Stress–energy tensor

Remarks

Example: dust of energy density $\rho \ge 0$ moving with four-velocity $U$ has $T = \rho \, U^\flat \otimes U^\flat$, where $U^\flat = g(U, \cdot)$. For the observer $u = U$, $T(u, u) = \rho \, g(U, U)^2 = \rho$, and $T(u, e) = 0$ and $T(e, e') = 0$, since $g(U, e) = 0$: no momentum flow and no stress in the rest frame. A perfect fluid with pressure $p$ has $T = (\rho + p) U^\flat \otimes U^\flat + p \, g$. Non-example: an antisymmetric tensor such as the electromagnetic tensor $F$ is not a stress–energy tensor, since a stress–energy tensor must be symmetric. The electromagnetic field has the stress–energy tensor $F_{\mu\alpha}F_\nu{}^\alpha - \frac14 g_{\mu\nu} F_{\alpha\beta}F^{\alpha\beta}$, up to a constant factor. In an inertial frame of special relativity, the components $T^{00}$, $T^{0i}$ and $T^{ij}$ are the energy density, the energy flux and the stress tensor.