A step in the derivation of the Euler-Lagrange equations using Hamilton’s Principle

classical-mechanicslagrangian-formalismvariational-calculusvariational-principle

I am going through the derivation of the Euler-Lagrange equations from Hamilton's principle following Landau and Lifshitz Volume 1. We start by writing the variation in the action as,

$$\delta S = \delta\int_{t_1}^{t_2}L(q,\dot{q},t)dt\ $$ as $q \to q+\delta q.$

In the next step, the variation in $L$ is written as $$\delta L= \frac{\partial{L}}{\partial{q}}\delta q+\frac{\partial{L}}{\partial{\dot{q}}}\delta \dot{q}.$$ Shouldn't there be a term $\frac{\partial{L}}{\partial{t}}\delta t$ as well since the Lagrangian $L$ is a function of $(q,\dot{q},t)$?

Best Answer

  1. No, in the principle of stationary action/Hamilton's principle we are only varying the generalized coordinates $q$; not the time-parameter $t$, i.e. $\delta t=0$. Similarly, we are not varying the endpoints $t_1$ and $t_2$.

  2. However be aware that there do exist other variational principles, such as e.g. Maupertuis' principle, where $\delta t\neq 0$.

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