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Is redshift dependent simply upon the relative velocity and gravitational potential of transmitter and receiver?

Or is it path dependent; say the receiver runs away for billions of years then changes to the same velocity as the transmitter... or for any reason the space time in the path becomes locally stretched for a large portion of the path.

And whichever...why? And what is the evidence either way?

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  • $\begingroup$ The redshift is not path dependent, because... there is no redshift. It is a widespread misconception. You get a redshift only when you measure the energy of the photon at the emission and absorption in different frames, which is clearly wrong. However, if you properly measure the energy of the photon always in the same frame (e.g. either emission or absorption, but not in both), then the energy of the photon does not change during travel. And since there is no redshift, your question is not well defined. $\endgroup$ – safesphere Jan 29 at 3:54
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    $\begingroup$ @safesphere That's just avoiding the question. The point is that there is no inertial frame containing both the Earth and a very distant star. Furthermore, even if there were such a frame, there is nothing "wrong" with doing a measurement in one frame or another. In any case, the experimenter turns on an apparatus and it spits out a number. Who are you to say that number is "wrong"? $\endgroup$ – knzhou Jan 29 at 10:28
  • $\begingroup$ @knzhou You don't need an inertial frame that you are describing. And you don't measure the same value in different frames expecting the same result. A photon energy, as measured in one and the same frame, does not change during the flight, no matter what the path is. So if you make a common mistake by measuring the energy at emission in the emission frame and then measuring the energy at absorption in the absorption frame to see the "redshift", this redshift does not depend on the path, because it is an artifact of your measurement that depends on the frames, but not on the path. $\endgroup$ – safesphere Jan 29 at 10:45
  • $\begingroup$ @safesphere Again, this is an unusual definition of "wrong", and it's precisely the opposite of the spirit of relativity. Would you also say that length contraction does not occur, because all lengths except the proper length are the "wrong" length? The photon does not have an "actual" energy, we know even from Newtonian mechanics that energy depends on the frame, and each inertial frame is as good as any other. $\endgroup$ – knzhou Jan 29 at 10:51
  • $\begingroup$ Sorry, I dot understand your objection. Let me give an example. Say, I am on Earth and you are on a distant orbit. I send you a beam of light and you see it gravitationally redshifted. A typical interpretation is that the photons lose energy while ascending. This is incorrect. In your frame, that are emitted already "redshifted" and don't change the energy while ascending whether they go straight of circle around a black hole on their way to you. Photons don't change energy in flight, as you observe them in flight in your frame. They are emitted with the same energy as they are absorbed. $\endgroup$ – safesphere Jan 29 at 11:00
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Redshift is path-dependent, but the issue is that in most situations of real physical interest, the photons either arrive through a single path, or the effect is absolutely negligible.

Let me state some mathematics to explain this. The frequency of a photon with respect to coordinate time $t=x^0$ is proportional to the zeroth component of its wave-vector, we may even set it equal, $k_0 = f$. Now the zeroth component will evolve as $$ \frac{d k_0}{d t} = -\frac{u^t}{2} g^{\mu\nu}_{\;\;,t} v_\mu v_\nu \,,$$ where $v^\mu =d x^\mu / d t$ and $u^t = dt/d\tau$ (also remember that $v_\mu = g_{\mu\nu} v^\nu$ and $d x^0/d t = dt/dt = 1$). Of course, if $g^{\mu\nu}_{\;\;,t} \approx 0$, the redshift of $k_0$ will depend only on your four-velocity with respect to the static frame; this can be interpreted in terms of a gravitational potential, velocity etc. On the other hand, if the cumulative effect of $g^{\mu\nu}_{\;\;,t}$ cannot be neglected, then talking about gravitational potentials etc. looses meaning.

Now, when $g^{\mu\nu}_{\;\;,t}$ is not negligible, you can be an observer static with respect to $t$ and let an emitter emit two photons of the same frequency in their rest-frame. The first photon goes straight to you, and the second photon goes through a different path which can be arranged either through a gravitational lens or a mirror deflecting it. Generally speaking, the two photons will go through completely different $g^{\mu\nu}$ thus shifting its $k_0$ to completely different values before arriving to you. In other words, generally redshift is very much path-dependent.

