# Integration of equation of motion to obtain invariant angular momentum

Studying the equations of motion in a central-symmetric potential, we obtain:

$m ( r \ddot \theta + 2 \dot r \dot \theta ) = 0 \tag{1}$

From that my book concludes that the angular momentum ($m r^2 \dot \theta$) is constant:

$\frac{d}{d t} (m r^2 \dot \theta ) = 0 \tag{2}$

This is true, as differentiation shows:

$$\frac{d}{d t} (m r^2 \dot \theta ) = m ( r^2 \ddot \theta + 2 r \dot r \dot \theta )$$

This is equal to (1) after division by $r$. How could I have discovered that starting from (1) itself? Is there any way to see that multiplication by $r$ makes the integral more easily solvable?

Clarification: What I'm after is the integration, not the conversation law. How do I know that I should multiply by $r$ when I obtain an integration such as $\int r \ddot \theta + 2 \dot r \dot \theta dt$ ? Is there any guiding principle or do you just play around with the equation and hope for the best?

• Can you clarify in what way the integral can be solved more simply by multiplying by $r$? I see that this revised integrand is simpler to integrate, but adding $r$ also changes the integral. – Hrodelbert Mar 12 '15 at 12:26
• How did people in the 1700's (before Lagrange/Noether) figure out that there's a conserved quantity hiding in (1)? Did they just play around with the equation and realize that if they multiply by $r$, they get the time derivative of another quantity? Or is there a more general method to go from "oh, this is always zero" to "that thing never changes"? – Christian Aichinger Mar 12 '15 at 13:26
• I have updated my answer, hopefully to contain some things that might help you. I think at least that the more general method you ask for is either the construction of Noether charges or the general theory of Integrable systems. Of course Noether's theorem relies on the Lagrangian formalism. I am not sure if I can help you any further than this. – Hrodelbert Mar 12 '15 at 13:36