 # Lagrange formalism electrodynamics

Hello everybody,

it would be very nice if someone could help me here, please.

I am investigating the laplace equation.

\begin{equation} \begin{split} Laplace \phi &= 0 \text{ in Q}\\\ \phi &= 0 \text{ at } \Gamma_1\\\ \gamma \frac{\partial \phi}{\partial n} &= 0 \text{ at } \Gamma_2 \text{ and } \Gamma_3\\\ \gamma \frac{\partial \phi}{\partial n} &= - \frac{1}{|\Gamma_4|}*\dot{Q_2}(t)*\int \phi dx \text{ at } \Gamma_4. \end{split} \end{equation}

after a few calculations with the electrical potential $$\phi$$ as the test function, this results in the weak formulation

\begin{equation} \int_{Q} \gamma |\nabla \phi|^2dx - \frac{1}{|\Gamma_4|}\dot{Q}_2(t)\int_{\Gamma_4}\phi(x)dx = 0 \text{ for all } v \in V. \end{equation}

I HAVE to use the Newmark-beta-method. But this method needs the second order.

Also if i am using the classical approach function

\phi \approx \tilde{\phi} = \sum_{I=1}^{N} c_i(x) \phi_i(x)\\

\nabla \phi \approx \tilde{\nabla\phi} = \sum_{I=1}^{N} c_i(x) \nabla \phi_i(x)

Then i get

\gamma[\sum_{I=1}^{N} c_i(x) \int \nabla \phi_i \nabla \phi_j dx] = \frac{1}{|\Gamma_4|} I_2(t) \int_{\Gamma_4} \phi_j (x).

So one can see that this still doesn’t fit and i think it is the wrong way.

Ok. Then i tried the Lagrange Formalism from mechanics applied to electrodynamics with the lagragian

L(x(t), \dot{x}(t), t) = \underbrace{\frac{m}{2}|\dot{x}|^2}_{\text{kinetic energy}} - \underbrace{e\phi(x,t) + e\dot{x} * A(x,t)}_{\text{potential energy}}.

Then doing the Lagrange Formalism i get

m|\ddot{x}| = eE(x,t) + \underbrace{e\dot{x} \frac{\partial A(x,t)}{\partial x} - \frac{1}{e} \frac{\partial}{\partial x} A(x,t) \dot{x}}_{e(\dot{x} \times \underbrace{\nabla \times A(x,t))}_{B(x,t)}}.

Please excuse the lack of vector arrows. I hope it’s clear anyway.

So great. This would be nice, but i really don’t know how to handle A and if this approach is ok or not.

Thank you and kindly greetings

In a couple of examples i have seen

A_z = TrialFunction(V)


for the magnetical potential. But in my case the TrialFunction is already the electrical potential. Hello again.

Ok, i could use a mixed function space and apply
(phi, A_z) = TrialFunction(V).