Stress–energy tensor

Components

Because the stress–energy tensor is of order 2, its components can be displayed in 4 × 4 matrix form: $${\displaystyle T^{\mu \nu }=\left(\begin{array}{cccc}{T}^{00}& T^{01}& T^{02}& T^{03}\\ T^{10}& T^{11}& T^{12}& T^{13}\\ T^{20}& T^{21}& T^{22}& T^{23}\\ T^{30}& T^{31}& T^{32}& T^{33}\end{array}\right),}$$ where the indices μ and ν take on the values 0, 1, 2, 3. Each component of the stress–energy tensor has a direct physical interpretation.

In the following, k and ℓ range from 1 through 3.

1. The time–time component is the density of relativistic mass, i.e., the energy density divided by the speed of light squared. This component is $${\displaystyle T^{00}=\rho ,}$$ where $\rho$ is the relativistic mass density.
2. The areal density rate of relativistic mass flux across the x^k surface is equal to the kth component of momentum density, $${\displaystyle T^{0k}=T^{k0}.}$$
3. The components $${\displaystyle T^{k\ell }}$$ represent the areal flux density rate of the kth component of momentum across the x^ℓ surface (compare with the Cauchy stress tensor). In particular, $${\displaystyle T^{kk}}$$ (not summed) represents normal stress in the kth co-ordinate direction (k = 1, 2, 3), which may be thought of as the pressure in the direction indexed by k. The remaining components (⁠${\displaystyle k\ne \ell }$⁠) $${\displaystyle T^{k\ell }}$$ represent shear stress.

In solid state physics and fluid mechanics, the stress tensor is defined to be the spatial components of the stress–energy tensor in the proper frame of reference. In other words, the stress–energy tensor in engineering differs from the relativistic stress–energy tensor by a momentum-convective term.

Covariant and mixed forms

Most of this article works with the contravariant form, T^μν of the stress–energy tensor. However, it is often convenient to work with the covariant form, $${\displaystyle T_{\mu \nu }=T^{\alpha \beta }{g}_{\alpha \mu }{g}_{\beta \nu },}$$ or the mixed form, $${\displaystyle T^{\mu }{}_{\nu }=T^{\mu \alpha }{g}_{\alpha \nu }.}$$

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

Conservation law

In special relativity

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, $${\displaystyle 0=T^{\mu \nu }{}_{;\nu }\equiv {\mathrm{\nabla }}_{\nu }{T}^{\mu \nu }.}$$

The integral form of the non-covariant formulation is $${\displaystyle 0={\int }_{\mathrm{\partial }N}{T}^{\mu \nu }{\mathrm{d}}^3{s}_{\nu }}$$ where N is any compact four-dimensional region of spacetime; $\mathrm{\partial }N$ is its boundary, a three-dimensional hypersurface; and ${\mathrm{d}}^3{s}_{\nu }$ is an element of the boundary regarded as the outward pointing normal.

In flat spacetime and using linear coordinates, if one combines this with the symmetry of the stress–energy tensor, one can show that angular momentum is also conserved: $${\displaystyle 0=(x^{\alpha }{T}^{\mu \nu }-x^{\mu }{T}^{\alpha \nu }{)}_{,\nu }.}$$

In general relativity

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 $${\displaystyle 0=\mathrm{div}T=T^{\mu \nu }{}_{;\nu }={\mathrm{\nabla }}_{\nu }{T}^{\mu \nu }=T^{\mu \nu }{}_{,\nu }+{\mathrm{\Gamma }}^{\mu }{}_{\sigma \nu }{T}^{\sigma \nu }+{\mathrm{\Gamma }}^{\nu }{}_{\sigma \nu }{T}^{\mu \sigma }}$$ where ${\mathrm{\Gamma }}^{\mu }{}_{\sigma \nu }$ is the Christoffel symbol, which is the gravitational force field.

