# Aufgaben:Problem 8

## Contents

Let $$R > 0$$. On $$\mathbb{R}^2$$, consider the metric $$g(u,v) = \frac{4R^4}{(R^2+u^2+v^2)^2} \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix},\ (u,v) \in \mathbb{R}^2$$

a) Write down the Hamiltonian equations describing the geodesic curves.

b) Let $$C_1 = \frac{4R^4}{(R^2+u^2+v^2)^2}(2uv\dot{u} +(R^2-u^2+v^2)\dot{v})$$ $$C_2 = \frac{4R^4}{(R^2+u^2+v^2)^2}(2uv\dot{v} +(R^2+u^2-v^2)\dot{u})$$ $$C_3 = \frac{4R^4}{(R^2+u^2+v^2)^2}(v\dot{u} - u\dot{v})$$ Show that $$C_1, C_2, C_3$$ are conserved and that

$$-C_1 u + C_2 v +C_3(u^2+v^2-R^2) = 0 \forall (u,v)\in \mathbb{R}^2$$ c) Show that the energy h satisfies $$h \geq 0$$ and discuss the case $$h = 0$$.

d) Describe the geodesics when $$h > 0$$ depending on the values of $$C_1, C_2, C_3$$.

## Solution

### Solution (a)

$$g_{ij}(u,v) = \frac{4R^4}{(R^2+u^2+v^2)^2} \delta_{ij}$$

because $$\big(g^{ij}(u,v)\big) = \big(g_{ij}(u,v)\big)^{-1}$$ $$\Rightarrow g^{ij}(u,v) = \frac{(R^2+u^2+v^2)^2}{4R^4} \delta^{ij}$$

let $$x = (u,v)$$ and $$\xi = (\xi_1, \xi_2)$$, the hamiltonian is defined as $$h(x, \xi) = \frac{1}{2} g^{ij}(x)\xi_i \xi_j$$

$$\Rightarrow h(x, \xi) = \frac{(R^2+u^2+v^2)^2}{8R^4} (\xi_1^2 + \xi_2^2)$$

now we can compute the Hamiltonian equations:

$$\dot{u} = \frac{\partial h}{\partial \xi_1} = \frac{(R^2+u^2+v^2)^2}{4R^4} \xi_1$$ $$\dot{v} = \frac{\partial h}{\partial \xi_2} = \frac{(R^2+u^2+v^2)^2}{4R^4} \xi_2$$

$$\dot{\xi_1} = -\frac{\partial h}{\partial u} = -\frac{(R^2+u^2+v^2)}{2R^4} (\xi_1^2 + \xi_2^2) u$$

$$\dot{\xi_2} = -\frac{\partial h}{\partial v} = -\frac{(R^2+u^2+v^2)}{2R^4} (\xi_1^2 + \xi_2^2) v$$

we can also observe that:

$$\dot{\xi_2}u = \dot{\xi_1}v$$ and $$\dot{u}\xi_2 = \dot{v}\xi_1$$

### Solution (b)

$$C_1 = \frac{4R^4}{(R^2+u^2+v^2)^2}(2uv\dot{u} +(R^2-u^2+v^2)\dot{v})$$ using (a) $$= 2uv\xi_1 +(R^2 -u^2 +v^2)\xi_2 = 2uv\xi_1 +R^2\xi_2 -u^2\xi_2 +v^2\xi_2$$

if $$C_1$$ is conserved, its time derivative should be zero

$$\frac{dC_1}{dt} = 2\dot{u}v\xi_1 + 2u\dot{v}\xi_1 + 2uv\dot{\xi_1}+R^2\dot{\xi_2} - 2u\dot{u}\xi_2 - u^2\dot{\xi_2}+ 2v\dot{v}\xi_2 + v^2\dot{\xi_2}$$ using $$\dot{v}\xi_1= \dot{u}\xi_2$$ on the second term, the second and fifth term cancel: $$= 2\dot{u}v\xi_1 +2uv\dot{\xi_1}+R^2\dot{\xi_2} - u^2\dot{\xi_2}+ 2v\dot{v}\xi_2 + v^2\dot{\xi_2}$$ using $$v\dot{\xi_1} = u\dot{\xi_2}$$ on the second term: $$= 2\dot{u}v\xi_1 +2u^2\dot{\xi_2}+R^2\dot{\xi_2} - u^2\dot{\xi_2}+ 2v\dot{v}\xi_2 + v^2\dot{\xi_2}$$ $$= \dot{\xi_2}(R^2 + u^2+ v^2) + 2\dot{u}v\xi_1 + 2v\dot{v}\xi_2$$ using the Hamiltonian equations: $$= -\frac{(R^2+u^2+v^2)^2}{2R^4}(\xi_1^2 + \xi_2^2) v + 2v\frac{(R^2+u^2+v^2)^2}{4R^4} \xi_1^2 + 2v\frac{(R^2+u^2+v^2)^2}{4R^4} \xi_2^2 =0$$

