Question
Consider R³ as a manifold with the flat Euclidean metric, and coordinates {x, y, z}. Introduce spherical polar coordinates {r, θ, ϕ} related to {x, y, z} by
x = rsinθcosϕ (1)
y = rsinθsinϕ (2)
z = rcosθ (3)
so that the metric takes the form
ds2 = dr2 + r2dθ2 + r2sin2θdϕ2 (4)
(a) A particle moves along a parameterised curve given by
x(λ) = cosλ , y(λ) = sinλ , z(λ) = λ (5)
Express the path of the curve in the {r, θ, ϕ} system.
(b) Calculate the components of the tangent vector to the curve in both the Cartesian and spherical polar coordinates.
Answers
It is fairly obvious that the curve is a helix. It has unit radius, the distance between each rung is 2π and so it goes up at an angle of 45°. It is shown below, compressed in the Z direction.
The helix, with some vector components |
Calculating the components of the tangent vector d/dλ in polar coordinates was non-trivial for me. It involved the chain, the quotient and the cos-1 rules of differentiation. I checked them by using the tensor transformation law from Cartesian to spherical polar components. The law is
In this case b…z and α...ω indices disappear, so it became a bit simpler but there were still almost 50 equations to step through and I had to add the tan-1 rule of differentiation to my armoury. I discovered that this method gave a different result for dθ/dλ component of my tangent vector. The possibility of errors had became enormous. Symbolab, a wonderful differential equation calculator, came to my rescue and I used it to check everything. It discovered a couple of minor errors in my 50 equation epic, but eventually pinned down the error to the first calculation of dθ/dλ from the equation of the curve in the {r, θ, ϕ} system. It just goes to show how important it is to have an 'independent' check.
More Gains
1) I have also now realised why the metric is sometimes written in the form like
ds2 = dr2 + r2dθ2 + r2sin2θdϕ2
and sometimes as a matrix. And how to get from one to the other.
2) In oder to draw the helix, I developed a spreadsheet to draw the 3-D curve. I am inspired to do something more general to draw any 3-D curves from a specified perspective.
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