Suppose I paramaterize my curve in one coordinate system, how do I specify it in a different coordiante system?

Question

Suppose some curve $\vec r (t) = (x,y,z)$ is given in one coordinate system, how do I compute it in a different coordinate system?

For example, say the helix $\vec r(t) = (\cos t, \sin t, t)$ in Cartesian coordinates, how do I transform this to say spherical coordiantes ($r = (x^2 + y^2 + z^2)^{1/2}$, $\theta = \arctan (y/x) $, $\phi = \arccos (z/r)$ ), or any other coordinate system?

Answer 1

Well I think you mostly gave your own answer.

We already have $(x,y,z)=(\cos(t),\sin(t),t)$. Now we just use the formulas you gave: $$r=(x^2+y^2+z^2)^{1/2}=(\cos^2(t)+\sin^2(t)+t^2)^{1/2}=(1+t^2)^{1/2}$$ $$\theta=\arctan(y/x)=\arctan(\sin(t)/\cos(t))=\arctan(\tan(t))=t$$ $$\phi=\arccos(z/r)=\arccos\left(\frac{t}{(1+t^2)^{1/2}}\right)$$

As long as you have the rule for converting between coordinates, you can apply it to the parameterization. The same things works in any coordinate system. (Can you convert to cylindrical now?)

Answer 2

Simply substitute the expressions for $r, \theta, \phi$ in $x, y, z$: \begin{align} r(t) &= [x(t)^2 + y(t)^2 + z(t)^2]^{1 / 2} = [\cos^2 t + \sin^2 t + t^2]^{1 / 2} = \sqrt{1 + t^2} \\ \theta(t) &= \arctan \frac{y(t)}{x(t)} = \arctan \frac{\sin t}{\cos t} = \arctan \tan t \\ \phi(t) &= \arccos \frac{z(t)}{r(t)} = \arccos \frac{t}{\sqrt{1 + t^2}} \end{align}

One must interpret these formulas carefully: The formula for $\theta$ in terms of $x, y$ only holds for $x > 0$, so we can only expect our formula for $\theta$ to hold for some interval on which $x(t)$ is always positive. Indeed, our formula for $\theta(t)$ is discontinuous at half-integer multiples of $\pi$ (corresponding to points when $x = 0$). If we restrict attention to the interval $-\frac{\pi}{2} < t < \frac{\pi}{2}$, for example, we can simplify our formula for $\theta(t)$ to $\theta(t) = t$. (There is no analogous issue with the formula for $\theta(t)$, by the way: Since $0 \leq t^2 < 1 + t^2$, $0 \leq \frac{t}{\sqrt{1 + t^2}} < 1$, and hence $\phi(t)$ is continuous.)

On the other hand, using our formula $\theta(t) = t$, which a priori only holds on the mentioned interval, and substituting $(r(t), \theta(t), \phi(t))$ into the formulas for expressing rectangular coordinates as functions of spherical coordinates, we see that equality really holds for all $t$, not just $t$ in the given interval. The price one pays for this convenience is that $(r, \theta, \phi)$ are no long strictly coordinates, as many triples, namely $(r, \theta + 2 \pi k, \phi)$, $k \in \Bbb Z$, represent the same points.