Consequently, if ${\xi }^{\mu }$ is any Killing vector field, then the conservation law associated with the symmetry generated by the Killing vector field may be expressed as $${\displaystyle 0={\mathrm{\nabla }}_{\nu }\left({\xi }^{\mu }{T}^{\nu }{}_{\mu }\right)=\frac{1}{\sqrt{-g}}{\mathrm{\partial }}_{\nu }\left(\sqrt{-g}{\xi }^{\mu }{T}_{\mu }^{\nu }\right)}$$

The integral form of this is $${\displaystyle 0={\int }_{\mathrm{\partial }N}{\xi }^{\mu }{T}^{\nu }{}_{\mu }\sqrt{-g}{\mathrm{d}}^3{s}_{\nu }.}$$

In special relativity

In special relativity, the stress–energy tensor contains information about the energy and momentum densities of a given system, in addition to the momentum and energy flux densities.

Given a Lagrangian density $\mathcal{L}$ that is a function of a set of fields ${\varphi }_{\alpha }$ and their derivatives, but explicitly not of any of the spacetime coordinates, we can construct the canonical stress–energy tensor by looking at the total derivative with respect to one of the generalized coordinates of the system. So, with our condition $${\displaystyle \frac{\mathrm{\partial }\mathcal{L}}{\mathrm{\partial }{x}^{\nu }}=0}$$

By using the chain rule, we then have $${\displaystyle \frac{d\mathcal{L}}{d{x}^{\nu }}=d_{\nu }\mathcal{L}=\frac{\mathrm{\partial }\mathcal{L}}{\mathrm{\partial }({\mathrm{\partial }}_{\mu }{\varphi }_{\alpha })}\frac{\mathrm{\partial }({\mathrm{\partial }}_{\mu }{\varphi }_{\alpha })}{\mathrm{\partial }{x}^{\nu }}+\frac{\mathrm{\partial }\mathcal{L}}{\mathrm{\partial }{\varphi }_{\alpha }}\frac{\mathrm{\partial }{\varphi }_{\alpha }}{\mathrm{\partial }{x}^{\nu }}}$$

Written in useful shorthand, $${\displaystyle d_{\nu }\mathcal{L}=\frac{\mathrm{\partial }\mathcal{L}}{\mathrm{\partial }({\mathrm{\partial }}_{\mu }{\varphi }_{\alpha })}{\mathrm{\partial }}_{\nu }{\mathrm{\partial }}_{\mu }{\varphi }_{\alpha }+\frac{\mathrm{\partial }\mathcal{L}}{\mathrm{\partial }{\varphi }_{\alpha }}{\mathrm{\partial }}_{\nu }{\varphi }_{\alpha }}$$

Then, we can use the Euler–Lagrange Equation: $${\displaystyle {\mathrm{\partial }}_{\mu }\left(\frac{\mathrm{\partial }\mathcal{L}}{\mathrm{\partial }({\mathrm{\partial }}_{\mu }{\varphi }_{\alpha })}\right)=\frac{\mathrm{\partial }\mathcal{L}}{\mathrm{\partial }{\varphi }_{\alpha }}}$$

And then use the fact that partial derivatives commute so that we now have $${\displaystyle d_{\nu }\mathcal{L}=\frac{\mathrm{\partial }\mathcal{L}}{\mathrm{\partial }({\mathrm{\partial }}_{\mu }{\varphi }_{\alpha })}{\mathrm{\partial }}_{\mu }{\mathrm{\partial }}_{\nu }{\varphi }_{\alpha }+{\mathrm{\partial }}_{\mu }\left(\frac{\mathrm{\partial }\mathcal{L}}{\mathrm{\partial }({\mathrm{\partial }}_{\mu }{\varphi }_{\alpha })}\right){\mathrm{\partial }}_{\nu }{\varphi }_{\alpha }}$$

We can recognize the right hand side as a product rule. Writing it as the derivative of a product of functions tells us that $${\displaystyle d_{\nu }\mathcal{L}={\mathrm{\partial }}_{\mu }\left[\frac{\mathrm{\partial }\mathcal{L}}{\mathrm{\partial }({\mathrm{\partial }}_{\mu }{\varphi }_{\alpha })}{\mathrm{\partial }}_{\nu }{\varphi }_{\alpha }\right]}$$

