I have a simple question regarding an example presented by Leonard Susskind and George Hrabovsky in their book on Classical Mechanics The Theoretical Minimum. In page 151, they state:

"If there is no explicit time dependence in the Lagrangian, then the energy $H$ is conserved. If, however, the Lagrangian is explicitly time-dependent, then the Hamiltonian is not conserved."

An example is given by the authors: suppose that a charged particle is moving between the plates of a capacitor with potential difference $\epsilon x$. If the field $\epsilon$ is constant, the Lagrangian is written as $$L= \frac{1}2m\dot x^2 + \epsilon x.$$ In this case, the energy is conserved. If the field $\epsilon$ is not constant (i.e. the capacitor is charging), the Lagrangian has an explicit time dependence and it is written as $$L= \frac{1}2m\dot x^2 + \epsilon(t) x.$$

My question: why the first Lagrangian doesn't have an explicit time dependence? Don't we have an explicit time dependence through $\dot x$? Even if $\dot x$ is constant in this case, isn't $\dot x$ generally explicitly time-dependent?


2 Answers 2


Explicit dependence would mean $\partial_tL\ne0$. Note that $\partial_t\dot{x}=0$.


In the Lagrangian formalism, the Lagrangian is a function $L(x, \dot{x}, t)$. The notation $\frac{\partial L}{\partial t}$ means nothing but "the partial derivative of L with respect to its third argument". The partial derivative notation always means differentiating a function with respect to a certain "argument slot", regardless of what you put in the slot afterwards. (Personally, I have been confused about this for several years before realising how to think about it.)


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