# Equivalence of two definitions of the stress-energy tensor in general relativity

I'm new in general relativity and I readed these two definitions of stress–energy tensor.

The stress–energy tensor is defined as the tensor $$T_{αβ}$$ of order two that gives the flux of the $$α$$th component of the momentum vector across a surface with constant $$x^β$$ coordinate.

2. In the same page in "Hilbert stress–energy tensor" paragraph we have:

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

Is the first definition equals to the second? If it is so, I would like to see a proof. If it isn't so, why two different stress-energy tensors?

• The first definition depends on which notion of momentum is used. Interestingly, the canonical SEM tensor and the Hilbert SEM tensor are only the same for scalar matter, cf. Energy-Momentum Tensor in QFT vs. GR and links therein. Mar 17, 2018 at 15:32
• I saw the link but I don't know the current Noether, I don't see the first definition (that by flux) in that link. Can you post a proof that they are equal? Thank you.
– asv
Mar 17, 2018 at 22:12

What you are essentially curious about is the equivalence between the canonical and Hilbert stress-energy tensors.

It also occurs to me that the concept of the canonical tensor can be made clearer. I won't do many exact proofs, but I'll try to make things "believable" for you.

Conservation laws:

To understand the canonical tensor, you need to understand conservation laws. In classical mechanics (nonrelativistic), $Q$ is a conserved charge, if $Q$ is a function $$Q:T\mathcal C\rightarrow \mathbb R,\ (q,\dot q)\mapsto Q(q,\dot q)$$ on phase space (in this case, velocity phase space, but you can also do it via momentum phase space in Hamiltonian mechanics), satisfying the following:

If $q(t)$ is a trajectory that satisfies the equations of motion, then along these trajectories $$\frac{d}{dt}Q(q(t),\dot q(t))=0.$$

In classical field theory, a conservation law is given by a continuity equation. If $\rho$ is a scalar field and $\mathbf j$ is a vector field, then a continuity equation is of the form $$\frac{\partial\rho}{\partial t}+\boldsymbol\nabla\cdot\mathbf j=0,$$ where $\rho$ is the density of some quantity, and $\mathbf j$ is the current density or flux density of the same quantity.

This conservation law is local, and can be turned into a "global" conservation law of the form of a classical mechanical law. Let us define $$Q(t)=\int d^3x\ \rho(\mathbf x,t),$$ where the integral extends over all of space at time $t$. The time derivative of $Q$ is then $$\frac{dQ}{dt}=\int d^3x\frac{\partial\rho}{\partial t}=-\int d^3x\ \boldsymbol{\nabla}\cdot\mathbf j=-\int_{\partial\mathbb R^3}d\boldsymbol{\sigma}\cdot\mathbf j,$$ where at the last equality, Gauss' theorem has been used. If the current density $\mathbf j$ is localized and does not extend into infinity, then the boundary term vanishes, hence we have $dQ/dt=0$.

In relativistic field theory (with $c=1$), a continuity equation is of the form $$\partial_\mu j^\mu=0,$$ where $(j^\mu)=(\rho,\mathbf j)$. This is exactly the same thing as before, the only additional information is that the density $\rho$ and the current $\mathbf j$ need to form a 4-vector together to be Lorentz-covariant.

Energy and momentum:

Consider some arbitrary field. The field has a total energy $E$. This energy is highly nonlocal, it is basically the energy of the entire field, extending over all of space. We expect that we can assign to the total energy a local density $\rho$ and a current density $\mathbf S$, satisfying $$E=\int d^3x\ \rho(\mathbf x,t)$$ and $$\frac{\partial\rho}{\partial t}+\boldsymbol\nabla\cdot\mathbf S=0.$$

The field also has a total momentum $\mathbf P$ with cartesian components $P^i$. The $i$th component has a momentum density $\pi^i$, which satisfies $$P^i=\int d^3x\ \pi^i(\mathbf x,t),$$ and a current density $\boldsymbol\sigma^i$, which together obey the continuity equation $$\frac{\partial\pi^j}{\partial t}+\boldsymbol\nabla\cdot\boldsymbol{\sigma}^j=0.$$ In components, this is $$\frac{\partial \pi^j}{\partial t}+\frac{\partial\sigma^{ij}}{\partial x^i}=0,$$ where $\sigma$ is a second order tensor, called the stress tensor.

