Doppler & Gravitational Redshift: Non-Inertial Observer in SR?

[gptejms]

Oxymoron,in your calculation of dz/dtau &. dt/dtau are you doing this:-starting from the Rindler metric find out the christoffel symbols and use those in the geodesic equation.If yes,pl. write out the details of your calculation i.e. christoffel symbols etc.

Have you shown that the relativistic doppler effect formula follows from the GR calcution during the period of acceleration?

 

[bernhard.rothenstein]

In Special Relativity,

If an inertial photon emitter, A, emits photons every second and an observer, B (moving away from A at a constant velocity v), intercepts these photons and instantaneously reflects them back to A then we would expect to see the Doppler redshift effect.

My question I would like to pose is this: suppose that B is non-inertial, suppose that B is accelerating away from A (where A remains inertial). If B is accelerating then we could also say that B is inertial in a gravitational field and therefore we should see the Gravitational redshift effect.

Does the accelerating observer still see regular doppler effect? Is the expected gravitational redshift just the Doppler effect in disguise? Is there gravitational redshift at all?

PS. I understand that SR does not work for curved spaces and just because I said that B expects to see grav. redshift does not mean I think there is curvature. What gravitational effects B sees would be, I guess, pseudo-gravitational.

The paper
Frequency measurements by uniformly accelerating observers
R.Neutze and W.Moreau
Physics Letters A 179 (1993) 389,390
offers an answer to your questions.

 

[vanesch]

Oxymoron,in your calculation of dz/dtau &. dt/dtau are you doing this:-starting from the Rindler metric find out the christoffel symbols and use those in the geodesic equation.If yes,pl. write out the details of your calculation i.e. christoffel symbols etc.

If it can be of any help, I worked out with Mathematica the equations of motion for a geodesic in a general diagonal metric (of which the Rindler metric is a special case), with the connection coefficients (called gammanormal) and all that.
You can find it as the attached document "geodesic1.nb" in
this post:
https://www.physicsforums.com/showpost.php?p=994335&postcount=138

(where I had to do it for another reason, which was a counter argument to Hossenfelder's anti-gravity)

Small detail: I write the metric as: ds^2 = g11 dt^2 + g22 dx^2 +...
note that there is no minus sign before g11, so it has to be included in the definition of g11 itself.

 

[pervect]

For the line element

-(1+gz)dt^2 + dx^2 + dy^2 + dz^2

(one form of the rindler metric)

it's fairly easy to compute that the only non-zero Christoffel symbols

\Gamma_{ztt} = -\Gamma_{tzt} = -\Gamma_{ttz} = g(1+gz)

from the defintion of the Christoffel symbols, and the fact that only g_{tt} has any non-zero derivatives. (This uses the usual form, where the last two indexes commute).

see for example
http://en.wikipedia.org/wiki/Christoffel_symbols<br /> G_{ijk} = \frac{1}{2} \left( - \partial g_{jk}/ \partial x^i + \partial g_{ik} / \partial x^j + \partial g_{jk} \partial x^i \right)<br />

(This is the usual form where symmetric in the last two indices.) Usually it is desired to raise the first index, though.

Raising the index with g^{ij} (easy to calculate and do) then gives:

<br /> \Gamma^z{}_{tt} = (1+gz)g \hspace{.5 in} \Gamma^t{}_{zt} = \Gamma^t{}_{tz} = \frac{g}{1+gz}<br />

 

[vanesch]

For the line element

-(1+gz)dt^2 + dx^2 + dy^2 + dz^2

(one form of the rindler metric)

it's fairly easy to compute that the only non-zero Christoffel symbols

\Gamma_{ztt} = -\Gamma_{tzt} = -\Gamma_{ttz} = g(1+gz)

from the defintion of the Christoffel symbols, and the fact that only g_{tt} has any non-zero derivatives. (This uses the usual form, where the last two indexes commute).

see for example
http://en.wikipedia.org/wiki/Christoffel_symbols


<br /> G_{ijk} = \frac{1}{2} \left( - \partial g_{jk}/ \partial x^i + \partial g_{ik} / \partial x^j + \partial g_{jk} \partial x^i \right)<br />

(This is the usual form where symmetric in the last two indices.) Usually it is desired to raise the first index, though.

Raising the index with g^{ij} (easy to calculate and do) then gives:

<br /> \Gamma^z{}_{tt} = (1+gz)g \hspace{.5 in} \Gamma^t{}_{zt} = \Gamma^t{}_{tz} = \frac{g}{1+gz}<br />

Indeed, that's what I have too.

For fun, I attach a quite general mathematica notebook that does this kind of calculations...

 

[pervect]

I tend to use GRtensorII for this sort of stuff - it's very nice. I gather it has a mathematica version too, GRTensorM. There's also GRtensorJ, a "teaching" version.

http://grtensor.phy.queensu.ca/
http://grtensor.org/teaching/

On the totally free side, there is Maxima. (GRtensorII itself is free, but it requires either Maple or Mathematica, depending on the version). Maxima is a stand-alone totally free program based on Macsyma. It's not nearly as nice as GRTensorII though it does have some limited tensor-handling capability built in.

http://maxima.sourceforge.net/

Some other possibilites are mentioned at
http://math.ucr.edu/home/baez/RelWWW/software.html

I suppose I should add that there are a couple of different ways of odering Christoffel symbols, GRtensorII uses the "wrong" one. It's easily fixed with a one-line defintion, but a new user might not notice this.

 

[gptejms]

Thank you all for your answers!
I had asked another question in the the same post which is "Have you shown that the relativistic doppler effect formula follows from the GR calcution during the period of acceleration?"Hope you guys will write out the calculation for this too.

 

[pervect]

BTW, the above Christoffel symbols give the following geodesic equations for a worldline paramaterized by proper time tau, i.e. 4 functions

t(tau),x(tau),y(tau),z(tau).

<br /> \frac{d^2 t}{d \tau^2} + \frac{2g}{1+gz} \frac{dt}{d\tau} \frac{dz}{d\tau} = 0<br />

<br /> \frac{d^2 z}{d \tau^2} + g(1 + gz) \left( \frac{dt}{d\tau} \right)^2 = 0<br />

<br /> \frac{d^2 x}{d \tau^2} = \frac{d^2 y}{d \tau^2} = 0<br />

If one wants to solve them, it is worth noting that the first of these equations is equivalent to

<br /> \frac{d}{d\tau} \left( \left( (1+gz(\tau))^2 \right) \frac{dt(\tau)}{d\tau} \right) = 0<br />

as applying the chain rule will show

which means that (1+gz)^2 dt/d\tau is a constant,

This is a result of the fact that E_0 is constant because the unit t vector is a Killing vector.

This observation makes solving the above equations a lot easier, dt/dtau can be eliminated, leaving only one equation for z:

P_x and P_y are also constant, of course it is obvious that if d^2x/d\tau^2=0, dx/d\tau = constant.

[add-correct]
The above equations imply that g_tt dt^2 + g_zz dz^2 = constant, but they don't specify the value of the constant.

Depending on whether one wants null, spacelike, or timelike geodesics, one must make the above constant equal to 0, +1, or -1.

The solution for null geodesics turns out to be:

z = -1/g + sqrt(2 K \tau/g)
t = ln(\tau)/2g