Why doesnt a photon escape from a black hole?

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SUMMARY

The discussion focuses on the inability of photons emitted from within the Schwarzschild radius (r < a) of a black hole to escape to a distant observer. The trajectory of a photon is defined by the equation dot{r}^2 = c^2 lambda^2 - (1 - a/r)(J^2/r^2), where J represents angular momentum and lambda is defined as lambda = dot{t}(1 - a/r). When the radial coordinate r is less than the Schwarzschild radius a, the expression for dot{r} becomes imaginary, confirming that light cannot escape from this region, as discussed in Dirac's General Relativity text.

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  • Understanding of General Relativity principles
  • Familiarity with Schwarzschild geometry
  • Knowledge of photon trajectory equations
  • Basic grasp of angular momentum in relativistic contexts
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stunner5000pt
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Explain why elight emitted at a point inside the Scwarzschild radius (r<a) cannot be seen by a distant observer. (by one at r-> infinity)
well we know the trajectory of a photon is given by


[tex]\dot{r}^2 = c^2 \lambda^2 - \left( 1 - \frac{a}{r}\right) \frac{J^2}{r^2}[/tex]

where J repsents the angular momentum in units of mass
lambda is [tex]\lambda = \dot{t} \left( 1 - \frac{a}{r} \right)[/tex]

so when r < a does the expression for r dot become imaginary? Is that what i am aiming to prove?
 
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See the argument given by Dirac in its GR book when discussing black holes.

Daniel.
 

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