📢Ditch the Messy Math The Elegant Tensor Secret to Centrifugal Force

The fictitious centrifugal force, initially defined by the vector triple product $\vec{F}{c}=m \vec{\omega} \times(\vec{\omega} \times \vec{x})$, can be elegantly transformed into a linear tensor expression, $F{c}^i=T^{i j} x^j$. By applying the standard vector triple product identity, the force is expanded to $\vec{F}_c=m\left[\vec{\omega}(\vec{\omega} \cdot \vec{x})-\omega^2 \vec{x}\right]$, which allows for the direct identification of the components of the tensor as $T^{i j}=m\left(\omega^i \omega^j-\omega^2 \delta^{i j}\right)$, where $\omega^2=\omega^k \omega^k$ and $\delta^{i j}$ is the Kronecker delta. This tensor $T^{i j}$ is symmetric and depends quadratically on the angular velocity $\vec{\omega}$, effectively encoding the force's linear dependence on the position vector $\vec{x}$. Physically, the centrifugal force vanishes if and only if the displacement vector $\vec{x}$ is parallel to the angular velocity $\vec{\omega}$, meaning the particle lies along the axis of rotation.

Tensor Form of the Centrifugal Force in Rotating Frames | Proof and Derivationviadean on Notion