I Path independence redshift photon

[wnvl2]

Is the redshift of a photon between two events independent of the path? I assume for this question that the the cosmological constant is zero.

Chatgpt writes: "In a gravitational field, the redshift can depend on the path taken by the photon. For example, in a Schwarzschild metric (which describes the spacetime around a non-rotating massive object), the gravitational redshift of a photon emitted from a point in the gravitational field to a point at infinity is path-independent. However, if you consider paths that pass close to the massive object (e.g., a photon that grazes the surface), the effective potential and spacetime curvature affect the redshift differently."

Is that last sentence correct for non rotating black holes? I understand that in case of a Kerr (Newman) metric a photon can gain energy. Are there other examples? Can this happen in case of a Schwarzschild metric?

 

[PeterDonis]

Is the redshift of a photon between two events independent of the path?

Generally speaking, no.

I assume for this question that the the cosmological constant is zero.

Ok, but you should also state up front what spacetime geometry you are talking about. You seem to be talking about black holes.

Chatgpt

Is not an acceptable source. You need to be looking at textbooks or peer-reviewed papers. Sean Carroll's online GR lecture notes are free:

https://www.preposterousuniverse.com/grnotes/

I understand that in case of a Kerr (Newman) metric a photon can gain energy.

Where do you get this understanding from?

 

[PeterDonis]

Is the redshift of a photon between two events independent of the path?

To expand somewhat on the answer at the top of my previous post:

First, "photon" is the wrong term. We're in the relativity forum, not the QM forum, and "photon" is a quantum term (and even in QM it doesn't mean what most people think it means). A better term would be "light pulse". Or, if we're talking about paths in spacetime, "null path".

Second, if "event" means what it usually means in relativity, i.e., a point in spacetime (not just a point in space), there might not be a null path between two events at all. There only will be if the events are null separated. And in such cases, there is typically only one null path between the events anyway, so the question is moot.

But I suspect what you actually mean is something more along the lines of: if we pick two points in space (one of which might be at infinity), and look at different possible ways a light pulse could travel between those two points in space, then in general, no, the redshift will not be the same for those different ways the light pulse can travel, because the different ways will pass through different regions with different spacetime curvatures.

 

[wnvl2]

I read most of Schutz and I know the lecture notes from Carroll. It is not that my understanding of general relativity is just based on chatgtp.

I want to leave the spacetime geometry open. Limitations are that there is no cosmological constant and that the metric is stationary. The question is in fact: 'For what kind of spacetime geometries is 'the redshift of a photon between two events independent of the path'. If I get a metric, how can I see that the redshift will be (in)dependent on path?

I assume that for a Schwarzschild metric it is independent of path, but I am not 100% sure.
And I assume that for a Kerr metric it is not independent of the path. Is it related to off diagonals from the time in the metric the cannot be made zero?

 

[wnvl2]

Second, if "event" means what it usually means in relativity, i.e., a point in spacetime (not just a point in space), there might not be a null path between two events at all. There only will be if the events are null separated. And in such cases, there is typically only one null path between the events anyway, so the question is moot.

Can gravitational lensing serve as an example of a phenomenon where multiple paths are possible?

How can I know in such a case that the frequencies from both paths will be the same?

 

[Ibix]

If the metric is stationary there's a timelike Killing field and the inner product of the tangent vector of a geodesic and a Killing field is a constant of the motion. So two light pulses emitted at the same event with the same energy as measured by a Killing observer will have the same energy as measured by a Killing observer at another event. Energy is conserved, in other words.

Gravitational lenses are not necessarily stationary spacetimes. If a particular one is, the above argument applies. If it's not, the red shift can be path dependant.

I think presence or absence of a cosmological constant is irrelevant, except that the decision removes some stationary spacetimes from consideration

 

[PeterDonis]

I want to leave the spacetime geometry open.

Not very:

Limitations are that there is no cosmological constant and that the metric is stationary.

That is a pretty strict constraint on the geometry. And note that you seem to be implicitly considering a further constraint, namely a vacuum spacetime, which is an even stricter constraint.

I assume that for a Schwarzschild metric it is independent of path,

Again, you need to be very careful about what "path" means. In most cases (see further comment below) there is only one null path through spacetime that is possible. Which makes the whole question moot.

Can gravitational lensing serve as an example of a phenomenon where multiple paths are possible?

If you mean paths through spacetime, yes. Although requiring both lensing paths to be between the same two events in spacetime is also a pretty strict constraint: in a case where lensing is possible, there will often be multiple paths between two points in space, but the travel times along those spatial paths will be different, so the spacetime paths will not be between the same pair of events.

How can I know in such a case that the frequencies from both paths will be the same?

In general you can't. It will depend on the specifics of the lensing situation. In many cases I would not expect the shifts to be the same for multiple paths.

 

[wnvl2]

In general you can't. It will depend on the specifics of the lensing situation. In many cases I would not expect the shifts to be the same for multiple paths.

Do you mean that in most cases of lensing, the metric is not stationary, which results in the shifts generally not being the same? Otherwise, if the metric is stationary, wouldn’t that contradict what Ibix writes?

