Stress–energy tensor

Most of this article works with the contravariant form, T^μν of the stress–energy tensor. However, it is often necessary to work with the covariant form,

${\displaystyle T_{\mu \nu }=T^{\alpha \beta }g_{\alpha \mu }g_{\beta \nu },}$

or the mixed form,

${\displaystyle T^{\mu }{}_{\nu }=T^{\mu \alpha }g_{\alpha \nu },}$

or as a mixed tensor density

${\displaystyle {\mathfrak {T}}^{\mu }{}_{\nu }=T^{\mu }{}_{\nu }{\sqrt {-g}}\,.}$

This article uses the spacelike sign convention (−+++) for the metric signature.

The stress–energy tensor is the conserved Noether current associated with spacetime translations.

The divergence of the non-gravitational stress–energy is zero. In other words, non-gravitational energy and momentum are conserved,

${\displaystyle 0=T^{\mu \nu }{}_{;\nu }=\nabla _{\nu }T^{\mu \nu }{}.\!}$

When gravity is negligible and using a Cartesian coordinate system for spacetime, this may be expressed in terms of partial derivatives as

${\displaystyle 0=T^{\mu \nu }{}_{,\nu }=\partial _{\nu }T^{\mu \nu }.\!}$

The integral form of this is

${\displaystyle 0=\int _{\partial N}T^{\mu \nu }\mathrm {d} ^{3}s_{\nu }\!}$

where N is any compact four-dimensional region of spacetime; ${\displaystyle \partial N}$ is its boundary, a three-dimensional hypersurface; and ${\displaystyle \mathrm {d} ^{3}s_{\nu }}$ is an element of the boundary regarded as the outward pointing normal.

In flat spacetime and using Cartesian coordinates, if one combines this with the symmetry of the stress–energy tensor, one can show that angular momentum is also conserved:

${\displaystyle 0=(x^{\alpha }T^{\mu \nu }-x^{\mu }T^{\alpha \nu })_{,\nu }.\!}$

When gravity is non-negligible or when using arbitrary coordinate systems, the divergence of the stress–energy still vanishes. But in this case, a coordinate-free definition of the divergence is used which incorporates the covariant derivative

${\displaystyle 0=\operatorname {div} T=T^{\mu \nu }{}_{;\nu }=\nabla _{\nu }T^{\mu \nu }=T^{\mu \nu }{}_{,\nu }+\Gamma ^{\mu }{}_{\sigma \nu }T^{\sigma \nu }+\Gamma ^{\nu }{}_{\sigma \nu }T^{\mu \sigma }}$

where ${\displaystyle \Gamma ^{\mu }{}_{\sigma \nu }}$ is the Christoffel symbol which is the gravitational force field.

Consequently, if ${\displaystyle \xi ^{\mu }}$ is any Killing vector field, then the conservation law associated with the symmetry generated by the Killing vector field may be expressed as

${\displaystyle 0=\nabla _{\nu }(\xi ^{\mu }T_{\mu }^{\nu })={\frac {1}{\sqrt {-g}}}\partial _{\nu }({\sqrt {-g}}\ \xi ^{\mu }T_{\mu }^{\nu })}$

The integral form of this is

${\displaystyle 0=\int _{\partial N}{\sqrt {-g}}\ \xi ^{\mu }T_{\mu }^{\nu }\ \mathrm {d} ^{3}s_{\nu }=\int _{\partial N}\xi ^{\mu }{\mathfrak {T}}_{\mu }^{\nu }\ \mathrm {d} ^{3}s_{\nu }}$

Note that the trace is defined to be ${\displaystyle T_{\mu }^{\mu }}$. Note that

${\displaystyle T_{\mu }^{\mu }=T^{\mu \nu }g_{\nu \mu }.}$

When we use our formula for the stress–energy tensor found above,

${\displaystyle T_{\mu }^{\mu }={\frac {\partial {\mathcal {L}}}{\partial (\partial _{\mu }\phi _{\alpha })}}g_{\mu \nu }\partial ^{\nu }\phi _{\alpha }-g_{\mu \nu }g^{\mu \nu }{\mathcal {L}}.}$

