Thursday 7 March 2019

homework and exercises - Photon on null geodesic


If given an FRW metric


$ds^2 = -dt^2 + a^2(t)[dx^2+dy^2+dz^2]$


and for the trajectory followed by a photon (null geodesic; $ds^2=0$) with affine parameter $\lambda$, know that


$g_{\mu\nu}\,\frac{dx^{\mu}}{d\lambda}\,\frac{dx^{\nu}}{d\lambda} = 0.$


How does one find


$\frac{dt}{d\lambda} = \frac{w0}{a(t)}$?


Already found the nonzero Christoffel coeffs and the remaining geodesic equations,



$ 0 = \frac{d^2x^i}{d\lambda^2} + \Gamma^i_{ti}\,\frac{dt}{d\lambda}\,\frac{dx^i}{d\lambda} $ and $ 0 = \frac{d^2t}{d\lambda^2} + \Gamma^t_{ii}\,\frac{dx^i}{d\lambda}\,\frac{dx^i}{d\lambda} $


with $ \Gamma^i_{ti} = \frac{1}{a(t)}\,\frac{da(t)}{dt} $ and $ \Gamma^t_{ii} = a(t)\,\frac{da(t)}{dt} $


I expect it to be pretty simple (yet not seeing it) and that the null geodesic in the $x$-direction come from something like (?):


$ ds^2 = -(dt^2)+a^2(t)\,dx^2 = 0 \rightarrow \frac{dt}{a(t)} = \pm dx $


I also believe that $dx/d\lambda = c$ if the affine parameter is related by $\lambda = t$.


This is where I am confused and need it to move on to look at the cosmological redshift and derive the ratio of emitted and observed energies of a photon at $t1$ and $t2$.


Anything to straighten me out would be great!



Answer



I'll use dots for derivative with respect to affine parameter. The FRW metric has Killing vectors $\partial_x, \partial_y, \partial_z$ each of which leads to a conservation equation: \begin{align} c_x &= \dot x\cdot\partial_x = a^2\dot x \\ c_y &= \dot y\cdot\partial_y = a^2\dot y \\ c_z &= \dot z\cdot\partial_z = a^2\dot z \end{align} which implies $$ a^4(\dot x^2 + \dot y^2 + \dot z^2) = w_0^2, \qquad w_0^2 \equiv c_x^2 +c_y^2 +c_z^2 $$ The null geodesic equation can, as you basically point out, be written $$ \dot t^2 = a^2 (\dot x^2 + \dot y^2 + \dot z^2) $$ multiplying both sides by $a^2$ and using the conservation condition derived above gives $$ a^2\dot t^2 = w_0^2 $$ which is essentially what you were looking for.


Unsolicited Advice (which has helped me enormously in research by the way)



When solving for geodesics, learn to love conserved quantities, and therefore Killing vectors!!!


No comments:

Post a Comment

Understanding Stagnation point in pitot fluid

What is stagnation point in fluid mechanics. At the open end of the pitot tube the velocity of the fluid becomes zero.But that should result...