Gravity exerted by a fast moving object versus stationary object?

[PAllen]

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.

 

[Trenton]

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.

 

[PeterDonis]

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.

 

[Bill_K]

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.

 

[PeterDonis]

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 let's 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?

 

[WannabeNewton]

According to Wald (page 127) "For nonstationary configurations, however, there is no known analog in general relativity of the Newtonian potential".

 

[PAllen]

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.

 

[bcrelling]

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?

 

[PAllen]

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.

 

[PeterDonis]

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.

 

[PAllen]

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.]

 

[PeterDonis]

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.

 

[Agerhell]

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?

 

[PeterDonis]

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.

 

[bcrelling]

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 exert a constant force on a black hole (assuming the source of the laser maintains constant distance from the BH)?

 

[PeterDonis]

light has momentum and would exert a constant force on a black hole

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]

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 black hole 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?

 

[PeterDonis]

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.

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.

 

[WannabeNewton]

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.

 

[jartsa]

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.

 

[bcrelling]

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.

Perhaps this is too deep for me, so I don't expect you to reiterate everything again as I probably won't get it. However could you just tell me what happens in these situations?

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?

(in both of these situations consider that the rocket and BH and a third party observer are stationary with respect to each other at t = 0, and that the forces are measured at the first instant t = 0, and that we are viewing from the frame of the third party observer who is stationary and at a finitite distance.)

 

[jartsa]

Let's consider a still standing rock and a black hole that approaches from the right. After the "collision" the rock will be traveling to the left, and it will have gained momentum to the right.

(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.)

This is a bit counter intuitive, so it's better that I also tell what kind of changes happen to the black hole: The velocity to the left decreases, and the momentum to the left increases.

Does anybody disagree with any of this?

 

[PeterDonis]

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?

What do you mean by "fires its thrust into a black hole"? If you mean the rocket exhaust falls through the hole's horizon in its entirety, then if the rocket thrust is the same for both cases then each rocket should experience the same force pushing it away from the hole, yes.

Btw, I should clarify one thing: whether the rocket actually increases its distance from the hole (strictly speaking, from the hole's horizon) depends on its initial distance and the amount of rocket thrust. If the thrust is too small compared to the hole's gravity, the rocket itself (not just its exhaust) will fall in. There is a particular amount of thrust for any given distance from the hole that will cause the rocket to "hover", staying at the same distance; an amount of thrust greater than that will cause its distance to increase. In all these cases, the direction of thrust (i.e., the direction it pushes the rocket) is away from the hole; but its magnitude affects what happens.

Also, one other thing: you may be visualizing the rocket thrust as "pushing" on the hole, the way it would push on the Earth, for example, if the rocket was launching itself from the Earth. But the Earth has a surface to push on; the hole does not. The rocket exhaust that falls into the hole, as I think I said before, eventually hits the singularity at r = 0 and disappears. This makes any analysis based on conservation laws more complicated. (Another complication is that the singularity at r = 0 is spacelike, so it is really an "instant of time", not a "place in space". But I'll stop now, since I already piled on enough in my last post. )

 

[PeterDonis]

After the "collision" the rock will be traveling to the left, and it will have gained momentum to the right.

These two statements contradict each other. The rock can't have momentum to the right if it is moving to the left.

(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.)

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? You need to keep in mind that momentum is frame-dependent; it's not an absolute quantity.

Does anybody disagree with any of this?

Yes. See above.

 

[jartsa]

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?

The rest frame of the rock that has not been disturbed yet.
Let's say we give a large neutron star a large momentum x.

And then we give a small object on the surface of the large neutron star a large momentum -2*x.

When we are giving the neutron star the large momentum, the coordinate velocity of it becomes nearly c.

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.

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, and the system has momentum into the opposite direction, which is the direction that we made the small object to move.We are far away from the neutron star when we are doing these momentum adjustment operations, we are using remote controlled robots.

 

[PeterDonis]

The rest frame of the rock that has not been disturbed yet.

This frame is not an inertial frame, and it's also not a frame that matches up with the symmetry of the spacetime (since the hole itself is not at rest in it), so it will not work like you are thinking it will. In particular, I don't think you can assume that momentum is conserved in this frame. See further comments below.

When we are giving the neutron star the large momentum, the coordinate velocity of it becomes nearly c.

In an appropriately chosen frame, yes. However, as noted above, this frame will not work like either an inertial frame, or like the normal Schwarzschild coordinates. See below.

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.

You can't assume this, because that statement is only true in Schwarzschild coordinates, and, as noted above, you're not using Schwarzschild coordinates. (Also, you're ignoring the effect of the neutron star's gravity on the motion of the small object; it will slow down, meaning its momentum will not be constant.)

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

The neutron star is moving in that direction, and the small object is moving in the opposite direction. Which direction the "system" is moving in depends on how you define the "system" and its motion. How are you doing that?

and the system has momentum into the opposite direction, which is the direction that we made the small object to move.

As above, that depends on how you define the "system" and its motion. But however you do it, you can't have the system moving in one direction but having momentum in the opposite direction. That implies that you can't assume that the momentum of the system is just the sum of the momenta of the individual parts; i.e., you can't assume that momentum is conserved in the frame you are using.

In short, I think that you can't just assume things work the way your intuition says they do in the frame you are using; I think you actually need to do the math. That will be difficult since the frame you are using is not either a standard inertial frame or standard Schwarzschild coordinates, as noted above.

We are far away from the neutron star when we are doing these momentum adjustment operations, we are using remote controlled robots.

What difference does that make?

 

[jartsa]

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?

(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)

 

[PeterDonis]

(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)

Coordinate velocities are not physical velocities. Also, even if we assume for the sake of argument that these coordinate velocities work like velocities in an inertial frame, those velocities don't add linearly; light emitted in the rightward direction by an object moving at 0.999999999999c to the left still moves at c to the right.

Also, as I've already noted, the coordinates you are using are not the coordinates in which statements like "the coordinate velocity of light is slower near the gravitating body" are even known to be true. You can't assume things will work in these coordinates like you think they will. You need to actually do the math.

 

[bcrelling]

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?)

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)?

(Just ignore me if I'm only scratching the surface lols!)

 

[jartsa]

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)?

Well ok, yes.


Now let's consider a light beam and a black hole approaching each other, using the same idea of non-problematic momentum:

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)

Velocity change of the black hole = momentum change / mass