Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 3 / 37D / b / i

Table of contents
- Solution i

Solution

Put

ρ = r^{2} + a^{2}, ϖ = ρ sin θ, z = r cos θ .

Since

x^{2} + y^{2} = ϖ^{2}

, the Euclidean spatial line element is

d x^{2} + d y^{2} + d z^{2} = d ϖ^{2} + ϖ^{2} d ϕ^{2} + d z^{2} .

Now

d ϖ = \frac{r}{ρ} sin θ d r + ρ cos θ d θ, d z = cos θ d r - r sin θ d θ .

The cross terms cancel. With

Σ = r^{2} + a^{2} cos^{2} θ, Δ = r^{2} + a^{2},

we obtain the oblate spheroidal coordinates line element

d x^{2} + d y^{2} + d z^{2} = \frac{Σ}{Δ} d r^{2} + Σ d θ^{2} + Δ sin^{2} θ d ϕ^{2} .

Adding the time coordinate gives

d s^{2} = - d t^{2} + \frac{r ^{2} + a ^{2} cos ^{2} θ}{r ^{2} + a ^{2}} d r^{2} + (r^{2} + a^{2} cos^{2} θ) d θ^{2} + (r^{2} + a^{2}) sin^{2} θ d ϕ^{2} .

 Ancestors (11)

 View article source

 Discussion (0)

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.

  See all articles in the same topic Create my own version