An alternative derivation to find the radius of the Photon Sphere of a Black Hole

Question

I know that the radius of the photon sphere of a Black Hole is derived by solving the Einstein field equations and it represents the smallest gravitational stable orbit around the singularity. My question is that is there any way that I can derive the value for the radius of the photon sphere using Newtonian Mechanics rather than solving the Einstein field equations?

Answer 1

The simple answer is that no Newtonian mechanics cannot predict the location of the photon sphere. In Newtonian mechanics we calculate the orbital velocity by equating the gravitational acceleration to the centripetal acceleration i.e.

$$ \frac{GM}{r^2} = \frac{v^2}{r} $$

And that gives us the Newtonian orbital speed as:

$$ v = \sqrt{\frac{GM}{r}} $$

Putting $v=c$ to locate the photon sphere then gives:

$$ r_\text{photon} = \frac{GM}{c^2} = \tfrac{1}{2} r_s $$

which is wrong.

However we can use Newtonian mechanics with a minor correction. In the corotating frame of the object in orbit the object is accelerated inwards by the gravitational force, and it has a fictitious centrifugal force pushing it outwards. So there is a net force (I'll write this as a force per unit mass to keep the mass of the object out of the equations):

$$ F_\text{net} = \frac{GM}{r^2} - r\omega^2 $$

And if we integrate the force with respect to $r$ we get an effective potential.

$$ V_{eff}(r) = -\frac{GM}{r} + \frac{L^2}{2r^2} \tag{1} $$

where $L$ is the angular momentum, $L = I\omega = mr^2\omega = mrv$. An object in a stable circular orbit sits at the minimum of this potential. I won't do the calculation, but if you find the minimum of the effective potential you recover the Newtonian expression for the orbital speed.

What general relativity does is modify this effective potential. In fact we get different effective potentials for photons and for massive objects, which is why photons and massive objects have different closest stable orbits at $\frac{3}{2}r_s$ and $3r_s$ respectively. The effective potential for photons is:

$$ V_{eff}(r) = \frac{L^2}{2r^2} - \frac{GML^2}{c^2r^3} \tag{2} $$

You have to use GR to get this correction to the effective potential, but assuming you're willing to accept this then the rest is just the usual Newtonian mechanics approach. We differentiate equation (2) and set it equal to zero to find the extremum, and this gives:

$$ -\frac{2L^2}{2r^3} + \frac{3GML^2}{c^2r^4} = 0 $$

And rearranging gives:

$$ r = \frac{3GM}{c^2} = \tfrac{3}{2} r_s $$

which is the correct result.

Answer 2

Centrifugal force of mass m at radius r Fc = m ω2 r = m ω2 r2/r Velocity v v = ω r , where v is virtual velocity Fc = m v2 / r (1) Gravity g at r for g = G Mbh / r2 (2) Gravity force on mass m at r for mass = Mbh Fg = G Mbh / r2 At radius r Fc = Fg Due to space dilation Ld at r, virtual velocity v = C/Ld equating (1) and (2) GM/r^2 = ω2 r = (v r)^2/r = (C/Ld)^2*r^2/r = C^2 r/Ld^2 After manipulation of sides 2GM/C^2 = 2 r/Ld^2 But 2GM/C^2 = Rsh Rsh =2 r/Ld^2 (3) Space dilation Ld Ld=1/(1-Rsh/r)^0.5 (4) Substituting (4) Ld in (3) we get Rsh = 2 r(1-Rsh/r) Rsh = 2 r – 2 Rsh r = 3 Rsh/2 r =1.5 Rsh

QED