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 '18 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 '18 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 '18 at 17:50
  • $\begingroup$ That’s for deriving the general equations of motion from $\delta \int Ldt =0$ $\endgroup$ – DS08 Jul 13 '18 at 18:06

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 '18 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 '18 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 '18 at 19:45

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.