4 ms·

Is there an equivalent of Tidal Heating [1] taking place between the two black holes? It would extract kinetic energy and put it into.. heating the black holes.

by mpreda 2y ago

Is there an equivalent of Tidal Heating [1] taking place between the two black holes? It would extract kinetic energy and put it into.. heating the black holes.. whatever that may mean. Assuming there is movement and friction in the core of a black hole. [1] https://en.wikipedia.org/wiki/Tidal_heating https://en.wikipedia.org/wiki/Tidal_heating

isoprophlex 2y ago
A black hole only has mass, angular momentum, and charge. So you can only deal with those three... not deform and knead a physical mass of planet as in tidal heating.
- orlp 2y ago
  This is often stated but it literally can't be true. It's true for stationary solutions to the relevant equations, but we're very much not living in a stationary world. Here's a simple thought experiment disproving your claim. A person hovers just above the origin of a supermassive black hole. They chuck a massively charged object into the black hole. If what you said is true they should observe the charge instantly being transported to the singularity, since a black hole can't have any attributes such as where charge is distributed within the horizon. Where it gets impossible is that someone very far away around the same supermassive black hole could observe a small charge increment. They in turn could chuck charged stuff into the black hole and now you've got faster than lightspeed communication.
  - elashri 2y ago
    > above the origin of a supermassive black hole I spent couple of minutes trying to understand your thought experiment and was puzzled why can't I understand it. It seems that it is probably because I don't understand what you mean by "just above the origin of supermassive black hole"? I feel that you have something interesting to say but not clear.
    - orlp 2y ago
      Mental typo, I meant horizon, not origin.
      - elashri 2y ago
        The no-hair theorem (scientific term for what you are descriping) essentially states that black holes can be completely described by only three externally observable classical properties: mass, charge, and angular momentum. The internal distribution of matter or charge within the black hole is irrelevant from the outside once it crosses the event horizon. The charge of an object thrown into a black hole doesn't need to "instantaneously" reach the singularity. In fact, it doesn't even "travel" in the sense that the observer would see a time delay based on where the charge is inside the horizon. The EM field generated by this charged object can still be observed outside the horizon before the object crosses the horizon. In classical GR, the exterior of a charged black hole is governed by the Reissner-Nordström metric [1], and it already includes the influence of charge. Once the object enters the horizon, the outside observer will perceive the black hole's external field to have changed. The field adjustment is instant for the outside observer because, from their point of view, the object never actually crosses the event horizon (due to infinite time dilation at the horizon from their perspective). The charge appears to have been "absorbed" by the black hole when the object is still outside the horizon. So no-hair theorem doesn't imply that properties such as charge or angular momentum must be "smeared" instantaneously to the singularity inside the black hole. The theorem only describes how these properties manifest externally, not their behavior inside the horizon. Once a charge object passes the event horizon, the information about its charge is not causally connected to any observer inside the black hole (except in QM contexts, where the information paradox becomes relevant). However, the charges affects the external metric of the black hole immediately and completely as seen by external observers. Also it doesn't say that the charge or mass needs to be uniformly distributed or behave in any specific way inside the event horizon. It only states that from the outside, black holes look like point-like objects characterized by mass, charge, and angular momentum. The specifics of how charge is distributed inside the black hole’s event horizon aren't visible to outside observers and therefore don't affect the validity of the theorem. [1] https://en.wikipedia.org/wiki/Reissner%E2%80%93Nordstr%C3%B6m_metric https://en.wikipedia.org/wiki/Reissner%E2%80%93Nordstr%C3%B6...
  - saagarjha 2y ago
    You can’t actually observe objects going into a black hole. They just become redshifted and fall slower and slower from your perspective towards the event horizon.
  - isoprophlex 2y ago
    The singularity is not the same thing as its event horizon. That said, it's the no hair theorem. It could of course still be wrong...
  - ben_w 2y ago
    The object will be observed asymptotically decelerating as it approaches the event horizon, never gets observed crossing by a distant observer. IIRC the event horizon expands to encompass the object just before it gets to the original horizon, due to the Schwarzschild formula. Also IIRC, the electric field is blurred out by the geodesics by this point, as if it were from the interior. But that's based on what I've heard, I have yet to derive useful results from the Einstein field equations, even though I think I should give it a go and I can follow them well enough to code a simple simulation…
gizmo686 2y ago
Gravitational waves have a similar effect of sapping the kinetic energy of orbiting black holes. Instead of heating the black holes, the energy is emitted as gravitational radiation. Having said that, now that we have actual observations of the gravitational radiation emmited from merging black holes, we can be pretty confident in our understanding of the magnitude of this energy loss. And that understanding is that it does not become significant until well into the last parsec.
- vlovich123 2y ago
  But however insignificant it is, wouldn’t it always be there sapping angular momentum & over a long period of time drag any two objects closer and closer together? What would be the repulsive force counteracting this?
  - Salgat 2y ago
    So the question isn't how they merge, but how they do it so quickly.
    - antognini 2y ago
      Yes, dynamical friction can "quickly" (on the order of 100 Myr) bring the two black holes to a distance of ~1 pc to each other. Gravitational wave radiation can "quickly" cause a merger if they are within a bit less than ~0.1 pc of each other. But how they go from 1 to 0.1 pc on a timescale of ~100 Myr is unknown.
  - ardel95 2y ago
    It would be. Which is why any pair of orbiting bodies will eventually collide. It’s just that for black holes this effect is insignificant (a merger would take much longer than the age of the Universe) until they get close to each other, much closer than 1 parsec.
- Iwan-Zotow 2y ago
  It was observed long time ago, Nobel 1993

by mpreda 2y ago

isoprophlex 2y ago

A black hole only has mass, angular momentum, and charge. So you can only deal with those three... not deform and knead a physical mass of planet as in tidal heating.

