# Time dilation in the field interpretation of GR

1. Mar 1, 2015

### Sonderval

IIUC, there are two different interpretations of GR - either as curvature of space time (as in Misner Thorne Wheeler) or as a (spin-2)field that influences all matter (as in the books by Weinberg or in the Feynman Lectures on Gravitation) and that leads to the underlying Minkowski metric being inobservable.

In the field interpretation, I can understand that fields do affect the distances between objects ("rulers"). I am trying to find a good heuristic or intuitive explanation or example, how the running of clocks is affected in the field interpretation. (There is a calculation in http://arXiv.org/abs/astro-ph/0006423v1 for the example of a H-atom, but this is rather un-intuitive and I seem not to be able to see the physics behind it.)

As a simple example, consider a particle in a constant magnetic field. This could be sued as a clock by measuring the cyclotron frequency
$$f_0=qB/2 \pi m$$

In a gravitational field, the frequency should become smaller so the clock runs slower than the far-away clock. However, I think m inside the gravitational field should be reduced (because of a gain in potential energy) by a factor (1-gR/c²) - because the energy needed to transport a particle from R to infinity is mgR - , which results in an increase of the frequency.

Probably, the magnetic field also has to be transformed (the paper cited above shows how the vacuum susceptibility is affected), but again the physics behind that is not too clear to me.

So is there any simple way to explain the time dilation in the field interpretation for this (or another) example?

2. Mar 1, 2015

### Staff: Mentor

Does that mean you already have one for the curvature of spacetime interpretation? If so, what is it? And why can't it be adapted to the field interpretation? The math of both is exactly the same.

Meaning, the frequency of the clock as seen by the far-away clock. Yes, that frequency becomes smaller.

To clarify terminology, the system deep in the gravitational field has less potential energy, not more; potential energy is negative (at least, it is with the usual normalization that has it going to zero at infinity). See below.

No, it isn't, because g is not constant; it varies with radius. The correct formula is $U = - G M / R$ for an object at radius $R$. (Note that this happens to be the same as the Newtonian formula; this is a special property of the particular spacetime we are considering, and does not happen in most other cases.)

Also, there is a square root in the time dilation factor; it is $\sqrt{1 + 2 U / c^2} = \sqrt{1 - 2 G M / c^2 R}$.

However, if you want to view this as a reduction in "mass", the "mass" that is reduced is not the $m$ that appears in the formula you gave; it's the total energy of the system, including the electron's kinetic and electromagnetic potential energy and the energy stored in the magnetic field. That is what gets reduced by the time dilation factor I just gave. So to figure out how that comes about mathematically, you would need to look at the formula for the total energy and see what would change by being in a gravitational field.

Also, you should be aware that this idea of "energy being reduced" when an object is deep in a gravitational field is not the only possible interpretation of the effect of the field; another interpretation is that the field affects the propagation of light, so light that is emitted by the clock deep in the gravity well has the same frequency when it is emitted as for any other clock of the same construction; but as the light travels outward from the clock to infinity, it loses energy by climbing against the field, so when it arrives at infinity, it has a lower frequency.

3. Mar 3, 2015

### harrylin

Just as a side note, that other interpretation is in disagreement with Einstein and his argumentation. As GR models light and radio waves as waves, I cannot see how the other interpretation could possibly be viable. Okun publiished a paper about the impossibility of that other interpretation, but perhaps a concrete example is more useful.
Say you listen to a radio station at 100 MHz when in Spacelab (simplified case of Earth "in rest"). Then:
1. the time delay must remain constant;the fact that the field affects the speed of light can not change the constant time delay.
2. As no wave crests can get lost or appear from nowhere, the frequency cannot change when "in flight" in vacuum.

If you know a loophole in that very basic argument of Einstein (1911), please present it.

4. Mar 3, 2015

### Staff: Mentor

First, I didn't say the field affects the speed of light. I said it affects the energy of the light; the light has to lose energy as it gains altitude. Otherwise you could build a perpetual motion machine. Einstein gave a simple argument showing that. The field does affect the coordinate speed of light, but that is irrelevant to the argument about the change in light frequency, since the change in coordinate speed affects all light rays equally.

