyle \nabla_{j}T^{ij}=0 \ \ \ \ \ (1)$

We can show that the geodesic equation actually follows from this conservation condition. For the case of ‘dust’ (a fluid whose constituent particles are locally at rest with one another), the stress-energy tensor is

$\displaystyle T^{ij}=\rho_{0}u^{i}u^{j} \ \ \ \ \ (2)$

where ${\rho_{0}}$ is the dust’s density in its own rest frame and ${u^{i}}$ is its four-velocity measured in the observer’s frame. In this case

 $\displaystyle \nabla_{j}T^{ij}$ $\displaystyle =$ $\displaystyle \nabla_{j}\left(\rho_{0}u^{i}u^{j}\right)\ \ \ \ \ (3)$ $\displaystyle 0$ $\displaystyle =$ $\displaystyle u^{i}\nabla_{j}\left(\rho_{0}u^{j}\right)+\rho_{0}u^{j}\nabla_{j}u^{i} \ \ \ \ \ (4)$

Note that ${\rho_{0}}$ is not necessarily a constant so its gradient will, in general be non-zero.

From the equation

$\displaystyle g_{ij}u^{i}u^{j}=-1 \ \ \ \ \ (5)$

we get

 $\displaystyle \nabla_{k}\left(g_{ij}u^{i}u^{j}\right)$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (6)$ $\displaystyle g_{ij}u^{i}\nabla_{k}u^{j}+g_{ij}u^{j}\nabla_{k}u^{i}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (7)$ $\displaystyle 2g_{ij}u^{i}\nabla_{k}u^{j}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (8)$ $\displaystyle g_{ij}u^{i}\nabla_{k}u^{j}$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (9)$

The second line follows from the fact that the absolute gradient of the metric tensor is zero (so there’s no term ${u^{i}u^{j}\nabla_{k}g_{ij}}$ in the product rule expansion). The third line comes from swapping the bound indices ${i}$ and ${j}$ in the second term in line 2, and then using the symmetry of the metric tensor (${g_{ij}=g_{ji}}$).

Returning to 4, we can multiply through by ${g_{il}u^{l}}$ and get

 $\displaystyle g_{il}u^{l}u^{i}\nabla_{j}\left(\rho_{0}u^{j}\right)+\rho_{0}g_{il}u^{l}u^{j}\nabla_{j}u^{i}$ $\displaystyle =$ $\displaystyle -\nabla_{j}\left(\rho_{0}u^{j}\right)+\rho_{0}u^{j}\left(g_{il}u^{l}\nabla_{j}u^{i}\right)\ \ \ \ \ (10)$ $\displaystyle 0$ $\displaystyle =$ $\displaystyle -\nabla_{j}\left(\rho_{0}u^{j}\right) \ \ \ \ \ (11)$

where we used 5 on the first term on the LHS of the first line, and 9 on the second term of the RHS of the first line (with indices suitably relabelled). Therefore

$\displaystyle \nabla_{j}\left(\rho_{0}u^{j}\right)=0 \ \ \ \ \ (12)$

Substituting this back into 4 we get

$\displaystyle u^{j}\nabla_{j}u^{i}=0 \ \ \ \ \ (13)$

The absolute gradient of a four-vector can be written in terms of Christoffel symbols as

$\displaystyle \nabla_{j}u^{i}=\partial_{j}u^{i}+\Gamma_{kj}^{i}u^{k} \ \ \ \ \ (14)$

so we get

 $\displaystyle u^{j}\nabla_{j}u^{i}$ $\displaystyle =$ $\displaystyle u^{j}\left(\partial_{j}u^{i}+\Gamma_{kj}^{i}u^{k}\right)\ \ \ \ \ (15)$ $\displaystyle 0$ $\displaystyle =$ $\displaystyle \frac{\partial x^{j}}{\partial\tau}\frac{\partial u^{i}}{\partial x^{j}}+\Gamma_{kj}^{i}u^{k}u^{j}\ \ \ \ \ (16)$ $\displaystyle 0$ $\displaystyle =$ $\displaystyle \frac{du^{i}}{d\tau}+\Gamma_{kj}^{i}u^{k}u^{j}\ \ \ \ \ (17)$ $\displaystyle 0$ $\displaystyle =$ $\displaystyle \frac{d^{2}x^{i}}{d\tau^{2}}+\Gamma_{kj}^{i}\frac{dx^{j}}{d\tau}\frac{dx^{k}}{d\tau} \ \ \ \ \ (18)$

This is just the geodesic equation, so we see that (for dust, anyway; the result is generally true for fluids but is harder to prove) conservation of four-momentum implies the geodesic equation.

# Vacuum stress-energy and the cosmological constant

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 21, Problem 21.9.

