Does "adding a constant to the potential energy" imply "adding a total time derivative to the Lagrangian." 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.
 A: 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.
A: 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.
