An identity for the Lorentz cross product

Question

I need some help finding the error in my proof for an identity for the Lorentz cross product.

For $x=(x_1,x_2,x_3)^T, y=(y_1,y_2,y_3)^T \ \in R^{2,1}$ the Lorentz cross product is defined as

det[x,y,z]=$\langle\,x \times y,z\rangle$

for all $z \in R^{2,1}$ where $\langle\,\cdot, \cdot\rangle$ is the Lorentz scalar product of $R^{2,1}$.

In coordinates we have

$x \times y=\begin{pmatrix}x_2y_3-y_2x_3\\y_1x_3-x_1y_3\\x_1y_2-y_1x_2\end{pmatrix}$.

Now let

$a=(a_1,a_2,a_3)^T, b=(b_1,b_2,b_3)^T, c=(c_1,c_2,c_3)^T \in R^{2,1}$. I want to prove the identity

$a \times (b \times c)=c\langle\,a, b\rangle-b\langle\,a,c\rangle$

I did it in coordinates. We have

$b \times c=\begin{pmatrix}b_2c_3-c_2b_3\\c_1b_3-b_1c_3\\b_1c_2-c_1b_2\end{pmatrix}$

and

$a \times (b \times c)=\begin{pmatrix}a_2(b_1c_2-c_1b_2)-(c_1b_3-b_1c_3)a_3\\(b_2c_3-c_2b_3)a_3-a_1(b_1c_2-c_1b_2)\\a_1(c_1b_3-b_1c_3)-(b_2c_3-c_2b_3)a_2\end{pmatrix}$

$=\begin{pmatrix}-c_1b_2a_2-c_1b_3a_3+b_1a_2c_2+b_1a_3c_3\\-c_2a_1b_1-c_2a_3b_3+b_2a_1c_1+b_2a_3c_3\\-c_3a_1b_3-c_3a_2b_2+b_3a_1c_1+b_3a_2c_2\end{pmatrix}$

We also have

$c\langle\,a, b\rangle-b\langle\,a, c\rangle$= \begin{pmatrix}c_1a_2b_2-c_1a_3b_3-b_1a_2c_2+b_1a_3c_3\\c_2a_1b_1-c_2a_3b_3-b_2a_1c_1-b_2a_3c_3\\c_3a_1b_1+c_3a_2b_2-b_3a_1c_1-b_3a_2c_2\end{pmatrix}

So far the equations do not coincide. It seems that I have made a mistake but I don't see where. Any help would be appreciated.

Answer 1

The problem is with your formula for Lorentz cross product. It should be: $x \times y=\begin{pmatrix}x_2y_3-y_2x_3\\y_1x_3-x_1y_3\\y_1x_2- x_1y_2\end{pmatrix}$.

With this formula everything fits.

Answer 2

The correct identity is $$a\times (b\times c)=b\langle\,a,c\rangle-c\langle\,a, b\rangle.$$The right-hand-side is negative of $c\langle\,a, b\rangle-b\langle\,a,c\rangle$.

You calculated $a\times (b\times c)$ wrongly. I calculated it as $$\begin{pmatrix}-c_1b_2a_2-c_1b_3a_3+b_1a_2c_2+b_1a_3c_3\\-c_2a_1b_1-c_2a_3b_3+b_2a_1c_1+b_2a_3c_3\\-c_3a_1\color{red}{b_1}-c_3a_2b_2+b_3a_1c_1+b_3a_2c_2\end{pmatrix}$$ The highlighted symbol is where you calculated wrong. Double check that part.

You also calculated $c\langle\,a, b\rangle-b\langle\,a,c\rangle$ wrongly. I calculated it as $$\begin{pmatrix}c_1a_2b_2\color{red}+c_1a_3b_3-b_1a_2c_2\color{red}-b_1a_3c_3\\c_2a_1b_1\color{red}+c_2a_3b_3-b_2a_1c_1-b_2a_3c_3\\c_3a_1b_1+c_3a_2b_2-b_3a_1c_1-b_3a_2c_2\end{pmatrix}$$

Again, the highlighted part is where you were wrong. Double check those.