# Relativity of potetial energy

1. Jan 26, 2010

### Naty1

from THE RIDDLE OF GRAVITATION,1992, Peter Bergmann (a student of Einstein)

In response to a post in another thread sometime ago I mentioned that KE is relative and potential energy is not, based on the above. There was some disagreement which I never pursued and so I'd like to hear views regarding the "absolute" nature of potential energy now.

In thinking about it more, it would seem maybe if an observer is racing past an elevated object above earth, say an apple on a table, and determines the PE of the apple, he will observe the table height as reduced...so maybe it's not absolute??

Last edited: Jan 26, 2010
2. Jan 26, 2010

### Meir Achuz

Potential energy can transform either as a Lorentz scalar or the 4th component of a four vector. For some reason, Peter must have been referring only to the scalar PE.

3. Jan 26, 2010

### bcrowell

Staff Emeritus
In the apple example, taken in the context of GR, I think it's a little difficult to give a good answer. The reason is that due to the equivalence principle, GR doesn't allow us to associate a local energy density with the gravitational field.

If we take the apple example in the context of Newtonian mechanics, then there is no length dilation, so gravitational PE is frame-independent.

In the context of electromagnetism+SR, the concept of potential energy is basically defined as the integrated energy density of the EM fields. This is frame-dependent. E.g., a Doppler-shifted light wave has a different energy.

4. Jan 26, 2010

### Jonathan Scott

I'd say the reason for this, at least in a semi-Newtonian approximation, is that the effect of a potential is multiplicative; it has a similar effect to multiplying the energy and momentum by (1-Gm/rc^2) or more generally by (1+Phi) where Phi is the Newtonian potential. For something with rest mass, this is like multiplying the rest mass by the same factor, and the potential energy is the resulting change in the rest mass.

5. Jan 26, 2010

### Naty1

Would someone explain this comment?? Thanks.

Last edited: Jan 26, 2010
6. Jan 26, 2010

### Naty1

also,
I'm unsure how to interpret the above because isn't CHARGE a frame independent scalar?? Or are there complications with that simple statement?? I guess this would be a point charge, not charge density, which due to length contraction would not be the same for all observers.

7. Jan 26, 2010

### bcrowell

Staff Emeritus
In Newtonian gravity, the energy density of the gravitational field is
[tex]
-\frac{1}{8\pi G}g^2 \qquad ,
[/itex]
and the gravitational field g is frame-independent. In GR, the gravitational field g is not really a well-defined thing. By the equivalence principle, you can always pick any point and make g=0 there by adopting a free-falling frame of reference. There isn;'t even a gravitational field tensor. The closest thing to a gravitational field in GR is the Christoffel symbol, and it doesn't transform as a tensor.

Charge is frame-independent, but E and B are frame-dependent.

8. Jan 26, 2010

### bcrowell

Staff Emeritus
Hmm...but in the Newtonian limit you recover the newtonian additivity of the potentials, so if source 1 makes Phi1, and source 2 makes Phi2, then the combination of sources makes Phi1+Phi2. Or, when you say "semi-Newtonian approximation," do you mean something else? I guess if "semi-" means that you want to include the lowest-order nonvanising relativistic corrections, then you could use the Einstein-Infeld-Hoffman expansion, in which the interaction is described as a velocity-dependent instantaneous interaction. The velocity-dependence might make Phi frame-dependent.

Last edited: Jan 26, 2010
9. Jan 26, 2010

### Altabeh

Let's say the potential energy is really frame-independent. Since kinetic energy is frame-dependent and so is the total energy, doesn't this mean that we are violating the conservation law of energy? This is because KE would transform into PE, say, if the table was launched at 2 PM and reached the maximum height of about 25000 feet above the surface of Earth by 3 PM thus PE = the total energy which is frame-dependent but, on the other hand, PE isn't frame-dependent so where's the lost energy? Besides, I think we don't consider the fact that the quantity 'height' is frame-dependent so is PE.

The potential energy varies depending on the reference frame we choose to measure the system because it simply involves a frame-dependent quantitiy. In the long run, I'd be so tough to say Bergmann's claim is a misleading or maybe a historical error.

AB

10. Jan 26, 2010

### Jonathan Scott

I'm assuming a weak enough field that (1 + Phi1)/(1 + Phi2) = (1 + Phi1 - Phi2), so the multiplicative ratio between two potentials has the same effect as the Newtonian difference.

When you compare the effective potentials (time rate) between two locations at rest and then transform to comparing them between the same two locations in motion, the ratio remains the same.

11. Jan 26, 2010

### Jonathan Scott

The definition of potential energy is tricky in relativistic gravity, and especially in GR, but I think that the usual informal interpretation for an object with rest mass is that it is the energy of the extra rest mass due to the potential difference between the locations being compared. This isn't 100% compatible with the Newtonian terminology about energy conservation, but then Newtonian terminology doesn't really mix with GR. For a moving object, this interpretation would mean that the potential energy has its own very tiny bit of kinetic energy! If you want something which should be conserved, you'd need to apply the potential factor (-Gm/rc^2 or whatever) to the total energy rather than the rest mass.

12. Jan 26, 2010

### Altabeh

At least here we know that the negative gradient of the gravitational potential energy U near the Earth's surface equals the Newtonian gravitational force which if we are in SR, then it is frame-dependent so I'd assume that the other side of equation must hold this frame-dependency due to having a meaningful comparison between PE and NGF when discussing their behavior wrt to reference frames in which they are measured.

I can feel that Jonathan you are, too, uncomfortable with what Bergmann claims just because you are so aware of that r in the denominator of NGF and of course its dependency to the reference frame it is measured in.

AB

13. Jan 27, 2010

### Naty1

bcrowell:
Both answers in #7 are very helpful; I'm apparently not struggling for a physical interpretation that is clearly obvious to all.

bcrowell (#7) posts
I understand I think, to the extent it seems to be somewhat of a dilemma, and that seems even crazier than the fact that magnetic and electric fields "transform" among frames.

The idea that an electric charge behaves one way and it's field(s) a different way in relativity appears utterly inconsistent. I'm guessing this results at least in part from the fact the relativity is a continuous formulation and a point charge is an idealized separate formulation.

But its no more a paradox than GR and QM not meshing so I guess we have gone about as far as we can right now.

14. Jan 27, 2010

### Naty1

There is another thread I read here in the last few days that pointed out that PE is posited to reside in the gravitational field between the masses and varies as the separation varies.

anyway, I appreciate the comments from all. Thanks.

15. Jan 27, 2010

### Jonathan Scott

If you try to formulate a Newtonian theory to work just like electrostatics, you get negative energy in the field (because a system has less energy when the masses are closer together). If however you include the known GR effect of time dilation, you get that potential energy would affect mass/energy directly, but by twice the expected amount. This would however still work out exactly right if there is an equal positive energy density in the field (just switching the sign).