According to the book Relativity, Thermodynamic and Cosmology by Tolman, the general relativistic analogue of the first law of thermodynamics is given by:

$$\frac{\partial I^\mu_\nu}{\partial x^\nu}-\frac{1}{2} I^{\alpha\beta}\frac{\partial g_{\alpha\beta}}{\partial x^\mu}=0$$ $$(T^{\mu\nu})_{;\nu}=0$$

where $I$ is the energy-momentum tensor density $$-8\pi I_{\mu\nu}(dg^{\mu\nu})= R_{\mu\nu}d(g^{\mu\nu}\sqrt{-g}).$$

What I do not understand from these equations is: what happens if there is energy loss, for example emission of radiation?

Does this form already includes this case?


1 Answer 1


First of all, as mentioned in the book you cited, $(T^{\mu\nu})_\nu=\nabla_\nu T^{\mu\nu}=0$ is nothing more than the conservation of energy-momentum. One could see it as "the general relativistic analog of the first law of thermodynamics" but I don't think this really adds any meaning to it.

To answer your question, no, these equations do not include the case of energy loss, or to better phrase it, they include it but do nothing to explain it.
The reason why is that in flat space $(\nabla\rightarrow\partial)$ the conservation of a tensor current $\partial_\nu T^{\mu\nu}=0$ can be seen as the generalization of the conservation of the more familiar vector current $\partial_\nu j^{\nu}=0$, meaning that they both lead to the conservation of a "charge" (energy, momentum, and angular momentum in the tensor case, and electric charge in the vector case). In curved space though, while the vector current still leads to the conservation of a charge (using the concepts of codifferential and generalized Stokes' Theorem) the tensor current has an extra term, dependent on the Christoffel's symbols, which means that there isn't a properly defined "charge", and so energy (in the non-gravitational sense) could be lost (we know that it isn't and that it just becomes gravitational energy, but gravitational energy isn't present in the energy-momentum tensor).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.