Second, when you say "the time delay must remain constant", it is; that is, the coordinate time delay between two wave crests is constant. See below.

And it doesn't; that is, the coordinate frequency does not change. What Einstein left out of this argument was that coordinate frequency and proper frequency (i.e., the frequency measured by a particular observer who is at rest at a particular altitude) need not be the same. This would not be fully understood until the Schwarzschild solution expressed in Schwarzschild coordinates was fully understood, which arguably did not happen until the 1960s.

The physical argument in favor of the "field affects the light frequency" interpretation is simple. Take two identical light sources (say atoms making the same energy level transition); put one at infinity and the second deep down in a gravity well. Measure the frequency of the light emitted locally--i.e., using a device at rest right next to each atom. Both measurements give the same result. This indicates that, locally, both light rays have the same energy. So when the light from the source deep in the gravity well reaches infinity, and is redshifted, it seems obvious that the change must be due to something that happened to the light as it traveled; if it were due to some change in the source, then the local measurement would have shown a different result, but it didn't.

If you really want to make an argument against the "field changes the light frequency" interpretation, you can, by invoking the fact that the "frequency" of the light is not a property only of the light: it's a property of the interaction between the light and the measuring device. The frequency of light observed by an observer with 4-velocity $u$ is just $p \cdot u$, where $p$ is the 4-momentum of the light. Since the light travels on a null geodesic, its 4-momentum is parallel transported along its worldline, which is as close as we can get in a curved spacetime to "doesn't change". So if its observed frequency changes, what must have changed is the 4-velocity of the observer, relative to the 4-velocity of the source. In other words, the observer at infinity must have a 4-velocity that "points in a different direction" in spacetime, relative to the 4-velocity of the source--even though the two are at rest relative to each other. Note that this interpretation also falsifies the "field changes the object's energy" interpretation: the locally measured energy of the object deep in the gravity well does not change (and this can be verified by measurement as well). The only thing the field changes is the "direction" in spacetime of the 4-velocity of an observer at rest.

Really, though, all of this is just semantics. The math is the same regardless of what interpretation you put on it. The reason we have different interpretations is as heuristics: to help us develop intuition for the physics. Different heuristics may work better in different situations. No one interpretation is "right"; as long as they all use the same math underneath, they can all be useful.

5. Mar 4, 2015

### harrylin

I meant that the time delay between emission and detection must remain constant as for example determined by the ground station or the receiving station. And note that in a discussion based on Einstein's 1911 argument any term refers to coordinate x and not proper x. In that context, "the light has to lose energy as it gains altitude" is erroneous or misleading, as also clarified by the fact that the (coordinate) frequency does not change.
See above; while nothing was "left out", that fact immediately followed from the theory.
That's easy to debunk: according to Einstein's GR the resonance frequency of everything - including any light emitter and any detector's reference frequency - is equally reduced in a gravity well. As a result no difference can be detected locally, but the emitted light will be redshifted as measured with a detector at infinity. Your example can therefore not differentiate which interpretation is correct. However, you ducked my example which I still think does disprove that alternative interpretation, and much better than any arguing in the style of Okun.

The calculation as performed by for example the ground station is straightforward: I think that the carrier frequency cannot be reduced in flight, it's a mathematical impossibility due to the continuity requirement of signal propagation. For example, assume that for 10 s the carrier wave is emitted at 100 MHz and it contains a 1kHz sound signal. For simplicity, say that it is received starting for example 1 s later at 99 MHz due to "something happening to the light" that results in a "change of frequency as it travels" (ignore the fact that according to GR accepted wave theory, this is impossible in vacuum; the reason will immediately become clear next!).
If I correctly understand the alternative interpretation's calculation, the 1kHz sound must now arrive as a 990 Hz sound as reckoned by the ground station; so that the whole sound pattern of 10'000 cycles will take roughly 10.1 s to reach the detector! Consequently the last second of radio emission will have been waiting in space for about 0.1 second.

