# No redshift in a freely falling frame

1. Jun 14, 2012

### vin300

Why is there no redshift in a freely falling frame? The photon in a freely falling frame also rises in the gravitational field, so isn't it supposed to be redshifted?

2. Jun 14, 2012

### Staff: Mentor

From the viewpoint of an observer at rest in the freely falling frame, there is no "gravitational field". He is weightless, feeling no force, so to him, physics looks the same as in free space with no gravity. (We are assuming the freely falling frame is small enough that tidal effects are negligible.) Photons in free space with no gravity do not redshift.

From the viewpoint of an observer at rest in the gravitational field, the freely falling frame is accelerating downward. Suppose a photon is emitted upward towards a freely falling observer some distance above, who is at rest in the gravitational field at the instant the photon is emitted. By the time the photon reaches the observer, it will have redshifted, but the observer will have picked up just enough downward velocity so that when the observer receives the photon, there will be a Doppler blueshift that exactly cancels the gravitational redshift.

3. Jun 14, 2012

### TSny

As you can see from PeterDonis' excellent answer, "redshift" (wavelength) is not a property belonging to a photon itself. It is a joint property of a photon and an observer (detector).

4. Jun 14, 2012

### Naty1

Does the PeterDonis description also apply to the deBroglie wavelength of a free falling matter particle?

I'm thinking of the Tamara Davis article I posted where she notes such a deBroglie wavelength of a matter particle redshifts in the same proportion in cosmological expansion as the wavelength [redshift] of a photon.

I'm wondering if that analogy applies to gravitational redshift as well.

I'm having trouble understanding " but the observer will have picked up just enough downward velocity so that when the observer receives the photon..."

In what sense is the 'at rest' observer accelerating downward??

5. Jun 14, 2012

### Staff: Mentor

I would say yes, because the de Broglie formula is basically just the rest mass m > 0 version of the formula for light. I can't think of any literature offhand where I've seen it discussed, though.

In the sentence you quoted, "observer" refers to a freely falling observer. I was trying to describe how an observer static in the field would view the observations made by a freely falling observer. The "acceleration" of the free-falling observer is only coordinate acceleration, true, but from the viewpoint of the static observer it still changes the relative motion of the two during the time of flight of the photon.

6. Jun 14, 2012

### Staff: Mentor

Hmm... this does bring up a good point relative to Naty1's question which I just posted an answer to. I need to qualify my answer somewhat.

In the case of a photon in free space, its velocity is constant; so the "dispersion relation" between frequency and wavelength is also constant (they are both the same in relativistic units where c = 1). Another way of saying this is that a Lorentz transformation on a photon does not change its speed; it just changes its frequency/wavelength, via the Doppler shift.

In the case of a massive particle, things are more complicated. The particle's velocity is not constant, since a Lorentz transformation changes it, so its dispersion relation is not constant either. That complicates the relationship between de Broglie frequency/wavelength and the particle's kinematics--energy and momentum. Falling or rising in a gravity field still has an effect, but I'm not sure it can be described as simply as the effect on a photon can be.

7. Jun 14, 2012

### TSny

See attachment for experimental observation of effect of gravity on "matter waves" of neutrons. You can get the gist of the article from reading the first few paragraphs and the last few paragraphs.

[I have never tried attaching a file before, so I hope the attachment works. If not, you can view the article here

http://www.atomwave.org/rmparticle/ao%20refs/aifm%20refs%20sorted%20by%20topic/inertial%20sensing%20refs/gravity/COW75%20neutron%20gravity.pdf [Broken]

]

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• ###### Neutron Interferometry in Gravity Field.pdf
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8. Jun 14, 2012

### Staff: Mentor

By the way, vin300, in case you want to search on your own, the principle that PeterDonis described is called the "Equivalence Principle" and is one of the key concepts behind General Relativity.

9. Jun 14, 2012

### yuiop

Just to be clear, the "no redshift" condition only occurs if the source and the observer are both free falling and reasonably close to each other. If the source is stationary relative to the gravitational field, then a stationary observer lower down in the gravitational well, will see the light as blue shifted and a free falling observer lower down will see the same light as red shifted.

