I'm reading Lanczos Variational Principles of Mechanics p.124, and following a discussion of how for scleronomic systems we get

$$\sum_{i=1}^{n} p_i\dot q_i - L = const.\tag{53.12}$$

For rheonomic systems it's stated that

$$\delta L=dL-\frac{\partial L}{\partial t}dt = \epsilon\left(\dot L -\frac{\partial L}{\partial t}\right)\tag{53.22}$$ where $\epsilon=dt$, which leads to $$\left[\sum_{i=1}^{n} p_i\dot q_i - L\right]^{t_2}_{t_1} = -\int^{t_2}_{t_1} \frac{\partial L}{\partial t} dt\tag{53.23}$$

However, when I do the variation

$$\delta\int_{t_1}^{t_2} L~dt= \epsilon\int_{t_1}^{t_2} \left(\dot L -\frac{\partial L}{\partial t}\right)dt = \epsilon L|_{t_1}^{t_2} - \int_{t_1}^{t_2}\frac{\partial L}{\partial t}dt$$

I'm getting an extra $\epsilon L|_{t_1}^{t_2}$ term? Any insight on what missing would be greatly appreciated!

  • $\begingroup$ Related: physics.stackexchange.com/q/94381/2451 and links therein. $\endgroup$
    – Qmechanic
    Jul 13, 2018 at 17:45
  • $\begingroup$ That question is for scleronomic (time-independent) systems where we get conservation of energy. This question is about deriving the the rheonomic equivalent $\endgroup$
    – DS08
    Jul 13, 2018 at 17:48
  • $\begingroup$ Arn't we supposed to make $\epsilon$ depend on $t$ and to vanish at the endpoints? $\endgroup$
    – mike stone
    Jul 13, 2018 at 17:50
  • $\begingroup$ That’s for deriving the general equations of motion from $\delta \int Ldt =0$ $\endgroup$
    – DS08
    Jul 13, 2018 at 18:06

1 Answer 1


Well, Lanczos uses the infinitesimal transformations

$$ t^{\prime} - t ~=:~\delta t ~=~0, \qquad \text{(no horizontal variation)}\tag{A''}$$ $$ q^{\prime i}(t) - q^i(t)~=:~\delta_0 q^i ~=~\epsilon\dot{q}, \qquad \text{(vertical variation)}\tag{B''}$$ $$ q^{\prime i}(t^{\prime}) - q^i(t)~=:~\delta q^i ~=~\epsilon\dot{q}. \qquad \text{(full variation)},\tag{C''} $$

cf. eq. (53.1). It is explained in Section V in my Phys.SE answer here that

$$ d(p_i\epsilon\dot{q}^i)~=~d(p_i\delta_0q^i)~\approx~\delta_0 L ~=~\frac{\partial L}{\partial q^i }\delta_0 q^i + \frac{\partial L}{\partial \dot{q}^i }\delta_0 \dot{q}^i ~=~\epsilon\frac{\partial L}{\partial q^i }\dot{q}^i + \epsilon\frac{\partial L}{\partial \dot{q}^i } \ddot{q}^i ~=~ \epsilon\frac{dL}{dt}-\epsilon\frac{\partial L}{\partial t}. \tag{D''} $$

Noether's theorem then yields that the would-be bare Noether current, full Noether current, and conservation law are $$j~=~p_i\dot{q}^i,$$ $$J~=~p_i\dot{q}^i-L,$$ and $$ \frac{dJ}{dt}~\approx~-\frac{\partial L}{\partial t},$$ respectively,

  • $\begingroup$ So effectively the $\epsilon L|_{t_1}^{t_2}$ terms just vanish? $\endgroup$
    – DS08
    Jul 13, 2018 at 19:10
  • $\begingroup$ No, that term (which btw is already present in the scleronomic case) causes the full Noether current to be different from the bare Noether current. $\endgroup$
    – Qmechanic
    Jul 13, 2018 at 19:17
  • $\begingroup$ Ohh I see what’s going on. The left side is equal to $\sum p_i \dot q_i |_{t1}^{t2} $so I really just had to add those terms to the left side. This makes sense now. Thanks! $\endgroup$
    – DS08
    Jul 13, 2018 at 19:45

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