4 ms·

Would our brains even be able to detect the moments the earth entered one? We'd probably get a little warning from amateur astronomers as they freaked out mont

by ck2 11y ago

Would our brains even be able to detect the moments the earth entered one? We'd probably get a little warning from amateur astronomers as they freaked out months before it happens.

huuu 11y ago
What if we entered one and came out on the other side without even noticing something happened?
- sunshinerag 11y ago
  What if that already happened/happening/happenwhen
- gonzo41 11y ago
  there is no other side. its not a physical hole. its a gravity hole. Think of a moon that drags you onto it. You can't ever have enough power to reverse moving towards it. Light isn't quick enough to leave the pull. thus it is dark.
  - nonbel 11y ago
    >"there is no other side. its not a physical hole." Isn't a black hole more like an edge of the universe? My understanding is that you end up with a Zeno's paradox situation where no matter how close you get you can never actually reach it.
    - xtacy 11y ago
      Not quite -- in your reference frame, you cross the blackhole's event horizon and go inside, but an observer (who is outside the event horizon) will only see you reach the event horizon and slowly "fade away."
      - nonbel 11y ago
        Thanks, so what is the deal with the four-velocity always equaling c^2? Ie (maybe it is totally wrong): c^2 = (v_t)^2 - (v_x)^2 - (v_y)^2 - (v_z)^2 where v_i indicates speed along that dimension. I thought this was why all the weird time dilation effects occur, velocity along time v_t ->0 as velocity through space increases. You seem to be saying that from the traveler's perspective they are not accelerating faster and faster towards the black hole. As I said, maybe my understanding is just totally incorrect here. Edit: Actually, I was really confused (this is all based on looking it up the other day). According to the above, v_t would need to increase as eg v_x increases. This is just wrong, but I'll leave it here in case someone wants to explain it correctly.
        raattgift 11y ago
        This is asking an essential Special Relativity (SR) question into a conversation about an object where gravitation is not just relevant but central, so I'll also give you an a modern Special Relativity answer cast in such a way to follow with a couple of comments about "c" in General Relativity. Below you'll note that I'll treat GR strictly as geometry. Your equation is some algebraic mangling of the Minkowski line-element, which we can write as ds^2 = -cdt^2 + dx^2 + dy^2 + dz^2 in Cartesian coordinates with a -+++ metric signature. That's the interval in flat spacetime (aka Minkowski spacetime, although in both cases sometimes "time" is omitted when a reader will not be confused). The conversion constant "c" there is the sole free parameter of the Poincare group, which is the local isometry group of Minkowski spacetime (i.e., it applies each point in spacetime), which includes translations, Lorentz boosts and rotations on the three spacelike axes (x, y, and z in the form above) and a unidirectional translation on the timelike axis (t). The parameter corresponds (in SR) to the speed of a massless particle, and light is assumed to be massless. It's a postulate of Special Relativity that all non-gravitational physics is invariant under the Poincare group, and that's baked into the Standard Model, for instance. However, in GR "c" corresponds to the surface of a nonempty, convex, open cone of tangent vectors at some point p on the (curved) manifold; light at point p is constrained to that surface exactly, and massive particles at point p stay inside that boundary. In GR we can only meaningfully compare speed and velocity locally, where that means either in the limit as spacetime intervals go to zero, or equivalently in the local section of the fibre bundle, because the very definition of spacetime curvature means the parallel transport of one vector to another for comparison purposes is path-dependent, and that applies on all four axes, so we cannot even meaningfully compare two clocks (with which we might measure relative velocity) unless they are at the same point in spacetime. However, we can work things out so that a small (but not exactly zero-sized) region of spacetime is treated as flat, and then use Special Relativity or even Newtonian mechanics to discuss the matter content in that region, right down to comparing velocities. This "flattening" can be done quite successfully in a number of ways, and can even be done in a region with significant curvature by using the formalisms of semiclassical gravity. The cost is in artifacts introduced into the non-gravitational content of the region of non-negligbly curved spacetime, most noticeably in terms of differences in particle count and even the interpretation of matter field excitations as particles. This mathematical "flattening away" of the curvature returns us from a causal cone built on a hyperbolization of a series of first-order quasilinear partial differential equations (the Einstein Field Equations) to the Poincare group applying at every point, and thus we return to "c" as being mathematically special, rather than relating to one of perhaps many causal cones (since one can have many hyperbolizations, and thus many timelike-spacelike conversion constants although so far there is only evidence for one).
  - huuu 11y ago
    Of course I was half joking. It's not about if light is quick enough to leave the pull. It's about the light not being fast enough to be stopped by the mass. If light is accelerating towards the center fast enough so the mass won't be able to stop it the light will go though it. But ofcourse that means that light should go faster than light.
ceejayoz 11y ago
The stars would start disappearing and being distorted millennia before, I'd imagine.
imglorp 11y ago
We'd see it coming long before. They emit radiation at many frequencies, plus all the stars orbiting them (ie, a galaxy) indicate the mass of what they're orbiting. So if you happen to see any galaxies coming towards us in the next billion years (aside from Andromeda) do let the world know.
- jfoster 11y ago
  Are there any risks associated with them? For instance, a burst of radiation that obliterates us? Is that possible?
  - ianlevesque 11y ago
    No, but definitely don't look up gamma ray bursts.

