1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Why don't black holes disconnect themselves from the universe?

  1. Apr 18, 2010 #1
    Gravitons are considered to be the mediating particles of gravitational force just as photons are considered to be the mediating particles of electromagnetic force. Both have zero rest mass. If extreme gravity of black holes does not allow photons of light to escape, why does it not do the same with gravitons?

    In other words : Observations of the cosmos show that black holes exert tremendous influence on the nearby stars to the extent of deforming them. It means in spite of extreme gravity, gravitons are able to ‘operate’ between a black hole and a nearby star. Why doesn’t the black hole ‘suck’ the gravitons just like photons and disappear/disconnect from the universe?

    Thanks in advance.
  2. jcsd
  3. Apr 18, 2010 #2


    User Avatar
    Science Advisor
    Gold Member

    Gravitons (if they exist*) propagate a change in gravity. No gravitons, or gravity waves, are required to move simply for gravity to exist.

    *I'm no expert, but I think that gravitons have neither been proven nor disproven to exist. Gravity waves are, however, a necessary part of general relativity, and propagate at the speed of light when there is a change of gravity.
  4. Apr 18, 2010 #3


    User Avatar
    Science Advisor
    Gold Member
    2017 Award

    I think you are 'over egging' the effect of black holes in general.
    You can expect black holes with masses the same as small stars or less. They will have no greater effect on a nearby star than an equivalent mass in the same place. I'm not sure what "observations" you refer to but they can't apply to all black holes.

    I think the way you are extending what you have found out about black holes into such a dramatic scenario is a bit speculative. Have you a serious reference to this?
  5. Apr 19, 2010 #4
  6. Apr 19, 2010 #5


    User Avatar
    Science Advisor
    Gold Member

    This question goes to the heart of a theory of quantum gravity, which has yet to be solved. What we do know is gravitational waves exist and gravity has properties that distinguishes it from photons. For another less than satisfactory explanation see http://imagine.gsfc.nasa.gov/docs/ask_astro/answers/980601a.html
  7. Apr 19, 2010 #6


    User Avatar
    Science Advisor
    Gold Member
    2017 Award

    As I suspected, this link refers to Super Massive Black Holes. Not just any old black hole! If we had a black hole in place of our Sun, we'd go round in the same orbit as we do at the moment (just a bit more chilly).
  8. Apr 20, 2010 #7
    Your contention has raised a question in my mind.

    It's only the super massive black holes that 'tear apart' the nearby stars. Whatever was there before the formation of a massive black hole (an extremely big star etc.) would not have 'tore apart' the neighboring stars.

    It means when something turns into a black hole something special to its gravity happens. In other words, if our sun turns into a black hole, the earth would not revolve around it as it is doing at present. It may actually be sucked up by the black hole(of our sun).

    I leave full scope of my argument being wrong.
  9. Apr 20, 2010 #8


    User Avatar
    Science Advisor
    Gold Member
    2017 Award

    I suppose that would follow, unless there is some reason for a 'threshold' effect, for big enough black holes, which doesn't occur for 'normal' ones. My understanding stops way before that, though.:smile:
  10. Apr 20, 2010 #9
    No, nothing special happens...

    A supermassive black hole FORMS differently, and being "super massive" it has the capacity to "do more damage". That said, if you imagined an impossible star with the mass of HUUUUGE black hole, then allowed it collapse, you would still be able to orbit it at he same distance you would have orbited the original star.

    Of course, in real life that BH is spinning like mad, has a truly enormous Ergoregion (see my name), has a MASSIVE accretion disk (which is a pretty violent place), and is probably blasting radiation from its "poles" (see galactic jets).

    That being said, if I light a match, and you light 1 million matches it's still fire. A pound of C4 vs. a Ton of it... it's still the same deal.

    Finally, if you magically turned Sol into a black hole with PRECISELY the same mass (this doesn't happen in real life), and you don't have any transitional period... then yes, you would orbit it just as before. The changes would be: Loss of Solar Ejecta, light, heat, and we'd probably be bombarded by lethal radiation. That doesn't change the gravity however... but if you get too close.... then it changes.