However, our universe is approximately homogeneous and isotropic, and it turns out that in a perfectly homogeneous and isotropic universe any photon, even one sent zig-zagging through space by a system of mirrors, will end up with a shift of $k_0$ that depends only on the ratio of the scale-factors $a$ at the time of its emission and the time of its observation.

When we include overdensities that actually occur in our universe, this will generally introduce a noise to this perfectly path-independent relation. But if your photon flew at least a few Megaparsecs before getting to you and has not been strongly lensed on the way, you can really trust the redshift to indicate the era the photon comes from. In other words, in cosmology, redshift is mostly path-independent.

As for strong gravitational lensing, this will induce path dependence if the matter causing the lensing is moving with relativistic speeds with respect to the cosmological background, because otherwise $g^{\mu\nu}_{\;\;,t}$ is small (roughly $\sim V/c$ small, where $V$ is the typical velocity in the system). Matter configurations that exhibit such speeds are black-hole and/or neutron-star binaries, as well as various catastrophic events such as supernovas. The redshift of the light that passed through such systems would not be indicative of its original cosmological era.

However, I believe that observing this phenomenon is highly improbable (and I am quite sure it was never observed). As for experimental confirmation that the redshift was path-dependent: we would have to observe two photons arriving to us from the same source through two paths, and at least one of them passing through such a relativistic, typically extremely variable system! Such double images of the same source occur habitually in strong lensing, but detecting both of them with sufficient accuracy is even more of a lost case then detecting just one image. I.e., I believe there is no experimental or observational evidence of path dependency of redshift, and there will not be any in the near future.

As for the tests that of the statement that in quasi-stationary fields redshift is path-independent, one should simply refer to the usual experimental tests of gravitational redshift, because a path-dependent theory would necessarily predict deviations from relativistic predictions. On the other hand, I do not believe path-independence for photon paths of cosmological scales can be practically tested, since 1) most double images correspond to photons that do not travel on very different paths on cosmological scales, and 2) we would probably not be able to separate intrinsic variation of redshift within the emitting source (or various other obscuration etc.) from the actual path-dependence of the redshift.

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  • $\begingroup$ I'm not able to retrieve your first equation in what I know. It looks like a geodesic equation, but I never saw it with a derivative wrt coordinate time. Would you give some help? $\endgroup$ – Elio Fabri Jan 29 at 20:03
  • $\begingroup$ @ElioFabri Well, that is because it was wrong, thank you for pointing it out :) $\endgroup$ – Void Jan 30 at 8:43
  • $\begingroup$ I believe what you are saying is that the redshift is path independent in static cases (which was my point in the comments above), but not in dynamic cases. For example, if a photon is reflected off a moving mirror, the frequency would obviously change. However, this is not due to the path per se, but exclusively due to the energy exchange between the photon and the mirror. If I cross a street on a walkway fine, but next time cross away from the walkway and get hit by a car, then well, yeah, the result would be very different, but the car would also suffer some damage due to the interaction. $\endgroup$ – safesphere Jan 30 at 10:30
  • $\begingroup$ @safesphere Yes, $k_0$ is conserved in stationary space-times. As for the moving mirror, this is an interesting point I missed when thinking about this. The mirror would also have to move at relativistic speeds in the cosmological frame for this effect so be significatn, though, so in some sense this is analogous to the gravitational lens. $\endgroup$ – Void Jan 30 at 19:16
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The (completely general) formula for red shift in GR is $$ \frac{\omega_B}{\omega_A} = \frac{ p_\mu(B) v^\mu(B) }{ p_\mu(A) u^\mu(A) } $$ where $p$ is the 4-momentum of the photon traveling from A to B, $u$ is the 4-velocity of the emitter who observes $\omega_A$ and $v$ is the 4-velocity of the detector who observers $\omega_B$. In this result, $u$ and $v$ have nothing to do with the photon's path (except they are evaluated at A, B), but $p$ has everything to do with the photon's path: it is the tangent 4-vector to the worldline.

