Energy & momentum in non-Minkowski coordinates

[DrGreg]

In Special Relativity (flat spacetime) we can define a contravariant 4-momentum vector as

$$P^{\alpha} = m\frac{dX^{\alpha}}{d\tau}$$​

for a particle of mass m where $X^{\alpha}$ is the "coordinate vector" of a particle. Or as

$$\textbf{P} = (h\nu, \hbar \textbf{k})$$​

for a photon (in Minkowski coordinates).

Better still, we can define a covariant 4-momentum covector as

$$P_{\alpha} = g_{\alpha\beta}P^{\beta}$$​

We can then define energy relative to a frame as the time component (rescaled into the correct units) of the vector, and momentum relative to a frame as the space component (similarly rescaled) of the vector. In Minkowski coordinates, it doesn't really matter whether we use contravariant or covariant 4-momentum; after rescaling you get the same answer either way.

However, in non-Minkowski coordinates, the two methods do give different answers. Does it matter which choice we make? And if so, why?

If we choose Rindler coordinates (coordinates in which a Born-rigid accelerating rocket is stationary), does the 4-momentum vector or covector naturally give rise to something resembling Newtonian potential energy?

Finally, can any of the above be adapted to make sense in general relativity (curved spacetime). I'm aware of the standard Physics FAQ (Is Energy Conserved in General Relativity?) which says that, in general, there is no global conservation of energy, only local. But it also says

"In certain special cases, energy conservation works out with fewer caveats. The two main examples are static spacetimes and asymptotically flat spacetimes."

..."The Schwarzschild metric is both static and asymptotically flat, and energy conservation holds without major pitfalls."​

So, for example, does it make any sense to talk about potential energy in Schwarzschild coordinates?

 

[atyy]

It has to do with a Killing vector, and contravariant/covariant is not important. See snoopies622's #8 and George Jones's #12 at https://www.physicsforums.com/showthread.php?t=263012.

 

[JesseM]

Just addressing the last part of your post, Wheeler and Taylor talk about an "effective potential" function for the Schwarzschild metric in chapter 4 of https://www.amazon.com/dp/020138423X/?tag=pfamazon01-20...just looking at what they say in the summary on p. 4-20, they tell us:

The Principle of Extremal Aging and the Schwarzschild metric combine to give us two constants of the motion--angular momentum L and energy E:

$$\frac{L}{m} = R^2 \frac{d\phi}{d\tau}$$

$$\frac{E}{m} = (1 - \frac{2M}{R})\frac{dt}{d\tau}$$

Note the wristwatch time increment $$d\tau$$ in both of these equations.

Then they say that when you solve the equations of motion for an orbiting (or falling) satellite of mass m in Schwarzschild coordinates, the following relation involving the energy E and the "effective potential" V tells you the radial velocity at any moment:

$$(\frac{dR}{d\tau})^2 = (\frac{E}{m})^2 - (\frac{V}{m})^2$$

where the squared effective potential is defined as

$$(\frac{V(R)}{m})^2 = (1 - \frac{2M}{R})[1 + \frac{(L/m)^2}{R^2}]$$

They mentioned earlier that you can also define a similar sort of effective potential for a satellite in Newtonian gravity, although in this case the equation would be $$\frac{V(R)}{m} = - \frac{M}{R} + \frac{(L/m)^2}{2R^2}$$. They say that the factor (L/m)^2/2R^2 is a "repulsive" factor due to angular momentum, and explain:

What is the source of the repulsive term in the effective potential?

An effective potential--Newtonian equation [28] on page 4-12 and relativistic equation [32] on page 4-18--looks beyond azimuthal motion and focuses on progress along one dimension: the radial dimension. But the single r-dimension fails to represent reality. As the pole rider pictured in Figure 6 (a person riding on a pole which is rotating around one end) climbs inward along the pole, he is also sweeping around in a circle. The pole-rider experiences an outward force, sometimes called a centrifugal force or centrifugal pseudo force, resulting form the circular orbit of his noninertial rotating reference frame. This outward force is the source of the repulsive term in the effective potential.

Can you give me a more concrete feeling for the repulsive term in the effective potential?

Yes. Think of yourself clinging to a lightweight pole pivoted at one end and rotating freely in a horizontal plane (Figure 6). The only force on you toward the center is the force of your hands grasping the pole. It is hard enough to hang on when you revolve far from the pivot; you feel pulled outward by an inertial "centrifugal force" because you are in a rotating reference frame. Now you grasp the pole tenaciously and pull yourself inward, hand over hand. You find that the rate of rotation increases--conservation of angular momentum! The effective outward force on you increases sharply, for two reasons. First, you are closer to the pivot and therefore travel in a tighter circle. Second, you move at greater speed around the smaller circle. The closer to the center you manage to drag yourself along the pole, the more savagely the repulsive centrifugal potential tears at your grip. Try as you will, there is a minimum radius beyond which you are not strong enough to haul yourself inward. Desperately you hang on at that innermost point. Soon your head swims, your hands tire, and you slip outward along the pole, leaving the pole, and plunge into the net held by your laughing companions. You have experienced personally the ferocious repulsive term in the effective potential.

 

[atyy]

I think as long as the spacetime is "static", there is a Newtonian potential:
http://books.google.com/books?id=0J_dwCmQThgC&printsec=frontcover#PPA122,M1
http://books.google.com/books?id=QagG_KI7Ll8C&printsec=frontcover#PPA72,M1