The energy-momentum tensor¶

Dust, fluid, and flux¶

A fluid is a material which flows, that is, has forces perpendicular to some imaginary surface which are much greater than the forces parallel to it. A perfect fluid is the limit where the substance has pressure but zero stress.

Dust is an idealised form of matter consisting of a collection of non-interacting particles which are not moving relative to one-another, and so the collection has zero pressure. Thus there is a momentarily comoving reference frame (MCRF) with respect to which all of the particles have zero velocity. If all of the particles have the same rest mass, \(m\), but the cloud of dust may have a varying mass density, \(n\).

Transforming to a frame movig at a velocity \(\vec{v}\) to the MCRF then the stationary volume element \(\Delta x \Delta y \Delta z\) will be Lorentz contracted to \(\Delta x' \Delta y' \Delta z' = (\Delta x / \gamma) \Delta y \Delta z\), for relative motion along the \(x\)-axis, increasing the number density to \(n \gamma\), producing a flux through the area \(\Delta y \Delta z\). All the particles pass through \(\Delta y' \Delta z'\) in a time \(\Delta t'\) for \(\Delta x' = v \Delta t'\), so the total number of particles is

This produces an \(x\)-directed flux,

or, defining a flux vector, \(\vec{N}\), and letting \(\vec{U} = (\gamma, \gamma v^x, \gamma v^y, \gamma v^z)\) be the velocity 4-vector,

In the MCRF \(\vec{U} = (1, \vec{0})\), so \(g(\vec{U}, \vec{U}) = -1\) so

The components of the flux vector \(\vec{N}\) in the frame are then

Any function \(\phi(t,x,y,z)\) over spacetime defines a constant surface, and its gradient \(\of{\dd}{\phi}\) defines a normal to the surface. The unit normal gradient is defined

Contracting this with the flux vector gives the flux across the corresponding surface.

The energy-momentum tensor¶

Energy and mass are interconvertible (special relativity); for a dust particle of mass \(m\) the energy density of the dust (in the MCRF) is \(E = mn\). In a moving frame the number density becomes \(n \gamma\), and the energy of each particle is \(\gamma m\), giving a total energy of \(\gamma^2 mn\) in a moving frame. The \(\gamma^2\) term cannot be generated by a simple Lorentz boost, so we require something of higher order. Forming the \((2,0)\)-tensor

with \(\rho = mn\) the mass density of the dust. This is the energy-momentum (stress-energy) tensor. The components of the tensor can be retrieved by contracting it with basis one-forms, \(\of{\omega}^{\alpha} = \of{\dd}x^{\alpha}\),

(2)¶\[\]

The \(\mathbf{00}\) component, \(T^{00}\) is the energy (flow of zeroth component of momentum across surface of constant time). The \(\mathbf{0i}\) component, \(T^{0i} = \gamma m \times \gamma n v^i\) is the energy flux across a surface of constant \(x^i\). The \(\mathbf{i0}\) component, \(T^{i0} = p^i \times N^0 = m \gamma v^i \times \gamma n\) is the flux of the \(i\)th component of momentum across a surface of constant time into the future. This is the \(i\)th component of momentum density. This is the same as the energy flux, so

The \(\mathbf{ij}\) component is the flux of \(i\)-momentum across a surface of constant \(x^j\), and has the dimensions of a pressure.

In general \(\ten{T}\) is symmetric, \(T^{\alpha \beta} = T^{\beta \alpha}\).

In a perfect fluid there is no preferred direction, so the spatial part of \(\ten{T}\) is proportional to the spatial part of the metric, and there is no momentum transport perpendicular to the surface of a fluid element. Thus

for a perfect fluid. Hence,

Dust has no pressure, so in the MCRF it has

The final property is the conservation law; if energy is conserved then the energy-momentum entering an arbitrary volume must be equal to that leaving it. Thus

That is

Similarly