The Einstein equation is

$\displaystyle R^{ij}=8\pi G\left(T^{ij}-\frac{1}{2}g^{ij}T\right)+\Lambda g^{ij} \ \ \ \ \ (1)$

Up to now, we’ve usually taken ${\Lambda=0}$, since we know from the Newtonian limit that ${\Lambda}$ must be very small. If ${\Lambda\ne0}$, the Newtonian limit becomes

$\displaystyle \nabla^{2}\Phi=4\pi G\rho-\Lambda \ \ \ \ \ (2)$

so ${\Lambda}$ acts as a negative mass density, that is, it adds a repulsive term into the gravitational force. Einstein originally introduced it to counter the attractive force of gravity on a cosmological scale, since at the time it was believed that the universe was static (neither expanding nor contracting) and if gravity were purely attractive, the universe would be contracting.

At the moment, the universe is believed to be expanding so ${\Lambda}$ is believed to be non-zero and positive, although still small enough that its effects are not noticeable on the scale of the solar system (or indeed on a galactic scale). Because of this, ${\Lambda}$ is called the cosmological constant.

We can include ${\Lambda}$ within the stress-energy tensor by defining a vacuum stress-energy as

$\displaystyle T_{vac}^{ij}=-\frac{\Lambda}{8\pi G}g^{ij} \ \ \ \ \ (3)$

We can define a vacuum stress-energy scalar:

 $\displaystyle T_{vac}$ $\displaystyle \equiv$ $\displaystyle g_{ij}T_{vac}^{ij}\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{\Lambda}{8\pi G}g_{ij}g^{ij}\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{4\Lambda}{8\pi G} \ \ \ \ \ (6)$

Therefore

 $\displaystyle T_{vac}^{ij}-\frac{1}{2}g^{ij}T_{vac}$ $\displaystyle =$ $\displaystyle -\frac{\Lambda}{8\pi G}g^{ij}+\frac{2\Lambda}{8\pi G}g^{ij}\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\Lambda}{8\pi G}g^{ij} \ \ \ \ \ (8)$

and we can write 1 as

 $\displaystyle R^{ij}$ $\displaystyle =$ $\displaystyle 8\pi G\left(T^{ij}-\frac{1}{2}g^{ij}T+T_{vac}^{ij}-\frac{1}{2}g^{ij}T_{vac}\right)\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle \equiv$ $\displaystyle 8\pi G\left(T_{all}^{ij}-\frac{1}{2}g^{ij}T_{all}\right) \ \ \ \ \ (10)$

where ${T_{all}^{ij}}$ includes the stress-energy from the mass-energy density and the vacuum.

The dominant energy condition is a constraint placed on the stress-energy tensor so that observers in any local orthogonal frame will measure the fluid’s speed to be less than the speed of light. The condition is that if ${a^{i}}$ is any four-vector that is causal, that is, it satisfies the conditions

 $\displaystyle \mathbf{a}\cdot\mathbf{a}$ $\displaystyle \le$ $\displaystyle 0\ \ \ \ \ (11)$ $\displaystyle a^{t}$ $\displaystyle >$ $\displaystyle 0 \ \ \ \ \ (12)$

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

$\displaystyle b^{i}=-T^{ij}g_{jk}a^{k} \ \ \ \ \ (13)$

then ${\mathbf{b}}$ is also a causal four-vector. For the vacuum stress-energy this condition says

 $\displaystyle b^{i}$ $\displaystyle =$ $\displaystyle -T_{vac}^{ij}g_{jk}a^{k}\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\Lambda}{8\pi G}g^{ij}g_{jk}a^{k}\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\Lambda}{8\pi G}\delta_{\; k}^{i}a^{k}\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\Lambda}{8\pi G}a^{i} \ \ \ \ \ (17)$

That is, ${b^{i}}$ is just a positive (if ${\Lambda>0}$) constant multiplied by ${a^{i}}$, so if ${a^{i}}$ is causal, then ${b^{i}}$ must also be causal. Thus ${T_{vac}^{ij}}$ satisfies the dominant energy condition.

# Einstein equation on the surface of a sphere

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 21, Problem 21.7.

According to the Einstein equation, the Riemann tensor in 2D must be zero in empty space, implying that gravitational fields cannot exist in 2D. Another consequence of the Einstein equation is that the stress-energy must be zero on the surface of a sphere. That is, even though a 2D surface is manifestly curved, the curvature is not the result of any mass or energy. This is another example of how general relativity breaks down in two dimensions.

The Einstein equation is

$\displaystyle R^{ij}=\kappa\left(T^{ij}-\frac{1}{2}g^{ij}T\right)+\Lambda g^{ij} \ \ \ \ \ (1)$

$\displaystyle R^{ij}=\left[\begin{array}{cc} \frac{1}{r^{4}} & 0\\ 0 & \frac{1}{r^{4}\sin^{2}\theta} \end{array}\right] \ \ \ \ \ (2)$

The metric for a sphere is (in both forms):

 $\displaystyle g^{ij}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{cc} \frac{1}{r^{2}} & 0\\ 0 & \frac{1}{r^{2}\sin^{2}\theta} \end{array}\right]\ \ \ \ \ (3)$ $\displaystyle g_{ij}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{cc} r^{2} & 0\\ 0 & r^{2}\sin^{2}\theta \end{array}\right] \ \ \ \ \ (4)$

Since the off-diagonal elements of ${g^{ij}}$ and ${R^{ij}}$ are all zero, 1 tells us that

$\displaystyle T^{\theta\phi}=T^{\phi\theta}=0 \ \ \ \ \ (5)$

To deal with the diagonal elements, we first need the stress-energy scalar.

