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Maxwell-Chern-Simons action

  1. Nov 4, 2014 #1
    I have the Chern Simons action, and I've found the equations of motion ##\epsilon^{\mu\nu\rho}F_{\nu\rho}=0##. A problem I was looking at said show that the e.o.m. is "gauge equivalent to the trivial solution". I don't understand what this means. Obviously the e.o.m. is manifestly gauge invariant, so I don't know what else to do.

    Also I'm only beginning to look at gauge symmetries but is this the general idea; if the equations of motion are invariant under ##A_\mu \rightarrow A_\mu + \partial_\mu \Lambda##, where ##\Lambda## is a function of the space-time coordinates, then we're free to impose whatever gauge condition we like, such as Lorenz gauge or Coulomb gauge, without changing the physics of the system?
     
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  3. Nov 5, 2014 #2
  4. Nov 7, 2014 #3

    julian

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    Using the identity (just edited identity - there was a typo)

    [itex]\epsilon_{\mu \nu' \rho'} \epsilon^{\mu \nu \rho} = \delta_{\nu'}^\nu \delta_{\rho'}^\rho - \delta_{\nu'}^\rho \delta_{\rho'}^\nu[/itex]

    the e.o.m, [itex]\epsilon^{\mu \nu \rho} F_{\nu \rho} = 0[/itex], are equivalent to

    [itex]F_{\mu \nu} = 0[/itex]

    where we have used [itex]F_{\mu \nu} = - F_{\nu \mu}[/itex]. So this is the trivial solution - nothing happens, the [itex]F_{\mu \nu}[/itex] are zero. And as you already know the [itex]F_{\mu \nu}[/itex] are invariant under a gauge transformation

    [itex]A_\mu \mapsto A_\mu + \partial_\mu \Lambda[/itex]

    viz

    [itex]F_{\mu \nu} \mapsto \partial_\mu (A_\nu + \partial_\nu \Lambda) - \partial_\nu (A_\mu + \partial_\mu \Lambda) = \partial_\mu A_\nu - \partial_\nu A_\mu = F_{\mu \nu}[/itex].

    I don't know if that is what they want for their question. Usually you just say the solutions are just pure gauge, i.e.

    [itex]A_\mu = \partial_\mu \Lambda[/itex]

    or "flat connections".

    The situation becomes more interesting when:

    (i) when you take space-time to have a non-trivial topology
    (ii) coupling to dynamical matter fields. Here the e.o.m. are [itex]k \epsilon^{\mu \nu \rho} F_{\nu \rho} = J^\mu[/itex] where [itex]J^\mu[/itex] are source fields due to matter - the e.o.m. are equivalent to

    [itex]F_{\mu \nu} = {1 \over k} \epsilon_{\mu \nu \rho} J^\rho[/itex]

    by the above identity.

    (iii) coupling to a Maxwell term (from the title of the thread you already probably know this)


    Gauges: So talking about 3+1 space-time Maxwell's equations now. You know how the [itex]F^{\mu \nu}[/itex] give you the electric and magnetic fields? ([itex]F^{0i} = E^i[/itex] and [itex]F^{ij} = \epsilon^{ijk} B^k[/itex]). These are your observables - what you measure, and are unchanged by a gauge transformations, as demonstrated above. So essentially the physics is the same in all gauges. There are many gauges that you can choose, each perhaps better adapted to the situation.

    First let us ask the mathematical question of whether it is possible for such a gauge condition to be imposed at all. Write [itex]A^\mu = (\phi , \vec{A})[/itex]. Let us look at this question with respect to the Coulomb gauge...The question is given a gauge potential [itex]A_\mu[/itex] is it possible to find a function [itex]\Lambda[/itex] such that the new gauge potential [itex]A_\mu' = A_\mu + \partial_\mu \Lambda[/itex] satisfies [itex]\vec{\nabla} \cdot \vec{A}' = 0[/itex]. To find out we write:

    [itex]\vec{\nabla} \cdot \vec{A}' = \vec{\nabla} \cdot \vec{A} + \nabla^2 \Lambda = 0[/itex]

    which leads to the question, given [itex]\vec{\nabla} \cdot \vec{A}[/itex], can we solve

    [itex]\nabla^2 \Lambda = - (\vec{\nabla} \cdot \vec{A})[/itex]

    for [itex]\Lambda[/itex]. And the answer is yes as this is just Poisson's equation, which has solution

    [itex]\Lambda (\vec{r}) = {1 \over 4 \pi} \int {(\vec{\nabla} \cdot \vec{A}) (\vec{r}_1) \over |\vec{r} - \vec{r}_1|} d
    \vec{r}_1[/itex].

    So we have proved that such a choice is always possible...Now what physical situation is it useful to employ this gauge?

