Can Gravitons Escape Black Holes?

In summary, it is my understanding that gravitons are the elementary particles thought to transmist gravity. They must be massless and therefore travel at the speed of light. With this in mind, how can black holes gravitationally effect other bodies as the gravitons emitted by the black hole would not be able to escape?I asked the same question a while ago on another forum, and I don't understand how the black hole could effect anything as the gravitons emitted by the black hole would not be able to escape.
  • #1
eep
227
0
It is my understanding that gravitons are the elementary particles thought to transmist gravity. They must be massless and therefore travel at the speed of light. With this in mind, how can black holes gravitationally effect other bodies as the gravitons emitted by the black hole would not be able to escape?
 
Physics news on Phys.org
  • #3
Gravitons are noting more than an unproven hypothesis. It is very likely that they do not exist. If they would exist, one would encounter an absurd situation: Mass is composed of particles , held together by forces. The energy contained in the fields has also mass. So, every virtual particle (responsible for instance for the electromagnetic force) within the mass must contribute to the total mass. This means that there should for every virtual particle also exist a virtual graviton and since "the gravitational field" is also a source of gravitation (Einsteins Field Equations are nonlinear) it means that every virtual gravitaton must have a corresponding graviton to be able to feel the gravitational force. And these in turn must always have gravitons such that they are also subject to gravitation. So, one can continue this up to infinity, meaning that the whole space must be completely filled with particles, so that it would resemble more a solid than a vacuum.
In short, the concept of gravitons should best be abandoned.
 
  • #4
Brinx said:
I asked the same question a while ago on another forum.
Here's a link with an answer:
http://www.math.ucr.edu/home/baez/physics/Relativity/BlackHoles/black_gravity.html

"...In addition, however, virtual particles aren't confined to the interiors of light cones: they can go faster than light! Consequently the event horizon, which is really just a surface that moves at the speed of light, presents no barrier..."

I don't get this however to escape from the singularity virtual particles must have an infinite speed
 
  • #5
As far as I know gravity acts from the surface of a mass (assuming it's spherical) a bit like gauss's law of charge, everything inside cancells out.

Your graviton only then emerges from the surface of the black hole so it doesn't need to escape from a singularity at infinite speed. What speed it needs to travel at I don't know. Anything within the event horizon can't get back out (unless you travel >c) but this isn't nessesarily the surface of the object so hmmm... This is more me thinking aloud rather than a proper answer!

After reading that page I'm still confused. I thought gravity was information and therefore limited to the speed of light. I guess it doesn't work like that then.

So using GR how does the information travel fast enough to escape and tell things outside the black hole to move towards it?
 
  • #6
Unsing GR the field is alredy there
using QM...there are problems...
 
  • #7
notknowing said:
Gravitons are noting more than an unproven hypothesis. It is very likely that they do not exist. If they would exist, one would encounter an absurd situation: Mass is composed of particles , held together by forces. The energy contained in the fields has also mass. So, every virtual particle (responsible for instance for the electromagnetic force) within the mass must contribute to the total mass. This means that there should for every virtual particle also exist a virtual graviton and since "the gravitational field" is also a source of gravitation (Einsteins Field Equations are nonlinear) it means that every virtual gravitaton must have a corresponding graviton to be able to feel the gravitational force. And these in turn must always have gravitons such that they are also subject to gravitation. So, one can continue this up to infinity, meaning that the whole space must be completely filled with particles, so that it would resemble more a solid than a vacuum.
In short, the concept of gravitons should best be abandoned.
Are this mean one uncompatibility of general relativity and quantum mecahnics? How precisely?
 
  • #8
notknowing said:
This means that there should for every virtual particle also exist a virtual graviton and since "the gravitational field" is also a source of gravitation (Einsteins Field Equations are nonlinear) it means that every virtual gravitaton must have a corresponding graviton to be able to feel the gravitational force.

Not if the graviton mass is = 0 ; because it would not attract ...or not?
 