    As terrible as the "rubber sheet" analogy is, in this case it's useful. The incline does become more extreme as you approach the BH... the difference is that eventually escape becomes impossible. The absolute range of the "dent" in the sheet never changes, only its geometry.
  11. Apr 20, 2010 #10


    User Avatar
    Science Advisor

    I feel it necessary to spell out explicitly that the gravity of the object DOES change were it to transform from a sun to a black hole. What Frame Dragger is saying (and is commonly quoted) is that this change is negligibly small at the distance of the Earth's orbit. It is not as if there is some magical cutoff beyond which there is no difference, it is simply that we can safely ignore any changes in gravity at this distance. This is what we mean by the gravity changes (appreciably) when you get "too close". Of course, this "too close" limit is defined by how precisely you are making your measurements.
  12. Apr 20, 2010 #11
    All true, but to be fair I was presenting an impossible scenario in which our sun intantly and without emission or perturbation, becomes a BH. Obviously the real thing is quite violent, and I have doubts as to how well we'd orbit a BH that was... well... Frame Dragging.

    Bottom line, sure, changing the geometry is going to have an effect, but that's why the caveat is, "from a stable orbit". This also does nothing to distinguish a supermassive BH, from a stellar-mass BH. Yes, you get a mushroom cloud from the MOAB, but not from lighting a puddle of gasoline... still.. the basics are unchanged.
  13. Apr 20, 2010 #12
    The question asked by Deepak is a very good one, and one that I have asked myself. I suppose the obvious answer is that gravitons are not affected by gravity because their exchange is responsible for the gravitational force (in accordance with a quantum theory of gravity, which of course we don't have). Truly, though, I don't know the answer, even though I'm a physicist, and I wish one of the high-powered physicists on PF woud supply an answer, or at least speculate about it. The replies to the question I have viewed so far seem to skirt the issue.
  14. Apr 20, 2010 #13
    Welcome to PF!

    Which question posed? There have, in fact, been several iterations now. Anyway, what branch of physics are you involved in, and how do you feel that people here, myself included have skirted this issue?

    For a first post, you certainly don't mince words, but you also haven't offered much in the way of detail for your critique. It's a bit unusual to join a site, and your first post dismisses all present and asks for a PF Mentor. A less charitable person than myself would be suspicious that you are not new here at all. :smile:
  15. Apr 21, 2010 #14
  16. Apr 21, 2010 #15
    Thanks for your detailed answer.
  17. Apr 21, 2010 #16
    My pleasure.

    @crager: Hmmm... my first reaction is: "who the hell knows?!". I still find the notion of gravitons + geometric gravity = Headache. I'd love to know the answer however.
  18. Apr 21, 2010 #17


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    The OP asked - I think - the following question:

    If a BH is called a BH because "not even photons can escape from it", then how come that gravitons can escape from it ?

    What has been discussed in this thread is that, indeed, at a sufficient distance from a BH, gravity acts in the same way as if the mass making up the BH was made up by a less dense, non-BH object. So yes, the earth will orbit the sun-turned-into-a-BH in the same way as when the sun was still there in its normal, "diluted" form. But the question is WHY ?

    In general relativity, that's fairly straightforward to show. But that is a classical theory. But in the quantum theory of gravity which we don't have yet, the question can be posed of how the heck do the virtual gravitons get out. They have to be virtual because, as someone noticed, real gravitons are supposed to describe quanta of gravitational waves. In fact, gravitons are the quanta of a *linearized*, "weak-field" theory of gravity waves (and are purely hypothetical). We don't know what is supposed to get out of a BH in a quantum context. (for sure, *I* don't know it :smile: ).

    BTW, you can also ask: how does an electric field get out ? Because a BH CAN be electrically charged a priori. And even though *real* photons won't get out (light will not get out), the Coulomb field (and so the virtual photons which make up the static coulomb interaction) do "get out".
  19. Apr 22, 2010 #18
    Do magnetic fields interact with gravitational fields . Can you bend a magnetic field with a gravitational field .
  20. Apr 22, 2010 #19
    Sure, just consider a gravitational lens. Light = EM radiation after all.
  21. Apr 22, 2010 #20
    so then gravitons interact with magnetism .
  22. Apr 22, 2010 #21
    Gravitons are hypothetical, but my understanding is that they would be the quanta of spacetime, and its geometry. It doesn't interact with EM beyond the usual GR fashion. I don't believe even hypothetical gravitons could interact with photons, but I admit that this is "beyond the standard model", and beyond me :redface:
  23. Apr 23, 2010 #22
    i guess what do you mean by interact , if photons are affected by gravity does this mean they are interacting with gravitons , or is the incorrect , and if it interacts with photons this doesn't mean it interacts with the EM field.
  24. Apr 23, 2010 #23
    Hmmmm... I'm really not sure. By interact I mean that say, Light, follows a path determined by the SET, which in turn is describing spacetime curvature (gravity). If gravitons are a quanta of that, photons wouldn't interact (participate in any force but gravity) with Gravitons. Remember, gravitons are theoretically massless spin-2 particles... so they shouldn't participate in ANY interactions (as we normally think of that). Then again, if they're determining the geometry of spacetime, wich EM radiation follows... ???
  25. Apr 23, 2010 #24
    interesting , if the graviton had a gravitational field , does this mean that it's own field can emit its own particles , and to conserve energy would the graviton go down in frequency .
  26. Apr 23, 2010 #25


