Kinetic energy and curved space-time

[jfy4]

Hi,

I was wondering if the non-relativistic kinetic energy (or the total energy) of a body in a curved space-time can still be written as

\frac{\mathbf{p}^2}{2m}

As an example, is the kinetic energy \frac{\mathbf{p}^2}{2m} in the Schwarzschild metric for a non-relativistic particle?

Thanks a lot.

 

[LAHLH]

In relativity the energy observed by an observer with four velocity U^{\mu} of a particle with four momentum p^{\mu} is E=-p_{\mu}U^{\mu}. Note that this is a Lorentz covariant scalar, anyone in whatever Lorentz frame can calculate this energy and get the same result to find out what the energy some observer measures some particle as.

If you are in Minkowski space and go into the Momentarily Co-moving Reference Frame (MCRF) you would find that U^{\mu}=(1,0,0,0) and P_{\mu}=(-E, p_x,p_y,p_z) so you see why the above definition makes sense.

Note that this is just the energy from rest mass and motion (not gravitational potential), and E=m_0 for an observer co-moving with a particle of invariant mass m_0. Boosting to another frame of course (and restoring factors of c^2) then E=\gamma m_0 c^2=\tfrac{m_0 c^2}{\sqrt{1-v^2/c^2} }=mc^2+mv^2/2+...

There is another notion of energy in a stationary spacetime (like Schw), where you have a timelike KV,K. This time translational symmetry leads to a conserved quantity: E=-K_{\mu} dx^{\mu}/d\lambda and this definition includes not only rest mass and motional energy, but also gravitational potential energy. I'm not sure exactly why this contains potential too, I just know it does, maybe someone else could explain why this is the case?

 

[Sam Gralla]

Well, the E associated with the killing field is constant for a geodesic. So, if you imagine radial infall, where you pick up kinetic energy, you have to be compensating with loss of gravitational potential energy. In this sense E includes gravitational potential energy. However, I think it's better to think of it as your ordinary energy as measured at infinity (where the "gravitational potential energy" is zero).

 

[jfy4]

This is my attempt at a solution, I would appreciate if you guys checked my physics/math

Start with E=-p_{a}u_{obs}^{a}

then I pulled the u_{obs} from the Hartle's gravity book for a stationary observer

u_{obs}^{a}=\left( (1-\frac{2M}{R})^{-\frac{1}{2}},0,0,0)\right)

and the momentum is p^{a}=m(u^t,\vec{u})

Now the scalar product is

-mg_{tt}(u^t)(u_{obs}^{t})=m\sqrt{1-\frac{2M}{R}}\gamma

Then the kinetic energy is this minus the rest mass, (if you haven't noticed, its natural units...). We can then expand gamma to first order for low velocity limit and we get

KE=\left(m+\frac{m\vec{v}^2}{2}\right)\sqrt{1-\frac{2M}{R}}-m=m\left(\sqrt{1-\frac{2M}{R}}-1\right)+\sqrt{1-\frac{2M}{R}}\frac{\vec{p}^2}{2m}

This is my answer.

 

[bcrowell]

Well, the E associated with the killing field is constant for a geodesic. So, if you imagine radial infall, where you pick up kinetic energy, you have to be compensating with loss of gravitational potential energy. In this sense E includes gravitational potential energy. However, I think it's better to think of it as your ordinary energy as measured at infinity (where the "gravitational potential energy" is zero).

This issue isn't so much why the conserved quantity includes both potential and kinetic energy, it's why there is not even any uniquely natural way to split the energy into potential and kinetic parts.

One thing to consider is the following. The conserved quantity associated with a Killing vector \xi_a is v^a\xi_a, and this is a scalar, so for the Schwarzschild spacetime, the conserved quantities E and L are coordinate-independent. This is unlike the situation in Newtonian mechanics, where global frames of reference exist, and energy and angular momentum are frame-dependent. If we were to define kinetic and potential energies K and U, however, then they would clearly be coordinate-dependent. This would make them in some sense less interesting than E and L. For example, if you adopt a coordinate system that is rotating relative to the Schwarzschild coordinates, E and L will have the same values, but a particle with a coordinate velocity of zero in the Schwarzschild coordinates could have a nonzero coordinate velocity in the rotating ones.

Another general consideration applies to spacetimes that are stationary but (unlike the Schwarzschild spacetime) not static. Since it's stationary, there is a conserved quantity associated with the timelike Killing vector, and we can interpret that as the energy per unit mass of a test particle. But the condition for a metric to be describable in terms of a single scalar potential is that it be static. Since our spacetime isn't static, there is no useful way to define a potential, and therefore it would be surprising if there was some useful way to define the potential energy of a test particle.