 

[PeterDonis]

If the metric is stationary there's a timelike Killing field and the inner product of the tangent vector of a geodesic and a Killing field is a constant of the motion.

Yes, the usual name for this constant of the motion is "energy at infinity". There are some technicalities involved here as well, about how to precisely define which Killing field is the one to use, in cases (such as the Kerr-Newman family of black hole spacetimes) where there are other Killing fields present as well

two light pulses emitted at the same event with the same energy as measured by a Killing observer will have the same energy as measured by a Killing observer at another event.

Yes, this is true, but there's a more general statement that's true as well, if the spacetime is stationary. The two light pulses don't have to be emitted and received at the same spacetime events. They only have to be emitted and received at the same Killing worldlines (i.e., integral curves of the timelike Killing field). Or, if we use those Killing worldlines to label "points in space", the light pulses only have to be emitted and received at the same points in space. This covers cases like gravitational lensing where the travel times are different along different paths.

Even this case is still restrictive, though, because most observation points move in space in the sense of their worldlines not being Killing worldlines. For example, if we consider the Earth as a test mass moving in the spacetime geometry of the Sun, and we approximate that geometry as Schwarzschild, the Earth is moving with respect to the Killing worldlines of that Schwarzschild geometry. So if two light pulses from the same object that are gravitationally lensed by the Sun both arrive at the Earth, but at different times, we can no longer use the argument above to say that the frequency shift of both pulses is the same, because the reception events aren't on the same Killing worldline. (Similar remarks apply if the emitting object is also moving with respect to the Killing worldlines: the two light pulses traveling different paths might not have been emitted at the same time.)

And of course the same applies for gravitational lensing by, say, distant black holes, because both the Earth and the emitting object are extremely likely to be moving with respect to the Killing worldlines of the black hole's spacetime geometry.

 

[PeterDonis]

Do you mean that in most cases of lensing, the metric is not stationary

In some cases, it might not be, but that's not my main concern.

if the metric is stationary, wouldn’t that contradict what Ibix writes?

No. It's just that what @Ibix wrote has much more limited application than you might think. See my post #9.

 

[Ibix]

In summary, stationary spacetimes define a timelike Killing field and you may therefore foliate them by picking a spacelike surface and translating it along the Killing field. The geometry of all surfaces generated by this translation is identical to the starting surface. That is, there's a time-independent definition of space, and you can talk sensibly about "path through space" by projecting a worldline along the Killing field onto one such spatial plane. Red shift is independent of path through space in this sense.

However, most observers and most sources won't be stationary in space in this sense, so you will typically see some time evolution of redshift as one or both parties move through space and hence you may simultaneously see multiple differently red shifted images of the same object. This isn't due to path dependence of red shift though, but rather due to the fact that the light arriving simultaneously by different paths was emitted at different times and hence emitted at different places. There may also be different kinematic Doppler of the source, and your own motion through space may induce different kinematic Doppler in the received pulses.

And in reality lens spacetimes may not be stationary.

 

[Ibix]

There are some technicalities involved here as well, about how to precisely define which Killing field is the one to use, in cases (such as the Kerr-Newman family of black hole spacetimes) where there are other Killing fields present as well

Are there multiple timelike Killing fields in these spacetimes?

 

[Orodruin]

But I suspect what you actually mean is something more along the lines of: if we pick two points in space (one of which might be at infinity), and look at different possible ways a light pulse could travel between those two points in space, then in general, no, the redshift will not be the same for those different ways the light pulse can travel, because the different ways will pass through different regions with different spacetime curvatures.

I don't obtain that. Perhaps you can find a flaw in the following reasoning:

In order to define "points in space" at all we need a stationary spacetime, i.e., a timelike Killing field $K$. Once we have that, the only reasonable definition of "gravitational redshift" is that offered by considering stationary observers, i.e., observers whose world lines are integral curves of $K$. Taking the 4-frequency of the light pulse as $N$, the frequency seen by a stationary observer is $g(V,N) = g(K,N)/\sqrt{g(K,K)}$ (where $V = K/\sqrt{g(K,K)}$ is the 4-velocity of the observer). Referring to the emitter with sub-index 0 and the observer with sub-index 1, the emitted frequency is $f_0 = g(K_0,N_0)/\sqrt{g(K_0,K_0)}$ and the observed frequency is $f_1 = g(K_1,N_1)/\sqrt{g(K_1,K_1)}$.

Now, a light pulse follows a null geodesic and since $K$ is a Killing field and $N$ the tangent of the geodesic, $g(K,N)$ is constant along that geodesic and therefore $g(K_0,N_0) = g(K_1,N_1)$. Consequently$$
\frac{f_1}{f_0} = \frac{g(K_1,N_1)}{\sqrt{g(K_1,K_1)}} \frac{\sqrt{g(K_0,K_0)}}{g(K_0,N_0)} = \sqrt{\frac{g(K_1,K_1)}{g(K_0,K_0)}}$$
This depends only on the ratio of the norms of the Killing field at observer and emitter and not on the particular geodesic chosen.

