There is no Transverse Gravitatonal Redshift

[codelieb]

There is no "Transverse Gravitatonal Redshift"

There has been a lot of noise on this forum lately about A. Mayer's so-called "Transverse Gravitational Redshift." It is not difficult to prove that no such effect exists, yet I have not seen anyone post a calculation to prove it. I have therefore posted a Mathematica notebook at http://www.feynmanlectures.info/notgr/notgr.nb" showing definitively that there is no "Transverse Gravitational Redshift."

Here is an outline of the calculation:

I consider two inertial frames of reference: a frame co-moving with the rocket at the moment of transmission with origin at the transmitter at the time of transmission ("transmitter's frame"), and a frame co-moving with the rocket at the moment of reception with origin at the receiver at the moment of reception ("receiver's frame"). **I make all calculations in the transmitter's frame.** (One could make the calculations from any inertial frame and get the same result, but the transmitter's frame is convenient for this purpose.) To find the radius vector from the spatial origin of the transmitter's frame to that of the receiver's frame at an arbitrary time, and to find their relative velocity at an arbitrary time I use the formulas for relativistic constant acceleration (see http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html" ) to show that that light transmitted from one side of the accelerating rocket at frequency w0 is received at the other side of the rocket at frequency w0 (thus, no redshift).

For those of you who do not have Mathematica, you can download the free Mathematica Player from http://www.wolfram.com/products/player/" , which comes with the Mathematica engine, so you can examine my equations and run the calculations for yourselves.

Mike Gottlieb
Physics Department
California Institute of Technology
http://www.feynmanlectures.info"

 

[codelieb]

I have received email from A. Mayer expressing some concerns over the underlying assumptions he infers, and thus the validity, of the calculation I have posted at

http://www.feynmanlectures.info/notgr/lgr.nb"

which I claim shows, by direct calculation based on well-known and well-verified relativistic formulas, that there is no such thing as "transverse gravitational redshift."

To address Mr. Mayer's concerns (and anyone else who might share similar concerns) I have posted a similar calculation of the "longitudinal gravitational redshift" that does occur when the transmitter is located in the tail of the rocket and the receiver is in the nose directly above it:

http://www.feynmanlectures.info/notgr/lgr.nb"

The result of this calculation for the longitudinal redshift is exactly the well-known experimentally verified redshift w0(1-gh/c^2) that occurs when light transmitted at frequency w0 in a uniform gravitational field with acceleration g, goes from a region of lower to higher potential, parallel to the potential's gradient, a distance h. (See: The Feynman Lectures on Physics Volume II Section 42-6 "The speed of clocks in a gravitational field.")

Both calculations use the exact same formulas for relativistic constant acceleration (see http://math.ucr.edu/home/baez/physic...SR/rocket.html ), for the relativistic Doppler effect (see http:////en.wikipedia.org/wiki/Relat...Doppler_effect), and for simple geometry. The two calculations differ only in the parameters necessary to model the different locations of the transmitter and the receiver in the two (different) cases, and besides these modeling parameters, the two calculations are 100% identical. Simple symmetry considerations imply that if one is right, they are both right.

[See original post regarding free Mathematica Player.]

Mike Gottlieb
Physics Department
California Institute of Technology
http://www.feynmanlectures.info"

 

[Mentz114]

Prof. Gottlieb,

thanks for this. I have played your notebook and I have to say I have a slight worry about the straight line cdt. For some reason I have a feeling there will be aberration and the angle of reception is not what you've calculated. Or am I thinking of freely-falling frames ?

M

 

[codelieb]

M,

I have a slight worry about the straight line c dt. For some reason I have a feeling there will be aberration and the angle of reception is not what you've calculated. Or am I thinking of freely-falling frames ?

I can't quite say what you are thinking of (since I am not telepathic :-), but I will try to answer your concern:

Please recall that all my calculations are made in an inertial frame co-moving with the rocket at the moment of transmission, whose origin (event) is the location of the transmitter at the moment of transmission ("transmitter's frame").

In the transmitter's frame, the radiation from the transmitter (which I imagine as a photo flashbulb), is an electromagnetic field, radiating outward from the origin in all directions. The wavefront is spherical and its radius advances at the velocity of light 'c'. The 'dt' in my calculation is defined to be the time (in the transmitter's frame) that it takes the light to get from the transmitter to the receiver, which equals the time required for the wavefront to advance from the point of transmission to the point of reception (in the transmitter's frame), equal to 'c dt'.

The dotted line in my figure marked "c dt" is not meant as a representation of the radiation, per se (which is not 1-dimensional). However, if you prefer to think of light (from a flashbulb) as particles traversing empty space, then (in any inertial frame) they advance isotropically in straight lines, and 'c dt' is the distance they travel in the time dt.

Mike

 

[Mentz114]

Mike,
I'm still not convinced. I think your result holds for infinitesmal dy , dt but you have to integrate to get the effect for finite dy, dt.