    The Coulomb gauge leads to the Poission equation [itex]\nabla^2 \phi = - {\rho \over \epsilon_0}[/itex], as in electrostatics, and the vector potential, [itex]\vec{A}[/itex], can be shown to satisfy a wave equation with the divergenceless component of [itex]\vec{J}[/itex] as its source

    [itex]\nabla^2 \vec{A} - {1 \over c^2} {\partial^2 \vec{A} \over \partial t^2} = - \mu_0 \vec{J} + {1 \over c^2} \nabla {\partial \phi \over \partial t} = - \mu_0 \vec{J}_t[/itex].

    So it will have a simple form when no sources are present at all. More importantly, this gauge is useful and simple for far field radiation problems. The asymptotic behaviour of [itex]\phi[/itex] is as [itex]{Q \over r}[/itex] (where [itex]Q[/itex] is the total charge), and because its gradient behaves as:

    [itex]- \vec{\nabla} \phi \sim {Q \over r^3} \vec{r}[/itex]

    we can neglect [itex]\phi[/itex] in the exact formula [itex]\vec{E} = - \vec{\nabla} \phi - {\partial \vec{A} \over \partial t}[/itex] when calculating radiation fields and the equations become:

    [itex]\vec{E} \sim - {\partial \vec{A} \over \partial t}[/itex]

    [itex]\vec{B} = \vec{\nabla} \times \vec{A}[/itex]

    for far fields.

    The Lorentz gauge is often considered as it fits in well with considerations of special relativity.
     
    Last edited: Nov 7, 2014
  5. Nov 7, 2014 #4

    julian

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    Let me be a bit pedagogical about Maxwell's equations, gauge potentials and gauge invariant field equations...

    Maxwell's equations are

    [itex]\vec{\nabla} \cdot \vec{E} = \rho[/itex]
    [itex]\vec{\nabla} \times \vec{B} - {\partial \vec{E} \over \partial t} = \vec{J}[/itex]

    where [itex]\rho[/itex] is the charge density and [itex]\vec{J}[/itex] the current density. The last two Maxwell's equations are

    [itex]\vec{\nabla} \times \vec{E} + {\partial \vec{B} \over \partial t} = 0[/itex]
    [itex]\vec{\nabla} \cdot \vec{B} = 0[/itex]

    The last two equations can be solved by writing fields in terms of a scalar potential and a vector potential:

    [itex]\vec{E} = - \vec{\nabla} \phi - {\partial \vec{A} \over \partial t}[/itex]
    [itex]\vec{B} = \vec{\nabla} \times \vec{A}[/itex].

    The potentials uniquely determine the fields, but the fields do not uniquely determine the potentials - we can make the changes:

    [itex]\phi' = \phi + {\partial \Lambda \over \partial t}[/itex]
    [itex]\vec{A}' = \vec{A} - \vec{\nabla} \Lambda[/itex]

    without effecting [itex]\vec{E}[/itex] and [itex]\vec{B}[/itex].

    There is an elegant relativistic notation: the gauge field is

    [itex]A^\mu = (\phi , \vec{A})[/itex]

    and the above mentioned `changes' can be written

    [itex]A_\mu' = A_\mu + \partial_\mu \Lambda[/itex]

    and the field strength tensor is

    [itex]F^{\mu \nu} = \partial^\mu A^\nu - \partial^\nu A^\mu[/itex]

    and in components:

    [itex]F^{0i} = E^i[/itex],
    [itex]F^{ij} = \epsilon^{ijk} B^k[/itex].

    The first two Maxwell's equations can be written as:

    [itex]\partial_\nu F^{\mu \nu} = J^ \mu[/itex]

    where [itex]J^\mu = (\rho , \vec{J})[/itex] is the charge-current density 4-vector. The last two Maxwell's equations can be written as:

    [itex]\epsilon_{\mu \nu \rho \sigma} \partial^\rho F^{\mu \nu} = 0[/itex]

    which are automatically satisfied. THESE are your gauge invariant field equations! However it is often useful to work with the gauge potential instead and then find the electric and magnetic fields by taking derivatives. If we do work with e.o.m. for the gauge potential we must impose gauge constraints in order to eliminate gauge degrees of freedom...
     
    Last edited: Nov 7, 2014
  6. Dec 20, 2014 #5
    Wow, what an answer! Thank's a lot!

    Re: gauge equivalent to trivial. I asked my professor what it meant and he said "We think about equations of motion as equations for ##A_{\mu}##. Gauge equivalent to trivial means - solution for ##A_{\mu}## is gauge equivalent to zero." I'm not entirely sure how to show that.

    Also I added a Maxwell term to the Lagrangian and found the equations of motion ##\partial_\mu F^{\mu\nu} + ke^2\epsilon^{\nu\alpha\beta}F_{\alpha\beta}=0##. I need to show that ##F_{\mu\nu}## satisfies the Klein Kordon equation. I introduced the dual field strength but all I end up with is ##(\partial_\mu \epsilon^{\nu\mu\lambda}+2ke^2g^{\nu\lambda})\tilde{F}_\lambda##. The obvious idea would be to multiply by the thing in brackets with a minus in the middle but that's also not getting me anywhere.
     
