Problem 14 - Classical Mechanics

Question

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 \(F=-q \Lambda\).)

Step-by-Step Solution

Answer

Yes, two Lagrangian functions that differ by a total time derivative of a function F dependent on generalized coordinates yield the same equations of motion.

Step 1: Hamilton's Principle

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.

Step 2: New Lagrangian function

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.

Step 3: Calculate the variation for the new Lagrangian

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: \(\delta\int_{t_1}^{t_2} L'(q, \dot{q}, t)\,dt = \delta\int_{t_1}^{t_2} L(q, \dot{q}, t)\,dt + \delta\int_{t_1}^{t_2} \frac{dF}{dt}\,dt\)

Step 4: Examine the second term

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: \(\delta\int_{t_1}^{t_2} \frac{dF}{dt}\,dt = 0\)

Step 5: Show that both Lagrangians give the same equations of motion

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.

Step 6: Gauge transformation for a charged particle

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.