Gravity exerted by a fast moving object versus stationary object?

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PAllen, yes I had considered this precise scenario as it is the standard early life of a star. The initialy diffuse cloud contracts under gravity, heats up and radiates away energy. But during this process no nuclear reactions are involved and so the mass of the gas itself is unchanged. The higher temparature though, as this is energy, increases the mass via mass-energy equivilence. Also needless to say, there is mass enegy equivelence of the radiated energy.

Seen from infinity, the mass of the initial cold cloud would be just the mass of the gas. But as cloud condensed and got hotter it would gain mass - in spite of radiating away mass in the form of energy. Even if the proto star has insufficent mass to become a star and instead becomes a Jupiter like object which (after billions of years) cools to the initial temperature of the initial difuse cloud and thus the same mass of the initial cloud, the millions of tonnes of light are still out there contributing to the total mass of the universe.

The proto star process, unless the initial gravitational energy has a mass energy equivelence in it's own right, would seem to have created mass, literally out of thin air! And yet as I point out, how can it have if this depends on a future unknown state?

I should probably say at this point, what I was contemplating when I came up with this. I was playing around with models of the universe and trying to work out what factors contributed to total gravity. Clearly in this context if gravitational energy is a factor it has greater propensity towards absurdity than it does in the example of the proto star.

But this is too simple a concept to have been overlooked. There must be good textbooks out there that explain what can and can't be considered to contribute gravity to the universe.
 
  • #27
PAllen
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PAllen, yes I had considered this precise scenario as it is the standard early life of a star. The initialy diffuse cloud contracts under gravity, heats up and radiates away energy. But during this process no nuclear reactions are involved and so the mass of the gas itself is unchanged. The higher temparature though, as this is energy, increases the mass via mass-energy equivilence. Also needless to say, there is mass enegy equivelence of the radiated energy.
there that explain what can and can't be considered to contribute gravity to the universe.
This may be your core misunderstanding. Heat, and all forms of energy, contribute to gravitational mass = inertial mass (principle of equivalence). Radiation of infrared, or whatever, lead to decrease of mass. Presence or absence of nuclear reactions is irrelevant.

See this classic paper:

http://arxiv.org/abs/gr-qc/9909014
.....

But this is too simple a concept to have been overlooked. There must be good textbooks out there that explain what can and can't be considered to contribute gravity to the universe.
All energy and momentum and pressure and rest mass contribute to the stress energy tensor which is the source of gravity in GR (not all are additive; mass is an extremely complex concept in GR; however all of these contribute to the source term of the equations of GR).

The up shot is, if you imagine a cloud collapsing without radiating, its mass does not change at all. The rest mass of particles counts for less because it is a lower potential; however, the KE as heat exactly compensates. In GR this is an exact consequence of Birkhoff's theorem if you assume spherically symmetric collapse. The collapsed cloud weighs less only as it radiates energy.
 
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  • #28
PeterDonis
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Seen from infinity, the mass of the initial cold cloud would be just the mass of the gas. But as cloud condensed and got hotter it would gain mass - in spite of radiating away mass in the form of energy.
I think you're mixing up distinct processes, and also your usage of the word "mass" is getting you into difficulties, because you're conflating different meanings of that term. Let me try to re-describe the cloud collapse scenario without using these terms at all, and see if that helps. (Also note that I'm not going to use the term "gravitational potential energy" at all; that concept can be helpful but it can also lead to confusion, like the concept of "mass" does.) [Edit: I see pervect said the same thing as I'm going to say, but much more briefly.]

Start with a cloud of gas that is (a) spherically symmetric, and (b) entirely at rest at some instant of time. Suppose we measure the mass of this cloud by putting a test object into a circular orbit about it at some large distance, and measuring both the distance and the orbital period, and applying Kepler's Third Law. This will yield some number M.

Now we wait a while, and the cloud starts to collapse. We'll assume that the collapse is also spherically symmetric, so any given particle of gas within the cloud only moves radially. After the cloud has been collapsing for some time, so individual gas particles within the cloud have a significant inward radial velocity, but *before* the cloud has radiated any energy away, we measure the mass again using the above method. What result will we get?

The answer is that we will still get M. This follows easily from Birkhoff's Theorem, which guarantees that the spacetime metric outside a spherically symmetric mass distribution is independent of whatever is happening inside the mass. Another way to see why we still get M is to note that, contrary to what I think your intuition is telling you, the kinetic energy that is gained by the particles of gas in the cloud as it collapses does *not* increase the cloud's externally measured mass. The reason is that, as pervect pointed out, the externally measured mass of the cloud is determined by its stress-energy tensor, not the relativistic mass of its individual particles. The SET of the cloud does change as the cloud collapses, but it changes in a way that leaves the externally measured mass of the cloud the same.

