
#1
Sep304, 12:34 AM

P: 35

I'm trying to find the line element in spherical coordinates as well as a velocity element. I know that they are (ds)^2=(dr)^2+r^2(sin(theta))^2(dtheta)^2+r^2(dphi)^2 and sqrt[(dr/dt)^2+r^2(sin theta)^2(dtheta/dt)^2+r^2(dphi/dt)^2].
I know that this should be a quick and easy problem, but I simply can not figure it out. I would really appreciate some help on this one. 



#2
Sep304, 01:18 AM

Sci Advisor
HW Helper
P: 3,149

Start from Cartesian coordinates in which [itex]ds^2 = dx^2+dy^2+dz^2[/itex] then calculate the differentials dx, dy and dz using:
[tex]x = r \sin \theta \cos \phi[/tex] [tex]y = r \sin \theta \sin \phi[/tex] [tex]z = r \cos \theta[/tex] Substitute for dx, dy and dz in [itex]ds^2 = dx^2+dy^2+dz^2[/itex] and after a bit of algebra you should get the desired result. 



#3
Sep304, 10:45 AM

Sci Advisor
HW Helper
PF Gold
P: 4,108

To find the velocity vector, write the position vector as [itex] r(t) \hat r [/itex].
Then [itex]v(t)=\frac{d}{dt} \left[ r(t) \hat r \right] [/itex]. Use the product rule. You'll have to compute [itex]\frac{d}{dt} \hat r [/itex], where [tex]\hat r= \sin\theta\cos\phi \hat\imath + \sin\theta\sin\phi \hat\jmath + \cos\theta \hat k[/tex]. To simplify what you get, you might find it useful to know that [tex]\hat \theta= \cos\theta\cos\phi \hat\imath + \cos\theta\sin\phi \hat\jmath  \sin\theta \hat k[/tex] and [tex]\hat \phi= \sin\phi \hat\imath + \cos\phi \hat\jmath [/tex] You can derive these expressions for the sphericalpolar unit vectors if you calculate the vectorial element [tex]d \vec s = (dx)\hat \imath + (dy)\hat \jmath + (dz)\hat k [/tex] using Tide's expressions for x, y, and z. [The strategy is to group the terms in [itex] dr[/itex], [itex]d\theta[/itex], and [itex]d\phi[/itex].] 


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