In relativity, energy and momentum are unified into a single 4-vector, the 4-momentum: $$(p^\mu)=(E,\mathbf p),$$ so we have the unified relation $$P^\mu=\int d^3x\ \pi^\mu,$$ and the continuity equation is $$\frac{\partial\pi^\nu}{\partial t}+\frac{\partial\sigma^{i\nu}}{\partial x^i}=0.$$ This is relativistically invariant, if we can write it as $$0=\partial_\mu T^{\mu\nu},$$ where $T^{\mu\nu}$ is a Lorentz-tensor called the stress-energy (or energy-momentum) tensor, whose components have the following meaning: $$T^{00}\text{ - Energy density} \\ T^{0i}\text{ - Momentum density} \\ T^{i0}\text{ - Energy flux density/current density} \\ T^{ij}\text{ - Stress tensor/i-th component of the flux density of the j-th momentum component}.$$

As you can see, the flux interpretation appears here properly.

Noether's theorem:

I want to be brief here, since this post is already overlong. The following should be clear from a course on classical mechanics:

To every infinitesimal symmetry of a physical system, there is a corresponding conserved charge. This is Noether's theorem. In mathematical form, the variation $\delta q^i$ is an infinitesimal quasi-symmetry of the Lagrangian $L$, if under the variation, it changes into a total derivative: $$\delta L=\frac{d}{dt}K$$ for some phase-space function $K$.

Then, the charge $Q$ given by $$Q=\frac{\partial L}{\partial\dot q^i}\delta q^i-K$$ is conserved, (has $dQ/dt=0$), provided the equations of motions hold.

It is customary in classical mechanics to derive the following:

• The total energy $E$ is conserved, if the system is invariant under time translations.

• The total momentum $\mathbf p$ is conserved if the system is invariant under spatial translations.

In field theory, we use a Lagrangian density $\mathcal L$ to formulate theories. There is a corresponding result, also called Noether's theorem. Let the dynamical fields be denoted by $\phi^a(\mathbf x,t)$. The variation $\delta\phi^a$ is a quasi-symmetry of the system if the Lagrangian changes as $\delta\mathcal L=\partial_\mu K^\mu,$ where $K^\mu$ is some 4-vector field. For such a quasi-symmetry, the current $$j^\mu=\frac{\partial\mathcal L}{\partial\partial_\mu\phi^a}\delta\phi^a-K^\mu$$ satisfies $$\partial_\mu j^\mu=0,$$ provided the equations of motion hold.

If the Lagrangian has no explicit dependance on the spacetime coordinates, then the spacetime translation $x^\mu\mapsto x^\mu+a^\mu$ for some constant 4-vector $a^\mu$ is a quasi-symmetry.

Applying Noether's theorem gives that $$j^\mu=T^\mu_{\ \nu}a^\nu$$ is conserved, where $$T^\mu_{\ \ \nu}=\frac{\partial\mathcal L}{\partial\partial_\mu\phi^a}\partial_\nu\phi^a-\delta^\mu_\nu\mathcal L$$ is the canonical stress-energy tensor.

If $a^\mu$ was truly arbitrary, then $T$ itself is conserved in the sense that $\partial_\mu T^{\mu\nu}=0$. We can identify $T$ with the tensor field discussed before, because, it is a conserved current associated with spacetime-translations, so it expresses the energy-momentum 4-current, which is precisely the same thing we considered in a more heuristic manner before.

Equivalent currents:

Assume that $j^\mu$ is a 4-current. The globally conserved charge is $$Q=\int d^3x\ j^0.$$ If $\Sigma^{\mu\nu}$ is an antisymmetric tensor, then $$\tilde{j}^\mu=j^\mu+\partial_\nu\Sigma^{\mu\nu}$$ generates the same charge $Q$, provided that $\Sigma$ falls off at infinity. Feel free to check it!

This is relevant, because in general, the canonical tensor $T^{\mu\nu}$ fails to have some good properties. It will

• not, in general be symmetric, yet conservation of angular momentum would have that (look it up);

• not, in general be gauge-invariant for gauge theories, which is a big no-no;

• not, in general be traceless for conformally invariant fields, which if once again a faux-pas.

