Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Curvature ~ dT/ds or dn/ds?

  1. Jun 29, 2010 #1
    curvature ~ dT/ds or dn/ds??

    In Riemannian Geometry: A Beginner's Guide on pg. 5, Chap. 2, the curvature, [tex]\kappa[/tex], is defined as the rate of change of the tangent vector, T, along the curve, x(t), where 't' is a parameter along the curve. (A physical analogy is the tangential velocity and centripetal acceleration in circular motion.)

    [tex]\overrightarrow{\kappa} \equiv \f{d\overrightarrow{T}}/{ds}[/tex]

    However, then the author says that the magnitude of the curvature is equal to the magnitude of the rate of change of the normal.

    [tex]\kappa \doteq |\f{d\overrightarrow{n}}/{ds}|[/tex]

    I don't understand how it can be both.

    Souldn't [tex]\kappa \propto |\overrightarrow{n}|[/tex] ?

    IOW, the magnitude of a vector, in this case the curvature, is the magnitude of the rate of change of the tangential vector. No?

    Thanks in advance.
     
    Last edited: Jun 30, 2010
  2. jcsd
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?
Draft saved Draft deleted



Similar Discussions: Curvature ~ dT/ds or dn/ds?
  1. Show that ds^2 = ds'^2 (Replies: 14)

Loading...