diff --git a/src/2_coordinates.md b/src/2_coordinates.md index d55bb139b04b6838423f5d964972b27af3d0f8ed..dedcd244f491d549dcf73a56c10d12e8972629da 100644 --- a/src/2_coordinates.md +++ b/src/2_coordinates.md @@ -264,10 +264,10 @@ We want to find the length of the (small) red segment $d s$. By inspecting the figure, we see that the horizontal (i.e. parallel to the $xy$-plane) segment $d l$ is perpendicular to the vertical segment $dz$. Using for $d l$ the length we obtained before for a line -segment in the $xy$ plane, expressed in polar coordinates, we +segment in the $xy$ plane expressed in polar coordinates, we immediately find: $$d s^2 = d l^2 + d z^2 = d r^2 + r^2 d \varphi^2 + d z^2.$$ -The volume element is consequently given as +The volume element is consequently given as: !!! info "Volume element in cylindrical coordinates" $$dV = r dr d\varphi dz.$$ @@ -281,27 +281,34 @@ and (2) the direction of the line connecting the origin to our point. The specification of this direction can be identified with a point on a sphere which is centered at the origin: - - -The position of a point on the sphere is specified using the two angles -$\theta$ and $\phi$ indicated in the figure. +<figure markdown> +  + <figcaption>The position of a point on the sphere is specified using the radius $r$ and two angles +$\theta$ and $\phi</figcaption> +</figure> !!! warning - Note that in mathematics, often the angles are labelled the other way - round: there, $\phi$ is used for the angle between a line running from - the origin o the point of interest and the $z$-axis, and $\theta$ for + In mathematics, the angles are often labelled the other way + around: there, $\phi$ is used for the angle between a line running from + the origin to the point of interest and the $z$-axis, and $\theta$ for the angle of the projection of that line with the $x$-axis. The - convention used here is custom in physics. - -The relation between Cartesian and coordinates is defined by -$$x = r \cos \varphi \sin \vartheta$$ -$$y = r \sin\varphi \sin \vartheta$$ $$z = r \cos\vartheta$$ The inverse -transformation is easy to find: $$r = \sqrt{x^2+y^2+z^2}$$ -$$\theta = \arccos(z/\sqrt{x^2+y^2+z^2})$$ -$$\phi = \begin{cases} \arctan(y/x) &{\rm for ~} x>0; \\ - \pi + \arctan(y/x) & {\rm for ~} x<0 {\rm ~ and ~} y>0;\\ - -\pi + \arctan(y/x) &{\rm ~ for ~} x<0 {\rm ~ and ~} y<0. - \end{cases}$$ These relations can be derived from the following figure: + convention used here is customary in physics. + +The relation between Cartesian and spherical coordinates is defined by: +!!! info "The relation between Cartesian and spherical coordinates" + $$x = r \cos \varphi \sin \vartheta$$ + $$y = r \sin\varphi \sin \vartheta$$ $$z = r \cos\vartheta$$ + +The inverse transformation is easy to find: +!!! info "The inverse relatuion between Cartesian and spherical coordinates" + $$r = \sqrt{x^2+y^2+z^2}$$ + $$\theta = \arccos(z/\sqrt{x^2+y^2+z^2})$$ + $$\phi = \begin{cases} \arctan(y/x) &{\rm for ~} x>0; \\ + \pi + \arctan(y/x) & {\rm for ~} x<0 {\rm ~ and ~} y>0;\\ + -\pi + \arctan(y/x) &{\rm ~ for ~} x<0 {\rm ~ and ~} y<0. + \end{cases}$$ + +These relations can be derived from the following figure: 