On top of that, this would also imply that the gravitational field does not affect clock frequency, contrary to standard GR and clock retardation in other experiments.
With this I get back on topic: in the field interpretation of GR as formulated by Einstein, gravitational time dilation implies that the gravitational field reduces resonance frequencies so that clocks tick slower and emitted light is red shifted.
Usually it's just semantics, but it looks as if this one is more than just semantics... except if the calculation for the alternative interpretation doesn't lead to error. But that was exactly my point: I still cannot see how the interpretation that something happens to the radio wave frequency between emission and reception can give the right answer in my example!

Last edited: Mar 4, 2015
6. Mar 4, 2015

### Staff: Mentor

This isn't an observable; the observer at the detection site can't observe the time of emission, and the observer at the emission site can't observe the time of detection. Each one can communicate their observations to the other, but in order to compare them, they need to first synchronize clocks, since neither one can make both observations directly.

But coordinates are not physical. Einstein wasn't fully clear about this in 1911 because he hadn't fully developed GR yet. In later writings he is quite clear that coordinates have no physical meaning. So while I agree that "light loses energy as it gains altitude" might be misleading if we are talking only about coordinate frequency, I don't care because I'm not interested in coordinates; I'm interested in the physics, and the coordinates by themselves don't tell you about the physics. You need to compute invariants.

No, this is wrong. Please note that I specifically said the frequency as measured by a detector at rest right next to the atom. In other words, proper frequency, not coordinate frequency. The proper frequency is unchanged, and that is exactly what GR predicts. The coordinate frequency is changed, but coordinates are not physical. I am being careful to specify physical observables, not coordinates.

I am not trying to show which interpretation is "correct". I am trying to show that no interpretation is "correct"; all of them are just interpretations, and which one is "correct" is a physically meaningless question. The physics is contained in the observables, and all the interpretations give the same observables. The things you are talking about are not observables; they are coordinate quantities that have no physical meaning. (Note that the same is true in your Spacelab example: you are talking about "frequency" and "time delay" as if those were absolute quantities, but the actual observed frequencies and time delays are not; only the coordinate frequencies and time delays are. But those aren't what are actually observed. Look at the actually observed frequencies and time delays and you will see that they change.)

7. Mar 4, 2015

### harrylin

They may synchronize clocks with for example GPS. Indeed coordinates are are mere instruments, like the rulers (and clocks) from which they were derived. Just as you, I'm only interested in the physics. Now, as GPS works with coordinates you could reason that it's meaningless while I reason that it's both useful and meaningful. In order to correctly understand the physics one has to understand what happens both concerning invariants and measurements with coordinate systems.
With "they change" you surely mean that "they differ"; I suspect that this last one is just semantics. According to GR all the locally measured frequencies and time delays are constant for the simplified case considered here - both in the ground station and Spacelab. But, once more, if it's all just semantics, then my example is excellent to clarify what you exactly meant with the alternative interpretation.

So, please correct my simple calculation example and show how it can work with the following alternative interpretation, as you phrased it (of course you are welcome to correct yourself; I just try to get this other explanation crystal clear here):

- "energy [is not] reduced" when an object is deep in a gravitational field. Instead:
- the field affects the propagation of light / the energy of the light;
- as the light travels outward from the clock to infinity, it loses energy by climbing against the field, so when it arrives at infinity, it has a lower frequency

8. Mar 4, 2015

### Staff: Mentor

"Working with" coordinates is not the same as viewing coordinates as having physical meaning. Coordinates are perfectly fine as intermediate numbers in a calculation, which GPS uses them for. They are also perfectly fine as arbitrary labels attached to physical locations: that's what the latitude and longitude you get from GPS are (arbitrary labels attached to locations on the Earth). But if GPS says I am at North latitude 38 degrees 15 minutes 24 seconds, and West longitude 85 degrees 32 minutes 27 seconds, that doesn't mean those numbers describe anything physical at that location. It just means they're a label I can use to refer to it.

