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

• Prof. Hawking
In summary: I think)In summary, the photon sphere 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
Prof. Hawking

## 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

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

#### Attachments

• 9783642395956-c2.pdf
789.5 KB · Views: 343

## 1. Why is the photon sphere outside the event horizon?

The photon sphere is a region around a black hole where photons can orbit the black hole in a circular path. It is located just outside the event horizon, which is the point of no return for anything that enters the black hole. This is because the intense gravitational pull of the black hole causes the fabric of space-time to curve, creating a circular path for photons to orbit.

## 2. Can anything escape from the photon sphere?

While photons can orbit the black hole at the photon sphere, they cannot escape from it. This is because the gravitational pull at the photon sphere is so strong that the escape velocity, the minimum speed required to escape the gravitational pull, is equal to the speed of light. Since photons travel at the speed of light, they do not have enough velocity to escape the photon sphere.

## 3. Is the photon sphere the same for all black holes?

No, the size of the photon sphere depends on the size and mass of the black hole. Smaller black holes have smaller photon spheres, while larger black holes have larger photon spheres. Additionally, the spin and charge of a black hole can also affect the size and shape of the photon sphere.

## 4. What happens to objects that enter the photon sphere?

If an object enters the photon sphere, it will be pulled towards the black hole and eventually cross the event horizon. Once inside the event horizon, the object cannot escape from the black hole. However, if the object has enough velocity, it can escape the gravitational pull of the black hole and not cross the event horizon.

## 5. How is the photon sphere related to the concept of an "observable universe"?

The photon sphere is not directly related to the observable universe. The observable universe is the portion of the universe that we can see and observe, while the photon sphere is a specific region around a black hole. However, the concept of an observable universe can be applied to the photon sphere in the sense that anything within the photon sphere cannot be observed from the outside since even light cannot escape from it.

Replies
57
Views
2K
Replies
51
Views
2K
Replies
7
Views
331
Replies
3
Views
1K
Replies
7
Views
464
Replies
6
Views
2K
Replies
21
Views
2K
Replies
11
Views
1K
Replies
1
Views
1K
Replies
5
Views
824