 $\displaystyle T$ $\displaystyle =$ $\displaystyle g_{ij}T^{ij}\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle r^{2}T^{\theta\theta}+r^{2}\sin^{2}\theta T^{\phi\phi} \ \ \ \ \ (7)$

We have

 $\displaystyle R^{\theta\theta}$ $\displaystyle =$ $\displaystyle \frac{1}{r^{4}}\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \kappa\left(T^{\theta\theta}-\frac{1}{2r^{2}}T\right)+\frac{\Lambda}{r^{2}}\ \ \ \ \ (9)$ $\displaystyle R^{\phi\phi}$ $\displaystyle =$ $\displaystyle \frac{1}{r^{4}\sin^{2}\theta}\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \kappa\left(T^{\phi\phi}-\frac{1}{2r^{2}\sin^{2}\theta}T\right)+\frac{\Lambda}{r^{2}\sin^{2}\theta} \ \ \ \ \ (11)$

Combining these we get

 $\displaystyle \frac{R^{\theta\theta}}{\sin^{2}\theta}-R^{\phi\phi}$ $\displaystyle =$ $\displaystyle \kappa\left(\frac{T^{\theta\theta}}{\sin^{2}\theta}-T^{\phi\phi}\right)=0\ \ \ \ \ (12)$ $\displaystyle \frac{T^{\theta\theta}}{\sin^{2}\theta}-T^{\phi\phi}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (13)$ $\displaystyle T^{\theta\theta}$ $\displaystyle =$ $\displaystyle T^{\phi\phi}\sin^{2}\theta\ \ \ \ \ (14)$ $\displaystyle T$ $\displaystyle =$ $\displaystyle 2r^{2}\sin^{2}\theta T^{\phi\phi} \ \ \ \ \ (15)$

Therefore

 $\displaystyle T^{\theta\theta}-\frac{1}{2r^{2}}T$ $\displaystyle =$ $\displaystyle T^{\phi\phi}-\frac{1}{2r^{2}\sin^{2}\theta}T=0 \ \ \ \ \ (16)$

holds identically. Thus the stress-energy contribution to the Einstein equation is always zero on a sphere (although the stress-energy tensor may have two non-zero components, these two components always combine to give zero contribution to the Einstein equation).

# Maxwell stress tensor

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Problem 8.3.

We’ve seen the stress-energy tensor in the context of mass in general relativity and also had a brief look at the corresponding tensor in electromagnetism. Here we’ll look at a purely classical, non-relativistic form of the tensor in electromagnetism. In doing so, we’ll look only at the spatial components of the tensor, so it becomes a ${3\times3}$ matrix.

The derivation starts with a calculation of the total force due to electromagnetic fields on the charges and currents within some volume ${\mathcal{V}}$. From the Lorentz force law, we have

 $\displaystyle \mathbf{F}$ $\displaystyle =$ $\displaystyle \int_{\mathcal{V}}\rho\left(\mathbf{E}+\mathbf{v}\times\mathbf{B}\right)d^{3}\mathbf{r}\ \ \ \ \ (1)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int_{\mathcal{V}}\left(\rho\mathbf{E}+\mathbf{J}\times\mathbf{B}\right)d^{3}\mathbf{r} \ \ \ \ \ (2)$

We can think of the integrand as a force density, or force per unit volume ${\mathbf{f}}$:

$\displaystyle \mathbf{f}\equiv\rho\mathbf{E}+\mathbf{J}\times\mathbf{B} \ \ \ \ \ (3)$

We can express this entirely in terms of fields by using Maxwell’s equations:

 $\displaystyle \rho$ $\displaystyle =$ $\displaystyle \epsilon_{0}\nabla\cdot\mathbf{E}\ \ \ \ \ (4)$ $\displaystyle \mathbf{J}$ $\displaystyle =$ $\displaystyle \frac{1}{\mu_{0}}\nabla\times\mathbf{B}-\epsilon_{0}\frac{\partial\mathbf{E}}{\partial t} \ \ \ \ \ (5)$

So we get

$\displaystyle \mathbf{f}=\left(\epsilon_{0}\nabla\cdot\mathbf{E}\right)\mathbf{E}+\left[\frac{1}{\mu_{0}}\nabla\times\mathbf{B}-\epsilon_{0}\frac{\partial\mathbf{E}}{\partial t}\right]\times\mathbf{B} \ \ \ \ \ (6)$

