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I'm having some difficulties understanding the cross product in a left-handed coordinate system. I want to compute $\hat{i} \times \hat{j}$ for both systems in the picture (the first one is right-handed and the second is left-handed).

For the first system (right handed), I'm calculating the cross product this way and I correctly obtain $\hat{i} \times \hat{j} = \hat{k}$. But when I try to apply the same method to the second case, since $\hat{i}$ is still on the $x$ directon and $\hat{j}$ on the $y$ direction, I'm getting the same result, which is not correct since this is a left-handed system. What am I missing here?

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I'm getting the same result, which is not correct since this is a left-handed system. What am I missing here?

You are not missing anything. It is correct, you should be getting the same result. You have changed both the handed ness of the coordinate system and the roles of the vectors. Those two changes cancel out so that in the end you get the same result. To make this easier to see let’s use $’$ symbols to indicate things in the left handed coordinate system. So $\hat i’$ is the i vector in the left-handed system and $\times’$ is the left handed cross product.

Now, we can use the fact that $\hat i=\hat j’$, $\hat j=\hat i’$, and $\hat k=\hat k’$ to show $$\hat i \times \hat j = \hat k$$ $$\hat j’ \times \hat i’ = \hat k’$$ $$\hat i’ \times \hat j’ = -\hat k’$$ $$\hat i’ \times’ \hat j’ = \hat k’$$

So we see that your result is correct. You changed the handedness of the coordinate system and you changed the definitions of your axes and basis vectors. Each of those changes introduced a minus sign which canceled each other out.

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In a left-handed system (which is the one on the right), the relation that connects your basis vectors $\textbf{e}_1, \textbf{e}_2, \textbf{e}_3$ (that signify $\textbf{i}, \textbf{j}$ and $\textbf{k}$ respectively) is:

$$\textbf{e}_i \times \textbf{e}_j = \sum_{k=1}^3\epsilon_{ijk} \textbf{e}_k$$

where $\epsilon_{ijk}$ is the $\textbf{Levi - Civita Symbol}$, defined (in this case) as:

$$ \epsilon_{ijk} = \begin{cases} -1 & \text{if} \ (i,j,k) = (1,2,3), (2,3,1) \text{ or } (3,1,2) \\ +1 & \text{if} \ (i,j,k) = (3,2,1), (1,3,2) \text{ or } (2,1,3) \\ 0 & \text{if} \ i=j, j=k \text{ or } k=i \end{cases} $$ You can test each case individually. Now that you know how your basis vectors interact, you can calculate the cross product of two abstract vectors: $$\textbf{a} = a_1\textbf{e}_1 + a_2\textbf{e}_2 + a_3\textbf{e}_3 \text{ and } \textbf{b} = b_1 \textbf{e}_1 + b_2 \textbf{e}_2 + b_3 \textbf{e}_3 $$

Since $\textbf{e}_i \times \textbf{e}_i = 0 \ \forall i \in \{1,2,3\}$. $\textbf{a} \times \textbf{b}$ simplifies quickly to:

$$\textbf{a} \times \textbf{b} = a_1b_2(\textbf{e}_1 \times \textbf{e}_2) + a_1b_3(\textbf{e}_1 \times \textbf{e}_3) + a_2b_1(\textbf{e}_2 \times \textbf{e}_1)+a_2b_3(\textbf{e}_2\times\textbf{e}_3) + a_3b_1(\textbf{e}_3 \times \textbf{e}_1) + $$ $$+a_3b_2(\textbf{e}_3 \times \textbf{e}_2) = $$ $$a_1b_2(-\textbf{e}_3) + a_1b_3\textbf{e}_2 + a_2b_1\textbf{e}_3 + a_2b_3\textbf{e}_1+a_3b_1 (-\textbf{e}_2) + a_3b_2\textbf{e}_1 = $$ $$(a_2b_3-a_3b_2)\textbf{e}_1 + (a_1b_3 - a_3b_1)\textbf{e}_2 + (a_2b_1-a_1b_2)\textbf{e}_3$$ And in the diagram's notation: $$\textbf{a} \times \textbf{b} = (a_2b_3 -a_3b_2)\textbf{i} +(a_1b_3-a_3b_1)\textbf{j}+(a_2b_1-a_1b_2)\textbf{k}$$ Hope that helped

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In a left-handed system, positive rotation is clockwise about the axis of rotation. Did you take this into account properly? And did you use left hand? It seems to work for me.

The method you sited is for a right-handed coordinate system. Its discussed in your reference but a specific method is not given for a left-handed system.

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  • $\begingroup$ What would I have to change in the determinant method in order to make it work with left-handed systems? $\endgroup$ – Tails_97 Oct 4 '19 at 22:19
  • $\begingroup$ i just basing that on what was written in your reference. I've never used a left-handed coordinate system so I am fairly clueless how it works with one. If you reread the first section of your reference it discusses handedness. $\endgroup$ – jmh Oct 4 '19 at 23:36
  • $\begingroup$ Another thing is i x j = k. But in left-hnded system i x j = -k. Not sure of the implications of that. Mabe you need j x i instead. $\endgroup$ – jmh Oct 4 '19 at 23:39

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