# Polar form of Newton's Second Law

Can someone please show the steps for deriving the total energy of a particle from the Polar form of Newton's Second Law?

The total energy is $$T+V$$, with $$T$$ kinetic and $$V$$ potential, and by energy conservation $$0=\dot{T}+\dot{V}$$. Taking $$T$$ to depend only on $$x$$ and $$V$$ only on $$\dot{x}$$, $$0=\ddot{x}\cdot\nabla_\dot{x}T+\dot{x}\cdot\nabla_xV$$. We want this to be equivalent to $$0=m\ddot{x}+\nabla_xV$$, so $$\nabla_\dot{x}T=m\dot{x}$$. The convention that $$T(0)=0$$ gives $$T=\frac{m\dot{x}^2}{2}$$.

This argument doesn't commit us to Cartesian coordinates; scalar products, including squared lengths, are unchanged when we switch to polar coordinates. In three dimensions, $$T=\frac{m}{2}\left(\dot{r}^2+r^2\dot{\theta}^2+r^2\sin^2\theta\dot{\phi}^2\right).$$The last term is deleted in two dimensions, to which we typically switch if radial forces' angular momentum conservation implies the motion is confined to a plane.