Understanding the Four Momentum of a Photon in General Relativity

[thepopasmurf]

Hi, I'm trying to reconcile some general relativity stuff and I'm getting confused on a few topics.
Would someone be able to explain the properties of the four momentum of a photon? The way I understand it is that you take the four wave-vector of the photon and multiply it by $\hbar$. The four wave-vector is simply $\omega / c$ as the zeroth compent and the 3-wave-vector as the spatial components. Right?

Now onto the four-momentum properties, how does the four-momentum of a photon change as it travels through spacetime? Are any components constant?

Thanks, I'll probably have follow up questions but I'd like to start with these

 

[atyy]

What do you mean by photon? A massless classical particle or light ray or a classical electromagnetic wave or a quantized electromagnetic wave? What do you mean change as it travels through spacetime? For example, are you talking a about a plane wave or a general solution of Maxwell's equations?

 

[thepopasmurf]

Ah, let me clarify myself. By photon I mean any particle that has zero mass therefore travels at the speed of light. Basically I would like to know the properties of the four-momentum of massless particles.

 

[PAllen]

Ah, let me clarify myself. By photon I mean any particle that has zero mass therefore travels at the speed of light. Basically I would like to know the properties of the four-momentum of massless particles.

Well, its a null vector (norm is 0). Otherwise, it can be anything. Expressed in any orthonormal basis, time component will be energy, rest momentum and in appropriate units p=E/c.

Not sure what else you are asking.

 

[thepopasmurf]

Well, I have some notes that say that if the metric is independent of 't', then the time component is the same at emission and reception of the photon. Why is this? Is this true of the time component of any null vector? What happens to the spatial components?

 

[PAllen]

Well, I have some notes that say that if the metric is independent of 't', then the time component is the same at emission and reception of the photon. Why is this? Is this true of the time component of any null vector? What happens to the spatial components?

That can't be true as stated. The time component, among other things, depends on the tangent vector (4-velocity) of the receiver. In this formalism, that's how doppler is represented. Suppose, as a hypothesis, it is meant that if the metric is static then you are considering only static emitter and absorber. Then, it is still not true. Consider the Schwarzschild geometry: you have gravitational redshift between static observers, so both time and spatial components change between emitter and absorber (as expressed in each one's local orthonormal basis).

 

[thepopasmurf]

But for an observer at infinity, do the components remain the same? As in if we remain in the frame of an observer at infinity, do the time and spatial components of the four momentum change as the photon travels between emitter and receiver?

 

[PAllen]

But for an observer at infinity, do the components remain the same? As in if we remain in the frame of an observer at infinity, do the time and spatial components of the four momentum change as the photon travels between emitter and receiver?

Yes, an observer at infinity sees maximum redshift for a photon emitted at smaller r coordinate. Any null 4-momentum would have the same character (assuming the null 4-momentum undergoes parallel transport along a null worldline): both time and spatial components would be smaller expressed in the frame of the observer at infinity than in the frame of the emitter.

 

[Elroch]

thepopasmurf, you can't really simply think of components "remaining the same" as they are only defined in local co-ordinate frames which are far from unique. Even the frequency depends on which Lorentz frame you use.

You implicitly assumed you had a photon with a fixed energy and momentum, which would mean a plane wave in flat space-time. But in GR, you are in curved space time, so there is no such thing as a simple plane wave.

You need to define a wave function on curved space time. On this curved space time, you need to define operators for momentum and energy etc. These could be defined in a local frame and then extended to the curved space time using parallel transport. But I believe this extension may be path dependent, so it could be tricky. It's also worth noting single photons can require a large domain: a photon wave function can even pass on both sides of a black hole in gravitational lensing!

I am not very familiar with this stuff, but perhaps someone else can elaborate.

 

[thepopasmurf]

thepopasmurf, you can't really simply think of components "remaining the same" as they are only defined in local co-ordinate frames which are far from unique. Even the frequency depends on which Lorentz frame you use.

That's a good point. But if I remain in a single frame of reference then the four-momentum of the photon remains the same along its path right? PAllen's post above seems to indicate that this is so. For now I would like to simplify the photon to just being a point particle with frequency and zero mass.

 

[PAllen]

That's a good point. But if I remain in a single frame of reference then the four-momentum of the photon remains the same along its path right? PAllen's post above seems to indicate that this is so. For now I would like to simplify the photon to just being a point particle with frequency and zero mass.

No, I said it is not so, in any case you mention. I described how even for two static observers, one at infinity, it is not true.

However, looking at how you asked the question, my "Yes" should have been "no". But the rest of my answer (in #8) should have made the meaning clear. The following should have left no doubt: "both time and spatial components would be smaller expressed in the frame of the observer at infinity than in the frame of the emitter"

 

[George Jones]

thepopasmurf, for questions in your two threads on this, I think you should read Appendix 9A General approach to gravitational redshifts from General Relativity: An Introduction for Physicists, by Hobson, Efstathiou, and Lasenby.

 

[thepopasmurf]

Sorry I guess I misunderstood. Can I clarify a few points that you made?

In post #6 you say that

both time and spatial components change between emitter and absorber (as expressed in each one's local orthonormal basis)

.
Is this saying that both emitter and absorber see the components change independently of each other ,i.e. the emitter tracking the particle would see the components change along the particle's path?
Or
Are you saying that the momentum of the particle as measured by the emitter at emission is different from the momentum that the absorber measures at absorption.

I agree with the latter statement.

In post #8, can you clarify what you mean by character? I think I interpreted that to mean the same. Also is this post describing the case where the observer at infinity receives the photon because that is not what I meant. I'm using the observer at infinity to mean an independent set of coordinates to describe the situation.

Thanks for being patient with me, I've been having a touch time wrapping my head around some of these concepts.

@George Jones: Ok, I'll have a look for that, thanks.

 

[PAllen]

Sorry I guess I misunderstood. Can I clarify a few points that you made?

In post #6 you say that .
Is this saying that both emitter and absorber see the components change independently of each other ,i.e. the emitter tracking the particle would see the components change along the particle's path?
Or
Are you saying that the momentum of the particle as measured by the emitter at emission is different from the momentum that the absorber measures at absorption.

I agree with the latter statement.

The latter is what I meant. The former is not meaningful. You can't measure the frequency of a distant photon.

In post #8, can you clarify what you mean by character? I think I interpreted that to mean the same. Also is this post describing the case where the observer at infinity receives the photon because that is not what I meant. I'm using the observer at infinity to mean an independent set of coordinates to describe the situation.

I meant that the behavior of null 4-momentum vectors and null paths is not tied to light or photons. SR and GR would treat any massless particle the same way. Here is the complete sentence: "Any null 4-momentum would have the same character (assuming the null 4-momentum undergoes parallel transport along a null worldline): both time and spatial components would be smaller expressed in the frame of the observer at infinity than in the frame of the emitter". The part after the colon is the characteristic that would apply to any null 4-momentum vector.

I am only interested in observer as an instrument that takes measurements. Coordinate representations, by themselves, don't tell you anything. Only using them in conjunction with a metric (and on instrument or observer world line) to derive invariant quantities gives you observables in GR. I can design coordinates for a given spacetime geometry such that the components of a parallel transported null 4-momentum oscillate. This still wouldn't change physics, because the dot product with orthonormal basis vectors for an observer at some point on the path would not change - the metric representation in such fishy coordinates would preserve the invariance.

Thanks for being patient with me, I've been having a touch time wrapping my head around some of these concepts.

@George Jones: Ok, I'll have a look for that, thanks.