# Is energy conserved in general relativity? Does $\nabla_aT^{ab}_{\rm matter}=0$ represent the conservation of energy and momentum?

1. For example, the radiation dominated cosmology, the energy density of radiation is proportional to $a^{-4}$ and the volume is proportional to $a^3$, where $a$ is the scale factor. So the total energy of radiation is propotional to $a^{-1}$. So where is the loss of energy of radiation? Is it because the gravitational field has the energy?

2. Does $\nabla_aT^{ab}_{\rm matter}=0$ represent the conservation of energy and momentum of matter field in GR?

• In a word, yes. – user21433 Apr 22 '14 at 14:23
• @SeanD But how can you explain the radiation dominated cosmology? It is a very easy solution of GR. What's wrong with it? – 346699 Apr 22 '14 at 14:28
• possible duplicate of Energy conservation in General Relativity – Kyle Kanos Apr 22 '14 at 14:31
• In a word, no. - – Philip Gibbs - inactive Oct 27 '14 at 9:33

After taking a look at the answer by Jim, I am not sure in my knowledge at the moment. However, lets try to figure out the details. I claim, that energy-momentum tensor of matter in GR is not conserved by itself since matter always interacts with gravitational field and the total energy should be taken into account instead.

Vanishing of the covariant divergence $\nabla_a T^{ab}=0$ exactly reflects this feature. Consider this equation integrated over a 4-dimensional volume \begin{aligned} 0=&\int_V d^4x \sqrt{-g}\;\nabla_a T^{ab}=\int_V d^4x \nabla_a(\sqrt{-g}\; T^{ab})=\int_V d^4x \partial_a(\sqrt{-g}\; T^{ab})+\ldots\\ =&\int_{\partial V} d\Sigma_a T^{ab}+\ldots, \end{aligned} where dots represent whatever Christoffel symbols appear there. Hence, we see, that if we choose the surface $\partial V$ in the usual way when only its $x^0=const$ parts contribute to the integral we will get the usual conservation law deformed by the connection terms $$0=P^b(x^0=t_2)-P^b(x^0=t_1)+\dots.$$ So, since the difference of 4-momentum at different time does not vanish, the energy momentum tensor of matter is not conserved by itself.

However, if we take into account the contribution from gravity that is calculated in the usual way $$T^{grav}_{ab}=\frac{1}{\sqrt{-g}}\frac{\delta S_{EH}}{\delta g^{ab}}=R_{ab}-\frac12g_{ab}R,$$ we see the well known feature of GR that the total energy-momentum tensor vanishes due to the Einstein equations $$T_{ab}+T_{ab}^{grav}=0.$$ That is kind of obvious, because the total energy momentum tensor is obtained by variation of the total action wrt $g_{ab}$ and therefore gives exactly EOM for gravitational field with a source.

It seems, that the loss of energy of radiation goes into energy of the gravitational field. In addition, you may read the 2 volume of Landau-Lofshitz to find out how people define energy-momentum pseudo-tensor of gravitational field that does not cancel the matter energy-momentum tensor and therefore is more useful for some applications.

• Thanks. I'm a beginer of GR, but I listened that the energy of gravtiy is not local and have no density of energy. So why is there a energy-momentum tensor of gravity? – 346699 Apr 24 '14 at 18:12
• I can always formally apply the rule $T_{ab}=(-g)^{-1/2}\delta S/\delta g^{ab}$ to the Einstein-Hilbert action. The point is, that this rule always gives the symmetric energy-momentum tensor for matter fields. However, for gravity it obviously gives just EOM and hence vanishes. That is to say that energy-momentum tensor for gravity is not defined. At least in this simple way. – Edvard Apr 24 '14 at 20:40
• It is right that the equation $\nabla_aT^{ab}=0$ only describes the energy of the matter fields and you need to include gravity to get the full conservation law. However the full expression for energy including gravity is not given by simply varying the metric as you suggest. The required expression from Noether's theorem is different and does not give something equal to zero. The Landau-Lifshitz method is one valid approach but there are better ways that do not need pseudo-tensors. – Philip Gibbs - inactive Oct 27 '14 at 9:29
• That's true, I agree. I meant the usual prescription, that suggests to vary by metric. Certainly, there should be a way to get something non-trivial instead of $T^{ab}_{full}=0$, however I know only the LL approach. – Edvard Oct 28 '14 at 11:19

The energy of radiation falls off like $a^{-1}$ because space is expanding. As space expands, the peaks of an electromagnetic wave expand with it, which makes them get farther apart. This means the wavelength of radiation increases as space expands, thus the frequency decreases. Since energy is $E=h\nu$, if the frequency decreases proportional to $a^{-1}$, then the energy also falls off like $a^{-1}$. This is called cosmological redshift.

Also, as mentioned in the comments, $\nabla_{\mu}T^{\mu}_{~\nu}=0$ means that the energy-momentum tensor is conserved. It is the GR equivalent of the conservation of energy and the conservation of momentum laws.

• I disagree on the conservation argument for $\nabla T$. See my answer. – Rexcirus Dec 17 '14 at 20:26
• @Rexcirus I'm saying that $\nabla T$ is the GR equivalent; the analog of the classical energy-momentum conservation laws. I don't want that interpreted as me saying it means non-gravitational energy-momentum conservation laws are valid, which they aren't globally. – Jim Dec 17 '14 at 22:16

$\nabla T=0$ is not a conservation law. You are not considering the energy of the gravitational field in this way.

If you do the calculation you find something like $\partial_{\nu} (\sqrt{-g} T^{\mu \nu}) \neq 0$, so you can't define a conserved charge like $P^{\mu}= \int \sqrt{-g} d^4x T^{\mu0}$.

To account for the gravitational scalar field you could construct a pseudotensor, but this is not satisfactory. Anyway, in some cases you have symmetries that allow you to have conserved quantities.

For the energy loss due to cosmological redshift see here: redshift

EDIT: I add something due to comments. Going from curved space to a local inertial frame of course you have $\nabla T=0 \rightarrow \partial T=0$ and the latter IS a conservation law. But not in general. In a generic dynamical spacetime the energy is indeed not conserved.

References: Hartle, Gravity, pag. 482 cap. 22 (Local Cons. of En-Momentum in Curved Space). See in particular the example on FRW cosmology that nicely answer the initial question.

• I've looked through a few books just now (Crowell's GR, Franklin's Advanced Mech & GR, and Wald's GR) and all three say that $\nabla_\nu T^{\mu\nu}$ is a conservation law. Now, all three do state that this would be a local conservation law and not a global one. Perhaps this is the point you are trying to make? – Kyle Kanos Dec 17 '14 at 20:49
• The stress-energy tensor is conserved. $\nabla_{\nu}T^{\mu\nu}=0$ The non-gravitational energy and momentum are not always conserved globally in general relativity, which corresponds to a non-zero covariant divergence of the stress-energy pseudotensor. And what Kyle refers to is that there is always a way to express this pseudotensor such that it is conserved locally. This means non-gravitational energy-momentum conservation laws apply locally. But $T^{\mu\nu}$ (the tensor, not the pseudotensor) includes gravitational energy-momentum and is still conserved globally. – Jim Dec 17 '14 at 22:13