 

[Orodruin]

Even this case is still restrictive, though, because most observation points move in space in the sense of their worldlines not being Killing worldlines.

Ok, I realise now I just wrote into math what @Ibix said by words. My point is that motion in space (as defined by the Killing field considered) seems to be excluded by requiring the observations to take place the same point in space (again, as defined by the Killing field). The frequency becomes the same regardless of the path. That you will have an additional local Doppler shift depending on which direction you move relative to the stationary observer should not come as any surprise. We don't even need GR for that.

 

[Ibix]

In order to define "points in space" at all we need a stationary spacetime,

I don't think that's quite true in general - we talk about FLRW co-moving observers as being at points in space and there's no timelike Killing field in that case. I think we get away with it there because the three spacelike Killing fields pick out a notion of space at one time.

Note that the OP didn't clearly specify that he was only interested in stationary spacetimes until the post after the one you quoted. I suspect @PeterDonis had a slightly more general spacetime in mind when he wrote that.

 

[Orodruin]

I don't think that's quite true in general - we talk about FLRW co-moving observers as being at points in space and there's no timelike Killing field in that case. I think we get away with it there because the three spacelike Killing fields pick out a notion of space at one time.

Yes, this is true. FLRW spacetimes get away thanks to the spatial symmetries, but there are then other symmetries at play - namely the full isometry of space. This picks out a family of timelike world lines that are everywhere orthogonal to the spatial directions.

Note that the OP didn't clearly specify that he was only interested in stationary spacetimes until the post after the one you quoted. I suspect @PeterDonis had a slightly more general spacetime in mind when he wrote that.

I think similar arguments with apply to the FLRW setting, but using the spacelike Killing fields rather than the timelike ones. The difference being that the ratio now depends on the cosmological time of emission/reception. On the other hand, due to the spatial symmetries in the FLRW spacetime, there will be only one path allowed (barring shenanigans with closed universes ...)

 

[PeterDonis]

Are there multiple timelike Killing fields in these spacetimes?

Yes. There is no Killing field that is timelike everywhere, but there are regions of these spacetimes in which multiple timelike Killing fields exist. This is because a linear combination of Killing fields with constant coefficients is also a Killing field.

To pick out the timelike Killing field with respect to which "energy at infinity" is defined, you have to pick out the unique Killing field that is timelike at infinity. That's the one you are most likely thinking of. But it is not the only Killing field that is timelike in a finite region of the spacetime.

 

[PeterDonis]

motion in space (as defined by the Killing field considered) seems to be excluded by requiring the observations to take place the same point in space (again, as defined by the Killing field).

Yes. My point was that that requirement excludes, for example, observations made by us on Earth at different times, since we do not remain at the same point in space with respect to any applicable Killing field, even if we approximate the fields of the sun and every black hole that we've observed gravitational lensing with as stationary.

 

[Orodruin]

Yes. My point was that that requirement excludes, for example, observations made by us on Earth at different times, since we do not remain at the same point in space with respect to any applicable Killing field, even if we approximate the fields of the sun and every black hole that we've observed gravitational lensing with as stationary.

Sure, but there is no deep insight from that in my opinion as it is related to the stationary path independent shift through a couple of local Doppler shifts. The corresponding effect is there in SR too: Does Doppler shift depend on the state of motion? A: Yes it does.

 

[pervect]

Is the redshift of a photon between two events independent of the path? I assume for this question that the the cosmological constant is zero.

The question needs a bit of sharpening up, as I think others have commented. The frequency of a photon (or , since GR is a classical theory, a pulse of coherent energy) is simply not a property of an event and a photon.

Example: Observers A and B are at the same event with different velocities. They observe the photon/pulse of energy to have different frequencies due to doppler shift.

If we replace "event" by "observed by an observer with some 4-velocity", then there is an easy mathematical answer in the special case of a stationary metric, which I think is your concern. (That's asuming I'm guessing your intent0.

But the tricky part is formulating the question correctly, frequency is simply not a property of a photon at an "event". If you reformulate the question, we can give you an invariant property along the photon's path which may answer your underlying question and concerns about the existence of a conserved energy. The short version is that in a stationary (or static) metric, there is a concept of a conserved energy, which I think is what your question is about.

I will drop a few hints of entities that will be useful. A photon or pulse doesn't have an invariant frequency, but it has a four-frequency https://en.wikipedia.org/wiki/Four-frequency. An "event" doesn't have a four velocity, but if you know the four-velocity of some observer, and the four-frequency of a photon/pulse, you can find the frequency that observer measures for the "photon".

The second hint. (It may even be an answer and not a hint). A stationary or static metric inherently has a time-like Killing vector field. Along any geodesic, the inner product of the 4-frequency of the "photon" and the time-like Killing vector is constant (conserved) along the geodesic.

For particles, we often use the invariance of the inner product of the energy-momentum 4-vector and the Killing vector field to get a conserved energy-at-infinity of a particle. We're essentially doing the same thing with the "photon", but we use the four-frequency in this application instead of the 4-momentum.

If you are interested in cases other than a stationary metric, though, you'll run into the usual difficulties with GR and energy conservation.