Commit 6b7e9ee6 authored by Michael Wimmer's avatar Michael Wimmer

add more text

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......@@ -67,19 +67,19 @@ $(r,\varphi)$ indicated. From this, we can see that the *Cartesian*
coordinates $(x,y)$ of the point are related to the polar ones as
$$x = r \cos\varphi;$$
$$y = r \sin \varphi.$$
$$\begin{equation} x = r \cos\varphi; \end{equation}$$
$$\begin{equation} y = r \sin \varphi.\end{equation}$$
The inverse relation is given as
$$r=\sqrt{x^2 + y^2};$$
$$\begin{equation} r=\sqrt{x^2 + y^2}; \label{rxy}\end{equation}$$
$$\begin{equation} \varphi=\begin{cases}
\arctan(y/x) & \text{$x>0$,}\\
\pi + \arctan(y/x) & \text{$x<0$ and $y>0$,}\\
-\pi + \arctan(y/x) & \text{$x<0$ and $y<0$.}
\end{cases} \label{phixy}\end{equation}$$
The last formula for $\varphi$ warrants a closer explanation: It is easy
to see that $\tan(\varphi)=y/x$ - but this is not a unique relation, due to
......@@ -133,8 +133,8 @@ 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,
which is indeed the area of a circle with radius 0.
......@@ -143,10 +143,10 @@ understood: the area swept by an angle difference $d\varphi$
Often, in physics important equations involve derivatives given in terms
of Cartesian coordinates. One prominent example are equations of the form
$$\left(\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2)\right)
$$\left(\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2}\right)
f(x, y) = \ldots.$$
The derivative operator $\left(\frac{\partial^2}{\partial x^2} +
\frac{\partial^2}{\partial y^2)\right)$ is so common it has its own name:
\frac{\partial^2}{\partial y^2}\right)$ is so common it has its own name:
the Laplacian (here for two-dimensional space).
Such an equation is universal, but for particular situations it might be
......@@ -162,9 +162,17 @@ involving the chain rule for a function of several variables.
Let $f$ be a function of $n$ variables: $f(y_1, y_2, \ldots, y_n)$,
as well as $g_i(x_1, x_2, \ldots, x_n)$ for $i=1,2,\ldots, n$. Then
$$\frac{\partial}{\partial{x_i} = \sum_{j=1}^n
$$\frac{\partial}{\partial x_i} = \sum_{j=1}^n
\fac{partial f}{\partial y_j} \frac{\partial g_j}{\partial x_i}$$
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
we look at this expression, we need to understand what it *means* to take the derivative
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}.
# Cylindrical coordinates
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