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.

  • 3
    $\begingroup$ \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} $\endgroup$
    – Eli
    Nov 13, 2022 at 22:07

1 Answer 1


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). $$\frac{df}{dt}=\frac{d\phi}{dt}\frac{df}{d\phi}$$ $$L=r^2\frac{d\phi}{dt}\implies\frac{d\phi}{dt}=\frac{L}{r^2}$$


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