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Invariance of EM Lagrangian

  1. Apr 11, 2016 #1

    ShayanJ

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

    Show that in order for the free Lagrangian to be invariant when ## A^\mu ## is transformed by a transformation U, it has to transform as below:
    ## A'^{\mu}=\frac i g (\partial^\mu U) U^{-1}+U A^\mu U^{-1} ##

    2. Relevant equations

    The wording of the problem is a bit awkward I think, but it seems it means the free EM Lagrangian!

    ## L_{free \ EM}=-\frac 1 4 F_{\mu \nu} F^{\mu \nu} ##
    ## F_{\mu \nu}=\partial_\mu A_\nu - \partial_\nu A_\mu ##

    3. The attempt at a solution

    I've never seen it like this before. It was always figuring out the transformation rule for ## A^\mu ## somehow that it leaves another field's Lagrangian invariant!
    The only method that comes into my mind is substituting ## A'^\mu ## in ## F'_{\mu \nu} F'^{\mu \nu} ## and expanding it. But it doesn't show any sign of convinient cancellations early enough and it seems I have to expand it all the way. But it would be a giant with 100 terms in it! So I think there has to be a better way for doing this. But I just can't figure it out!

    Any small hint is appreciated.
    Thanks
     
  2. jcsd
  3. Apr 11, 2016 #2
    I think you are dealing with scalar O(n) gauge theory. The Lagrangian density can be expressed as L=1/2(∂μΦ)TμΦ- 1/2m2ΦTΦ where Φ is a vector of fields. Now, the Lagrangian is invariant under the transformation Φ' |→ UΦ whenever U is a constant matrix belonging to the nxn orthogonal group O(n). Also, (∂μΦ) |→ (∂μΦ)' = U(∂μΦ) and this expresses global symmetry. However, if you require local invariance the U will not pass through the derivative when U=U(x); obviously ∂μ(UΦ) ≠ U(∂μΦ). I suggest that you introduce the covariant derivative Dμ = ∂μ + igAμ, where g is a coupling constant and A(x) is the gauge field. The required result should immediately follow after a simple calculation.
     
    Last edited: Apr 11, 2016
  4. Apr 11, 2016 #3

    ShayanJ

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    Thanks for the comment but I know that and the problem seems to require that only the EM Lagrangian gets involved.
     
  5. Apr 11, 2016 #4
    I think that "free Lagrangian" in the problem statement refers to no interactions between the scalar fields. I think you are making a mistake in identifying the vector potential of classical EM with the gauge field of QFT.
     
  6. Apr 11, 2016 #5

    ShayanJ

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    The vector potential of classical EM is the gauge field of QFT if the gauge group is U(1)!
    And without mentioning any specific Lagrangian, it could be a complex scalar field or a spinor field!
    And actually it was a problem on a test. I asked the professor which Lagrangian he meant, is it the EM Lagrangian? He said its obvious! He's really bad at teaching!:frown:
     
  7. Apr 12, 2016 #6

    CAF123

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    Hi Shyan,
    I am still learning this as well so take what I say with a pinch of salt :) From the structure of the transformation law for ##A_{\mu}##, I would guess the lagrangian is pure Yang-Mills, i.e $$\mathcal L = -\frac{1}{4} F_{\mu \nu}^a F^{\mu \nu, a}$$ with gauge group SU(N). So the ##F_{\mu \nu}## has an additional piece in its structure.
    Possibly?
     
  8. Apr 12, 2016 #7

    ShayanJ

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    Thanks for the comment but no. If you set ## U=e^{i\alpha(x)} ## in the transformation law given in the first post, you see its exactly a U(1) gauge transformation.

    I'm starting to think that the professor just confused me with the wording of the problem and his unclear answer to my question and I simply should have used a complex scalar field or a Dirac field.

    It would be good if I had some confirmation from science advisors whether people actually check the transformation rule given in the first post using only the free EM Lagrangian or not!(@vanhees71 , @strangerep ,@samalkhaiat ,@Demystifier , and any other advisors...you have no comment?)
     
