# Energy of a particle

1. Jan 16, 2005

### hellfire

In the first chapters of Schutz's "A first course in general relativity", the energy of a particle is defined to be E = p0 (e.g. in 2.20). But in the chapter 7.4 discussing "conserved quantities" in curved spacetimes, the energy is defined to be E = - p0 = - g0upu. After reading 7.4 it is not clear for me why p0 is used for the definition of E instead of p0, as done in the first chapters of the book. Any help would be appreciated.

2. Jan 16, 2005

### Garth

Edit Added for rigour: The following is all measured in the rest frame of the coordinate system.
Ua = [1,0,0,0]
Otherwise E = - paUa

It is confusing because in GR there is no consistent definition of energy. This is because in general energy is not conserved.

The energy of a particle is only conserved if there is a time-like killing vector (static field) and the energy of a gravitating mass and its gravitational field can be only properly defined at an asymptotic null-infinity.

Basically you have to reach a region where the GR metric becomes Minkowskian - i.e. a SR metric - before you can proceed.

p0 is a natural candidate for the energy of a particle as it is the time component of the 4-momentum, however it is not conserved even if there is a time-like killing vector.

p0 is not an obvious choice until you realise that it is conserved because U0 is a killing vector in a static field and
p0 = mU0.

But note that if there is another moving mass nearby the field is not static and this definition falls apart. It won't even do for the Earth's field because of the presence of the Moon, let alone the Sun!

Note The theory of Self Creation Cosmology addresses this problem and defines the energy of a particle as [p0p0]1/2, which in SCC is conserved.

I hope this helps.

Garth

Last edited: Jan 17, 2005
3. Jan 16, 2005

### pmb_phy

The first section that you refer to is about SR where, in the coordinates he uses (Minkowski coordinates) p0 = p[suo]0[/sup] so there is no difference and typically people think of the time component as energy (although they shouldn't imho).

As far as why p[suo]0[/sup] is energy please see - http://www.geocities.com/physics_world/gr/conserved_quantities.htm

Pete

4. Jan 16, 2005

### pervect

Staff Emeritus
For an object in geodesic motion, pa ka or pakb will be a conserved quantity of the motion if and only if ka (in the first case) or ka (in the second) is a Killing vector.

In the exterior region of the Schwarzschild metric, it turns out that the Killing vector ka is a unit time like vector, i.e. ka = (1,0,0,0) regardless of the coordinates. This is the underlying reason that P0 is a conserved quantity. It's a particular feature of the Schwarzschild metric that ka = (1,0,0,0) is a Killing vector, while ka = (1,0,0,0) is NOT a Killing vector. (Well, actually, there are a fair number of metrics that have the property that ka = (1,0,0,0) is a Killing vector - any static metric where the gij are not functions of time will have this property, at least in the vacuum region of the metric.)

Because it's a conserved quantity in the Schwarzschild metric, P0 is often called the "energy" of the particle, though some authors are more careful and call it the "energy parameter". The important thing to know is that this is a conserved quantity for a particle in geodesic motion in the Schwarzschild metric (or in general in a static metric). It is important to also realize that one can NOT determine the total energy of a system simply by summing up P0.

In general spacetimes, P0 is not necessarily conserved - in fact, the constant density solution for a spherically symmetric star deosn't allow P0 as a conserved quantity in the interior region of the star. This last observation comes from my own calculations, but one can confirm that (1,0,0,0) is not a Killing vector in the interior region with Killing's equation.

5. Jan 18, 2005

### pmb_phy

. On that point it is important to note that even in SR P0 is not always conserved. If a charged particle is in an Electrtic field then P0 its not conserved.

Pete

6. Jan 18, 2005

### Garth

Pete - Is not pervect refering to P0 not being conserved when no work is done on or by the particle?

This occurs in a general non-static space-time metric, such as the Earth's space-time distorted by the Moon and Sun.

Garth

7. Jan 18, 2005

### pmb_phy

When the gravitational force does work on the particle in a static gravitational field then P0 is conserved.

Pete

8. Jan 18, 2005

### Garth

Pete - True, but where do we find a static gravitational field?

Garth

Last edited: Jan 18, 2005
9. Jan 18, 2005

### pmb_phy

You're sitting in one.

Pete

10. Jan 18, 2005

### Garth

Pete - No I'm not - I'm sitting on the Earth, orbiting the Sun, orbiting the Milky Way Galaxy, and with the Moon orbiting it. I see tides rise and fall twice a day. This space-time gravitational field is only approximately static - which in my book is not static!

Garth

11. Jan 18, 2005

### pervect

Staff Emeritus
I hope this discussion has been helpful to hellfire, and that we're not over-Killing the thread :-)

12. Jan 18, 2005

### pmb_phy

Do you know what a static g-field is?

We're speaking of concepts and not actual situations. If you want to have it your way and be exact then we can't describe anything exactly.

Pete

13. Jan 18, 2005

### hellfire

It is helpful, thank you. The first three posts are a clear answer to my question, but of course you are free to discuss this - or related aspects - more; I will keep my attention...

14. Jan 19, 2005

### Garth

Yes - and as the Earth's gravitational potentials are perturbed by the presence of the Moon's and Sun's gravitational fields, I'm not sitting in one!
Pete - we have had this discussion before. Energy is only conserved in GR under very special circumstances, which do not exist in the real universe. Here we live in a dynamical universe not a static one, therefore we had better accommodate ourselves to the realisation Einstein and others had about GR, and proved by Emmy Noether, that GR is an example of an improper energy theorem and therefore energy is not in general conserved.

Garth

Last edited: Jan 19, 2005
15. Jan 19, 2005

### pervect

Staff Emeritus
We've had this dicussion before? Really? Oh yeah, I guess we have :-)
(kidding).

