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The variation of a scalar field (from Ryder's QFT book)

  1. Jul 20, 2013 #1
    Hello!
    Im currently reading Ryder's QFT book and am confused with the variation of a scalarfield.
    He writes that the variation can be done in two ways,
    [tex]
    \phi(x) \rightarrow \phi'(x) = \phi(x) + \delta \phi(x)
    [/tex]
    and
    [tex]
    x^\mu \rightarrow x'^\mu = x^\mu + \delta x^\mu.
    [/tex]
    This seems reasonable. But later, in order to proceed with Noether's theorem he states that the variation is of the form (eq 3.22):
    [tex]
    \Delta x^\mu = X^\mu_\nu \Delta \omega^\nu,
    [/tex]
    and
    [tex]
    \Delta \phi = \Phi_\mu \delta \omega^\mu.
    [/tex]

    I dont know how to interpret this. What is a function over spacetime? Is 'X' our generators for the group of transformations?
    If someone feels like expanding on this transformation I would be very happy. I can see that we are describing the total variation in the last equations but why dont we just vary the scalarfield with another scalar function? It seems to just complicate things this way. :(

    Thanks for your time guys and girls!
     
  2. jcsd
  3. Jul 20, 2013 #2

    samalkhaiat

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  4. Jul 21, 2013 #3
    I'm sorry samalkhaiat. It seems to be to much for me to get into. More than I hoped. Is there really not any 'easy' intuitional explanation for this transformation that could help me? The other way to transform the scalar field seems so easy to picture.
     
  5. Jul 21, 2013 #4
    If both the matrix X and the infinitesimal paramter '\delta \omega' are functions over spacetime our choice of these objects are certainly not unique...
     
  6. Jul 21, 2013 #5
    As far as I can tell, the [itex]X^\mu_\nu[/itex] term does indeed represent the infinitesimal generators of the transformation group--by plugging in the appropriate value for [itex]\omega^\nu[/itex], we can vary the coordinates in whichever way we want to. Only [itex]X^\mu_\nu[/itex] is a function over spacetime, though--[itex]\omega^\nu[/itex] is just a single tensor that controls which combination of generators we're interested in.

    The reason you set the transformation up this way, instead of just starting with the total scalar variation, is that you will eventually be interested in what happens when you vary the coordinates in specific directions. For instance, when you vary the coordinates in a purely time-like direction, Noether's Theorem gives you the conservation of energy, and when you vary it in a space-like direction, you get the conservation of momentum.

    So once you churn through the machinery of Noether's Theorem, you get a conserved quantity with an extra tensor index. By contracting a specific direction into that quantity, you can get either energy or momentum out, or any combination thereof. The key is that the derivation is done in a relativistically invariant way. If you tried to vary in specific directions up front, you could get out the various components of the conserved quantities, but it would be hard to assemble them into an invariant four-vector. By doing it this way, it all happens automatically.
     
    Last edited: Jul 21, 2013
  7. Jul 21, 2013 #6

    Avodyne

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    IMO varying the coords as well as the fields is unnecessarily confusing. See the derivation of Noether's theorem in Srednicki's text.
     
  8. Jul 22, 2013 #7
    Thanks for your help! I think I get it now. Ryder however is not a book for beginners. I would say that he makes all the misstakes that are usually confunsing when trying to learn QFT. Understanding field quantization form this book alone would be awful.
     
  9. Jul 22, 2013 #8

    vanhees71

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    It's easier to deal with "external symmetries" first, but varying only the fields is not sufficient to understand Poincare invariance, which is the most fundamental symmetry principle underlying relativistic QFT.
     
  10. Jul 22, 2013 #9

    Avodyne

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    You can do poincare symmetry without explicit coord changes; for translation symmetry, simply define ##\delta\varphi(x)=a^\mu\partial_\mu\varphi(x)## where ##a^\mu## is an infinitesimal translation parameter. Then take into account that the variation of the lagrangian density is a total divergence (rather than zero). See Srednicki's text for details.
     
