# Photon energy conservation

1. Jul 3, 2010

### TrickyDicky

Is photon energy conserved for comoving observers in redshifted light from receding galaxies?
And in the case of gravitational redshift?

Thanks

2. Jul 4, 2010

### nutgeb

Photon energy is strictly conserved, if you look at it from the correct perspective. If you consider galaxies to be moving through space (as opposed to space itself expanding), then redshift is merely an accumulation of tiny Doppler shifts along a photon's trajectory through space. Doppler shifts do not constitute a loss of energy per se, rather the redshift occurs as the photon loses momentum relative to the observer, or its wavelength physically stretches as each wavecrest takes longer to arrive, as it chases and enters successive reference frames that are receding ever faster away from the emitter.

The cosmological redshift is a function primarily of location -- that is, the proper distance traveled away from the emitter, which in turn determines the changes in its proper velocity (c + H * D) it experiences in each successive reference frame. (H is the then-current Hubble rate and D is proper distance from the emitter). The photon of course always travels at a local (peculiar) speed of c, but its proper velocity relative to the emitter progressively increases with distance due to Hubble's Law. Note that at large cosmic distances, the accumulation of tiny Doppler shifts along the photon's trajectory yields a much different total redshift figure than would a single end-to-end Doppler shift velocity calculation. The accumulation of redshifts along the trajectory is incremental because the photon must adjust its proper velocity (relative to the emitter) to maintain a local speed of c in each successive recession frame it travels through, even though the photon is not absorbed in any of those frames until it reaches the ultimate observer.

The basic concept of cosmological redshift as Doppler shift, and the implications for conservation of energy, are the subject of the cover article of the current issue of Scientific American, authored by Tamara Davis. You can read it at www.sciam.com. As she points out, even in the recent past many authors have written as if the cosmological redshift somehow violates energy conservation, and they often resort to unhelpful platitudes such as "General Relativity does not require energy to be conserved in an expanding universe". But really it is just a failure to take the Doppler velocity differential of the observer and emitter fully into account. The photon's momentum relative to its emitter (redshift) decreases with distance; but its proper velocity relative to its emitter increases with distance, and in exactly the same proportion. Therefore at each location along the photon's trajectory, multiplying momentum and proper velocity together always yields a constant. That constant relates the photon's energy conservation back to the emitter's reference frame. An increase in proper velocity doesn't come as a free lunch; the price paid for it is the loss of momentum observed as redshift.

You also asked about gravitational redshift. From the perspective of the observer, gravitation causes blueshift, not redshift, as the photon is accelerated toward the observer by the sphere of cosmic mass energy defined by the photon's radius from the observer at each instant in time. However, if redshift is treated as Doppler shift in the FRW metric, then gravitational blueshift is already subsumed in the Doppler shift, so it is not a separate factor in the calculation. The effect of gravitation is reflected in the decrease of the Hubble rate over time. As a result of this decrease, the photon's accumulated proper velocity (H * D) increase relative to the emitter is less than it would have been if gravity weren't progressively slowing down the Hubble rate. Since energy conservation is maintained by the Doppler shift formula, the gravitational contribution is already captured in the overall energy conservation.

But the mathematical role of gravity is coordinate-specific. If Schwarzschild coordinates are used instead of FRW coordinates, then gravitational blueshift becomes a discrete element of the redshift calculation, and it is multiplied by an element equivalent to the accumulated Special Relativistic Doppler shift. In Schwarzschild coordinates, all of the cosmic mass in the sphere defined by the emitter's radius from the observer is considered to be concentrated at the observer's location. Therefore clocks run slower at the observer than at the emitter, and the observer sees more wavecrests arriving per beat of slower local clock time (i.e., higher frequency), which is observed as blueshift. The total energy of a system comprised of a Schwarzschild mass and a photon (or any object) freefalling radially toward it is strictly conserved.

Last edited: Jul 5, 2010
3. Jul 6, 2010

### TrickyDicky

From this energy conservation considerations, can we properly say that in every instance, a redshift (either cosmological, gravitational, Doppler in case you distinguish it from the cosmological) of incoming light implies that the observer local clock goes faster than at the source and therefore time dilation is directly linked to redshift?.( that can be observed too in the lightcurves of supernovae Ia, I think)

4. Jul 6, 2010

### nutgeb

That's a tricky question to answer. In general I would say no, but like so many aspects of relativity, the answer is dependent on which coordinates you choose.

