Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Can a photon through some process be made to orbit a celestial or any other object?

Two follow-up questions.

  1. Can this orbit be described as the photon crossing its own path.

  2. Will this wave-function be effected by positive interference. To the effect of increasing frequency?

share|improve this question
    
The first follow up is weird--- the photon orbits in a finite time, so it is never intersecting it's past. The last question is also weird, there can be interference between the orbiting "wavefunction" (just electromagnetic field in this case, don't think of wavefunctions for a photon), but the frequency isn't changing by interference, it is just interfering. –  Ron Maimon Jun 27 '12 at 4:15
    
The prevalent theory is wave-particle-duality. Wave and particle at the same time this means you can not just use the description that suits your method that is not positivism. –  Argus Jun 27 '12 at 4:39
    
@RonMaimon: as you previously stated inside a blackhole it is space-like so if the particle has momentum and no mass it is at every point in its radial time-like orbit at the same time. –  Argus Jun 27 '12 at 4:42
    
The reason I say don't use particles is because it is confusing not because it is incorrect. The wavefunction of a photon is not so simple to describe, it is not like the wavefunction of a nonrelativistic electron. The orbit is outside the black hole, the spacelike business of the t coordinate is inside the black hole. Even in the interior, there aren't closed loops in time. –  Ron Maimon Jun 27 '12 at 6:16
    
Closed loops in time is an opinion LQG might disagree. OK can a photon obsorb itself? –  Argus Jun 27 '12 at 6:19
add comment

3 Answers 3

up vote 3 down vote accepted

In short, yes. But there are 2 caveats for the orbit for a massless particle around a spherical body. Both can be seen from the following plot, borrowed form "Spacetime and Geometry" by Carroll pg.211:

enter image description here

i) There is only one possible orbit for $r = 3 GM = \frac{3 R_{s}}{2}$ where $R_s$ is the Schwarzschild radius. This orbital radius could very well be within the radius of the body being orbited, hence the particle may not even be able to get this trajectory in the first place.

ii) The orbit is unstable.

EDIT: As for the updated, additional questions:

First of all, the above analysis was done for a classical particle, so there is no notion of waves or interference there. In this sense, the photon will cross its own path the same way my dog crosses his own path when he runs in a circle.

If you really want to treat the photon quantum mechanically, I can only take a stab at it under some idealized circumstances. In one idealization we could take the photon state is very localized (high probability of detecting it in small range, virtually zero everywhere else). We can send this photon once around in the orbit - in this case it will act a lot like a classical particle in that it won't interfere at all with the 'tail' of the wavefunction from the prior pass. (That we can set up states like this is merely an intuitive notion on my part since I know we can set up single photon states and there is little chance of detecting them at say, Alpha Centauri, so localized states of a single photon seem to be possible). Now for the other idealization, we could take a state that isn't very localized at all, say a plane wave directed tangentially to the orbit. Now that the wave function isn't localized it's going to be able to 'feel' all around the space and I think its going to want to 'fall off' the trajectory where it can be at a state of lower energy. That is, over a long period of time its going to sense the instability and have a very small probability of being detected at along the radius at $ 3 G M$ rendering the question sort of moot. One thing you could do is stand on one side of the body being orbited and shoot a photon towards it and have your buddy on the other side detect it, which is pretty much the double slit experiment and will result in interference (the 2 paths are going left or right of the body being orbited). (I am more posting this part of my answer as a guess to see what other people think and less out of certainty).

I want to emphasize point (ii) above - something that is unstable is a bit like having a needle balanced on its tip - sure you can do it for a split second but the slightest movement will make it fall over so really no matter how many special circumstances you want to invent or questions you want to ask about the photon orbiting its just not going to stay there very long.

share|improve this answer
    
Really good about point two since this double slit like path intersects itself even if only once can this photon absorb and re-emit itself. Got the "huge" stretch in this "perfect" scenario assuming we use the half orbit of the quantized wave and the photon re-emits itself into a "perfect" orbit is there a limit to the attainable frequency (takeing into account this is extremely extremely theoretical) slowly building a picture in my head thanks to the very intuitive answers –  Argus Jun 27 '12 at 16:08
add comment

The short answer is yes, in the vicinity of the event horizon of a black hole, there are photon orbits. For the static black hole, this orbit is the boundary between stable and unstable orbits at $r = 3M$. For stationary black holes, it's more complicated.

UPDATE:

With regards to the follow-up questions:

(1) Photons follow null geodesics of the spacetime. The photon orbit at $r = 3M$ for a static hole is a closed helical null geodesic but, as another answer provides, the orbit is unstable.

(2) Do photons have wave-functions?

share|improve this answer
    
Can this orbit be so explained as the photon intersects its own path? –  Argus Jun 27 '12 at 1:56
    
The explanation of the orbit is in the geometry of the spacetime in the vicinity of the hole and the explanation of the geometry is in the equations of GTR. –  Alfred Centauri Jun 27 '12 at 2:07
    
A propagating em field does as this is the classification of a photon I assumed one implies the other –  Argus Jun 27 '12 at 3:04
    
Let us find out. –  Argus Jun 27 '12 at 3:05
1  
A photon is as relativistic as any particle can be; it is intrinsically relativistic since, in Newtonian mechanics, massless particles simply don't exist (no momentum, no inertia, no energy, etc.). So, I'm just not quite understanding what your grasping to ask and understand here. –  Alfred Centauri Jun 27 '12 at 3:28
show 3 more comments

Photons are electrons with a positive charge and can be influenced by magnetism which can push or pull them in a certain direction. To have affect of creating an orbit, the mass's diameter has to be greater than the distance the photon travels per second, and the orbit will be the diameter of the mass + distance between the photon and the mass x 2 x 3.14159.

share|improve this answer
2  
According to the standard model Electrons have mass, photons do not. Photons do not carry a charge. Maybe you are thinking of a positron? –  RedGrittyBrick Oct 5 '12 at 11:02
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.