Gravity exerted by a fast moving object versus stationary object?

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  • #51
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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.)
 
  • #52
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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 momemtum to the left increases.

Does anybody disagree with any of this?
 
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  • #53
PeterDonis
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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. :wink:)
 
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PeterDonis
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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.
 
  • #55
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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.
 
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PeterDonis
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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?
 
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  • #57
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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)
 
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PeterDonis
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(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.
 
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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!)
 
  • #60
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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
 
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PeterDonis
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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)
Once again, you are using a coordinate chart in which you can't assume that conservation of momentum works. Unless you have done the math, you can't just help yourself to the above statements. A coordinate chart in which a black hole is moving does *not* work like a standard inertial frame in SR, nor does it work like standard Schwarzschild coordinates in GR.

Velocity change of the black hole = momentum change / mass
Even if the coordinates you are using did work like standard coordinates, this would not be true; relativistic momentum is not mass times velocity. You really need to do some math on this problem.
 
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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.
ok on the first part, but the rest depends on your definition of gravity; such an increase in KE is not sourced from the Einstein stress energy momentum tensor...the source of gravity as usually defined....which is in the rest frame of the mass.

Hmmm... a faster moving object, with more KE, say going past a large gravitating source such as a planet, would be deflected LESS than a similar object with lower velocity....???

Anyway, if the objects were two space ships, each would measure the planet as having the different KE relative to them.....
 

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