How to find Kinetic Energy in General Relativity?

Question

I am trying to determine what exactly the kinetic energy would be in General relativity and to my surprise, I am much more confused than I thought I would be. So for special relativity, the four momentum $\mathbf{P}$ is defined as:

$$\mathbf{P} = \left( \frac{E}{c},\mathbf{p} \right)$$

where $\mathbf{p}$ is the 3 momentum (spatial components) and $E/c$ is the total energy divided by the speed of light. Therefore we have:

$$\langle \mathbf{P},\mathbf{P} \rangle = \eta_{\mu \nu}P^{\mu}P^{\nu} = -\left( \frac{E}{c} \right)^2 +p^2 = -(m_0c)^2$$

$$E = m_0 c^2\sqrt{1+\left(\frac{p}{m_0 c}\right)^2}$$

where $m_0$ is the rest mass. So to obtain the kinetic energy $E_k$, I think one can simply subtract away the rest mass energy.

$$E_k = m_0 c^2\left(\sqrt{1+\left(\frac{p}{m_0 c}\right)^2}-1\right)$$

So if all of this is correct, I should be able to go through a similar approach with general relativity via the correspondence principle. The energy should be related to the momentum inner product:

$$\langle \mathbf{P},\mathbf{P} \rangle = g_{\mu \nu}P^{\mu}P^{\nu} = -(m_0c)^2$$

The issue is, I don't know what the four momentum would be in general relativity. If I remember correctly, I don't think it equals the same thing as the special relativistic case, because the time component isn't related directly to the energy anymore (is this correct??).

Let us suppose that I have solved the geodesic equation, and have the explicit solution for all of the coordinates $x^{\mu}$. I should be able to find the four velocity $u^{\mu}$ (using a proper time affine parametrization $\tau$):

$$u^{\mu} = \frac{\partial x^{\mu}}{\partial \tau}$$

So the four momentum should be related to this. However, the four momentum must include the mass. In special relativity, the momentum is given by:

$$p^{\mu} = m u^{\mu} = m_0 \gamma \cdot (c,\mathbf{v})$$

where $\mathbf{v}$ is the 3-velocity without including the gamma factor. The gamma factor ultimately comes from flat space metric and so is not appropriate here. What would be the equivalent in general relativity? And how would I relate this to the energy?

A guess of mine is perhaps since my coordinates include the contribution from the curved metric $g^{\mu\nu}$ intrinsic to them from solving the geodesic equation (or any other equation of motion that includes the metric), one might be able to say the four momentum is:

$$p^{\mu} = m_0 u^{\mu}$$

where $m_0$ is the rest mass? If not, if I have explicit forms of $x^{\mu}$ and $u^{\mu}$, how can I find the momentum and therefore the kinetic energy? Hope this makes sense...

Answer 1

The definition of momentum is the same in GR but you need to use the metric $g_{uv}$ to calculate the magnitude and momentum might not be conserved.

$m_0 \frac{dx_i}{d\tau} = m_0 \frac{dx_i}{dt} \frac{dt}{d\tau}$

Following A First Course in General Relativity by Bernard Schutz. Page 190.

In GR,

$m \frac{p_{\small \beta}}{d\tau} = \frac{1}{2} g_{\mu \nu,\small\beta} p^\mu p^\nu $

If all the components of $g_{\mu\nu}$ are independent of $x^{\small \beta}$ for some fixed index $\small \beta$, then $p_{\small \beta}$ is a constant along any particle's trajectory.

For example, in a stationary gravitational field, where there is a coordinate system with time independent metric components, $p_0$ is conserved.

For a weak gravitational field,

$ds^2= -(1+2\phi)dt^2 +(1-2\phi)(dx^2 + dy^2 + dz^2 )$

where $\phi$ is the Newtonian gravitational potential.

$\vec{p} \space \cdot \vec{p} = -m^2 = g_{\mu\nu} p^\mu p^\nu = -(1+2\phi)({p^0})^2 +(1-2\phi)((p^x )^2 + (p^y)^2 + (p^z)^2 )$

$-p^0 \approx m + m \phi +\frac{\vec{p}^2}{2m}=\text{rest mass} + \text{gravitational potential}+\text{kinetic energy}$

$p^0$ is conserved along the particles trajectory. This generalizes the Newtonian concept of conservation of energy.

Answer 2

This is no different in GR than in SR. Vectors like the momentum four-vector live in the tangent space at a particular spacetime point, which means they act just like you expect from SR.

I don't think it equals the same thing as the special relativistic case, because the time component isn't related directly to the energy anymore (is this correct??).

No, it's the same as in SR. You can define a local frame of reference, which defines a unit vector in the time direction.

The gamma factor ultimately comes from flat space metric and so is not appropriate here.

No, the gamma doesn't have anything to do with the metric. If you choose appropriate local Minkowski coordinates, then the metric looks exactly like the metric in SR, diag(1,-1,-1,-1). The gamma just occurs because you're normalizing the velocity vector.

Since you mention a geodesic, it may be that you're thinking of the case where the spacetime has a timelike Killing vector, and then there is a conserved energy. That's a different thing. Most spacetimes don't have a timelike Killing vector, and then there is no conserved energy that you can define for test particles. But that has nothing to do with the general ideas of how you define the energy and the momentum four-vector, which are the same in all cases.

It's also not true that the momentum four-vector is defined from a normalized velocity vector. That wouldn't work for massless particles (in either SR or GR). The only real things you can say in general are that the mass is the norm of the momentum vector, and the momentum vector has to be parallel to the (normalized or unnormalized) velocity vector (or else it would violate rotational symmetry).