  • #9
As far as I have understood from my query on that other forum, changes in gravity (i.e. changes in curvature of spacetime), that are for example caused by changes in position of matter, are thought to propagate at the speed of light. The curvature that is already 'in place' doesn't need to be propagated, apparently: it's already a local property of spacetime. So during the collapse of an object to a black hole, all locations outside the (future) event horizon are 'updated' regarding the new situation up until the actual formation of that event horizon, after which the curvature of spacetime outside the event horizon doesn't change anymore - at least not as a consequence of the configuration of collapsing matter inside the event horizon. How spacetime curvature can remain in one configuration without it being 'in touch' with the source of that curvature (the matter making up the black hole) is beyond me.
 
  • #10
Born2Perform said:
Not if the graviton mass is = 0 ; because it would not attract ...or not?


No, the graviton (non existing) mass should be different from zero because otherwise it would not feel gravity and the Einstein field equations imply that gravity (curvature) can be the source of gravity (curvature). Of course, the gravitons rest mass could still be zero (similar to the zero rest mass of a photon but the non-zero energy of the photon).
 
  • #11
exponent137 said:
Are this mean one uncompatibility of general relativity and quantum mecahnics? How precisely?


No, they are not incompatible. The incompatibility arises just because of a wrong approach, there is no fundamental compatibility. Gravitation should not be treated in the same way as the electromagnetic force. The key to understanding gravity lies in the quantum vacuum which theoreticians unfortunately usually (try to) throw out of their calculations (using so-called renormalisation).
 
  • #12
Brinx said:
I asked the same question a while ago on another forum. Here's a link with an answer:

http://www.math.ucr.edu/home/baez/physics/Relativity/BlackHoles/black_gravity.html

I haven't found a more elaborate explanation anywhere else, yet...

Yep - that's the standard FAQ answer.

Several important things to note:

1) "Gravitons" aren't the usual approach to GR at all, though some approaches can derive an equivalent theory to GR from "gravitions".

2) One must distinguish between virtual gravitons and real gravitons, just as one must distinguish between virtual photons and real photons. The question (and most people who ask the question) unfortunately does not / do not make this important distniction. One can use the same argument to ask "How can a black hole be charged, if virtual photons can't escape it?"
 
Last edited:
  • #13
notknowing said:
No, they are not incompatible. The incompatibility arises just because of a wrong approach, there is no fundamental compatibility. Gravitation should not be treated in the same way as the electromagnetic force. The key to understanding gravity lies in the quantum vacuum which theoreticians unfortunately usually (try to) throw out of their calculations (using so-called renormalisation).

Do you have a peer-reviewed reference for this statement?

Highly speculative personal theories generally aren't welcome here on PF, see the posting guidelines.
 
Last edited:
  • #14
Yes, I have a peer-reviewed reference for this statement.
You can find it on http://home.online.no/~avannieu/darkmatter/
The paper is :
R. Van Nieuwenhove, Quantum Gravity : a Hypothesis, Europhysics Letters, 17 (1), pp. 1-4 (1992)

PS : I have seen much more speculative theories on this site without any perr-reviewed reference (in fact maybe as much as 99 %)
 
Last edited by a moderator:
  • #15
I'm getting confused here. Surely there can be no messenger-particle hail of "gravitons" flooding through all space and matter to create the effect of gravity? Which means gravitons are better thought of as something like curvature calculation units? Can anybody correct this or give me a concept of what the graviton is supposed to be? Real or virtual?
 
  • #16
Cornell's FAQ on black holes says the opposite thing of what you all say:
http://curious.astro.cornell.edu/question.php?number=264

Cornell university FAQ said:
"Virtual particles are essentially allowed to do anything at all short of violating causality during their lifetimes. They necessarily violate conservation of energy just by existing, and they're also allowed to violate many other physical laws before they disappear. One of these is the restriction on traveling slower than the speed of light"

this guy says that gravitons excapes from the black holes because their speed is >c ;
i know FAQs are made to let normal people understand but after this i am a bit confused..
 
