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I 4 vector upper and lower indices

  1. Oct 12, 2016 #1
    I'm working through some intro QFT using Peskin accompanied by David Tong's notes, and have a question over notation. From Peskin I have:

    p_\mu p^\mu=g^{\mu\nu}p^\mu p^\nu=E^2-|\mathbf{p}|^2
    [/tex] with
    \partial_\mu=\frac{\partial}{\partial x^\mu}=\bigg(\frac{\partial}{\partial x^0},\mathbf{\nabla}\bigg)

    Does this mean that:

    \partial^\mu=\frac{\partial}{\partial x_\mu}=\bigg(\frac{\partial}{\partial x^0},-\mathbf{\nabla}\bigg)

    and if so, is there any reason why the upper/lower index flips when expressing a derivative compared with when writing just a normal vector? It's a bit of a pain when you're starting out so I'm guessing there must be a good reason for it that emerges later.

    I'd also like someone to just confirm that I've taken this derivative properly (might seem a bit laboured but I want to make triple sure I've got the notation correct right away):

    If we have a Lagrangian density of


    the derivative with respect to dphi is:

    \frac{\partial L}{\partial(\partial_\mu\phi)}=\frac{\partial}{\partial_\mu}\bigg(\frac{1}{2}\partial^\mu\phi\partial_\mu\phi\bigg)=\frac{\partial}{\partial_\mu}\bigg(\frac{1}{2}g^{\mu\nu}\partial_\mu\phi\partial_\mu\phi\bigg)=\frac{1}{2}g^{\mu\nu}\frac{\partial}{\partial_\mu}\bigg(\partial_\mu\phi\partial_\mu\phi\bigg)=\frac{1}{2}g^{\mu\nu}\frac{\partial}{\partial_\mu}(\partial_\mu\phi)^2


    Thanks as always to you good folk.
  2. jcsd
  3. Oct 12, 2016 #2
    The answer is correct - but your working needs some work - you are introducing too many repeated indices that mess up with the summation convention and confuse you. It is probably a fortunate coincidence in this case that your sloppy working produces the correct result!

    In particular,
    [tex]g^{\mu \nu} \partial_\mu = \partial^\nu \neq \partial^\mu[/tex]

    It is good bookkeeping practice to avoid conflicts between indices that are not related - i.e. you should rename the dummy indexes so they do not coincide. As an illustration of the correct way you should do this calculation:
    [tex]\frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi)} = \frac{\partial}{\partial (\partial_\mu \phi)} \left[\frac{1}{2} \partial^{\alpha} \phi\,\partial_{\alpha} \phi\right] = \frac{1}{2} \frac{\partial}{\partial (\partial_\mu \phi)} \left[g^{\alpha \beta} \partial_{\beta} \phi\,\partial_{\alpha} \phi \right]
    = \frac{1}{2} g^{\alpha \beta} \left[\delta^{\mu}_{\,\beta} \, \partial_{\alpha} \phi + \partial_{\beta} \phi\, \delta^{\mu}_{\,\alpha}\right]\\
    = \frac{1}{2} \left[ g^{\alpha \mu}\, \partial_{\alpha} \phi + g^{\mu \beta}\, \partial_{\beta} \phi \right] = \partial^{\mu} \phi
    This may seem awfully tedious, and as you get more familiar, there is a tendency to simply skip steps, but for more complicated scenarios, it is extremely important that we keep track of the indices very carefully.
  4. Oct 26, 2016 #3
    Thanks for this - its kind of bizarre that I have never come across this kind of thing explicitly in my courses, and I'm finding it really hard to find clear online material on it as well.
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