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WannabeNewton
Science Advisor
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According to Wald (page 127) "For nonstationary configurations, however, there is no known analog in general relativity of the Newtonian potential".
PAllen said:Another thing worth looking at is the Komar mass.
See: http://en.wikipedia.org/wiki/Komar_mass
for an introduction. A key point is the term Kdv = √gtt dv. This means the contribution of locally measured mass to mass measured at a distance is decreased by the gravitational redshift factor. This is proportional 'surface gravity'. Thus, a collection of pieces of matter with some total mass at infinite separation, when brought together in a collection, will have their contribution reduced proportionally to surface gravity. The more compact the object, the more the surface gravity, therefore the larger redshift factor and the smaller the contribution.
bcrelling said:Would this reduction in mass for an object moved closer to a gravitational source exactly equal the relatavistic mass gained through increasing its velocity by falling the same distance?
bcrelling said:Would this reduction in mass for an object moved closer to a gravitational source exactly equal the relatavistic mass gained through increasing its velocity by falling the same distance?
PAllen said:The general effect it shows (decrease in contribution to gravitational mass for a collection of bodies closer together) is true for the collapsing case
PeterDonis said:More precisely, it's true if the collapsing case includes radiating away energy as the object collapses; the decrease in gravitational mass is equal to the energy radiated away. If you look at a collapse with no energy radiated away, the gravitational mass of the system does not change. See above.
PAllen said:What I am referring to is the idea that the local rest mass of infalling bodies contributes less (to gravitational mass measured at a distance), but their momentum increases; since both contribute to ADM mass, it remains unchanged during collpase, even 'before' radiation. While I know that you can't cleanly separate these, you can motivate that there are competing effects that balance.
Agerhell said:Now imagine you have an object moving "inwards" radially fast in a spherically symmetric gravitational field from a black hole. Then initially the black hole and the inwards moving object will be accelerated towards each other.
Agerhell said:At some point in time there will be no acceleration at all and when the inwards moving object comes really close to the Schwarzschild radius of the black hole (or perhaps sooner if it is moving fast?) it will decelerate.
Agerhell said:At this stage the black hole must be pushed back from instead of accelerated towards the inwards moving object...
Agerhell said:This is like if the mass of the inwardsmoving object was negative...
Agerhell said:What do you mean by "mass" in general relativity, is it "E/c^2", or is it resistance to acceleration when a force is acting on an object or is it something ells?
PeterDonis said:I'm not sure how one would exert a force on a black hole, so I'm not sure how a definition of mass as "resistance to force" (which is, strictly speaking, a definition of inertial mass, not gravitational mass) would apply to a black hole.
bcrelling said:light has momentum and would exert a constant force on a black hole
PeterDonis said:Would it? How would it exert the force? The black hole is vacuum; there's nothing there for the light to hit. It would just fall through the horizon and get destroyed in the singularity.
bcrelling said:I gather that a photon has momentum(p = h λ).
If a black hole consumes a photon surely it must take on the photon's momentum otherwise conservation of momentum is violated.
PeterDonis said:Interesting question. First, note that if this argument is valid, it will hold for any object falling into the hole, not just a photon. (Although there is a key difference between a photon and a massive object--see below.)
However, as it stands, it can't be valid, because momentum is frame-dependent, and conservation laws can't be frame-dependent. So whatever is conserved, it can't be momentum by itself; at the very least, we have to pick a particular frame in which to analyze the problem.
If you look at the actual math of freely falling objects in the gravitational field of a black hole (or indeed any spherically symmetric mass), there are two constants of the motion, energy at infinity and angular (not linear) momentum. To simplify things we can restrict ourselves to scenarios where the object is moving purely radially, so its angular momentum is zero. But even then, we have only energy at infinity as a constant of the motion (i.e., conserved quantity), and energy at infinity is evaluated in the frame in which the black hole is at rest. In that frame, the hole does not gain any momentum, by definition; it only gains energy--the energy it gains is equal to the energy at infinity of the object that falls in, so total energy in this frame is conserved.
Now what about momentum in the black hole's rest frame? If we drop in a massive object, we can drop it in "from rest at infinity", i.e., with zero initial momentum in the black hole's rest frame; in this case, total momentum is obviously conserved in the hole's rest frame, since the hole's final momentum in that frame is zero.
However, we can't drop a photon into the hole from rest at infinity, because a photon can never be at rest; this is the key difference I referred to above. What this tells me is that we can't just consider the photon in isolation; we have to consider its source as well, including the source's momentum. First take the simple case of a source that is at rest at infinity. Then when the source shoots the photon at the black hole, it recoils, so the total momentum at infinity of source + photon remains zero in the hole's rest frame. Thus, total momentum in that frame remains conserved. It is true that the source can increase the distance between itself and the hole, giving the appearance of a repulsive force exerted on the hole, by repeatedly firing photons in this way; but that is just another way of saying that we can make a photon rocket.
If the photon source has some nonzero original momentum in the black hole's rest frame, then a similar analysis to the above indicates that that will be the final total momentum of the entire system in that frame.
bcrelling said:A conventional rocket fires its thrust into a black hole- do they both experience the same repulsive force?
A "photon rocket" fires its thrust into a black hole- do they both experience the same repulsive force?
jartsa said:After the "collision" the rock will be traveling to the left, and it will have gained momentum to the right.
jartsa said:(the rock will be gaining momentum all the time that the black hole is exerting a pulling force on it, which is a long time, therefore the momentum is huge, and towards the black hole.)
jartsa said:Does anybody disagree with any of this?
The rest frame of the rock that has not been disturbed yet.PeterDonis said:In a properly chosen frame, such as the black hole's rest frame, yes. But you started out saying that the rock was at rest and the hole was moving. What frame do you think you're doing your analysis in?
jartsa said:The rest frame of the rock that has not been disturbed yet.
jartsa said:When we are giving the neutron star the large momentum, the coordinate velocity of it becomes nearly c.
jartsa said:When we are giving the small object a large momentum, the coordinate velocity of it becomes not as large as in the previous case, because coordinate speed of light is smaller on the surface of the neutron star.
jartsa said:So the system consisting of the neutron star and the small object is moving into one direction, which is the direction that we made the neutron star to move
jartsa said:and the system has momentum into the opposite direction, which is the direction that we made the small object to move.
jartsa said:We are far away from the neutron star when we are doing these momentum adjustment operations, we are using remote controlled robots.
jartsa said:(the coordinate velocity of the neutron star and the coordinate velocities of the photons are not very different, because the local speed of light is low)
jartsa said:Here's yet another thought experiment:
Let's say we are observing, from far away, a neutron star moving very fast to the left. On the neutron star there's a light bulb shining light all around.
Now we ask the following question:
What is the direction of the momentum of a photon that is moving to the right, away from the light bulb, but whose position is changing so that it is more and more to the left from us as time passes?)
bcrelling said:Surely there would be a redshift, and due to p=hf, the momentum would be less(due to decreased frequency) but still in same direction(to the right)?
jartsa said:When approaching:
Objects gain momentum towards each other.
When the light has disappeared:
The momentum of the black hole =
The momentum of the black hole at time t + the momentum of the light at time t
(t can be chosen freely)
jartsa said:Velocity change of the black hole = momentum change / mass
Quote by dipole
I think "relativistic mass" is a concept which should be avoided. It's better to think in terms of energy and to understand that energy is a source of gravitation, so an object moving with a lot of kinetic energy is going to have a stronger gravitational field.