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Spherical coordinates length from differential length

  1. Mar 25, 2014 #1
    is it logical to ask this question in Spherical coordinates:

    Using the differential length dl , find the length where r=1 0<Θ<∏/4 ∏/2< θ <∏/4 where Θ is the azimuthal angle.

    What I mean by ∏/2< θ <∏/4 is that the line is a "diagonal" line which has an ascention of ∏/4 from the xy plane. I dont know how else to write it.

    Is the differential length dl = sqrt( (rdΘ)^2 + (dθ)^2 ) ?
     
    Last edited: Mar 25, 2014
  2. jcsd
  3. Mar 25, 2014 #2

    Simon Bridge

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    Suspect not - but I'm not sure what you are asking.

    if ##\vec r(t)=(r(t), \theta(t),\phi(t))^t## parameterzes a curve in spherical-polar coordinates, and ##dl## is the length element along that curve in the direction of increasing parameter ##t##, then ##dl## can be expressed in terms of the curve ##\vec r(t)## and the spherical-polar unit vectors.

    See:
    http://home.comcast.net/~szemengtan/ClassicalMechanics/SingleParticle.pdf [Broken]
    section 1.5 onwards.
     
    Last edited by a moderator: May 6, 2017
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