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Raising and Lowering Indices and expansion

  1. Feb 12, 2010 #1

    Not sure if this is the best place to post this question but..

    Why can we expand [tex](\partial_\mu \phi)^2[/tex] in this way:

    [tex](\partial_\mu \phi)^2=(\partial_\mu \phi)(\partial^\mu \phi)[/tex]

    I mean [anything]^2 should equal [anything]*[anything] - why have be raised one of the indices above?

  2. jcsd
  3. Feb 12, 2010 #2
    Because you are contracting the two indices using the metric, which usually comes down to the Minkowski metric in QFT. What it really means is
    (\partial_\mu \phi)^2=(\partial_\mu \phi)(\partial^\mu \phi) = \eta^{\mu\nu} (\partial_\mu \phi)(\partial_\nu\phi)
    The Minkowski metric [itex]\eta^{\mu\nu}[/itex] is sometimes defined as [itex]\eta^{00}=-1[/itex], and [itex]\eta^{ii}= 1[/itex] for the spatial indices. Plugging this back in gives a minus
    (\partial_\mu \phi)^2 = -\partial_0\phi\partial_0\phi + \partial_x\phi\partial_x\phi = -(\partial_0\phi)^2 + (\partial_x\phi)^2
    This is really different from "just taking the square".

    Squaring an expression which contains an index [tex]\mu[/tex] means contracting over that index using the metric. It's not the same as "expanding the square".
  4. Feb 12, 2010 #3
    So its simply a question of definition - how the square of something is defined, in QFT.

    Great explanation xepma - thanks:)
  5. Feb 12, 2010 #4
    can you (or anyone else who's happy to) tell me why we can raise and lower matrices in this way:

    [tex]F_{\mu \nu}=g_{\mu \rho}g_{\nu \sigma} F^{\rho \sigma}[/tex]

    How does the above expression come about?

  6. Feb 13, 2010 #5
    This is about the calculation of tensors.
    For the practical use, we can neglect the abstract definition of dual space, linear functional and so on...
    And this is just a useful language in physics.

    Consider this, in special relativity, some quantities are expressed in terms of so-called 4-vectors, [tex] v^\mu[/tex], which transform under the Lorentz transformation as [tex] v^\mu \rightarrow v'^\mu = \Lambda^\mu{}_\nu v^\nu [/tex], where [tex]\Lambda^{\mu}{}_\nu[/tex] is the Lorentz transformation matrix and the Einstein summation convention is used.

    The Einstein summation convention is that, whenever we meet two objects, with one upper index and one lower index, we must sum over the index with possible range of the indices. For example, [tex]a^\mu b_{\mu} = a^0b_0 + a^1b_1 + a^2b_2 + a^3b_3[/tex]

    So what is the object with one lower index? It's also a vector, which is defined via the introduction of metric tensor. We define the metric tensor as a diagonal matrix, [tex]\eta_{\mu_\nu} \equiv \text{diag}(-1,1,1,1) [/tex]. In this way, the lower-index vector is defined as [tex] v_\mu \equiv \eta_{\mu\nu}v^\nu [/tex].
    Define [tex]\eta^{\mu\nu}[/tex] (lets call it metric tensor too) as the inverse matrix of the metric tensor [tex]\eta_{\mu\nu}[/tex], we see that the metric tensor can be used to pull indices up and down. For example, [tex] a^\mu = \eta^{\mu\nu}a_{\nu}[/tex]

    The definition of Lorentz transformation is a linear homogeneous transformation such that the object like [tex]a^\mu b_\mu[/tex] is a scalar under Lorentz transformation.

    So, your quantity [tex] \partial_\mu \phi \partial^\mu\phi [/tex] is a scalar, so that it can be put in the Lagrangian as the kinetic energy of a scalar field [tex]\phi[/tex].
  7. Feb 13, 2010 #6
    Thanks Ismaeli:)

    So again, it's just the way a subscripted tensor is defined in QFT...
  8. Feb 14, 2010 #7
    It's not in any way specific to QFT. These conventions show up all over the place in relativity. Since you're talking about relativistic QFT, it is natural that they show up there as well.

    What should probably be made clear is that the notation you're asking about is simply a conventional shorthand. Fundamentally, the quantity of interest is [itex](\partial_\mu\phi)(\partial^\mu\phi)[/itex]. But, physicists are lazy and would rather write [itex](\partial_\mu\phi)^2[/itex] a whole bunch of times than the full expression. And, since it's generally know that any term in the Lagrangian must be a scalar, there's no confusion about how the two indices must relate.
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