I am solving the vector equation:

$$\vec \nabla P(r,\phi) = \vec f(r,\phi)$$

where $\vec f$ is conservative, in polar coordinates.

Am I allowed to the following?

$$\partial_r P= f_r$$ $$\partial_\phi P=r f_\phi$$

Or, if I am to solve the line integral, using the following fact

$$P(r,\phi)-P(r_0,\phi_0)=\int_\mathcal{C} \vec f(r,\phi)\cdot \vec dl$$

with $\mathcal{C}$ any path from $(r_0,\phi_0)$ to $(r,\phi)$, I could go along the direction $\hat r$ and write:

$$P=\int_{r_0}^r f_r(r,\phi) dr$$

or not?


The gradient operator in polar coordinates is given by

$$\nabla =\hat r \frac{\partial }{\partial r}+\hat \theta \frac{1}{r}\frac{\partial }{\partial \theta}$$

Since the gradient of a scalar field is a conservative, then the line integral between two points is not affected by the path taken between points. Therefore, one can write

$$\begin{align} P(\vec r_2)-P(\vec r_1)&=\int_{\vec r_1}^{\vec r_2} \nabla P \cdot d\vec \ell\tag 1 \end{align}$$

The points $\vec r_1$ and $\vec r_2$ are not in general collinear. In other words, the unit vectors $\hat r_1 =\hat x \cos \theta_1+\hat y \cos \theta_1$ and $\hat r_2 =\hat x \cos \theta_2+\hat y \cos \theta_2$ with $\theta_1\ne \theta_2$ point in different directions.

Therefore, in general the line integral in $(1)$ is not equivalent to a path integral along a constant $\theta$ direction.

  • $\begingroup$ This does not answer the question, since this seems in agreement with the first derivation, isn't it? $\endgroup$ – usumdelphini Sep 14 '15 at 14:50
  • $\begingroup$ @usumdelphini I edited to expand on the reasoning. Please let me know how I can improve my answer. I really want to give you the best answer I can. $\endgroup$ – Mark Viola Sep 14 '15 at 15:04

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