Use Hamilton's principle to show that if \(F\) is any function of the generalized co-ordinates, then the Lagrangian functions \(L\) and \(L+\mathrm{d} F / \mathrm{d} t\) must yield the same equations of motion. Hence show that the equations of motion of a charged particle in an electromagnetic field are unaffected by the 'gauge transformation' (A.42). (Hint: Take \(F=-q \Lambda\).)

Hamilton's Principle states that the integral of the Lagrangian L, over the time interval from \(t_1\) to \(t_2\), is stationary (minimum or no change at all) for the true path followed by the system: \(\delta\int_{t_1}^{t_2} L(q, \dot{q}, t)\,dt = 0\) Here, \(q(t)\) are the generalized coordinates, and \(\delta\) is the variation in the path followed by the system.

Let's consider two Lagrangian functions \(L\) and \(L'\) where \(L'=L+\frac{dF}{dt}\), where F is a function of the generalized coordinates q.

As a next step, let's evaluate the variation of the integral of the new Lagrangian over the same time interval from \(t_1\) to \(t_2\). Since we're interested in checking if the change in the Lagrangians affects the equations of motion, let's compute \(\delta\int_{t_1}^{t_2} L'(q, \dot{q}, t)\,dt\): \(\delta\int_{t_1}^{t_2} L'(q, \dot{q}, t)\,dt = \delta\int_{t_1}^{t_2} (L + \frac{dF}{dt})(q, \dot{q}, t)\,dt\) Using the linearity property of the \(\delta\) operator, we can rewrite this expression as: \(\delta\int_{t_1}^{t_2} L'(q, \dot{q}, t)\,dt = \delta\int_{t_1}^{t_2} L(q, \dot{q}, t)\,dt + \delta\int_{t_1}^{t_2} \frac{dF}{dt}\,dt\)

Let's focus on the second term: \(\delta\int_{t_1}^{t_2} \frac{dF}{dt}\,dt = \int_{t_1}^{t_2} \frac{d}{dt}(\delta F)\,dt\) Here, we've taken the total time derivative out of the variation because the derivative and variation operators commute. Also note that since F is a function of q only, its variation is \(\delta F(q,t)\). Now apply the fundamental theorem of calculus: \(\int_{t_1}^{t_2} \frac{d}{dt}(\delta F)\,dt = \delta F(t_2) - \delta F(t_1)\) Since the choice of the path has no effect on the initial and final states (i.e., the variation of F is zero at the boundary points \(t_1\) and \(t_2\)), both terms on the right-hand side are zero, and hence: \(\delta\int_{t_1}^{t_2} \frac{dF}{dt}\,dt = 0\)

Now we return to the expression for the variation of the integral of the new Lagrangian. From Step 4, we have shown that the second term vanishes, which implies that: \(\delta\int_{t_1}^{t_2} L'(q, \dot{q}, t)\,dt = \delta\int_{t_1}^{t_2} L(q, \dot{q}, t)\,dt\) This means that both Lagrangians L and L' give rise to the same equations of motion, since the stationary condition (Hamilton's Principle) remains unchanged.

Now we are asked to show that the equations of motion of a charged particle in an electromagnetic field are unaffected by the gauge transformation (A.42). The gauge transformation is given by: \(A' = A - \nabla\Lambda\) Since we're considering the case where \(F = -q\Lambda\), we can write the total time derivative of F as: \(\frac{dF}{dt} = -q\frac{d\Lambda}{dt}\) We already proved that the Lagrangian functions L and L' (where \(L'=L+\frac{dF}{dt}\)) must yield the same equations of motion. Therefore, under the gauge transformation (A.42), the equations of motion of a charged particle in an electromagnetic field remain unchanged. By following the steps above, we have demonstrated that two Lagrangian functions, differing by a total time derivative of a function F dependent on generalized coordinates, yield the same equations of motion. We have also shown that the equations of motion of a charged particle in an electromagnetic field are unaffected by the gauge transformation.