Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.


share|cite|improve this question
up vote 9 down vote accepted

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: $$\frac{\partial}{\partial x}\arctan x = \frac{1}{x^2+1}$$

share|cite|improve this answer
luksen: yes, eventually I understand. Thank you. – Peter4075 Jun 20 '11 at 18:03
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. – 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:
$$r\;d\theta \;\;=\;\; - \frac{y\;dx}{r} + \frac{x\;dy}{r},$$ $$d\theta \;\;=\;\; - \frac{y\;dx}{r^2} + \frac{x\;dy}{r^2}.$$

share|cite|improve this answer
I confess at my level I find luksen's answer a bit clearer but thanks anyway. – Peter4075 Jun 22 '11 at 19:36
Well, I hate differentiating inverse trig functions. – Carl Brannen Jun 22 '11 at 20:35

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.