B Solving an Integral on a Spherical Surface - Tips

[victorneto]

Hello.

I ask for solution help from the integral below, where y and x represent angles in a metric of a spherical, 2-D surface. He was studying how to obtain the geodesic curves on the spherical surface, the sphere of radius r = 1, to simplify. The integral is the end result. It is enough, now, to integrate, to obtain the equation of the geodesics in such surface. It happens that I could not solve the integral, for which I used all the methods I know (there must be some that I do not know ...), with no results. I then used Mathematica (Wolfram Alpha), but the answer is a very complex, unintelligible equation. At least for me.$$Δy=∫1/Sin[x]*Sqrt[k^-2*Sin[x]^2-1]$$

Could anyone indicate any tips?

 

[vanhees71]

Could you clearly define the integral you want to solve? It's hard to make sense of your formula above.

It's of course not very smart to use spherical coordinates to get the geodesics on a sphere. Cartesian coordinates with the obvious holonomous constraint are much more convenient (which admittedly is unintuitive since spherical coordinates seem "natural" to describe a sphere, which is why they are called "spherical" in the first place ;-))).

The geodesics on the sphere follow from the Lagrangian
$$L=\frac{1}{2} \dot{\vec{x}}^2 - \frac{\lambda}{2} (\vec{x}^2-1),$$
where $\vec{x} \in \mathbb{R}^3$, and $\lambda$ is the Lagrange parameter for the constraint, $\vec{x}^2=1$.

The canonical momenta are
$$\vec{p}=\frac{\partial L}{\partial \dot{\vec{x}}}=\dot{\vec{x}}$$
and the geodesic is thus given by
$$\dot{\vec{x}}=\ddot{\vec{x}}=\frac{\partial L}{\partial \vec{x}}=-\lambda \vec{x}. \qquad (1)$$
To determine $\lambda$ we take the 2nd deriviative of the contraint equation
$$\vec{x}^2=1 \; \Rightarrow \; \vec{x} \cdot \dot{\vec{x}}=0 \; \Rightarrow \; \dot{\vec{x}}^2 + \vec{x} \cdot \ddot{\vec{x}}=0.$$
With the equation of motion (1) we get
$$\dot{\vec{x}}^2-\lambda \vec{x}^2=\dot{\vec{x}}^2-1=0. \qquad (2)$$
Further we have the "Hamiltonian" of the variational principle as a conserved quantity since the Lagrangian doesn't depend explicitly on "time":
$$H=\vec{p} \cdot \dot{\vec{x}}-L=\frac{1}{2} \dot{\vec{x}}^2 + \frac{\lambda}{2} (\vec{x}^2-1) = \frac{1}{2} \dot{\vec{x}}^2=\text{const}.$$
We can choose the constant to be $1$ by just rescaling the independent parameter $t$. BTW: This is the advantage of the "square form" of the Lagrangian for the geodesics, because it always automatically makes the curve parameter an affine parameter for the geodesic. So we have $\dot{\vec{x}}^2=1$ and thus using this in (2) gives $\lambda=1$. Plugging this into (1) finally gives
$$\ddot{\vec{x}}=-\vec{x}.$$
The general solution of this obviously is
$$\vec{x}(t)=\vec{A} \cos t + \vec{B} \sin t.$$
The constraint tells us that
$$\vec{x}^2=1 \; \Rightarrow \; \dot{\vec{x}} \cdot \vec{x}=0 \; \Rightarrow \; (\vec{A} \cos t +\vec{B} \sin t)(-\vec{A} \sin t + \vec{B} \cos t) = (\vec{B}^2- \vec{A}^2) \sin t \cos t + \vec{A} \cdot \vec{B} (\cos^2 t - \sin^2 t)=\frac{1}{2} (\vec{B}^2-\vec{A}^2) \sin(2 t) + \vec{A} \cdot \vec{B} \cos(2 t)=0.$$
Setting $t=0$ implies $\vec{A} \cdot \vec{B}=0$. Setting $t=\pi/4$ yields $|\vec{A}|=|\vec{B}|$.

This means that $\vec{A}$ and $\vec{B}$ are two perpendicular vectors of equal length $\ell$. Defining thus $\vec{e}_1=\vec{A}/\ell$ and $\vec{e}_2 = \vec{B}/\ell$ and $\vec{e}_3=\vec{e}_1 \times \vec{e}_2$ one has a Cartesian coodinate system, where the geodesic reads
$$\vec{x}=\ell \begin{pmatrix} \cos t \\ \sin t \\ 0 \end{pmatrix}.$$
Employing once more the constraint $\vec{x}^2=1$ gives $\ell 1$, and indeed the geodesic is a unit circle around the origin, i.e., a great circle on the sphere.