Question

Prove that the spherical polar co-ordinate system is orthogonal.

Answer 1

hey yaa...***

Cylindrical coordinate system is orthogonal :

Cartesian coordinate system is length based, since dx, dy, dz are all lengths. However, in other curvilinear coordinate systems, such as cylindrical and spherical coordinate systems, some differential changes are not length based, such as dθ, dφ. Suppose we define a new orthogonal coordinate system, such as spherical coordinates defined by

x=rsinθcosϕ,y=rsinθsinϕ,z=rcosθ.x=rsin⁡θcos⁡ϕ,y=rsin⁡θsin⁡ϕ,z=rcos⁡θ.

Is there an efficient and general way to prove that the coordinate system is orthogonal, other than by writing out the off-diagonal metric coefficients and finding that they are all zero? Of course it's not that bad for spherical coordinates, but for (say) toroidal or bispherical it gets messy. Let's say you have a system of coordinates u,v,wu,v,w with tangent vectors eu,ev,eweu,ev,ew. Given the relationship between these coordintaes and Cartesian coordinates, you can express eueu as a linear combination of ex,ey,ezex,ey,ez, and the same for the others.

You can use a wedge product to determine the size of the parallelepiped formed from these vectors, and you should be able to express that wedge product in terms of the Cartesian basis vectors. That is, you should be able to find

eu∧ev∧ew=α(ex∧ey∧ez)eu∧ev∧ew=α(ex∧ey∧ez)

for some number αα.

In the general case, the magnitude of αα is bounded as follows:

|α|≤|eu||ev||ew||α|≤|eu||ev||ew|

The bound is saturated when the vectors are orthogonal. This allows you to compute only the diagonal metric components and a wedge product (equivalent to computing a determinant). Debatable whether this is "easier" than computing all the metric components, though.

As such its orthogonal...

•••••hope it helpss•••••