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

[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).

$$\frac{d^2 t}{d \tau^2} + \frac{2g}{1+gz} \frac{dt}{d\tau} \frac{dz}{d\tau} = 0$$

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

$$\frac{d^2 x}{d \tau^2} = \frac{d^2 y}{d \tau^2} = 0$$

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

$$\frac{d}{d\tau} \left( \left( (1+gz(\tau))^2 \right) \frac{dt(\tau)}{d\tau} \right) = 0$$

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