Lagrangian method for an LC-Circuit

[RickRazor]

In the paper http://physics.unipune.ernet.in/~phyed/26.2/File5.pdf, the author solves the LC-circuit using Euler-Lagrange equation. She assumes that the Lagrangian function for the circuit is $$L=T-V$$ where
$$T=L\dot q^2/2$$ is the kinetic energy part $$V=q^2 / 2C$$ is the potential energy.She then goes on to say that $$
\frac{d}{dt} \left ( \frac{\partial L}{\partial \dot q} \right ) - \frac{\partial L}{\partial q} = 0
$$
is the equation of motion. I have a couple of questions:1) Is Hamilton's principle applicable here? If yes, why is the kinetic energy $$
\dfrac{{L\dot q^2}}{2}
?$$
2) Inductors also store potential energy temporarily in the magnetic field. Why don't we consider that term?

 

[Orodruin]

Inductors also store potential energy temporarily in the magnetic field. Why don't we consider that term?

Because it is not potential energy in the sense of the Lagrangian method. The dynamic variable is the charge $q$ and the energy stored in the inductor depends on its time derivative $\dot q$. This makes the energy stored in the inductor "kinetic" energy.

 

[vanhees71]

You can use the Hamilton principle for any differential equation (be it an ordinary or partial one), if you find a Lagrangian, whose Euler-Lagrange equation gives this differential equation.

That you don't have electromagnetic fields here explicitly is due to the quasi-stationary approximation for compact circuits made. "Compact" here means that the entire setup is small compared to the typical wavelengths of the charge-current distributions and the em. field. Then you can eliminate the field by introducing the usual characteristics of the elements (resistance for resistors, capacitance for capacitors, mutual and self-inductivities of coils).

Indeed in your case the Euler-Lagrange equations give (I write $\tilde{L}$ for the Lagrangian, because otherwise there's a clash of notation since we also use $L$ for the inductivity):
$$\frac{\mathrm{d}}{\mathrm{d} t} \frac{\partial \tilde{L}}{\partial \dot{q}}=L \ddot{q}=\frac{\partial \tilde{L}}{\partial q}=-\frac{1}{C} q$$,
describing the charge $q$ at the capacitor. Indeed
$$L \ddot{q}=-\frac{q}{C}$$
is the differential equation you also get from Faraday's Law and charge conservation (leading to Kirchhoff's Laws for circuits). So your Lagrangian just gives the right equation, and that's why you can use it to analyze it in terms of analytical mechanics. E.g., since $\tilde{L}$ is not explitly time dependent, the Hamiltonian
$$H=p \dot{q}-\tilde{L}, \quad p = \frac{\partial \tilde{L}}{\partial \dot{q}}=L \dot{q}$$
is conserved: $H=E=\text{const}$.