Anyway, to make my usual comments, there are at least two different but very closely related standard notions of energy in GR (Bondi energy and ADM energy), both of which are derived from the notion of asymptotically flat space-time. There is a detailed discussion of this in Wald's "General Relativity", and I'm sure a search of the forum will turn up a lot of past discussion on this topic as well.

Since the universe appears to be expanding, one can indeed raise significant questions about whether or not asymptotically flat space-time actually exists in the real universe.

As an alternative of modifying GR so that energy is conserved (Garth's Self-Creation cosmology), it's worthwhile asking the question of whehter the universe really does conserve energy on a cosmological scale. An alternate solution to the problem is to say that energy conservation is only approximate - that in cases where one can ignore cosmological expansion, one can come up with a conserved energy based on approximating the universe as being asymptotically flat, but that in other cases, where cosmological expansion is significant and asymptotic flatness breaks down, energy conservation also breaks down.

16. Jan 19, 2005

### pmb_phy

So you agree that nothing in the universe can be described precisely and that when we speak of physics here we know that and are not worried about it and are usually speaking about ideal situations. E.g. If, relative to ther surface of the earth, I drive my car 50 mph in a straight line for one hour then I will have traveled 50 miles.

We understand that this is not literally possible because of miniscule variations in various variables. But we can say these things and talk about it and not have to worry that rather than 50 miles I might actually have traveled 50 miles and 1 nanometer.

We know you care about minor variations like that. However the situations we discuss here we almost never care about them.

Pete

17. Jan 19, 2005

### Garth

Pete - The perturbations of the Moon on the Earth's spherically symmetric and 'approximately static' gravitational potentials cause significant tides under which the Earth rotates so that a lot of energy can be extracted. Hardly just a "nanometer" or two.

But also the Earth is in an elliptical orbit around the Sun. As the Sun's dimensionless gravitational potential at the Earth is about 10-8, the six-monthly variation in its potential is about +/-10-10. Measured from the Earth the Sun's total energy varies, and is not conserved, by about 1 part in 10-10, which considering the Sun is quite a lot!

Garth

Last edited: Jan 20, 2005
18. Jan 19, 2005

### pervect

Staff Emeritus
To focus my earlier response a bit, I'll add a few more comments to the specific problem raised - that of the Earth, sun, & moon. If we isolate the solar system from the rest of the galaxy, by assuming that the solar system is embedded in an asymptotically flat space time, we can use the definitions of energy I mentioned earlier to come up with a theoretical conserved energy for the solar system (or the simpler 3-body problem of just the Earth, Moon & sun if one desires).

One does have to take into account the gravitational radiation that the Earth, moon, and other planets are emitting to come up with a truly conserved energy with this approach.

Actually carrying out the calculations would be extremely difficult, as I don't believe there are any analytical solutions to even the two-body problem in GR, much less the three-body problem. Approximate numerical solutions do exist, using the Newtonian solutions for the orbits would be an excellent starting approximation for further numerical refinement.

One will undoubtedly find many physical effects which are hard to model and a lot more significant than the effects predicted by GR in the issue of accurately modelling even the three-body system of the Earth, moon, & sun as they actually exist. If we replace the Earth moon & sun by black holes of the same mass, the GR analysis would be a lot simpler - this may or may not be regarded as cheating. An actual analysis that would represnet the actual Earth, moon & sun rather than their idealizations would have to include such diverse factors as:

the solar wind and light pressure from the Sun's radiation, the magnetic field of the Earth and it's interaction with the solar wind

the evolution of the sun as it burns it's nuclear fuel (which effects the other two factors above)

tidal interactions due to the extended nature of the planets & sun

so it would be a very difficult problem.

19. Jan 20, 2005

### Garth

It would indeed be a difficult problem to accurately model these effects, however that is not what I see the issue here to be.

The Sun losing mass through solar wind or losing energy through electromagnetic radiation, or the system losing energy through gravitational radiation can all be accommodated by the conservation of energy. Energy lost on one side of the equation is energy gained somewhere else.

My point is the principle of relativity and the equivalence of different inertial frames of measurement.

For example as the Earth approaches the Sun moving from aphelion to perihelion the red shift of the Sun's radiation decreases by about 1 part in 10-10. Therefore, other things remaining equal, as measured in the Earth's frame of reference the Sun would appear to heat up by that amount. The amount of heat energy stored by the Sun would appear to have increased as measured from the inertial frame of reference of the Earth and not the Sun. And this is only a tiny part of the problem.

To simplify; consider a massive gravitating body, A with mass M, alone in space, static, with no radiation, solar wind, gravitational wave or other loses of mass or energy. Consider a point mass, a test particle, falling from rest at infinity onto A. Because A has a static gravitational field there is a time-like Killing vector, A's 4-velocity U0, and the total energy of the test particle with 4-velocity one-form V0, here defined by the 'energy parameter' P0 = mV0, is conserved. Conserving the energy parameter and comparing it at infinity where v = 0 with its value at altitude r one obtains
v = [2GM/r]1/2 :- as in Newton where KE + PE = constant.

However from the test particles point of view, in its freely falling inertial frame of reference, the metric components of A's gravitational field are not time independent, they change with time as the particle descends. Therefore there is no time-like Killing vector and A's total energy is not conserved in that freely falling frame of reference.

We note that the desire to conserve energy may lead us to choose A's centre of mass as a preferred frame of reference but this would not be consistent with the principles of GR, instead it would be a Machian approach to gravitational theory which is the route Self Creation Cosmology has taken.

Last edited: Jan 20, 2005
20. Jan 20, 2005

### pmb_phy

Do you understand my point?

Pete