Now we wait a while longer, and the cloud starts to radiate energy away as it collapses. After some more time has passed, the cloud has radiated away energy E (as measured by collecting the radiation very, very far away). We measure the mass again at this point using the above method. Now we will get M - E as the result; the energy radiated away is exactly balanced by a reduction in the measured mass of the cloud itself. So energy is not created out of nothing.
 
  • #29
PAllen
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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.
 
  • #30
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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.
Is this analogous to the effect of "nuclear binding energy" where an atom has less mass than the sum of it's separate subcomponents(when apart)?
 
  • #31
PAllen
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Is this analogous to the effect of "nuclear binding energy" where an atom has less mass than the sum of it's separate subcomponents(when apart)?
Yes, it is analogous.
 
  • #32
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Peter Donis et al - This has put my mind at rest. Pervect was aluding to the same but he is far more fluent in math than I am so he doesn't need the descriptions as much as I do! I knew that gravity was the result of the stress-energy tensor but I make a lot of mistakes when I try to construct it. Presumably this is why PE is by convention negative? I also like the analogy to the nuclear binding energy.

On the original issue of gravity exerted by a relativistic mass, what is the plausibilty of the following?

In the big bang only a percentage of mass would become matter and probably most of it is in the form of photons and a large proportion of these are at the 'edge' traveling outwards. Assuming a sphere and noting that gravity normally cancels out inside spheres; given that the source of the gravity is itself traveling out at c, there would be a net field acting to accelerate outwards and strongest nearer the edge, as the cancelling gravity can no longer reach you in time! I think the result would be a collapsing gravity field and so an expansion of space.

This conjecture if it is in any way viable, has advantages over the concept of 'dark energy' forcing galaxies apart - which is for me, is no more comfortable than the concept of proto stars creating mass/energy out of thin air.
 
  • #33
PeterDonis
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On the original issue of gravity exerted by a relativistic mass, what is the plausibilty of the following?
Not good, because all the discussion we've been having in this thread only holds in the case of a static gravitational field--i.e., a single isolated massive object. Concepts like "gravitational potential energy" don't work at all in the non-static case, such as our universe, which is expanding. The stress-energy tensor still acts as the source of gravity, but the specific solution of the Einstein Field Equation that describes the universe is very different from the one that describes an isolated gravitating mass.
 
  • #34
Bill_K
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Not good, because all the discussion we've been having in this thread only holds in the case of a static gravitational field--i.e., a single isolated massive object. Concepts like "gravitational potential energy" don't work at all in the non-static case, such as our universe, which is expanding.
Isn't this overstating it a bit, Peter? The concept of gravitational potential energy certainly exists in the weak field case. And as far as I know, for strong fields, asymptotic flatness alone is a sufficient condition.
 
  • #35
PeterDonis
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The concept of gravitational potential energy certainly exists in the weak field case.
Even for a non-stationary weak field? For example, for an expanding universe with a very, very small energy density?

And as far as I know, for strong fields, asymptotic flatness alone is a sufficient condition.
Even if the spacetime is not stationary? Asymptotic flatness lets you define a notion of "infinity" where the potential energy can go to zero, but how does that help to define a potential energy in the strong field region?
 
  • #36
WannabeNewton
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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".
 
  • #37
PAllen
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Asymptotic flatness alone is all you need to define global conserved quantities (energy, momentum, angular momentum), via the ADM approach. However, I don't see how that translates to something resembling a potential without much more restrictive assumptions. In other words, what I've read agrees with WannabeNewton's Wald quote.
 
  • #38
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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.
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?
 
  • #39
PAllen
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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?
Relativistic mass is an old fashioned concept from SR, altogether inapplicable to GR. What you can say is that via conservation of ADM mass, imass contribution reduction from gravitational binding, contribution from increased momentum of bodies coalescing, and radiation (both EM and GW) balance out. Unfortunately, your attempts to to separate these out into simple, additive quantities is doomed to fail in GR.

I gave the Komar mass reference as a general aid to understanding. The Komar mass integral is not applicable to a collapsing body. The general effect it shows (decrease in contribution to gravitational mass for a collection of bodies closer together) is true for the collapsing case, but the exact integral defining Komar mass is not actually applicable.
 
  • #40
PeterDonis
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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?
This is not a good way of putting it, IMO, but it is true that if we take a system that starts at rest at some initial radius and let it collapse to a smaller radius, and do not allow any energy to radiate away, the total gravitational mass of the system is unchanged. One way of looking at this is to note that the individual parts of the system will not be at rest at the smaller radius; they will be moving inward. So the increase in kinetic energy of the parts of the system exactly compensates for the decrease in gravitational potential energy, so that the total energy (and hence the total gravitational mass) is unchanged. (Of course, as I said, this only holds as long as no energy is radiated away during the collapse.)

