Talk:Spherical coordinate system/Archive

The description of spherical coordinates is wrong. You interchange theta with phi


 * This is a long-standing debate/controversy/argument/cacophony of accusations. Mathematicians (American, mostly) usually do it with phi coming down from the z-axis, while everyone else usually has this be theta. My own personal preference is for the latter. Revolver


 * Not wrong, just unconventional. I've looked at three sources and two very authoratitive ones (Borowski and Borwein's Reference Dictionary of Mathematics, Collins and MIT Press's Encyclopedic Dictionary of Mathematics) have phi coming down from the z-axis.  That said, I was taught (at school and university in the UK) that phi came down from the z-axis, so it might indeed be a UK/US thing as Revolver says. -- Paul G 15:05, 22 Dec 2003 (UTC)


 * More a mathematicians/physicists thing :-)

I'm not sure this is right at first glance: the formulae do not appear to match the description in words: phi and theta appear to be opposite ways round. Specifically, the TeX equations seem to be written for azimuth = &theta; Lots of checking needed here. -- The Anome 15:15, 22 Dec 2003 (UTC)


 * I'm not sure what an azimuth and colatitude are supposed to be (even after looking at their articles), but in the equations, &phi; only needs to range from 0 to &#960;, and &theta; should be given a full 2&#960; range... Can't really fix it myself, as I don't understand all the words, perhaps someone should just delete all the text, as math is easier to understand than text anyway... &#922;&#963;&#965;&#960; Cyp  15:35, 22 Dec 2003 (UTC)


 * I get

\begin{vmatrix}d\rho\\d\phi\\d\theta\end{vmatrix} =\begin{vmatrix} \frac{x}{\rho}&\frac{y}{\rho}&\frac{z}{\rho}\\ \frac{xz}{\rho^2\sqrt{x^2+y^2}}&\frac{yz}{\rho^2\sqrt{x^2+y^2}}&\frac{-x^2-y^2}{\rho^2\sqrt{x^2+y^2}}\\ \frac{-y}{x^2+y^2}&\frac{x}{x^2+y^2}&0 \end{vmatrix}\cdot \begin{vmatrix}dx\\dy\\dz\end{vmatrix} $$
 * Someone please check this.
 * Rasmus Faber 15:41, 22 Dec 2003 (UTC)


 * Worked out one of the middle row again, and got something different than the first time, so your version is probably more accurate. Have to get a blank piece of paper to calculate on, rather than scribbling all over the place... &#922;&#963;&#965;&#960; Cyp  16:08, 22 Dec 2003 (UTC)
 * Ok, worked it out more carefully, and I get the same as the above now.
 * Tak, fordi du fandt fejlen. &#922;&#963;&#965;&#960; Cyp  16:40, 22 Dec 2003 (UTC)