One can use this method to understand the physics, certainly. But I strongly disagree that one has to. The physics can be understood without ever talking about coordinates at all. Read the second to last paragraph of my post #4. I draw a physical conclusion there without ever even talking about coordinates.

Unless I'm misunderstanding the scenario, this is simply wrong; you are basically denying that gravitational redshift exists. My understanding of the scenario is: we have a ground station at rest on the Earth (for this scenario we can ignore the Earth's rotation), and Spacelab "hovering" at some large altitude directly over the ground station, so the two are at rest relative to each other. The ground station sends a radio signal to Spacelab; the frequency $\omega_e$ of the signal on emission is measured locally at the ground station, and the frequency $\omega_r$ of the signal on reception is measured locally by Spacelab. Then we have $\omega_r < \omega_e$, both as a prediction of GR and as an experimentally measured fact (the Pound-Rebka experiment).

Now, if by "locally measured frequency" you meant that Spacelab has an identical radio transmitter to the ground station's, and it locally measures the emitted frequency of its own transmitter (not the frequency of the signal received from the ground station), then yes, this locally measured frequency will be $\omega_e$. But if Spacelab uses its transmitted to send a signal to the ground station, and the ground station locally measures the frequency of the signal it receives, it will measure $\omega_r > \omega_e$. So the frequency received is still different from the frequency emitted, as measured locally.

Let me repeat this key point: all the interpretations use the same math. All the interpretations make the same predictions for all observables. The difference is only in what words in ordinary language are used to describe what the (same) math is saying and what the (same) observed results "mean".

9. Mar 5, 2015

### harrylin

I doubt that anyone on this forum views coordinates as having physical meaning!

As I tried to explain how gravitational redshift exists in what I view as the standard field interpretation, you definitely misunderstood what I wrote. Anyway, I was trying to understand what you wrote, so let's look at your following clarification:
So far so good; up to that point there is no distinction possible between the two interpretations. As I illustrated, the problem arises with the continuity condition if one takes the alternative interpretation as phrased by you at face value.
With the alternative interpretation as presented by you taken literally, I found a result that disagrees with observations and common sense. As you could not answer how that alternative interpretation can work in my example, most probably it is not an alternative interpretation at all; just semantics that for me is too ordinary and potentially misleading, in the sense that I illustrated. Thus I'm satisfied with your clarifications! I hope that it was also helpful for Sonderval.
I note that he/she did not reply further; so I wait for his/her comments

Last edited: Mar 5, 2015
10. Mar 5, 2015

### Sonderval

@harrylin+PeterDonis
Sorry for not reacting -somehow my mail notofication did not work and I assumed that the thread is dead...
Thanks for the clarifications - however, I'm not sure I really get your argument at #3, harrylin.

11. Mar 5, 2015

### Staff: Mentor

In other words, you used different math, which is wrong. If you came up with different math by "taking literally" what you thought I was presenting as the alternative interpretation, then you definitely misunderstood something.

Because I couldn't understand how you were coming up with different math, when I specifically said the alternative interpretation uses the same math. If the ordinary language description appears to be leading you to different math, then that means you disregard the ordinary language description; it doesn't mean you change the math. The math is what comes directly from the theory; the ordinary language description is always going to be less reliable than the math, simply because of the limitations of ordinary language. That's why we use math in physics to do the actual work.

12. Mar 9, 2015

### harrylin

A radio signal is propagated as a wave according to GR field theory. According to the second interpretation something happens to radio waves in transit when changing height. However, a radio wave that is sent upwards over a fixed distance but "the field affects the propagation of light / the energy of the light", can -as far as I know- only reduce in energy (and thus in frequency) by slowing down more and more. That is contrary to field theory, illogical and contrary to observation. Inversely, signals that are sent down would even arrive earlier and earlier, rapidly being received before being sent! I therefore asked peterdonis to explain how he understands that alternative interpretation by means of that example. But it now appears to be just different (and in my opinion, poor) semantics for in fact the same interpretation.

Last edited: Mar 9, 2015
13. Mar 9, 2015

### Sonderval

Sorry, I still don't get it - "slowing down more and more" in what coordinate system or frame? Wouldn't it be clearer to say that the wavelength (and thus the frequency) is affected by the field, but the speed is always c (to a fixed observer at given height)?

