When the photon is born on the horizon .bang?

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When a photon is generated at the event horizon of a black hole, its fate depends on its position relative to the horizon. A photon strictly at or outside the event horizon can escape, while one inside cannot. The event horizon acts as a null hypersurface, meaning a photon on the horizon can only remain there and cannot escape. The discussion highlights the complexities of photon behavior in extreme gravitational fields. Understanding these dynamics is crucial for comprehending black hole physics.
Giulio B.
when the photon is born on the horizon...bang?

2 electrons are traveling around a black hole, after some time they meet themselves exactly at the horizon of the hole, generating a photon.

if for fate the particle at the moment of be born has a perpendicular speed (opposed relative the center of the black hole), what happends to the photon?

i'm not sure it falls down neither escape, if we suppose it to be at the matematical confine of the horizon , but it must anyway maintain his c speed...:bugeye:
 
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A photon strictly at or outside the event horizon can escape; a photon inside the event horizon cannot.

- Warren
 
chroot said:
A photon strictly at or outside the event horizon can escape

No, the event horizon is a null hypersurface, so, if a photon is on the horizon, the best that it can do (in terms of "escaping") is to stay on the horizon.

Regards,
George
 
In an inertial frame of reference (IFR), there are two fixed points, A and B, which share an entangled state $$ \frac{1}{\sqrt{2}}(|0>_A|1>_B+|1>_A|0>_B) $$ At point A, a measurement is made. The state then collapses to $$ |a>_A|b>_B, \{a,b\}=\{0,1\} $$ We assume that A has the state ##|a>_A## and B has ##|b>_B## simultaneously, i.e., when their synchronized clocks both read time T However, in other inertial frames, due to the relativity of simultaneity, the moment when B has ##|b>_B##...

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