It is known that when the hamiltonian is time independent, it also does not vary with time.

That is, $\frac{\partial \mathcal{H}}{\partial {t}}=0$ implies $\frac{\mathrm{d} \mathcal{H}}{\mathrm{d} {t}}=0$ on solutions of the hamiltonian equations. This is an easy calculation using the hamiltonian equations and the chain rule: $\frac{\mathrm{d} \mathcal{H}}{\mathrm{d} {t}}= \frac{\partial \mathcal{H}}{\partial {p}} \frac{\mathrm{d} \mathcal{q}}{\mathrm{d} {t}} + \frac{\partial \mathcal{H}}{\partial {q}} \frac{\mathrm{d} \mathcal{q}}{\mathrm{d} {t}}= \frac{\partial \mathcal{H}}{\partial {p}} (-\frac{\partial \mathcal{H}}{\partial {q}} ) + \frac{\partial \mathcal{H}}{\partial {q}} (-\frac{\partial \mathcal{H}}{\partial {p}} )=0$.

Is the same true for the Lagrangian?

That is, assuming $L(q, \dot q )$, is it true that $\frac{dL}{dt}=0$ on solutions satisfying the Euler-Lagrange equations?

The canonical example of the Lagrangian is $L=T-V=\frac 12 m \sum \dot q_i ^2-V(q)$, which we naturally look at first.

In this case: $\frac{dL}{dt}= \sum_i \frac{\partial \mathcal{L}}{\partial {q_i}} \frac{\mathrm{d} \mathcal{q_i}}{\mathrm{d} {t}} + \sum_i \frac{\partial \mathcal{L}}{\partial {\dot q_i}} \frac{\mathrm{d} \mathcal{\dot q_i}}{\mathrm{d} {t}}= m \sum _i \dot q_i \ddot q_i- \sum \frac{\partial V}{\partial {q_i}} \dot q_i= m \sum _i \dot q_i \ddot q_i+ \sum_i F_i \dot q_i = \sum (m v_i a_i +F_i v_i)$.

How can I see in a concrete example what this is equal to? (And in particular, whether this is zero?)

Please don't use any 'empirical' facts, only the mathematical formulation.


2 Answers 2


For any system in which you expect energy transfer between kinetic and potential energy the Lagrangian won't be a constant of motion. To take a concrete example, consider a one-dimensional system corresponding to a falling body under (some approximation of) gravity:

$$ L(q, \dot{q}) = \frac{1}{2}m \dot{q}^2 - m g q. $$

The equations of motion can be easily solved to result in

$$ q(t) = A_0 + A_1t - \frac{g}{2} t^2. $$

Consider for simplicity the initial conditions $A_0 = A_1 = 0$. Then $q(t) = -\frac{g}{2} t^2$ and while the sum of the kinetic and potential energy (corresponding to the Hamiltonian)

$$ \frac{1}{2}m \dot{q}^2 + mgq = \frac{1}{2}m (-gt)^2 - \frac{mg^2}{2}t^2 \equiv 0 $$

is constant, the difference between the kinetic and the potential energy (which is the Lagrangian) is not:

$$ L(q(t),\dot{q}(t)) = \frac{1}{2}m \dot{q}^2 - mgq = \frac{1}{2}m (-gt)^2 + \frac{mg^2}{2}t^2 = mg^2t^2.$$


Given that $m_ia_i=F_i$, you get $2\sum_iv_iF_i=0\implies F_i=0\implies V=\text{const}.$

  • $\begingroup$ Why $v \cdot F =0 $ implies $F=0$? (Isn't it even false in circular motion?) $\endgroup$
    – Emolga
    Nov 21, 2015 at 19:12
  • 1
    $\begingroup$ @Leullame: this should be true for every choice of initial conditions, so for every $v$. $\endgroup$
    – enzotib
    Nov 21, 2015 at 19:31
  • $\begingroup$ What about lagrangians that are not of the form $L=T-V$? For example, I think that whenever the lagrangian is positively homogeneous, it is a constant of motion. This is simply because $\mathcal H = (m-1) L$, where m is the order of the homogeneity. $\endgroup$
    – Emolga
    Nov 21, 2015 at 22:18
  • $\begingroup$ @Leullame: given that $\partial\mathcal{H}/\partial{t}=-\partial{L}/\partial{t}$, this could be possible only for $m=0$. $\endgroup$
    – enzotib
    Nov 22, 2015 at 10:11

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