Energy-Momentum Tensor

Throughout this chapter, I will be using the terms density and flux. For 4D density will be a value per unit volume at constant time. For example the number density (e.g. the number of particles):

N^t = \frac{\#}{\Delta x \Delta y \Delta z},

while the flux is some value passing through a unit surface in unit time while the third coordinate is held constant:

\begin{align*} N^x &= \frac{\#}{\Delta t \Delta y \Delta z}, \\ N^y &= \frac{\#}{\Delta t \Delta x \Delta z}, \\ N^z &= \frac{\#}{\Delta t \Delta x \Delta y}. \end{align*}

This can be visualized in 2D:

\begin{align*} N^t = \frac{\#}{\Delta x} = 2, \\ N^x = \frac{\#}{\Delta t} = 1. \end{align*}

Consider the Galilean transform:

Number flux and density with Galilean transform

\begin{align*} \tilde{N}^t = \frac{\#}{\Delta \tilde{x}} = 2, \\ \tilde{N}^x = \frac{\#}{\Delta \tilde{t}} = 0. \end{align*}

and the Lorentz transform:

Number flux and density with Lorentz transform

\begin{align*} \tilde{N}^t = \frac{\#}{\Delta \tilde{x}} = 1, \\ \tilde{N}^x = \frac{\#}{\Delta \tilde{t}} = 0. \end{align*}

Recall the initial definition:

\begin{align*} N^t &= \frac{\#}{\Delta x \Delta y \Delta z}, \\ N^x &= \frac{\#}{\Delta t \Delta y \Delta z}, \\ N^y &= \frac{\#}{\Delta t \Delta x \Delta z}, \\ N^z &= \frac{\#}{\Delta t \Delta x \Delta y}. \end{align*}

In Galilean relativity, the $N^t$ component is the rest number density:

n = N^t = \frac{\#}{\Delta x \Delta y \Delta z}.

And use it in the definition of the spatial components:

\begin{align*} N^x &= \frac{\#}{\Delta t \Delta y \Delta z} = \frac{\Delta x}{\Delta t} \frac{\#}{\Delta x \Delta y \Delta z} = u^x n, \\ N^y &= \frac{\#}{\Delta t \Delta x \Delta z} = \frac{\Delta y}{\Delta t} \frac{\#}{\Delta x \Delta y \Delta z} = u^y n, \\ N^z &= \frac{\#}{\Delta t \Delta x \Delta y} = \frac{\Delta z}{\Delta t} \frac{\#}{\Delta x \Delta y \Delta z} = u^z n, \end{align*}

or in vector notation:

\boldsymbol{N} = n \begin{bmatrix} 1 \\ u^x \\ u^y \\ u^z \end{bmatrix}.

In special relativity, with the Lorentz transformation, the vector

\boldsymbol{N} = n \left(\gamma \begin{bmatrix} 1 \\ u^x \\ u^y \\ u^z \end{bmatrix}\right),

where the vector in brackets is the four-velocity:

\boldsymbol{N} = n \boldsymbol{U}.

This number flux four-vector describes the total number (of partiacles, for example) in a space time box of size 1. If we want to describe the four-momentum in a box of size 1, another four-vector won't work. We need a 4x4 tensor - the energy momentum tensor.

Energy-Momentum Tensor

T^{\alpha \beta} = \begin{bmatrix} T^{tt} & T^{tx} & T^{ty} & T^{tz} \\ T^{xt} & T^{xx} & T^{xy} & T^{xz} \\ T^{yt} & T^{yx} & T^{yy} & T^{yz} \\ T^{zt} & T^{zx} & T^{zy} & T^{zz} \end{bmatrix}

The energy-momentum tensor component $T^{\alpha \beta}$ is the flux of the $P^{\alpha}$ component in a unit volume of constant $\beta$ . Consider the following illustration:

where the purple vectors are the four-momentum vectors. In the image above, the components are:

\boldsymbol{P} = P^t \boldsymbol{e_t} + P^x \boldsymbol{e_x} = 0.2 \boldsymbol{e_t} + 0.1 \boldsymbol{e_x}.