Last edited by a moderator:
  • #17
This is really a difficult subject. One is really considering the graviton as a virtual particle with zero rest mass (to ensure a force with infinite range). The graviton (if it would exist) should however have some mass (since it not as rest) but some people think that this can not have an effect on the rest of the universe because it exists for a too short time. This argument would hold for virtual particles which are not massless (such as W's) but I don't see how the argument could hold for gravitons. If you think of the mass of an atom than you find that most of the particle (if not all) consists of fields or virtual particles interacting between its constituents (for instance virtual gluons between the quarks, virtual photons between quarks, electrons, etc.). Maybe quarks are composed of other sub-particles with again virtual particles between these, etc. So, in the end most of matter seems to be in the form of virtual particles. If one then claims that these virtual particles do not contribute to the mass of the particle than I ask, what is then making up its mass ? Thus, coming back to the graviton now, it looks plausible that if it would exist, that it would feel gravitation and hence one encounters the implausible situation I mentioned in a previous message (space completely filled with gravitons).
 
  • #18
Born2Perform said:
Cornell's FAQ on black holes says the opposite thing of what you all say:
http://curious.astro.cornell.edu/question.php?number=264



this guy says that gravitons excapes from the black holes because their speed is >c ;
i know FAQs are made to let normal people understand but after this i am a bit confused..

The sci.physics faq says much the same thing:

In addition, however, virtual particles aren't confined to the interiors of light cones: they can go faster than light! Consequently the event horizon, which is really just a surface that moves at the speed of light, presents no barrier.

I couldn't use these virtual photons after falling into the hole to communicate with you outside the hole; nor could I escape from the hole by somehow turning myself into virtual particles. The reason is that virtual particles don't carry any information outside the light cone. See the physics FAQ on virtual particles for details.

Of course, before that, the sci.physics.faq points out, as I also like to point out, that GR is not actually formulated in terms of a particle or field model. GR is formulated in terms of curvature, instead. (Certain non-standard formulations of GR in terms of spin 2 particles do exist, however - references on request, or search the forum).

There is no doubt that virtual particles are weird. I personally think that they are more trouble than they are worth, but many people seem to find the idea of forces carried by "particles" attractive, no matter how weird the "particles" act. I personally suspect that most people who are fond of virtual particles mistakenly think they act like normal particles, which is why I start out by pointing out that forces are not carried by real particles.

Since GR isn't actually formulated in terms of "particles" at all, I would suggest that the quantum physics forum is the best place to ask about the weird and counter-intuitive properties of virtual particles, or to complain about how weird they are :-).
 
Last edited by a moderator:
  • #19
Farsight said:
I'm getting confused here. Surely there can be no messenger-particle hail of "gravitons" flooding through all space and matter to create the effect of gravity? Which means gravitons are better thought of as something like curvature calculation units? Can anybody correct this or give me a concept of what the graviton is supposed to be? Real or virtual?

Gravitons, if they exist, would come in both "real" and "virtual" flavors, just as photons do.

Real gravitons would be quantized packets of gravitational radiation, just like real photons are packets of quantized electromagnetic radiation. Real gravitions, like real photons would carry information. Real gravitions, like real photons, could not escape from a black hole.

Note that since we can't detect gravitational radiation at all (hopefully that will change soon)m, we are a VERY long way from being able to detect whether or not gravitational radiation is quantized. Qunatitiation of gravitational radiation is entirely theoretical at this point.

Virtual gravitons would be responsible for the attractive force between masses, just as virtual photons are responsible for the attractive and repulsive forces bedtween charges.

Virtual gravitons would share the many oddities of virtual photons.

Note that while neither real photons nor real gravitions could escape a black hole, both virtual photons and virtual gravitons could (in principle, anyway), as a black hole has both a mass and a charge.
 
  • #20
Thanks for your answers, everyone. I don't really know anything about virtual particles besides the fact that virtual photons are responsible for the electromagnetic force. So the gravitons reponsible for the gravitational force would be virtual gravitons, and therefore can escape due to their ability to travel greater than the speed of light. Hrm.
 
  • #21
eep said:
So the gravitons reponsible for the gravitational force would be virtual gravitons, and therefore can escape due to their ability to travel greater than the speed of light.

uhm...better if this idea does not remain in your head because most of people who talked here are sceptical about this...what i understood from pervect & co. is that GR is not a particle theory so we won't have the answer without a QG
 
  • #22
Born2Perform said:
uhm...better if this idea does not remain in your head because most of people who talked here are sceptical about this...what i understood from pervect & co. is that GR is not a particle theory so we won't have the answer without a QG

Yes. Excellent point.

