Unit vectors in Spherical Coordinates

**Starproj** · Jun30-09, 03:43 PM

Does anyone know a good sight that explains, step-by-step, how to derive unit vectors in spherical coordinates? I am at that unfortunate place where I have been looking at it for so long I know the answer from sheer memorization, but don't understand the derivation. From the definitions I am looking at, each unit vector is a derivative of the vector r wrt the unit vector in question divided by the absolute value of the same derivative (as described at http://mathworld.wolfram.com/SphericalCoordinates.html). It is the denominator that is throwing me off.

Can anyone help before I lose my mind?

Thanks!

**Nick Bruno** · Jun30-09, 06:04 PM

hi, the denominator is not the absolute value. Its the norm... or in other words the magnitude, or in other words the vector norm, or in other words the L2-norm if u wanna talk about linear algebra.

This means you take the square root of each component squared (added together)

http://mathworld.wolfram.com/Norm.html

In this case, the norm of the position vector must be 1... since the derivative of the vector wrt the variables in question is the unit vector.

**danong** · Jul1-09, 02:29 AM

hi there,
actually the vector of spherical coordinate depens on which parameter u wish to represent,
for example in euclid, or in riemann the presentation of the vector of a particular coord system would be different,
i suggest u pick up a fundamental study in position vector in euclid before this,

the equation in mathforums at (71) onward are the one who is using the spherical coordinate system to derive, in this case i,j,k component is presented as r, detta, sigma components respectively.
It's only a matter of how you select and define your components.

**arildno** · Jul1-09, 07:09 AM

Ok, this is fairly trivial.

Assume that some vector $LaTeX Code: \\vec{u}$ (dependent on some independent variables) has unit size irrespective of the values of the independent variables, i.e:
$LaTeX Code: \\vec{u}^{2}=1(1)$

Then, labeling an independent variable as $LaTeX Code: x_{i}$ , we get by differentiating (1) wrt. to that variable:
$LaTeX Code: 2\\frac{\\partial\\vec{u}}{\\partial{x}_{i}}\\cdot\\vec{ u}=0$ , i.e, the derivatives of the unit vector are orthogonal to it!

Thus, starting out with the radial vector,
$LaTeX Code: \\vec{i}_{r}=\\sin\\phi\\cos\\theta\\vec{i}+\\sin\\phi\\sin \\theta\\vec{j}+\\cos\\phi\\vec{k}$ , we perform the two differentiations here:
$LaTeX Code: \\frac{\\partial\\vec{i}_{r}}{\\partial\\phi}=\\cos\\phi\\ cos\\theta\\vec{i}+\\cos\\phi\\sin\\theta\\vec{j}-\\sin\\phi\\vec{k}=\\vec{i}_{\\phi}$
and:
$LaTeX Code: \\frac{\\partial\\vec{i}_{r}}{\\partial\\theta}=\\sin\\ph i(-\\sin\\theta\\vec{i}+\\cos\\theta\\vec{j})=\\sin\\phi\\vec{ i}_{\\theta}$

where the appropriate forms of the unit vectors $LaTeX Code: \\vec{i}_{\\phi},\\vec{i}_{\\theta}$ have been indicated.

**Starproj** · Jul1-09, 11:15 AM

Originally Posted by Nick Bruno

hi, the denominator is not the absolute value. Its the norm... or in other words the magnitude, or in other words the vector norm, or in other words the L2-norm if u wanna talk about linear algebra.

This means you take the square root of each component squared (added together)

http://mathworld.wolfram.com/Norm.html

In this case, the norm of the position vector must be 1... since the derivative of the vector wrt the variables in question is the unit vector.

Thanks for your input - I get it now! I grinded through the math with a few simple trig substitutions and got the answers provided.

I appreciate you taking the time to relpy.

**danong** · Jul1-09, 11:13 PM

lol, hey do not ignore the contribution from arildno,
as i said you need to have some fundamental concept before coming to this, and arildno is giving you the real deriving of it.