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A Differentiation of a product of 4-gradients wrt a 4-gradient

  1. Apr 19, 2016 #1
    I know that ##\frac{\partial}{\partial (\partial_{\mu}\phi)} \big( \partial_{\mu} \phi\ \partial^{\mu} \phi \big) = \partial_{\mu} \phi##.

    Now, I need to prove this to myself.

    So, here goes nothing.

    ##\frac{\partial}{\partial (\partial_{\mu}\phi)} \big( \partial_{\mu} \phi\ \partial^{\mu} \phi \big)##
    ## = \frac{\partial}{\partial (\partial_{\mu}\phi)} \big( \eta^{\mu\nu}\partial_{\mu} \phi\ \partial_{\nu} \phi \big)##
    ##= \eta^{\mu\nu}\ \partial_{\nu} \phi + \eta_{\mu\nu} \eta^{\mu\nu}\ \partial_{\mu} \phi##,

    where I first differentiated the factor ##\partial_{\mu}\phi## with respect to ##\partial_{\mu}\phi## and then I differentiated the factor ##\partial_{\nu}\phi## with respect to ##\partial_{\mu}\phi##.

    Am I correct so far?
     
  2. jcsd
  3. Apr 20, 2016 #2
    Your indices in the second term don't match with those in the first so that's an indication something is wrong.

    Try using the following form

    $$\frac{\partial}{\partial (\partial_{\alpha}\phi)} \big( \partial_{\mu} \phi\ \partial^{\mu} \phi \big)$$

    This helps you avoid the problem where you have ill-defined expressions like ##\eta_{\mu\nu}\eta^{\mu\nu}\partial_\mu\phi##.
    The latter can mean two things
    $$\eta_{\mu\nu}\left(\eta^{\mu\nu}\partial_\mu\phi\right)=\eta_{\mu\nu}\partial^\nu\phi=\partial_\mu\phi$$
    or it can mean (using ##\eta_{\mu\nu}\eta^{\mu\nu}=D## with D the number of spacetime dimensions)
    $$\left(\eta_{\mu\nu}\eta^{\mu\nu}\right)\partial_\mu\phi=4\partial_\mu\phi$$
     
  4. Apr 21, 2016 #3
    Firstly, I need to say that ##
    \frac{\partial}{\partial (\partial_{\mu}\phi)} \big( \frac{1}{2}\partial_{\mu} \phi\ \partial^{\mu} \phi \big) = \partial^{\mu} \phi##.
    I made a mistake in my first post of placing the ##\mu## index on the RHS downstairs, instead of upstairs. I also made a mistake of forgetting the factor of ##\frac{1}{2}##.

    Now,

    ##\frac{\partial}{\partial (\partial_{\alpha}\phi)} \big( \frac{1}{2}\partial_{\mu} \phi\ \partial^{\mu} \phi \big)##
    ##=\frac{1}{2}\eta^{\mu\nu}\frac{\partial}{\partial (\partial_{\alpha}\phi)} \big( \partial_{\mu} \phi\ \partial_{\nu} \phi \big)##
    ##=\frac{1}{2}\eta^{\mu\nu}(\eta_{\alpha\mu}\partial_{\nu}\phi+\eta_{\alpha\nu}\partial_{\mu}\phi)##
    ##=\frac{1}{2}(\eta_{\alpha\mu}\eta^{\mu\nu}\partial_{\nu}\phi+\eta_{\alpha\nu}\eta^{\nu\mu}\partial_{\mu}\phi)##
    ##=\eta_{\alpha\mu}\eta^{\mu\nu}\partial_{\nu}\phi##
    ##=\delta^{\nu}_{\alpha}\partial_{\nu}\phi##
    ##=\partial_{\alpha}\phi##

    Am I correct?
     