This is also mentioned in the FAQ(s)

Gravitons don't exist in general relativity, because GR is not a quantum theory. They might be part of a theory of quantum gravity when it is completely developed, but even then it might not be best to describe gravitational attraction as produced by virtual gravitons. See the physics FAQ on virtual particles for a discussion of this.

From the virtual particle FAQ:

Quantum gravity is not yet a complete, established theory, so gravitons are still speculative. It is also unlikely that individual gravitons will be detected any time in the near future.

Furthermore, it is not at all clear that it will be useful to think of gravitational "forces," such as the one that sticks you to the Earth's surface, as mediated by virtual gravitons. The notion of virtual particles mediating static forces comes from perturbation theory, and if there is one thing we know about quantum gravity, it's that the usual way of doing perturbation theory doesn't work.

<...>

Other approaches to quantum gravity are needed, and they might not describe static fields with virtual gravitons.
 
  • #23
pervect said:
One can use the same argument to ask "How can a black hole be charged, if virtual photons can't escape it?"

I thought that stuff that fell into a black hole never actually crosses the event horizon, at least in the Schwarzschild metric with respect to the point of view of an observer outside the hole.

Along this line, I've been trying to program the Cambridge geomtric algebra version of GR into an applet. They apparently give orbits that are identical to that of GR, but time is mixed up with radius (i.e. the metric is not diagonal) in a way that allows stuff to fall across the event horizon in finite coordinate time. Does this make any sense to you? Here's a reference to the paper from which I am using the equations:
http://www.mrao.cam.ac.uk/~clifford/publications/abstracts/grav_gauge.html

Note that there is what I found to be a confusing typo where they repeatedly write something like [tex]f_1g_2 - g_2f_1[/tex] where they meant [tex]f_1g_2 - g_1f_2[/tex] as f and g are scalar functions.

Carl
 
  • #24
CarlB said:
I thought that stuff that fell into a black hole never actually crosses the event horizon, at least in the Schwarzschild metric with respect to the point of view of an observer outside the hole.

Yes, that's true, from that particular POV. That might even be the right POV to use if one was using something like the Lienard-Wiechart potentials to calculate the charge of a black hole. (I'm not actually positive that these generalize to curved space-time offhand, however).

From the POV of someone falling into the black hole it takes only a finite amount of time to cross the event horizon, of course.

Along this line, I've been trying to program the Cambridge geomtric algebra version of GR into an applet.

Carl

I can't say that I really understand this paper at this time, so I'm not in a position to help with writing an applet for it.

I can possibly provide some general guidance with the GR part of the problem, though, and address some other concerns.

One general concern I have about the applet is coordinate system compatibility. One can make the same solution look different by using different coordinates. Some effort is going to have to be made to avoid this sort of false comparison, to use compatible coordiantes.

I would suggest using Schwarzschild coordinates, where the radial variable is defined such that the circumference is always 2*pi*r. This alows theta and phi to be defined in the usual manner (2Pi radians convers the entire surface).

This coordinate defintion is possible when the theory in question defines circumference. A fair comparison requires that the circumference be the same circumference as would be measured by a standard SI ruler. I'll talk about my concerns about rulers later.

Going back to the paper:

I have seen other formulations of GR in terms of a flat space-time, however, most notably

http://xxx.lanl.gov/abs/astro-ph/0006423

The above formulation, however, is torsion-free and (claims to be) equivalent to standard GR. It sounds to me like the Cambridge authors are not aware of this approach, at least I don't see it cited. There are some differences, but a lot of similarities.

The Cambridge paper says:

At the very least, this means that physicists have the choice of formulating gravity in terms of space-time curvature or in terms of forces in flat space-time.

The Straumann paper says

Although this field theoretic approach, which has been advocated repeatedly by a number of authors, starts with a spin-2 theory on Minkowski spacetime, it turns out in the end that the flat metric is actually unobservable, and that the physical metric is curved and dynamical.

What I find attractive about the Straumann paper is that it goes on to re-examine what physics looks like in terms of "standard" rulers, recognizing that it takes a non-standard ruler to make curved space-time flat.