  11. Jul 22, 2013 #10

    samalkhaiat

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    Ok, let [itex]G[/itex] be a space-time group and [itex]\mathcal{ L }[/itex] its Lie algebra. The group can be formally realized as a group of coordinate transformations:
    [tex]\bar{ x }^{ \mu } = g^{ \mu }{}_{ \nu } ( \omega ) \ x^{ \nu } . \ \ (1)[/tex]
    By expanding [itex]g[/itex] infinitesimally around the identity:
    [tex]g^{ \mu }{}_{ \nu } = \delta^{ \mu }_{ \nu } + \left( \omega \cdot \Sigma \right)^{ \mu }_{ \nu } ,[/tex]
    with [itex]\Sigma[/itex] being the infinitesimal generators of [itex]\mathcal{ L }[/itex], Eq(1) becomes,
    [tex]\bar{ x }^{ \mu } = x^{ \mu } + \omega^{ \alpha } \left( \Sigma_{ \alpha } \right)^{ \mu }_{ \nu } \ x^{ \nu } ,[/tex]
    or
    [tex]\delta x^{ \mu } = \omega^{ \alpha } \left( \Sigma_{ \alpha } \right)^{ \mu }_{ \nu } \ x^{ \nu } , \ \ \ (2)[/tex]
    where [itex]\alpha = 1, 2 , …,[/itex] dim([itex]\mathcal{ L }[/itex]). If [itex]G[/itex] is the Poincare’ group, then
    [tex]
    \omega^{ \alpha } ( \Sigma_{ \alpha } )^{ \mu }_{ \nu } \equiv a^{ \sigma } P_{ \sigma } \delta^{ \mu }_{ \nu } + \frac{ 1 }{ 2 } \omega^{ \rho \sigma } ( \Sigma_{ \rho \sigma } )^{ \mu }_{ \nu } ,
    [/tex]
    where the Lorentz generators are given by
    [tex]
    \left( \Sigma_{ \rho \sigma } \right)^{ \mu }_{ \nu } = \eta_{ \rho \nu } \ \delta^{ \mu }_{ \sigma } - \eta_{ \sigma \nu } \ \delta^{ \mu }_{ \rho } .
    [/tex]
    If you compare Eq(2) with that of Ryder’s, you see that he buries the generators in his (stupid) symbol [itex]X^{ \mu }_{ \nu }[/itex]. Of course this is wrong because one of the indices HAS to run over the whole dimension of [itex]\mathcal{ L }[/itex] ( I denoted it by [itex]\alpha[/itex]) and the other index runs over the space-time dimensions. So, to construct the correct version of his symbol you can write Eq(2) as
    [tex]\delta x^{ \mu } = \omega^{ \alpha } X^{ \mu }_{ \alpha } ( x ) ,[/tex]
    where, as a function of x,
    [tex]X^{ \mu }_{ \alpha } ( x ) \equiv \left( \Sigma_{ \alpha } \right)^{ \mu }{}_{ \nu } \ x^{ \nu } .[/tex]
    Ok, let us now talk about fields, [itex]\phi^{ r } ( x )[/itex], [itex]r = 1,2,..n[/itex].
    They transform by finite dimensional ([itex]n \times n[/itex] matrix) representation, [itex]D^{ r }_{ s } ( g )[/itex], of the group [itex]G[/itex]:
    [tex]\bar{ \phi }^{ r } ( \bar{ x } ) = D^{ r }{}_{ s } \ \phi^{ s } ( x ) . \ \ \ (3)[/tex]
    Again, we expand the matrix [itex]D ( g )[/itex] to first order in the parameters, [itex]\omega^{ \alpha }[/itex], as
    [tex]D^{ r }_{ s } = \delta^{ r }_{ s } + \omega^{ \alpha } \left( \Sigma_{ \alpha } \right)^{ r }_{ s } .[/tex]
    Inserting this in Eq(3), we get
    [tex]\Delta \phi^{ r } ( x ) \equiv \bar{ \phi }^{ r } ( \bar{ x } ) - \phi^{ r } ( x ) = \omega^{ \alpha } ( \Sigma_{ \alpha } )^{ r }_{ s } \ \phi^{ s } ( x ) .[/tex]
    Or, if you define (as Ryder “does” ) the functions
    [tex]\Phi^{ r }_{ \alpha } ( \phi ) \equiv \left( \Sigma_{ \alpha } \right)^{ r }_{ s } \ \phi^{ s } ( x ) ,[/tex]
    then you can write the infinitesimal transformations of the fields as
    [tex]\Delta \phi^{ r } ( x ) = \omega^{ \alpha } \ \Phi^{ r }_{ \alpha } ( \phi ) .[/tex]
    Notice that the index [itex]r = 1,2,…,n[/itex] is a representation index (n is the number of components of the field), while [itex]\alpha[/itex], as before, runs over the dimension of the Lie algebra [itex]\mathcal{ L }[/itex]. A scalar field, by definition, transforms by the identity element of the group [itex]G[/itex], [itex]D^{ r }_{ s } = \delta^{ r }_{ s }[/itex], the so-called trivial representation. So, for scalar fields, we have
    [tex]\Delta \phi^{ r } ( x ) = 0, \ \forall r .[/tex]
    The last thing I want to mention is that we can define the variation in the functional form of the fields as follow. First, expand the transformed field to first order in [itex]\delta x^{ \mu }[/itex]
    [tex]\bar{ \phi }^{ r } ( \bar{ x } ) = \bar{ \phi }^{ r } ( x ) + \delta x^{ \mu } \partial_{ \mu } \phi^{ r } ( x ) ,[/tex]
    then, you find
    [tex]
    \delta \phi^{ r } ( x ) \equiv \bar{ \phi }^{ r } ( x ) - \phi^{ r } ( x ) = - \delta x^{ \mu } \partial_{ \mu } \phi^{ r } + \omega^{ \alpha } ( \Sigma_{ \alpha } )^{ r }_{ s } \ \phi^{ s } .
    [/tex]

    Sam
     
  12. Jul 22, 2013 #11
    Peskin and Schroeder has a fairly straightforward definition of both Noether's Theorem, as well as a couple specific applications like internal symmetries and spacetime translations. Check out the end of section 2.2 (if you don't have a copy, Amazon's preview includes those pages).
     
  13. Jul 22, 2013 #12
    I think it's okay to just vary the fields, and show that it leads a change in the Lagrangian equal to a 4-divergence.

    However, while somewhat mathematically less confusing, I think physically it's odd.

    Physically when translating you are changing the coordinates and the fields, and not just the fields.

    So saying that the action is invariant under both coordinate and field change makes more physical sense, but can be slightly more confusing when you encounter it for the first time (I've seen textbook authors get it wrong - I mean they obviously get the end result correct because everyone knows the end result, but wrong reasoning: but as long as they make an even number of sign errors, they'll get the right result lol)
     
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