2
$\begingroup$

I am trying to derive

$$\mathrm dx\ \mathrm dy = r\,\mathrm dr\ \mathrm d\phi.$$

I start with the following ansatz:

$$\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} r\cos\phi \\ r\sin\phi \end{pmatrix}.$$

The change $\mathrm dx$ in $x$ is equal to:

$$\mathrm dx = \frac{\partial x}{\partial r}\ \mathrm dr + \frac{\partial x}{\partial \phi}\ \mathrm d\phi$$

and similarly for $y$. This gives

$$\mathrm dx\ \mathrm dy = \left(\frac{\partial x}{\partial r} \frac{\partial y}{\partial \phi} + \frac{\partial y}{\partial r} \frac{\partial x}{\partial \phi}\right) \mathrm dr\ \mathrm d\phi = r(\cos^2\phi - \sin^2\phi)\ \mathrm dr\ \mathrm d\phi \quad\text{(ignoring terms like $(\mathrm dr)^2$ and $(\mathrm d\phi)^2$)}$$

This is evidently incorrect and differs from the Jacobian determinant (only) by the minus sign:

$$\left| \frac{\partial (x,y)}{\partial (r,\phi)} \right|\mathrm dr\ \mathrm d\phi = \left| \frac{\partial x}{\partial r} \frac{\partial y}{\partial \phi} - \frac{\partial y}{\partial r} \frac{\partial x}{\partial \phi}\right| \mathrm dr\ \mathrm d\phi.$$

Where is my ansatz incorrect? How do I derive the Jacobian determinant?

$\endgroup$
4
  • 2
    $\begingroup$ I think you are referring to the same problem here. $\endgroup$ Oct 24 '18 at 11:33
  • $\begingroup$ I think the issue is with directions. From what you are doing, you are assuming $dx$ is in the $\hat{x}$ direction and $dy$ in the $\hat{y}$ direction. Therefore you are assuming that the area element, by computing $dx dy$ is a square. In reality, we having a mapping from 1 two-dimensional space to another $G: (r,\theta) \mapsto (x,y)$. So as you vary $r$, $(x,y)$ will move in some random direction and as you vary theta with $r$ fixed, $(x,y)$ move in some other random direction $\endgroup$
    – DWade64
    Oct 24 '18 at 13:09
  • $\begingroup$ If you have a function $f$ defined in the xy plane, then $dA$ or $dx dy$ stands symbolically for an area element of the domain. If you have a function $g$ in the $r\theta$ plane, then $dA$ or $drd\theta$ would stand symbolically for an area element of $g$'s domain. If you want to think of the dx in dx dy as a differential, then what function is it a differential of? (you can only take differentials and derivatives of functions). $dx$ would stand for the differential of the identity function for $x$, and $dy$ would be the differential for $\endgroup$
    – DWade64
    Oct 24 '18 at 13:25
  • $\begingroup$ the identity function for $y$. $dr$ would be the differential for the identity function of $r$ $\endgroup$
    – DWade64
    Oct 24 '18 at 13:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.