When you have a space-time, and a metric, curvature follows unavoidably from the curvature tensor. If you have only a space-time (and haven't picked a metric yet), different metrics might make it curved or flat.

I'm not quite sure I understand at this point how (or even if) Lorentz invariance is kept when one starts changing the defintions of standard rulers & clocks, however.

One of several things that I'm not sure about with respect to the Cambridge paper is what concept (if any) the theory has that represents a standard SI ruler. Without this information, it's hard to get actual physical predictions out of the theory, because physically we'd make measurements with standard SI rulers and standard SI clocks.
 
  • #25
CarlB said:
I thought that stuff that fell into a black hole never actually crosses the event horizon, at least in the Schwarzschild metric with respect to the point of view of an observer outside the hole.

Along this line, I've been trying to program the Cambridge geomtric algebra version of GR into an applet. They apparently give orbits that are identical to that of GR, but time is mixed up with radius (i.e. the metric is not diagonal) in a way that allows stuff to fall across the event horizon in finite coordinate time. Does this make any sense to you?

As far as I can tell, the Cambridge group's theory makes exactly the same predictions as standard general relativity about what a distant observer will see as an object falls into a Schwarzschild black hole. If I remember correctly, I got Lasenby to admit as much during a presentation he gave at a conference in '95.

One has to very careful when using coordinates to make *physical* predictions. The coordinates that the Cambridgers use may not be well known, but they certainly exist in standard GR, and they certainly have been used before in standard GR.

These coordinates are called Painleve-Gullstrand coordinates in the GR literature, and a good reference is the book https://www.amazon.com/gp/product/0521830915/?tag=pfamazon01-20 (5.1.4) by Eric Poisson. Also, on the internet, Chris Hillman has promoted the use of these coordinates.

An an expended, updated version of the Cambridgers work is given http://www.arxiv.org/abs/gr-qc/0405033".

Lasenby said:
A crucial question to address is whether any experimental tests are likely to distinguish between general relativity and GTG in the immediate future. The biggest differences between general relativity and GTG to emerge to date lie in the treatment of black hole singularities [55, 58], but these are unlikely to be testable for some considerable time! A more promising area is the link between gravity and quantum spin. GTG makes a clear prediction for the type and magnitude of this interaction, whereas it is not uniquely picked out in general spin-torsion theories, or in more general Poincar´e gauge theory. Any experiment measuring this interaction would therefore provide a clear test of GTG. A partial exploration of the effects of spin interactions in GTG is contained in [49].

Again, they don't mention any observable differences for things falling into Schwarzschild black hole, and the coordinates that they use (for this) can also be used in standard GR.
 
Last edited by a moderator:
  • #26
Thanks for the response pervect. I guess I'm not at all clear on virtual particles.
 
  • #27
pervect said:
One general concern I have about the applet is coordinate system compatibility. One can make the same solution look different by using different coordinates. Some effort is going to have to be made to avoid this sort of false comparison, to use compatible coordiantes.

It turns out that the Cambridge and Schwarzschild solutions use the same coordinates, other than coordinate time. So the orbits are identical. Where the GR particles get stuck on the event horizon, the gauge particles go through to the singularity. I've sort of modeled this in a fake way (i.e. Newtonian particles that get stuck on a fake event horizon) to give an idea what happens between regular GR and gauge gravity:
http://www.gaugegravity.com/testapplet/SweetGravity.html
But note that the above is wrong in that the gauge particles, instead of going on to the singularity, climb back out of the "event horizon". Hopefully I'll get the simulation with real physics running today or tomorrow.

Now there is a difference in coordinate time at a fixed radius from the black hole. So to make the two theories compatible in that way, I am saving the initial conditions and correcting the gauge theory to match the GR for passage of time for a particle fixed in that manner.

