I'm reading (I'm trying to read) Schutz's "A first course in general relativity" (1985). On page 126 he mentions that a small change in angle theta in polar coordinates is given by:

small change in theta - polar coordinates

I can't see why this is. Can anyone please explain.



The polar angle is given by

$$\theta = \arctan(y/x)$$

the total derivative ("a small change in theta") of theta is given by

$$d\theta = \frac{\partial \theta}{\partial x}dx + \frac{\partial \theta}{\partial y}dy$$

this works out to $$\frac{\partial \theta}{\partial x} = \frac{-y}{x^2+y^2} = \frac{-y}{r^2}$$

$$\frac{\partial \theta}{\partial y} = \frac{x}{x^2+y^2} = \frac{x}{r^2}$$

therfore the result is as given in the book.

Edit: in case it is not clear: $$\frac{\partial}{\partial x}\arctan x = \frac{1}{x^2+1}$$

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    $\begingroup$ Wishing to avoid inv trig like Carl, I've always preferred to think of it as differentiating $\tan\theta = y/x$ instead. Since $x = r\cos\theta$, the LHS is $(\sec^2\theta)d\theta = d\theta(r/x)^2$, and RHS is $(xdy-ydx)/x^2$ by quotient rule. Hence the book's result: $r^2d\theta = xdy-ydx$, rearranged. $\endgroup$ – Stan Liou Aug 13 '11 at 23:42

I'll give what I think is a more beautiful explanation. First of all, calculate $dr$ in terms of $dx$ and $dy$:
$$r^2 = x^2 + y^2$$ $$2r\;dr = 2x\;dx + 2y\;dy$$ $$dr = \frac{x\;dx}{r} + \frac{y\;dy}{r}$$

With regard to the above, note that $(x/r,y/r)$ is a unit vector in the $dr$ direction. Since $d\theta$ is perpendicular to $dr$; we first look for a unit vector that is perpendicular to $(x/r, y/r)$. The answer is clearly $(y/r,-x/r)$ or its negative. To get the right sign, note that when $y=0$, we are on the x-axis and we want $d\theta$ to point in the $+y$ direction so we use $(-y/r,x/r)$.

And we need an overall scale. The circumference of the unit circle is $2\pi$ and this is what you get from integrating $d\theta$ from $0$ to $2\pi$. For larger radii, we need to multiply by the radius. So we get:
$$r\;d\theta \;\;=\;\; - \frac{y\;dx}{r} + \frac{x\;dy}{r},$$ $$d\theta \;\;=\;\; - \frac{y\;dx}{r^2} + \frac{x\;dy}{r^2}.$$

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  • $\begingroup$ I confess at my level I find luksen's answer a bit clearer but thanks anyway. $\endgroup$ – Peter4075 Jun 22 '11 at 19:36
  • $\begingroup$ Well, I hate differentiating inverse trig functions. $\endgroup$ – Carl Brannen Jun 22 '11 at 20:35

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