2
$\begingroup$

I can't wrap my head around the result of the cross product of two vectors in spherical coordinates. Is it a vector or something that I can represent geometrically?

For example, given two vectors in spherical coordinates:

$\overrightarrow a = (a,\theta,\phi)$

$\overrightarrow b = (b,0,0)$

Taking the cross product $[\overrightarrow a \times \overrightarrow b] = -ab\sin \theta \overrightarrow{e_\phi}$

Ok, so where does this vector start and where does it point to? Is it a "different" kind of vector? Does it point anywhere?

$\overrightarrow a$ and $\overrightarrow b$ point away from the origin and start on the origin. Where does the resulting vector of their cross product point to?

If $ab\sin \theta$ is the vector modulus and $-\overrightarrow{e_\phi}$ is the direction it is pointing, does it mean that it starts in any point of the sphere of radius $ab\sin \theta$ and points in the $-\overrightarrow{e_\phi}$ direction?

In that case, if I do the cross product of $\overrightarrow a$ and the resulting vector, what will be the $\theta$ angle between both vectors??? It can be anything!!

This final question appeared to me when studying and finding in my textbook something similar to this:

$[\overrightarrow a \times \overrightarrow b] = -ab\sin \theta \overrightarrow{e_\phi}$

$[\overrightarrow a\times[\overrightarrow a \times \overrightarrow b]] = a^2b\sin \theta \overrightarrow{e_\theta}$

I understand the first cross product but cannot find how to compute the second. $\theta$ between $\overrightarrow a$ and the resulting vector is 90º? How?

EDIT: Now that I think about it, I am satisfied if someone helps me understand the meaning of a vector of type $a\overrightarrow{e_\theta}$ for example.

$\endgroup$
4
  • $\begingroup$ possible duplicate of What is the general formula for calculating dot and cross products in spherical coordinates? $\endgroup$ Jul 12, 2015 at 17:45
  • $\begingroup$ That accepted answer to that question does not derive an expression for the cross product, but does give some insights of how to get it. Also remember that the cross product between two vectors will be perpendicular to both vectors, so for your example this implies the resulting $\phi$ has to be equal to $\frac{\pi}{2}$ in order to be perpendicular to $\vec{b}$. $\endgroup$ Jul 12, 2015 at 17:48
  • $\begingroup$ I skimmed through that topic before but didn't find a good answer. Since you pointed that it could be a dupplicate I payed much more attention this time and found an answer that partially explains what I want math.stackexchange.com/a/243303/254072. So thank you. $\endgroup$
    – rmarques
    Jul 12, 2015 at 17:55
  • $\begingroup$ Ah... facepalms... of course it has to be perpendicular... Thank you so much @fibonatic, that helps with the missing sine on the second formula. $\endgroup$
    – rmarques
    Jul 12, 2015 at 17:58

3 Answers 3

4
$\begingroup$

Note: one need to be careful when working in spherical coordinates, as there are two conventions used for the notation of the polar and azimuthal angles. See https://en.wikipedia.org/wiki/Spherical_coordinate_system and the 2 pictures near the top. In this answer I am using the physicist's convention, since it complies better with your own notation.

To clear the first point of confusion: All vectors start at the origin. They can be added together visually by placing them head to tail, but you should always think of them as having the tail at the origin and the head pointing in whatever direction it's supposed to.

The cross product in rectangular coordinates and the cross product in spherical coordinates are the same thing. The only difference is the way we represent them in formulas.

Notation also seems to be confusing you. The basis in spherical coordinates works differently from the basis in rectangular coordinates. See http://mathworld.wolfram.com/SphericalCoordinates.html and scroll down to the listing of unit vectors for a complete account. Caution: This article uses the mathematician's notation. You will have to swap the roles of $\theta$ and $\phi$ in your head, or just write things down to keep track. e.g. $\hat{\phi}$ there is what we'll think of as $\vec{e}_\theta$.

