# Why is black hole photon sphere outside the event horizon?

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1. Jul 8, 2015

### Prof. Hawking

1. The problem statement, all variables and given/known data
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.

2. Relevant 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)

3. The attempt at a solution

2. Jul 9, 2015

### Simon Bridge

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.

3. Jul 9, 2015

### Prof. Hawking

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)

4. Jul 9, 2015

### stevebd1

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-

$$v_s=\sqrt{\frac{Gm}{r\left(1-\frac{2Gm}{c^2r}\right)}}$$

5. Jul 10, 2015

### Prof. Hawking

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

6. Jul 12, 2015

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