The reason for doing this is so that the two theories should match orbits. That is, for two test bodies starting at the same point and falling towards the black hole and then escaping back to the same radius (properly precessed), the simulation should show the two test bodies on the same orbit, but with gauge particle getting ahead on falling, and with the GR particle catching up on rising back out. But doing it this way will preserve collisions (i.e. events).

pervect said:
I have seen other formulations of GR in terms of a flat space-time, however, most notably

http://xxx.lanl.gov/abs/astro-ph/0006423

I'm not a gravity guy, but I think that the above is well known in the industry. I think it's mentioned in some of the other gauge gravity papers, particularly a paper by David Hestenes.

pervect said:
One of several things that I'm not sure about with respect to the Cambridge paper is what concept (if any) the theory has that represents a standard SI ruler. Without this information, it's hard to get actual physical predictions out of the theory, because physically we'd make measurements with standard SI rulers and standard SI clocks.

It's basically standard GR, but with the coordinate system chosen to be flat. If you wanted to choose some other coordinate system you could do it that way too. I believe that the same group has done that, with the curved coordinates used for representing the curvature of the universe as a whole. This seems sort of natural to me, that curvature in spacetime should be something that is seen only in vast vast distances.

I've got spotty internet right now. I should have the GR and gauge gravity simulations up tonight.

Carl
 
Last edited by a moderator:
  • #28
George Jones said:
As far as I can tell, the Cambridge group's theory makes exactly the same predictions as standard general relativity about what a distant observer will see as an object falls into a Schwarzschild black hole. If I remember correctly, I got Lasenby to admit as much during a presentation he gave at a conference in '95.

From having gone through the effort of programming it into the computer, I also believe that this is true. Thanks for the reference to Painleve-Gullstrand coordinates, and to Chris Hillman on the internet.

To me, the two important features of the Cambridge approach are (1) there is a very natural geometric algebra type Dirac equation buried in there, and (2) it's on flat space. These make it a natural gravity to combine with particle theory, which is my interest.

Carl
 
  • #29
CarlB said:
It turns out that the Cambridge and Schwarzschild solutions use the same coordinates, other than coordinate time.

So do Painleve and Schwarzschild coordinates. Basically a certain time transformation maps one into the other. Time and space are not at right angles in the Painleve-Gullstrand coordinate system however, something you were complaining about earlier. But the PG coordinates are finite at the event horizon.

Take a look at, for instance, https://www.physicsforums.com/showthread.php?t=126307

Unfortunately I have more work to do to sort out all the sign issues with the lograthmic conversion of the arctanh() function, they drive Maple nuts :-(.

[add]OK, I think this is all sorted out now, and it seems to match up well with the online homework solution

http://www.physics.umd.edu/grt/taj/776b/hw1soln.pdf

So the orbits are identical. Where the GR particles get stuck on the event horizon, the gauge particles go through to the singularity.

I think from this remark that you are taking the coordinates too seriously. A change of coordinates never affects the physics. The particles are still stuck on the horizon from the POV of the obsever at infinity, and the particles were NEVER stuck on the horizon from their own POV.

When we change the coordiantes, we do not change the particles orbit at all. We change only the description of the particle's orbit, we do not change anything physical.

An example might help.

Suppose we consider the coordinate system of an observer with a constant 1 light year/year^2 acceleration - a rindler observer, in a spaceship heading away from Earth with a constant acceleration.

The rindler observer, in his spaceship, obsevers an event horizon 1 light year behind him, the so-called rindler horizon. This is very similar to the event horizon of a black hole.

[add]
The metric for the accelerating spaceship , assumed to be accelerating in the z direction, is

-(1+gz)dt^2 + dx^2 + dy^2 + dz^2

You can see that at gz=-1, the cofficient of dt^2 becomes zero, just as it does in the Schwarzschild coordinate system. This marks the "event horizon" of the Rindler obsever, a plane located at z = -1.

[end add]

The rindler observer never sees the Earth as being more than 1 year old, because light emitted after that date never catches up with him as long as he continues to accelerate. (See anything on "hyperbolic motion" for why this is true, I think there is even something in Wiki about this).

The rindler observer then concludes (if he takes coordiantes too seriously) that the Earth never gets older than 1 year, and that the Earth never actually passes through the rindler horizon, 1 light year behind him.

The Earth observer does not observer anything unusual at 1 year, and cheerfully passes through "the horizon" (which doesn't even exist in his coordinate system) with no difficulty whatsoever.

Thinking that the Earth never gets more than 1 year old because the rindler obsever in the acclerating spaceship doesn't see it happen is a lot like thinking that the particle never falls into the event horizon because the outside observer doesn't see it.

