Proving Geodesics in Hyperbolic Geometry

[ChrisJ]

Given $ds^2 = y^{-2}(dx^2 + dy^2)$, I am trying to prove that a demicircle centred on the x-axis, written parametrically as $x=r\cos\theta + x_0$ and $y= r \sin \theta$ are geodesics. Where $r$ is constant and $\theta \in (0,\pi)$

I have already found the general form of the geodesic equation for this metric by finding the christoffell symbols,

<br /> 0 = \frac{d^2x}{ds^2} - \frac{2}{y}\frac{dx}{ds}\frac{dy}{ds} \\<br /> 0 = \frac{d^2y}{ds^2} - \frac{1}{y} \left( \frac{dy}{ds} \right)^2 + \frac{1}{y}\left( \frac{dx}{ds} \right)^2<br />Then I was not exactly sure where to go from here, but just started spitballing,

<br /> dx = -r \sin \theta d\theta\\<br /> dy = r \cos \theta d\theta\\<br /> ds^2 = \left( \frac{1}{r^2 \sin^2 \theta } \right) \left( -r^2 \sin^2 \theta d\theta^2 + r^2 \cos^2 \theta d \theta^2 \right) \\<br /> ds^2 = (\cot^2 \theta- 1 ) d\theta^2 \\<br /> ds = \sqrt{\cot^2 \theta-1} d\theta<br />

Am I even on the right track here? Any help/advice is much appreciated.

 

[sweet springs]

Take good care of sign of $dy^2$. The fifth line is ds^2=cosec^2\theta d\theta<br /> ^2

 

[ChrisJ]

Take good care of sign of $dy^2$. The fifth line is ds^2=cosec^2\theta d\theta<br /> ^2

Sorry, I cannot see that?

What I did was this,
<br /> ds^2 = \left( \frac{1}{r^2 \sin^2 \theta } \right) \left( -r^2 \sin^2 \theta d\theta^2 + r^2 \cos^2 \theta d \theta^2 \right) \\<br /> ds^2 = \frac{-r^2 \sin^2 \theta d\theta^2}{r^2 \sin^2 \theta} + \frac{r^2 \cos^2 \theta d \theta^2}{r^2 \sin^2 \theta} \\<br /> ds^2 = -1 d\theta^2 + \cot^2 d \theta^2 \\<br /> ds^2 = (\cot^2 \theta - 1) d \theta^2<br />

 

[Orodruin]

What is your idea behind computing $ds$ in terms of $d\theta$? How is it going to help you find out whether or not your curves are geodesics? (I am not saying it doesn't, just asking for your rationale.)

I also agree with @sweet springs , you need to be more careful with your signs. Note that $(-1)^2 = 1$ and not $-1$.

 

[ChrisJ]

What is your idea behind computing $ds$ in terms of $d\theta$? How is it going to help you find out whether or not your curves are geodesics? (I am not saying it doesn't, just asking for your rationale.)

I also agree with @sweet springs , you need to be more careful with your signs. Note that $(-1)^2 = 1$ and not $-1$.

The idea was to find $ds^2=$...$d\theta^2$ to then find the non-vansihing christoffel symbols of metric in terms of $\theta$ to plug into new geodesic equation.

This is what I have done so far now, leaving where last post left off and fixing mistake

<br /> ds^2 = \left( \frac{1}{r^2 \sin^2 \theta } \right) \left( r^2 \sin^2 \theta d\theta^2 + r^2 \cos^2 \theta d \theta^2 \right) \\<br /> ds^2 = d\theta^2 + \cot^2 d \theta^2 \\<br /> ds= \sqrt{(\cot^2 \theta + 1)} d \theta \\<br /> ds = \csc \theta d\theta \\<br /> ds^2 = \csc^2 d\theta^2<br /> <br />

Then I can write a new tensor metric as such

<br /> <br /> g_{\alpha\beta} = \begin{pmatrix} 0 &amp; 0 \\ 0 &amp; \csc^2 \theta \end{pmatrix} \\<br /> g^{\alpha\beta} = \begin{pmatrix} 0 &amp; 0 \\ 0 &amp; \sin^2 \theta \end{pmatrix} \\<br /> \\<br /> \Gamma^{\theta}_{\theta\theta} = \frac{1}{2} \sin^2 \theta \left( - \frac{\partial}{\partial \theta} \csc^2 \theta \right) \\<br /> \Gamma^{\theta}_{\theta\theta} = \sin^2 \theta \cot^2 \theta \csc^2 \theta = \cot^2 \theta \\<br />

