1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Lagrangian of a circuit

  1. Apr 16, 2008 #1


    User Avatar

    Hello folks,

    I could really do with a few hints with this. (As soon as possible!)

    1. The problem statement, all variables and given/known data

    For a repeated line of cells, two 'equations of motion' can be written:

    [tex]C_{n}\dot{U_{n+1}} = I_{n}-I_{n+1}[/tex]
    [tex]L_{n}\dot{I_{n}} = U_{n} - U_{n+1}[/tex]

    where [tex]C_{n}[/tex] is the capacitance, [tex]U_{n+1}[/tex] the voltage after the nth cell, [tex]I_{n}-I_{n+1}[/tex] the charging current, [tex]L_{n}[/tex] the inductance.

    Work out the Lagrangian that generated these equations.

    You should find that

    [tex]E = \sum_{n}\left( (1/2)\L_{n}\dot{Q_{n}}^{2} + (1/2)C_{n}U_{n+1}^{2} \right)[/tex]


    [tex]H = (1/2)\sum_{n}\left( \frac{P_{n}^{2}}{L_n}} + \frac{(Q_{n+1}-Q_{n})^{2}}{C_{n}} \right)[/tex]

    2. Relevant equations

    Note that
    [tex]Q_{n} = -\int I_{n} dt => C_{n}U_{n+1} = Q_{n+1} - Q_{n}[/tex]

    3. The attempt at a solution

    I have suggested a Lagrangian of

    [tex]L = (1/2)L_{n}\dot{Q_{n}}^{2} + Q_{n}(U_{n+1}-U_{n}) + (1/2)C_{n}\dot{U_{n+1}^{2}} + U_{n+1}(I_{n}-I_{n+1})[/tex]

    (which can be turned into a sum for all the repeated circuits)

    but it doesn't seem convincing, despite the fact that you can recover the original 'equations of motion' using Euler-Lagrange equations (using Q and U).

    I'm not sure how the energy is being derived from the Lag. in this case (in mechanics, it was always the case that L = T - U, hence E = T + U), and it seems clear that you aren't going to get either that expression for the energy or that Hamiltonian from my guess-work Lagrangian.

    Any suggestions?

    Many thanks!
  2. jcsd
  3. Apr 16, 2008 #2
    I don't think Ulf would like this any more than Chris.
  4. Apr 16, 2008 #3


    User Avatar
    Science Advisor
    Homework Helper

  5. Apr 17, 2008 #4


    User Avatar

    I think Anony. is under the impression that, being part of a set Q, this isn't up for discussion. According to the tutor I asked, we *can* discuss these Qs with other Physicists, exchange ideas, argue, etc. What we're not allowed to do is just copy someone's answer (which PhysicsForums also prohibits).
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Lagrangian of a circuit
  1. Problem on Lagrangian (Replies: 5)

  2. QFT lagrangian (Replies: 1)