We are going to look into angular momentum soon. Before that, let's review the vector cross product.
The definition of the cross product is as follow
$ \begin{eqnarray*} \vec{x} \times \vec{y} &=& \left|\begin{array}{ccc}i & j & k\\x_1 & x_2 & x_3\\y_1 & y_2 & y_3\end{array}\right| \\ &=& (x_2 y_3 - x_3 y_2) e_1 + (x_3 y_1 - x_1 y_3 ) e_2 + (x_1 y_2 - x_2 y_1) e_3 \end{eqnarray*} $
In this form, we can easily seen we can write it as a summation of these indices
$ \begin{eqnarray*} \vec{x} \times \vec{y} &=& \sum\limits_{i,j,k = 0}^{3}{f(i, j, k)x_i y_j e_k} \\ \end{eqnarray*} $
It is easy to verify that this is true, therefore we have $ \vec{x} \times \vec{y} = \sum\limits_{i,j,k = 0}^{3}{\epsilon_{ijk}x_iy_je_k} $, where $ \epsilon_{ijk} $ is the Levi-Civita symbol. We can even simplify it to just $ \epsilon_{ijk}x_iy_je_k $, using the Einstein's summation notation.
As a simple sum, differentiate it will be easy. Suppose we have the vectors as a function of time, then we have
$ \begin{eqnarray*} \frac{d}{dt} \vec{x} \times \vec{y} &=& \frac{d}{dt}(\epsilon_{ijk}x_i(t)y_j(t)e_k) \\ &=& \epsilon_{ijk}(x_i'(t)y_j(t) + x_i(t)y_j'(t))e_k \\ &=& \epsilon_{ijk}x_i'(t)y_j(t)e_k + \epsilon_{ijk}x_i(t)y_j'(t)e_k \\ &=& \vec{x'} \times \vec{y} + \vec{x} \times \vec{y'} \end{eqnarray*} $
The definition of the cross product is as follow
$ \begin{eqnarray*} \vec{x} \times \vec{y} &=& \left|\begin{array}{ccc}i & j & k\\x_1 & x_2 & x_3\\y_1 & y_2 & y_3\end{array}\right| \\ &=& (x_2 y_3 - x_3 y_2) e_1 + (x_3 y_1 - x_1 y_3 ) e_2 + (x_1 y_2 - x_2 y_1) e_3 \end{eqnarray*} $
In this form, we can easily seen we can write it as a summation of these indices
$ \begin{eqnarray*} \vec{x} \times \vec{y} &=& \sum\limits_{i,j,k = 0}^{3}{f(i, j, k)x_i y_j e_k} \\ \end{eqnarray*} $
| i | j | k | f(i,j,k) |
| 1 | 2 | 3 | 1 |
| 1 | 3 | 2 | -1 |
| 2 | 1 | 3 | -1 |
| 2 | 3 | 1 | 1 |
| 3 | 1 | 2 | 1 |
| 3 | 2 | 1 | -1 |
| otherwise | 0 | ||
It is easy to verify that this is true, therefore we have $ \vec{x} \times \vec{y} = \sum\limits_{i,j,k = 0}^{3}{\epsilon_{ijk}x_iy_je_k} $, where $ \epsilon_{ijk} $ is the Levi-Civita symbol. We can even simplify it to just $ \epsilon_{ijk}x_iy_je_k $, using the Einstein's summation notation.
As a simple sum, differentiate it will be easy. Suppose we have the vectors as a function of time, then we have
$ \begin{eqnarray*} \frac{d}{dt} \vec{x} \times \vec{y} &=& \frac{d}{dt}(\epsilon_{ijk}x_i(t)y_j(t)e_k) \\ &=& \epsilon_{ijk}(x_i'(t)y_j(t) + x_i(t)y_j'(t))e_k \\ &=& \epsilon_{ijk}x_i'(t)y_j(t)e_k + \epsilon_{ijk}x_i(t)y_j'(t)e_k \\ &=& \vec{x'} \times \vec{y} + \vec{x} \times \vec{y'} \end{eqnarray*} $
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