# Do black holes violate conservation of momentum?

#### glugglug

##### Diamond Member
Say you have a black hole with a net postiive charge.

A proton approaches the singularity and the electrical field from the proton repels it. But the electrical field from inside the black hole itself can not escape due to gravity, so the outside proton is unaffected.

#### DrPizza

##### Administrator Elite Member Goat Whisperer
I don't think there's any violation of the conservation of momentum. There's just no way to measure the change in momentum of the black hole on such a minute scale. Perhaps this will bother you: as I jump up the impulse given to me is equal to the impulse the earth experiences (but opposite in directions, obviously)... I go up and the earth goes down, each with the same change in momentum. And after I leave the surface, I exert a force on the earth, equal to the earth's gravity's force on me. The earth accelerates toward me as I accelerate toward the earth. Both have the same force acting on them, so F=ma. For me, it's F = little mass, big a. For the earth, it's the same F = BIGGGG mass, littttttle a.

#### DrPizza

##### Administrator Elite Member Goat Whisperer
Oh, forgot to mention, as I jump up, someone on the other side of the earth with the same mass as me jumps up at the exact same time, thus everything cancels out

#### glugglug

##### Diamond Member
The black hole would accelerate, perhaps even measurably so. It doesn't need to weigh all that much, it could even be below 1g if composed of a few protons orbiting each other at relativistic speeds.

What I want to know is, since the electrical field from the protons inside the black hole can not escape, are protons outside the black hole repelled by the electrical field or are they completely unaffected, violating conservation of mass since the black hole is accelerating away from their electrical field and they are not being pushed in the opposite direction?

#### Shalmanese

##### Platinum Member
My guess is that black holes are always electrically neutral since electric fields are mediated by photons. Thus, a proton would be unaffected by a black hole.

#### CQuinn

##### Golden Member

(1) Doesn't your question ignore the "existence" of the Event Horizon?

That boundary effectively places the singularity within its own universe.

(2) Aren't you assuming that the laws of physics apply the same close to the singularity as they do
on the upside of the Event Horizon?

#### Jeff7

##### Lifer
Originally posted by: glugglug
Say you have a black hole with a net postiive charge.

A proton approaches the singularity and the electrical field from the proton repels it. But the electrical field from inside the black hole itself can not escape due to gravity, so the outside proton is unaffected.

Does anyone really know what the charge of a black hole is? Can it even have a charge if gravity lets nothing escape?

Originally posted by: Shalmanese
My guess is that black holes are always electrically neutral since electric fields are mediated by photons. Thus, a proton would be unaffected by a black hole.

Electrons also carry charge.
The proton would be affected by a black hole, just likely not electrically. Since the proton does carry mass, it would be drawn by gravity towards the black hole, and cross the event horizon. Would this change the charge of the black hole? Possibly. Once we understand what happens inside a black hole (or rather, if), then that could be answered.

#### MrDudeMan

##### Lifer
as a proton approachs a black hole, it tries to escape and ends up taking a curved instead of direct shot. near the event horizon, it splits into two different particles. its called a positron, the anti-matter form of a proton. part of it escapes due to the angle after the split, and the other half is pushed the other direction, or toward the singularity.

therefore, i have no idea how to answer your question, but i thought id give you this tidbit of information.

#### glugglug

##### Diamond Member
Originally posted by: MrDudeMan
as a proton approachs a black hole, it tries to escape and ends up taking a curved instead of direct shot. near the event horizon, it splits into two different particles. its called a positron, the anti-matter form of a proton. part of it escapes due to the angle after the split, and the other half is pushed the other direction, or toward the singularity.

therefore, i have no idea how to answer your question, but i thought id give you this tidbit of information.

Um.... a positron is the anti-matter form of an electron, and I'm fairly sure nothing with mass escapes, but that's not the question.

