Take derivative to a cross product of two vectors with respect to the position vector [closed]

Question

I'm doing classical mechanics about Lagrange formulation and confused about something about vector differentiation.The Lagrangian is given: $$\mathcal{L}=\frac{m}{2}(\dot{\vec{R}}+\vec{\Omega} \times \vec{R})^2-V(\vec{R})=\frac{m}{2}\dot{\vec{R}}^2+m\dot{\vec{R}} \cdot (\vec{\Omega }\times \vec{R})+\frac{m}{2}(\vec{\Omega} \times \vec{R})^2-V(\vec{R})$$

Then the equation of motion of given by Euler-Lagrange equation: $$\frac{d}{dt}(\frac{\partial \mathcal{L}}{\partial \dot{\vec{R}}})=\frac{\partial \mathcal{L}}{\partial \vec{R}}$$

The left hand side is easy to evaluate. In order to evaluate the right hand side, rewrite the relevant term(the second and the third term) using the cyclic property of the triple product as $$m\vec{R}\cdot(\dot{\vec{R}}\times \vec{\Omega})+\frac{m}{2}(\vec{\Omega} \times \vec{R})\cdot(\vec{\Omega} \times \vec{R})$$

The derivative of the first term with respect to $\vec{R}$ is straightforward and for the second term we get $$\frac{m}{2} \frac{d}{d \vec{R}}(\vec{\Omega} \times \vec{R})\cdot(\vec{\Omega} \times \vec{R})=m(\frac{d}{d \vec{R}}(\vec{\Omega} \times \vec{R}))\cdot(\vec{\Omega} \times \vec{R})$$

Then, using the triple product rule $$=m({\frac{d}{d\vec{R}}(\vec{R})})\cdot ((\vec{\Omega} \times \vec{R})\times \vec{\Omega})=m(\vec{\Omega} \times \vec{R})\times \vec{\Omega}$$

I don't understand the last step. Specifically, I don't know how to apply triple product rule in the last step. I'm wondering whether I can write: $$m(\frac{d}{d \vec{R}}(\vec{\Omega} \times \vec{R}))\cdot(\vec{\Omega} \times \vec{R})=m(\vec{\Omega}\times\frac{d}{d\vec{R}}(\vec{R}))\cdot(\vec{\Omega} \times \vec{R})$$

Then apply triple product rule.

However, is the derivative term, $\frac{d}{d\vec{R}}(\vec{R})$, a unit matrix? How to cross product a vector with a matrix?

Answer 1

I would suggest making use of the famous identity

$$(\vec{\Omega}\times\vec{R})^2=\vec{\Omega}^2\vec{R}^2-(\vec{\Omega}\cdot\vec{R})^2$$

Proof (for the record):

$$(\vec{\Omega}\times\vec{R})\cdot(\vec{\Omega}\times\vec{R})=-\big(\vec{\Omega}\times(\vec{\Omega}\times\vec{R})\big)\cdot\vec{R}$$

The first parenthesis is

$$(\vec{\Omega}\cdot\vec{R})\vec{\Omega}-\vec{\Omega}^2\vec{R},$$

QED

Answer 2

HINT: If you want to do this without index notation, try rewriting the term $(\vec{\Omega} \times \vec{R})\cdot(\vec{\Omega} \times \vec{R})$ using the cyclicity of the triple product and the $BAC$–$CAB$ rule before taking the derivative. It should be possible to put this expression into a form that does not involve cross products at all, and in this form it is easier to see how to take its derivative.