What is the Purpose of the Geodesic Equation in General Relativity?

[space-time]

I started studying the geodesic equation:

∂²x^μ/∂s² = - Γ^μ_ab(∂x^a/∂s)(∂x^b/∂s)

where the term s is proper time according to the wiki(https://en.wikipedia.org/wiki/Geodesics_in_general_relativity).
The 2nd derivative on the left side of the equation is the acceleration in the x^μ direction.

Now my question is:

When I solve the geodesic equation, what exactly am I trying to solve for? Am I solving for s (or rather am I supposed to plug in information into the equation in order to derive s)? If not s, then what exactly am I solving for? If you already have your Christoffel symbols derived, then you should have already long had your x^μ functions derived (which is why I don't think you are solving for x^μ).

To be honest, this equation seems more to me like the Euler - Lagrange equations in the sense that, the main purpose doesn't seem to be to solve the equation, but more so to plug in known information in order to derive equations of motion.

 

[PeterDonis]

When I solve the geodesic equation, what exactly am I trying to solve for?

The function $x^{\mu} (s)$ that describes a geodesic worldline. Since it's a second order differential equation, you will need to specify two initial conditions in order to obtain a unique solution--usually these are $x^{\mu}$ and $dx^{\mu} / ds$ at $s = 0$.

If you already have your Christoffel symbols derived, then you should have already long had your xμ functions derived

No. The Christoffel symbols are derivatives of the metric, not of $x^{\mu}$. The coordinates $x^{\mu}$ here are four functions of the parameter $s$; they tell you the coordinates of some object whose worldline you want to describe at each value of $s$.

 

[Orodruin]

To add to what Peter already said, I think it should be pointed out that $s$ may be any affine parameter, not necessarily the proper time (although it is an obvious choice for time-like geodesics). For example, for light-like geodesics, the geodesic equations look exactly the same, but $s$ can no longer be identified with (a multiple of) the proper time, since the length of light-like geodesics is zero.

 

[space-time]

The function $x^{\mu} (s)$ that describes a geodesic worldline. Since it's a second order differential equation, you will need to specify two initial conditions in order to obtain a unique solution--usually these are $x^{\mu}$ and $dx^{\mu} / ds$ at $s = 0$.
No. The Christoffel symbols are derivatives of the metric, not of $x^{\mu}$. The coordinates $x^{\mu}$ here are four functions of the parameter $s$; they tell you the coordinates of some object whose worldline you want to describe at each value of $s$.

I see. Well then here is a question:

The right side of the equation invokes Einstein's summation convention where the derivatives of the $x^{\mu}$ functions (with respect to s) are taken multiple times for all values of μ. Now, if I don't know what $x^{\mu}$ is from the beginning, then how am I supposed to differentiate functions like x¹ or x² with respect to s? Here's an example of what I mean:

The Morris Thorne wormhole metric has Christoffel symbols Γ^μ_ab as follows:Γ¹₂₂ and Γ¹₃₃ are non zero. These are the only non-zero Christoffel symbols where μ= 1.

As a result of this, when I set μ= 1 (which means I am solving for x¹) in the geodesic equation, the equation turns out to look like this:

∂²x¹/∂s = [- Γ¹₂₂ * (∂x²/∂s) * (∂x²/∂s) ] + [- Γ¹₃₃ * (∂x³/∂s) * (∂x³/∂s) ]

(That is what the summation on the right of the equation ends up being).

As you can see, the summation contains derivatives of x² and x³ despite the fact that I'm solving for x¹. If I don't know any of the $x^{\mu}$ terms to begin with, then how am I supposed to take said derivatives of x² and x³?

 

[Orodruin]

As you can see, the summation contains derivatives of x2 and x3 despite the fact that I'm solving for x1.

Why do you think this is a problem? It is a set of coupled differential equations which have to be solved together.

 

[Nugatory]

Now, if I don't know what $x^{\mu}$ is from the beginning, then how am I supposed to differentiate functions like x¹ or x² with respect to s?

That's what makes solving systems of coupled differential equations so much fun. You have to make an educated guess as to what all the functions $x^\mu(s)$ might look like, plug these into the equations, see if they work, adjust and try again if it doesn't.

 

[Orodruin]

That's what makes solving systems of coupled differential equations so much fun. You have to make an educated guess as to what all the functions $x^\mu(s)$ might look like, plug these into the equations, see if they work, adjust and try again if it doesn't.

This sounds a bit like solving coupled differential equations is based on guesswork. I think it should be mentioned that there will often be first integrals based on symmetries which may often be used to integrate some of the equations and by using the result the other equations are simplified. Of course, this is not particular to relativity, but appears already in classical mechanics, eg, a charged particle moving in a constant magnetic field.

 

[m4r35n357]

When I solve the geodesic equation, what exactly am I trying to solve for?

Look at the derivation of equation (3) in this article. It neglects phi and theta, but the geodesic equations are written in full in terms of t and r. The rest of the article is fundamentally useful too, but goes beyond your actual question.

 

[Orodruin]

Look at the derivation of equation (3) in this article. It neglects phi and theta, but the geodesic equations are written in full in terms of t and r. The rest of the article is fundamentally useful too, but goes beyond your actual question.

The article you link to is dealing with the Schwarzschild metric. This is not applicable to the metric which the OP is studying.

 

[m4r35n357]

The article you link to is dealing with the Schwarzschild metric. This is not applicable to the metric which the OP is studying.

The OP did not quote a metric, it was a general question regarding the meaning of the geodesic equation. I maintain that I have answered the OP correctly with an illustrative example.