We now need to do a bit of vector calculus gymnastics. From the product rule

$\displaystyle \frac{\partial}{\partial t}\left(\mathbf{E}\times\mathbf{B}\right)=\frac{\partial\mathbf{E}}{\partial t}\times\mathbf{B}+\mathbf{E}\times\frac{\partial\mathbf{B}}{\partial t} \ \ \ \ \ (7)$

$\displaystyle \frac{\partial\mathbf{B}}{\partial t}=-\nabla\times\mathbf{E} \ \ \ \ \ (8)$

Combining these two we get

 $\displaystyle \frac{\partial\mathbf{E}}{\partial t}\times\mathbf{B}$ $\displaystyle =$ $\displaystyle \frac{\partial}{\partial t}\left(\mathbf{E}\times\mathbf{B}\right)-\mathbf{E}\times\frac{\partial\mathbf{B}}{\partial t}\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\partial}{\partial t}\left(\mathbf{E}\times\mathbf{B}\right)+\mathbf{E}\times\left(\nabla\times\mathbf{E}\right) \ \ \ \ \ (10)$

We can insert this into 6 and while we’re at it, we can add on a term ${\frac{1}{\mu_{0}}\left(\nabla\cdot\mathbf{B}\right)\mathbf{B}}$. This is always zero because ${\nabla\cdot\mathbf{B}=0}$, but it gives the equation a symmetry that will be useful in a minute. We get for the force density:

 $\displaystyle \mathbf{f}$ $\displaystyle =$ $\displaystyle \epsilon_{0}\left(\nabla\cdot\mathbf{E}\right)\mathbf{E}+\frac{1}{\mu_{0}}\left(\nabla\cdot\mathbf{B}\right)\mathbf{B}+\frac{1}{\mu_{0}}\left(\nabla\times\mathbf{B}\right)\times\mathbf{B}-\epsilon_{0}\frac{\partial\mathbf{E}}{\partial t}\times\mathbf{B}\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \epsilon_{0}\left(\nabla\cdot\mathbf{E}\right)\mathbf{E}+\frac{1}{\mu_{0}}\left(\nabla\cdot\mathbf{B}\right)\mathbf{B}+\frac{1}{\mu_{0}}\left(\nabla\times\mathbf{B}\right)\times\mathbf{B}-\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle$ $\displaystyle \epsilon_{0}\frac{\partial}{\partial t}\left(\mathbf{E}\times\mathbf{B}\right)-\epsilon_{0}\mathbf{E}\times\left(\nabla\times\mathbf{E}\right) \ \ \ \ \ (13)$

Now another identity from vector calculus says

$\displaystyle \nabla\left(\mathbf{A}\cdot\mathbf{B}\right)=\mathbf{A}\times\left(\nabla\times\mathbf{B}\right)+\mathbf{B}\times\left(\nabla\times\mathbf{A}\right)+\left(\mathbf{A}\cdot\nabla\right)\mathbf{B}+\left(\mathbf{B}\cdot\nabla\right)\mathbf{A} \ \ \ \ \ (14)$

If ${\mathbf{A}=\mathbf{B}=\mathbf{E}}$, we get

$\displaystyle \nabla\left(E^{2}\right)=2\mathbf{E}\times\left(\nabla\times\mathbf{E}\right)+2\left(\mathbf{E}\cdot\nabla\right)\mathbf{E} \ \ \ \ \ (15)$

so

 $\displaystyle \mathbf{E}\times\left(\nabla\times\mathbf{E}\right)$ $\displaystyle =$ $\displaystyle \frac{1}{2}\nabla\left(E^{2}\right)-\left(\mathbf{E}\cdot\nabla\right)\mathbf{E}\ \ \ \ \ (16)$ $\displaystyle \mathbf{B}\times\left(\nabla\times\mathbf{B}\right)$ $\displaystyle =$ $\displaystyle \frac{1}{2}\nabla\left(B^{2}\right)-\left(\mathbf{B}\cdot\nabla\right)\mathbf{B} \ \ \ \ \ (17)$

