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I often read,that if the lagrangian $L=p\dot{q}-H$ of a pair of coordinates in phase space $(q,p)$ and $P\dot{Q}- K $, for some new pair of coordinates $(Q,P)$ only differ by a total time derivative $dF/dt$, the Variation of the action functional doesn't differ too, which implies that the canonical equations for the new coordinates hold.

My question now is: if the action functional $S(q,p,t)=S'(Q,P,t) + c$ where $c= \int (dF/dt )dt $ is the integrated total time derivative of $F$ (which disappears in the Variation); why then is the variation of both sides equal? I mean surely the Variation of the $(q,p)$ coordinates would be the same for both sides, but why should the Variation of the $(Q,P)$ coordinates on the right hand side be the same as the Variation of the $(q,p)$ coordinates on the left hand side?

Qmechanic
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jonb25
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  • Because the new variables $Z^J\equiv (Q^j,P_j)$ are a fixed function (say, $f$) of the old variables $z^I\equiv (q^i,p_i)$ and $t$ (and vice versa). – Qmechanic Jul 21 '18 at 10:06
  • ok thank you for the answer! is it so, because that implies that Q(q+ ɛu(t),p+ ɛv(t),t) = Q+ ɛx(t) for some arbitrary functions u,v so that x(t) is arbitrary too? So that with q(t, ɛ)=q + ɛu(t) , p(t, ɛ) = p + ɛv(t) : (dS(Q(q(t, ɛ),p(t, ɛ),t),P(q(t, ɛ),p(t, ɛ),t),t)/dɛ) ist the same as : ( dS(Q(t, ɛ),P(t, ɛ),t)/d ɛ) for ɛ = 0 ? (sorry for my ugly writing, Im truly new to this site and can't yet write with mathematica or so ) – jonb25 Jul 21 '18 at 10:20
  • We are varying both $Z^J$ and $z^I$, but not independently: We are not varying their inter-relational function $f$. – Qmechanic Jul 21 '18 at 10:42
  • ok I think Im overlooking something there. I don't really understand how you vary Z^J and z^I at the same time if they are dependent on each other (as you say with the function f) ? Doesn't varying one implie how the other coordinates change? – jonb25 Jul 21 '18 at 10:53
  • oh ok I see, so if we have function S(Q,P,t) and Q,P are again in a fixed relation to (q,p) (with f), this implies that varying Q,P is the same as varying q,p for S ? – jonb25 Jul 21 '18 at 10:59
  • Yes, up to temporal boundary terms. – Qmechanic Jul 21 '18 at 11:08

1 Answers1

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It is important to stress that the $2n$ new variables $Z^J\equiv (Q^j,P_j)$ and the $2n$ old variables $z^I\equiv (q^i,p_i)$ are not $4n$ but only $2n$ independent variables.

We assume that the new variables $Z^J$ are a function of the old variables $z^I$ and time $t$ (and vice versa), and that the Jacobian matrix $\partial Z^J / \partial z^I$ is invertible.

In particular an infinitesimal variation $\delta z$ of $z$ induces a corresponding infinitesimal variation $\delta Z$ of $Z$.

See also this related Phys.SE post.

Qmechanic
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