I am confused about energy and momentum conservation, time and space translation symmetry, and the continuity equation.

Suppose we have a mass $m$ in inertial space far from any gravitational fields. Suppose the mass has an initial velocity $v$; thus, it has an initial momentum $p=mv$ and an initial kinetic energy of $T=\frac12mv^2$.

My understanding of the continuity equation is we imagine a closed surface this mass passes through and we say momentum is conserved if the momentum of $m$ exiting the imaginary surface is the same as the momentum of m entering the surface. This comes about by space translation symmetry.

Now, suppose during the time $m$ is within the closed surface, an impulse is applied to the mass, so that when it exits the surface, it has a greater momentum. Do we say the momentum is still conserved because we include the impulse that acted on the mass during its transit within the closed surface?

Also, the same question applies to the kinetic energy the mass gains by work being done on it by a force $f$ and displacement $d$ within the closed surface. We still say energy is conserved because work was done on the mass while it was in transit within the closed surface. Correct?

Now, suppose the mass $m$ starts out in a gravitational field and free falls to the earth. During its fall, it also passes through an imaginary closed surface. The momentum and the kinetic energy of the mass will have increased from its initial values prior to entering the closed surface. But, relativity states gravity is not a force, but a curving of space-time. While the mass accelerates through the imaginary closed surface, it gains energy and momentum even though there is no force or "weight" acting on it.

Is it correct to say that because there was no time or space translation symmetry within this closed surface-- spacetime is curved-- that energy and momentum is not conserved? Or, do we think of a real force acting on the mass due to the gravitational field so that we can assert energy and momentum is conserved locally?

  • $\begingroup$ In regards to your first question (is momentum still conserved), see this answer of mine to a very similar question. $\endgroup$
    – Kyle Kanos
    Commented Jul 15, 2014 at 2:56

1 Answer 1


The continuity equation (without sources) is usually written as follows

$$\partial_t \rho + \nabla \cdot \mathbf{j} = 0$$

If you identify $\rho$ as the mass density, integrate over some volume $V$ and use the divergence theorem you get the result that you mention in your question. Namely, the change in mass in $V$ equals the amount of mass flowing through the boundary.

If you have some source (or sink) of mass, you can add terms to the right-hand side.

$$\partial_t \rho + \nabla \cdot \mathbf{j} = \mathcal{S}$$

In GR, conservation of energy and momentum is expressed using the stress-energy tensor $T^{\mu\nu}$

$$\nabla_\mu T^{\mu\nu} = 0$$

where we're taking the covariant derivative to account for the fact that our spacetime can be curved.

Now we can interpret this equation in the same manner as we did for the continuity equation. If we write the covariant derivative in a coordinate basis it becomes easier to compare to the continuity equation. The above equation becomes

$$\partial_t (\sqrt{-g}\, T^t{}_\nu) + \partial_i (\sqrt{-g}\, T^i{}_\nu) = \sqrt{-g}\, T^\kappa{}_\lambda \Gamma^\lambda_{\nu\kappa}$$

where $g$ is the determinant of the metric and $\Gamma^\lambda_{\nu\kappa}$ are the Christoffel symbols.

Comparing this to the continuity equation we see that the term on the right-hand side looks like a source term. This term arises purely because of the geometry of our spacetime.

So the lack of translation invariance in curved spacetime gives rise to non-zero source terms in the conservation equations.


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