    Last edited: Apr 21, 2016
  5. Apr 21, 2016 #4
    Why do you use that
    $$
    \frac{\partial\left(\partial_\mu\phi\right)}{\partial\left(\partial_\alpha\phi\right)}=\eta_{\mu\alpha}
    $$

    A hint that something is wrong, is that the indices don't correspond. You can check this by performing a transformation ##x^\mu\to x^{\prime\mu}## you should find that there is an upper and a lower index. [*]

    It's equal to ##\delta^\alpha_\mu\,\,\,\left(=\eta^\alpha_{\,\,\, \mu}\right)## as far as I can tell.

    [*]: Oddly it seems that this would turn out correct if you contract the metric so I'm starting to doubt myself.
     
  6. Apr 21, 2016 #5
    Ok, let me rework my answer using your suggestion.

    ##\frac{\partial}{\partial (\partial_{\alpha}\phi)} \big( \frac{1}{2}\partial_{\mu} \phi\ \partial^{\mu} \phi \big)##
    ##=\frac{1}{2}\eta^{\mu\nu}\frac{\partial}{\partial (\partial_{\alpha}\phi)} \big( \partial_{\mu} \phi\ \partial_{\nu} \phi \big)##
    ##=\frac{1}{2}\eta^{\mu\nu}({\eta^{\alpha}}_{\mu}\partial_{\nu}\phi+{\eta^{\alpha}}_{\nu}\partial_{\mu}\phi)##
    ##=\frac{1}{2}({\eta^{\alpha}}_{\mu}\eta^{\mu\nu}\partial_{\nu}\phi+{\eta^{\alpha}}_{\nu}\eta^{\nu\mu}\partial_{\mu}\phi)##
    ##={\eta^{\alpha}}_{\mu}\eta^{\mu\nu}\partial_{\nu}\phi##
    ##=\eta^{\alpha\nu}\partial_{\nu}\phi##
    ##=\partial^{\alpha}\phi##

    It appears that in the answer, the index ##\alpha## should be upstairs, not downstairs. I made this mistake in my previous post.
     
  7. Apr 21, 2016 #6
    Try to see why the ##\alpha## index is upstairs and also why you can write ##\eta^\alpha_{\,\,\,\mu}=\delta^\alpha_\mu## where delta is the kronecker delta.

    Other than that it is correct.
     
  8. Apr 21, 2016 #7
    Shouldn't the index ##\alpha## be upstairs because we are differentiating the product of an upstairs index and a downstairs index with respect to a downstairs index?

    I think ##\eta^\alpha_{\,\,\,\mu}=\delta^\alpha_\mu## because ##\eta^\alpha_{\,\,\, \mu} = \eta^{\alpha\nu}\eta_{\nu\mu} = \delta^\alpha_\mu##.

    Am I correct?
     
  9. Apr 21, 2016 #8
    The latter part is perfect.

    The first part becomes clearest when explicitly changing coordinates In other words how does the

    $$
    \partial_\mu\phi(x)=\frac{\partial\phi(x)}{\partial x^\mu}
    $$

    Apply a coordinate transformation ##x^\mu\to x^{\prime\mu}## and look at the way the Jacobian shows up.
    The Jacobian is given by (or "upside down" doesn't matter too much since we look at invertible transformations)

    $$
    \frac{\partial x^{\prime\mu}}{\partial x^{\mu}}
    $$
     
  10. Apr 22, 2016 #9
    Under the coordinate transformation ##x^{\mu}\rightarrow x'^{\mu}={\Lambda^{\mu}}_{\nu}x^{\nu}##,

    ##\partial_{\mu}\phi(x)~=~\frac{\partial\phi(x)}{\partial x^{\mu}}~##

    ##\rightarrow \frac{\partial\phi(\Lambda^{-1}x)}{\partial x^{\mu}}=\frac{\partial (\Lambda^{-1}x)^{\nu}}{\partial x^{\mu}}\frac{\partial\phi(\Lambda^{-1}x)}{\partial (\Lambda^{-1}x)^{\nu}}=\frac{\partial}{\partial x^{\mu}}\Big( {(\Lambda^{-1})^{\nu}}_{\rho}x^{\rho} \Big)(\partial_{\nu}\phi)(\Lambda^{-1}x)={(\Lambda^{-1})^{\nu}}_{\rho}\delta^{\rho}_{\mu}(\partial_{\nu}\phi)(\Lambda^{-1}x)={(\Lambda^{-1})^{\nu}}_{\mu}(\partial_{\nu}\phi)(\Lambda^{-1}x)##.

