# Photon in a circular orbit

1. Oct 27, 2012

### lailola

Let be a photon in a circular orbit (r=3M). I want to know the period measured by a stationary observer at this radius.

Because we are working with a photon we cannot 'talk' about proper time, that's why I don't understand how can I calculate this.

If I had a massive particle I could do it (just finding an expression for $\frac{d\phi}{d\tau}$ and then calculating $\Delta\tau$ for $\Delta\phi=2\pi$.

Thanks for any help

2. Oct 27, 2012

### PAllen

A stationary observer had constant r. You have to specify which stationary observer you mean. Suppose one at the same r as the light is orbiting. Then, if you can figure out that a the light path will go through (r,0) and (r,t1), then just integrate proper time along t for this r. This gives you exactly what a stationary observer will measure on their watch for one period of the light.

3. Oct 27, 2012

### lailola

But there's no proper time. We have a photon, instead of proper time we have to use an affine parameter λ, for example.

4. Oct 27, 2012

### PAllen

You are computing proper time along the stationary observer's world line, not along the photon path. Actually, you don't need to integrate since gtt doesn't depend on t. You just figure t coordinate difference for one period of the light path, and multiply by √gtt(r), for the given r. This will give you proper time for the orbit measured by a stationary observer (for whom r, theta, phi are fixed).

5. Oct 27, 2012

### lightarrow

You think it's not correct to divide the circular path in n straight segments of lenght h = 2pi r/n and then sending n to infinite?

6. Oct 27, 2012

### PAllen

Do you know how to find theta(t) for the photon orbit, where t is coordinate time? If so, the rest is trivial. If not, then ask about that.

7. Oct 27, 2012

### lailola

I think you mean: $\frac{dt}{d\tau}=g_{tt}^{-1/2}\rightarrow \Delta\tau=\Delta t (1-2M/r)$ where t is the time measured by an observer at infinity. And I get that expression from the expression of the Schwarzschild metric.

But in this case what I have is (see image). That's why I cannot see how I could get the upper expression for photons.

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Last edited: Oct 27, 2012
8. Oct 27, 2012

### PAllen

One issue is the orbit. If you take as given that you have the right radius for a circular photon orbit (if you want to see the derivation of that, ask about that), then the path of the light is given by just setting proper interval differential to zero:

0 = gtt c^ dt^2 - r^2 d$\theta$^2

From this you can trivially get d$\theta$/dt. Then 2$\pi$ / <angular speed> is coordinate time for the period. Then proper time along static observer's world line is just √gtt times this amount.

9. Oct 27, 2012

### lailola

Ok, I think now I understand you. Setting d$r/d\lambda$ in the equation in the figure I can get the expression that you've writen. Then, to calculate the proper time measured by the observer I use the Schw. metric again, but this time d$\tau$ is not zero cause, as you said before, now we are considerating the observator's world line. As the observator is stationary all terms are zero but dt and I can calculate the proper time as a function of t. Is it right?

Last edited: Oct 27, 2012
10. Oct 27, 2012

### PAllen

You got it.

11. Oct 27, 2012

### A.T.

Can't you just use the fact that stationary observers at r=3M will measure the local speed of the orbiting photon as c (or 1), according to their proper times? So the proper time they measure would be simply 6πM?

12. Oct 27, 2012

### PAllen

This isn't strictly a local measurement of lightspeed (it is a complete orbit). Stationary observers near an EH do not necessarily measure c for light speed over some distance (this has been discussed in a number of threads here, a year or so ago). Note that stationary observers are not inertial observers.

Meanwhile, the method I outlines seems quite trivial to me. It then arrives at 6 pi M, which is not a trivial fact. I guess you could justify by an argument like the light remains all at the same potential, as does the observer, so the speed must be the same a local speed of light, over the whole orbit. But then all of this needs to be justified - versus a small amount of algebra to get the result directly from the metric.

