3 ms·

Yah, I errored in thinking 3 beams would negate that, but it wouldn't, it would be just below, as you say. But you have another issue: Even if you are just bel

by ars 3y ago

Yah, I errored in thinking 3 beams would negate that, but it wouldn't, it would be just below, as you say. But you have another issue: Even if you are just below the speed of light, most of the mass would become relativistic mass (and relativistic momentum), with almost no rest mass. But there's a postulate that only rest mass can make a black hole, and relativistic mass doesn't count. (Because otherwise you could travel fast and see inside the black hole.) So we are left with a contradiction.

btilly 3y ago
The better demonstration of that postulate is that it is only a black hole if it is a black hole in all reference frames. So if it isn't a black hole in its own reference frame, it isn't a black hole. As for the effect of moving masses, that kind of "drags" space-time along with it. And rotating masses do as well. In the case of rotating masses that's called https://en.wikipedia.org/wiki/Lense%E2%80%93Thirring_precession https://en.wikipedia.org/wiki/Lense%E2%80%93Thirring_precess... and was what Gravity Probe B was measuring. As for the light, you can solve the mystery as follows. Go to the reference frame in which the light is just all pouring into one spot. In that spot photons are meeting photons and creating a sea of pairs of particles - and now it is obvious that a black hole could be made.
- ars 3y ago
  Photons do not create particles in a vacuum - they need to interact with matter to do that - one of the reasons being in order to conserve momentum, some of which goes to the matter. So at that meeting spot they would go right through each other.
  - btilly 3y ago
    No, they do. When a particle meets its antiparticle, you get 2 energetic photons. All such interactions are perfectly reversible. So when 2 energetic photons meet, you can get a particle and antiparticle. Both energy and momentum are conserved in this process.
    - ars 3y ago
      When they meet they produce photons traveling in opposite directions, so there is no net momentum. If there was incoming momentum from the particle (they they move both forward and toward each other) then the two emitted photons will also be traveling forward (i.e. at an angle toward each other). But here we explicitly placed the photons to be traveling in the same direction, so there is some net momentum that can not be carried by the particles. i.e. particle anti-particle annihilation will never produce two photons that are both traveling straight forward. So the event is not reversible.
      - btilly 3y ago
        No, we explicitly placed the photons to be traveling in different directions so that they could meet. If they meet, and they meet with sufficient energy, they can create particle anti-particle pairs. You can read more about this at https://en.wikipedia.org/wiki/Two-photon_physics https://en.wikipedia.org/wiki/Two-photon_physics.
        ars 3y ago
        The angle they are traveling is one that's impossible to reverse - in annihilation it's impossible for the two photons to be emitted at such a close angle. I read about the two photon physics - and the two photons can scatter off of each other, but they can't produce particles except in very high electromagnetic fields, which isn't the case here.
        codethief 3y ago
        I'm really not following you. If two photons meet at an angle – any angle –, there will be a rest frame in which they're meeting from exactly opposite (180°) directions. They can absolutely produce electron positron pairs if their energy is high enough. The process γ γ′ → e+ e− is a classic calculation in QED, and although it's never been observed directly for two single/isolated photons[0], there is no good reason to believe QED should be wrong in this particular case and, indeed, all evidence we have is pointing towards this process being perfectly possible[1]. [0]: https://en.m.wikipedia.org/wiki/Breit%E2%80%93Wheeler_process https://en.m.wikipedia.org/wiki/Breit%E2%80%93Wheeler_proces... [1]: https://www.energy.gov/science/np/articles/making-matter-collisions-light https://www.energy.gov/science/np/articles/making-matter-col...
codethief 3y ago
> But there's a postulate that only rest mass can make a black hole, and relativistic mass doesn't count. That's news to me. Could you provide a reference? "Relativistic mass" is just rest mass + kinetic energy, and so yes, it should influence the energy-momentum tensor like everything else. > (Because otherwise you could travel fast and see inside the black hole.) How so?
- btilly 3y ago
  I explain it in https://news.ycombinator.com/item?id=37344926 https://news.ycombinator.com/item?id=37344926. Something can't become a black hole just because it is moving fast. If it is a black hole, it is a black hole in all reference frames. Including the one where it is just sitting there and the rest of the universe is going the other way.
  - codethief 3y ago
    Of course the (non-)existence of a black hole is a coordinate-/observer-independent statement (unless we talk only about apparent horizons), but the full energy ("relativistic mass") of a moving object will still show up in the energy-momentum tensor as seen from an outside (non-moving) observer. For instance, if we take a hot gas, we absolutely have to account for its temperature & pressure (the kinetic energy & momentum of its constituents) in the energy-momentum tensor. To see that this is a necessity, even for a single particle moving at relativistic speeds, consider energy conservation: If you accelerate a particle, you have to expend energy and that energy needs to come from somewhere (say nuclear fission). As a consequence, this energy will already show up in the energy-momentum tensor before acceleration (in the fission case: as rest mass of your isotopes) and, by local energy-momentum conservation (= 4-divergence of the tensor being zero), it then must also show up in the tensor afterwards. Whether or not that energy can lead to a black hole is a different question. For a single massive object flying at relativistic speeds in an otherwise empty universe: Probably not, because one can simply transform that "relativistic mass" away by switching coordinates, as you say. For several objects moving relative to each other at high speeds, or even a single particle moving relative to the "rest of the universe", it's not so simple, though – you won't find a coordinate system in which all terms in the energy-momentum tensor are small (let alone zero) everywhere.

