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Topological sigma model, Euler Lagrange equations

  1. May 29, 2012 #1
    1. The problem statement, all variables and given/known data
    My question refers to the paper "Topological Sigma Models" by Edward Witten, which is available on the web after a quick google search. I am not allowed to include links in my posts, yet. I want to know how to get from equation (2.14) to (2.15).
    We consider a theory of maps from a Riemann surface [itex]\Sigma[/itex] with complex structure [itex]\varepsilon[/itex] to a Riemannian manifold [itex]M[/itex] with an almost complex structure [itex]J[/itex]. [itex]h[/itex] is the metric on [itex]\Sigma[/itex], [itex]g[/itex] the metric on [itex]M[/itex].
    The map [itex]\phi:\Sigma \to M[/itex] is locally described by functions [itex]u^i(\sigma)[/itex]. [itex]H^{\alpha i}[/itex] is a commuting field ([itex]\alpha = 1,2[/itex] is the tangent index to [itex]\Sigma[/itex] and [itex]i=1,\ldots,n[/itex] runs over a basis of [itex]\phi^*(T)[/itex], which is the pullback of the tangent bundle of [itex]T[/itex] to [itex]M[/itex]). [itex]H^{\alpha i}[/itex] obeys [itex]H^{\alpha i}=\varepsilon^\alpha{}_\beta {J^i}{}_j H^{\beta j}[/itex].
    We consider the Lagrangian [itex]\mathcal{L}=\int d^2\sigma(-\frac{1}{4}H^{\alpha i}H_{\alpha i} + H^\alpha_i \partial_\alpha u^i + (\text{terms independent of H}))[/itex].
    [itex]H[/itex] is a non-propagating field, since the Lagrangian does not depend on its derivative.
    Using the Euler-Lagrange equations I want to show: [itex]H^i_\alpha=\partial_\alpha u^i \epsilon_{\alpha\beta}J^i{}_j\partial^\beta u^j[/itex]

    2. Relevant equations
    [itex]\mathcal{L}=\int d^2\sigma(-\frac{1}{4}H^{\alpha i}H_{\alpha i} + H^\alpha_i \partial_\alpha u^i + (\text{terms independent of H}))[/itex]
    [itex]H^{\alpha i}=\varepsilon^\alpha{}_\beta {J^i}{}_j H^{\beta j}[/itex]
    [itex]\frac{\partial \mathcal{L}}{\partial H^\alpha_i(\sigma)}=0[/itex]

    3. The attempt at a solution
    Since [itex]\mathcal{L}[/itex] does not depend on the derivative of [itex]H[/itex], the Euler Lagrange equations simply state [itex]\frac{\partial \mathcal{L}}{\partial H^\alpha_i(\sigma)}=0[/itex]. I tried to evaluate this:
    [itex]\frac{\partial}{\partial H^\alpha_i(\sigma)}\left(\int d^2s(-\frac{1}{4}H^{\beta j}(s)H_{\beta j}(s) + H^\beta_j(s) \partial_\beta u^j(s))\right)[/itex]
    [itex]=\frac{\partial}{\partial H^\alpha_i(\sigma)}\left(\int d^2s(-\frac{1}{4}h_{\beta\gamma}g^{jk}H^{\beta}_k(s) H^{\gamma}_j(s)+ \epsilon^\beta{}_\gamma J_j{}^k H^\gamma_k(s) \partial_\beta u^j(s))\right)[/itex] where the second equation of the realtions above is used
    [itex]=-\frac{1}{4}h_{\beta\gamma}g^{jk}(h^{\beta\alpha} g_{ik} H^{\gamma}_{j}(\sigma) + H^{\beta}_{k}(\sigma) h^{\alpha\gamma} g_{jk}) + \varepsilon^\beta{}_\gamma J_j{}^k h^{\alpha\gamma} g_{ik} \partial_\beta u^{j}(\sigma)[/itex]
    [itex]=-\frac{1}{2}H^\alpha_i+\varepsilon^{\beta\alpha}J_{ji}\partial_\beta u^j(\sigma)[/itex]

    Unfortunately, that is not really close to the expression that I am looking for. Can someone find mistakes? I appreciate any help.

    Best regards, physicus
     
    Last edited: May 29, 2012
  2. jcsd
  3. May 29, 2012 #2

    fzero

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    Use the constraint to write

    $$-\frac{1}{4} H^{\alpha i}H_{\alpha i} + H_{\alpha i} \partial^\alpha u^i
    =- \frac{1}{4} H^{\alpha i}H_{\alpha i} + \frac{1}{2} H_{\alpha i} \partial^\alpha u^i
    - \frac{1}{2} \epsilon_{\beta\alpha} {J^i}_jH_{\alpha i} \partial^\beta u^j .$$

    Note that the sign of the 3rd term changes when you transpose the indices on ##\epsilon_{\beta\alpha}## after computing the variation.
     
  4. May 29, 2012 #3
    Thanks a lot, that was very helpful. I got it now.
     
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