Why is black hole photon sphere outside the event horizon?

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SUMMARY

The discussion centers on the concept of the photon sphere of a black hole, which is located 1.5 times the distance from the singularity compared to the event horizon. The minimum velocity required for a stable orbit is defined by the equation v=sqrt((2GM)/r), while escape velocity is given by v=sqrt((GM)/r). Participants clarified that the photon sphere is indeed the minimum orbital distance, and confusion arose regarding its position relative to the event horizon. Reliable resources for further exploration of these concepts include specific online physics sources that provide equations relevant to orbits in curved spacetime.

PREREQUISITES
  • Understanding of general relativity and black hole physics
  • Familiarity with gravitational equations, specifically escape velocity and orbital velocity
  • Knowledge of Schwarzschild spacetime metrics
  • Basic mathematical skills for manipulating equations
NEXT STEPS
  • Study the derivation of the Schwarzschild metric in general relativity
  • Learn about the implications of the photon sphere in astrophysics
  • Explore the concept of stable orbits in curved spacetime
  • Review online resources for advanced physics equations, such as those found on Fourmilab and university physics papers
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Astronomy students, physicists, and researchers interested in black hole dynamics and gravitational physics will benefit from this discussion.

Prof. Hawking
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Homework Statement


I am preparing a report on black holes and I recently learned about a phenomenon I was previously unaware of: the photon sphere of a black hole. While reading an article on said occurrence (I have now confirmed this on multiple sources) the photon sphere which is the minimum distance anything (in this case light) can orbit from a black hole and the photon sphere is the lower bound where only something going at the speed of light (photons) can orbit.

However the distance from the singularity for this boundary is 1.5 times as far as the event horizon is from the center of the black hole. From what I understand the equation for the velocity required to orbit an orbit an object stabely is v=sqrt((2GM)/r) where G is the gravitational constant M is the mass of the object being orbited and r is the distance from the center of mass of the object ( I found this equation on physicsforums here https://www.physicsforums.com/threa...-speed-for-putting-an-object-in-orbit.299929/). However the formula for escape velocity (the event horizon in the case of a black hole) is v=sqrt((GM)/r).

These formulas would show that to escape from the gravity of a body takes much more energy than to orbit it from the same distance however the speed of light is not enough to orbit outside the minimum escapable distance at the speed of light if this distance is inside the photon sphere so I am rather confused.

Homework Equations


what I found to be the formula for minimum velocity to be in a stable orbit v=sqrt((2GM)/r)
escape velocity v=sqrt((GM)/r)

The Attempt at a Solution

 
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These formulas would show that to escape from the gravity of a body takes much more energy than to orbit it from the same distance...
It is easier to say at a set distance from a body by moving around it than it is to get away completely, just like it is often easier to walk around a solitary mountain than to climb it.

... however the speed of light is not enough to orbit outside the minimum escapable distance at the speed of light if this distance is inside the photon sphere so I am rather confused.
But didn't you just finish explaining that the photon sphere is at the minimum orbit distance?
Therefore the answer is that it is a logical contradiction in the sentence.
 
Simon Bridge said:
It is easier to say at a set distance from a body by moving around it than it is to get away completely, just like it is often easier to walk around a solitary mountain than to climb it.

But didn't you just finish explaining that the photon sphere is at the minimum orbit distance?
Therefore the answer is that it is a logical contradiction in the sentence.
What I don't understand is why that minimum orbital distance is farther away than the minimum escapable distance. It seems like the photon sphere would be inside the event horizon because you don't need to be as far away where the gravity is weaker because it takes less energy to orbit

Also to your second comment I was talking about a photon being unable to orbit between the event horizon and the photon sphere (sorry for confusion)
 
Prof. Hawking said:
what I found to be the formula for minimum velocity to be in a stable orbit v=sqrt((2GM)/r)
escape velocity v=sqrt((GM)/r)

You've essentially got these the wrong way round. Escape velocity is v_e=\sqrt(2Gm/r) and the Newtonian equation for stable orbit is v_s=\sqrt(Gm/r) though in GR in Schwarzschild spacetime (i.e. static black hole), the equation for a stable orbit is-

v_s=\sqrt{\frac{Gm}{r\left(1-\frac{2Gm}{c^2r}\right)}}
 
stevebd1 said:
You've essentially got these the wrong way round. Escape velocity is v_e=\sqrt(2Gm/r) and the Newtonian equation for stable orbit is v_s=\sqrt(Gm/r) though in GR in Schwarzschild spacetime (i.e. static black hole), the equation for a stable orbit is-

v_s=\sqrt{\frac{Gm}{r\left(1-\frac{2Gm}{c^2r}\right)}}
Thank you Steve I realize now that I had the formulas backwards so now this makes perfect sense, also thanks for the more applicable formula for stable orbit,

where did you find this formula as I didn't come across it in any of my googling, is there a reliable online source where I could easily access all these types of less common physics formulas because that would be very helpful in the future
 

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