I think that is how astronomers hunt for these elusive objects using radio telescope and also gravitational lensing, I'm wondering how these photons lose their energy while encountering black hole or some massive objects and I remember energy cannot be created or destroy so what gives?
Asked
Active
Viewed 388 times
1 Answers
-4
When a photon escapes a black hole, why is it red shifted?
It isn't actually red-shifted. There is no magical mechanism by which the ascending photon loses energy. Gravitational redshift is where the photon is emitted at a lower energy. See Einstein talking about it here:
"An atom absorbs or emits light at a frequency which is dependent on the potential of the gravitational field in which it is situated".
If you have two identical atoms at two different altitudes, you will measure the light from the lower atom to have a lower frequency. But the frequency didn't reduce as the photon ascended, the frequency was already lower when it was emitted.
I think that is how astronomers hunt for these elusive objects using radio telescope and also gravitational lensing, I'm wondering how these photons lose their energy while encountering black hole or some massive objects and I remember energy cannot be created or destroyed so what gives?
Energy cannot be created or destroyed, and ascending photons don't lose any energy. In similar vein descending photons don't gain any energy. If you send a 511keV photon into a black hole, the black hole mass increase is 511Kev/c². When you fall down to a lower altitude, some of your mass-energy is converted into kinetic energy, which is typically dissipated. You're then left with a mass deficit. So you've lost energy. So it looks like the photon has gained it. You measure the photon frequency to be higher because you and your clocks go slower when you're lower, but the photon freqency hasn't really changed, you have. It's similar for a photon in space. If you accelerate towards it, it doesn't change, you do.
John Duffield
- 11,381