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.)


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.