by ck2 11y ago

Would our brains even be able to detect the moments the earth entered one? We'd probably get a little warning from amateur astronomers as they freaked out months before it happens.

huuu 11y ago

What if we entered one and came out on the other side without even noticing something happened?

sunshinerag 11y ago
What if that already happened/happening/happenwhen
gonzo41 11y ago
there is no other side. its not a physical hole. its a gravity hole. Think of a moon that drags you onto it. You can't ever have enough power to reverse moving towards it. Light isn't quick enough to leave the pull. thus it is dark.
- nonbel 11y ago
  >"there is no other side. its not a physical hole." Isn't a black hole more like an edge of the universe? My understanding is that you end up with a Zeno's paradox situation where no matter how close you get you can never actually reach it.
  - xtacy 11y ago
    Not quite -- in your reference frame, you cross the blackhole's event horizon and go inside, but an observer (who is outside the event horizon) will only see you reach the event horizon and slowly "fade away."
    - nonbel 11y ago
      Thanks, so what is the deal with the four-velocity always equaling c^2? Ie (maybe it is totally wrong): c^2 = (v_t)^2 - (v_x)^2 - (v_y)^2 - (v_z)^2 where v_i indicates speed along that dimension. I thought this was why all the weird time dilation effects occur, velocity along time v_t ->0 as velocity through space increases. You seem to be saying that from the traveler's perspective they are not accelerating faster and faster towards the black hole. As I said, maybe my understanding is just totally incorrect here. Edit: Actually, I was really confused (this is all based on looking it up the other day). According to the above, v_t would need to increase as eg v_x increases. This is just wrong, but I'll leave it here in case someone wants to explain it correctly.
      - raattgift 11y ago
        This is asking an essential Special Relativity (SR) question into a conversation about an object where gravitation is not just relevant but central, so I'll also give you an a modern Special Relativity answer cast in such a way to follow with a couple of comments about "c" in General Relativity. Below you'll note that I'll treat GR strictly as geometry. Your equation is some algebraic mangling of the Minkowski line-element, which we can write as ds^2 = -cdt^2 + dx^2 + dy^2 + dz^2 in Cartesian coordinates with a -+++ metric signature. That's the interval in flat spacetime (aka Minkowski spacetime, although in both cases sometimes "time" is omitted when a reader will not be confused). The conversion constant "c" there is the sole free parameter of the Poincare group, which is the local isometry group of Minkowski spacetime (i.e., it applies each point in spacetime), which includes translations, Lorentz boosts and rotations on the three spacelike axes (x, y, and z in the form above) and a unidirectional translation on the timelike axis (t). The parameter corresponds (in SR) to the speed of a massless particle, and light is assumed to be massless. It's a postulate of Special Relativity that all non-gravitational physics is invariant under the Poincare group, and that's baked into the Standard Model, for instance. However, in GR "c" corresponds to the surface of a nonempty, convex, open cone of tangent vectors at some point p on the (curved) manifold; light at point p is constrained to that surface exactly, and massive particles at point p stay inside that boundary. In GR we can only meaningfully compare speed and velocity locally, where that means either in the limit as spacetime intervals go to zero, or equivalently in the local section of the fibre bundle, because the very definition of spacetime curvature means the parallel transport of one vector to another for comparison purposes is path-dependent, and that applies on all four axes, so we cannot even meaningfully compare two clocks (with which we might measure relative velocity) unless they are at the same point in spacetime. However, we can work things out so that a small (but not exactly zero-sized) region of spacetime is treated as flat, and then use Special Relativity or even Newtonian mechanics to discuss the matter content in that region, right down to comparing velocities. This "flattening" can be done quite successfully in a number of ways, and can even be done in a region with significant curvature by using the formalisms of semiclassical gravity. The cost is in artifacts introduced into the non-gravitational content of the region of non-negligbly curved spacetime, most noticeably in terms of differences in particle count and even the interpretation of matter field excitations as particles. This mathematical "flattening away" of the curvature returns us from a causal cone built on a hyperbolization of a series of first-order quasilinear partial differential equations (the Einstein Field Equations) to the Poincare group applying at every point, and thus we return to "c" as being mathematically special, rather than relating to one of perhaps many causal cones (since one can have many hyperbolizations, and thus many timelike-spacelike conversion constants although so far there is only evidence for one).
- huuu 11y ago
  Of course I was half joking. It's not about if light is quick enough to leave the pull. It's about the light not being fast enough to be stopped by the mass. If light is accelerating towards the center fast enough so the mass won't be able to stop it the light will go though it. But ofcourse that means that light should go faster than light.

ceejayoz 11y ago

The stars would start disappearing and being distorted millennia before, I'd imagine.

imglorp 11y ago

We'd see it coming long before. They emit radiation at many frequencies, plus all the stars orbiting them (ie, a galaxy) indicate the mass of what they're orbiting. So if you happen to see any galaxies coming towards us in the next billion years (aside from Andromeda) do let the world know.

jfoster 11y ago
Are there any risks associated with them? For instance, a burst of radiation that obliterates us? Is that possible?
- ianlevesque 11y ago
  No, but definitely don't look up gamma ray bursts.