Now, in flat space, one can write $d_{\nu }\mathcal{L}={\mathrm{\partial }}_{\mu }[{\delta }_{\nu }^{\mu }\mathcal{L}]$. Doing this and moving it to the other side of the equation tells us that $${\displaystyle {\mathrm{\partial }}_{\mu }\left[\frac{\mathrm{\partial }\mathcal{L}}{\mathrm{\partial }({\mathrm{\partial }}_{\mu }{\varphi }_{\alpha })}{\mathrm{\partial }}_{\nu }{\varphi }_{\alpha }\right]-{\mathrm{\partial }}_{\mu }\left({\delta }_{\nu }^{\mu }\mathcal{L}\right)=0}$$

And upon regrouping terms, $${\displaystyle {\mathrm{\partial }}_{\mu }\left[\frac{\mathrm{\partial }\mathcal{L}}{\mathrm{\partial }({\mathrm{\partial }}_{\mu }{\varphi }_{\alpha })}{\mathrm{\partial }}_{\nu }{\varphi }_{\alpha }-{\delta }_{\nu }^{\mu }\mathcal{L}\right]=0}$$

This is to say that the divergence of the tensor in the brackets is 0. Indeed, with this, we define the stress–energy tensor: $${\displaystyle T^{\mu }{}_{\nu }\equiv \frac{\mathrm{\partial }\mathcal{L}}{\mathrm{\partial }({\mathrm{\partial }}_{\mu }{\varphi }_{\alpha })}{\mathrm{\partial }}_{\nu }{\varphi }_{\alpha }-{\delta }_{\nu }^{\mu }\mathcal{L}}$$

By construction it has the property that $${\displaystyle {\mathrm{\partial }}_{\mu }{T}^{\mu }{}_{\nu }=0}$$

Note that this divergenceless property of this tensor is equivalent to four continuity equations. That is, fields have at least four sets of quantities that obey the continuity equation. As an example, it can be seen that $T^0{}_0$ is the energy density of the system and that it is thus possible to obtain the Hamiltonian density from the stress–energy tensor.

Indeed, since this is the case, observing that ${\mathrm{\partial }}_{\mu }{T}^{\mu }{}_0=0$, we then have $${\displaystyle \frac{\mathrm{\partial }\mathcal{H}}{\mathrm{\partial }t}+\mathrm{\nabla }\cdot \left(\frac{\mathrm{\partial }\mathcal{L}}{\mathrm{\partial }\mathrm{\nabla }{\varphi }_{\alpha }}{\dot{\varphi }}_{\alpha }\right)=0}$$

We can then conclude that the terms of $\frac{\mathrm{\partial }\mathcal{L}}{\mathrm{\partial }\mathrm{\nabla }{\varphi }_{\alpha }}{\dot{\varphi }}_{\alpha }$ represent the energy flux density of the system.

Trace

The trace of the stress–energy tensor is defined to be ⁠${\displaystyle T^{\mu }{}_{\mu }}$⁠, so $${\displaystyle T^{\mu }{}_{\mu }=\frac{\mathrm{\partial }\mathcal{L}}{\mathrm{\partial }({\mathrm{\partial }}_{\mu }{\varphi }_{\alpha })}{\mathrm{\partial }}_{\mu }{\varphi }_{\alpha }-{\delta }_{\mu }^{\mu }\mathcal{L}.}$$

Since ⁠${\displaystyle {\delta }_{\mu }^{\mu }=4}$⁠, $${\displaystyle T^{\mu }{}_{\mu }=\frac{\mathrm{\partial }\mathcal{L}}{\mathrm{\partial }({\mathrm{\partial }}_{\mu }{\varphi }_{\alpha })}{\mathrm{\partial }}_{\mu }{\varphi }_{\alpha }-4\mathcal{L}.}$$

In general relativity

In general relativity, the symmetric stress–energy tensor acts as the source of spacetime curvature, and is the current density associated with gauge transformations of gravity which are general curvilinear coordinate transformations. (If there is torsion, then the tensor is no longer symmetric. This corresponds to the case with a nonzero spin tensor in Einstein–Cartan gravity theory.)

