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Thread: Calculus of Variations - Noether's Theorem for fields

  1. #1
    May 2008
    Melbourne Australia

    Calculus of Variations - Noether's Theorem for fields

    I am working on the attached problem from the book "Emmy Noether's Wonderful Theorem". I believe I can answer parts a,b,c and d, but I would like help with part e.

    Now if my answer's to parts a-d are not correct then that might explain why I can't do part e. So I will put some of my working below.

    Part a)

    A Lagrangian density function that gives the Poisson equation is:

    $\displaystyle \mathcal{L} = \frac 12 \phi_i \phi^i-\phi \rho$ for i = 1,2,3

    Part b)

    $\displaystyle p^\mu=\frac{\partial \mathcal{L}}{\partial \phi_\mu}$


    $\displaystyle p^0=0$ and $\displaystyle p^i = \phi^i = \phi_i$

    Part C)

    $\displaystyle \mathcal{H}_\rho ^{.v}=\phi_\rho p^v-\delta _\rho ^v \mathcal{L}$


    $\displaystyle \mathcal{H}_i ^{.0}= 0$, $\displaystyle \mathcal{H}_0 ^0= \mathcal{L}$ and

    $\displaystyle \mathcal{H}_j ^k=\phi_j p^k-\delta _j ^k \mathcal{L}=E_jE_k-\delta_k^jk(\frac 12 \phi_s \phi^s-\phi \rho)=E_jE_k-\delta_k^j(\frac 12 E_s E^s-\phi \rho) $

    $\displaystyle \mathcal{H}_j ^k=E_jE_k-\delta_k^j\eta +\delta_k^j \rho \phi = T_{jk}+\delta_{jk} \rho \phi$

    Notice that this is slightly different to what I was asked to show. I think I am right and the text has a typo. Partly because my version looks more like an energy term.

    Part D)
    The generators are [0 y -x 0]'. But I choose to make the problem a little more general. I choose to make the rotation about an axis $\displaystyle \hat n $, then the rotation around z is a special case and the generator becomes $\displaystyle \tau^\rho = [0, \hat n \times \underline r]'$

    Now the Rund Trautman identity is:
    $\displaystyle \bigtriangledown \cdot F=\frac{\partial \mathcal{L}}{\partial \phi} \zeta + \frac{\partial \mathcal{L}}{\partial \phi_\rho} (\partial_\rho \zeta)+(\partial_\rho\mathcal{L})\tau^\rho-\mathcal{H}_\rho^v(\partial_v\tau^\rho)$

    Taking care of the zero generators this simplifies to:

    $\displaystyle \bigtriangledown \cdot F=(\partial_j\mathcal{L})\tau^j-\mathcal{H}_j^i(\partial_i\tau^j)$

    Now $\displaystyle \partial_j\tau^j=0$ unless rho=1 and v=2 or rho = 2 and v=1 and tau is equal to 1 and -1 respectively. So the Hamiltonian term disappears and Rund Trautman becomes

    $\displaystyle \bigtriangledown \cdot F=(\partial_j\mathcal{L})\tau^j=-\bigtriangledown(\rho \phi)\cdot (\hat n \times \underline r)=-\bigtriangledown(\rho \phi)\cdot (\hat n \times \underline r)-\rho\phi \bigtriangledown \cdot (\hat n \times \underline r)$ because derivatives of a constant vector are zero


    $\displaystyle \bigtriangledown \cdot F=-\bigtriangledown \cdot [\phi \rho(\hat n \times \underline r)]$


    $\displaystyle F=-\phi \rho(\hat n \times \underline r)$

    Part E)
    Noether's Theorem with (zeta = 0) simplifies to:

    $\displaystyle j^j=-\mathcal{H}_i^{.j}\tau^i-F^j=-(E_jE_i-\delta_{ij}\eta+\delta_{ij}\rho\phi)(\hat n \times \underline r)_i-F^j$ and $\displaystyle \partial_\rho j^\rho=0$


    $\displaystyle \underline j=-[E(E\cdot\hta n \times \underline r)] + [\eta(\hat n \times \underline r)] -[\rho\phi(\hat n \times \underline r)]-F$

    Cancelling the last two terms:
    $\displaystyle \bigtriangledown \cdot \underline j=-\bigtriangledown\cdot[E(E\cdot\hta n \times \underline r)] + \bigtriangledown\cdot[\eta(\hat n \times \underline r)]$

    From here I have a few ideas but have not been able to make anything work. My ideas include:
    1. Take the divergence of j then get all of the terms into a form of a vector dotproduct with n. The n can then be dropped as it is an arbitrary direction.
    2. Integrate the divergence of j over a sphere with a radius that approaches infinity this integral will be a constant or zero. Taking the time derivative of this integral will get rid of any constant, then I can use the divergence theorem to write:

    $\displaystyle \int _S \frac d{dt}(j.\hat r) dA$=0

    But I haven't been able to make either of these work.
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