# Defining negative energy of gravitational field

1. Jan 31, 2014

### Herbascious J

I have been learning that gravity has a negative energy associated with it. I've heard this stated a couple different ways, but I would like to understand a distinction. One line of thought declares that the potential energy of an object within a gravitational field is negative. Another line of thought states that the energy of the gravitational field itself is negative. I'm curious, is there a difference between these two ideas? It makes sense that an object has a negative energy associated with it because of its position within an already existing gravitational field. However, according to GR, and the idea of bent space-time, does the field itself some how contain negative energy by the nature of it's warp, independently from the objects that may be near it? I hope this makes sense. It seems in one description objects have an associated negative energy relative to the field. In another description the field itself is created from, or holds, negative energy. Perhaps this is simply a misunderstanding and they are one and the same.

Last edited: Jan 31, 2014
2. Jan 31, 2014

### WannabeNewton

Keep in mind that the energy of the gravitational field is much more difficult a concept in GR than it is in Newtonian gravity. The concept of a gravitational potential energy doesn't even make sense for non-stationary space-times. Regardless, see here: http://en.wikipedia.org/wiki/Positive_mass_theorem

3. Jan 31, 2014

### Herbascious J

I think this outlines part of the confusion I've been having. I was under the impression that gravity according to GR wasn't really a force acting on objects at all, but instead, objects simply coasting along warped geodesics, and experiencing no force of acceleration. To me this sounds like the only energy involved is kinetic, which remains constant. So with this idea, I've always imagined the energy of the field was in it's warping, independently of the objects. Somehow the field is warped holding a kind of energy. Is this not accurate?

4. Jan 31, 2014

### WannabeNewton

Yes, that's true of freely falling particles at least.

Let's be careful here. We don't want to confuse the energy of test particles in a gravitational field with the gravitational energy density of the gravitational field itself.

The phrase "energy of test particles" is itself contentious because there are at least two major notions of "energy of test particles". There's the expression $E = -p_{\mu}u^{\mu}$ where $p^{\mu}$ is the particle's 4-momentum and $u^{\mu}$ the 4-velocity of the observer making the energy measurement; this expression is valid for all space-times and is the same as the expression in SR for energy of a particle as measured by an observer. However $E$ does not include gravitational potential energy and in general we cannot even define a gravitational potential energy unless space-time is stationary, meaning it possesses a time-translation symmetry.

We can describe this time-translation symmetry by a vector field $\xi^{\mu}$ which we call a time-like Killing field and all it does is carry all fields in space-time along the flow of time in a way such that the metric tensor $g_{\mu\nu}$ remains invariant along this flow. We can then define another energy expression by $\tilde{E} = -p_{\mu}\xi^{\mu}$. If the particle is freely falling then $\tilde{E}$ corresponds to the total conserved energy along the geodesic describing the particle. If space-time is asymptotically flat, meaning the gravitational field dies off as you approach spatial infinity, then $\tilde{E}$ corresponds to the energy of the particle as measured by an inertial observer at infinity. $\tilde{E}$ includes the gravitational potential energy but in a non-trivial way in general because the gravitational potential is defined as $\phi = \frac{1}{2}\ln (-\xi_{\mu}\xi^{\mu})$.

However let's consider the special case of an observer who is carried along by the flow of $\xi^{\mu}$ so that said observer has a 4-momentum $p^{\mu} = m(- g_{00})^{-1/2}\xi^{\mu}$; this is an observer who is at rest in the gravitational field; then $\tilde{E} = m(-g_{00})^{1/2}$. Furthermore imagine we're in Schwarzschild space-time wherein $g_{00} = -(1 - \frac{2M}{r})$ hence $\phi = \frac{1}{2}\ln(1 - \frac{2M}{r})$.

If $r >> 2M$ then we can expand this to 1st order in $\frac{2M}{r}$ to get $\phi = -\frac{M}{r} + O((\frac{2M}{r})^{2})$. This should be familiar to you as the Newtonian gravitational potential. Note then that $\tilde{E} = m(1 + \phi) + O((\frac{2M}{r})^{2})$; if we plug back in the factors of $c$ and $G$ by dimensional analysis then $\tilde{E} = mc^2 + m\phi$. So for an observer at rest in the gravitational field far away from the source, $\tilde{E}$ is just the rest energy plus the gravitational potential energy.

