Difference between revisions of "Aufgaben:Problem 8"

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(Created page with "==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}^...")
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==Task==
  
 +
Let \(R < 0\). On \(\mathbb{R}^2\), consider the metric
 +
$$ g(u,v) = \frac{4R^4}{(R^2+u^2+v^2)^2} \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix},\ (u,v) \in \mathbb{R}^2$$
  
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''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 \( \frac{d}{dx} ln(x) \vert_{x=1} = -x\dfrac{(\ln(1)-\ln(1-\frac{x}{k}))}{\frac{x}{k}} \).
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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 \(C_1, C_2, C_3\).
  
−
and then use Bernoulli-L'Hopital to get:
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==Solution==
  
−
$$ \lim_{k \rightarrow \infty} \frac{\ln(1-\frac{x}{k})}{\frac{1}{k}} = \lim_{k \rightarrow \infty}  \frac{\frac{x}{k^2}}{(\frac{x}{k} - 1) \frac{1}{k^2}} = -x $$
+
(a)
 +
$$ g_{ij}(u,v) = \frac{4R^4}{(R^2+u^2+v^2)^2} \delta_{ij}$$
  
−
Second thing: Nevermind, confused myself...
+
because \( \big(g^{ij}(u,v)\big) =  \big(g_{ij}(u,v)\big)^{-1}\)
 +
$$ \Rightarrow g^{ij}(u,v)  = \frac{(R^2+u^2+v^2)^2}{4R^4} \delta_{ij}$$
  
−
Greetings
+
let \( x = (u,v)\) and \( \xi = (\xi_1, \xi_2)\), the hamiltonian is defined as
−
A.
+
$$h(x, \xi) = \frac{1}{2} g^{ij}(x)\xi_i \xi_j $$
  
−
''I didn't write that: I meant that the limit of the different quotient is the derivative. I was a bit inconsistent with the variable x (now it's fixed, I've used a y). If you prefer Bernoulli it's the same for me. Nick''
+
$$ \Rightarrow h(x, \xi) = \frac{(R^2+u^2+v^2)^2}{8R^4} (\xi_1^2 + \xi_2^2)$$
  
−
Okay, I see what you meant now. I personally find it a bit confusing. But I guess it's a matter of taste.
+
now we can compute the Hamiltonian equations:
  
−
A.
+
$$ \dot{u} = \frac{\partial h}{\partial \xi_1} = \frac{(R^2+u^2+v^2)^2}{4R^4} \xi_1$$
 +
$$ \dot{v} = \frac{\partial h}{\partial \xi_2} = \frac{(R^2+u^2+v^2)^2}{4R^4} \xi_2$$
  
−
To Part b)
+
$$ \dot{\xi_1} = -\frac{\partial h}{\partial u} = -\frac{(R^2+u^2+v^2)}{2R^4} (\xi_1^2 + \xi_2^2) u$$
  
−
Question:
+
$$ \dot{\xi_2} = -\frac{\partial h}{\partial v} = -\frac{(R^2+u^2+v^2)}{2R^4} (\xi_1^2 + \xi_2^2) v$$
  
−
Since we had an whole exercise sheet on Lebesgue-convergence  and  i would assume they would ask of us to find  an  upper bound  function  g  to prove that we're allowed  to use that theorem. Why don't you do that? (It might be obvious but those just learning it by heart won't notice.)
+
we can also observe that:
  
−
Answer:
+
$$ \dot{\xi_2}u = \dot{\xi_1}v$$
 +
and
 +
$$ \dot{u}\xi_2 = \dot{v}\xi_1$$
  
−
The upper bound function is for dominated convergence, not for monotone convergence. But I agree it is not so clearly stated, I'll fix it, thanks
 
  
−
Again to part b)
 
−
As I remember (and according to Wikipedia http://de.wikipedia.org/wiki/Satz_von_der_monotonen_Konvergenz) the function to which \(f_k\) converges, has to be \(\in L^1 \). That should maybe shown too?
 
  
−
[[User:Trubo-Warrior|Trubo-Warrior]] ([[User talk:Trubo-Warrior|talk]]) 12:20, 5 January 2015 (CET)
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(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
  
−
''Since we all like alternative versions here on the wiki, and I've encountered some confusion among my friends about the limit-thing we discussed above, I felt so free to put the alternative version in. Best, A.''
+
$$\frac{\partial C_1}{\partial t} = 2\dot{u}v\xi_1 + 2u\dot{v}\xi_1 + 2uv\dot{\xi_1}+R^2\dot{\xi_2} - 2u\dot{u}\xi_2 - u^2\dot{\xi_2}+ 2v\dot{v}\xi_2 + v^2\dot{\xi_2}$$
 +
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 for the alternative version: yes, it was a good idea, so everybody can choose the one she/he prefers.
+
now for \( C_2 \)
−
For \( L^1 \): it is, since \( e^{-x} \) is in Schwartz space (over positive real numbers), but I've added it to clarify. I didn't have this requirement in the formulation of the exercise class, but I think you're right.
+
−
Cheers, '[[User:Nick|Nick]] ([[User talk:Nick|talk]]) 17:24, 12 January 2015 (CET)'
+
  
