6 ms·

Could someone please explain to me in plain English, how is it that the black holes "emit radiation at many frequencies" and, at the same time, "the light is un

• badlogic 11y ago
  As surrounding matter falls into the black hole, it heats up and emits radiation. There's also Hawking radiation, but that's not detectable.
  • alphydan 11y ago
    To be slightly more specific: Astronomers think there is a black hole when you see gas moving very very fast towards an invisible and massive object. That gas heats up from the friction of its particles moving rapidly towards the black hole. This heated gas emits X or Gamma radiation when it travels that fast. However, once it goes past the event horizon nothing else is seen. So you have something very heavy which is attracting objects and accelerating the surrounding gas. At the very "edge", sometimes pairs of particles are formed out of the vacuum (for example an electron and an anti-electron (positron)). Hawking asked: What if one particle leaves away from the horizon, but the other one leaves towards the center of the black hole? We would see an object that is sending particles from its event horizon. So in some sense particles would escape from the edge of the black hole. That's Hawking radiation (in simplified terms).
    • pflanze 11y ago
      Hmm, so there is this thing where physical theories seem to predict that each piece of matter should be complemented with an according piece of anti-matter (because those then make themselves "whole" by turning into pure energy), but we don't know where the anti-matter is. Your description made me think, could it all be in black holes?
      • Natanael_L 11y ago
        Antimatter is still regular matter, but with the exact opposite charge to the common particles we keep seeing. So technically yes, it could all have disappeared into black holes, but it wouldn't really be antimatter anymore after entering the singularity due to being effectively broken down into pure energy/matter.
        • danbruc 11y ago
          I don't think many - if any - physicist believe that there is a real singularity within a black hole. The existence of singularities in the solutions of our equations more likely just tells us that the equations are just good approximations that break down at the center of black holes.
        • lisivka 11y ago
          What will happen when black hole will merge with antimatter black hole of same mass?
          • Natanael_L 11y ago
            It is no longer antimatter after entering the black hole (edit: once reaching the presumed singularity, but either way you still won't see any difference from the outside). The mass is the same, only the charge is different. They'd merge normally. You might be thinking of negative matter, in which case they'd effectively cancel out, erase each other.
            • danbruc 11y ago
              Nothing strange happens when something falls into a black hole. We could right now be beyond the event horizon of a newly forming black hole. Unless a black hole is particularly small, in which case there will be notable tidal forces near the event horizon, the event horizon of a black hole is a pretty ordinary place hardly distinguishable from flat space. So nothing happens to matter or antimatter when it falls into a black hole.
    • ars 11y ago
      Can you explain the friction? To have friction you have to have two things moving at different speeds rubbing on each other. But infalling gas all moves in the same direction.
      • maaku 11y ago
        (1) Gas is composed of particles that are not moving in the same direction. (2) Electromagnetic waves come from the movement of an electric charge. Even a single molecule falling into an event horizon is a moving charge (plasma) or dipole (gas).
        • deleted 11y ago
          [deleted]
      • semi-extrinsic 11y ago
        The infalling gas will, in general, be rotating like water around a drain. The rotation leads to a shear (technical word for fluid rubbing against itself), and since the gas has non-zero viscosity (technical word for fluid friction coefficient), shear leads to heat production. In the purely non-rotating case, there is another source of heat as well: compression. As gas from far away moves closer to the black hole, the distance between gas molecules decreases, and the pressure increases. This causes the temperature to increase. Just feel the outside of a bicycle pump after quickly inflating two full tires.
      • raattgift 11y ago
        It's not really so much friction as collision, which in post-classical terms is objects coming close enough together for (mainly electromagnetic) quantum interactions to dominate. The objects around a black hole are on different intersecting geodesics; "collisions" result in energy exchanges that boost the particles onto different geodesics, typically that wind closer to the horizon. Near the horizon inverse Compton scattering becomes important, too: high energy photons emitted by matter-matter collisions end up colliding with charged massive particles with a transfer of momentum from the latter to the photon. The latter's change of momentum leaves it on a geodesic still closer to the horizon, while the photon is boosted to high energy (perhaps escaping to infinity as an X-Ray).
    • sullyj3 11y ago
      I don't understand hawking radiation. People say it would eventually evaporate the black hole, but wouldn't the particle falling in increase its mass, whilst the emitted anti-particle would have no effect?
      • Retric 11y ago
        First it only applies to rotating black holes. Second it's particle and virtual particle not anti particle.
        • danbruc 11y ago
          I don't think that this is true, non-rotating black holes can also emit Hawking radiation. Besides that, there are probably no non-rotating black holes, having an angular momentum of (exactly) zero is a pretty improbable situation and a single particle falling not exactly radially into a non-rotating black hole would turn it into a rotating black hole albeit with a really tiny angular momentum.
          • Retric 11y ago
            I don't see why it would not work on stationary black holes, but I keep seeing it in reference to rotating black holes. Though, as you say all black holes rotate.
            • danbruc 11y ago
              According to Wikipedia Hawking started his work after another physicist pointed out that rotating black holes should emit radiation. Maybe that stuck somehow. Hawking's work followed his visit to Moscow in 1973 where the Soviet scientists Yakov Zeldovich and Alexei Starobinsky showed him that, according to the quantum mechanical uncertainty principle, rotating black holes should create and emit particles.