However, as PAllen says, viewing the change in kinetic energy as an increase in "relativistic mass" is not really applicable in GR; it's better to just look at it as a change in energy. Also, viewing the change in gravitational potential energy as a "reduction in mass" is not really a good way to think of it either. One way of seeing why is to imagine an observer who is free-falling inward along with one of the individual parts of the collapsing object. Suppose this observer measures the mass/energy of the part he is falling next to. He will measure it to have its normal rest mass (i.e., the rest mass it would have if it were far away from all gravitating bodies), and to have zero kinetic energy (because it's at rest relative to him). In other words, the fact that it is falling inward, and that it is in a region of lower gravitational potential energy, will make no difference to his measurements.

Also, the above assumes that we even have a valid notion of "gravitational potential energy", which is only true in a very restricted set of spacetimes. (Whether or not a non-stationary spacetime like that of a collapsing object is even in this restricted set is an interesting question; I think it is if the collapse is spherically symmetric and asymptotically flat, but others might disagree.)

The general effect it shows (decrease in contribution to gravitational mass for a collection of bodies closer together) is true for the collapsing case
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.
 
  • #41
PAllen
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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.
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.

[edit: and once you consider radiation, the ADM mass (which includes all the radiation at infinity) still remains unchanged, but the Bondi mass (which does not include the radiation) decreases.]
 
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  • #42
PeterDonis
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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.
Ah, ok, that's basically what I was saying in the first part of post #40.
 
  • #43
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Returning to the title of the thread: "Gravity exerted by a fast moving object versus stationary object".

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 inwardsmoving object will be accelerated towards each other. At some point in time there will be no acceleration at all and when the inwardsmoving object comes really close to the Schwarzschild radius of the black hole (or perhaps sooner if it is moving fast?) it will decelerate. At this stage the black hole must be pushed back from instead of accelerated towards the inwardsmoving object...

This is like if the mass of the inwardsmoving object was negative...

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?
 
  • #44
PeterDonis
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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.
In the sense of coordinate acceleration, in Schwarzschild coordinates, yes. See below.

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.
In the sense of coordinate acceleration, in Schwarzschild coordinates, yes. But that doesn't mean what you think it means. See below.

At this stage the black hole must be pushed back from instead of accelerated towards the inwards moving object...
No, this is not correct. The inwards moving object still falls into the hole, because Schwarzschild coordinates get more and more distorted as you get closer to the horizon, so they don't give a good description of what's actually happening physically. Google on Painleve coordinates for a coordinate chart that gives a much better description of an object falling into the hole; in these coordinates the infalling object continues to accelerate all the way down.

This is like if the mass of the inwardsmoving object was negative...
No, it isn't. See above.

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?
I would say in general "something else", but it really depends on the particular spacetime you're considering and what you want the "mass" to tell you. For a black hole spacetime, or indeed for a general stationary isolated gravitating object, there are at least three possible definitions for "mass": the ADM mass, the Bondi mass, and the Komar mass. For a classical Schwarzschild black hole they are all equal, but they can be different for other kinds of gravitating objects.

Also, for an isolated object one can assign a 4-momentum vector whose length is its mass (I think the ADM mass is the one this strictly applies to), so one could say that M = E / c^2 if one sets up asymptotic inertial coordinates "at infinity" such that the object's 4-momentum components in those coordinates are (E, 0, 0, 0).

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.
 
  • #45
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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.
Couldn't this be done with a laser, as light has momentum and would excert a constant force on a blackhole (assuming the source of the laser maintains constant distance from the BH)?
 
  • #46
PeterDonis
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light has momentum and would excert a constant force on a blackhole
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.
 
  • #47
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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.
I gather that a photon has momentum(p = h λ).
If a blackhole consumes a photon surely it must take on the photon's momentum otherwise conservation of momentum is violated.
And if that stream of momentum packets is constant it would have the effect of a constant force.

How else could momentum be conserved?
 
  • #48
PeterDonis
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I gather that a photon has momentum(p = h λ).
If a blackhole consumes a photon surely it must take on the photon's momentum otherwise conservation of momentum is violated.
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.
 
  • #49
WannabeNewton
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You have to be somewhat careful with that bcrelling. If you allow for a particle with even extremely small non-zero angular momentum to fall into a Schwarzschild black hole, you are perturbing the solution from Schwarzschild to Kerr. If the particle falls in with no angular momentum (so radially) then an effective conservation law of importance is the Schwarzschild black hole analogue of the first law of thermodynamics, ##\delta M = \frac{\kappa}{8\pi}\delta A## where ##\kappa## given by ##\kappa^{2} = -\frac{1}{2}\nabla^{a}\xi^{b}\nabla_{a}\xi_{b}## is the surface gravity across the horizon of the black hole (here ##\xi^{a}## is the killing vector field normal to the horizon which for the Schwarzschild black hole coincides with the stationary killing vector field) and ##A## is the surface area of the horizon.
 
  • #50
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If I have understood correctly a method to accelerate a black hole is of interest here.

Here's my suggestion: Let's put a satellite on a highly eccentric orbit around the black hole.

The satellite deforms the event horizon at one side of the black hole, so that the black hole propels itself with its non-symmetric Hawking radiation.
 
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