Using the raising and lowering properties of the metric, and the fact that ${\displaystyle g^{\mu \nu }g_{\mu \alpha }=\delta _{\alpha }^{\nu }}$,

${\displaystyle T_{\mu }^{\mu }={\frac {\partial {\mathcal {L}}}{\partial (\partial _{\mu }\phi _{\alpha })}}\partial _{\mu }\phi _{\alpha }-\delta _{\mu }^{\mu }{\mathcal {L}}.}$

Since ${\displaystyle \delta _{\mu }^{\mu }=4}$, so that we can conclude that

${\displaystyle T_{\mu }^{\mu }={\frac {\partial {\mathcal {L}}}{\partial (\partial _{\mu }\phi _{\alpha })}}\partial _{\mu }\phi _{\alpha }-4{\mathcal {L}}.}$

In general relativity, the stress tensor is studied in the context of the Einstein field equations which are often written as

${\displaystyle R_{\mu \nu }-{\tfrac {1}{2}}R\,g_{\mu \nu }+\Lambda g_{\mu \nu }={8\pi G \over c^{4}}T_{\mu \nu },}$

where ${\displaystyle R_{\mu \nu }}$ is the Ricci tensor, ${\displaystyle R}$ is the Ricci scalar (the tensor contraction of the Ricci tensor), ${\displaystyle g_{\mu \nu }\,}$ is the metric tensor, Λ is the cosmological constant (negligible at the scale of a galaxy or smaller), and ${\displaystyle G}$ is the universal gravitational constant.

In special relativity, the stress–energy of a non-interacting particle with rest mass m and trajectory ${\displaystyle \mathbf {x} _{\text{p}}(t)}$ is:

${\displaystyle T^{\alpha \beta }(\mathbf {x} ,t)={\frac {m\,v^{\alpha }(t)v^{\beta }(t)}{\sqrt {1-(v/c)^{2}}}}\;\,\delta (\mathbf {x} -\mathbf {x} _{\text{p}}(t))={\frac {E}{c^{2}}}\;v^{\alpha }(t)v^{\beta }(t)\;\,\delta (\mathbf {x} -\mathbf {x} _{\text{p}}(t))}$

where ${\displaystyle (v^{\alpha })_{\alpha =0,1,2,3}\!}$ is the velocity vector (which should not be confused with four-velocity, since it is missing a ${\displaystyle \gamma }$)

${\displaystyle (v^{\alpha })_{\alpha =0,1,2,3}=\left(1,{\frac {d\mathbf {x} _{\text{p}}}{dt}}(t)\right)\,,}$

${\displaystyle \delta }$ is the Dirac delta function and ${\displaystyle E={\sqrt {p^{2}c^{2}+m^{2}c^{4}}}}$ is the energy of the particle.

For a perfect fluid in thermodynamic equilibrium, the stress–energy tensor takes on a particularly simple form

${\displaystyle T^{\alpha \beta }\,=\left(\rho +{p \over c^{2}}\right)u^{\alpha }u^{\beta }+pg^{\alpha \beta }}$

where ${\displaystyle \rho }$ is the mass–energy density (kilograms per cubic meter), ${\displaystyle p}$ is the hydrostatic pressure (pascals), ${\displaystyle u^{\alpha }}$ is the fluid's four velocity, and ${\displaystyle g^{\alpha \beta }}$ is the reciprocal of the metric tensor. Therefore, the trace is given by

${\displaystyle T_{\,\alpha }^{\alpha }=g_{\alpha \beta }T^{\beta \alpha }=3p-\rho c^{2}\,.}$

The four-velocity satisfies

${\displaystyle u^{\alpha }u^{\beta }g_{\alpha \beta }=-c^{2}\,.}$

In an inertial frame of reference comoving with the fluid, better known as the fluid's proper frame of reference, the four velocity is

${\displaystyle (u^{\alpha })_{\alpha =0,1,2,3}=(1,0,0,0)\,,}$

the reciprocal of the metric tensor is simply

${\displaystyle (g^{\alpha \beta })_{\alpha ,\beta =0,1,2,3}\,=\left({\begin{matrix}-{\frac {1}{c^{2}}}&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{matrix}}\right)\,}$

and the stress–energy tensor is a diagonal matrix

${\displaystyle (T^{\alpha \beta })_{\alpha ,\beta =0,1,2,3}=\left({\begin{matrix}\rho &0&0&0\\0&p&0&0\\0&0&p&0\\0&0&0&p\end{matrix}}\right).}$