  9. Apr 13, 2016 #8

    samalkhaiat

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    Shyan, I rarely visit the section of the forums. So, please notify me by PM in the future.
    Okay, I think your instructor has messed the problem up very badly. As it stands, the wording of the problem makes no sense at all. I believe that he had the following problem in mined, but failed to formulate it properly.
    The problem: Let [itex]\mathcal{L}(\Phi , \partial \Phi)[/itex] be a free generic matter field theory. Assume that [itex]\mathcal{L}(\Phi , \partial \Phi) = \mathcal{L}(\Phi^{'} , \partial \Phi^{'})[/itex] is invariant under the following (global) transformations [tex]\Phi^{'} = U \Phi , \ \ \ \partial_{\mu}\Phi^{'} = U \partial_{\mu}\Phi ,[/tex] where [itex]U[/itex] is a constant (matrix).
    Show that demanding invariance under local transformation (achieved by letting [itex]U \to U(x)[/itex]) forces you to introduce a (matrix-valued) vector field [itex]A_{\mu}[/itex], which couples to the matter field [itex]\Phi[/itex] and transforms according to [tex]A^{'}_{\mu} = i (\partial_{\mu}U) U^{-1} + U A_{\mu} U^{-1} . \ \ \ \ \ \ (1)[/tex]
    Can you do this problem?
    Remark: (i) If [itex]A_{\mu}[/itex] is a number (not matrix), i.e., if the gauge group is [itex]U(1)[/itex], then the second term in the transformation law is [tex]U A_{\mu} U^{-1} = A_{\mu} U U^{-1} = A_{\mu} .[/tex] In this case, you can easily show that the field tensor [itex]F_{\mu \nu} = \partial_{\mu}A_{\nu} - \partial_{\nu}A_{\mu}[/itex] is invariant under the [itex]U(1)[/itex] gauge transformation [tex]A^{'}_{\mu} = i (\partial_{\mu}U) U^{-1} + A_{\mu} . \ \ \ \ \ \ (2)[/tex] This means that you need to add the gauge invariant Maxwell’s Lagrangian [itex]- \frac{1}{4}F^{2}[/itex], to the gauge invariant matter field Lagrangian that you obtained by introducing the gauge field [itex]A_{\mu}[/itex]. This leads you to the QED Lagrangian: [tex]\mathcal{L}_{QED} = \mathcal{L}( \Phi , D_{\mu}\Phi ) - \frac{1}{4} F_{\mu \nu} F^{\mu \nu} .[/tex]
    (ii) If [itex]A_{\mu}[/itex] is a matrix-valued vector field, i.e., if [itex]U(x)[/itex] belongs to a non-abelian gauge group, then the matrix-valued field tensor is given by
    [tex][D_{\mu} , D_{\nu}] \Phi \equiv i F_{\mu \nu}\Phi , \ \ \ \ \ (3)[/tex]
    [tex]F_{\mu \nu} = \partial_{\mu}A_{\nu} - \partial_{\nu}A_{\mu} + [A_{\mu} , A_{\nu}].[/tex]
    In this case, you can show (using 3) that the field tensor transforms according to
    [tex]F^{'}_{\mu \nu} = U F_{\mu \nu} U^{-1}.[/tex]
    Thus, to construct a gauge-invariant Lagrangian for the gauge fields you need to trace over the group index: [tex]\mathcal{L}_{YM} = - \frac{1}{4} \mbox{Tr} ( F_{\mu \nu} F^{\mu \nu}) .[/tex] In this case, the resulting gauge-invariant interacting theory is described by the following total Lagrangian
    [tex]\hat{\mathcal{L}} = \mathcal{L}( \Phi , D_{\mu}\Phi ) - \frac{1}{4} \mbox{Tr} ( F_{\mu \nu} F^{\mu \nu}) .[/tex]
     
  10. Apr 13, 2016 #9

    ShayanJ

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    Thanks Sam, that's very kind of you.
    I think...but I'm not getting the right result!

    At first I note that for the local transformation, we'll have
    ## \mathcal L(U\Phi,\partial (U \Phi))=\mathcal L(U\Phi,(\partial U)\Phi+U\partial \Phi) ##,
    this can't be equal to ## \mathcal L(\Phi,\partial \Phi) ## for a general U.

    In order to achieve invariance, I consider the Lagrangian
    ## \mathcal L(\Phi,\partial \Phi+A \Phi)=L(\Phi,\partial \Phi)+\frac{\partial \mathcal L}{\partial(\partial_\mu \Phi)} A_\mu \Phi ##
    instead. (Actually expanding the Lagrangian and retaining only the first term here doesn't seem quite right to me because it assumes that ## A_\mu ## are small!)

    Then I apply a local transformation U to the fields to get:
    ## \mathcal L(U\Phi,(\partial U)\Phi+U \partial \Phi+A' U \Phi)=\mathcal L(U\Phi,U\partial \Phi)+\frac{\partial \mathcal L}{\partial(\partial_\mu \Phi)} \left[ (\partial_\mu U)+A'_\mu U \right] \Phi ##
    So for the new Lagrangian to be invariant under a local transformation U, we should have:
    ## A_\mu=(\partial_\mu U)+A'_\mu U ##
    But this is not correct! I can't find out what went wrong!
     
