0
$\begingroup$

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.

$\endgroup$
1
  • 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

1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.