Difference between revisions of "Aufgaben:Problem 8"

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==Task==
  
 +
Let \(R > 0\). On \(\mathbb{R}^2\), consider the metric
 +
$$ g(u,v) = \frac{4R^4}{(R^2+u^2+v^2)^2} \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix},\ (u,v) \in \mathbb{R}^2$$
  
−
''Concerning part a)''
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a) Write down the Hamiltonian equations describing the geodesic curves.
  
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First thing: It is kind of unclear to me how \( \frac{d}{dx} ln(x) \vert_{x=1} = -x\dfrac{(\ln(1)-\ln(1-\frac{x}{k}))}{\frac{x}{k}} \). ''No, I meant that the limit of the fraction is the derivative (it's just the definition of the derivative as limit of the difference quotient), but if you prefer Bernoulli it's the same for me.''
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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
  
−
Wouldn't it be much easier to just:
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$$ -C_1 u + C_2 v +C_3(u^2+v^2-R^2) = 0 \forall (u,v)\in \mathbb{R}^2$$
 +
c) Show that the energy h satisfies \(h \geq 0\) and discuss the case \(h = 0\).
  
−
$$ \lim_{k \rightarrow \infty} \ln((1-\frac{x}{k})^{k}) = \lim_{k \rightarrow \infty}  k\ln(1-\frac{x}{k}) = \lim_{k \rightarrow \infty}  \frac{\ln(1-\frac{x}{k})}{\frac{1}{k}} $$
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d) Describe the geodesics when \(h > 0\) depending on the values of \(C_1, C_2, C_3\).
  
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and then use Bernoulli-L'Hopital to get:
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==Solution==
  
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$$ \lim_{k \rightarrow \infty}  \frac{\ln(1-\frac{x}{k})}{\frac{1}{k}} = \lim_{k \rightarrow \infty}  \frac{\frac{x}{k^2}}{(\frac{x}{k} - 1) \frac{1}{k^2}} = -x $$
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===Solution (a)===
  
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Second thing: Nevermind, confused myself...
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$$ g_{ij}(u,v)  = \frac{4R^4}{(R^2+u^2+v^2)^2} \delta_{ij}$$
  
−
Greetings
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because \( \big(g^{ij}(u,v)\big) =  \big(g_{ij}(u,v)\big)^{-1}\)
−
A.
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$$ \Rightarrow g^{ij}(u,v)  = \frac{(R^2+u^2+v^2)^2}{4R^4} \delta^{ij}$$
 +
 
 +
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
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$$ \dot{u}\xi_2 = \dot{v}\xi_1$$
 +
 
 +
 
 +
 
 +
===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.

Latest revision as of 06:10, 29 July 2015

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.

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