Part of the redshift in a "normal" Doppler shift in Minkowski coordinates is attributable to Special Relativistic (SR) time dilation, and part of it is not. At non-relativistic velocities, SR time dilation makes an insignificant contribution to the observed redshift. At relativistic velocities, SR time dilation dominates the redshift.

However, FRW coordinates are most frequently used in large-scale cosmology. As I explained, the cosmological redshift is interpreted as an accumulated Doppler shift in FRW coordinates. FRW coordinates are unique because, as a result of the homogeneity of the spatial distribution which is built into the metric, all comoving observers share the same cosmological time, i.e. the same elapsed time since the Big Bang. Their clocks all run at the same rate. As a result, a photon travelling between two comovers does not experience SR time dilation, regardless of the relative recesssion velocities of the emitter and observer. In FRW coordinates, the incremental redshifts are in the form of SR redshift with the SR time dilation element eliminated. (Although if the tiny accumulated links in a photon's trajectory are small enough, even if local space were considered to be Minkowski, and SR time dilation were included in the equation, it would mathematically drop out as the recession velocity change from link to link approaches zero.) You can think of FRW coordinates as being like Minkowski coordinates to which an inverse Lorentz transformation has already been applied, but with respect to comoving velocities only. By the way that's why both distant comoving recession velocities and distant photon proper velocities can exceed c in FRW coordinates, while local peculiar velocities cannot.

On the other hand, relativistic peculiar velocities (i.e., local velocity of an object relative to the local Hubble flow) can include significant amounts of SR time dilation in the redshift. This is because the inverse Lorentz transformation built into FRW coordinates provides a common clock rate only for purely comoving objects. The redshift contributed by peculiar motion is multiplied against the redshift caused by comoving motion, to yield the total cosmological redshift.

As I explained in my first post, there is no discrete element of gravitational time dilation in redshift in FRW coordinates, because the gravitational effect is indirect, subsumed in the Doppler shift equation. Since FRW comovers share a common clock rate, it is apparent that there can be no gravitational time dilation as between any comoving emitter and observer.

If Schwarzschild coordinates are used instead, then the redshift is comprised of one element of gravitational time dilation and one element of accumulated SR redshift, multiplied together. The SR redshift in Schwarzschild coordinates includes the normal SR time dilation.

If Minkowski coordinates are used, for example to model a small region with essentially flat spacetime (no cosmic gravity), then the redshift is purely SR, and SR time dilation occurs.

The redshift observed from purely comoving type 1a Supernovae does not include any SR time dilation in FRW coordinates. But if the supernova has a large peculiar velocity, that will contribute an element of SR redshift. Also keep in mind that for comoving supernova much closer than z~1, the difference between the SR and cosmological redshift calculation can become vanishingly small.

Some authors use the term "time dilation" to refer to the fact that the observed duration of a 1a Supernova cycle (from start to finish) is elongated in the same proportion as the light is redshifted. I think that terminology is unnecessarily confusing, because nothing akin to SR time dilation occurs in that case. What is really happening is that the train of photons becomes radially stretched apart during transit. This occurs because the photons at the front of the train must increase their proper velocity (relative to the observer) at a faster rate than photons at the rear of the train, as they chase and enter regions of slower Hubble velocity (H * D). (Slower because here we are considering the Hubble velocity from the observer's prespective, so it slows down as proper distance D from the observer decreases, thereby enabling the photon's proper velocity toward the observer to increase.) You can analogize it to an evenly spaced line of joggers running along at the same peculiar velocity, but who are running over a series of slowly moving sidewalks, where each successive moving sidewalk along the route is moving backward a little less fast than the previous sidewalk. The line of joggers will progressively stretch apart. This happens even though the sidewalks themselves don't stretch, and each jogger maintains the same peculiar speed relative to whichever sidewalk she is running on. The distance between photons becomes stretched in the same way that the distance between wavecrests stretches.

