Shortest arc between two points in polar coordinates

[mnb96]

Hello,
If we consider a Euclidean plane $$\mathbb{R}^2$$ with the ordinary inner product, and we "distort" it through a cartesian->polar transformation, how should I compute the shortest arc between two points $$(r,\theta)$$ and $$(r',\theta')$$ ?

 

[zhentil]

Hello,
If we consider a Euclidean plane $$\mathbb{R}^2$$ with the ordinary inner product, and we "distort" it through a cartesian->polar transformation, how should I compute the shortest arc between two points $$(r,\theta)$$ and $$(r',\theta')$$ ?

What metric are you using on the polar plane?

 

[mnb96]

I am using the metric I derived from the equations
$$x=r cos(\theta)$$
$$y=r sin(\theta)$$

From those I got:

$$M = diag(1,r^2)$$

 

[zhentil]

I guess I'm not sure what you're looking for. The shortest arc will be the image of an honest straight line under the isometry.

 

[mnb96]

Ok. I guess my original question was meaningless.
As far as I could understand, computing a shortest-arc length makes sense only on surfaces whose curvature changes locally. The $\mathbb{R}^2$ plane is flat, so the shortest arcs between two points are always straight lines.

Basically, all I have to do is to consider the straight line connecting the two points (in cartesian coordinates), and convert its parametric representation into polar coordinates.

Is this correct?

 

[zhentil]

Computing geodesics is the same no matter what the metric does. It's just particularly easy here.

So yes, you're correct.