orlp 2y ago
This is often stated but it literally can't be true. It's true for stationary solutions to the relevant equations, but we're very much not living in a stationary world. Here's a simple thought experiment disproving your claim. A person hovers just above the origin of a supermassive black hole. They chuck a massively charged object into the black hole. If what you said is true they should observe the charge instantly being transported to the singularity, since a black hole can't have any attributes such as where charge is distributed within the horizon. Where it gets impossible is that someone very far away around the same supermassive black hole could observe a small charge increment. They in turn could chuck charged stuff into the black hole and now you've got faster than lightspeed communication.
- elashri 2y ago
  > above the origin of a supermassive black hole I spent couple of minutes trying to understand your thought experiment and was puzzled why can't I understand it. It seems that it is probably because I don't understand what you mean by "just above the origin of supermassive black hole"? I feel that you have something interesting to say but not clear.
  - orlp 2y ago
    Mental typo, I meant horizon, not origin.
    - elashri 2y ago
      The no-hair theorem (scientific term for what you are descriping) essentially states that black holes can be completely described by only three externally observable classical properties: mass, charge, and angular momentum. The internal distribution of matter or charge within the black hole is irrelevant from the outside once it crosses the event horizon. The charge of an object thrown into a black hole doesn't need to "instantaneously" reach the singularity. In fact, it doesn't even "travel" in the sense that the observer would see a time delay based on where the charge is inside the horizon. The EM field generated by this charged object can still be observed outside the horizon before the object crosses the horizon. In classical GR, the exterior of a charged black hole is governed by the Reissner-Nordström metric [1], and it already includes the influence of charge. Once the object enters the horizon, the outside observer will perceive the black hole's external field to have changed. The field adjustment is instant for the outside observer because, from their point of view, the object never actually crosses the event horizon (due to infinite time dilation at the horizon from their perspective). The charge appears to have been "absorbed" by the black hole when the object is still outside the horizon. So no-hair theorem doesn't imply that properties such as charge or angular momentum must be "smeared" instantaneously to the singularity inside the black hole. The theorem only describes how these properties manifest externally, not their behavior inside the horizon. Once a charge object passes the event horizon, the information about its charge is not causally connected to any observer inside the black hole (except in QM contexts, where the information paradox becomes relevant). However, the charges affects the external metric of the black hole immediately and completely as seen by external observers. Also it doesn't say that the charge or mass needs to be uniformly distributed or behave in any specific way inside the event horizon. It only states that from the outside, black holes look like point-like objects characterized by mass, charge, and angular momentum. The specifics of how charge is distributed inside the black hole’s event horizon aren't visible to outside observers and therefore don't affect the validity of the theorem. [1] https://en.wikipedia.org/wiki/Reissner%E2%80%93Nordstr%C3%B6m_metric https://en.wikipedia.org/wiki/Reissner%E2%80%93Nordstr%C3%B6...
- saagarjha 2y ago
  You can’t actually observe objects going into a black hole. They just become redshifted and fall slower and slower from your perspective towards the event horizon.
- isoprophlex 2y ago
  The singularity is not the same thing as its event horizon. That said, it's the no hair theorem. It could of course still be wrong...
- ben_w 2y ago
  The object will be observed asymptotically decelerating as it approaches the event horizon, never gets observed crossing by a distant observer. IIRC the event horizon expands to encompass the object just before it gets to the original horizon, due to the Schwarzschild formula. Also IIRC, the electric field is blurred out by the geodesics by this point, as if it were from the interior. But that's based on what I've heard, I have yet to derive useful results from the Einstein field equations, even though I think I should give it a go and I can follow them well enough to code a simple simulation…

gizmo686 2y ago

Gravitational waves have a similar effect of sapping the kinetic energy of orbiting black holes. Instead of heating the black holes, the energy is emitted as gravitational radiation. Having said that, now that we have actual observations of the gravitational radiation emmited from merging black holes, we can be pretty confident in our understanding of the magnitude of this energy loss. And that understanding is that it does not become significant until well into the last parsec.

vlovich123 2y ago
But however insignificant it is, wouldn’t it always be there sapping angular momentum & over a long period of time drag any two objects closer and closer together? What would be the repulsive force counteracting this?
- Salgat 2y ago
  So the question isn't how they merge, but how they do it so quickly.
  - antognini 2y ago
    Yes, dynamical friction can "quickly" (on the order of 100 Myr) bring the two black holes to a distance of ~1 pc to each other. Gravitational wave radiation can "quickly" cause a merger if they are within a bit less than ~0.1 pc of each other. But how they go from 1 to 0.1 pc on a timescale of ~100 Myr is unknown.
- ardel95 2y ago
  It would be. Which is why any pair of orbiting bodies will eventually collide. It’s just that for black holes this effect is insignificant (a merger would take much longer than the age of the Universe) until they get close to each other, much closer than 1 parsec.
Iwan-Zotow 2y ago
It was observed long time ago, Nobel 1993