  7. Dec 21, 2014 #6

    julian

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    Start with your equation

    ##\epsilon^{\nu\mu\lambda} \partial_\mu \tilde{F}_\lambda+2ke^2g^{\nu\lambda} \tilde{F}_\lambda = 0## Eq.1

    and apply ##\epsilon_{\alpha \beta \nu} \partial^\alpha## to both sides:

    ## \epsilon_{\alpha \beta \nu} \epsilon^{\nu\mu\lambda} \partial^\alpha \partial_\mu \tilde{F}_\lambda + 2ke^2 g^{\nu\lambda} \epsilon_{\alpha \beta \nu} \partial^\alpha \tilde{F}_\lambda = 0 ##

    and then use the identity I quoted in my first post:

    ## (\delta^\mu_\alpha \delta^\lambda_\beta - \delta^\lambda_\alpha \delta^\mu_\beta) \partial_\mu \partial^\alpha \tilde{F}_\lambda + 2ke^2 g^{\nu\lambda} \epsilon_{\alpha \beta \nu} \partial^\alpha \tilde{F}_\lambda = 0 ##

    This becomes

    ##\partial_\alpha \partial^\alpha \tilde{F}_\beta - \partial_\beta \partial^\alpha \tilde{F}_\alpha+ (2ke^2) g_{\beta \rho} [- \epsilon^{\rho \alpha \lambda} \partial_\alpha \tilde{F}_\lambda] = 0 ## Eq.2

    where we have also done a simple rewriting of the third term. The second term in Eq.2 vanishes because ## \partial^\alpha \tilde{F}_\alpha = 0## as can be seen from:

    ##\partial_\mu \tilde{F}^\mu = {1 \over 2} \epsilon^{\mu \nu \rho} \partial_\mu F_{\nu \rho} = {1 \over 2} \epsilon^{\mu \nu \rho} \partial_\mu (\partial_\nu A_\rho - \partial_\rho A_\nu) = 0 ##.

    Also using Eq.1 in the third term of Eq.2, Eq.2 then reads:

    ## \partial_\alpha \partial^\alpha \tilde{F}_\beta + (2ke^2) g_{\beta \rho} [(2ke^2) g^{\rho \lambda} \tilde{F}_\lambda] = 0 ##

    which easily simplifies to

    ## \big[ \partial_\alpha \partial^\alpha + (2ke^2)^2 \big] \tilde{F}_\beta = 0 ##.
     
    Last edited: Dec 21, 2014
  8. Dec 21, 2014 #7

    julian

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    And to show that ##F^{\mu \nu}## satisfies the Klein-Gordon equation write the last equation of the previous post as:

    ##{1 \over 2} \epsilon_{\beta \rho \sigma} \big[ \partial_\alpha \partial^\alpha + (2ke^2)^2 \big] F^{\rho \sigma} = 0##

    contract with ##\epsilon^{\beta \mu \nu}##, use that identity again and that ##F^{\mu \nu} = - F^{\nu \mu}##.
     
  9. Dec 21, 2014 #8
    Fantastic, thank you very much :D
     
  10. Dec 21, 2014 #9
    I'm now trying to derive the Hamiltonian. The paper I'm reading says I start by rewriting the Lagrangian as
    ##\mathcal{L}=\frac{1}{2e^2}E_i^2-\frac{1}{2e^2}B^2 +
    k\epsilon^{ij}\dot{A}_iA_j + 2kA_0B##

    I have most of this, but instead of ##2kA_0B## I have
    ##kA_0B + k\epsilon^{ij}A_i\partial_j A_0##

    I'm guessing the two terms are equal since that would give me what I need but I don't know how to show it.

    Edit:
     
    Last edited: Dec 21, 2014
  11. Dec 21, 2014 #10
    I rewrote the last term as
    ##\partial_j(\epsilon^{ij}A_iA_0)-\epsilon^{ij}\partial_jA_iA_0##

    Relabling indices in the second term here and swapping on the epsilon gives me what I want, but I still have an extra term of ##-\epsilon^{ij}\partial_i(A_jA_0)##

    Since it's a total derivative and hence wont change the equations of motion I'm thinking they might have dropped it? I can't see any other reason for it disappearing.
     
    Last edited: Dec 21, 2014
  12. Dec 22, 2014 #11

    julian

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    When you get a term like this you convert it into surface term. Generally speaking with action principles, such terms may vanish by boundary conditions, but if they don't then the action is not extremum. In order to cancel out the surface integral, and so obtain an action that is stationary for solutions of the e.o.m. in the bulk under all variations (and not just those variations that vanish for the surface term), you have to add a boundary term to the action principle.

    How surface terms are treated in Chern-Simons theory is subtle and can lead to important boundary effects. I think from a previous thread you know that the gauge-invariance of Chern-Simons theory is only guaranteed if a boundary term vanishes. This has implications for current (non-)conservation on the boundary. I don't know enough about it to say much more right now.

    What paper are you looking at?
     
    Last edited: Dec 22, 2014
  13. Dec 22, 2014 #12
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