  11. Apr 14, 2016 #10

    samalkhaiat

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    No, you can not expand the Lagrangian. Expansion requires infinitesimal transformation, but you are dealing with finite transformation [itex]U(x)[/itex].
    Note that, by hypothesis, [itex]\mathcal{L}(\Phi , \partial \Phi)[/itex] is invariant under global (constant [itex]U[/itex]) transformation, because both arguments of [itex]\mathcal{L}[/itex] transform in the same way (i.e., covariantly): [itex]\Phi \to U\Phi[/itex] and [itex]\partial \Phi \to U \partial \Phi[/itex]. Now, keep this note in mined because we will need it bellow.
    When we make the transformation local by letting [itex]U \to U(x)[/itex], we find [itex]\Phi \to U(x) \Phi[/itex], but covariance is lost because [tex]\partial_{\mu}\Phi^{'} = U \partial_{\mu}\Phi + (\partial_{\mu} U) \Phi . \ \ \ \ \ (1)[/tex] This means that the original Lagrangian is no longer invariant under the local [itex]U(x)[/itex] transformation. So, to achieve invariance, our Lagrangian must be in the form [itex]\mathcal{L}(\Phi , \Psi_{\mu})[/itex], with both arguments [itex]\Phi[/itex] and [itex]\Psi_{\mu}[/itex] having the same transformation law under [itex]U(x)[/itex]. Thus, we only need to find the object [itex]\Psi_{\mu}[/itex] which transforms according to [tex]\Psi^{'}_{\mu} = U(x) \Psi_{\mu} . \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)[/tex]
    Once an expression for [itex]\Psi_{\mu}[/itex] is found, equation (2) will, therefore, insure the invariance of [itex]\mathcal{L}(\Phi , \Psi_{\mu})[/itex] under the local transformation [itex]U(x)[/itex].
    Now, (i) Since we need to recover the original theory in the limit [itex]\Psi_{\mu} \to \partial_{\mu}\Phi[/itex], the object [itex]\Psi_{\mu}[/itex] must depend linearly on [itex]\partial_{\mu}\Phi[/itex]. (ii) Since [itex]\Psi_{\mu}[/itex] and [itex]\partial_{\mu}\Phi[/itex] both carry Lorentz index, the new ingredient in the theory must also carry a Lorentz index. However, taking a Lorentz vector [itex]V_{\mu}[/itex] and setting [itex]\Psi_{\mu} = \partial_{\mu}\Phi + V_{\mu}[/itex] is not correct, because [itex]\partial_{\mu}\Phi[/itex] need not be a Lorentz vector. Indeed, [itex]\Phi[/itex] is an arbitrary matter field, it could itself be a vector field. So, the simplest choice which keeps Lorentz happy is achieved by taking
    [tex]\Psi_{\mu} = \partial_{\mu}\Phi + i A_{\mu}\Phi \equiv D_{\mu}\Phi . \ \ \ \ (3)[/tex] Thus, the field [itex]A_{\mu}[/itex] is the only new ingredient in the theory. Now, from the local invariance condition eq(2), we can easily deduce the transformation law for the new field, just substitute (3) in (2)
    [tex]\left( \partial_{\mu} + i A^{'}_{\mu} \right) U \Phi = U \left( \partial_{\mu}\Phi + i A_{\mu}\Phi \right) .[/tex]
    Okay, I leave you to expand the left-hand-side and obtain the required local transformation law for [itex]A_{\mu}[/itex]. Notice that in the new theory, the field [itex]A_{\mu}[/itex] is coupled to the matter field [itex]\Phi[/itex]. This means that imposing local invariance on a free globally invariant matter field theory leads to interacting field theory.
    If you like to expand your knowledge and see how this is done in details, see the PDF below.
     

    Attached Files:

    Last edited: Apr 14, 2016
  12. Apr 15, 2016 #11

    ShayanJ

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    Thanks for the explanations Sam.

    Your PDF is nice, but I just have a question, In section 2, you try to prove ## R_{\mu\nu}=R_{\nu \mu} ##. This is trivially true for the spatial components. But about ## R_{j0}=R_{0j} ##, ##R_{j0}## is zero, but ## R_{0j}=\int d^3 x \partial_j F(x) ## which you don't prove to be zero! But I really doubt that the integral over all space of the derivative of any given function is zero!

    The name of the PDF file contains "Ch5" which suggests its the chapter 5 of something. A series of lecture notes of yours maybe?
     
  13. Apr 15, 2016 #12

    samalkhaiat

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    I did. If you go down two lines, you see the boundary condition on [itex]F(x)[/itex].
    It is not any function.
    The question was the following: What boundary condition the function [itex]F(x)[/itex] must satisfy in order for [itex]\int d \sigma_{\mu} \ \partial_{\nu} F(x) = \int d \sigma_{\nu} \ \partial_{\mu} F(x) [/itex] to be identically true?
    Now, [tex]R_{0j} - R_{j0}= \int_{V} d^{3}x \ \partial_{j}F(x) .[/tex]
    By the divergence theorem
    [tex]\int_{V} d^{3}x \ \partial_{j}F(x) = \int_{S^{2}} dS \ \hat{n}_{j} \ F(x) ,[/tex]
    we have the following behaviour
    [tex]R_{0j} - R_{j0} \sim 4\pi | \vec{x}|^{2} F(x) |_{|\vec{x}| \to \infty} .[/tex]
    This, of course vanishes when [itex]F(x)[/itex] dies off faster that [itex]\frac{1}{| \vec{x}|^{2}}[/itex].

    Actually, except for minor improvement, I made the material in that PDF when I was in grad school (22 years ago!). But, yes they are part of huge set of notes I delivered at various places on symmetries in classical and quantum field theories. The plan was (and still is) to organize them into a textbook, but unfortunately I haven’t got time to typeset them all yet.
     
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