Is relative mass making gravitational field?


by marshallaw
Tags: field, gravitational, mass, relative
marshallaw
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#1
Dec6-12, 11:09 PM
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Hello
When you have something and it gains a lot of mass m due to its high kinetic energy, so it gains a lot of relative energy. So, every object has its own gravitational field. So is the RELATIVE mass making a gravitatinal field?
mathematically if g=GM/Rˇ2 does work, where you MUST add relative mass
Well this would be very amazing, strange, bizzare if this was true
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bcrowell
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Dec6-12, 11:36 PM
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If I'm understanding correctly, then you're using the terms "relative energy" and "relative mass" to mean what is referred to in standard terminology as mass-energy. In standard (modern) terminology, "mass" is absolute, not relative. What we now call mass is what people used to refer to as rest mass.

The source of the gravitational field is the stress-energy tensor, not mass-energy. You can sometimes get away with estimating gravitational effects by plugging mass-energy into Newton's law of gravity, but it won't always work. For example, parallel beams of light exert zero gravitational force on one another, but antiparallel ones exert nonzero force.
PAllen
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Dec7-12, 12:37 AM
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A next level of approximation is invariant mass. It is often accurate up to the point where self gravitation is important, for simple systems. For example it 'explains' the anti-parallel vs. parallel beams. In the former, the invariant mass of the system adds the beam energy because the momentum cancels. In the parallel case, the invariant mass is zero. Similarly, the KE of an isolated body (no matter how fast it is moving in some frame) has no relevance for any curvature invariant produced by it. However, as part of confined system of bodies (e.g. particles in a box), the invariant mass of the system will tend to include much of the KE of each body due to momentum cancellation. Just don't take this too far - the real source, as mentioned, is the stress energy tensor.

tom.stoer
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Dec7-12, 02:21 AM
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Is relative mass making gravitational field?


this question could be studied using the Aichelburg–Sexl ultraboost metric http://en.wikipedia.org/wiki/Aichelb...exl_ultraboost
haushofer
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Dec7-12, 07:02 AM
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I would like to nominate that as the metric with the most beautiful name.
pervect
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Dec7-12, 09:48 AM
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I tend to view energy (sometimes called relativistic mass) as a better approximation for gravity in most circumstances than invariant mass.

Examples where this approach works reasonably well.

particles in a box moving thermally
velocities induced by a relativistic flyby.

In fact, the velocities induced by a relativistic flyby increase MORE rapidly than the approximation usig energy- see http://dx.doi.org/10.1119/1.14280 .

So using invariant mass to compute velocities due to a relativistic flyby is a very bad approximation, it's off by a factor of 2*gamma, whereas the using energy is only off by a factor of 2:1 or so.

But as everyone has pointed out, the actual situation is more complex, it's really the stress energy tensor that causes gravity, and if you want to get truly accurate answers that's what you need to use.

Which leads to the other reason I like to use energy. It's easier to justify as an approximation - you're approximating the stress-energy tensor by just the energy part, and throwing away the others (the momentum part, and the stress part). Invariant mass isn't part of the stress energy tensor the way energy is.

The other thing that's worth mentioning is that F=GmM/r^2 doesn't work at all for relativistically moving bodies. The "field" from a moving body isn't at all spherically symmetric.

Some insight into this can be gained by considering the electric field of a rapidly moving body. A search should find a lot of posts on this topic.
Naty1
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Dec7-12, 10:29 AM
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Marshallaw.....WOW you got great replies here,,,,,sometimes an initially insigthful answer spawns others.....
As usual when members of this group reply I have some recollecting to do. If you are
that way, check out the first section here:
http://en.wikipedia.org/wiki/Invariant_mass
for a bit of background.....especially 'frames'.

The source of the gravitational field is the stress-energy tensor, not mass-energy.
took me by surprise when I first read it several years ago. Later I realized that it meant the mass-energy was only a piece of the source....as pervect illustrated while I was typing!
PAllen
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Dec7-12, 10:45 AM
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Quote Quote by pervect View Post
I tend to view energy (sometimes called relativistic mass) as a better approximation for gravity in most circumstances than invariant mass.

Examples where this approach works reasonably well.

particles in a box moving thermally
velocities induced by a relativistic flyby.

In fact, the velocities induced by a relativistic flyby increase MORE rapidly than the approximation usig energy- see http://dx.doi.org/10.1119/1.14280 .

So using invariant mass to compute velocities due to a relativistic flyby is a very bad approximation, it's off by a factor of 2*gamma, whereas the using energy is only off by a factor of 2:1 or so.

But as everyone has pointed out, the actual situation is more complex, it's really the stress energy tensor that causes gravity, and if you want to get truly accurate answers that's what you need to use.

Which leads to the other reason I like to use energy. It's easier to justify as an approximation - you're approximating the stress-energy tensor by just the energy part, and throwing away the others (the momentum part, and the stress part). Invariant mass isn't part of the stress energy tensor the way energy is.

The other thing that's worth mentioning is that F=GmM/r^2 doesn't work at all for relativistically moving bodies. The "field" from a moving body isn't at all spherically symmetric.

Some insight into this can be gained by considering the electric field of a rapidly moving body. A search should find a lot of posts on this topic.
I was talking about estimating curvature invariants not deflection of fly by trajectories. For that purpose, there is s a class of simple systems (non-interacting particles, including minimal mutual gravitational influence per particle, small-ish total mass), such that ADM=BONDI mass (since no radiation) = invariant mass. This means the total curvature is due to the invariant mass. This approximation remains good for highly relativistic speeds between the particles, and for high COM speed (given all the other conditions).

Also, the invariant mass approximation helps explain why motion relative to an observer has no tendency to produce a black hole; and why parallel motion of two bodies or beams is different in an invariant way from anti-parallel motion, in terms of the effective curvature mass.
PAllen
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Dec7-12, 11:22 AM
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One more point on the difference between modeling fly by deflection versus curvature invariants:

Imagine series of identical asteroids fired to pass e.g. 10,000 km passed a some observer maintaining fixed position relative to distant stars. The first projectile is a .1, then .5., then .9c, then .999c, etc. The 'static' observer measures a peak Kretschmann invariant (the simplest curvature invariant that isn't zero for a vacuum in GR). The peak Kretschmann invariant for each projectile will be the same.

[Edit: a few further observations:
- The above statement is true despite that peak thrust needed for observer to retain 'static' position is greater for each successive projectile.

- Imagine two asteroids going by at the same speed in tandem past hypothetical 'static' observer; versus two asteroids going by, coming from opposite directions, same speed, same closest approach to each other and to observer as tandem case. I believe:

- the peak Kretschmann invariant measured by observer will be greater for the opposing motion fly by.
- the peak force to hold position will be similar for the two scenarios (unlike the Kretschmann measurement)

Thus, invariant mass is a better first approximation for curvature invariants (e.g. black hole formation) or estimating ADM/Bondi mass. Total energy is a better first approximation for estimating peak attractive force.

]


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