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Gravitons are the application of the methods of quantum field theory to the *linearized* version of general relativity (that is, a kind of weak-field limit). When you take general relativity, and you apply a series development of the metric and limit yourself to first order, you find the linearised Einstein equations.

    For short: absence of gravity implies that the metric is the Minkowski metric of special relativity (which can be seen as a diagonal matrix with elements (1,-1,-1,-1) on the diagonal in an orthonormal coordinate system). Gravity manifests itself by a deviation of the metric from this Minkowski (or "flat") metric, and the Einstein equations give you the link with the matter and energy distribution. In a certain way, the Einstein equations are to the metric, what the Maxwell equations are to the E and B fields: the Maxwell equations give you the link between the E and B fields and the distribution (and motion) of charges. The Einstein equations do the same for the metric and the distribution (and motion) of matter and energy.

    However, there's a big caveat: the Einstein equations are non-linear, as where the Maxwell equations are linear. This means that if you double all charges and currents, you will double the E and B fields. Not so for gravity. This is why Einstein's theory is so rich in strange solutions (and so difficult to handle).

    But, for WEAK gravity, we know that the gravitational response is linear (twice the mass gives you twice the gravitational pull in Newton's theory of gravity). So if we do a series expansion of the metric tensor around the flat value (Minkowski tensor), and limit ourselves to small deviations and first order, we can linearise Einstein's equations, and we get something that has some similitude to the Maxwell equations...
    Except that our thing is now a second-order tensor, and not a (first-order) vector (E-field, say).

    These are the linearised Einstein equations. If you push it somewhat, you can get out Newtonian gravity (a bit like you can get out Coulomb electrostatics out of Maxwell's equations). But you get also some wave equations, similar to the equations of electromagnetic waves.

    And if you apply quantum field theory bluntly to those wave equations, well, you get a quantum field theory that looks somewhat like quantum electrodynamics. Except that you get a different "mediating particle", not a photon, but a "graviton". Because the field quantity is a 2nd order tensor (the metric tensor, or better, the deviation of the metric tensor from the flat metric tensor), our "graviton" will be a spin-2 particle. Because in EM, the field is a vector (say, E-field), the photon is a spin-1 particle.

    So, "by imitation" of what happens to the quantum version of electromagnetism, one can quantize the linearised version of Einstein's equations, and one then finds a particle, called the "graviton".

    Except that the theory is full of difficulties. There where a trick in electromagnetism, called, renormalization, allows one to get rid of all infinities that devellop during the calculations, this trick doesn't work for the graviton. And we remain with the difficulty that we patched quantum theory upon a *linearised* version of Einstein's equations. So it is not really clear what the graviton is supposed to mean.

    Now, there's a lot more to say about this, but that's beyond my own knowledge.

    As to "gravitational self-coupling", well, because of this linearised version of the equations on which one applied quantum field theory, there is no direct graviton-graviton interaction, as there is also no direct photon-photon interaction (there are indirect interactions, through pair creation, but this is even out of scope for gravitons, as higher-order diagrams diverge because of the lack of renormalisation).
    I think that for the same reason, there are no direct photon-graviton interactions, but I might be wrong here.

    But self-particle interactions ARE considered in other theories. Non-abelian gauge theories, such as QCD, do have gluon-gluon interactions. That is because the equations one starts with are non-linear (but the gauge stuff saves us there).

    Does the graviton have "its own gravitational field" ? Well, because we linearised that away, no. Like the photon doesn't have its own charge (but there it is believed to be exact).
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook