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I was reading about the applications of Euler-Lagrange equation in Mathematics for Physics by Stone, Goldbart; they were showing the application of the variational principle for the central force problem $F = -\partial_r V(r)$ where $V$ is the scalar potential; then the authors concluded $\dot{(mr^2\dot \theta)} = 0$ which implies $l =mr^2\dot \theta = \textrm{const.} $

However, then they wrote:

Warning: We might realize, without having gone to the trouble of deriving it from the Lagrange equations, that rotational invariance guarantees that the angular momentum $l = mr^2\dot\theta $ is constant. Having done so, it is almost irresistible to try to short-circuit some of the labour by plugging this prior knowledge into $$L =\frac12 m\left(\dot r^2 + r^2\dot \theta^2\right) - V (r) \tag{1.53}$$

so as to eliminate the variable $\dot \theta$ in favour of the constant $l$. If we try this we get $$L \stackrel{?}{\to}\frac12 mr^2 +\frac{l^2}{2mr^2} - V (r). \tag{1.54}$$

We can now directly write down the Lagrange equation $r$, which is $$m\ddot r + \frac{l^2}{mr^3} \stackrel{?}{=}-\frac{\partial V}{\partial r} \tag{1.55}$$

Unfortunately this has the wrong sign before the $l^2/mr^3$ term! The lesson is that we must be very careful in using consequences of a variational principle to modify the principle.

Indeed the sign is wrong; there should be '$-$' and not '$+$' in $(1.55)$. However, I'm not getting why it yielded the wrong result - the wrong sign.

After all, we have used the correct conclusion that came from the variational principle itself; still a wrong sign appeared. Why is it so?

What did the authors meant by the last line?

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When we take the partial derivative of $$\frac12 m\left(\dot r^2 + r^2\dot \theta^2\right)$$ with respect to $r$, that means changing $r$ while keeping $\theta$ as it was. Accordingly, we find the change of kinetic energy is balanced by the derivative of potential $V(r)$ with respect to $r$.

But if we first place $\dot \theta = C/r$ and then differentiate with respect to $r$, this is no longer a partial derivative with respect to $r$; this is a directional derivative in some direction that involves changing both $r$ and $\theta$. This sort of variation involves modifying $\dot\theta$ (velocity, not trajectory), and it's hard to say what it even represents in physical terms. In mathematical terms, there is no reason to expect this directional derivative to be the same as the partial derivative with respect to $r$.

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  • $\begingroup$ Thanks for the response; +1 for it sounds reasonable. I'll however not accept it now; when I resume this topic again, I'll ping you and tick it. $\endgroup$ – user142971 Dec 26 '16 at 9:07

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