Putting this into 12 we get

 $\displaystyle \mathbf{f}$ $\displaystyle =$ $\displaystyle \epsilon_{0}\left(\nabla\cdot\mathbf{E}\right)\mathbf{E}+\frac{1}{\mu_{0}}\left(\nabla\cdot\mathbf{B}\right)\mathbf{B}-\epsilon_{0}\frac{\partial}{\partial t}\left(\mathbf{E}\times\mathbf{B}\right)-\ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle$ $\displaystyle \frac{1}{\mu_{0}}\mathbf{B}\times\left(\nabla\times\mathbf{B}\right)-\epsilon_{0}\mathbf{E}\times\left(\nabla\times\mathbf{E}\right)\nonumber$ $\displaystyle$ $\displaystyle =$ $\displaystyle \epsilon_{0}\left(\nabla\cdot\mathbf{E}\right)\mathbf{E}+\frac{1}{\mu_{0}}\left(\nabla\cdot\mathbf{B}\right)\mathbf{B}-\epsilon_{0}\frac{\partial}{\partial t}\left(\mathbf{E}\times\mathbf{B}\right)-\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle$ $\displaystyle \frac{1}{2}\nabla\left(\epsilon_{0}E^{2}+\frac{1}{\mu_{0}}B^{2}\right)+\epsilon_{0}\left(\mathbf{E}\cdot\nabla\right)\mathbf{E}+\frac{1}{\mu_{0}}\left(\mathbf{B}\cdot\nabla\right)\mathbf{B}\nonumber$ $\displaystyle$ $\displaystyle =$ $\displaystyle \epsilon_{0}\left[\left(\nabla\cdot\mathbf{E}\right)\mathbf{E}+\left(\mathbf{E}\cdot\nabla\right)\mathbf{E}\right]+\frac{1}{\mu_{0}}\left[\left(\nabla\cdot\mathbf{B}\right)\mathbf{B}+\left(\mathbf{B}\cdot\nabla\right)\mathbf{B}\right]-\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle$ $\displaystyle \frac{1}{2}\nabla\left(\epsilon_{0}E^{2}+\frac{1}{\mu_{0}}B^{2}\right)-\epsilon_{0}\frac{\partial}{\partial t}\left(\mathbf{E}\times\mathbf{B}\right)\nonumber$

It might not seem that we’re making any progress, since the equations just get longer with each alteration. However, we can now introduce the Maxwell stress tensor ${\overleftrightarrow{\mathbf{T}}}$ which is a ${3\times3}$ matrix with components defined by

$\displaystyle \boxed{T_{ij}\equiv\epsilon_{0}\left(E_{i}E_{j}-\frac{1}{2}\delta_{ij}E^{2}\right)+\frac{1}{\mu_{0}}\left(B_{i}B_{j}-\frac{1}{2}\delta_{ij}B^{2}\right)} \ \ \ \ \ (21)$

Note that the tensor is symmetric: ${T_{ij}=T_{ji}}$. If we define the scalar product of the tensor with an ordinary vector to be another vector:

$\displaystyle \left[\mathbf{a}\cdot\overleftrightarrow{\mathbf{T}}\right]_{j}=\sum_{i}a_{i}T_{ij} \ \ \ \ \ (22)$

where the subscript ${j}$ indicates the ${j}$th component of the resulting vector, then the divergence is

 $\displaystyle \left[\nabla\cdot\overleftrightarrow{\mathbf{T}}\right]_{j}$ $\displaystyle =$ $\displaystyle \sum_{i}\partial_{i}T_{ij}\ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \epsilon_{0}\sum_{i}\left(\left(\partial_{i}E_{i}\right)E_{j}+E_{i}\left(\partial_{i}E_{j}\right)-\frac{1}{2}\delta_{ij}\partial_{i}E^{2}\right)+\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle$ $\displaystyle \frac{1}{\mu_{0}}\sum_{i}\left(\left(\partial_{i}B_{i}\right)B_{j}+B_{i}\left(\partial_{i}B_{j}\right)-\frac{1}{2}\delta_{ij}\partial_{i}B^{2}\right)\nonumber$ $\displaystyle$ $\displaystyle =$ $\displaystyle \epsilon_{0}\left(\left(\nabla\cdot\mathbf{E}\right)E_{j}+\left(\mathbf{E}\cdot\nabla\right)E_{j}-\frac{1}{2}\partial_{j}E^{2}\right)+\ \ \ \ \ (25)$ $\displaystyle$ $\displaystyle$ $\displaystyle \frac{1}{\mu_{0}}\left(\left(\nabla\cdot\mathbf{B}\right)B_{j}+\left(\mathbf{B}\cdot\nabla\right)B_{j}-\frac{1}{2}\partial_{j}B^{2}\right)\nonumber$

Comparing this with 20, we see that we can write ${\mathbf{f}}$ in terms of ${\overleftrightarrow{\mathbf{T}}}$ and the Poynting vector as

$\displaystyle \boxed{\mathbf{f}=\nabla\cdot\overleftrightarrow{\mathbf{T}}-\epsilon_{0}\mu_{0}\frac{\partial\mathbf{S}}{\partial t}} \ \ \ \ \ (26)$

The total force on the volume is then

 $\displaystyle \mathbf{F}$ $\displaystyle =$ $\displaystyle \int_{\mathcal{V}}\mathbf{f}d^{3}\mathbf{r}\ \ \ \ \ (27)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int_{\mathcal{V}}\left(\nabla\cdot\overleftrightarrow{\mathbf{T}}-\epsilon_{0}\mu_{0}\frac{\partial\mathbf{S}}{\partial t}\right)d^{3}\mathbf{r} \ \ \ \ \ (28)$

From the formula 23 for the divergence, we can see that the vector resulting from the divergence has as its components the divergences of each column of ${\overleftrightarrow{\mathbf{T}}}$. Therefore we can apply the divergence theorem to the first term in the integrand to get

$\displaystyle \boxed{\mathbf{F}=\int_{\mathcal{S}}\overleftrightarrow{\mathbf{T}}\cdot d\mathbf{a}-\epsilon_{0}\mu_{0}\frac{\partial}{\partial t}\int_{\mathcal{V}}\mathbf{S}d^{3}\mathbf{r}} \ \ \ \ \ (29)$

where ${\mathcal{S}}$ is any surface that encloses only the charges and currents within ${\mathcal{V}}$.