              • Retric 11y ago
                Ahh, ok thanks that's the connection I was missing.
• qrendel 11y ago
  IANAP(hysicist), but afaik it's due to pair production of photons at the boundary of the black hole. One photon falls into the black hole, the other escapes in the opposite direction, conserving momentum. The pair production of photons is stimulated by the black hole itself through <some complicated interaction>, thereby causing the black hole to lose energy over time as the escaping photons are lost, which are called Hawking radiation. The many frequencies occur because the pair production produces pairs of photons at different energies/frequencies.
  • MichaelApproved 11y ago
    No, we can't detect Hawking radiation, yet. It's too faint. http://physics.stackexchange.com/questions/48032/how-would-you-detect-hawking-radiation http://physics.stackexchange.com/questions/48032/how-would-y... > Emitted Radiation When material falls into a black hole from a companion star, it gets heated to millions of degrees Kelvin and accelerated. The superheated materials emit X-rays, which can be detected by X-ray telescopes such as the orbiting Chandra X-ray Observatory. > In addition to X-rays, black holes can also eject materials at high speeds to form jets. Many galaxies have been observed with such jets. Currently, it is thought that these galaxies have supermassive black holes (billions of solar masses) at their centers that produce the jets as well as strong radio emissions. http://science.howstuffworks.com/dictionary/astronomy-terms/black-hole3.htm http://science.howstuffworks.com/dictionary/astronomy-terms/...
    • qrendel 11y ago
      I thought the question was about radiation emitted by the black hole itself, i.e. Hawking radiation, not radiation emitted by matter outside but near the black hole, i.e. accretion disk radiation. It's not really contradictory that stuff outside the event horizon can still escape.
      • MichaelApproved 11y ago
        You are correct, I was mistaken.
• danbruc 11y ago
  There are several effects. Accelerating black holes, or actually any accelerated matter, emits gravitational waves. This are disturbances of space-time itself and unlike electromagnetic waves mediated by photons moving through space-time. At least in a classical picture of gravity, not sure how that would work out within quantum gravity when gravitational waves are mediated by gravitons. But I guess you can just think of the gravitational waves being emitted at the event horizon. Then there is electromagnetic radiation emitted from matter falling towards the black hole but still outside of the event horizon. The gravitational forces of the black hole heat up this matter to the point that it emits electromagnetic radiation in a wide spectrum. Finally there is Hawking radiation which is a suspected electromagnetic radiation due to quantum effects near the event horizon. The associated black-body temperature is tiny, two billionth of a Kelvin for a 30 solar mass black hole and decreasing with increasing mass. The temperature of the cosmic microwave background with 2.7 Kelvin is in comparison gigantic and therefore black holes absorb much more cosmic microwave background radiation than they emit Hawking radiation. The universe has to become a lot cooler before Hawking radiation can start to evaporate black holes.
  • raattgift 11y ago
    Gravitational waves (GWs) propagate a _change_ in the gravitational field. Consequently, GWs are only shed by accelerating matter where the change in motion is not rotationally or spherically symmetric. Spherically symmetric gravitational collapses (of dust or gas, for example) are certainly systems of accelerated matter, but do not shed gravitational waves. Rotating spherical bodies also do not shed GWs. To first order, this means the rotation of stars and planets do not shed GWs, although a planet revolving around the star in an essentially Kepler orbit will shed GWs, as will a pair of stars in close orbit around each other. GWs solve the apparent removal of momentum-energy from a system where there is no removal of mass-energy by non-gravitational radiation. For most systems the energy loss is negligible. That's why the first GW detection involved a late stage inspiral of two compact massive objects with no way to ditch their angular momentum electromagnetically or via neutrino (or other dark matter) emission. GWs themselves have spin 2 symmetry; in theories with gravitons, those are spin-2 gauge bosons, typically massless (because they are long range), and you get large numbers of them forming the classical GWs, just as you will find large numbers of photons forming a bright flicker of light, although you can detect tiny numbers of photons because electromagnetism is relatively strong -- tiny numbers of gravitons are not going to be detectable by humanity any time soon. It's not curvature (and certainly not gravitational forces) heating up accretion disks. The matter in the disks move on intersecting geodesics and occasionally collide. Such collisions tend to ditch the particles' momentum into emitted photons. Hawking radiation is an effect of accelerated observers seeing more (and more energetic) particles than non-accelerated ones, and gravitational collapses producing an acceleration between past and future observers. Where a past observer sees a tiny number of low-frequency particles in the "vacuum" outside the collapsing object, a future observer (after the horizon has formed) will see a larger number of higher frequency ones. The acceleration is curvature dependent, and curvature just outside the horizon depends on black hole mass -- more massive black holes have less curvature just outside the horizon (which is further away from the gravitational singularity or whatever is at the centre of a black hole). These particles near but outside the horizon can escape, which takes mass-energy away from the dynamical spacetime just outside the horizon. That in turn causes the horizon to retract and that exposes stronger curvature. That in turn produces a stronger acceleration for further future observers, and the process continues -- that's black hole evaporation. Hawking radiation is always emitted, for SMBHs that only grow by incidental interception of CMB radiation, and even for stellar mass black holes. Infalling matter simply replaces (and then some) the outgoing "fossil" infrared radiation from the space just outside the black hole as the horizon was forming.