On top of all this, we also have the energy/mass of space-time itself. This is what you are referring to. In Newtonian mechanics this is just the gravitational energy density of a gravitational field but the concept of a gravitational energy density is made complicated in GR by the equivalence principle (particularly the lack of gauge invariance). We instead work with global quantities like the Komar energy or ADM energy. See here for more: http://en.wikipedia.org/wiki/Mass_in_general_relativity#Types_of_mass_in_general_relativity

Last edited: Jan 31, 2014
5. Jan 31, 2014

### Herbascious J

Ok, so the energy/mass is the energy of the field itself. This sounds familiar, I remember something about the energy of gravity itself adding to the gravitational field in GR, in a kind cascading effect that should be taken into account. But this sounds to me like the energy is positive and not negative. The article about positive mass theorem seemed to indicate the energy of the field was also positive. I recently heard a lecture by Alan Guth that seemed to state that energy was negative and balanced out positive energy in the universe. It was a comment made in passing, but I'm curious how that plays into this discussion. Thank you for the excellent resources and response.

6. Jan 31, 2014

### WannabeNewton

Right, just like how an electromagnetic field itself has energy (in classical field theory we can easily write down the energy of the electromagnetic field in the most general context-for the gravitational field in GR we can only define its energy under special circumstances).

What may not be familiar to you is the concept of both the electromagnetic field and the gravitational field having momentum and angular momentum as well.

7. Feb 1, 2014

### Herbascious J

Just out of curiosity; when you say that a field has momentum and angular momentum, does that mean if they spin, then somehow the field will want to continue spinning momentarily if the source of the field is suddenly stopped from spinning? Or does that mean the field has angular momentum in the same way that a fundamental paricle has integer spin, like a quantum effect?

8. Feb 1, 2014

### WannabeNewton

See page 4 of this document (example 8.4-the Feynman cylinder paradox): http://oldwww.phys.washington.edu/users/vladi/phys543/2011PHYS543finalWsolutions.pdf [Broken]

We're talking purely classical fields here. Classical fields can have angular momentum too. The Feynman cylinder paradox is an example of electromagnetic field angular momentum. We also have gravitational field angular momentum in GR. For example Kerr space-time has an angular momentum $J = Ma$. The angular momentum of a stationary space-time is what causes for example the Lense-Thirring precession ("frame dragging") of gyroscopes at rest in the gravitational field.

The flow lines of the vector field $\xi^{\mu}$ defined in post #4 start twisting around one another because of the angular momentum of the space-time and the precession of a gyroscope at rest in the gravitational field can be directly related back to this twisting, which is codified by the vector field $\omega^i = \epsilon^{ijk}\partial_j \xi_k$.

Quantum fields have associated angular momentum operators which decompose into orbital and spin parts. The spin part $S^i$ is non-trivial to describe in the general context so I'll leave that alone. The orbital part is just $L^i = \epsilon^{ijk}x^j \partial^k$. Both of these act on states created from vacuum by the creation operators associated with the field.

Last edited by a moderator: May 6, 2017
9. Feb 1, 2014

### Jilang

So would it be incorrect to associate the angular momentum with the state rather than the field itself?

10. Feb 1, 2014

### WannabeNewton

Remember what the multi-particle states from vacuum represent conceptually in this case: mode excitations of a given field. In other words you shouldn't think of the two as independent/separate entities. The mode excitations are after all part of the field. It isn't as black and white when compared to the case of classical fields.

11. Feb 1, 2014

### Jilang

Thanks very much, that makes it most clear. I am still unclear about the OP though; is the energy of a static gravitational field positive or negative?

12. Feb 1, 2014

### WannabeNewton

13. Feb 1, 2014

### Jilang

Thanks for that. The link in post #2 ( if I'm reading right) would suggest that in Newtonian mechanics the energy of the field is negative, but in GR it's zero or perhaps having no single definition of the concept. Is that correct?

14. Feb 2, 2014

### pervect

Staff Emeritus
Newtonian mechanics has gravity as instantaneous interaction at a distance, rather than a classical field like E&M. One has to take into account the gravitational binding energy, but I don't think there is any clear way of localizing it as a classical field, at least not that I can recall reading about. It's hard to localize something that manifests as instantaneous action at a distance, so I suspect there isn't any way to localize it.

In GR, there isn't any general notion of the energy of the field, though there's a well defined notion of the Lagrangian density. Also, in a static geometry, you can find textbook statements that the mass of a system is lower than the sum of its parts, with attribution to the difference to "the gravitational binding energy". But you still can't localize the binding energy to any specific spot. MTW has a chapter on why this is impossible, as I recall.

So the net result is that there is some notion that gravitational binding energy exists in both static GR and Newtonian mechanics, but no clear prescription with either theory on exactly how this energy could be given a location.

Also, unless you have a static geometry, or an asymptotically flat space-time, GR doesn't have any well defined notion of energy.