Basically, your simulation is about coordinates, about how quantities that are infinite in one coordinate system are finite in another. Gauge gravity doesn't have anything to do with the issue at all, and I think your approach is going to mislead poor newbies.

It's basically standard GR, but with the coordinate system chosen to be flat. If you wanted to choose some other coordinate system you could do it that way too.

A picky point here. Just changing the coordinates is not enough to change curved space-time into flat. (Coordinates never affect the physics! Coordinates are just labels on a map, and the map is not the territory.) You have to go further, and change the metric as well, i.e. the defintion of length and time.
 
Last edited:
  • #30
pervect said:
Time and space are not at right angles in the Painleve-Gullstrand coordinate system however, something you were complaining about earlier.

I never complained about this, I said it was a very good thing.

pervect said:
Unfortunately I have more work to do to sort out all the sign issues with the lograthmic conversion of the arctanh() function, they drive Maple nuts :-(.

I ran into a bit of a problem, maybe related. I tried to do the calculation in proper time and then I have to keep track of coordinate time in order. But I get [tex]dt/ds = (E + \sqrt{r/2}dr/ds)/(1 - 2/r)[/tex] which goes to 0/0 at r=2. So before I can have particles pass the event horizon, I have to do the division.

pervect said:
I think from this remark that you are taking the coordinates too seriously. A change of coordinates never affects the physics. The particles are still stuck on the horizon from the POV of the obsever at infinity, and the particles were NEVER stuck on the horizon from their own POV.

It's good for you to point this out for the newbies. Of course I know it.

pervect said:
Basically, your simulation is about coordinates, about how quantities that are infinite in one coordinate system are finite in another. Gauge gravity doesn't have anything to do with the issue at all, and I think your approach is going to mislead poor newbies.

I agree. My interest in gauge gravity has nothing to do with the convenient coordinates, it comes from particle physics.

pervect said:
A picky point here. Just changing the coordinates is not enough to change curved space-time into flat.

Good for you to point this out for the newbies. I call the gauge gravity theory "flat" because Doran, Gull and Lasenby at Cambridge, and Hestenes at Arizona State all call it "flat". In addition to out numbering me 4 to 1, they also have published their papers in the peer reviewed literature, have good academic positions, have written quite a number of papers on gravitation, and like I said before, I admire their theory for what it implies regarding elementary particles.

By the way, I'm working on the GR simulation first and it is not pretty. I've got precession, but it doesn't look right to me. And for another thing, the orbits far from the black hole don't match between Newton and Einstein which is a very bad thing. I'm also out of time to work on it for the next day at least.

I'm going to work on the 0/0 problem (by fixing the equations). Maybe that will give some correction where I've got a minus sign or something somewhere.

Carl
 
Last edited:
  • #31
CarlB said:
I ran into a bit of a problem, maybe related. I tried to do the calculation in proper time and then I have to keep track of coordinate time in order. But I get [tex]dt/ds = (E + \sqrt{r/2}dr/ds)/(1 - 2/r)[/tex] which goes to 0/0 at r=2. So before I can have particles pass the event horizon, I have to do the division.

I've finished updating the Painleve-Gullstrand thread, having resolved the sign issues and also having found a stable online reference that's likely not to disappear in a few weeks that gets the same results that I do.

I'm not sure what approach you are using, but you might want to check that some of the conseved quantities for orbital motion are actually being conserved by your integrator (unless you are already using the conservation of these quanties to get your solution).

You might find http://www.fourmilab.ch/gravitation/orbits/

helpful, in that it calculates GR orbits, and has some discussion of the formulas as well.

You can also download the source code for their applet at the bottom of the page, too.

Note that for the Schwarzschild metric, the conserved quantities are

E0 = (dt/dtau) (1-2M/r)
L = r^2 (dphi/dtau)

As I mentioned, in the PG thread, for the PG metric, the expression for the conseved E0 changes to

E0 = (1-2*M/r)*(dt/dtau) - sqrt(2M/r) * (dr/dtau)

(The expression for L remains unchanged).

This is different by a sign convention from the conserved quantity I mentioned in the PG thread, however it uses the same sign convention as the Schwarzschild energy.