Last edited: Oct 27, 2012
13. Oct 29, 2012

### lightarrow

It's not a SR question? Then where is the issue?

14. Oct 29, 2012

### PAllen

It is obviously not an SR issue. Photons don't orbit BH in SR. The method I describe will work for an elliptical photon orbit, and leads to the understanding of how the circular case simplifies due to the fixed radius.

Note that for a finite chord segment, you must integrate the metric to get proper distance; then you would find that light would not follow that path anyway in this metric. So your literally described procedure [post #5] cannot be carried out in a valid way.

Last edited: Oct 29, 2012
15. Oct 29, 2012

### pervect

Staff Emeritus
I was just going to point that out to you, but you already know it. So, you calculate the photon trajectories using the affine parametrization using the geodesic equations (or more simply by using the metric and setting the interval to zero). The period of the orbit is the length of the timelike worldline of the static observer, not the length of the worldline of the photon, which would be zero.

16. Oct 31, 2012

### lightarrow

The OP didn't write about photons orbiting around a BH, if he intended that I couldn't know it...

17. Oct 31, 2012

### jnorman

i was taught that free photons have no specific location, and thus, would have no identifiable orbital parameters. i do not understand the discussion here. can someone please clarify what i am missing here? thanks.

18. Oct 31, 2012

### PAllen

This is a pure classical treatment. Photon is a stand in for 'massless particle following a null path'. This is done all the time in SR and GR. It really would be a small pulse of light treated classically. However, the difference is immaterial unless the wavelength corresponding to the photon energy is significant compared to curvature. In that case, I am not sure there is any full theory - just a bunch semi-classical heuristics.

19. Oct 31, 2012

### PAllen

Well, if you had a little GR background you would know immediately. The OP said r=3M. Well, 3M is the radius at which light orbits a spherically symmetric BH of mass M. Then, everything between OP and me made clear that's what he/she was talking about (use of geodesic equation, discussion of metric coefficients, etc.).

20. Oct 31, 2012

### Dickfore

Sure there is, Maxwell's equations in curved spacetime. In them, you use the covariant derivative.
$$F_{\mu \nu} \equiv D_{\mu} A_{\nu} - D_{\nu} A_{\mu} = \partial_{\mu} A_{\nu} - \partial_{\nu} A_{\mu}$$
In the absense of external currents and charges, the Maxwell's equations read:
$$D_{\nu} F^{\mu \nu} = 0$$
This is nontrivial as it contains the Christtoffell symbols

Last edited: Oct 31, 2012
21. Oct 31, 2012

### PAllen

The question asked and my response concerned quantum behavior of photons. What does the above have to do with it? [My phrase: the difference is imaterial - referred to classical vs quantum treatment] I thought it obvious I was referring to the case where the wavelength is comparable to analogs of curvature radius, so flat space QFT would break down. I am not aware any current approach used in this realm is considered theoretically sound or complete.

22. Oct 31, 2012

### lightarrow

Sorry, I didn't know it, indeed (I thought 3M was 3 metres ).
Anyway, "spherically symmetric" means also "non-rotating"? If it rotates, the Lense-Thirring effect gives different values for light speed in the two tangent directions, as measured from a stationary observer at that radius?

23. Oct 31, 2012

### PAllen

A rotating BH cannot be spherically symmetric in the technical sense. It is axially symmetric.

24. Oct 31, 2012

### Dickfore

If the wavelength is comparable to the curvature, then this is the low wavelength limit if I'm not mistaken. In it, quantum effects are negligible. What becomes important is the scattering and interference of wavefront from the inhomogeneities of the gravitational field. This is different from the geometric optics limit (which is formally valid in the zero wavelength limit), when the em wave travels along a trajectory a massless particle would take.

25. Oct 31, 2012

### PAllen

Ah, you are right, as I phrased it. I was trying to come up with a simple description of when, near a black hole, a classical treatment (whether using curved space EM, or geometrical optics) would be significantly inaccurate. I blew it trying to be too general and simple..