Last edited: Jul 6, 2010
5. Jul 6, 2010

### TrickyDicky

Ok, so I understand we can observe redshifts that are not associated to time dilation.
Can you provide me with some specific example?
Thanks

6. Jul 6, 2010

### JustinLevy

To expand on nutgeb's comments, here's what we'd get for the doppler frequency shift if it was due "only" to time dilation (in flat spacetime inertial coordinate systems):
$$f = f_0 \frac{1}{\sqrt{1 - v^2/c^2}}$$
where f_0 is the frequency in the rest frame of the source, and f is the frequency in an inertial frame which the source is moving at speed v.

This clearly doesn't match the relativistic doppler formula. What nutgeb was referring to is that the doppler formula includes a newtonian classical sense of "running towards" or "running away" from the source will increase or decrease the rate at which we come across "pings" (or light wave crests) from the source moving through space. We could actually describe the situation from the rest frame of the source, and an observer rushing towards the source will hit the pings more rapidly than an observer rushing away from the source (due to just simple classical reasons that their relative velocity between the pings is different). Then you can get the full relativistic formula by adjusting this result by the time dilation factor to relate time in the source's inertial frame, to the time as measured by the observer.

Putting this together gives the actual relativistic doppler formula:
$$f = (1 - v/c) f_0 \frac{1}{\sqrt{1 - v^2/c^2}} = f_0 \sqrt{\frac{1-v/c}{1+v/c}}$$

---
To get back to your original question on cosmological redshift, note that it is possible to strictly say energy-momentum is conserved locally in GR, but it becomes difficult to discuss whether it is conserved in a volume. Only in situations which the spacetime has time invariance can we get around this (for example Schwarzschild, Kerr, Minkowski, etc solutions). So for Schwarzschild we can say a bit easier that energy is conserved in gravitational redshift. FRW cosmological solution doesn't have this time symmetry, so we need to be careful when discussing what we mean by "energy conservation" in the first place.

One way around this is to use pseudo-tensors to describe energy-momentum. This frustratingly breaks the beautiful coordinate invariance of GR, but this was Einstein's approach since he couldn't find another way. This led to bizarre arguments over whether gravity waves were "real" or just "coordinate artifacts". After some simplistic arguments by Feynman, the consensus has settled on gravity waves being real/physical in GR. Pseudo-tensors allow a means to discuss the energy in more complicated situations, even dynamical spacetimes that include gravity waves. From this point of view, we can state more clearly: yes, energy is conserved in cosmological redshift.

The issues of energy in GR can be subtle. And it is quite (very) possible some of the subtlety still escapes me. So hopefully some more knowledge members can comment, as I'd like to understand this better myself.

Last edited: Jul 6, 2010
7. Jul 6, 2010

### marcus

TD, the only reason I'm responding is that it doesn't seem to me that anyone has yet given you a simple direct answer to your actual question.

I think you are asking about the photon energy as it would be measured by successive comoving observers along the path. The answer is clearly NO. The energy of the photon does not remain constant. It decreases in proportion as the wavelength increases. Each successive observer would see it as less.

Let me make sure we mean the same thing by "comoving". What I mean by a comoving observer is one who is at rest relative to the CMB---at rest with respect to the expansion process (the rate looks the same in all directions.) Is that what you mean?

It seems to me that others may have been addressing other questions--such as the following: How does it look from the standpoint of the emitter? Suppose we include in the photon's speed an extra contribution from expansion of distances, so its speed away from the emitter can be greater than c---will this in some sense compensate for the decline in momentum? If you define a special "energy" that incorporates the photon's superluminal "proper speed" away from the emitter, can you get this special energy (seen from the emitter's standpoint) to remain constant? Those seem like fun questions, but AFAICS not exactly the question you posed at the start. Please let me know if I'm wrong.

Last edited: Jul 6, 2010
8. Jul 6, 2010

### nutgeb

Marcus, I think your answer, while superficially accurate in a certain narrow sense, misses the point of Tamara Davis' article in Scientific American. Although an observer receiving light from a moving emitter will measure and calculate that the photon has less energy than when it left the emitter, that does not represent a true loss of energy for the system as a whole, only a shift from the perspective of one participant to the perspective of another participant. After all, how can the observer talk about the supposedly higher energy of the photon when it left the emitter, except by shifting his own perspective to the emitter's reference frame?

