# When is the Lagrangian a constant of motion?

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.

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 $2\sum_iv_iF_i=0\implies F_i=0\implies V=\text{const}.$

• Why $v \cdot F =0$ implies $F=0$? (Isn't it even false in circular motion?) – Emolga Nov 21 '15 at 19:12
• @Leullame: this should be true for every choice of initial conditions, so for every $v$. – enzotib Nov 21 '15 at 19:31
• 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. – Emolga Nov 21 '15 at 22:18
• @Leullame: given that $\partial\mathcal{H}/\partial{t}=-\partial{L}/\partial{t}$, this could be possible only for $m=0$. – enzotib Nov 22 '15 at 10:11