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Suppose, this is the line element of a spherically symmetric FLRW metric, $$ ds^2 = -[1 + 2ψ(t,x)]dt^2 + a^2(t) [1 - 2ψ(t,x)]dx^2 $$ and the geodesic equation is, $$ \frac{d^2x^α}{dλ^2} = - Γ_{βγ}^α \frac{dx^β}{dλ} \frac{dx^γ}{dλ} $$ where λ is the affine parameter. Again, for null geodesic we can write

$$ ds^2 = 0 $$ Now, if I want to solve this null geodesic equation I know that I need to convert the four 2nd order differential equation into eight first order equation first and then easily anyone can solve this by using any mathematical tools. Actually, my concern is about the initial conditions, how can I set initial conditions to solve this null geodesic equation, i.e. what will be the initial conditions here for x, y, z and for the four velocities, \$$ \frac{dt}{dλ}, \frac{dx}{dλ}, \frac{dy}{dλ}, \frac{dz}{dλ}$ respectively, suppose, the initial condition for time is today that means I want to integrate the geodesic equation from today i.e, the redshift is zero.

Actually I have asked the same question here, please see the link

but did not get the answer, that's why I'm asking here again.

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  • $\begingroup$ I think this place would fit your question. $\endgroup$ – Max0815 Feb 25 at 4:39
  • $\begingroup$ This has been cross-posted in Physics as well as in Astronomy where an answer has been posted. $\endgroup$ – uhoh Feb 25 at 7:08
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First of all, notice that a geodetic preserve the metric of the tangent vector, since:

$\nabla\left\langle \dot{\gamma}(t),\dot{\gamma}(t)\right\rangle=2 \left\langle\nabla\left(\dot{\gamma}(t)\right),\gamma(t)\right\rangle=0$

Thus, if you want a null geodesic, it is enough for the initial vector to be null, and the condition (that is, for the geodesic to be light-like) will follow. we are imposing one condition, which is going to give a relation between x and t (notice that, since you are talking about squared differentials, the solutions is not going to be unique, but you will have to decide their signs).

The geodesic then follows from the point you choose in the spacetime from which to start, in this case from today, on a certain point.

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  • $\begingroup$ Thanks for your answer. Can you please explain me a bit about the term ''uniqueness'' to solve geodesic equation. $\endgroup$ – Photon Mar 3 at 23:03
  • $\begingroup$ @Photon there is not only one vector with norm $0$. In fact, it is true for every vector for which $c^2t^2-x^2=0$. The solution is thus not unique. In fact, you could have a geodesic pointing toward the past or toward the future, that is your choice. Using this equation, you are considering only light-like geodesic. Using the assumption that the geodesic starts from today and I suppose points to the future, the geodesic (that is unique only if you specify the starting point and the starting velocity) will follow $\endgroup$ – Gabriele Cassese Mar 4 at 6:34
  • $\begingroup$ Thanks again. Yes, that is my concern, actually. How can I specify the starting position and the starting velocity as well, is there any formula or any other procedure? $\endgroup$ – Photon Mar 4 at 6:44
  • $\begingroup$ @Photon You just have to choose them and then use them as initial condition in your differential equation $\endgroup$ – Gabriele Cassese Mar 4 at 6:47

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