10. Jun 14, 2012

### pervect

Staff Emeritus
Another way of saying this is that there is no redshift to first order. If you plot redshift vs distance, for example, using a reasonably intuitive notions of redshift and distance (this somewhat ambiguious description is defined exactly by using fermi-normal coordinates for the distances and the metric coeffficeint g_00 for the redshift), you'll see that at the origin the curve is a horizontal line, a curve having zero slope, but due to second and higher-order terms the redshift becomes non-zero as you get further away from the origin of the coordinate system.

Last edited: Jun 14, 2012
11. Jun 15, 2012

### vin300

Can it be shown mathematically?

12. Jun 15, 2012

### pervect

Staff Emeritus
What I said can be shown mathematically. If you have or can find MTW's Gravitation, see pg 332, eq 13.73

It gives the Fermi normal metric for a non-rotating observer with zero propagation to accurate to second order The expression for g_00 is

$$g_{00} \approx -1 -R_{\hat{0}\hat{l}\hat{0}\hat{m}}x^l x^m$$

where R is the Riemann curvature tensor in the frame-field basis of the falling observer.

It's mentioned elsewhere in MTW that the components of the curvature tensor $R_{ijkl}$ for the Schwarzschild metric near a single massive body aren't affected by motion towards or away from the body. Furthermore, the radial component is given by $R_{0r0r}$ = -2GM/r^3[/itex], equal to the Newtonian tidal force. The transverse components are half that and of the opposite sign (and are also equal to the Newtonian tidal force). The off diagonal terms R_{0i0j} for i not equal to j are zero.

The biggest issues are 1) knowing that the metric g_00 represents time dilation and 2) thinking of Fermi-normal coordinates as representing "an intuitive notion of distance". The last is probably the most problematic - though they are a pretty straightforward generalization of coordinates from the flat space-time to curved. The biggest issue is that there is more than one way to make the transition.

Online, http://arxiv.org/pdf/gr-qc/0010096v1.pdf and http://arxiv.org/pdf/0901.4465.pdf may be helpful

The later reference also provides some justification for thinking of the Fermi Normal coordinates as having a "physical" meaning (though I prefer to say intuitive).

Last edited: Jun 15, 2012
13. Jun 15, 2012

### Naty1

seemed clear to me after this morning coffee!! Maybe I was drunk yesterday!!??

I should have posted that with my question: I have been puzzling about exactly that since I read the Tamara Davis article. She does not explain her conclusion just states it.

I think you are right: it can't be as simple. If a free falling photon exhibits one redshift, how could a matter particle exhibit the same redshift in the same spacetime? The changing speed of the matter particle does not affect the speed of the emitted deBroglie wavelength but it must affect it's redshift relative to a more rapidly receding photon.

Last edited: Jun 15, 2012
14. Jun 15, 2012

### Naty1

oops now I am not so sure:

I came across this from Jonathan Scott, which I also thought correct:

so I have more thinking to do.

edit: Upon reconsideration I think the above from Scott is incorrect regarding frequency: frequency corresponds to the KE of light so it should change as a photon moves in a non uniform static gravitational field. So that description does NOT seem to answer our question....

I'll leave this here in case others find that 'logic' of value....or in case I'm wrong again!

Last edited: Jun 15, 2012
15. Jun 15, 2012

### GAsahi

It is considerably more complicated than the above. The first half of your statement is correct, the second one is not. Proof

$$d \tau^2=(1-r_s/r)dt^2-dr^2/(1-r_s/r)$$

1. If the two observer and the source are stationary, then $dr=0$ at radial coordinates $r_1<r_2$ , you can write:

$$d \tau_1^2=(1-r_s/r_1)dt^2$$
$$d \tau_2^2=(1-r_s/r_2)dt^2$$

It follows that :

$$\frac{d \tau_1}{d \tau_2}=\sqrt {\frac{1-r_s/r_1}{1-r_s/r_2}}$$

Since $r_1<r_2$ it follows
$$d \tau_1< d \tau_2$$
i.e $$f_1>f_2$$ (blueshift)

