Commit be7850ec authored by Michael Wimmer's avatar Michael Wimmer

first finished draft

parent 6b7e9ee6
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......@@ -133,7 +133,7 @@ understood: the area swept by an angle difference $d\varphi$
We find:
\int_0^{2\pi} d\varphi \int_0^r_0 r dr =
\int_0^{2\pi} d\varphi \int_0^{r_0} r dr =\\
2\pi \int_0^r_0 r dr = 2 \pi \frac{1}{2} r_0^2 = \pi r_0^2,
......@@ -163,7 +163,7 @@ involving the chain rule for a function of several variables.
as well as $g_i(x_1, x_2, \ldots, x_n)$ for $i=1,2,\ldots, n$. Then
$$\frac{\partial}{\partial x_i} = \sum_{j=1}^n
\fac{partial f}{\partial y_j} \frac{\partial g_j}{\partial x_i}$$
\frac{partial f}{\partial y_j} \frac{\partial g_j}{\partial x_i}$$
We start by replacing the function $f(x, y)$ by a function in polar coordinates
$f(r, \varphi)$, and ask what is $\frac{\partial}{\partial x} f(r, \varphi)$. When
......@@ -172,7 +172,40 @@ of a function of $r, \varphi$ in terms of $x$?
For this, we need to realize that there are relations between the coordinate systems.
In particular, $r = r(x,y)$ and $\varphi = \varphi(x, y)$ as defined in equations
\ref{rxy} and \ref{phixy}.
\ref{rxy} and \ref{phixy}. In fact, we have been rather sloppy in our notation above,
as the functions $f(x,y)$ and $f(r, \varphi)$ do not mean that I substitute $x=r$
and $y=\varphi$! It is more precise to state that there are two diferent
functions $f_\text{cart}(x,y)$ and $f_\text{polar}(r, \varphi)$ that are equivalent,
in the sense that
$$f_\text{cart}(x, y) = f_\text{polar}(r(x,y), \varphi(x,y))$$
In physics, we usually never write this down explicitly, but we are aware that these
are two different functions from the fact that they use different coordinates.
With this information, we can now apply the chain rule:
$$ \frac{partial}{\partial x} f(r, \varphi) =
\frac{\partial f}{\partial r} \frac{partial r(x, y)}{\partial x} +
\frac{\partial f}{\partial \varphi} \frac{\partial \varphi(x,y)}{\partial x}
and it is now a matter of (tedious) calculus to arrive at the right result.
This is the task of exercises 3 and 4, which finally compute the Laplacian
in polar coordinates.
!!! warning Inverse function theorem
In this calculation one might be tempted to use the inverse
function theorem to compute derivatives like
$\frac{\partial \varphi}{\partial x}$ from the much simpler
$\frac{\partial x}{\partial \varphi}$. Note though that here we
are dealing with functions depending on several variables, so the
*Jacobian* has to be used (see [Wikipedia]( A direct calculation is in this particular case more easy.
Note that this procedure also carries over to other coordinate systems,
although the calculations can become quite tedious. In these cases,
it's usually best to look up the correct form.
# Cylindrical coordinates
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