Then new geodesic equation becomes, noting that since $ds=\csc \theta d\theta$ that $d\theta = \sin \theta ds$...
<br /> 0= \frac{d^2\theta}{ds^2} - \Gamma^{\theta}_{\theta\theta} \left( \frac{d\theta}{ds} \right)^2 \\<br /> 0=\frac{d}{ds}\left( \sin \theta \right) - \left( \cot^2 \theta \right) \left( \sin^2 \theta \right) \\<br /> 0=\cos \theta \sin \theta - \cos^2 \theta<br />

Which does not equal zero. Any help is really appreciated. Thanks.

 

[Orodruin]

The idea was to find ds2=ds2=ds^2=...dθ2dθ2d\theta^2 to then find the non-vansihing christoffel symbols of metric in terms of θθ\theta to plug into new geodesic equation.

This will not work. All you are doing is getting an induced metric on the circle. Your manifold has two coordinates and the metric is non-degenerate.

You need to find the Christoffel symbols in your original coordinate system (or another coordinate system on your original manifold) and show that the curve is a geodesic. Note that a curve can be a non-affinely parametrised geodesic, which means that it will not satisfy the geodesic equation directly, but can be made to do so by a reparametrisation.

 

[ChrisJ]

This will not work. All you are doing is getting an induced metric on the circle. Your manifold has two coordinates and the metric is non-degenerate.

You need to find the Christoffel symbols in your original coordinate system (or another coordinate system on your original manifold) and show that the curve is a geodesic. Note that a curve can be a non-affinely parametrised geodesic, which means that it will not satisfy the geodesic equation directly, but can be made to do so by a reparametrisation.

Oh, bugger. Ok thanks.

Well the geodesic equation, given the chrsitoffel symbols to the original coordinate system is as follows
<br /> 0 = \frac{d^2x}{ds^2} - \frac{2}{y}\frac{dx}{ds}\frac{dy}{ds} \\<br /> 0 = \frac{d^2y}{ds^2} - \frac{1}{y} \left( \frac{dy}{ds} \right)^2 + \frac{1}{y}\left( \frac{dx}{ds} \right)^2<br />

are you saying that I need to work the following out?
<br /> 0 = \frac{d}{ds} \left( \frac{d}{ds} \left( r \cos \theta + x_0 \right) \right) - \frac{2}{r \sin \theta} \left( \frac{d}{ds} (r \cos \theta + x_0) \right) \left( \frac{d}{ds}r\sin \theta \right) \\<br /> 0 = \frac{d}{ds}\left(\frac{d}{ds} r \sin \theta \right) - \frac{1}{r\sin \theta} \left( \frac{d}{ds} r\sin \theta \right)^2 + \frac{1}{r \sin \theta} \left(\frac{d}{ds} (r \cos \theta + x_0) \right)^2<br />

 

[Orodruin]

are you saying that I need to work the following out?

Yes. This should not be too difficult seeing that there is only one variable (the curve parameter $\theta$). Do not forget that $\theta$ is not necessarily an affine parameter.

 

[ChrisJ]

Yes. This should not be too difficult seeing that there is only one variable (the curve parameter $\theta$). Do not forget that $\theta$ is not necessarily an affine parameter.

ok its been a while since I've done implicit differentiation, but am I correct in thinking that, for example $\frac{d}{ds} r \sin \theta = r \cos \theta \frac{d\theta}{ds} = r \cos \theta \sin \theta$

Since in my wrong method above I found that $\frac{d\theta}{ds}= \sin \theta$

 

[Orodruin]

ok its been a while since I've done implicit differentiation, but am I correct in thinking that, for example $\frac{d}{ds} r \sin \theta = r \cos \theta \frac{d\theta}{ds} = r \cos \theta \sin \theta$

Since in my wrong method above I found that $\frac{d\theta}{ds}= \sin \theta$

Yes, this would amount to implicitly reparametrising the curve with the path length $s$.