The Hilbert stress–energy tensor of a source-free electromagnetic field is

${\displaystyle T^{\mu \nu }={\frac {1}{\mu _{0}}}\left(F^{\mu \alpha }g_{\alpha \beta }F^{\nu \beta }-{\frac {1}{4}}g^{\mu \nu }F_{\delta \gamma }F^{\delta \gamma }\right)}$

where ${\displaystyle F_{\mu \nu }}$ is the electromagnetic field tensor.

The stress–energy tensor for a complex scalar field ${\displaystyle \phi }$ which satisfies the Klein–Gordon equation is

${\displaystyle T^{\mu \nu }={\frac {\hbar ^{2}}{m}}(g^{\mu \alpha }g^{\nu \beta }+g^{\mu \beta }g^{\nu \alpha }-g^{\mu \nu }g^{\alpha \beta })\partial _{\alpha }{\bar {\phi }}\partial _{\beta }\phi -g^{\mu \nu }mc^{2}{\bar {\phi }}\phi ,}$

and when the metric is flat (Minkowski in Cartesian Coordinates) its components work out to be:

${\displaystyle {\begin{aligned}T^{00}&={\frac {\hbar ^{2}}{mc^{4}}}\left(\partial _{0}{\bar {\phi }}\partial _{0}\phi +c^{2}\partial _{k}{\bar {\phi }}\partial _{k}\phi \right)+m{\bar {\phi }}\phi ,\\T^{0i}=T^{i0}&=-{\frac {\hbar ^{2}}{mc^{2}}}\left(\partial _{0}{\bar {\phi }}\partial _{i}\phi +\partial _{i}{\bar {\phi }}\partial _{0}\phi \right),\ \mathrm {and} \\T^{ij}&={\frac {\hbar ^{2}}{m}}\left(\partial _{i}{\bar {\phi }}\partial _{j}\phi +\partial _{j}{\bar {\phi }}\partial _{i}\phi \right)-\delta _{ij}\left({\frac {\hbar ^{2}}{m}}\eta ^{\alpha \beta }\partial _{\alpha }{\bar {\phi }}\partial _{\beta }\phi +mc^{2}{\bar {\phi }}\phi \right).\end{aligned}}}$

The Hilbert stress–energy tensor is defined as the functional derivative

${\displaystyle T_{\mu \nu }={\frac {-2}{\sqrt {-g}}}{\frac {\delta S_{\mathrm {matter} }}{\delta g^{\mu \nu }}}={\frac {-2}{\sqrt {-g}}}{\frac {\partial ({\sqrt {-g}}{\mathcal {L}}_{\mathrm {matter} })}{\partial g^{\mu \nu }}}=-2{\frac {\partial {\mathcal {L}}_{\mathrm {matter} }}{\partial g^{\mu \nu }}}+g_{\mu \nu }{\mathcal {L}}_{\mathrm {matter} },}$

where ${\displaystyle S_{\mathrm {matter} }}$ is the nongravitational part of the action, ${\displaystyle {\mathcal {L}}_{\mathrm {matter} }}$ is the nongravitational part of the Lagrangian density, and the Euler-Lagrange equation has been used. This is symmetric and gauge-invariant. See Einstein–Hilbert action for more information.

Noether's theorem implies that there is a conserved current associated with translations through space and time. This is called the canonical stress–energy tensor. Generally, this is not symmetric and if we have some gauge theory, it may not be gauge invariant because space-dependent gauge transformations do not commute with spatial translations.

In general relativity, the translations are with respect to the coordinate system and as such, do not transform covariantly. See the section below on the gravitational stress–energy pseudo-tensor.

In the presence of spin or other intrinsic angular momentum, the canonical Noether stress energy tensor fails to be symmetric. The Belinfante–Rosenfeld stress energy tensor is constructed from the canonical stress–energy tensor and the spin current in such a way as to be symmetric and still conserved. In general relativity, this modified tensor agrees with the Hilbert stress–energy tensor.