Solving Line & Velocity Elements in Spherical Coordinates

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SUMMARY

The discussion focuses on deriving the line element and velocity element in spherical coordinates. The line element is defined as (ds)^2=(dr)^2+r^2(sin(theta))^2(dtheta)^2+r^2(dphi)^2, while the velocity element is expressed as sqrt[(dr/dt)^2+r^2(sin theta)^2(dtheta/dt)^2+r^2(dphi/dt)^2]. The transformation from Cartesian coordinates involves substituting x, y, and z with their spherical equivalents and applying the product rule to find the velocity vector. Key unit vectors in spherical coordinates, such as \hat \theta and \hat \phi, are also derived for simplification.

PREREQUISITES
  • Understanding of spherical coordinates and their mathematical representation
  • Familiarity with vector calculus and differentiation
  • Knowledge of unit vectors in three-dimensional space
  • Proficiency in algebraic manipulation and substitution techniques
NEXT STEPS
  • Study the derivation of the line element in different coordinate systems
  • Learn about the product rule in vector calculus
  • Explore the applications of spherical coordinates in physics and engineering
  • Investigate the properties of unit vectors in various coordinate systems
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Students and professionals in physics, mathematics, and engineering who are working with spherical coordinates and need to understand the derivation of line and velocity elements in this context.

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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.
 
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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.
 
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 spherical-polar 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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