# Differential form relation from conserved angular momentum

In Binney and Tremaine chapter 3 equation 3.9 they use the fact that in a spherically symmetric potential you get from the Euler Lagrange equations that angular momentum is conserved $$r^2 \dot{\phi}=L$$

then it is said that this relation follows $$\frac{d}{dt} = \frac{L}{r^2}\frac{d}{d\phi}$$ but I do not see how.

I tried to derive it the following way $$r^2 \frac{d\phi}{dt}=L$$ $$r^2 \frac{d\phi}{dt} \frac{d}{d \phi}=L \frac{d}{d \phi}$$ $$\frac{d}{dt} = \frac{L}{r^2}\frac{d}{d\phi}$$

But I do not understand what multiplying by $$\frac{d}{d\phi}$$ means and why it is legal. Since we have not acted the derivative on anything yet, I cannot use the limit definition to view it as a fraction. the lone $$d$$ in the numerator is concerning me.

• \begin{aligned}\dfrac{du}{dt}=\dfrac{du}{d\phi }\dfrac{d\phi }{dt}\\ \dfrac{d\phi }{dt}=\dfrac{L}{r^{2}}\\ \dfrac{d}{dt}=\dfrac{L}{r^{2}} \dfrac{d}{d\phi }\end{aligned}
– Eli
Nov 13, 2022 at 22:07

The relation: $$\frac{d}{dt}=\frac{L}{r^2}\frac{d}{d\phi}$$ is just an operator that satisfies any function $$f(\phi(t))$$. In other words, any function that can be expanded via multivariable chain rule. The equation is not something you can work with without inputting some scalar $$f$$. By default, it is written this way to make it clear that there's more than one satisfactory function (such as radius $$r(\phi(t))$$ in orbits). $$\frac{df}{dt}=\frac{d\phi}{dt}\frac{df}{d\phi}$$ $$L=r^2\frac{d\phi}{dt}\implies\frac{d\phi}{dt}=\frac{L}{r^2}$$