Aufgaben:Problem 8

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Task

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.