# Photon "escaping" from photon sphere in Schwarzchild space

• N00813
In summary, we can use the equations of motion and the condition for escape to infinity to find the angle at which a photon can escape to infinity from a general r between 2(mu) and 3(mu) near a Schwarzschild black hole.
N00813

## Homework Statement

Close to a Schwarzschild black hole, a photon is emitted between r = 2(mu) and 3(mu), where $\mu = \frac{GM}{c^2}$. The photon is emitted at an angle (alpha) to the radial direction. At r = 2(mu), the highest angle that the photon can escape at is (alpha) = 0; at r = 3(mu), the highest angle is (pi/2). For a general r between 2(mu) and 3(mu), what is the highest angle (alpha) in which the photon can escape to infinity?

## Homework Equations

Eqns of motion (EQM):
$(1-\frac{2\mu}{r}) \frac{dt}{d\sigma} = k$

$c^2 (1-\frac{2\mu}{r}) (\frac{dt}{d\sigma})^2 - (1-\frac{2\mu}{r})^(-1) (\frac{dr}{d\sigma})^2 -r^2 ((\frac{d\phi}{d\sigma})^2 =0$
$r^2 (\frac{d\phi}{d\sigma}) =h$

Impact parameter:
$b = \frac{h}{ck} = r sin (\phi)$

## The Attempt at a Solution

I guessed that the condition to escape was that $\frac{dr}{d\phi} =0$ at r = 3 mu to give b, and then using that to give (phi), but that didn't get the right result for r = 2(mu).
I think that I don't really know what the condition for escape to infinity is.

The condition for escape to infinity is that the photon's energy (or equivalently, its frequency) at infinity is greater than or equal to its energy at the emission point. This can be expressed mathematically as:

\frac{E_{\infty}}{E_{emission}} \geq 1

Using the equations of motion, we can express the photon's energy at infinity as:

E_{\infty} = c^2 (\frac{dt}{d\sigma})_{\infty}

where (\frac{dt}{d\sigma})_{\infty} is the time component of the photon's 4-momentum at infinity. Similarly, the energy at the emission point can be expressed as:

E_{emission} = c^2 (\frac{dt}{d\sigma})_{emission}

Using the equation for (\frac{dt}{d\sigma}), we can rewrite this as:

E_{emission} = c^2 \frac{k}{1-\frac{2\mu}{r}}

Now, we can use the condition for escape to infinity to solve for (alpha), the angle at which the photon is emitted. We can set E_{\infty} = E_{emission} and solve for (\frac{dt}{d\sigma})_{emission}. Then, we can use the equation for (\frac{dt}{d\sigma}) to solve for (alpha).

I hope this helps!

## 1. What is a photon sphere in Schwarzchild space?

A photon sphere in Schwarzchild space is a theoretical region around a non-rotating black hole where photons (particles of light) can orbit the black hole in a circular path. This region is located at a distance of 1.5 times the Schwarzschild radius, which is the distance from the center of the black hole where the escape velocity equals the speed of light.

## 2. How does a photon "escape" from the photon sphere in Schwarzchild space?

In order for a photon to escape from the photon sphere, it must possess enough energy to overcome the gravitational pull of the black hole. This can happen if the photon is either emitted with enough energy or acquires enough energy through a process known as gravitational redshift.

## 3. Can photons orbit a black hole indefinitely on the photon sphere?

Yes, photons can theoretically orbit the black hole indefinitely on the photon sphere as long as they do not interact with any other particles or objects. However, due to the presence of other matter and energy in the surrounding space, it is unlikely for a photon to remain in a stable orbit for a long period of time.

## 4. What is the significance of the photon sphere in Schwarzchild space?

The photon sphere is significant because it represents the closest distance that a particle can orbit a black hole without being pulled into the event horizon. It also plays a role in the appearance of black holes, as photons from the photon sphere can form the bright rings seen around black holes in images.

## 5. Can anything other than photons escape from the photon sphere in Schwarzchild space?

No, only particles that travel at the speed of light, such as photons, can escape from the photon sphere. Anything with mass, including other particles like protons or electrons, would require infinite energy to escape the gravitational pull of the black hole within the photon sphere.

Replies
10
Views
1K
Replies
5
Views
1K
Replies
11
Views
2K
Replies
3
Views
1K
Replies
4
Views
418
Replies
42
Views
4K
Replies
5
Views
2K
Replies
1
Views
1K