# Potential energy question

1. Dec 21, 2004

### Ivan Seeking

Staff Emeritus
I just made the following statement regarding two masses, M and m, in orbit about each other.

"Btw, IIRC [and I may not be... I will check], as the potential energy of the gravitational field of M, wrt to m, is converted to kinetic energy in m, the mass of M decreases according Einstein's mass/energy relationship. As m loses kinetic energy to potential energy the mass M increases. This accounts for the energy storage and exchange.

Am I screwing this up or does memory serve correctly? It makes sense but that is usually a bad sign.

2. Dec 21, 2004

### pmb_phy

What does "IIRC" mean?
This is not quite clear to me. What do you mean by "potential energy of the gravitational field of M, wrt to m". "With respec to m"????? The field itself does not have potential energy. There is a mutal potential energy between the two bodies.

Let's assume that you're refering to two bodies in orbit about each other for which the speed of each and the distance between them changes as a function of time. In such a case the kinetic energy and potential energy will change with time and (disgregarding energy carried away in gravitational waves) the total energy of the system will be constant. The proper mass of each body will remain constant.
I think you're screwing it up big time.

Pete

3. Dec 21, 2004

### Creator

In Newtonian orbital mechanics the masses stay constant. The energy shuttles back & forth from kinetic to potential energy based upon a change in radial distances and velocities.

Creator

Opps; looks like I stepped over Pete.

Last edited: Dec 21, 2004
4. Dec 21, 2004

### Ivan Seeking

Staff Emeritus
Eeeeek! This stemmed from a question about where the potential energy is stored. What am I forgetting here?

Last edited: Dec 21, 2004
5. Dec 21, 2004

### Ivan Seeking

Staff Emeritus
IF you think I am forgetting about the energy exchange between potential and kinetic energy, that's not what I'm saying.

6. Dec 21, 2004

### Ivan Seeking

Staff Emeritus
You seem to be using an elementary interpretation of potential energy.

Yes, no?

Oh yes, IIRC, If I recall correctly
Also, I meant the potential energy of the field at m, not wrt m.

Last edited: Dec 21, 2004
7. Dec 21, 2004

### pmb_phy

When I see people use the term "potential energy is located at.." I runaway and hide. :surprised

Pete

8. Dec 21, 2004

### Creator

You're forgetting about DISTANCE between the two masses. Potential energy, U, varies inversely proportional to distance and is given by definition as:

U = -GMm/R

And yes, it is elementary; it is, as I said, Newtonian mechanics. :tongue2:
Creator

(edited to add in neg. sign).

Last edited: Dec 21, 2004
9. Dec 21, 2004

### Ivan Seeking

Staff Emeritus
I think there is a more fundamental interpretation. This is always tough because I used to be relatively certain of what I know and what I don't, but I haven't studied some of this for so long now...scary....

Last edited: Dec 21, 2004
10. Dec 21, 2004

### Creator

At this point I think I'm going to take pmb's advice. :rofl:

11. Dec 21, 2004

### Chronos

Ivan, your memory is not so bad. Remember the Einstein box? Anyways, it goes like this. Suppose you fire a flash of light with an energy of 13.6ev at a hydrogen atom. This will knock the electron out of orbit and you will be left with a proton and an electron, which collectively have a total mass that is exactly 13.6ev more than a hydrogen atom. This difference is the potential energy residing in the separated particles. If you bring them back together, the reformed hydrogen atom will release this potential energy in another flash of light with an energy of 13.6ev. In the case of gravitational potential energy, the energy used to separate mass m from mass M is stored by both objects. This is, of course, offset by the energy used to move them apart. If you borrow all the energy used from M, its mass will decrease and the mass of m will increase. If you borrow equally from m and M, both will have the same mass after being separated as they did prior to separation. If you steal the energy from say the moon, m and M will both increase in mass while the moon looses mass.

12. Dec 21, 2004

### Ivan Seeking

Staff Emeritus
Wheeeewwwwwwww! Thanks Chronos. pmb_phy and Creator scared me for a minute there. :uhh:

...and as pmb_phy pointed, under GR there is [still] no energy in the field. Correct?

13. Dec 21, 2004

### pervect

Staff Emeritus
I would suggest reading the sci.physics.faq on energy conservation in general relativity located

here

To put it as simply as I can, there is a well defined concept of conservation of energy in general relativity which does however require specific boundary conditions ("asymptotically flat space-times").

However, when a system contains significant gravitational energy, it's usually not possible to localize it in GR. (The exception is a static space-time. Planets orbiting each other aren't static, though.).

