Is photon frequency intrinsic or observer-dependent?

[Roberto Pavani]

TL;DR

If frequency is just a projection $\nu = k^\mu u_\mu$, what distinguishes a radio photon from a gamma ray intrinsically?

In GR, the frequency of a photon is observer-dependent: $\nu = g_{\mu\nu} k^\mu u^\nu$, where $k^\mu$ is the photon 4-momentum and $u^\nu$ is the observer's 4-velocity.
Two observers in relative motion measure different frequencies from the same photon (gravitational/cosmological redshift, Doppler).
This means that "the frequency" is not (or not only) a property of the photon itself, but of the photon–observer pair.
My question: if we take this seriously, does it imply that all photons are in some sense "the same object" (null geodesics with $k^\mu k_\mu = 0$), and the only thing that distinguishes them observationally is the projection onto the observer's frame?
If so, what distinguishes a radio photon from a gamma ray photon intrinsically, as opposed to relationally?

 

[loop quantum gravity]

what's the definition of $k^\mu$ here? is it momentum up to constant hbar or something else? maybe a wave number?

 

[Matterwave]

Not sure why you bring GR into this and not sure why you bring photons into this.

The measured frequency of waves depend on relative motion of the emitter and the observer. This is just as true for light as it is for sound waves. You get a doppler effect for either.

all photons are in some sense "the same object" (null geodesics with $k^\mu k_\mu = 0$), and the only thing that distinguishes them observationally is the projection onto the observer's frame?

Another property that comes to mind is the photon's helicity. You're gonna have to clarify what you mean by "the same object". Do you just mean indistinguishability? I.e. that we can't apply Maxwell-Boltzmann statistics to photons?

If so, what distinguishes a radio photon from a gamma ray photon intrinsically, as opposed to relationally?

Nothing. The photon's frequency scales linearly to its energy $E=h\nu$ and energy is clearly frame dependent.

If you move very fast relative to the CMB for example, you'll see the CMB blue shifted in front of you and red shifted behind you. And in fact, we can use this to detect our motion through the CMB (we call this the "Solar Dipole").

See the multipole section of: https://en.wikipedia.org/wiki/Cosmic_microwave_background?hl=en-US

 

[Matterwave]

Just to give a 1-sentence answer: the frequency of light is observer dependent.

 

[Dale]

what distinguishes a radio photon from a gamma ray photon intrinsically, as opposed to relationally?

Nothing.

 

[Roberto Pavani]

Thanks for the replies. A follow-up: if frequency is observer-dependent, and $E = h\nu$, then the photon's energy is also observer-dependent.
Does this mean a photon doesn't "have" an energy in any intrinsic sense, only in relation to an observer?

 

[Sagittarius A-Star]

Thanks for the replies. A follow-up: if frequency is observer-dependent, and $E = h\nu$, then the photon's energy is also observer-dependent.
Does this mean a photon doesn't "have" an energy in any intrinsic sense, only in relation to an observer?

Yes, because it is massless.

 

[audiefoster28]

I'd push back a bit on the 'nothing' answer though. The direction of the 4-momentum through spacetime is intrinsic — it's literally the geodesic the photon follows. A gamma ray and radio photon have k^μ vectors pointing in completely different directions, even if both are null. That direction doesn't change depending on who's looking. Frequency, sure, that's observer-dependent. But the path through spacetime itself isn't.

 

[Nugatory]

I'd push back a bit on the 'nothing' answer though. The direction of the 4-momentum through spacetime is intrinsic — it's literally the geodesic the photon follows. A gamma ray and radio photon have k^μ vectors pointing in completely different directions, even if both are null. That direction doesn't change depending on who's looking. Frequency, sure, that's observer-dependent. But the path through spacetime itself isn't.

That's similar to saying that two bullets moving relative to one another will have different kinetic energies in different frames (which also can be described in terms of non-parallel four-momenta). They're different, but not in a way that we would generally consider to be "intrinisically" different - they're identical bullets with different past histories.

 

[PeterDonis]

does it imply that all photons are in some sense "the same object"

Since we are talking about relativity, the term "photon" is out of place; that's a quantum term. A better term might be "light pulse" or "laser pulse".

With that correction, the answer to your question is no. Each light pulse has a distinct worldline in spacetime and a distinct 4-momentum $k^\mu$ tangent to that worldline. The thing that originally determines $k^\mu$ is the process that emitted the light pulse; it then gets parallel transported along the light pulse's worldline. That is sufficient to determine its inner product with the 4-velocity of any timelike observer whose worldline intersects the light pulse's worldline, and that inner product is what determines the energy/frequency/momentum/wavelength that the observer measures for the light pulse.

 

[PeterDonis]

The direction of the 4-momentum through spacetime is intrinsic — it's literally the geodesic the photon follows.

No. The geodesic is a worldline. The 4-momentum is a 4-vector $k^\mu$ that is tangent to that worldline. And...

A gamma ray and radio photon have k^μ vectors pointing in completely different directions, even if both are null.

No, this is not necessarily true. Suppose two light pulses (I'm using that term instead of "photon" for reasons explained in my post #10 just now) are both emitted at the same event in spacetime and in the same direction in space. Then their worldlines are the same, but their 4-momentum vectors $k^\mu$ are not the same--not because they "point in different directions" but because they have different inner products with the emitter's 4-velocity, because of the different emission processes. We can express this as two invariants, using the 4-velocity $u^\mu$ of the emission worldline at the event of emission:

Gamma ray pulse: energy at the emitter $(k_\gamma)_\mu u^\mu$ might be, say, $10^7$ eV.

Radio pulse: energy at the emitter $(k_R)_\mu u^\mu$ might be, say, $10^{-4}$ eV.

Same worldline, same direction in space, but very different $k^\mu$.

 

[Roberto Pavani]

One more observation: if both photons from an $e^+e^-$ annihilation are null ($k^\mu k_\mu = 0$ each), but their sum is timelike:
$k^\mu_{\gamma_1} + k^\mu_{\gamma_2} = p^\mu_{\rm total}$
with $p^\mu p_\mu = -(2m_e c)^2 \neq 0$, then individually neither photon "has" a rest frame, but together they reconstruct the timelike direction (rest frame) of the event that created them.
So the information about the source's proper time is split between the two photon, neither carries it alone, but together they encode it fully.

 

[robphy]

One more observation: if both photons from an $e^+e^-$ annihilation are null ($k^\mu k_\mu = 0$ each), but their sum is timelike:
$k^\mu_{\gamma_1} + k^\mu_{\gamma_2} = p^\mu_{\rm total}$
with $p^\mu p_\mu = -(2m_e c)^2 \neq 0$, then individually neither photon "has" a rest frame, but together they reconstruct the timelike direction (rest frame) of the event that created them.
So the information about the source's proper time is split between the two photon, neither carries it alone, but together they encode it fully.

This is essentially the energy-momentum analogue of light-cone coordinates (up to conventional normalization), as if gotten by a radar measurement.

 

[Dale]

A gamma ray and radio photon have k^μ vectors pointing in completely different directions

For a radio wave and a gamma ray propagating in the same spatial direction, the $k^\mu$ vectors are pointing in the same direction in spacetime.