Question

1) In class, we showed that the Lagrangian was not unique: we could add xx to L and still get Newton's equations. Note that xx =-(2) (a) Let L be a Lagrangian for a system with n degrees of freedom satisfying Lagrange's equations. Show that we can generally add the total derivative of any function of coordinates and time to L: , which is a special case of the following: (q1, q2, ...,qn, t), and that Lagrange's equations will be unchanged dF (b) Show that we can add ff to L in the expression for the action S, and that the action remains stationary, implying Lagrange's equations for either L or L dt

Answer 1

For L,

$\frac{\partial L}{\partial q}-\frac{\mathrm{d} }{\mathrm{d} t}\left (\frac{\partial L}{\partial \dot q} \right )=0$

For L'

$\frac{\partial L'}{\partial q}-\frac{\mathrm{d} }{\mathrm{d} t}\left (\frac{\partial L'}{\partial \dot q} \right )=\frac{\partial L}{\partial q}+\frac{\partial \dot{F}}{\partial q}-\frac{\mathrm{d} }{\mathrm{d} t}\left (\frac{\partial L}{\partial \dot q} +\frac{\partial \dot{F}}{\partial \dot{q}}\right )$

$\frac{\partial L'}{\partial q}-\frac{\mathrm{d} }{\mathrm{d} t}\left (\frac{\partial L'}{\partial \dot q} \right )=\frac{\partial \dot{F}}{\partial q}-\frac{\mathrm{d} }{\mathrm{d} t}\left (\frac{\partial \dot{F}}{\partial \dot{q}}\right )$

Now note that

$\dot{F}=\frac{\partial F}{\partial t}+\frac{\partial F}{\partial q}\dot{q}$

so that

$\frac{\partial \dot{F}}{\partial q}=\frac{\partial^2 F}{\partial q\partial t}+\frac{\partial^2 F}{\partial q^2}\dot{q}$

and

$\frac{\mathrm{d} }{\mathrm{d} t}\left (\frac{\partial \dot{F}}{\partial \dot{q}} \right )=\frac{\mathrm{d} }{\mathrm{d} t}\left (\frac{\partial F}{\partial q} \right )=\frac{\partial^2 F}{\partial q\partial t}+\frac{\partial^2 F}{\partial q^2}\dot{q}$

Finally,

$\frac{\partial L'}{\partial q}-\frac{\mathrm{d} }{\mathrm{d} t}\left (\frac{\partial L'}{\partial \dot q} \right )=0$

(b)

$S'=\int_{t_1}^{t_2} L' dt=\int_{t_1}^{t_2} \left (L+\frac{\mathrm{d} F}{\mathrm{d} t} \right ) dt=S+F(t_2)-F(t_1)$

$\delta S'=\delta S$