−
At the very end, don't think f(x) is in Schwartz Space, because t can take other values than integrers. I would prove that f(x) is in L1 separately:
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$$ C_2 = \frac{4R^4}{(R^2+u^2+v^2)^2}(2uv\dot{v} +R^2\dot{u}+u^2\dot{u}-v^2\dot{u}) $$
−
$$ \int_0^\infty  \left| x^{t-1}e^{-x} \right| \, \mathrm{d}x  = \int_0^\infty  x^{t-1}e^{-x} \, \mathrm{d}x =  \int_0^1  x^{t-1}e^{-x} \, \mathrm{d}x + \int_1^\infty  x^{t-1}e^{-x} \, \mathrm{d}x $$
+
$$ = 2uv\xi_2 +R^2\xi_1 + u^2\xi_1 -v^2\xi_1$$
 +
$$\frac{\partial C_2}{\partial t} = 2\dot{u}v\xi_2 + 2u\dot{v}\xi_2 + 2uv\dot{\xi_2}+R^2\dot{\xi_1} + 2u\dot{u}\xi_1 + u^2\dot{\xi_1}- 2v\dot{v}\xi_1 - v^2\dot{\xi_1}$$
  
−
$$  \int_0^1 \!  x^{t-1}e^{-x} \, \mathrm{d}x \leq \int_0^1 \!  x^{t-1} \, \mathrm{d}x = \frac{1}{t}$$  
+
using \( \dot{u}\xi_2 = \dot{v}\xi_1\) on the first term, the first and second to last term cancel:
−
because \( t>0 \), the integrant is positive and \(e^{-x} \leq 1\).  and for the second integral, if \( 0 < t < 1 \)
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$$= 2u\dot{v}\xi_2 + 2uv\dot{\xi_2}+R^2\dot{\xi_1} + 2u\dot{u}\xi_1 + u^2\dot{\xi_1} - v^2\dot{\xi_1}$$
−
$$ \int_1^\infty  x^{t-1}e^{-x} \, \mathrm{d}x \leq \int_1^\infty e^{-x} = \frac{1}{e}$$
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using \(u\dot{\xi_2} = v\dot{\xi_1}\) on the second term:
−
and if \(t > 1 \) we can partially integrate until we get an integral of the above form multiplied by a constant (the additional terms don't go to zero but are less than infinity)
+
$$= 2u\dot{v}\xi_2 + 2v^2\dot{\xi_1}+ R^2\dot{\xi_1} + 2u\dot{u}\xi_1 + u^2\dot{\xi_1} - v^2\dot{\xi_1}$$
 +
$$ = \dot{\xi_1}(R^2 + u^2+  v^2) + 2u\dot{v}\xi_2 + 2v\dot{u}\xi_1 $$
 +
using again the Hamiltonian equations:
 +
$$ = -\frac{(R^2+u^2+v^2)^2}{2R^4}(\xi_1^2 + \xi_2^2) u + 2u\frac{(R^2+u^2+v^2)^2}{4R^4} \xi_2^2 + 2u\frac{(R^2+u^2+v^2)^2}{4R^4} \xi_1^2 =0$$
  
−
Which all together means that
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and for \( C_3 \)
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$$  \int_0^\infty  \left| x^{t-1}e^{-x} \right| \, \mathrm{d}x  < \infty$$
+
−
or maybe it is enough to say that the gamma function is analytic for \(t>0\) , but I wouldn't count on it
+
−
"[[User:Carl|Carl]] ([[User talk:Carl|talk]]) 16:31, 15 January 2015 (CET)"
+
  
−
I agree that the story with Schwartz space doesn't really make sense, since we haven't even defined it for sets other than R^n; despite that, the version of the monotone convergence theorem that I had on the notes actually agrees with Lebesgue monotone convergence theorem, which doesn't require f to be in L1 (in fact, it states that it follows from the rest: http://en.wikipedia.org/wiki/Monotone_convergence_theorem). In any case, thanks for the proof :)
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$$ C_3 = \frac{4R^4}{(R^2+u^2+v^2)^2}(v\dot{u} - u\dot{v}) = (v\xi_1 - u\xi_2)$$
−
'[[User:Nick|Nick]] ([[User talk:Nick|talk]]) 21:38, 15 January 2015 (CET)'
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$$ \frac{\partial C_3}{\partial t} = \dot{v}\xi_1 + v\dot{\xi_1} - \dot{u}\xi_2 - u\dot{\xi_2} = \dot{u}\xi_2 + v\dot{\xi_1} - \dot{u}\xi_2 - v\dot{\xi_1} = 0$$
 +
 
 +
Last we wantto 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.
 +
 
 +
(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.
 +
 
 +
(d)
 +
working on it.

Revision as of 21:09, 11 June 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

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


(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

$$\frac{\partial C_1}{\partial t} = 2\dot{u}v\xi_1 + 2u\dot{v}\xi_1 + 2uv\dot{\xi_1}+R^2\dot{\xi_2} - 2u\dot{u}\xi_2 - u^2\dot{\xi_2}+ 2v\dot{v}\xi_2 + v^2\dot{\xi_2}$$ 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: $$ = -\frac{(R^2+u^2+v^2)^2}{2R^4}(\xi_1^2 + \xi_2^2) v + 2v\frac{(R^2+u^2+v^2)^2}{4R^4} \xi_1^2 + 2v\frac{(R^2+u^2+v^2)^2}{4R^4} \xi_2^2 =0$$

now for \( C_2 \)

$$ C_2 = \frac{4R^4}{(R^2+u^2+v^2)^2}(2uv\dot{v} +R^2\dot{u}+u^2\dot{u}-v^2\dot{u}) $$ $$ = 2uv\xi_2 +R^2\xi_1 + u^2\xi_1 -v^2\xi_1$$ $$\frac{\partial C_2}{\partial t} = 2\dot{u}v\xi_2 + 2u\dot{v}\xi_2 + 2uv\dot{\xi_2}+R^2\dot{\xi_1} + 2u\dot{u}\xi_1 + u^2\dot{\xi_1}- 2v\dot{v}\xi_1 - v^2\dot{\xi_1}$$

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

and for \( C_3 \)

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

Last we wantto 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.

(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.

(d) working on it.

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