In lensing there is more than one path (null geodesic) between given events A and B. We are used to the fact that different paths arrive at B with different spatial directions; it should not surprise us that they can have different amounts of $p_0$ too, and therefore yes, redshift can depend on path. In order to convince you that this can happen, I'll give a rather impractical example, and leave it to others to find more practical ones.

Suppose that two paths go from A to B, and one of these paths crossed through a local gravitational wave at the phase of that wave where spacetime was locally expanding, but the other path did not. The tidal forces stretch the wavefronts of the lightwave apart from one another. But as the light wave passes out of this gravitational region, the tidal forces fall to zero so do not squeeze it back again. I think such a photon ends up with a different frequency to its 'twin' when they both arrive at B.

I'm pretty sure this same type of argument can also be made in a static spacetime, but as I say, I'll leave to others to comment on that.

Such effects are probably too small or too rare to be relevant to observational astrophysics/cosmology, but I guess you never know ....

P.S. on red shift vs. gravitational time dilation

I wrote this answer after reading some comments to the original question in which one person asserts there is no such thing as red shift (of the photon or lightwave itself) because the observed frequency change is wholly owing to a difference in reference frames adopted for the frequency measurements at A and B. This perspective can be useful to training our intuition about GR in some situations. If my high-up caesium clock is ticking faster than your low-down one, then I should not be surprised if microwaves emitted by your caesium atom arrive at me with a rather sluggish oscillation compared to my clock. However, since more generally the frequency change can depend on path, clearly this "it is only about the observers' frames" perspective does not capture the whole truth. So I think there is such a thing as red shift, but it shouldn't be too quickly asserted that the photon itself is changing; rather one should declare all the relevant quantities, including what local clock you are using.

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  • $\begingroup$ "I'm pretty sure this same type of argument can also be made in a static spacetime" - Not easily. The only way the energy of a photon can change in flight is if the photon interacts with something by exchanging energy with it (e.g. a moving mirror). It is simply energy conservation. To produce any shift, something must give extra energy to or take it from the photon. It may be possible to do both though to have the total energy conserved. 2 photons passing a Kerr black hole on the opposite sides may get the opposite shifts (let the experts correct me), but the black hole energy is conserved. $\endgroup$ – safesphere Jan 30 at 10:44
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Redshift is path-dependent. The photon starts off with some energy-momentum four-vector, and that four-vector gets parallel-transported along the geodesic that is the photon's trajectory. Parallel transport is path-dependent; that's the definition of curvature.

Is redshift dependent simply upon the relative velocity and gravitational potential of transmitter and receiver ?

Most spacetimes don't have a well-defined gravitational potential. We have a potential only in the case of a static spacetime. For example, cosmological spacetimes aren't static, so there is no potential associated with them. For this reason, it is not meaningful to try to break down cosmological redshifts into kinematic and gravitational parts. We can't even tell whether a distant galaxy is moving relative to us. We can say that it's moving away from us, or we can say equally well that the space in between is expanding.

And what is the evidence either way ?

Interesting question. Gravitational lensing is evidence that the parallel transport of a photon's four-momentum is path-dependent. I don't know of a direct empirical proof that the energy component is path-dependent. This kind of thing is hard to test because I don't think we have any test theory in which it's not true that the energy is path-dependent.

mmeent says:

No, redshift is in general not dependent on the path followed by the light ray. At least not any of the cases we would normally consider (I'll come back to the exception at the end). [...] As far as I can see, the only way that redshift can become path dependent is if the lightray interacts with a gravitational field in such a way that it produces (or absorbs) gravitational waves.

You don't need anything this complicated or special. A much simpler example is the following. In the Schwarzschild spacetime, let a source at rest relative to the gravitating object emit light at event P, and follow two rays that pass symmetrically around the object, reuniting at a point Q directly on the opposite side. An observer at Q will measure unequal redshifts with unit probability if his velocity vector is randomly chosen.