For a scalar theory, the canonical tensor is fine, but if you calculate it for the electromagnetic field, it will commit all three crimes I have outlined above.

However, it is possible to make a transformation of the kind I have outlined above, to produce an equivalent (in the sense that it generates the same global conserved charges) tensor, which satisfies all three properties. There is also a way to perform this systematically, which is called the Belinfante-Rosenfeld tensor (look it up!). It is actually the Belinfante-tensor, which the Hilbert tensor is equal to.

The Hilbert tensor:

Here I will attempt to motivate that the Hilbert tensor is the same as a tensor that is equivalent to the canonical tensor (equivalence in the sense that has been defined before).

The covariant conservation law for the Hilbert tensor can be attained from a Noether's theorem-like result (this is related to what is called Noether's second theorem), when applied to the diffeomorphism invariance (coordinate-invariance) of GR. Consider a matter field $\psi$ with generally covariant Lagrangian $\mathcal L_m$. The action is $$S[\psi,g]=\int d^4x\ \mathcal L_m.$$

An infinitesimal coordinate transformation acts on the fields via the Lie derivative, so if we have $x^\mu\mapsto x^\mu+\epsilon \xi^\mu(x)$ (with the vector field $\xi^\mu$ vanishing on the integration boundary), then we have $\delta\psi=\mathcal L_\xi\psi$ and $\delta g_{\mu\nu}=\mathcal L_\xi g_{\mu\nu}$.

The change in the action is $$\delta S=\int d^4x\left(\frac{\delta S}{\delta\psi}\delta\psi+\frac{\delta S}{\delta g_{\mu\nu}}\delta g_{\mu\nu}\right).$$

If the matter equations of motion are satisfied, then the first functional derivative (with respect to $\psi$) vanishes, so we are left with $$\delta S=\int d^4x\ \frac{\delta S}{\delta g_{\mu\nu}}\delta g_{\mu\nu}.$$

The action was completely independent of the coordinates, so the variation necessarily vanishes. The Lie derivative of the metric can be written as $$\mathcal L_\xi g_{\mu\nu}=\nabla_\mu\xi_\nu+\nabla_\nu\xi_\mu.$$

Let us define $$T^{\mu\nu}=\frac{2}{\sqrt{-g}}\frac{\delta S}{\delta g_{\mu\nu}}.$$ We have then $$\delta S=\int d^4x\sqrt{-g}\frac{1}{2}T^{\mu\nu}(\nabla_\mu\xi_\nu+\nabla_\nu\xi_\mu)=\int d^4x\sqrt{-g}\ T^{\mu\nu}\nabla_\mu\xi_\nu=0,$$ where I have utilized the fact that this $T$ is symmetric by design. Because $\xi$ vanishes on the boundary, we may rewrite this as $$0=\delta S=\int d^4x\sqrt{-g}\nabla_\mu T^{\mu\nu}\xi_\nu,$$ and since $\xi$ was arbitrary, we have $$\nabla_\mu T^{\mu\nu}=0.$$

Now to motivate equivalence with the canonical tensor. I won't prove it here, but it can be shown (this is not trivial - basically Noether's theorem in special relativity for arbitrary coordinate transformations is somewhat ambigous), that if, in SR, we consider a "spacetime-dependent translation" of the form $x^\mu\mapsto x^\mu+\xi^\mu(x)$ instead of $x^\mu+a^\mu$ (with $a$ a constant), the transformation will no longer be a symmetry, but the change in the action is given by $$\delta S=-\int d^4x\ T^{\mu\nu}\partial_\mu\xi_\nu,$$ where this $T$ is the canonical tensor.

This is not a symmetry, and $\delta S$ doesn't vanish, but if the equations of motion for the matter field holds, then the variation of the action vanishes for arbitrary variations, so we still have this expression vanish. Now we assume that $T$ is symmetric (remember we can modify $T$ in a way that this relation still remains true), then we have $$\delta S=-\int d^4x\ \frac{1}{2}T^{\mu\nu}(\partial_\mu\xi_\nu+\partial_\nu\xi_\mu).$$ This is the negative of the expression for the Hilbert tensor (in the flat spacetime limit, so $\nabla_\mu\rightarrow\partial_\mu$ and $\sqrt{-g}\rightarrow 1$).