Here is a working definition of the cross product:

The cross product of two vectors $\vec{a}$ and $\vec{b}$ is another vector, mutually perpendicular to $\vec{a}$ and $\vec{b}$, with the direction determined by the right-hand rule. The magnitude of $\vec{a}\times\vec{b}$ is given by the area of the parallelogram spanned by $\vec{a}$ and $\vec{b}$.

Note that this definition makes no reference to the choice of coordinate system; it works in rectangular and spherical coordinates.

Therefore, if $\vec{a} = (a,\theta,\phi)$ and $\vec{b} = (b,0,0)$ in spherical coordinates, to find the cross product we need to find the direction the vector is pointing and the magnitude from this information. The key to this is to find the angle between $\vec{a}$ and $\vec{b}$. Well, $\vec{b}$ points straight along the $z$-axis, so it suffices to find the angle of $\vec{a}$ relative to the $z$-axis, and if you draw a picture you should see that this angle is $\theta$. (It doesn't matter what $\phi$ is here.) The area of the parallelogram spanned by $\vec{a}$ and $\vec{b}$ is then $ab\sin\theta$. As for the direction, to be perpendicular to the $z$-axis the vector must lie in the $xy$-plane, so we need to find the direction in the $xy$-plane perpendicular to $(a,\theta,\phi)$. Then it suffices to find the direction perpendicular to $(1,\pi/2,\phi)$, which is of course $(1,\pi/2,\phi\pm\pi/2)$. The right-hand rule says we take the minus sign, so $$ \vec{a}\times\vec{b} = ab\sin\theta(1,\pi/2,\phi-\pi/2). $$ We can take the plus sign by pointing in the opposite direction, so $$ \vec{a}\times\vec{b} = -ab\sin\theta(1,\pi/2,\phi+\pi/2). $$ Finally, let's express the last vector in rectangular coordinates. It is $$ (1,\pi/2,\phi+\pi/2)_{\text{sph}} = (\sin(\pi/2)\cos(\phi+\pi/2),\sin(\phi+\pi/2)\sin(\pi/2), \cos(\pi/2))_{\text{rec}} = (-\sin\phi,\cos\phi,0)_{\text{rec}} = \vec{e}_{\phi} $$ according to the definition of unit vectors in spherical coordinates. (See the mathworld link above if you haven't already.) Therefore we finally obtain $$ \vec{a}\times\vec{b} = -ab\sin\theta\vec{e}_\phi $$ as claimed.

Now to compute $\vec{a}\times\vec{a}\times\vec{b}$, we know that $\vec{a}$ and $\vec{a}\times\vec{b}$ are perpendicular, so they span a rectangle. Therefore the area they span is the product of their lengths, or $a^2b\sin\theta$. (And there is a minus sign from the previous computation.) For the direction we need to compute $$ \frac{\vec{a}}{\|\vec{a}\|}\times\vec{e}_{\phi}. $$ $\vec{e}_\phi$ is the vector lying in the $xy$-plane perpendicular to $(1,\pi/2,\phi)_{\text{rec}}$. To be perpendicular to this and $\vec{a}$, we must find the vector in the plane perpendicular to $\vec{e}_\phi$ which, if you draw a picture, you'll see is the one containing $\vec{a}$ and the $z$-axis. Therefore to find the perpendicular we just have to shift $\theta$ by $-\pi/2$ (the sign comes from the right-hand rule). Therefore $$ \frac{\vec{a}}{\|\vec{a}\|}\times\vec{e}_{\phi} = (1,\theta-\pi/2,\phi)_{\text{sph}} $$ and if we work this out in rectangular coordinates, $$ (1,\theta-\pi/2,\phi)_{\text{sph}} = (\sin(\theta-\pi/2)\cos\phi,\sin(\theta-\pi/2)\sin\phi,\cos(\theta-\pi/2))_{\text{rec}} = (-\cos\theta\cos\phi,-\cos\theta\sin\phi,-\sin\theta)_{\text{rec}} = -\vec{e}_\theta. $$ So putting everything together, $$ \vec{a}\times\vec{a}\times\vec{b} = -a^2b\sin\theta(-\vec{e}_\theta) = a^2b\sin\theta\vec{e}_\theta. $$ This is never zero except in the following cases (which are not mutually exclusive): if $\vec{a}=0$, if $\vec{b}=0$, or if $\vec{a}$ and $\vec{b}$ point in the same or opposite directions (so that $\sin\theta=0$). Which you could guess from the original working definition of the cross product: since $\vec{a}\times\vec{b}$ always points perpendicular to $\vec{a}$, as long as we don't have some trivial case the vectors $\vec{a}$ and $\vec{a}\times\vec{b}$ will span a parallelogram of nonzero area.

