Reference: Moore, Thomas A., *A General Relativity Workbook*, University Science Books (2013) – Chapter 8; Problem 8.2.

We’ve seen that geodesic curves are found by maximizing the proper time between two events, where the proper time interval is given by

We can do this by solving the set of Lagrangian differential equations given by

where the Lagrangian function is given by

Here, the object’s world path is given in parametric form by . The metric can depend on the position, so it too is ultimately a function of the parameter , which plays the role of in 2. For this Lagrangian, the coordinates are and the generalized velocities are . We therefore have

This follows because does not depend on (it depends only on the coordinates, not the velocities), and also because .

The other derivative comes out to

We can get rid of in these two results by noticing from 1 that

We therefore get

Putting it all together using 2, we get

(To those following along in Moore’s book: his equation 8.11 is wrong, and should be the equation given here.) We can now eliminate the parameter by multiplying through by :

This is one form of the *geodesic equation*, which is a second order ordinary differential equation. (It’s an ordinary differential equation despite the appearance of the partial derivative because we will know the metric as a function of the coordinates, so this derivative will be known when we set out to solve the ODE.)

As an example of how we can use this equation to determine geodesics, we’ll look at the usual example of the curved 2-d surface of a sphere of radius . The metric for this space is, using the usual spherical coordinates and

The required derivatives of are

Then 15 becomes (using as the parameter in place of ) for :

And for :

The last equation can be integrated once to get

where is a constant with dimensions of length, so that is dimensionless. Substituting this into the other ODE gives

We’ve decoupled the equations, although this latest ODE isn’t exactly easy to solve. We can make a bit of progress by observing that in 2-d space, the infinitesimal interval is given by , so

Here, this gives us

Although we could integrate this directly (using software), the answer isn’t terribly illuminating. We can take a different approach by rearranging the equation to get

where . Now we can use the substitution with and we get

If we want the path length to be zero at , this corresponds to , so we take . This gives

Defining and squaring both sides, we get

Returning to 23, we can now eliminate , since

Integrating this using software gives the rather cryptic result

We can convert this into something more meaningful if we remember that , so . Also, since

the inverse functions are related by

Putting this together, we get

We now have equations giving and in terms of :

We know that the geodesics on a sphere should be arcs from great circles, that is, arcs from circles formed by the intersection of a plane containing the centre of the sphere with the sphere itself. Note that we’re looking for great circles that connect any two points on the sphere, so these circles need not go through the poles. We can define these circles by considering a plane with equation where is a constant, and its intersection with the sphere . All these planes will contain the axis, but we are free to define the axis pointing in any direction from the centre of the sphere so we haven’t really restricted the solution in any way.

The intersection of the plane and sphere is given by converting to spherical coordinates:

Thus the equation is the equation of a great circle that includes the intersection of the axis with the sphere. Does this agree with the solution from the geodesic equation above?

Conventional spherical coordinates requires along the axis and since we’re passing all our planes through that axis, we need to choose the constant above to match this. We’ve taken on the equator, which also intersects the axis, so we need to take as well. Therefore the geodesics are

We need to use a bit of trigonometric wizardry to convert these equations. First, , so

The LHS and RHS are equal if

That is, these are great circles if we identify

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Chris KranenbergThe expression k=√(1-a^2 ) leading to √(-1+a^2 )=ik should be

√(-1+a^2 )=-ik, and since 〖tanh〗^(-1) (x) is an odd function, the integrated result is

∅-∅_0=(k〖tanh〗^(-1) (-iktanu))/(-ik)=(-k〖tanh〗^(-1) (iktanu))/(-ik)=(〖tanh〗^(-1) (iktanu))/i where u=s/R.

A different approach that may be more direct is taking the hyperbolic tangent of each side giving

tanh(i(∅-∅_0 ))=iktanu

Now,

tanh�?�(ix)=(e^ix-e^(-ix))/(e^ix+e^(-ix) )=2isinx/2cosx=isinx/cosx=itanx

So,

itan(∅-∅_0 )=iktanu

tan�?�(∅-∅_0 )=ktanu

Also, the right side of the equation after “putting this all together, we get�? should be

∅-∅_0=1/i iarctan(ktan(s/R))

growescienceTaking in equation 43 wouldn’t make any difference since, as you say, arctanh is an odd function so

which is the same result that I use to get equation 48. [Unless I’m missing something?]

Chris KranenbergYes, Glenn, it is the same result but for the less experienced explicitly using sqrt(-1 + a^2) = -ik and envoking the odd function property can be more clear.

growescienceI guess what I’m not seeing is why it’s more natural to take in equation 43 rather than . To me the positive square root always seems the more natural choice unless there’s some reason the negative root is required.

Chris KranenbergIt can be viewed as starting with +sqrt(1 – a^2) = k then factoring out an i then multiplying both sides by -i giving +sqrt(-1 – a^2) = -ik. Your point is well taken. Thanks.

growescienceFixed the tan -> arctan in equation 48. Thanks.

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JoelHow do you get from (35) to (36)? That integral should yield arcsin, shouldn’t it?

gwrowePost authorThe integral can also be an arcsin, but the arctan is correct too. You can use the trig identity

to convert between them. [To prove this identity, just let .]