 

[ChrisJ]

Yes, this would amount to implicitly reparametrising the curve with the path length $s$.

Ok, I think I am having a bit of trouble here, does not seem to equal zero the way I'm doing it.

I found all the derivatives I need separately to make it clearer, then plug in at the end.
<br /> \frac{d}{ds} r \sin \theta = r \cos \theta \frac{d\theta}{ds} = r \cos \theta \sin \theta \\<br /> \frac{d}{ds}(r \cos \theta + x_0) =- r \sin \theta \frac{d\theta}{ds} = -r \sin^2 \theta \\<br /> \frac{d}{ds} \left( \frac{d}{ds} \left( r \cos \theta + x_0 \right) \right) = \frac{d}{ds} \left( r \sin^2 \theta \right) = (-r \cos \theta \sin \theta + r \sin\theta \cos \theta) \frac{d\theta}{ds} = 0 \\<br /> \frac{d}{ds}\left(\frac{d}{ds} r \sin \theta \right) = \frac{d}{ds}(r \cos \theta \sin \theta) = (r \sin^2 \theta - r \sin^2 \theta) \frac{d\theta}{ds} =0<br />

So both of the second derivative in the geodesic equations equal zero, and just need to focus on the single derivatives. So,
<br /> 0 = \frac{d}{ds} \left( \frac{d}{ds} \left( r \cos \theta + x_0 \right) \right) - \frac{2}{r \sin \theta} \left( \frac{d}{ds} (r \cos \theta + x_0) \right) \left( \frac{d}{ds}r\sin \theta \right) \\<br /> 0 = -\frac{2}{r\sin \theta} \left( -r \sin^2 \theta \right) \left( r \cos \theta \sin \theta \right) \\<br /> 0 = 2r \sin^2 \theta \cos \theta<br />

and the other one..
<br /> 0 = \frac{d}{ds}\left(\frac{d}{ds} r \sin \theta \right) - \frac{1}{r\sin \theta} \left( \frac{d}{ds} r\sin \theta \right)^2 + \frac{1}{r \sin \theta} \left(\frac{d}{ds} (r \cos \theta + x_0) \right)^2 \\<br /> 0 =- \frac{1}{r \sin\theta} \left( r \cos \theta \sin \theta \right)^2 + \frac{1}{r \sin\theta} ( -r \sin^2 \theta )^2 \\<br /> 0 = -r \cos^2 \theta \sin \theta + r \sin^3 \theta<br />

 

[sweet springs]

0=dds(sinθ)−(cot2θ)(sin2θ)
0=cosθsinθ−cosθ​

I do not follow physical meaning but there is a mistake in your calculation. From the top one
cos\theta(\frac{d\theta}{ds}-cos\theta)=0
ds=\frac{d\theta}{cos\theta}
s=\frac{1}{2}log|\frac{1+sin\theta}{1-sin\theta}|+C

 

[ChrisJ]

I do not follow physical meaning but there is a mistake in your calculation. From the top one
ds=\frac{d\theta}{cos\theta}

Where did you get that from? From post 5, it can be seen that

ds^2 = \csc^2 \theta d\theta^2 \\<br /> ds = \csc \theta d \theta \\<br /> ds = \frac{d\theta}{\sin \theta}<br />

 

[sweet springs]

I've got it from your
0=\frac{d}{ds}(sin\theta)-cot^2\theta sin^2\theta
Could you elaborate which equation is a right one?

 

[ChrisJ]

I've got it from your
0=\frac{d}{ds}(sin\theta)-cot^2\theta sin^2\theta
Could you elaborate which equation is a right one?

Oh right ok, there might be an error there somewhere but that is a wrong method anyway. As per post 6 onwards tried to go about the problem from a different angle as per Orodruin's suggestion. However, I am pretty certain that $\frac{d\theta}{ds} = \sin \theta$, which that bit is still needed in new method.

 

[sweet springs]

If you are certain about that formula why do not you integtrate immediately to get
s=\frac{1}{2}log|\frac{1-cos\theta}{1+cos\theta}|+C?