Using the above sign convention, when the velocity at infinity is zero, E0 = 1. It can be regarded as the energy per unit rest mass.

The above conserved quantites along with the metric equation give enough information to determine the orbits, i.e. you can derive the effective potential for the Schwazschild equation

(dr/dtau)^2 + V^2(L,r) = E^2

from the above conseved quantites of the Schwarzschild metric and the Schwarzschild metric itself with a little algebra and the fact that ds^2 = -dtau^2.

You can derive a similar equation for (dr/dtau) in the PG metric by the same means.
 
Last edited:
  • #32
Good simulation link, except his java crashes my windows. It uses the usual methods of deriving the orbits that is in MTW (i.e. based on the constants of motion).

A method of deriving the orbits more similar to what one can do with Newtonian gravity (i.e. straightforward integration), which I think may be better for the computer, is here:
http://sb635.mystarband.net/relat.htm
http://sb635.mystarband.net/cip.htm

The above was originally published in "Computers in Physics", then by the American Institute of Physics, and put on the web by Steve Bell.

As it turns out, the above equations exhibit the same 0/0 behavior that mine do. This is inevitable when you have a division by (1-2/R). But I find mine suspicious, so I'm going to compare them with the above, when I get some time.

Carl
 
Last edited by a moderator:
  • #33
You can convert the first-order set of equations using conseved quantites into a second order set by just setting the derivatives of the conserved quantites to zero.

I think I recall this helping to get past some "turning points", but it's been too long for me to be sure it really helped. I'm not sure if it will help with your problem at all.

This set of 2nd order equations is equivalent to the geodesic equations you get from the Christoffel symbols (it may take some algebraic manipulation to see the equivalence).

i.e.

d^2 x^i / dtau^2 + [itex]\Gamma^i{}_{jk}[/itex] (d x^j / dtau) (d x^k / dtau) = 0
 
  • #34
pervect said:
You can convert the first-order set of equations using conseved quantites into a second order set by just setting the derivatives of the conserved quantites to zero.

I think I recall this helping to get past some "turning points", but it's been too long for me to be sure it really helped. I'm not sure if it will help with your problem at all.

My intuition says that this is the way to do it. Break the 2nd order equations down to 1st order equations.

I've now got a first attempt at the simulation up here:
http://www.gaugegravity.com/testapplet/SweetGravity.html

It still has the division by zero problem, so I've set it for an initial condition that stays away from the event horizon. Also, in my equation [tex]dt/ds = (E + \sqrt{r/2}dr/ds)/(1 - 2/r)[/tex] I had the wrong sign. The correct equation, of course, is [tex]dt/ds = (E - \sqrt{r/2}dr/ds)/(1 - 2/r)[/tex]. The choice of the sign amounts to a sort of arrow of time for the black hole.

Carl
 
Last edited by a moderator:

FAQ: Can Gravitons Escape Black Holes?

What is a graviton?

A graviton is a hypothetical particle that is believed to be the carrier of the force of gravity. It is predicted by certain theories, such as string theory, but has not yet been observed or confirmed by experiments.

How do gravitons and black holes relate?

Gravitons are thought to play a role in the formation and behavior of black holes. They are believed to be responsible for the gravitational pull that causes matter to collapse and form a singularity at the center of a black hole.

Can gravitons escape from a black hole?

No, according to current theories, gravitons cannot escape from a black hole. This is because the escape velocity required to overcome the strong gravitational pull of a black hole is greater than the speed of light, which is the maximum speed at which particles can travel.

How are gravitons different from other particles?

Gravitons are unique in that they are the only particles that are predicted to have a spin of 2, whereas all other known particles have a spin of either 0, 1/2, or 1. This spin is what allows gravitons to mediate the force of gravity.

Are gravitons the only explanation for gravity?

No, there are other theories and explanations for gravity, such as Einstein's theory of general relativity. Gravitons are just one possible explanation and are still a topic of ongoing research and debate among scientists.

Similar threads

Replies
4
Views
1K
Replies
21
Views
2K
Replies
5
Views
1K
Replies
4
Views
945
Replies
2
Views
1K
Replies
11
Views
1K
Back
Top