Example We can revisit the problem of finding the magnetic force between the two halves of a spherical shell of surface charge density ${\sigma}$ rotating with angular velocity ${\boldsymbol{\omega}=\omega\hat{\mathbf{z}}}$. In our earlier solution we used the Biot-Savart law and integrated over each differential ring in the rotating sphere. Using the stress tensor, we can integrate over any volume that encloses the upper half of the sphere, so we can choose the half space consisting of all space above the ${xy}$ plane (we’re assuming that the centre of the sphere is at the origin, so the ${xy}$ plane contains the sphere’s equator). Since the distribution of charges and currents is finite, all fields will go to zero at infinity, so we need to integrate only over the ${xy}$ plane.

We saw earlier that the magnetic field inside and outside the sphere is

$\displaystyle \mathbf{B}=\begin{cases} \frac{2\mu_{0}R\omega\sigma}{3}\left(\cos\theta\hat{\mathbf{r}}-\sin\theta\hat{\boldsymbol{\theta}}\right)=\frac{2\mu_{0}R\omega\sigma}{3}\hat{\mathbf{z}} & rR \end{cases} \ \ \ \ \ (30)$

In the ${xy}$ plane, ${\theta=\pi/2}$ so the field is

$\displaystyle \mathbf{B}=\begin{cases} \frac{2\mu_{0}R\omega\sigma}{3}\hat{\mathbf{z}} & rR \end{cases} \ \ \ \ \ (31)$

Since we’re interested only in the magnetic field, we can ignore ${\mathbf{E}}$ here, although there is a repulsive force between the two hemispheres due to the electric field as well. Also, as the currents are steady, ${\partial\mathbf{S}/\partial t=0}$.

From the symmetry of the problem, the force is in the ${z}$ direction, so we need to work out only ${\left[\overleftrightarrow{\mathbf{T}}\cdot d\mathbf{a}\right]_{z}}$. We get ${T_{xz}=T_{yz}=0}$ because ${B_{x}=B_{y}=0}$ on the ${xy}$ plane, so we’re left with just ${T_{zz}}$:

$\displaystyle T_{zz}=\frac{1}{2\mu_{0}}B_{z}^{2}=\begin{cases} \frac{2}{9}\mu_{0}\sigma^{2}\omega^{2}R^{2} & rR \end{cases} \ \ \ \ \ (32)$

The total force is then (the minus sign is because ${T_{zz}>0}$ and ${d\mathbf{a}}$ points towards ${-z}$):

 $\displaystyle \mathbf{F}$ $\displaystyle =$ $\displaystyle \int_{\mathcal{S}}\overleftrightarrow{\mathbf{T}}\cdot d\mathbf{a}\ \ \ \ \ (33)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\hat{\mathbf{z}}\left[\frac{2}{9}\mu_{0}\sigma^{2}\omega^{2}R^{2}2\pi\int_{0}^{R}r\; dr+\frac{2\pi}{18}\mu_{0}\sigma^{2}\omega^{2}R^{8}\int_{R}^{\infty}\frac{r\; dr}{r^{6}}\right]\ \ \ \ \ (34)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\hat{\mathbf{z}}\left(\frac{2\pi}{9}\mu_{0}\sigma^{2}\omega^{2}R^{4}+\frac{\pi}{36}\mu_{0}\sigma^{2}\omega^{2}R^{4}\right)\ \ \ \ \ (35)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{\pi}{4}\mu_{0}\sigma^{2}\omega^{2}R^{4}\hat{\mathbf{z}} \ \ \ \ \ (36)$

This agrees with the result we got earlier using the Biot-Savart law.

# Stress-energy tensor: negative pressure revisited

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 20; 9.

Suppose we have a stress-energy tensor of the form

$\displaystyle T^{ij}=\left[\begin{array}{cccc} \rho & 0 & 0 & 0\\ 0 & \alpha\rho & 0 & 0\\ 0 & 0 & \alpha\rho & 0\\ 0 & 0 & 0 & \alpha\rho \end{array}\right] \ \ \ \ \ (1)$

where ${\rho}$ is the energy density of the perfect fluid in its rest frame and ${\alpha}$ is a scalar constant.