In general relativity, the partial derivatives used in special relativity are replaced by covariant derivatives. What this means is that the continuity equation no longer implies that the non-gravitational energy and momentum expressed by the tensor are absolutely conserved, i.e. the gravitational field can do work on matter and vice versa. In the classical limit of Newtonian gravity, this has a simple interpretation: kinetic energy is being exchanged with gravitational potential energy, which is not included in the tensor, and momentum is being transferred through the field to other bodies. In general relativity the Landau–Lifshitz pseudotensor is a unique way to define the gravitational field energy and momentum densities. Any such stress–energy pseudotensor can be made to vanish locally by a coordinate transformation.

In curved spacetime, the spacelike integral now depends on the spacelike slice, in general. There is in fact no way to define a global energy–momentum vector in a general curved spacetime.

Einstein field equations

In general relativity, the stress–energy tensor is studied in the context of the Einstein field equations which are often written as $${\displaystyle G_{\mu \nu }+\mathrm{\Lambda }{g}_{\mu \nu }=\kappa T_{\mu \nu },}$$ where $G_{\mu \nu }=R_{\mu \nu }-\frac{1}{2}R{g}_{\mu \nu }$ is the Einstein tensor, $R_{\mu \nu }$ is the Ricci tensor, $R=g^{\alpha \beta }{R}_{\alpha \beta }$ is the scalar curvature, $g_{\mu \nu }$ is the metric tensor, Λ is the cosmological constant (negligible at the scale of a galaxy or smaller), and $\kappa =8\pi G/c^4$ is the Einstein gravitational constant.

Stress–energy in special situations

Isolated particle

In special relativity, the stress–energy of a non-interacting particle with rest mass m and trajectory ${\mathbf{x}}_{\text{p}}(t)$ is: $${\displaystyle \begin{array}{cc}{T}^{\alpha \beta }(\mathbf{x},t)& =\frac{m{v}^{\alpha }(t)v^{\beta }(t)}{\sqrt{1-(v/c{)}^2}}\delta \left(\mathbf{x}-{\mathbf{x}}_{\text{p}}(t)\right)\\ & =\frac{E}{c^2}{v}^{\alpha }(t)v^{\beta }(t)\delta \left(\mathbf{x}-{\mathbf{x}}_{\text{p}}(t)\right)\end{array}}$$ where $v^{\alpha }$ is the velocity vector (which should not be confused with four-velocity, since it is missing a $\gamma$) $${\displaystyle v^{\alpha }=\left(1,\frac{d{\mathbf{x}}_{\text{p}}}{dt}(t)\right),}$$ $\delta$ is the Dirac delta function and $E=\sqrt{p^2{c}^2+m^2{c}^4}$ is the energy of the particle.

Written in the language of classical physics, the stress–energy tensor would be (relativistic mass, momentum, the dyadic product of momentum and velocity) $${\displaystyle \left(\frac{E}{c^2},\mathbf{p},\mathbf{p}\mathbf{v}\right).}$$

Stress–energy of a fluid in equilibrium

For a perfect fluid, the stress–energy tensor takes on the form $${\displaystyle T^{\alpha \beta }=\left(\rho +\frac{p}{c^2}\right)u^{\alpha }{u}^{\beta }+p{g}^{\alpha \beta }}$$ where $\rho$ is the mass density and $p$ is the isotropic pressure in the rest frame, $u^{\alpha }$ is the fluid's four-velocity, and $g^{\alpha \beta }$ is the matrix inverse 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 the fluid's proper frame of reference, the four-velocity is $${\displaystyle u^{\alpha }=(1,0,0,0),}$$

the matrix inverse of the metric tensor is simply $${\displaystyle g^{\alpha \beta }=\left(\begin{array}{cccc}-\frac{1}{c^2}& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right)}$$ and the stress–energy tensor is a diagonal matrix $${\displaystyle T^{\alpha \beta }=\left(\begin{array}{cccc}\rho & 0& 0& 0\\ 0& p& 0& 0\\ 0& 0& p& 0\\ 0& 0& 0& p\end{array}\right).}$$

