For any timelike curve $p_\mu$ in General Relativity (section 3 of this review), we can project this into its timelike and spacelike components. Further, these projections are associated with conservation laws.

The projection along the timelike component is obtained by projecting $p_\mu$ onto $u^\mu$, where $u^\mu$ is the 4-velocity. This projection captures what occurs in the rest frame of the observer as he moves along with the fluid.

The projection along the spacelike component is obtained by defining a 'projection tensor' as follows: $$h_{\mu\nu}=g_{\mu\nu}+u_\mu u_\nu$$ where the projection is taken onto the spacelike components.

Then, energy and momentum conservation can be obtained from the covariant energy-momentum conservation equation $\nabla_\mu T^{\mu\nu}=0$ as follows:

  • Energy conservation (timelike projection): $u_\nu\nabla_\mu T^{\mu\nu}=0$
  • Momentum conservation (spacelike projection): $h_{\mu\nu}\nabla_\mu T^{\mu\nu}=(g_{\mu\nu}+u_\mu u_\nu)\nabla_\mu T^{\mu\nu}=0$

My questions:

1. Why is energy conservation associated with timelike projection and momentum conservation associated with spacelike projection?

2. How do we derive the expression for the projection tensor? Any suggestions to suitable references would be sufficient.


First of all, do note that the momentum conservation equations should read $h_{\rho\nu} \nabla_\mu T^{\mu\nu}=0$, otherwise you would have a single equation.

The strict distinction between energy and momentum conservation is only valid in the Local Rest Frame, the frame of the fluid element with velocity $u^\mu$: for some other observer with velocity $v^\mu \neq u^\mu$ the equation $u_\nu \nabla_\mu T^{\mu\nu}$ will not represent only energy conservation.

Now, to answer question 1: the $\mu\nu$ component of the stress-energy tensor is the flow of the $\mu$-th component of four-momentum through a surface of constant $x^\nu$, so if we project using the vector $u_\nu$ we are getting something which, in the LRF, is conservation of four-momentum through surfaces of constant time. This, coupled with the fact that the time component of the four-momentum is the energy is the motivation for the name.

The reasoning for the momentum conservation equations is similar.

To convince yourself that this makes sense, you can look at the nonrelativistic limit of these equations: see the Theoretical Physics Reference. There, they use the fact that in the LRF $u^\mu = (1, 0,0,0)$ to write the energy conservation equation as $\nabla_\mu T^{\mu 0}$ and similarly for the others. There they derive Euler's equations for a perfect fluid, but a very similar line of reasoning applies to the general case.

As for question 2, very loosely following section 3 in Taub (1978): to see that $h_{\mu}^{\nu} = \delta_\mu^\nu + u_\nu u^\mu$ is an orthogonal projector onto the subspace orthogonal to $u^\mu$ we need to check that it is idempotent ($h^\mu_\rho h^\rho_\nu = h^\mu _\nu$), that it is orthogonal, and that its $\ker$ is just the span of $u^\mu$:

$$ h^\mu_\rho h^\rho_\nu = (\delta^\mu_\rho + u^\mu u_\rho)(\delta^\rho_\nu + u^\rho u_\nu) = \delta^\mu_\nu + 2 u^\mu u_\nu + u^\mu u_\rho u^\rho u_\nu = h^\mu _\nu $$

since $u^\rho u_\rho = -1$. Orthogonality can be seen immediately from the definition.

Also, $(\delta^\mu_\nu + u^\mu u_\nu) u^\nu = 0$ but for a vector $k^\mu$ with $k^\mu u_\mu = 0$ we have $ (\delta^\mu_\nu + u^\mu u_\nu) k^\nu = k^\mu$.

This proves the statement. To get the (0,2)-tensor it is just necessary to lower an index.

To see why one might define $h_{\mu\nu}$ this way you can start by looking at the projection tensor onto the time-like subspace, which will be proportional to $u_\mu u^\nu$: $\pi_\mu^\nu = \alpha u_\mu u^\nu$, and since we want it to leave $u^\nu$ unchanged it will have to be $\alpha = -1$.

Then, the way to find $h_{\mu\nu}$ is to write the equation

$$ \delta_\mu^\nu = h_\mu^\nu + \pi_\mu^\nu $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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