If an observer that moves with four-velocity ${\mathbf{u}_{obs}}$ relative to the fluid’s frame is to see the energy density in his frame as positive, this imposes a constraint on the possible values of ${\alpha}$. We can analyze this by looking at the situation in the observer’s local orthonormal frame (LOF). In this case, the global frame is the rest frame of the fluid. In the observer’s own local frame, because his metric is flat and the observer is not moving relative to himself, ${\mathbf{u}=\left[1,0,0,0\right]}$. That is, in the local frame, ${\mathbf{u}=\mathbf{o}_{t}}$. Therefore, ${\mathbf{u}_{obs}}$ in the global frame is the transformed version of ${\mathbf{o}_{t}}$ so in the global frame

 $\displaystyle \mathbf{o}_{t}$ $\displaystyle =$ $\displaystyle \mathbf{u}_{obs}\ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left[\gamma,\gamma v^{x},\gamma v^{y},\gamma v^{z}\right] \ \ \ \ \ (3)$

We can find the energy density seen by the moving observer by applying the transformation for the stress-energy tensor. The transformation of the energy density is

$\displaystyle T_{obs}^{tt}=\left(\rho+P\right)\gamma^{2}-P \ \ \ \ \ (4)$

with ${P=\alpha\rho}$, so we get

$\displaystyle T_{obs}^{tt}=\gamma^{2}\left(1+\alpha\right)\rho-\alpha\rho>0 \ \ \ \ \ (5)$

Since the energy density ${\rho}$ in the rest frame is known to be positive, we can cancel it off and get

$\displaystyle \alpha>-\frac{\gamma^{2}}{\gamma^{2}-1} \ \ \ \ \ (6)$

If we want this condition to be satisfied for all possible observers, then we must take the limit as the observer’s speed ${v\rightarrow1}$, or ${\gamma\rightarrow\infty}$, giving

$\displaystyle \alpha>-1 \ \ \ \ \ (7)$

A negative ${\alpha}$ admits the possibility of a negative pressure, which we’ve already seen can’t correspond to a perfect fluid, even though it gives an acceptable stress-energy tensor.

# 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.

# Stress-energy tensor in a local orthonormal frame

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 20; Problem 20.8.

Earlier we saw that it’s possible to define a local orthogonal frame (LOF) at any event in spacetime, with that frame embedded in a global system. Our earlier example defined a locally flat frame embedded within the Schwarzschild metric, but we can also use it to compare two inertial frames, with one frame moving with a four-velocity ${\mathbf{u}_{obs}}$ relative to the other.

As before, we define the observer’s local basis vectors as ${\mathbf{o}_{i}}$ with coordinates in the observer’s frame:

 $\displaystyle \mathbf{o}_{t}$ $\displaystyle =$ $\displaystyle \left[1,0,0,0\right]\ \ \ \ \ (1)$ $\displaystyle \mathbf{o}_{x}$ $\displaystyle =$ $\displaystyle \left[0,1,0,0\right]\ \ \ \ \ (2)$ $\displaystyle \mathbf{o}_{y}$ $\displaystyle =$ $\displaystyle \left[0,0,1,0\right]\ \ \ \ \ (3)$ $\displaystyle \mathbf{o}_{z}$ $\displaystyle =$ $\displaystyle \left[0,0,0,1\right] \ \ \ \ \ (4)$

To find the components of a four-vector ${A^{i}}$ in the observer’s frame, we can work out the scalar product in any frame and, since a scalar is invariant, it will have the same value in all frames. That is

$\displaystyle A_{obs}^{i}=\eta^{ij}g_{km}\left(\mathbf{o}_{j}\right)^{k}A^{m} \ \ \ \ \ (5)$

where ${g_{km}}$ is the metric in the global frame and ${\eta^{ij}}$ is the flat metric in the observer’s frame.

For a second-rank tensor that is the product of two four-vectors, that is

$\displaystyle C^{ij}=A^{i}B^{j} \ \ \ \ \ (6)$

we can just apply the transformation to each four-vector separately, so we get

 $\displaystyle C_{obs}^{ip}$ $\displaystyle =$ $\displaystyle \eta^{ij}g_{km}\left(\mathbf{o}_{j}\right)^{k}A^{m}\eta^{pq}g_{rs}\left(\mathbf{o}_{q}\right)^{r}B^{s}\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \eta^{ij}g_{km}\left(\mathbf{o}_{j}\right)^{k}\eta^{pq}g_{rs}\left(\mathbf{o}_{q}\right)^{r}C^{ms} \ \ \ \ \ (8)$

Although not all second-rank tensors are the product of two four-vectors, this seems a reasonable definition of the transformation.