The flux of the $t$ and $x$ components when $t$ is held constant is:

\begin{align*} T^{t t} &= 2 P^t = 0.4, \\ T^{x t} &= 2 P^x = 0.2, \end{align*}

The flux of the $t$ and $x$ components when $x$ is held constant is:

\begin{align*} T^{t x} &= P^t = 0.2, \\ T^{x x} &= P^x = 0.1, \end{align*}

and as seen on the $T^{tx}$ and $T^{xt}$ , the tensor is symmetric:

T^{\alpha \beta} = T^{\beta \alpha}.

Another way to look at it: if we take the four-momentum $\boldsymbol{P}$ and in a unit volume of constant $\beta$ coordinate, we will get a column of the energy-momentum tensor $T^{\alpha \beta}$ , where $\alpha$ are the components of the four-momentum.

The energy-momentum tensor takes two vectors as inputs: $T(\boldsymbol{V}, \boldsymbol{U})$ . The output is the four momentum in the $\boldsymbol{V}$ direction inside a unit 3D box perpendicular to $\boldsymbol{U}$ . Inputting the basis vectors gives us the components:

T(\boldsymbol{e_{\alpha}}, \boldsymbol{e_{\beta}}) = T^{\alpha \beta}.

The $P^t$ component is energy, this means that the $T^{tt}$ component is the energy density and $T^{ti}$ is the energy flux:

\begin{align*} T^{tt} &= \frac{P^t}{\Delta x \Delta y \Delta z} = \frac{E}{V}, \\ T^{tx} &= \frac{P^t}{\Delta t \Delta y \Delta z}, \\ T^{ty} &= \frac{P^t}{\Delta t \Delta x \Delta z}, \\ T^{tz} &= \frac{P^t}{\Delta t \Delta x \Delta y}. \end{align*}

The $T^{it}$ components are the momentum densities:

\begin{align*} T^{xt} &= \frac{P^x}{\Delta x \Delta y \Delta z} = \frac{P^x}{V}, \\ T^{yt} &= \frac{P^y}{\Delta x \Delta y \Delta z} = \frac{P^y}{V}, \\ T^{zt} &= \frac{P^z}{\Delta x \Delta y \Delta z} = \frac{P^z}{V}. \end{align*}

The $T^{ii}$ components are pressure (force per unit area):

\begin{align*} T^{xx} &= \frac{P^x}{\Delta t \Delta y \Delta z} = \frac{F^x}{\Delta y \Delta z}, \\ T^{yy} &= \frac{P^y}{\Delta t \Delta x \Delta z} = \frac{F^y}{\Delta x \Delta z}, \\ T^{zz} &= \frac{P^z}{\Delta t \Delta x \Delta y} = \frac{F^z}{\Delta x \Delta y}. \end{align*}

And the $T^{ij}$ components are the shear stress.

Conservation of Energy-Momentum

The conservation of energy-momentum may be summed by the following statement: if the momentum flows into the box (negative divergence), then the density has to increase and if the momentum flows out of the box (positive divergence), then the density has to decrease:

\partial_{\nu} T^{\mu \nu} = T^{\mu \nu}{}_{, \nu} = 0,

this only works in cartesian coordinates. For general coordinates, we have to consider the covariant derivative.

The covariant derivative of the energy-momentum tensor is as follows:

\begin{align*} \nabla_{\boldsymbol{e_{\sigma}}} T &= \nabla_{\boldsymbol{e_{\sigma}}} (T^{\mu \nu} \boldsymbol{e_{\mu}} \otimes \boldsymbol{e_{\nu}}) \\ &= \nabla_{\boldsymbol{e_{\sigma}}} (T^{\mu \nu}) \boldsymbol{e_{\mu}} \otimes \boldsymbol{e_{\nu}} + T^{\mu \nu} \nabla_{\boldsymbol{e_{\sigma}}} (\boldsymbol{e_{\mu}}) \otimes \boldsymbol{e_{\nu}} + T^{\mu \nu} \boldsymbol{e_{\mu}} \otimes \nabla_{\boldsymbol{e_{\sigma}}} (\boldsymbol{e_{\nu}}) \\ &= T^{\mu \nu}{}_{,\sigma} \boldsymbol{e_{\mu}} \otimes \boldsymbol{e_{\nu}} + T^{\mu \nu} \Gamma^{\rho}{}_{\sigma \mu} \boldsymbol{e_{\rho}} \otimes \boldsymbol{e_{\nu}} + T^{\mu \nu} \boldsymbol{e_{\mu}} \otimes \Gamma^{\rho}{}_{\sigma \nu} \boldsymbol{e_{\rho}} \\ &= T^{\mu \nu}{}_{,\sigma} \boldsymbol{e_{\mu}} \otimes \boldsymbol{e_{\nu}} + T^{\rho \nu} \Gamma^{\mu}{}_{\sigma \rho} \boldsymbol{e_{\mu}} \otimes \boldsymbol{e_{\nu}} + T^{\mu \rho} \boldsymbol{e_{\mu}} \otimes \Gamma^{\nu}{}_{\sigma \rho} \boldsymbol{e_{\nu}} \\ &= (T^{\mu \nu}{}_{,\sigma} + T^{\rho \nu} \Gamma^{\mu}{}_{\sigma \rho} + T^{\mu \rho} \Gamma^{\nu}{}_{\sigma \rho}) \boldsymbol{e_{\mu}} \otimes \boldsymbol{e_{\nu}} \\ &= (T^{\mu \nu}{}_{,\sigma} + T^{\rho \nu} \Gamma^{\mu}{}_{\sigma \rho} + T^{\mu \rho} \Gamma^{\nu}{}_{\sigma \rho}) \boldsymbol{e_{\mu}} \otimes \boldsymbol{e_{\nu}} \\ &= T^{\mu \nu}{}_{; \sigma} \boldsymbol{e_{\mu}} \otimes \boldsymbol{e_{\nu}}. \end{align*}

The proper conservation of four-momentum is:

T^{\mu \nu}{}_{; \nu} = 0.

Note: in general relativity, the above formula only applies locally.

Energy-Momentum Tensor of Dust

Dust is a collection of particles not exerting pressure on each other. Each particle has four-momentum $\boldsymbol{P}$ and the densities are described by the number-flux four-vector $\boldsymbol{N}$ :

T^{\mu \nu} = P^{\mu} N^{\nu}.

The four-momentum may be rewritten as the rest mass times the four-velocity:

\boldsymbol{P} = m \boldsymbol{U},

and the number-flux four-vector may be rewritten as the rest number density times the four-velocity:

\boldsymbol{N} = n \boldsymbol{U},

substituting into the above definition for the energy-momentum tensor:

T^{\mu \nu} = m n U^{\mu} U^{\nu}.

Multiplying the mass by the number density, we get the mass density $\rho$ :

T^{\mu \nu} = \rho U^{\mu} U^{\nu}.

In the rest dust's rest frame, the only non zero components are:

\begin{align*} P^t &= m, \\ N^t &= n, \end{align*}

making the only non zero component of the energy-momentum tensor equal to:

T^{00} = \rho,

or in SI units:

T^{00} = \rho c^2.

This is the energy density. It is related to the mass-energy equivalence:

E = mc^2.

Energy-Momentum Tensor of a Perfect Fluid

Perfect fluid means that the particles exert pressure $p$ equally in all directions. The energy-momentum tensor components are equal to:

T^{\mu \nu} = (\rho + p) U^{\mu} U^{\nu} + p g^{\mu \nu},

where $g^{\mu \nu}$ is the inverse metric tensor with space-positive signature.

In its rest frame, the energy-momentum tensor look like this:

T^{\mu \nu} = \begin{bmatrix} \rho & 0 & 0 & 0 \\ 0 & p & 0 & 0 \\ 0 & 0 & p & 0 \\ 0 & 0 & 0 & p \end{bmatrix}.