#### rjain

##### Golden Member
Tom Van Flandern's latest articles seem to indicate that EM must propagate at superluminal speeds. Both Maxwell's equations and Einstein's equations require this to be true. However, oscillations in the EM and Gravitiational fields travel at c.
I'm currently thinking of a theory where virtual particles are simply particles that are still coherent with the source particle. When decoherence occurs, the effects of that must be instantaneous, and decoherence imposes conservation laws on the possible results. The effects must be instantaneous because of issues like conservation of angular momentum.
However, this theory does predict that the propagation of changes of magnitude in the field strength will be constrained to c as that factor is mediated by the number of virtual particles in any region. Propagations of changes of direction will be instantaneous, mediated by decoherence.

#### glugglug

##### Diamond Member
Originally posted by: rjain
Tom Van Flandern's latest articles seem to indicate that EM must propagate at superluminal speeds. Both Maxwell's equations and Einstein's equations require this to be true. However, oscillations in the EM and Gravitiational fields travel at c.

Interesting idea, I am wondering how the distinction is made between EM and oscillations in EM. Can you post a link to this article?

#### rjain

##### Golden Member
Originally posted by: glugglug
Originally posted by: rjain
Tom Van Flandern's latest articles seem to indicate that EM must propagate at superluminal speeds. Both Maxwell's equations and Einstein's equations require this to be true. However, oscillations in the EM and Gravitiational fields travel at c.

Interesting idea, I am wondering how the distinction is made between EM and oscillations in EM. Can you post a link to this article?
http://www.metaresearch.org/cosmology/gravity/speed_limit.asp
http://www.dipmat.unipg.it/~bartocci/ep6/ep6-vanfl.htm

I'm currently trying to explain my theory to Tom on the metaresearch messageboard under the Gravity and Relativity section.

The distinction between EM and oscillations in the EM field is the same as the difference between the speed at which a taut string's atoms can travel and the speed at which vibrations on that string propagate.

#### Jeff7

##### Lifer
Originally posted by: rjain
Tom Van Flandern's latest articles seem to indicate that EM must propagate at superluminal speeds. Both Maxwell's equations and Einstein's equations require this to be true. However, oscillations in the EM and Gravitiational fields travel at c.
I'm currently thinking of a theory where virtual particles are simply particles that are still coherent with the source particle. When decoherence occurs, the effects of that must be instantaneous, and decoherence imposes conservation laws on the possible results. The effects must be instantaneous because of issues like conservation of angular momentum.
However, this theory does predict that the propagation of changes of magnitude in the field strength will be constrained to c as that factor is mediated by the number of virtual particles in any region. Propagations of changes of direction will be instantaneous, mediated by decoherence.

EM must propagate at superluminal speeds? Isn't light just EM radiation? Is that to say then that light travels faster than light?

#### rjain

##### Golden Member
Originally posted by: Jeff7
EM must propagate at superluminal speeds? Isn't light just EM radiation? Is that to say then that light travels faster than light?
The "speed of light" (c) is not the speed of propagation of changes in the EM field in response to movements of charges. It is the speed of propagation of electrical and magnetic field oscillations along the field.

#### tboneuls

##### Banned
Irrelevant. The mass of a black hole is so great that matter collapses onto itself, regardless of polarity. The strength of the electron field is not strong enough to overcome such gravitational forces. Additionally, there have been no black holes observed with positive charges, so as far as we know, such a thing does not exist.

#### s0upx0n

##### Junior Member
It seems that by Maxwells equations if the black hole has a net charge there would be an electric field around it, possibly caused by the particles emiting the relevent electromagnetic field carriers previous to falling past the event horizon. That assumes that maxwells equations work around a singularity, but I thought that black holes were supposed to be something other then 'bare' singularities, so this may work.

However I think that discounts the nature of the question, the proton is falling into the black hole and it interacts with the particles around it, just because an electromagnetic field can't penetrate from inside the event horizon doesn't mean that the proton can't effect other particles near it. Thus providing some change in the net center of mass of the black hole. Actually it seems that a black hole's charge would be somewhat similar to its surrounding environment, and I thought the bulk of the universe was pretty neutral, so overall it seems that the black hole's gravitational force would basically be negligable. Also as the proton enters the black hole it becomes part of it, so any interaction between those two shouldn't change the center of mass of the black hole/proton system.

Just my \$.02, I'm not a physicist.

