# Metric in polar coordinate derivation

1. Nov 17, 2015

### DiracPool

At time 1:11:20, Lenny introduces the metric for ordinary flat space in the hyperbolic version of polar coordinates? Is that what he is doing here?

d(tau)^2 = ρ^2 dω^2 - dρ^2.

He then goes on to say that this metric is the hyperbolic version of the same formula for Cartesian space, i. e.,

ρ^2 dθ^2 + dρ^2

My question is what does the Cartesian version equal? It cannot equal d(tau)^2. Does it equal dS^2? That is,

dS^2 = ρ^2 dθ^2 + dρ^2

And what is this telling us? Is it the formula for the distance between 2 points on a Cartesian plane in polar coordinates? I looked for a derivation of this formula online and on you-tube and couldn't find it. Can someone direct me to a "B-level" online link I can look at?

2. Nov 17, 2015

### PWiz

This is the Euclidean distance (line element) in polar coordinates.
I'll derive it for you.
Take the distance formula in 2D space in Cartesian coordinates: $ds^2=dx^2+dy^2$ (Pythagoras's theorem)
Use the coordinate conversions $x=\rho cos \theta$ and $y=\rho sin \theta$. You can differentiate these equations to obtain $dx= - \rho sin \theta d\theta + cos \theta d\rho$ and $dy = \rho cos \theta d\theta + sin\theta d\rho$. Just square these two and add them up to get $ds^2 = (\rho ^2 sin^2 \theta d\theta ^2 - 2 \rho cos\theta sin \theta d\theta d\rho + cos^2 \theta d\rho^2) + (\rho ^2 cos^2 \theta d\theta ^2 +2 \rho sin\theta cos\theta d\theta d\rho + sin^2 \theta d\rho ^2) = \rho^2 d\theta^2 + d\rho^2$

(Note that deriving the line element in spherical coordinates [and other 3D coordinates] can get really messy. The polar coordinates case just happens to be really easy.)

3. Nov 17, 2015

### DiracPool

Great. Thanks for that, that's exactly what I was looking for.