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  • $\begingroup$ Thanks for your interesting reply... One point that springs immediately though....If one cannot differentiate between space expansion and relative velocity how come so many papers can claim that Hubble flow is accelerating or that CMB dipole anisotropy is implicitly interpreted as purely space expansion ? $\endgroup$ – cumfy Jan 28 at 20:49
  • $\begingroup$ @cumfy: Hubble flow is accelerating This is just a statement that $\ddot{a}>0$. It doesn't have anything to do with the interpretation you refer to. or that CMB dipole anisotropy is implicitly interpreted as purely space expansion ? Not sure what you mean by this. You might want to ask it as a separate question. $\endgroup$ – Ben Crowell Jan 28 at 21:54
  • $\begingroup$ apologies if poorly worded. Many, many results in astrophysics are predicated on the assumption (contrary to your assertion) that Hubble flow and differential velocity are seperable and distinguishable. Since your answer appears to assume they are inseparable your answer appears to conflict with physics concensus. Eg Planck results of z=1100 is not interpreted as CMB photons source could be 1000 ly away but receding .999c, but according to you they could (theoretically?). $\endgroup$ – cumfy Jan 28 at 23:07
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    $\begingroup$ @cumfy: Sorry, but I'm not able to understand your comments. I suggest you write a separate question on this topic. In your most recent comment, you seem to be misinterpreting what I said and using it to reach conclusions that don't have much to do with what I said -- but I don't think a comment thread is a good place to clarify this kind of thing. $\endgroup$ – Ben Crowell Jan 29 at 1:44
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    $\begingroup$ At the very least this answer is misleading. In stationary spacetimes (i.e. spacetimes with a timelike killing vector) energy is a constant of motion of geodesics. Hence in gravitational lensing by a stationary lens (which is almost always a good approximation) the redshift does not depend on the path. (Note that due to differing Shapiro delay you could in principle get different redshifts from two lensed images because you are view the object at different times.) $\endgroup$ – mmeent Jan 30 at 10:02
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Is redshift dependent simply upon the relative velocity and gravitational potential of transmitter and receiver?

Neither. Relative velocity is well defined in flat- but not in curved spacetime. @Ben Crowell has already commented on the gravitational potential.

The cosmological redshift $z$ depends on the relative increase of the scale factor $a$ between emission and absorption and is given by $z=(a_{now}-a_{then})/a_{then}$, which can be understood as stretching of the photon's wavelength during the time it travels.

The "stretching" in this sense doesn't mean however that the space is stretched physically. Expansion of the universe doesn't mean more than growing distances between comoving objects. For more see the answer @safespere linked in his comment.

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  • $\begingroup$ "Expansion of the universe doesn't mean more than growing distances between comoving objects." This appears to say it is the same as objects moving apart after an explosion in flat static (i.e. Minkowski) spacetime, as in the Milne model. Milne model is an FLRW model, but a rather odd one with no gravitation. So I think I wouldn't want to assert "... doesn't mean more than ...". The "stretching" picture is useful and precise for calculating cosmological redshift. $\endgroup$ – Andrew Steane Jan 30 at 9:50
  • $\begingroup$ @AndrewSteane The expansion and moving apart are not equivalent in principle, because in the expansion distant galaxies can cross the horizon and disappear forever, but if they are simply moving apart they must stay in the light cone. So the existence of the horizon may prove thet space expands in FLRW. However models are possible with the space expansion without the horizon, so the absence of the horizon does not disprove the space expansion. $\endgroup$ – safesphere Jan 30 at 11:03
  • $\begingroup$ @Andrew Steane "This appears to say it is the same as objects moving apart after an explosion in flat static (i.e. Minkowski) spacetime, as in the Milne model." Yes indeed in this special case, as the empty FRW-model and the Milne model are mathematically equivalent. Let me know if you want a reference. $\endgroup$ – timm Jan 30 at 15:38
  • $\begingroup$ @timm thanks but it's ok I am familiar with this; my comment was intended merely to point out a mild disagreement with the phrase "doesn't mean more than", on the basis that the field equation is a statement about spacetime as well as matter and e-m waves etc. $\endgroup$ – Andrew Steane Jan 30 at 16:38
  • $\begingroup$ The "stretching" in this sense doesn't mean however that the space is stretched physically. Expansion of the universe doesn't mean more than growing distances between comoving objects. It seems like you want to insist on some distinction, but it's not clear what the distinction is or whether it corresponds to anything empirically meaningful. You can say that space is stretched physically, or you can say that space is not stretched physically. Neither statement makes a testable prediction, and neither corresponds very well to how GR actually describes a cosmological spacetime. $\endgroup$ – Ben Crowell Jan 30 at 17:49
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No, redshift is in general not dependent on the path followed by the light ray. At least not any of the cases we would normally consider (I'll come back to the exception at the end).

