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I have a couple of questions about light here, and sorry of they are silly..

So since anything that goes beyond the event horizon can't go out, so what if a light beam was pointed somewhere behind the event horizon to the outside, would light have a slower speed then get sucked back? and what if the light particles were somehow exactly on the event horizon, would the light stand still? And why can't light stand still? why does it always move at that really high speed?

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5 Answers 5

The most elegant way I've seen to describe this is described in the paper The river model of black holes. If we write the Schwarzschild metric in Gullstrand-Painlevé coordinates we get (in units where $c = G = 1$):

$$ ds^2 = -dt_{ff}^2 + \left(dr + \beta dt_{ff} \right)^2 + r^2 d\Omega^2 $$

where:

$$ \beta = \frac{2M}{r} $$

This looks like the Minkowski metric except that the radial coordinate is replaced by $dr + \beta dt_{ff}$. The parameter $\beta$ has the dimensions of velocity, and in fact it's equal to the Newtonian escape velocity. $\beta < 1$ outside the event horizon, $\beta = 1$ at the event horizon and $\beta > 1$ inside the event horizon.

The physical interpretation of this is that the radial coordinate is flowing inwards at a velocity of $\beta$. In effect space is flowing inwards into the black hole and carries observers along with it in the same way a river carries along observers floating on it - hence the name river model. A light beam moves with velocity $1$ with respect to the spacetime around it, so relative to the observer at infinity the net velocity of a radial outbound light beam is:

$$ v_{eff} = 1 - \beta $$

So outside the event horizon $v_{eff} > 0$ and the light beam moves outwards. At the event horizon $v_{eff} = 0$ and the light beam is frozen at the horizon unable to move outwards. Inside the event horizon $v_{eff} < 0$ i.e. even if you shine the light directly outwards it still moves inwards towards the singularity.

As it happens I have addressed this issue before, in the question Why is a black hole black?. However that was a more algebraic approach and I think the river model approach is far more intuitive (insofar as anything in GR can be intuitive!).

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What do you mean by light is frozen at event horizon? Isn't speed of light constant? –  Sachin Shekhar Apr 3 at 19:26
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@SachinShekhar: in GR the speed of light is locally constant. If you measure it at your location you'll always get the value $c$. However anywhere else it can differ from $c$. –  John Rennie Apr 3 at 19:41
    
+1 What a great paper, there are some wonderfully creative teachers around. –  WetSavannaAnimal aka Rod Vance Apr 3 at 22:06

so what if a light beam was pointed somewhere behind the event horizon to the outside,

Inside the horizon, the curvature of spacetime is such that the direction "to the outside" is the past time direction.

In other words, to go 'back' towards the horizon is as impossible as it is to go 'back' in time.

Indeed, for the same reason that we inexorably 'move forward' in time, once inside the horizon, one moves inexorably towards the singularity at the 'center' of the black hole.

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General relativity describes gravity as the curvature of space-time. So to understand how black holes work, we must have a basic idea of how space-time works.

In regular 3D space, the length $\Delta s$ between two points is $\Delta s^2 = \Delta x^2 + \Delta y^2 + \Delta z^2$. This is just the pythagorean theorem applied in 3D. But in space-time, where we treat time as another coordinate, the distance between two points in space-time is $\Delta s^2 = (c\Delta t)^2-\Delta x^2 - \Delta y^2 - \Delta z^2$. This means the distance depends also on the difference in time. Light travels on paths that have no distance through space-time, i.e. $\Delta s^2 = 0$. Particles with mass travel on paths that give $\Delta s^2 > 0$.

Now when we curve space-time by adding mass or energy, the paths that give no change in length change. When we're inside a black hole, there are no paths that begin inside the black hole and end outside that have distance greater than or equal to zero. This means all paths inside the black hole end inside the black hole. The only way out is of you can travel faster than light, which is currently considered physically impossible.

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Faster than light can also not help. You need to warp Spacetime. –  Sachin Shekhar Apr 3 at 10:58
    
@SachinShekhar - "It is interesting that tachyons may emerge from under the event horizon." - arxiv.org/pdf/1301.5428.pdf (final page). –  Nuclear_Wizard Apr 3 at 11:09
    
This is fictional. If you fall on to Black Hole, singularity would lie in your direct future. To get outside, you need to travel in past which is impossible for even tachyons.. –  Sachin Shekhar Apr 3 at 11:12
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But tachyons travel along spacelike geodesics, so I'm fairly certain they can escape a black hole. Isn't travelling into the past essentially what tachyons are doing by following spacelike geodesics? –  Nuclear_Wizard Apr 3 at 11:29
    
@Nuclear_Wizard Maybe that would be a good new question. Intuitively, your comment on spacelike geodesics makes sense, but my GR level is a reasonable overall grasp of Misner Thorne Wheeler and so the question is likely a bit beyond that. –  WetSavannaAnimal aka Rod Vance Apr 3 at 22:10
  • As you said, since nothing can go out from the event horizon, if light is sent from within the event horizon it does not escape.

    However, it is not true, that light will slow down and go back. In fact, inside the event horizon all accessible paths point towards the center of the Black Hole.

    This means, that, once inside a Black Hole, you will be pushed to its center (in a finite time). No one knows what happen when you touch the singularity since it is not described by General Relativity.

  • For light emitted in the event horizon, at least in the Schwarzschild metric, it will be kept in the horizon.

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Light would vent and I it would go into the black hole at the speed of light but to the person that poi nted the light to the black hole, the beam of light will seamed to stop at the event horizon.

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