2. If the source is moving, things get more complicated.

$$d \tau_1^2=(1-r_s/r_1)dt^2$$
$$d \tau_2^2=(1-r_s/r)dt^2-dr^2/(1-r_s/r)=(1-r_s/r)dt^2 (1-\frac{(dr/dt)^2}{(1-r_s/r)^2})$$

Therefore:

$$\frac{d \tau_1^2}{d \tau_2^2}=\frac{1-r_s/r_1}{1-r_s/r}\frac{1}{1-\frac{v^2}{(1-r_s/r)^2}}$$

The second fraction is always larger than 1.
The first fraction is sometimes greater than 1 (for $r_1>r$) or smaller than 1, (for $r_1<r$), so you can get either blue or redshift (you don't always get redshift).
For $v^2=(1-r_s/r)(r_s/r_1-r_s/r)$ there is no shift whatsoever.

Last edited: Jun 15, 2012
16. Jun 15, 2012

### Staff: Mentor

But "KE change" is not invariant, it's observer dependent; that was really Jonathan Scott's point. Whether or not the KE changes depends on who is measuring the change. The KE of freely falling light changes from the viewpoint of static observers; that's what is normally referred to as gravitational redshift/blueshift. But from the viewpoint of the freely falling photon itself, its energy does not change; more precisely, its 4-momentum does not change, it is parallel transported along the photon's worldline. The "change", from the photon's point of view, is in the 4-velocity vectors of static observers at different heights.

17. Jun 15, 2012

### vin300

I've found an explanation that violates the equivalence principle. An observer in a uniform gravitational field sees objects accelerating downwards at a constant g. An observer with a constant upward acceleration g also sees the same thing, but there's something that's not equivalent in the two cases. An observer in a gravitational field says that a falling object has an unchaging total energy, all the gain in its kinetic energy is due to equivalent loss of potential energy, but the accelerating observer doesn't say that its potential energy is decreasing. The man in the field says the energy of the object doesn't change, but the accelerating one says that the energy changes at a constant rate.

18. Jun 15, 2012

### Staff: Mentor

First of all, the above is irrelevant to the EP because it's not based on local observations. The difference between the two observers (in so far as it is a difference--see below) only becomes evident when they make observations over time.

Second, even leaving out the above, you're using two different definitions of energy for the two accelerated observers; that's where the apparent difference comes from. The reason you can tell it's only apparent is that it doesn't affect the results of any actual experiments. Whatever experiment the observer accelerating in free space tries to run to show that the free-falling object's energy "changes at a constant rate", the observer at rest in the gravity field will be able to run the same experiment and get the *same* result. Similarly, whatever experiment the observer at rest in the field tries to run to show that the free-falling object's total energy is unchanged, the observer accelerating in free space will be able to run and get the same result. So the apparent "difference" between the two is only a matter of the definition of "total energy"; it doesn't reflect any actual physics.

Btw, what I said in the last paragraph is *not* true in general in GR; it's a special feature of Schwarzschild spacetime (strictly speaking, of Schwarzschild spacetime outside the horizon) that this extended correspondence with Rindler observers in Minkowski spacetime, which is what justifies the statements I made above, can be drawn. That's why those statements are not about the EP; the EP is true in any spacetime, but only locally.

19. Jun 16, 2012

### DrGreg

Energy is always measured relative to an observer (or, more accurately, relative to some coordinate system). In those cases where it is possible to meaningfully define energy relative to a non-inertial observer so that energy is conserved, you need to define potential energy. So there is a "pseudo-gravitational" potential energy associated with Rindler coordinates, for example.

20. Jun 18, 2012

### stevendaryl

Staff Emeritus
I think that your answers are correct, but I think that the computation is not quite correct. The reason why is because what you have calculated is the ratio of clock rates as measured using Schwarzschild coordinates. Relative clock rates for spatially separated clocks is a coordinate-dependent quantity; a different coordinate system might give a different answer.

The quantity that is directly observable, and is independent of coordinate systems, is this:
Let a photon be emitted from one observer. Let f1 be its frequency, as measured by that observer. Let the photon travel to the other observer. Let f2 be the frequency of that same photon, as measured by the second observer. Then the measured redshift or blueshift would be f2/f1.

I don't know right off the bat if that gives the same answer as the ratio you computed, or not, but conceptually, they are different.