Let's see what happenened.

When we considered the variation of the matter action in GR, we had $$\delta S=\int d^4x\left( \frac{\delta S}{\delta\psi}\delta\psi+\frac{\delta S}{\delta g_{\mu\nu}}\delta g_{\mu\nu}\right),$$ where $\delta S/\delta g_{\mu\nu}$ was what gave the expression with the stress-energy tensor.

When we considered the variation of the matter action in SR, we did not vary the Minkowski metric $\eta_{\mu\nu}$. After all, that was a "fixed" object. So the reason we didn't have a symmetry is because we "forgot" that variation, therefore, the term $$-\int d^4x\frac{1}{2} T^{\mu\nu}(\partial_\mu\xi_\nu+\partial_\nu\xi_\mu)$$ is precisely the negative of the $\delta S/\delta g_{\mu\nu}$ term, what we'd need to "add" to the variation to get symmetry.

Hence, the two stress-energy tensors are equivalent, at least in the sense of equivalence defined before.

Edit: My last point isn't exactly clear, so let me rephrase it in a more precise way.

The matter action $S$ in special relativity is denoted as $S_{SR}$, and the (modified) canonical tensor as $T^{\mu\nu}_C$. We have obtained (or well, I have said, that we obtain) $$\delta S_{SR}=-\int d^4x\ \frac{1}{2}T^{\mu\nu}_C(\partial_\mu\xi_\nu+\partial_\nu\xi_\mu).$$

By contrast, if we also consider the metric a dynamical field, the action is denoted as $S_{GR}$, but I am considering variations around a flat background. We have obtained $$\delta S_{GR}=\int d^4x\left(\frac{\delta S}{\delta \psi}\delta\psi+\frac{\delta S}{\delta \eta_{\mu\nu}}\delta\eta_{\mu\nu}\right)=\int d^4x\ \frac{\delta S}{\delta\eta_{\mu\nu}}\delta\eta_{\mu\nu}=\int d^4x\frac{1}{2}T^{\mu\nu}_H(\partial_\mu\xi_\nu+\partial_\nu\xi_\mu)=0,$$ where $T^{\mu\nu}_H$ is the Hilbert tensor.

However the two variations differ by precisely the metric variation, so we have $$\delta S_{SR}=\delta S_{GR}-\int d^4x\frac{\delta S}{\delta\eta_{\mu\nu}}\delta\eta_{\mu\nu}=- \int d^4x\frac{\delta S}{\delta\eta_{\mu\nu}}\delta\eta_{\mu\nu},$$ as $\delta S_{GR}=0$, so if you insert the expressions here, you can see that we get $$-\int d^4x\frac{1}{2}T^{\mu\nu}_C(\partial_\mu\xi+\partial_\nu\xi_\mu)=-\int d^4x\frac{1}{2}T^{\mu\nu}_H(\partial_\mu\xi+\partial_\nu\xi_\mu),$$ hence $$T_C\approxeq T_H.$$

• For exapmple if I consider the stress–energy of a fluid in equilibrium $T^{\alpha \beta} \, = \left(\rho + {p \over c^2}\right)u^{\alpha}u^{\beta} + p g^{\alpha \beta}$ Do I obtain same result if I write the lagrangian density of the fluid and calculate: ${\displaystyle T_{\mu \nu }={\frac {-2}{\sqrt {-g}}}{\frac {\delta ({\mathcal {L}}_{\mathrm {matter} }{\sqrt {-g}})}{\delta g^{\mu \nu }}}=-2{\frac {\delta {\mathcal {L}}_{\mathrm {matter} }}{\delta g^{\mu \nu }}}+g_{\mu \nu }{\mathcal {L}}_{\mathrm {matter} }}$ ? Correct me if I am wrong
– asv
Apr 9, 2018 at 19:10
• I made the example of fluid in reference to this: en.wikipedia.org/wiki/Perfect_fluid because I saw books that prove the cosmological model starting by that tensor and not by lagrangian density
– asv
Apr 9, 2018 at 19:31