M

 

[codelieb]

I am not sure what you want me to integrate over...

However, if you agree that the receiver sees radiation at some constant frequency, that implies that the frequency is the same in any interval dy,dt. What I am calculating is the instantaneous time rate of change of the phase of the wave at the receiver at the moment of reception. This calculated frequency has to be the same at the next moment, the next, and the next, ad infinitum (unless you posit that the receiver sees radiation whose frequency varies with time).

 

[Mentz114]

Those rocket equations look like time-like geodesics of the Rindler space-time. If the equivalence principle let's us model the rockets as frames in that metric, the calculation is different. The rockets are experiencing proper acceleration so that doesn't sound too likely. I'll think about it.

 

[yuiop]

In the transverse situation, the receiver is effectively going away from the transmitter and there is a doppler effect that redshifts the apparent frequency of the signal. However the average velocity (and degree of time dilation) of the clock on the right between the reception of two crests is higher than the speed of the clock on the left at the time the signal was emitted. This blueshifts the signal. It seems reasonable that the two effects would cancel each other out, resulting in no transverse gravitational redshift. I will check it out in more detail later.

 

[codelieb]

kev,

You are correct that the non-relativistic Doppler formula predicts a transverse redshift. However, the relativistic Doppler formula used in my calculation differs from the non-relativistic formula precisely in that it takes time dilation into account.

Mike

 

[codelieb]

M,

You were right to be suspicious of my calculation because it contains two errors, one of them related to aberration, as you suspected. However, these two errors exactly cancel each other, so the result is the same: no "transverse gravitational redshift." A description of the errors and their corrections can be found here:

http://www.feynmanlectures.info/notgr/notgr_errata.pdf

I learned something interesting from these mistakes that I believe give a clue regarding the false calculation of "transverse gravitational redshift" in A. Mayer's ebook. I suspect he ignored aberration (equivalent to the orientation of the receiver, as described in the errata). If you do that then you get, for the angle between the velocity vector and the incoming light, theta' = Pi/2 - alpha, with alpha = ArcTan[dy/d], instead of what it should be, theta = Pi/2 + alpha. Using this (wrong) theta' in the calculation produces a fictitious redshift very reminiscent of Mayer's:

w = w0/[1+(dg)^2/c^4]. [FALSE]

Mayer's formula (given in Eq. (66) in his ebook) is:

w = w0 Sqrt[1-(dg)^2/c^4]. [MAYER]

Note that to the first order in the small term (dg)^2/c^4,

1/[1+(dg)^2/c^4] ~= 1 - (dg)^2/c^4

Thus Mayer's redshift factor and the more exactly calculated (but false) redshift factor above (derived by ignoring aberration) differ, to the first order, only by a square. The reason the factors are not identical may be because Mayer uses Newtonian approximations in his derivation: for the time it takes the light to go from transmitter to receiver (which he approximates as d/c), for the velocity gained by the rocket in that time (which he approximates as gd/c), and for the angle between the rocket's velocity and the light path (which he approximates as Pi/2).

Mike

 

[Mentz114]

Hi Mike,

I'm convinced. I think what worried me was that the transmitter would have to aim above the receiver to hit it, which has now been taken into account.

I haven't seen Mayer's calculation but if he's working in GR then it's a standard problem which I have failed to do many a time !

M

 

[timewarp]

I downloaded the notgr.nb and also the errata.pdf and checked it.
I think you have got the angle of reception still wrong.
It is not corrected for abberation due to the velocity at the receiver.
I understand that you did use the general doppler formula as e.g. given by wikipedia
http://en.wikipedia.org/wiki/Relativistic_Doppler_effect#Accelerated_motion
But you just plugged in the theta, but should have used
theta_rxv = Arccos[(Cos[theta]-dv)(1-dv Cos[theta]]
instead (all using your variable names)

I'd really like to correct the notebook, but I have no Mathematic available (only the player).
I assume that with this correction the effects will not cancel out.

General relativistic doppler effects are derived in detail here:
http://www.mathpages.com/rr/s2-04/2-04.htm

The effect due to abberation at the receiver is shown very clearly in
http://www.mathpages.com/rr/s2-05/2-05.htm
(look for the diagram looking like ">--")

 

[pervect]

It seems to me that any notion of "gravitational redshift", transverse or not, is going to be coordinate dependent.

So, I would expect it to be there sometimes, and not there at others, all depending on one's coordinate choice.

To take a specific example, if one looks at an accelerating rocket from an inertial frame, there is no gravitation or gravitational effects at all, and there are only doppler shifts..

But if you choose to go to the rindler frame, then there is an associated metric, which varies with height, giving rise to a "gravitational redshift" when one uses the Rindler coordinates.

So it's purely a matter of bookeeping in this case as to whether or not "gravitational redshift" exists at all. In Rindler coordinates it exists, and in inertial coordinates it does not.