Suppose that we now have a perfect fluid at rest in its own frame and an observer that moves with four-velocity ${\mathbf{u}_{obs}}$ relative to the fluid’s frame. In this case, the global frame is the rest frame of the fluid. In the observer’s own local frame, because his metric is flat and the observer is not moving relative to himself, ${\mathbf{u}=\left[1,0,0,0\right]}$. That is, in the local frame, ${\mathbf{u}=\mathbf{o}_{t}}$. Therefore, ${\mathbf{u}_{obs}}$ in the global frame is the transformed version of ${\mathbf{o}_{t}}$ so in the global frame

 $\displaystyle \mathbf{o}_{t}$ $\displaystyle =$ $\displaystyle \mathbf{u}_{obs}\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left[\gamma,\gamma v^{x},\gamma v^{y},\gamma v^{z}\right] \ \ \ \ \ (10)$

In the fluid’s frame, the stress-energy tensor is

$\displaystyle T^{ij}=\left[\begin{array}{cccc} \rho & 0 & 0 & 0\\ 0 & P & 0 & 0\\ 0 & 0 & P & 0\\ 0 & 0 & 0 & P \end{array}\right] \ \ \ \ \ (11)$

We can find the energy density seen by the moving observer by applying 8 with ${C^{ij}=T^{ij}}$. Since both frames are inertial, both metrics are the flat space metric, that is, ${g_{ij}=\eta_{ij}}$. We thus get

$\displaystyle T_{obs}^{tt}=\eta^{tj}\eta_{km}\left(\mathbf{o}_{j}\right)^{k}\eta^{tq}\eta_{rs}\left(\mathbf{o}_{q}\right)^{r}T^{ms} \ \ \ \ \ (12)$

Because ${\eta_{ij}}$ and ${T^{ms}}$ are diagonal, the only non-zero terms in the sum are with ${j=t}$, ${q=t}$, ${k=m}$, ${r=s}$ and ${m=s}$, so we get

 $\displaystyle T_{obs}^{tt}$ $\displaystyle =$ $\displaystyle \eta_{km}\left(\mathbf{o}_{t}\right)^{k}\eta_{rs}\left(\mathbf{o}_{t}\right)^{r}T^{ms}\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(\mathbf{o}_{t}\right)^{t}\left(\mathbf{o}_{t}\right)^{t}T^{tt}+\sum_{i=x,y,z}\left(\mathbf{o}_{t}\right)^{i}\left(\mathbf{o}_{t}\right)^{i}T^{ii}\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \rho\gamma^{2}+P\gamma^{2}v^{2}\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \rho\gamma^{2}+P\gamma^{2}\left(1-1+v^{2}\right)\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(\rho+P\right)\gamma^{2}+P\gamma^{2}\left(v^{2}-1\right)\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(\rho+P\right)\gamma^{2}-P \ \ \ \ \ (18)$

$\displaystyle T^{ij}=\left(\rho+P\right)u^{i}u^{j}+Pg^{ij} \ \ \ \ \ (19)$

so the ${tt}$ component is

 $\displaystyle T^{tt}$ $\displaystyle =$ $\displaystyle \left(\rho+P\right)u^{t}u^{t}+\eta^{tt}P\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(\rho+P\right)\gamma^{2}-P \ \ \ \ \ (21)$

so we get the same answer.

# Metric tensor as a stress-energy tensor

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 20; Problem 20.7.

A curious possibility for the stress-energy tensor is

$\displaystyle T^{ij}=-\Lambda g^{ij} \ \ \ \ \ (1)$

where ${\Lambda}$ is a positive constant and ${g^{ij}}$ is any metric tensor. In a local inertial frame (LIF), ${g^{ij}=\eta^{ij}}$ and

$\displaystyle T^{ij}=\Lambda\left[\begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & -1 & 0 & 0\\ 0 & 0 & -1 & 0\\ 0 & 0 & 0 & -1 \end{array}\right] \ \ \ \ \ (2)$

Note that this tensor satisfies the energy conservation equation ${\nabla_{i}T^{ij}=0}$ as the covariant derivative ${\nabla_{i}g^{ij}}$ is always zero.

Comparing this with the form of ${T^{ij}}$ for a perfect fluid in its LIF

$\displaystyle T^{ij}=\left[\begin{array}{cccc} \rho & 0 & 0 & 0\\ 0 & P & 0 & 0\\ 0 & 0 & P & 0\\ 0 & 0 & 0 & P \end{array}\right] \ \ \ \ \ (3)$

we see that the energy density is ${\rho=\Lambda>0}$ and the pressure is ${P=-\Lambda<0}$. Such a tensor cannot arise from a perfect fluid because for such a fluid, for example

$\displaystyle T^{xx}=\int\int\int N\left(p\right)\frac{\left(p^{x}\right)^{2}}{p^{t}V}dp^{x}dp^{y}dp^{z} \ \ \ \ \ (4)$

The integrand is instrinsically non-negative because ${N\left(p\right)}$ is the number of particles in volume ${V}$ with a magnitude of momentum ${p}$ and must be non-negative. The component ${p^{t}=\gamma m}$ for massive particles or ${p^{t}=E}$ for photons, but in either case it too is positive. The numerator ${\left(p^{x}\right)^{2}}$, being a square, is also non-negative. Thus for a perfect fluid in its LIF, ${T^{xx}\ge0}$.

Although this tensor cannot be that of a perfect fluid, it does play a role in general relativity as we’ll hopefully see a bit later.