Electromagnetic stress–energy tensor

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

Scalar field

The stress–energy tensor for a complex scalar field $\varphi$ that satisfies the Klein–Gordon equation is $${\displaystyle T^{\mu \nu }=\frac{{\hslash }^2}{m}\left(g^{\mu \alpha }{g}^{\nu \beta }+g^{\mu \beta }{g}^{\nu \alpha }-g^{\mu \nu }{g}^{\alpha \beta }\right){\mathrm{\partial }}_{\alpha }\overline{\varphi }{\mathrm{\partial }}_{\beta }\varphi -g^{\mu \nu }m{c}^2\overline{\varphi }\varphi ,}$$ and when the metric is flat (Minkowski in Cartesian coordinates) its components work out to be: $${\displaystyle \begin{array}{cc}{T}^{00}& =\frac{{\hslash }^2}{m{c}^4}\left({\mathrm{\partial }}_0\overline{\varphi }{\mathrm{\partial }}_0\varphi +c^2{\mathrm{\partial }}_k\overline{\varphi }{\mathrm{\partial }}_k\varphi \right)+m\overline{\varphi }\varphi ,\\ T^{0i}=T^{i0}& =-\frac{{\hslash }^2}{m{c}^2}\left({\mathrm{\partial }}_0\overline{\varphi }{\mathrm{\partial }}_i\varphi +{\mathrm{\partial }}_i\overline{\varphi }{\mathrm{\partial }}_0\varphi \right),\mathrm{a}\mathrm{n}\mathrm{d}\\ T^{ij}& =\frac{{\hslash }^2}{m}\left({\mathrm{\partial }}_i\overline{\varphi }{\mathrm{\partial }}_j\varphi +{\mathrm{\partial }}_j\overline{\varphi }{\mathrm{\partial }}_i\varphi \right)-{\delta }_{ij}\left(\frac{{\hslash }^2}{m}{\eta }^{\alpha \beta }{\mathrm{\partial }}_{\alpha }\overline{\varphi }{\mathrm{\partial }}_{\beta }\varphi +m{c}^2\overline{\varphi }\varphi \right).\end{array}}$$

Variant definitions of stress–energy

There are a number of inequivalent definitions of non-gravitational stress–energy:

Hilbert stress–energy tensor

The Hilbert stress–energy tensor is defined as the functional derivative $${\displaystyle T_{\mu \nu }=\frac{-2}{\sqrt{-g}}\frac{\delta S_{\mathrm{m}\mathrm{a}\mathrm{t}\mathrm{t}\mathrm{e}\mathrm{r}}}{\delta g^{\mu \nu }}=\frac{-2}{\sqrt{-g}}\frac{\mathrm{\partial }\left(\sqrt{-g}{\mathcal{L}}_{\mathrm{m}\mathrm{a}\mathrm{t}\mathrm{t}\mathrm{e}\mathrm{r}}\right)}{\mathrm{\partial }{g}^{\mu \nu }}=-2\frac{\mathrm{\partial }{\mathcal{L}}_{\mathrm{m}\mathrm{a}\mathrm{t}\mathrm{t}\mathrm{e}\mathrm{r}}}{\mathrm{\partial }{g}^{\mu \nu }}+g_{\mu \nu }{\mathcal{L}}_{\mathrm{m}\mathrm{a}\mathrm{t}\mathrm{t}\mathrm{e}\mathrm{r}},}$$ where $S_{\mathrm{m}\mathrm{a}\mathrm{t}\mathrm{t}\mathrm{e}\mathrm{r}}$ is the nongravitational part of the action, ${\mathcal{L}}_{\mathrm{m}\mathrm{a}\mathrm{t}\mathrm{t}\mathrm{e}\mathrm{r}}$ 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.

Canonical stress–energy tensor

Noether's theorem implies that there is a conserved current associated with translations through space and time; for details see the section above on the stress–energy tensor in special relativity. 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 pseudotensor.

Belinfante–Rosenfeld stress–energy 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.

Gravitational stress–energy

Since by the equivalence principle, gravitational stress–energy will always vanish locally at any chosen point in some chosen frame, gravitational stress–energy cannot be expressed as a non-zero tensor; instead, we have to use a pseudotensor.

In general relativity, there are many possible distinct definitions of the gravitational stress–energy–momentum pseudotensor. These include the Einstein pseudotensor and the Landau–Lifshitz pseudotensor. The Landau–Lifshitz pseudotensor can be reduced to zero at any event in spacetime by choosing an appropriate coordinate system.