As you can see, this whole business with spherical coordinates is rather cumbersome for the cross product. This is because the cross product is designed to respect linear combinations, and therefore it works well in a rectangular system where the local geometry looks the same everywhere. However, in curvilinear coordinates (like spherical), vectors do not add coordinatewise like in rectangular coordinates, so the cross product cannot use the special linear structure of $\mathbb{R}^3$. The easiest way to do things is to learn to break up vectors into the local basis for spherical coordinates, and learn the formulas for the cross product of the local orthonormal basis in spherical coordinates. Either that, or as Griffiths suggests in Introduction to Electrodynamics, just convert to rectangular coordinates and try to do most things there.

$\endgroup$
1
  • $\begingroup$ I am really thankful for you taking the time to write such a complete answer, it is exactly what I needed. Cheers! $\endgroup$
    – rmarques
    Jul 12, 2015 at 19:22
0
$\begingroup$

Generally speaking, I don't think it's helpful to think of vectors as "starting somewhere". It can be helpful e.g. when you're adding vectors; then you can imagine one "starting" where the other one "ends"; but more often than not I'd say it's best not to think in these terms.

In your case, if you insist on wanting to visualize the vector as "starting somewhere" and "pointing somewhere", note that the frame of unit vectors $\vec e_r$, $\vec e_\theta$ and $\vec e_\phi$ is a function of position. The $\vec e_\phi$ that you're getting as a cross product could more precisely be denoted as $\vec e_\phi(\vec a)$; it's the unit vector that points in the direction in which the position changes if you change $\phi$ at $\vec a$. Thus, if you really want to imagine it "starting somewhere" and "pointing somewhere", it would make sense to imagine it attached at $\vec a$ and pointing along the $\phi$ coordinate line (the line on which $r$ and $\theta$ are constant and $\phi$ changes).

Since it's a unit vector, any factors in front of it determine the length of the resulting vector by their magnitude and the orientation by their sign (towards increasing $\phi$: positive, towards decreasing $\phi$: negative).

$\endgroup$
0
$\begingroup$

Note that $[\overrightarrow a \times \overrightarrow b] \perp \overrightarrow a, \overrightarrow b$ so that $\overrightarrow a \times [\overrightarrow a \times \overrightarrow b]$ is a vector of magnitude $|\overrightarrow a| |\overrightarrow a \times \overrightarrow b| = a |\overrightarrow a \times \overrightarrow b|$ that is both $\perp \overrightarrow a$ and in the plane spanned by $\overrightarrow a, \overrightarrow b$.

Also noting that $\overrightarrow a = a \overrightarrow{e_r}$ and that $\overrightarrow{e_r} \times \overrightarrow{e_\phi} = -\overrightarrow{e_\theta}$ you can get the second formula from the first.

Another source of confusion is the difference between the systems of spherical coordinates used in physics versus those used in mathematics (e.g. see https://en.wikipedia.org/wiki/Spherical_coordinate_system) - I can infer that you are using the physics system where $\theta$ is the polar angle. In this case, $(\overrightarrow{e_r}, \overrightarrow{e_\theta}, \overrightarrow{e_\phi})$ is a right-handed triple.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .