Stress-energy tensor

Stress Energy Momentum Tensor

The stress-energy tensor for short of elements $T^{\mu \nu }$ contains information about the stress, pressure, energy density, and momentum density of the matter in the spacetime. For rectilinear inertial frame coordinates

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

is a statment of energy conservation. The closest thing to such a statment for general relativity where globally rectilinear inertial frames don't exist when there is Riemannian spacetime curvature present is

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

which is a statment of energy conservation for local free fall frames for which the Christoffel symbols vanish reducing it to the expression just above.

For such a rectilinear inertial frame the elements of the stress energy tensor have the following interpretations

$T^{00}$ is the coordinate frame energy density.

$T^{ii}$ is a flow of momentum per area in the $x^{i}$ direction or the pressure on a plane whose normal is in the $x^{i}$ direction.

$T^{ij}$ is the $x^{i}$ component of momentum per area in the $x^{j}$ direction or describes a shearing from stresses.

$T^{0i}$ is the volume density of the ith component of momentum flow.

Of Dust

Given that an observer local to and moving with an element of dust moving uniformly with the bits around it finds that its local energy density is $\rho _{0}c^{2}$ then the stress-energy tensor according to a frame for which it moves at 4-velocity $U^{\mu }$ is

$T^{\mu \nu }=\rho _{0}U^{\mu }U^{\nu }$

Of an Ideal Fluid

Given that an observer local to and moving with an element of fluid finds that its local energy density is $\rho _{0}c^{2}$ and that its local pressure is p, then the stress-energy tensor according to a frame for which it moves at 4-velocity $U^{\mu }$ and the contravariant metric tensor is $g^{\mu \nu }$is

$T^{\mu \nu }=\left(\rho _{0}+{\frac {p}{c^{2}}}\right)U^{\mu }U^{\nu }-g^{\mu \nu }p$

Of the Massless Electromagnetic Field

In Newtonian gravitation it is specifically the mass of a thing that gravitates. The discovery of the Higgs particle settles the question on whether the photon has any mass. Since it does not interact with the Higgs field, it does not. General relativity describes gravitation differently than Newtonian physics though. In General relativity/Einstein equations it is the macroscopic feild's stress-energy tensor that couples to the Einstein curvature tensor in the field equations, not the mass of the field's mediating particles. If a field of particles gives rise to a field and the stress-energy tensor of that field is $T^{\mu \nu }$ then the fields particles are massless iff $T_{\mu }^{\mu }=0$. The stress-energy tensor can still have gravitational effect, even though the field particles are massless because $T^{\mu \nu }$ can be nonzero even when $T_{\mu }^{\mu }=0$, and it is $T^{\mu \nu }$ that is the source term in the field equations, not $T_{\mu }^{\mu }$.

Given an electromagnetic field tensor of $F_{\mu \nu }$ the stress-energy tensor for the electromagnetic field is

$T^{\mu \nu }=\epsilon \left(F^{\mu \lambda }{F_{\lambda }}^{\nu }+{\frac {1}{4}}g^{\mu \nu }F_{\lambda \sigma }F^{\lambda \sigma }\right)$

where in "local Cartesian coordinates" the electromagnetic field tensor is given by

$\|F'_{\mu \nu }\|={\begin{bmatrix}0&-E_{x}&-E_{y}&-E_{z}\\E_{x}&0&cB_{z}&-cB_{y}\\E_{y}&-cB_{z}&0&cB_{x}\\E_{z}&cB_{y}&-cB_{x}&0\end{bmatrix}}$

And a test charge will experience a 4-force from it of

$F^{\lambda }=q{\frac {U^{\mu }}{c}}g_{\mu \nu }F^{\nu \lambda }$

In terms of E&B in Cartesian inertial frame coordinates for flat spacetime the stress-energy tensor of an electromagnetic field can be written out as

$T^{00}={\frac {1}{2}}\epsilon \left(E^{2}+c^{2}B^{2}\right)$

$T^{0i}=T^{i0}=\epsilon \left(\mathbf {E} \times c\mathbf {B} \right)^{i}$

$T^{ij}=-\epsilon \left(E^{i}E^{j}+c^{2}B^{i}B^{j}-{\frac {1}{2}}\delta ^{ij}\left(E^{2}+c^{2}B^{2}\right)\right)$