## No redshift in a freely falling frame

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?
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 Blog Entries: 9 Recognitions: Gold Member Science Advisor 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.
 Recognitions: Gold Member Homework Help 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).

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## No redshift in a freely falling frame

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??

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 Quote by 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 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.

 Quote by Naty1 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??
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.

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 Quote by 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).
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.

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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/a...%20gravity.pdf

]
Attached Files
 Neutron Interferometry in Gravity Field.pdf (371.9 KB, 8 views)
 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.
 Blog Entries: 6 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.
 Recognitions: Science Advisor 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.
 Can it be shown mathematically?

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 Quote by vin300 Can it be shown mathematically?
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).

 Nowadays the Fermi normal coordinates are usually - although improperly - called Fermi coordinates. In exper- imental gravitation, Fermi normal coordinates are a powerful tool used to describe various experiments: since the Fermi normal coordinates are Minkowskian to first order, the equations of physics in a Fermi normal frame are the ones of special relativity, plus corrections of higher order in the Fermi normal coordinates, therefore accounting for the gravitational field and its coupling to the inertial effects. Additionally, for small velocities v compared to light velocity c, the Fermi normal coordinates can be assimilated to the zeroth order in (v/c) to classical Galilean coordi- nates. They can be used to describe an apparatus in a “Newtonian” way (e.g. [1, 3, 8, 10]), or to interpret the outcome of an experiment (e.g. [11] and comment [21], [5, 6, 15, 17]). In these approaches, the Fermi normal coordinates are considered to have a physical meaning, coming from the principle of equivalence (see e.g. [18]), and an operational meaning: the Fermi normal frame can be realized with an ideal clock and a non extensible thread [29].

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

 In the case of a photon in free space, its velocity is constant. In the case of a massive particle, things are more complicated. The particle's velocity is not constant, since a Lorentz transformation changes
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'm not sure it can be described as simply as the effect on a photon can be.
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.

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 ...I'm not sure it can be described as simply as the effect on a photon can be. I think you are right:

oops now I am not so sure:

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

 Neither photons nor even massive objects change in frequency or energy when moving in a static gravitational field as observed by any one observer. In Newtonian terms this is because the combined effects of changes in kinetic energy (from motion) plus potential energy (from time dilation) result in constant total energy.
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!

 Quote by yuiop 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.
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.

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