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

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?

• Newtonian mechanics has no concept of black holes, yet alone photons following curved paths in gravitational fields. – StephenG Jul 15 '17 at 4:04
• @StephenG: in Newtonian mechanics the mass of an object doesn't affect its gravitational deflection, so photons are deflected by gravity in the same was as any massive object would be if travelling at $c$. – John Rennie Jul 15 '17 at 4:55
• My friend derived the value for the radius of the photon sphere using Newtonian mechanics rather than General Relativity and as far as I can remember he used normal circular orbital dynamics . The derivation yielded 2 values of radius one for the radius of the photon sphere and another being half the Schwarzschild radius. Would that make any sense as its derived using Newtonian Mechanics? – Munj Patel Jul 15 '17 at 6:05
• @john-rennie To apply Newtonian gravity in this way just isn't sensible. Newtonian gravity says there is no force on a zero mass object. And to apply the deflection formula requires you to cancel out a fictitious photon mass (which still has to be zero). I don't consider that a reasonable thing to do in this context. The fact the result is wrong tends to underline this. – StephenG Jul 15 '17 at 6:11
• @StephenG: see this answer of mine – John Rennie Jul 15 '17 at 6:13

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

• Hi, welcome to Physics SE! Please see how to format your equations using MathJax. – Nihar Karve Jan 23 at 7:17

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.

• This is mixing and matching GR and Newton, which is not what the question asks. It's not without interest, but it's not what was asked for. And at a basic level Newtonian gravity says there is no force between a massless photon and a massive object, so there can be no deflection predicted. My view is the rest is just playing with mathematics. I'd draw an analogy between this and the accidental "prediction" of the Schwarzschild radius using Newtonian gravity. – StephenG Jul 15 '17 at 6:27
• @StephenG: see this paper on the Arxiv. Newtonian gravity does predict gravitational deflection of light. This happens because the ratio $F/m$ remains sensible in the limit of $m \rightarrow 0$. – John Rennie Jul 15 '17 at 6:45
• @john-rennie If we apply Newtonian mechanics by considering a particle of mass 'm' thrown with an initial velocity 'c' m/s from inside the horizon , applying conservation of energy from that point to the maximum height reached ( here it's equal to the radius of the Black Hole ) yields the value for the radius. Only there it's assumed that the escape velocity on the Horizon is greater than the speed of light . Solving the value for radius yields 2 different escape velocities with which one value yeilds the radius of the photon sphere and the other half the Schwarzschild radius. Is it correct? – Munj Patel Jul 15 '17 at 6:56
• @MunjPatel: no it isn't. – John Rennie Jul 15 '17 at 7:00
• @MunjPatel: you don't give the details of the calculation, but Newtonian mechanics does not apply to a black hole so any calculation using it is invalid. However the $r$ coordinate in the Schwarzschild metric is defined in such a way that Newtonian calculations sometimes accidentally give the correct answer. See for example Deriving a Schwarzschild radius using relativistic mass. – John Rennie Jul 15 '17 at 7:25