On page 2 of "Mechanics" Landau & Lifshitz say that $q=q(t)$ is a function for which action is a minimum. Before this they say that at times $t_1$ and $t_2$, the system occupies coordinates $q^{(1)}$ and $q^{(2)}$, respectively. However the function for which action is a minimum, or $q$ is $\int\limits_{t_1}^{t_2} L(q,\dot{q},t)\,dt$, and not just the coordinates. Why do they use the same symbol $q$ twice? Another thing is that I don't understand when they say:

"Since, for $t_1$ and $t_2$, all the functions $q+{\delta}q$ must take the values $q^{(1)}$ and $q^{(2)}$ respectively, it follows that ${\delta}q(t_1)={\delta}q(t_2)=0$"

This would be understandable if $q$ was $\int\limits_{t_1}^{t_2} L(q,\dot{q},t)\,dt$, but later on they justify the fact that $[\frac{\partial{L}}{{\partial{\dot{q}}}}{\delta}q]^{t_2}_{t_1}=0$ by the above, once again treating $q$ as if it were coordinates. So what do the $q$ refer to after all?


This is a statement of the principle of minimal action. Let $$\int\limits_{t_0}^{t_f}\text{d}t L (q(t),\dot{q}(t),t) $$ be the energy functional ($L = E_{kin} - E_{pot}$). First note that the energy is function of coordinate ($q$) and speed ($\dot{q}$). Then consider a infinitesimal variation of the trajectory from point $q(t_0)$ and $q(t_f)$: $$q(t) = q_0(t) + \delta q(t)$$, the action becomes: $$S(q_0(t) + \delta q(t)) = S(q_0(t)) + \delta S = \int\limits_{t_0}^{t_f} \text{d}t L (q_0(t),\dot{q}_0(t),t) + \bigg (\frac{\delta L }{\delta q} - \partial_t \frac{\delta L }{\delta \dot{q}}\bigg ) \delta q $$ $q_0(t)$ is the trajectory with minimal action, if the variation of the action with respect to $\delta q(t)$ vanishes.

Now about your questions: the symbol $\dot{q}$ is the speed, and not the coordinate. It is the time derivative of the position. About the second question, you want to vary infinitesimally the trajectory between $q^{(1)}$ and $q^{(2)}$, but not $q^{(1)}$ and $q^{(2)}$ themselves. So you require $\delta q(t_1) = \delta q(t_2) = 0$.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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