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Cross product

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Spanned parallelogram.png

$$u \times v = <A,B,C> \times <D,E,F> = \det \left( \begin{array}{ccc} i & j & k \\ A & B & C \\ D & E & F \end{array} \right) $$


We generalize the last theorem below.

Theorem (Conservation of Angular Momentum). If $X$ is a trajectory of the discrete model with central force, the angular momentum, $$H_{n+1}=X_n\times X_{n+1}',$$ is constant. Conversely, if the angular momentum remains constant along each trajectory of a discrete model of order two, the model's force is a central force.

Proof. The proof follows the last one. For the model: $$X_{n+1}' '=f(X_n),$$ where $f$ is some function, we have $$X_n=X_{n-1}+ X_{n}'\Delta t $$ and $$X_{n+1}'=X_n'+X_n' ' \Delta t=X_n'+f(X_n)\Delta t.$$ We substitute the first into the following: $$\begin{array}{ll} H_{n}&=X_{n-1}\times X_{n}' \\ &=(X_n-X_n'\Delta t)\times X_{n}' \\ &=X_n\times X_{n}'-X_n'\Delta t\times X_{n}' \\ &=X_n\times X_{n}'. \end{array}$$ We substitute the second into the following: $$\begin{array}{ll} H_{n+1}&=X_n\times X_{n+1}' \\ &=X_n\times (X_n'+f(X_n)\Delta t) \\ &=X_n\times X_n'+X_n\times f(X_n)\Delta t . \end{array}$$ Therefore, $$H_{n+1}=H_n\ \Longleftrightarrow\ X_n\times f(X_n) =0\ \Longleftrightarrow\ X_n\ ||\ f(X_n).$$ $\blacksquare$