As $$C_1, C_2$$ and the hamilton equation are symmetric under permutation of $$u$$ and $$v$$, $$C_2$$ is also conserved.

and for $$C_3$$

$$C_3 = \frac{4R^4}{(R^2+u^2+v^2)^2}(v\dot{u} - u\dot{v}) = (v\xi_1 - u\xi_2)$$ $$\frac{dC_3}{dt} = \dot{v}\xi_1 + v\dot{\xi_1} - \dot{u}\xi_2 - u\dot{\xi_2} = \dot{u}\xi_2 + v\dot{\xi_1} - \dot{u}\xi_2 - v\dot{\xi_1} = 0$$

Last we want to prove the identity:

$$-C_1 u + C_2 v +C_3(u^2+v^2-R^2) = \frac{4R^4}{(R^2+u^2+v^2)^2}\Big(-(2uv\dot{u} +(R^2-u^2+v^2)\dot{v})u$$ $$+ (2uv\dot{v} +(R^2+u^2-v^2)\dot{u})v + (v\dot{u} - u\dot{v})(u^2+v^2-R^2)\Big)$$ $$= \frac{4R^4}{(R^2+u^2+v^2)^2}\Big( -2uv\dot{u}u -R^2\dot{v}u+u^2\dot{v}u-v^2\dot{v}u + 2uv\dot{v}v +R^2\dot{u}v+u^2\dot{u}v-v^2\dot{u}v$$ $$+ v\dot{u}u^2 + v\dot{u}v^2 - v\dot{u}R^2 - u\dot{v}u^2 - u\dot{v}v^2 + u\dot{v}R^2\Big) = 0$$ as all the terms in the second bracket cancel.

### Solution (c)

$$h(x, \xi) = \frac{(R^2+u^2+v^2)^2}{8R^4} (\xi_1^2 + \xi_2^2)$$ $$h \geq 0$$ simply because all the terms are squared. For h = 0 the last bracket has to be zero because $$R > 0$$ $$\Rightarrow \xi_1^2 = - \xi_2^2 \Rightarrow \xi_1 = \xi_2 = 0$$ from the Hamiltonian equation follows that u = const. , v = const. which corresponds to the case of a constant geodesic.

### Solution (d)

$$C_1 u - C_2 v = C_3 (u^2+v^2-R^2)$$ for $$C_3\neq0$$: $$\frac{C_1}{C_3}u-\frac{C_1}{C_3}v=u^2+v^2-R^2$$ This looks like a circle, and indeed completing the square yields: $$(u-\frac{C_1}{2C_3})^2 + (v+\frac{C_2}{2C_3})^2=\frac{C_1^2+C_2^2}{4C_3^2}+R^2$$ The geodesics in this case are circles centered at $$(\frac{C_1}{2C_3}, -\frac{C_2}{2C_3})$$ with radius $$\sqrt{\frac{C_1^2+C_2^2}{4C_3^2}+R^2}$$. The radius is always greater than zero because $$R>0$$

For $$C_3=0$$: The geodesics are lines in $$\mathbb{R}^2$$ parametrized by $$(C_2,C_1)t, t\in\mathbb{R}$$.

If $$C_1 = C_3=0$$, the equations do not give a unique solution for $$u$$ but $$v = 0$$

If $$C_2 = C_3=0$$, the equations do not give a unique solution for $$v$$ but $$u = 0$$

If $$C_1= C_2= C_3=0$$:

$$C_3 = \frac{4R^4}{(R^2+u^2+v^2)^2}(v\dot{u} - u\dot{v}) \Rightarrow v\dot{u} = u\dot{v}$$

$$0 = C_1 =\frac{4R^4}{(R^2+u^2+v^2)^2}(2uv\dot{u} +(R^2-u^2+v^2)\dot{v})$$

$$0 = 2uv\dot{u} +(R^2-u^2+v^2)\dot{v} = 2u^2\dot{v} +(R^2-u^2+v^2)\dot{v} = (R^2+u^2+v^2)\dot{v} \Rightarrow \dot{v} = 0$$

$$C_2 = 0 \Rightarrow \dot{u} = 0$$

So in this case we have a geodesic that is constant at a single point in space.