In the same way, a ball caught by a running receiver appears to have lost much of its flight energy (compared to when the ball was thrown), but from the correct perspective it hasn't lost any energy at all. Any concept of energy conservation is nonsensical if the observer is allowed to ignore the relative motion of the emitter, when the total system is comprised of an emitter, an observer, and object moving between them. Only the system as a whole is capable of energy conservation.

In the same way, energy conservation of the expanding universe as a total system can be discussed meaningfully only as the aggregation of a subset of all subsystems, each of which subsystems is comprised of an emitter, an observer, and a photon passing between them. In each such subsytem as a whole, photon energy is strictly conserved when Doppler shift occurs.

Last edited: Jul 6, 2010
9. Jul 6, 2010

### nutgeb

JustinLevy's answer explains very crisply what I was alluding to. Thanks for that!

10. Jul 7, 2010

### TrickyDicky

Thanks for the replies. I think I get the picture about the energy part, even if it looks more complex than I thought.
But I am now more specifically interested in the link between time dilation and redshift, that is why I asked for some example of observed redshift not associated with some quantifiable time dilation?
In other words, in as much as redshift is an observed change of frequency photons, and as long as those photons in their own frame of reference are not losing energy (there is no "tired light" mechanism or scattering or anything like that), can we say that the change in frequency observed in our detector with respect to the frequency at the source is basically relativistic time dilation?

From the wiki: "Time dilation can arise from (1) relative velocity of motion between the observers, and (2) difference in their distance from gravitational mass"

Perhaps the question seems so trivial, that you missed my point, or maybe it has been implicitly answered and I didn't get it. Nutgeb said redshift is not always associated to time dilation, thus my asking for specific examples.

11. Jul 7, 2010

### Ich

In http://www.astro.ucla.edu/~wright/cosmo_01.htm" [Broken], more than 99% of the observed redshift are not due to time dilation. Is that what you wanted to know?

Last edited by a moderator: May 4, 2017
12. Jul 7, 2010

### TrickyDicky

In fact, I never asked if redshift is due to time dilation (not talking about causality issues here), but rather whether we could say that whenever there is redshift in EM waves (no matter if gravitational, doppler relativistic, cosmological) there is also some (even if infinitesimal, not talking about the percentage of the total) time dilation.

If the answer is negative, there is something I'm not getting, so examples would be great.

BTW, where in the wright tutorial page linked do they mention time dilation? I missed it

Last edited by a moderator: May 4, 2017
13. Jul 7, 2010

### TrickyDicky

14. Jul 7, 2010

### Calimero

Tricky, I guess that you want to know if when we observe distant redshifted object, do we observe time going slower there? And answer is yes, but you have to keep in mind that it is pure observational effect. In the same way astronomer from earth observes redshift in distant galaxy, astronomer from distant galaxy would observe our galaxy redshifted. So, do clocks attached to two distant comoving galaxies show any time dilation would depend on manner in which those two clocks are brought together for comparing. If you do it in symmetrical fashion (both are accelerated towards each other at same rate) there would be no time dilation.

Note that it is fundamentally different from gravitational redshift, where clock left in lower potential would show time going slower comparing to a clock in higher potential, even if they are brought together for comparing through a network of observers which all agree on their acceleration and speed being equal while traveling towards each other.

As for conservation of photon energy, I would suggest you to read for starters http://blogs.discovermagazine.com/cosmicvariance/2010/02/22/energy-is-not-conserved/" [Broken], blog post by well known physicist Sean Carroll.

Last edited by a moderator: May 4, 2017
15. Jul 7, 2010

### Ich

Be aware that time dilation (at least by the definition I'd use) is generally coordinate-dependent, and so is the answer to your question.
For example, cosmological redshift, when described in cosmological coodinates, is 0% doppler (including time dilation), 0% gravitational, and 100% cosmological.
The very same redshift, when described in static coordinates, is x% doppler (including time dilation), (100-x)% gravitational, and 0% cosmological.

That said, in static coordinates, I can't conceive of an example of redshift without at least infinitesimal time dilation.

16. Jul 7, 2010

### TrickyDicky

Thanks everyone, I see this concepts from a wider perspective now.
As always, there are more facets to a question than one is able to see at first.

Regards