#### rjain

##### Golden Member
no known laws of physics forbid an electromagnetic field from passing through the event horizon any more than they forbid a gravitiational field from doing the same. This is merely a fallacy introduced by the geometric interpretation of GR, which presupposes that SR is correct. There is no evidence that SR is any better than LR, and the experiences from the GPS system indicate that LR handles the universe better than SR where the two differ.

Here's one example of how the geometric interpretation of GR is bogus. Put two objects in the same orbit around some large body, one a little behind the other. According to the geometric interpretation of GR, the orbit is the shortest path through space-time between the two objects, which is why there needs to be "no force" applied to either object to keep them in orbit. Now put a string between the two objects and keep it taut between them. Theoretically, the string should not affect the orbit of either object, as their relative positions remain constant. Since the string is taut, it should take the shortest line between the two objects. However, this line is not the same as the path of the orbit. One of these paths is the shortest line, and one of them isn't.

#### s0upx0n

##### Junior Member
I think you may be correct that an electromagnetic field could penetrate the event horizon. Since black holes can give off the (I think it is) Hawking radiation, that could easily propogate the field.

I thought that in such a situation with the string between two planets, the aparent contradiction between the two paths of shortest distance arises from one's view of the string. In viewing at one instant, one confines the view of the string to its 'spacial' shortest path, not its space-time shortest path which can be observed as one watches the string orbit with the two planets, and coincides with the type of path the planets take. These paths would be different because the planets are moving over time thus they experience the force due to the change in space over time, whereas the view of the string is at one instant and can only respond to how warped space is.

#### rjain

##### Golden Member
Originally posted by: s0upx0n
I think you may be correct that an electromagnetic field could penetrate the event horizon. Since black holes can give off the (I think it is) Hawking radiation, that could easily propogate the field.
Hawking radiation isn't a propagation of EM fields. It's the creation of a particle-antiparticle pair near the event horizon, followed by one of the two falling into the event horizon, but the other escaping out of it.
I thought that in such a situation with the string between two planets, the aparent contradiction between the two paths of shortest distance arises from one's view of the string. In viewing at one instant, one confines the view of the string to its 'spacial' shortest path, not its space-time shortest path which can be observed as one watches the string orbit with the two planets, and coincides with the type of path the planets take.
According to the geometrical interpretation of GR, space cannot exist without time. If SR is to be taken seriously, the path of the string cannot have anything to do with the timelike (instantaneous) path between the two objects, as that would violate causality.
These paths would be different because the planets are moving over time thus they experience the force due to the change in space over time, whereas the view of the string is at one instant and can only respond to how warped space is.
The motion of the planets in the geometric interpretation of GR isn't a result of any force. In fact, it's the result of there being no force applied. If the string responded to how warped space is, then it would follow the curved orbit.

#### s0upx0n

##### Junior Member
My apologies for being less then clear.

I introduced the Hawking radiation as something that could be effected by the charge of the black hole and then escape to affect other particles. Basically it was exteranneous. Also I was unclear in that when I said the force on the planets, I meant the aparent 'mg' force due to being in a frame of reference that is accelerated with respect to another.

However for the planets and the string, it is my understanding that since the planets in themselves curve space, and they change position with time, they have a path different from a 'massless object'. Thus the shortest line for the string is significantly different from shortest line for the planets.

#### rjain

##### Golden Member
Originally posted by: s0upx0nI was unclear in that when I said the force on the planets, I meant the aparent 'mg' force due to being in a frame of reference that is accelerated with respect to another.
Well, that's not really a force in the geometrical interpretation of GR. It's simply an object following a geodesic (shortest path) in the absense of external force.
However for the planets and the string, it is my understanding that since the planets in themselves curve space, and they change position with time, they have a path different from a 'massless object'. Thus the shortest line for the string is significantly different from shortest line for the planets.
Right, they curve space in the geometrical interpretation of GR. The string isn't massless anyway, nor does it need to be. By definiton, it's following the shortest path, as it is taut, yet that shortest path isn't the shortest path that the orbiting objects follow. So which one is it? I agree with the view that the geometrical interpretation of GR just confuses the issue, and, given the experimental results that the weak equivalence principle isn't true, that interpretation is just plain wrong.