Can Photon Have Orbital Angular Momentum?

[Physics4Funn]

TL;DR

If a photon could orbit, would it have orbital angular momentum?

This is a very special case.
In my 50 years studying physics I have never seen any discussion of photons having orbital angular momentum. Any angular momentum for photons in orbit around a black hole must be a GR question. I have not specialized in GR but I don’t recall any discussion of it.
I don't see how to put this into the energy-momentum tensor.
Can anyone help answer this question? Thanks
https://www.quora.com/Is-it-wrong-t...286203624&comment_id=123013387&comment_type=2

 

[PeroK]

Summary:: If a photon could orbit, would it have orbital angular momentum?

This is a very special case.
In my 50 years studying physics I have never seen any discussion of photons having orbital angular momentum. Any angular momentum for photons in orbit around a black hole must be a GR question. I have not specialized in GR but I don’t recall any discussion of it.
I don't see how to put this into the energy-momentum tensor.
Can anyone help answer this question? Thanks
https://www.quora.com/Is-it-wrong-t...286203624&comment_id=123013387&comment_type=2

Angular momentum does not require orbital motion. A particle moving in a straight line has angular momentum about any point not on its path. You can view conservation of linear momentum as a special case of conservation of angular momentum.

 

[PeterDonis]

Any angular momentum for photons in orbit around a black hole must be a GR question.

Not just in a closed orbit about a black hole; a photon flying by the Sun and having its path bent by the Sun's gravity has orbital angular momentum about the Sun (as would, say, an ordinary object like a comet flying in from interstellar space and flying back out again on an escape trajectory). And yes, GR handles this just fine. See below.

I don't see how to put this into the energy-momentum tensor.

It isn't in the stress-energy tensor; orbital angular momentum is not a property of a source of gravity (which is what the stress-energy tensor describes); it's a property of a test body on a trajectory in the spacetime around a source of gravity. More specifically, it is a constant of geodesic motion associated with rotational symmetry of the spacetime.

 

[Physics4Funn]

Angular momentum does not require orbital motion. A particle moving in a straight line has angular momentum about any point not on its path. You can view conservation of linear momentum as a special case of conservation of angular momentum.

Yes, for a massive particle. OK, so just use L = r x p for a photon and use the momentum of the photon. Thank you. Good answer.

 

[Physics4Funn]

Not just in a closed orbit about a black hole; a photon flying by the Sun and having its path bent by the Sun's gravity has orbital angular momentum about the Sun (as would, say, an ordinary object like a comet flying in from interstellar space and flying back out again on an escape trajectory). And yes, GR handles this just fine. See below.
It isn't in the stress-energy tensor; orbital angular momentum is not a property of a source of gravity (which is what the stress-energy tensor describes); it's a property of a test body on a trajectory in the spacetime around a source of gravity. More specifically, it is a constant of geodesic motion associated with rotational symmetry of the spacetime.

Yes. you are correct. How dense of me. Thank you.

 

[PeroK]

Yes, for a massive particle. OK, so just use L = r x p

Why not for a photon?

 

[Physics4Funn]

yes. thank you.

 

[pervect]

Once one decides that photons have linear momentum, in flat space-time the angular momentum about any point can be calculated from it's linear momentum. In 3-vector notation, it's $r \times p$. In four vector notation, angular momentum is a bi-vector rather than a vector, and it's given by $r$ ^ $p$ where "^" is the wedge product. In 4 dimensions, there is no cross product as there is in 3 dimensions. There are versions for the angular momentum density in flat space-time for distributed systems, too, that can be derived from the momentum-energy density, i.e. one can derive the angular momentum density about a specified point giventhe stress-energy tensor.

Curved space-time is trickier. At the most basic level, there are conserved quantites for particles (including 'photons') in the Schwarzschild metric following geodesics that are called angular momentum. MTW, for instance, talks about the orbits of 'photons' in the Schwarzschild geometry. I don't believe the general issues with angular momentum in curved space-time are fundamentally different than the issues with linear momentum in curved space-time, but I could be wrong about that.

 

[Physics4Funn]

Yes, that makes sense. Thank you.

 

[vanhees71]

A photon is a single-quantum Fock state of the electromagnetic field. You can expand the electromagnetic field in terms of vector spherical harmonics, the socalled multipole expansion. This provides a basis of single-photon Fock states which are simultaneous eigenstates or the photon's energy, $J$ and $M$, where $J \in \{1,2,\ldots \}$ is the angular-momentum quantum number, i.e., the eigenvalue of $\hat{\vec{J}}^2$ is $J(J+1) \hbar^2$ and $M$ provides the eigenvalue of $\hat{J}_3$, $M \hbar$.

Note that I talk only about total angular momentum $\vec{J}$, which is the only physical angular momentum which can be uniquely and gauge-invariantly defined for the electromagnetic field. Sometimes one expands the total angular momentum in an "orbital angular momentum" and a "spin angular momentum", which is however gauge dependent and has to be taken with some care. Nevertheless one talks about "photons with higher orbital angular momentum". It's just jargon, and you have to be careful concerning the physical interpretation. You are save when referring only to the total angular momentum and/or helicity.