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I am trying to understand why Taylor says in his Classical Mechanics text, "we can always subtract a constant from the potential without effecting any physics."

I assume "doesn't effect any physics" means the equations of motion are unaltered — as is the result of adding a total time derivative to a Lagrangian.

How are the two statements related? Explicitly please.

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Yes, indeed the equations of motion are unaltered; clearly we consider potentials of the form $V(\mathbf q)$, Hence, the only contribution to the Euler Lagrange equations done by the potential is in the term $\frac{\partial L}{\partial q}$, since a constant vanishes under a derivative; then adding a constant do not affect the equations of motion. More intuitively, it has to do with the fact that a force $\mathbf F$ acting on a particle due to a potential $V$ is given by $\mathbf F=\partial V/\partial q$, again, the first derivative does the trick.

Another way to think about it is that the Energy of a system, is the sum $T+V$ as functions of $(q,\dot q, t)$ in an extremal of the action functional. So that if we added a constant to the potential (from the beginning), we would add it to this sum in the entire path $q(t)$. Hence, we "kinda" decide with how much energy the system starts, the actual rule that governs the dynamics is the conservation of the energy; not the initial energy.

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  • $\begingroup$ This answer is very helpful! However, despite how clear you made it for $U(r)$ you beg the question, what about a time-dependent potential? We have them in QM all the time, no? Does Lagrangian mechanics not work well with that case, and we differ to Hamiltonian? $\endgroup$
    – Lopey Tall
    Jul 15 '20 at 21:41
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    $\begingroup$ Hamiltonian and Lagrangian mechanics are equivalent, in the sense that they provide the same equation of motions. We use Hamiltonian formulations, for it is easier in dealing with evolution of observables. For potentials depending on time, the deal is more or less the same, a constant vanishes under the integral, but the Energy is now not conserved (Noether theorem) $\endgroup$ Jul 15 '20 at 21:55
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You can add $\dot q=\frac{dq}{dt}$ to the Lagrangian without harm, and this has nothing to do with a potential.

Since the EOM are obtained from derivatives of the Lagrangian, adding any constant (to the potential energy or otherwise) will have no effect.

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  • $\begingroup$ Can you expand on adding q dot to the Lagrangian? My question is how that relates to "adding a constant to the potential." And are you letting the generic function (usually denoted $F(q_i,t)$ ) added to the Lagrangian be q? $\endgroup$
    – Lopey Tall
    Jul 15 '20 at 21:43
  • $\begingroup$ Adding any contant to $L$ will not change the equation of motion and this is not same as adding $dF/dt$ for any $F$. Adding a total derivative add a constant to the action, without changing the function which minimizes this action since this function is obtained by a functional derivative of $S$. An analogy: the functions $x^2$and $x^2+2$ both have the min at $x=0$. Moreover $dF/dt$ is eventually connected to generating functions for canonical transformations in the Hamiltonian formalism, and such transformations have no interpretation in terms of adding a constant potential. $\endgroup$ Jul 15 '20 at 22:45

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