I have recently started studying String Theory and this notion of variations has come up. Suppose that we have a Lagrangian $L$ such that the action of this Lagrangian is just $$S=\int dt L.$$ The variation of our action $\delta S$ is just $$\delta S=\int dt \delta L.$$ I have read on other posts that the variation is defined to be $$\delta f=\sum_i \frac{\partial f}{\partial x^i}\delta x^i,$$ which seems like an easy enough definition. Having this definition in mind, I proceded to come across an action $$S=\int dt \frac12m \dot X^2-V(X(t))$$ which implies our Lagrangian is $L(t,X, \dot X)$ which makes our first varition follow as $$\delta S=m \int dt\frac12(2 \dot X)\delta \dot X-\int dt \frac{\partial V}{\partial X} \delta X$$ $$=-\int dt \left(m \ddot X+\frac{\partial V}{\partial X}\right)\delta X.$$ My question is, did that variation of the action follow the definition listed above? That is $$\delta S=\int dt\frac{\partial L}{\partial t} \delta t+\frac{\partial L}{\partial X} \delta X+\frac{\partial L}{\partial \dot X}\delta \dot X,$$ where the $$\frac{\partial L}{\partial t} \delta t$$ term vanishes because there is no $t$ variable.
2 Answers
Yes, that happened.
I guess you meant $$ \delta f = \sum_i \frac{\partial f}{\partial x_i} \delta x_i $$ on your third equation. Also you've implicity fixed inital $t_0$ and final $t_1$, so that your action integral really is $$ S = \int_{t_0}^{t_1} dt L $$ and therefore, since the limits are fixed, variation "commutes" with integration:$$\delta \int_{t_0}^{t_1} dt L = \int_{t_0}^{t_1} dt \delta L $$ (you can check out some problems where the end intervals are not fixed for variational problems and some extra stuff is needed - see Elsgolc).
Also related: Is the principle of least action a boundary value or initial condition problem?
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If the Lagrangian only depends on time through $X$ or $\dot{X}$, then we say that the Lagrangian has implicit but not explicit time dependence. So in your example, we would write \begin{equation} L(X, \dot{X}) = \frac{1}{2} m \dot{X}^2 - V(X) \end{equation} even though $X=X(t)$ depends on time. Given this Lagrangian, your variation is completely fine; the equations of motion are \begin{equation} m \ddot{X} + V'(X) =0 \end{equation}
In order to have explicit time dependence, you would need to have time appear explicitly, not simply through $X$ or $\dot{X}$. For example: \begin{equation} L(X, \dot{X}, t) = \frac{1}{2} m \dot{X}^2 - V(X) + J(t) X \end{equation} for some function of time $J(t)$. In this case, the equation of motion would be \begin{equation} m \ddot{X} + V'(X) = J(t) \end{equation}
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