14. Mar 9, 2015

### Staff: Mentor

This is not correct in curved spacetime. In flat spacetime, with no dispersive medium present, you would be correct that a given EM wave, when viewed from a fixed inertial frame, cannot change in frequency. (Note that "cannot" here is absolute--it can't change frequency without changing speed, and it can't change speed because there is no dispersive medium present, so it can't change frequency, period. A dispersive medium would allow the wave to change speed, but the curved spacetime scenario we are discussing is in vacuum, so there's no need to add further complications.)

In curved spacetime, however, as I have already pointed out several times, there is no such thing as a "fixed inertial frame" from which to view the wave. Two observers, both at rest in the gravitational field at different altitudes, are not in the same "fixed inertial frame", so you cannot extend the flat spacetime conclusion of "no frequency change without speed change" to cover this case. The argument simply does not apply, because a condition on which it depends (having a single fixed inertial frame covering the entire experiment) is not satisfied.

Note that this, by itself, does not show that the radio wave can change frequency without changing speed; it only shows that the argument you are giving does not apply. That is, you cannot say, based on your argument, that the alternate interpretation I proposed (where the field affects the propagation of light) does not "make sense", because we are talking about curved spacetime and your argument cannot be applied in curved spacetime.

15. Mar 10, 2015

### harrylin

I assumed (please correct me if that was mistaken) that a frame fixed to the Earth was used in your discussion. Similarly either a non-local frame as used for GPS and that is fixed to the radio station or to the simplified stationary Spacelab can be used in my example: both reference systems work for the interpretation that you apparently went for. If I correctly understand your discussion, your interpretation corresponds perfectly to the fundamental conclusion that in "the gravitational field [..] the clock goes slowly when it is placed in the neighbourhood of ponderable masses. It follows from this that the spectral lines in the light coming to us from the surfaces of big stars should appear shifted towards the red end of the spectrum" - https://en.wikisource.org/wiki/The_Foundation_of_the_Generalised_Theory_of_Relativity

Concerning my question about the other interpretation, in wave theory and using a reference system as I mentioned, the number of cycles is conserved so that in stationary conditions the frequency cannot change as Einstein explained in 1911 (and also I showed here), even when the wavelength is a function of position. However, it is possible to phrase the same interpretation in a totally different way; and I could not make sense of the alternative interpretation that peterdonis introduced. I therefore asked him to clarify what he means with the alternative interpretation by matching his phrasings to my example in such a way that the correct calculation results (maybe you can do it?).

16. Mar 10, 2015

### harrylin

PS: just to be very clear, I talk here about the frequency of the light waves in transit as determined with a "stationary", non-local coordinate system like GPS and as discussed in the section of Einstein's description that I referred to. In such a reference system the speed of light is according to GR a function of place and direction. And just in case it is not clear to you how time dilation in Einstein's gravitational field interpretation works out correctly in my example (merely looking at frequency and conservation of cycles), I'll be happy to elaborate.

17. Mar 10, 2015

### Sonderval

@harrylin
Thanks for the link to the EInstein book, I'll have a closer look at §22 ASAP.
Probably I have no clue what a "stationary non-local" system is and how it is supposed to work - I still don't see how the number of cycles argument works. Let's say we have two observers at different heights; the lower one transmitting 100 light signals at a frequency of 1Hz. For the upper observer, the time elapsed will be longer than 100s, so she will see a lower frequency. The number of cycles is constant, but the frequency is not. How does this example work out in the "stationary non-local system"?

18. Mar 11, 2015

### harrylin

In the GPS system satellite clocks are tuned to be in phase with Earthbound clocks, so that their clock frequency does not correspond to the "local time" but to Earth time. That allows consistent measurements.

