Question about the vector cross product in spherical or cylindrical coordinates

[dyn]

Hi
If i calculate the vector product of a and b in cartesian coordinates i write it as a determinant with i , j , k in the top row. The 2nd row is the 3 components of a and the 3rd row is the components of b.
Does this work for sphericals or cylindricals eg . can i put e_r , e_θ , e_φ in the top row with the relevant components in the 2nd and 3rd rows ?
I know that for the formula for grad , div and curl in sphericals or cylindricals then scale factors appear but is the cross product of 2 vectors independent of scale factors ?
Thanks

 

[jedishrfu]

The simplest solution is to convert both vectors to cartesian, do the cross product and convert backup to spherical or cylindrical.

However, doing the cross product spherically or cylindrically directly boils down to find a vector that is perpendicular to both vectors following the right hand rule convention and recalling that the magnitude of the resultant vector is:

$\vec A \times \vec B = |A||B|sin(\theta)$

where $\theta$ is the angle between $\vec A$ and $\vec B$ and |A| and |B| are the magnitudes of the respective vectors $\vec A$ and $\vec B$.

There is an old thread on it here:

https://www.physicsforums.com/threa...ame-way-i-do-in-cartesian-coordinates.772165/

A more detailed answer may be found here:

https://math.stackexchange.com/questions/1358590/cross-product-spherical-coordinates

 

[pasmith]

Hi
If i calculate the vector product of a and b in cartesian coordinates i write it as a determinant with i , j , k in the top row. The 2nd row is the 3 components of a and the 3rd row is the components of b.
Does this work for sphericals or cylindricals eg . can i put e_r , e_θ , e_φ in the top row with the relevant components in the 2nd and 3rd rows ?
I know that for the formula for grad , div and curl in sphericals or cylindricals then scale factors appear but is the cross product of 2 vectors independent of scale factors ?
Thanks

THe determinant method works for any set of orthogonal unit basis vectors which form a right-handed triple, so if in your convention \mathbf{e}_r \times \mathbf{e}_\theta = \mathbf{e}_{\phi} then <br /> \mathbf{a} \times \mathbf{b} = \left| \begin{matrix}<br /> \mathbf{e}_{r} &amp; \mathbf{e}_{\theta} &amp; \mathbf{e}_{\phi} \\<br /> a_r &amp; a_\theta &amp; a_\phi \\<br /> b_r &amp; b_\theta &amp; b_\phi<br /> \end{matrix}\right|. Curl doesn't work this way because in polar coordinates the basis vectors are functions of position and not constants.

 

[dyn]

Thank you

 

[romsofia]

Since I've seen you post in some physics sections before, here is an excerpt on how to use the levi-cievta sign from notes i wrote years ago (since physicists love the following two ideas, it's worth getting use to it):

We will now introducte the Levi-Cievta symbol as follows: $$\epsilon_{ijk} = \begin{cases}
1 & 123, 312, 231 \\
-1 & 321, 132, 213 \\
0 & i=j=k
\end{cases} $$

Where the numbers represent the subscripts. So that means $\epsilon_{123} = 1$. So how do you keep track of which is which? When I was younger, it was introduced to be a permutation, and a bunch of other nomenclature that I couldn't quite grasp until I was older. So I'll give you the secret I used! I simply start with 123, and then bump up the numbers to get 312, then again to get 231. Then, to get the -1, you just start with 321, and do the same. This is essentially what a permutation is, but when you're seeing all this for the first time, and a new symbol, even the simple things seem hard! So, how exactly does this relate to cross product? Well, let me throw out this equation (Note I am using Einstein summation): $$\vec{A} \times \vec{B} = \epsilon_{ijk}A^jB^k\hat{e_i}$$ Pretty obvious now, right? Okay, if it isn't obvious how they're related, don't worry! $A^j$ $B^k$ are the components of our vector, and $\hat{e_i}$is our basis vector. Let's explicitly write out the sum to see if it will give us the same result as the cross product. \\

$$
\epsilon_{ijk}A^jB^k\hat{e_i} = $$
\begin{align*}
\epsilon_{111}A^1B^1\hat{e_1} + \epsilon_{112}A^1B^2\hat{e_1} + \epsilon_{113}A^1B^3\hat{e_1} + \\ \epsilon_{123}A^2B^3\hat{e_1} + \epsilon_{122}A^2B^2\hat{e_1} + \epsilon_{121}A^2B^1\hat{e_1} +\\
\epsilon_{131}A^3B^1\hat{e_1}+\epsilon_{132}A^3B^2\hat{e_1}+\epsilon_{133}A^3B^3\hat{e_1} +\\ \epsilon_{211}A^1B^1\hat{e_2}+ \epsilon_{212}A^1B^2\hat{e_2} + \epsilon_{213}A^1B^3\hat{e_2} + \\
\epsilon_{223}A^2B^3\hat{e_2} + \epsilon_{222}A^2B^2\hat{e_2} + \epsilon_{221}A^2B^1\hat{e_2} + \\
\epsilon_{231}A^3B^1\hat{e_2} + \epsilon_{232}A^3B^2\hat{e_2} + \epsilon_{233}A^3B^3\hat{e_2} +\\
\epsilon_{311}A^1B^1\hat{e_3} + \epsilon_{312}A^1B^2\hat{e_3} + \epsilon_{313}A^1B^3\hat{e_3} + \\
\epsilon_{323}A^2B^3\hat{e_3} + \epsilon_{322}A^2B^2\hat{e_3} + \epsilon_{321}A^2B^1\hat{e_3} +\\
\epsilon_{331}A^3B^1\hat{e_3} + \epsilon_{332}A^3B^2\hat{e_3} + \epsilon_{333}A^3B^3\hat{e_3}
\end{align*}

So, I know what you're thinking... 27 terms from such a simple line?! That's the power of the Einstein summation convention! However, I only wrote this out so you can get a feel for that power. In reality, we just use what the Levi-Civita symbol tells, that is any $i=j=k$ just go to zero, so our 27 equations zoom down to 6.
So we're left with
\begin{align*}
\epsilon_{ijk}A^jB^k\hat{e_i} =
\epsilon_{123}A^2B^3\hat{e_1} + \epsilon_{132}A^3B^2\hat{e_1} \\
\epsilon_{231}A^3B^1\hat{e_2} + \epsilon_{213}A^1B^3\hat{e_2} \\
\epsilon_{312}A^1B^2\hat{e_3} + \epsilon_{321}A^2B^1\hat{e_3}
\end{align*}
I've specifically put them in this order for a reason, we can now factor out the specific basis vector on each line and let's see what we get.
We now see that our equation becomes
\begin{align*}
\epsilon_{ijk}A^jB^k\hat{e_i} =
\hat{e_1}(A^2B^3 - A^3B^2) + \\
\hat{e_2}(A^3B^1 - A^1B^3)+ \\
\hat{e_3}(A^1B^2 - A^2B^1)
\end{align*}Which, as you can see, makes no mention of the particular coordinate system you're in. I do go into more details in my notes, but I think this is long enough for you to get the idea!

 

[jedishrfu]

Nice tensor exposition! However, it may be overkill for the OP as Tensor Analysis is seldom taught until needed in grad school for either Differential Geometry or General Relativity.