Why the Lagrangian doesn't have an explicit time dependence?

Question

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?

No, it’s implicitly time-dependent. Explicit means $t$ itself actually appearing in $L$. — G. Smith, Dec 01 '19 at 07:55
Can't we write $\dot x(t)$, and then $L$ becomes explicitly time-dependent? — Naps, Dec 01 '19 at 07:58
Possible duplicate: What is the difference between implicit, explicit, and total time dependence, e.g. $\frac{\partial \rho}{\partial t}$ and $\frac{d \rho} {dt}$? — Qmechanic, Dec 01 '19 at 08:00
You can write that, but it doesn’t make $t$ “explicit”. It’s just a function argument. — G. Smith, Dec 01 '19 at 08:00
It is however wrong that the Hamiltonian is necessarily the energy of the system in the reference frame where the Lagrangian is initially computed when the Lagrangian does not depend explicitly on time. — Valter Moretti, Dec 01 '19 at 10:12
By the way, it's a good opportunity to realize the importance of the Beltrami Identity. — Frobenius, Dec 01 '19 at 14:26

score 5 · Accepted Answer · answered Dec 01 '19 at 08:05

5

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

answered Dec 01 '19 at 08:05

J.G.

24,837

score 3 · Answer 2 · answered Dec 01 '19 at 09:54

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.)

Why the Lagrangian doesn't have an explicit time dependence?

2 Answers2