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Homework Help: The length of a path on a sphere (in spherical coordinates)

  1. Jun 19, 2013 #1
    So, I'm to show that in spherical coordinates, the length of a given path on a sphere of radius R is given by:
    L= R[itex]\int_{\theta_1}^{\theta_2} \sqrt{1+\sin^2(\theta) \phi'^2(\theta)}d\theta[/itex],
    where it is assumed [itex]\phi(\theta)[/itex], and start coordinates are [itex](\theta_1,\phi_1) [/itex]and [itex](\theta_2, \phi_2)[/itex].

    I've tried starting by letting ds be a short path segment, such that
    [itex]ds=R \sqrt{ d\theta^2+ d\phi^2} [/itex] and [itex]d\phi=\frac{d\phi}{d\theta}d\theta=\phi'(\theta) d\theta[/itex]

    But this is obviously wrong, since I'll be missing a factor of sin^2(θ). If I draw the situation, it's also clear that I'll need the factor.

    Is it just a sphercial coordinate concept I'm missing, or am I way off?
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Jun 19, 2013 #2
    If a curve is given by x(t), y(t), z(t), where t is a parameter, how is its length computed?
  4. Jun 19, 2013 #3
    I'd use


    So I tried inserting x, y and z as spherical coordinates, and having θ as the parameter, in the end I end up with [itex]\sqrt{R^2}[/itex]. That's after I take the derivative with respect to θ.

    I'm not sure I got your hint.
  5. Jun 19, 2013 #4
    ds^2 = dx^2 + dy^2 + dz^2
    To translate this into spherical coordinates in the form:
    ds^2 = a*dr^2 + b*dθ^2 + c*dϕ^2,
    you need the Jacobian matrix for the transformation, which will give you a, b and c. The result is rather well known, but if you're not familiar with the technique, i would recommend you look it up. Your problem is straightforward from there.
  6. Jun 19, 2013 #5
    Your approach is correct in principle, so there must be an error in details. Unless you show all the steps, I can't help.
  7. Jun 19, 2013 #6
    It's a matter of geometry. As you approach the poles, the lines of constant longitude get closer together, and as you approach the equator, they get further apart. You need to draw a diagram to see why geometrically there is a factor of sin2θ.
  8. Jun 20, 2013 #7
    I've looked it up, but it puzzles me how I go from [itex]ds^2=dx^2+dy^2+dz^2[/itex] when dx, dy and dz are added together. I mean, if it was [itex]ds^2=dx^2 dy^2 dz^2[/itex] it would be straight forward for me.

    abs[∂(x,y,z)/∂(r,θ,phi)] * dr dθ dphi

    Can I simply transform each on its own? dx, dy and dz?
  9. Jun 20, 2013 #8
    With [itex]\theta[/itex] as the parameter, I insert [itex]x=r\sin\theta\cos\phi[/itex], [itex]y=r\sin\theta\sin\phi[/itex] and [itex]z=r\cos\theta[/itex].

    This would give me

    From there I end up with


    Using identities it reduces to [itex]\sqrt{r^2}[/itex]..
  10. Jun 20, 2013 #9
    Do you understand that, for example, $$ \frac {d} {d\theta} \sin \theta \cos \phi = \cos \theta \cos \phi - \sin^2 \theta \phi'(\theta) $$
  11. Jun 20, 2013 #10
    No, I can't figure out the last term on the right.
  12. Jun 20, 2013 #11
    Voko, is your 2nd term correct? In my opinion it ought to be " -sin(theta)*sin(phi)*(dphi/dtheta)"
  13. Jun 20, 2013 #12
    Darn, brain still half asleep :)

    Indeed that should have been $$ \cos \theta \cos \phi - \sin \theta \sin \phi \phi'(\theta) $$
  14. Jun 20, 2013 #13
    I see what you did there. So I should treat [itex]\phi(\theta)[/itex] properly in my derivatives.
  15. Jun 20, 2013 #14
    Well, yes that was fairly easy. Thanks alot. :)
  16. Jun 20, 2013 #15
    Here's a simpler way. Draw a circle. Draw the z-axis vertically through the center of the circle. Pick a point P on the upper half of the circumference. Draw a radius from the center of the circle to the point on the circumference. Label the angle θ between the radius and the z axis. Draw a horizontal line from the point P to the z axis. The length of this line is R sin θ. So you see how this length is related to longitudinal distance around the z axis?
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