by ars 3y ago

btilly 3y ago

The better demonstration of that postulate is that it is only a black hole if it is a black hole in all reference frames. So if it isn't a black hole in its own reference frame, it isn't a black hole. As for the effect of moving masses, that kind of "drags" space-time along with it. And rotating masses do as well. In the case of rotating masses that's called https://en.wikipedia.org/wiki/Lense%E2%80%93Thirring_precession https://en.wikipedia.org/wiki/Lense%E2%80%93Thirring_precess... and was what Gravity Probe B was measuring. As for the light, you can solve the mystery as follows. Go to the reference frame in which the light is just all pouring into one spot. In that spot photons are meeting photons and creating a sea of pairs of particles - and now it is obvious that a black hole could be made.

ars 3y ago
Photons do not create particles in a vacuum - they need to interact with matter to do that - one of the reasons being in order to conserve momentum, some of which goes to the matter. So at that meeting spot they would go right through each other.
- btilly 3y ago
  No, they do. When a particle meets its antiparticle, you get 2 energetic photons. All such interactions are perfectly reversible. So when 2 energetic photons meet, you can get a particle and antiparticle. Both energy and momentum are conserved in this process.
  - ars 3y ago
    When they meet they produce photons traveling in opposite directions, so there is no net momentum. If there was incoming momentum from the particle (they they move both forward and toward each other) then the two emitted photons will also be traveling forward (i.e. at an angle toward each other). But here we explicitly placed the photons to be traveling in the same direction, so there is some net momentum that can not be carried by the particles. i.e. particle anti-particle annihilation will never produce two photons that are both traveling straight forward. So the event is not reversible.
    - btilly 3y ago
      No, we explicitly placed the photons to be traveling in different directions so that they could meet. If they meet, and they meet with sufficient energy, they can create particle anti-particle pairs. You can read more about this at https://en.wikipedia.org/wiki/Two-photon_physics https://en.wikipedia.org/wiki/Two-photon_physics.
      - ars 3y ago
        The angle they are traveling is one that's impossible to reverse - in annihilation it's impossible for the two photons to be emitted at such a close angle. I read about the two photon physics - and the two photons can scatter off of each other, but they can't produce particles except in very high electromagnetic fields, which isn't the case here.
        codethief 3y ago
        I'm really not following you. If two photons meet at an angle – any angle –, there will be a rest frame in which they're meeting from exactly opposite (180°) directions. They can absolutely produce electron positron pairs if their energy is high enough. The process γ γ′ → e+ e− is a classic calculation in QED, and although it's never been observed directly for two single/isolated photons[0], there is no good reason to believe QED should be wrong in this particular case and, indeed, all evidence we have is pointing towards this process being perfectly possible[1]. [0]: https://en.m.wikipedia.org/wiki/Breit%E2%80%93Wheeler_process https://en.m.wikipedia.org/wiki/Breit%E2%80%93Wheeler_proces... [1]: https://www.energy.gov/science/np/articles/making-matter-collisions-light https://www.energy.gov/science/np/articles/making-matter-col...

codethief 3y ago

> But there's a postulate that only rest mass can make a black hole, and relativistic mass doesn't count. That's news to me. Could you provide a reference? "Relativistic mass" is just rest mass + kinetic energy, and so yes, it should influence the energy-momentum tensor like everything else. > (Because otherwise you could travel fast and see inside the black hole.) How so?

btilly 3y ago
I explain it in https://news.ycombinator.com/item?id=37344926 https://news.ycombinator.com/item?id=37344926. Something can't become a black hole just because it is moving fast. If it is a black hole, it is a black hole in all reference frames. Including the one where it is just sitting there and the rest of the universe is going the other way.
- codethief 3y ago
  Of course the (non-)existence of a black hole is a coordinate-/observer-independent statement (unless we talk only about apparent horizons), but the full energy ("relativistic mass") of a moving object will still show up in the energy-momentum tensor as seen from an outside (non-moving) observer. For instance, if we take a hot gas, we absolutely have to account for its temperature & pressure (the kinetic energy & momentum of its constituents) in the energy-momentum tensor. To see that this is a necessity, even for a single particle moving at relativistic speeds, consider energy conservation: If you accelerate a particle, you have to expend energy and that energy needs to come from somewhere (say nuclear fission). As a consequence, this energy will already show up in the energy-momentum tensor before acceleration (in the fission case: as rest mass of your isotopes) and, by local energy-momentum conservation (= 4-divergence of the tensor being zero), it then must also show up in the tensor afterwards. Whether or not that energy can lead to a black hole is a different question. For a single massive object flying at relativistic speeds in an otherwise empty universe: Probably not, because one can simply transform that "relativistic mass" away by switching coordinates, as you say. For several objects moving relative to each other at high speeds, or even a single particle moving relative to the "rest of the universe", it's not so simple, though – you won't find a coordinate system in which all terms in the energy-momentum tensor are small (let alone zero) everywhere.