14. Dec 21, 2004

### pmb_phy

I'm still not sure what the question really is so don't give up. It could be that I'm just not understanding your question.

If you're asking if gravitational energy has a mass related to it then yes, that's quite true.

Pete

15. Dec 22, 2004

### Garth

Ivan Seeking "As m loses kinetic energy to potential energy the mass M increases. This accounts for the energy storage and exchange."
Pete "The proper mass of each body will remain constant."
Creator "In Newtonian orbital mechanics the masses stay constant"
Chronos "....collectively have a total mass that is exactly 13.6ev more than a hydrogen atom. This difference is the potential energy residing in the separated particles"
Pete, "If you're asking if gravitational energy has a mass related to it then yes, that's quite true"

Confusing isn't it?

In Newton and GR the mass of the bodies stays constant. In Newton energy is kept in a separate account, and KE + PE = const.

In SR the two accounts appear to merge E = mc2
But physicists like to keep them separate anyway and have invariant particle masses, so that energy has now to be kept in the field. Even if this means in the case of a bound system such as an atom then that field has to store negative energy.

In GR particle masses are constant but except in certain static fields energy goes all over the place, it is not locally conserved and the value of a particle's energy cannot be transported from one position to another in the presence of curvature unless there is a time-like killing vector, which generally does not exist.

Therefore the great classical separate principles of the conservation of energy and mass were magnificently united by SR only to be discarded by GR, except in some special contrived situations.

Perhaps GR needs to be modified to restore the situation? I won't bore you again with my solution!

Garth

Last edited: Dec 22, 2004
16. Dec 22, 2004

### Andrew Mason

If one views a system of two bodies (non-EM radiating massive bodies separated by a distance) as a black box (in the frame of reference of the center of mass of the box), the mass of that box should not change. It should not depend on whether the energies of the bodies inside are potential or kinetic (ie. as gravitational potential becomes kinetic energy as the separation decreases). According to Special Relativity, kinetic energy contributes $KE/c^2$ to the mass of the body. Therefore gravitational potential energy must contibute equally to the mass of the body. Each will contribute to gravitation of the system. So, gravitational potential energy has to comprise part of the mass or inertia of each body.

So it is not a question of a body's mass being converted to kinetic energy or stored as potential energy. Its mass, and therefore its gravitation, is constant. Its mass merely oscillates between rest and relativistic mass (in the frame of reference of the centre of mass of the system).

AM2

Last edited: Dec 22, 2004
17. Dec 22, 2004

### Garth

And what happens when the gravitational potential energy changes because the two bodies freely fall towards each other?

Garth

18. Dec 22, 2004

### Andrew Mason

The kinetic energy (= loss of PE) provides relativistic mass in an amount KE/c^2 (for v<<c), and corresponding gravitation.

AM2

19. Dec 23, 2004

### Garth

Throw a ball vertically. It begins with rest mass m and KE E, at the highest point it has 'rest' mass m' and zero KE. Does m = m' as Pete said "The proper mass of each body will remain constant" or not? If it does, as in GR, what has happened to the KE?

The total energy of the projectile is actually given by
E = - P0U0 ,
so time dilation g00 comes into play. From the frame of reference of the Centre of Mass of the Earth this compensates for the potential energy component. Energy is conserved because U0 is a killing vector as the metric components are time independent in such a static gravitational field.

However, from the point of view of the projectile the field is not static as in that frame of reference the metric components change with time (they change with position, which changes with time for a moving observer) and there is no killing vector. In GR energy is not conserved and cannot even be easily defined.

Alternatively the rest mass of the projectile might be defined to include PE, as I think you are suggesting Andrew, in which case
m = m0exp(Phi) where Phi is the dimensionless Newtonian potential, as 'rest' mass is defined in SCC.

Garth

Last edited: Dec 23, 2004
20. Dec 23, 2004

### pmb_phy

The mass of a particle is given by the time component of 4-momentum, i.e. $m = \mu dt/d\tau$ where $\mu$ is the proper mass of the particle. Proper mass is an intrinsic property of a particle and is not changed by speed of position in a gravitational field. However proper mass is different than rest mass. This requires explanation - Take as an example a particular kind of gravitational field for which g0k = 0. This is called a "time-orthogonal" gravitational field. In this case

$$m = m(v,\Phi) = \frac{\mu}{\sqrt{1 + 2\Phi/c^2 - \beta}}$$

where $\beta = v/c$. Rest mass is defined as

$$m_0 = m(0,\Phi)$$

The proper mass is related to the mass through

$$\mu = m(0,0)$$
Note: E is not related to the m above through E = mc2.

Pete

Last edited: Dec 23, 2004