Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

What's the gradient vector field for $u+iv= \mathrm{Log~} z$? I got:

$$\frac{\mathrm du}{\mathrm dr} = 1/r$$ $$\frac{\mathrm du}{\mathrm d\theta} = 0$$ and $$\frac{\mathrm dv}{\mathrm dr} = 0$$ $$\frac{\mathrm dv}{\mathrm d\theta} = 1$$ But the answer is $(1/r)ur$ and $(1/r)u\theta$

share|cite|improve this question

1 Answer 1

Your problem is with the definition of gradient vector in orthogonal curvilinear coordinates: you need to include the scale factors. Specifically, $\nabla f = \sum_j \frac{1}{h_j} \frac{\partial f}{\partial x_j} {\bf u}_j$ where $x_j$ are the coordinates, ${\bf u}_j$ is the unit vector in the direction of increasing $x_j$, and $h_j = \left| \frac{\partial {\bf r}}{\partial x_j}\right|$. In particular for polar coordinates $(r,\theta)$, $$\nabla f = \frac{\partial f}{\partial r} {\bf u}_r + \frac{1}{r} \frac{\partial f}{\partial \theta} {\bf u}_\theta$$

share|cite|improve this answer

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.