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For a problem I am given that we have a 2D space time with the metric

$$ds^2=-\rho^2d\alpha^2+d\rho^2 \tag{A}$$.

We are also given that

$$\rho^2\frac{d\alpha}{d\tau}=l \tag{B},$$

for which we have shown that it remains constant along the geodesic. Now we have to express $\rho$ as $\rho(\alpha)$ under the condition that $\rho(\alpha=0)=l$.

We are given the hint that we should use the measure for an infinitesimal displacement to obtain a differential equation for $\rho=\rho(\alpha)$.

I have tried several approaches, but I really do not see how this should be done. I thought I should use that $$\rho(\alpha)^2 d\alpha=ld\tau$$ and then integrate both sides, but as you don't know how $\rho$ depends on $\alpha$ I was not really able to get any further than this.

Any help would be much appreciated!

1 Answers 1

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$$\rho^2\frac{d\alpha}{d\tau}=l,$$ a constant, with $\rho$ as $\rho(\alpha)$ in polar coordinates and $\rho(\alpha=0)=l$ as initial value represents geodesics on any surface of revolution in cylindrical coordinates $ (\rho, \alpha, z) $ generated from arc length $\tau$ basis. It is also called Clairaut's Law.

$\rho =l$ represents minimum radial distance from axis of symmetry $\rho=0.$

If the meridian $r=f(z) $ is not given it is impossible determine the geodesic on the surface in a 3D situation.

But, it integrates to a straight line in the plane as follows:

$$ \rho^2 \frac{d \alpha}{d \tau} = l $$

Letting

$$ \sin \psi = \frac{\rho d \alpha} {d \tau} $$

we have Clairaut's Law

$$ \rho \sin \psi = l $$

In text-books of differential geometry angle $\psi $ is given between radial line and arc which in this case is a straight line.

EDIT 1:

I assume $ s= \tau $ and a Euclidean metric

$$ ds^2=d \tau^2 = +\rho^2d\alpha^2+d\rho^2 $$

for the answer:

A highly exaggerated differential triangle in the plane is sketched below:

enter image description here

Assuming Euclidean metric the following is valid:

$$ \rho \sin \psi = \rho^2 \frac{d\alpha}{d\tau} = l \tag1 $$

$$\frac {d\rho}{d\tau} = \cos \psi \tag2 $$

Eliminate $\tau$

$$ \frac {d \alpha}{l}= \frac{d \rho}{\rho\, \sqrt{\rho^2-l^2}} \tag3$$

integrates to

$$ \frac{\alpha}{l}= \sec^{-1} \frac{\rho}{l}\tag4$$

$$ \rho = l\, \sec{ (\alpha + \beta) } \tag5 $$

where $\beta $ is an arbitrary constant of rotation for all straight lines of minimum distance $l$ from origin.

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    I added some extra information given in the question about the metric of the 2D spacetime. Does that help with this? – 2017-02-26
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    Yes that was essential. But sign is wrong, please check it. Also why two symbols for $ \tau, s?$ I think you got it now. – 2017-02-26
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    Yes this is the way that it's given in my exercise, with a minus sign in front of the first term and a plus sign in front of the second term. – 2017-02-26
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    But I know it is wrong. It is the standard *Euclidean metric* in polar coordinates $ \rho, \alpha$ valid for a straight line. I cannot help you with a confirmed negative sign and that differential geometrical element definition. – 2017-02-26
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    It's perfectly right for a minkovskian (thus flat) space-time. $\tau$ is the proper time of a pointlike particle following the geodesic and used to parametrize the trajectory. – 2017-02-27
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    OK, If $\tau$ is proper time what is your other parameter ( other than $\rho$) in your flat Minkowski space that you have assumed? Is the geodesic just a 1- dimensional ray of light? – 2017-02-27
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    @RafaBudría May be we are referring to different spacelike and timelike geodesics. I have considered simple flat spatial geometry. – 2017-02-27
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    A photon cannot be parametrized by the proper time because for photons, the proper time is zero. $\tau$ is commonly understood as the proper time for a moving particle: $\mathrm d\tau=\mathrm ds$ for a moving particle (along a geodesic or not). This point is implicitly assumed in the question. $\alpha$ isn't properly an angle. – 2017-02-27
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    Hope someone else sheds a few photons on this topic. At least for now we could add some more useful tags, or move it to Physics SE... – 2017-02-27