Gravitational redshift from doppler shift

[TrickyDicky]

Hi, I'm interested in the derivation of the gravitational redshift formula from the Doppler shift in an accelerated frame formula (or viceversa), that are linked by the Equivalence principle, it should be pretty straightforward but I get stuck. Anyone can show me this or point me to a link with the derivation worked out?
Thanks

 

[Dale]

Start with a detector on the floor and an emitter on the ceiling. Determine how much time it takes for light to go from the ceiling to the floor and then convert that into a velocity. Finally, use that velocity in the usual expression for the Doppler effect.

 

[bcrowell]

http://www.lightandmatter.com/html_books/genrel/ch01/ch01.html#Section1.5

See section 1.5.5.

 

[George Jones]

The standard argument, which I am too lazy to type right now, is given, for example, on pages 125-127 of the book Gravitation: Foundations and Frontiers by T. Padmanabhan.

A Google search turned up

http://www.springerlink.com/content/k13204g473901v54/,

which I have yet to read, but it appears that this article gives some criticisms of the standard argument.

 

[Passionflower]

For an accelerating observer in a linear accelerated frame in flat spacetime the following applies:

An observer with a proper acceleration of g will measure the frequency of light emitted from an inertial source a distance x away shifted by (c=1):

<br /> -g \left( 1+gx \right) ^{-1} \left( \ln \left( 1-{\frac {g}{v}}<br /> \right) \right) ^{-1}<br />

Dividing this for two observers (x2 > x1) we get:

<br /> -{\frac {g \left( {\it x2}-{\it x1} \right) }{1+g{\it x2}}}<br />

Now you could compare this to a situation in a homogeneous gravitational field. The catch is that a homogeneous field in GR is not straightforward.

I suppose we could start writing the metric and then observe the coordinate behavior of light and a static observer in such a field. Issue is there is no vanishing field at infinity, we could of course consider the light sent from an accelerating source which accelerates exactly against the inertial acceleration of the field but that obviously leads to a circular reasoning, since we would apply the equivalence principle in such a case.

 

[TrickyDicky]

For an accelerating observer in a linear accelerated frame in flat spacetime the following applies:

An observer with a proper acceleration of g will measure the frequency of light emitted from an inertial source a distance x away shifted by (c=1):

<br /> -g \left( 1+gx \right) ^{-1} \left( \ln \left( 1-{\frac {g}{v}}<br /> \right) \right) ^{-1}<br />

Dividing this for two observers (x2 > x1) we get:

<br /> -{\frac {g \left( {\it x2}-{\it x1} \right) }{1+g{\it x2}}}<br />

Thanks for the formulas, unfortunately I can't follow that derivation, could you write the complete equations? The first formula would give the observed frequency, but how we relate it to the source frequency? The second one I don't know what it is or how is derived.

For an object with non-relativistic velocity isn't the right equation?:

f=f&#039;(1+\frac{gx}{c^2})=f&#039;(1+\frac{gt}{c})

with f=frecuencia observada
f'=frecuencia emitida por el objeto
g= constant accelaration of the reference frame of the observer
x=distance that the object travels towards the observer
t=x/c time it takes for the object to travel distance x at v<<c

Now you could compare this to a situation in a homogeneous gravitational field. The catch is that a homogeneous field in GR is not straightforward.

Well , right the only gravitational fields we know are not homogenous being usually originated in almost spherical sources and besides they are only constant at a constant distance from the center of the source, that's why the equivalence principle only holds locally, where the cuvature is negligible.

I suppose we could start writing the metric and then observe the coordinate behavior of light and a static observer in such a field. Issue is there is no vanishing field at infinity

Sure, instead we would have infinity populated by objects at c, actually this is the velocity space of the Minkowski spacetime, so the metric is already written. This just shows that the Equivalence principle is just the confirmation that locally, our universe is SR, and that curvature only appears when we consider bigger distances.

 

[Passionflower]

Thanks for the formulas, unfortunately I can't follow that derivation, could you write the complete equations? The first formula would give the observed frequency, but how we relate it to the source frequency? The second one I don't know what it is or how is derived.

You can enter the acceleration (g) and the source frequency (v) and the result of the formula represents the target frequency. As I wrote before the second is simply the division, e.g. ( the formula for (x2 - x1)/x1 )

For an object with non-relativistic velocity isn't the right equation?:

f=f&#039;(1+\frac{gx}{c^2})=f&#039;(1+\frac{gt}{c})

That is an approximation not an exact formula.

 

[Ich]

Have a look at the attachment in https://www.physicsforums.com/showpost.php?p=1769524&postcount=33". It shows a spacetime diagram of two consecutive light signals (blue and red) going up from one accererating observer to the next. The red signal takes longer than the blue one, you can get the time dilation formula if you figure out how much.