# Does a gravitational field have mass?

1. Jun 29, 2007

### duordi

I have been trying to conceptualize how a gravity field is formed.
If gravity was a wave that was emitting from the host mass then it should lag behind as a mass moves but a gravitational field seems to arrive with it host mass and the gravitational field is fully formed.

It is as if the gravitational field has an existence of its own.

That is the gravitational field is like a standing wave which has a velocity identical to its host mass.

If the gravitational field has energy it must also have a rest mass however small it may be.

The field should therefore follow a free floating path just as its host mass does.

I use this point of view to escape having to explain how the gravitational field (curved space time) develops instantaneously after the mass moves to a new location.

My line of reasoning is that the gravitational field has all the properties of mass and therefore a gravitational field and its mass will follow identical ( or almost identical paths ) arriving in a new location together and that the gravitational field does not have to be redeveloped from the host mass as a source at each location.

Is my line of reasoning correct or is there another view point I am not aware of?

I guess I have a problem with how gravity seems to transfer information instantaneously.

Duane Eddy

2. Jun 30, 2007

### pervect

Staff Emeritus
There is a certain amount of ambiguity in what the term "gravitational field" might mean. Probably the best way to answer this question is to keep things simple by talking about the Newtonian limit. If you have two masses, m1 and m2 (you measure them when they are not interacting), and you bring them close together so that they interact gravitationally, in the Newtonian limit the mass of the system is m1+m2 - $E_b$ where $E_b$ is the Newtonian gravitational binding energy of the system, i.e. the amount of energy it would take to separate the two masses to infinity.

3. Jun 30, 2007

### Mentz114

I believe it's generally accepted that nothing, even gravity can transfer information instantaneously. If a large mass appeared near the earth ( which couldn't happen) we wouldn't feel any gravitation from it until the deformation wave reached us. Similarly if the sun disappeared we wouldn't know about by any signal for about 9 minutes.

4. Jun 30, 2007

### pervect

Staff Emeritus
I didn't really address the "speed" aspect of the original question, which was a bit sprawly.

I'd recommend http://www.math.ucr.edu/home/baez/physics/Relativity/GR/grav_speed.html

for some introductory reading on that topic. When one applies the logic in the original post to electromagnetism (much easier than gravity), one gets the incorrect notion that the electric field should "lag behind" the position of a moving charge. As the FAQ explains, the conservation of momentum dictates that the electric field should and does point towards the instantaneous position of the charge, not the retarded position. This happens because of the conservation of momentum.

One does not derive the speed of light (or the speed of gravity) by looking at the "direction" that the field points in. Observing the direction of the static field doesn't give one any information about how the field propagates. Rather, one disturbs the field by making some sort of change (such as accelerating a charge), and then looks for a physical effect that one can measure. In the case of both light and gravity as it is described by GR, if you perturb a mass, the physical effects of that pertubation will travel outwards at 'c'.

5. Jun 30, 2007

### pervect

Staff Emeritus
There's another aspect of the above sprawling question that should probably be addressed. This is the issue of localizing energy in "the gravitational field".

It turns out not to be possible. See for instance MTW's "gravitation", chapter 20, section 4.

A detaield discussion of energy in GR gets rather technical. As far as online popular references go, besides the above, I'd suggest the sci.physics.faq

http://math.ucr.edu/home/baez/physics/Relativity/GR/energy_gr.html

A third aspect that probably deserves some mention is the relationship between energy and mass. Not only is the question of what constitutes the "gravitational field" a little ambiguous, there is some question as to what the original poster might mean by "mass". In SR, there are a couple of different sorts of mass - invariant mass, and "relativistic mass". In GR, there are even more types of "mass" - see the wikipedia article for more detail.

Last edited: Jun 30, 2007
6. Jun 30, 2007

### pmb_phy

pervect - Many times in the past you've mentioned the Komar mass. Do you know what the definition is? Do you have an intuitive feeling for what it is and means? I've always been curious on this but I can only find the term in Wald and Wald is not a text I've read much yet. Its on my list of things to do.

Thanks

Best regards

Pete

7. Jun 30, 2007

### duordi

Thanks for the answers and the references, espically the references.

I think my biggest problem is changing the Newtonian mind set.

I am finding it is an easy thing to say that I am going to stop thinking that way...

Accomplishing this intension is harder then I thought it would be.

There are just so many subtle assumptions that I make with out realizing it.

I went though an SR text book and worked the problems and was planning to go through a GR book but I am thinking now I should go through the SR and work the problems again.

Then I may be ready for the GR.

Duane Eddy

8. Jul 1, 2007

### pervect

Staff Emeritus
I've got a semi-intuitive definition that works for static systems. This means, for example, that it will work for a black hole, but not a rotating black hole. (It turns out that the Komar mass can be defined for a rotating black hole, but not by the method I'm going to describe below).

Because one assumes a static system, one can use the Newtonian idea of a gravitational force as the force per unit mass required to hold an object "in place".

Gauss's law would suggest that we try integrating the normal component of this force (the force required to hold a unit test mass in place which we can think of as the "gravitational force") over an enclosing surface (say a sphere) and that this integral should be a constant regardless of the size and shape of the enclosing surface.

When we try this ,though, we don't get a constant -unless we modify the formula by multiplying the force by a "red shift factor" to get the force at infinity. If we integrate this "force at infinity" by the normal area of the enclosing boundary (the area as measured by the frame-field of an observer located at the boundary) we do get a constant.

Because we normalize the "force at infinity" for a unit test mass, it's really an acceleration, but I think it reads better to write "force at infinity".

In coordinate dependent terms, the red-shift factor is just $\sqrt{g_{tt}}$, assuming that the metric has the properties that $g_{tt}$ = 1 at infinity and that $g_{tt}$ is not a function of time. The Schwarzschild metric satisfies these requirements, for example, so the red-shift factor for the Schwarzschild metric s just sqrt(1-2m/r).

The area of an enclosing sphere at a Scwharzschild at radius 4 is $4 \pi r^2$ the force required to hold an object in place is $$\frac{m}{r^2 \, \sqrt{1-2m/r}}$$, the force at infinity is $$\frac{m}{r^2}$$, so the integral of the force at infinity by the area of an enclosing sphere is just $4 \pi m$, independent of the radius r of the enclosing sphere. So the Komar mass can be defined as 1/4 Pi multiplied by the integral of (force-at-infinity) * (area).

For more info online see http://en.wikipedia.org/wiki/Komar_mass (which I wrote, for the original treatment there's always Wald).

The Komar mass can be re-written as a volume integral. Under suitably idealized conditions (isotropic pressure, a locally Minkowskian coordinate system, and geometric units) the contribution of a volume element dV to the Komar mass is (red shift factor)*(rho +3P)dV . Here rho is the density and P is the pressure as they appear in the stress-energy tensor.

Last edited: Jul 1, 2007
9. Jul 2, 2007

### pmb_phy

Oy! So much to learn and so little time to learn it. Ever get that feeling?

Pete