I recently noticed there's something that escaped me in Lagrangian mechanics. I recently browsed though the first volume of Landau and L., where it is explained that two systems have identical dynamics if their lagrangians differ only by a total differential to time of a function (because the action will be extremal for the same path in configuration space). But a bit later, they show that translation symmetry in space gives rise to conservation of linear momentum... by assuming that the Lagrangian is invariant under a space translation. Now, this I knew, of course. However, "translation symmetry" should only imply that under an infinitesimal translation, the lagrangian should at most change by a total time differential of a function (but that function can depend on the infinitesimal translation at hand). So, instead of dL = \partial L / \partial x_i dx_i = 0 and using the E-L equation: d/dt ( \partial L / \partial x_i' dx_i ) = 0 from which follows : d/dt p_i = 0 and hence momentum conservation we should write: dL = d F(x_i, d x_i,t) / dt = d/dt F_i(x_i,t) d x_i with F an arbitrary function. Now, this then leads to: d/dt (p_i - F_i(x_i,t) ) = 0 which means: p_i - F_i(x_i,t) is conserved but given that F_i is arbitrary, this doesn't mean anything anymore. Nevertheless, the dynamics of the system is invariant under translation here, in that the solution of the equations of motion will give you the same solution in the slightly translated coordinate system than in the original coordinates system. So why can we require that the Lagrangian itself should be invariant under translations (for instance) ? Shouldn't it simply be the action which needs to be invariant under the symmetry of the system ? What am I missing ? I checked Goldstein, but there too is assumed that the Lagrangian itself is invariant under the symmetry at hand.