In almost all cases that we would want to consider (e.g. gravitational lensing, redshift from a source near a black hole, etc.) the metric that we would want to consider is stationary, meaning that the metric has a time translation symmetry or in GR parlence, it has a timelike Killing vector field.

The energy of a lightray is the contraction of this timelike Killing vector with the four-momentum of the lightray. Such contractions is a constant of motion of a geodesic, i.e. it does not change along a geodesic. In particular, it does depend on the exact path taken by the lightray (or equivalently it does not depend on the particulars of the curvature along the objects path).

Of course, generally, the energy of the lightray relative to the timelike Killing vector is not the energy of the light ray in the frame of the emitter, nor that in the frame of the reciever. However, the relation between the three depends only on local quantities at the emitter and reciever (4-velocity of emitter/reciever, the local metric, etc.)

Now, what happens when we let go of the stationarity requirement. The most considered case would that of lightrays in an expanding universe described by an FRWL metric. In that case the metric does not have a time translation symmetry and energy is not necessarily conserved along a geodesic. In fact, the metric does have a preferred timelike vector field defined by the Hubble flow, and the energy of a lightray relative to that timelike vector field does change along the ray. This is the cosmological redshift.

However, by explicit calculation one easily finds that the total gravitational redshift between an emitter and reciever at rest with respect to the Hubble flow is given simply by the ratio of the scale factors at the time of emission and reception. (And for emitters/recievers not at rest with respect to the Hubble flow, the difference is simply the normal relativisitic doppler shift.) In particular, it again does not depend on the details of the path taken or the particulars of how the scale factor has evolved.

As far as I can see, the only way that redshift can become path dependent is if the lightray interacts with a gravitational field in such a way that it produces (or absorbs) gravitational waves. In such a case the amount of energy lost (or gained) certainly depends on the particulars of the chosen path. However, the factional change in energy would be exceptionally tiny in all remotely realistic scenario's I can think of.

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  • $\begingroup$ However, by explicit calculation one easily finds that the total gravitational redshift between an emitter and reciever at rest with respect to the Hubble flow is given simply by the ratio of the scale factors at the time of emission and reception. But this applies only to an FLRW spacetime, which is an idealized model with more symmetry than the real universe. $\endgroup$ – Ben Crowell Jan 30 at 17:54
  • $\begingroup$ As far as I can see, the only way that redshift can become path dependent is if the lightray interacts with a gravitational field in such a way that it produces (or absorbs) gravitational waves. I don't think anything so complicated or special is required. I'll add a simple example to my own answer. $\endgroup$ – Ben Crowell Jan 30 at 18:43
  • $\begingroup$ The cosmological redshift doesn't imply a loss of energy by photons. This is a myth based on using a wrong frame. The energy of a cosmological photon measured in the same frame doesn't change with the expansion of the universe. For example, if you observe a photon in the frame of the emitter, there is no redshift. Similarly, if two spaceships fly apart and observe each other's tail lights redshifted, this is just an effect of the frame, but no photons actually lose energy. $\endgroup$ – safesphere Jan 30 at 20:39
  • $\begingroup$ @safesphere In an expanding FRW universe there is no such thing as the "same frame" at different spacetime events. Hence a receiver can never observe a photon in "the frame of the emitter". There is however a prefered frame set by the Hubble flow. It is a simple statement of fact that the energy of a photon changes relatively to the frame set by the Hubble flow. $\endgroup$ – mmeent Jan 31 at 8:04
  • $\begingroup$ @mmeent I didn't refer to "a receiver observing a photon in the frame of the emitter". Instead I referred to "an emitter observing a photon in the frame of the emitter (self)". I realize such an observation may not be practically possible, but this is irrelevant, because, what stops you from theoretically calculating the energy of the photon in the frame of the emitter? And this energy conserves, at least in the uniform expansion. Thus I don't see how your objection is relevant to my comment. $\endgroup$ – safesphere Jan 31 at 8:13

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