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Metric in polar coordinate derivation

  1. Nov 17, 2015 #1
    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. jcsd
  3. Nov 17, 2015 #2
    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.)
  4. Nov 17, 2015 #3
    Great. Thanks for that, that's exactly what I was looking for.
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