In my example (why change the numbers?) the Earthbound radio station emits for 10 seconds a carrier wave containing a 1kHz signal, and the reception according to the Earth bound GPS system happens with a certain time delay. The time delay is necessarily constant because in this example the distance is held constant. The signal will in this case be detected as 1kHz for 10 seconds, and 10'000 cycles will be received in that time. If alternatively a non-adjusted atomic clock is used for reception, then (exaggerated) the signal will be detected as 10'000 cycles of (roughly) 990 Hz for 10.1 seconds of local "Spacelab time". As this takes place during 10 seconds of Earth time, everything works out nicely: the interpretation is that the radio frequency in transit (in near vacuum) cannot be affected by the gravitational field, but instead it is the atomic clock frequency that is affected by the Earth's gravitational field.

As a reminder, it remains unclear to me how this can work with a truly different interpretation (and not merely different phrasing of the same interpretation), according to which, as "the light travels outward [..], it loses energy by climbing against the field, so when it arrives [..], it has a lower frequency".

And note that it is again one step further to go to an interpretation about what the effect of the Earth's gravitational field is on mass, since E = m c2. I recall having seen a discussion about that on sci.physics.research some years ago, and I can try to find that back if you like.

Last edited: Mar 11, 2015
19. Mar 11, 2015

### Staff: Mentor

You left out a key assumption of this interpretation; see the bolded insert above.

"Loses energy" is equivalent to "the standard of time is changing". The standard of time for an object at rest at Spacelab's altitude is different from the standard of time for an object at rest on the Earth's surface. This change in the standard of time manifests itself as a loss of energy for anything that climbs from one altitude to the other.

The root of the problem here is that we have two intuitions that cannot both be satisfied in this scenario. We have the intuition that the radio frequency in transit through a vacuum does not change; this intuition is supported by the mathematical fact that the 4-momentum of a photon is parallel transported along its null geodesic worldline. But we also have the intuition that energy is conserved; everything else except a photon loses energy when climbing in a gravitational field, so photons must also lose energy when climbing, or else energy is not conserved. (Einstein used a simple thought experiment based on this idea to argue that there must be gravitational redshift.)

No single interpretation can preserve both of these intuitions, because "energy" and "frequency" are just two different names for the same thing (with different units). Your interpretation preserves intuition #1 but violates intuition #2; my interpretation preserves intuition #2 but violates intuition #1.

(More precisely, my interpretation evades intuition #1 by recognizing that "energy" and "frequency" are frame-dependent, and pointing out that there is not a single inertial frame covering both Spacelab and the Earth's surface: the 4-velocities of the two observers are not "parallel", so their inner product with the parallle transported 4-momentum of the photons is not the same. But that just means we're now violating intuition #3, which says that two observers at rest relative to each other should have "parallel" 4-velocities. Ultimately, the only way out of all this is to stop relying on intuitions and intrepretations and just look at the math, which is unambiguous.)

20. Mar 11, 2015

### harrylin

Instead I specified the GPS system, but it works equally well for a Spacelab observer in this example; and I also stressed that frequency is frame dependent.

As I stressed, the time standards of both the earth's radio station and Spacelab in this example are different but constant. And the "climbing" light waves do not have a valid time standard.
Strictly speaking GR does not use the photon concept; and in any case, light has no (rest) mass so intuition can point either way. Once more, this is why I here refused to argue with words or intuition; instead I asked you to match your phrasings with correct math like I did, as calculations clarify what is meant with words and drive home where wrong interpretations fail. As far as I am aware, by 1916 Einstein only maintained the conclusion corresponding to one interpretation, as I cited: redshift is due to time dilation of atomic processes - and my calculation example appears to be compatible with only that interpretation.

PS. I asked you this now at least three times, and this will be the last time: please back up and clarify your "other interpretation" by precisely matching the interpretation as you phrased it* with the full correct calculation of my example. That's the only way to avoid a useless discussion. Therefore I will not react anymore to another type of (non)answer (sorry!).

*"energy [is not] reduced" when an object is deep in a gravitational field. Instead:
"the field affects the propagation of light / the energy of the light";
"as the light travels outward from the clock to infinity, it loses energy by climbing against the field, so when it arrives at infinity, it has a lower frequency"

Last edited: Mar 11, 2015