    How does this help?
     
    Last edited: Apr 22, 2016
  11. Apr 22, 2016 #10
    Well, now use it in the derivative w.r.t. ##\partial_\mu\phi##.
    You should get something of the form
    $$
    \frac{\partial\left(\partial_\mu\phi\right)}{\partial\left(\partial_\alpha\phi\right)}\to \left(\Lambda^{-1}\right)^\nu_{\,\,\, \mu}\Lambda^\alpha_{\,\,\, \beta}\frac{\partial\left(\partial_\nu\phi(x^\prime)\right)}{\partial\left(\partial_\beta\phi(x^\prime)\right)}
    $$

    This means that the object transform as a (1,1)-tensor i.e. it has one upper and one lower index.

    This might help with the details https://www.physicsforums.com/threads/kronecker-delta-as-tensor-proof.320692/
     
  12. Apr 22, 2016 #11
    Ok. So, under the coordinate transformation ##x^{\mu} \rightarrow {\Lambda^{\mu}}_{\nu}x^{\nu}##,

    ##\frac{\partial(\partial_{\mu}\phi(x))}{\partial(\partial_{\alpha}\phi(x))} \rightarrow \frac{\partial({(\Lambda^{-1})^{\nu}}_{\mu}(\partial_{\nu}\phi)(\Lambda^{-1}x))}{\partial({(\Lambda^{-1})^{\beta}}_{\alpha}(\partial_{\beta}\phi)(\Lambda^{-1}x))}=\frac{\partial({(\Lambda^{-1})^{\nu}}_{\mu}(\partial_{\nu}\phi)(\Lambda^{-1}x))}{\partial({(\Lambda)_{\alpha}}^{\beta}(\partial_{\beta}\phi)(\Lambda^{-1}x))}##

    How should I now take ##{(\Lambda)_{\alpha}}^{\beta}## to the numerator?
     
  13. Apr 22, 2016 #12
    You can take ##\Lambda## outside of the derivatives. But I suggest you look at a text on (special) relativity.

    I suppose you're studying relativistic field theory? This means that knowing how to quickly read and interpret indices (upper/lower) can help you focus on the physics content instead of the manipulations of expressions.
     
  14. Apr 28, 2016 #13
    Thanks for the reply.

    Let me finish the steps of my derivation:

    ##\frac{\partial({(\Lambda^{-1})^{\nu}}_{\mu}(\partial_{\nu}\phi)(\Lambda^{-1}x))}{\partial({(\Lambda)_{\alpha}}^{\beta}(\partial_{\beta}\phi)(\Lambda^{-1}x))}=\frac{\partial((\partial_{\beta}\phi)(\Lambda^{-1}x))}{\partial({(\Lambda)_{\alpha}}^{\beta}(\partial_{\beta}\phi)(\Lambda^{-1}x))}\frac{\partial({(\Lambda^{-1})^{\nu}}_{\mu}(\partial_{\nu}\phi)(\Lambda^{-1}x))}{\partial((\partial_{\beta}\phi)(\Lambda^{-1}x))}={(\Lambda^{-1})^{\nu}}_{\mu}{(\Lambda^{-1})^{\beta}}_{\alpha}\frac{\partial((\partial_{\nu}\phi)(\Lambda^{-1}x))}{\partial((\partial_{\beta}\phi)(\Lambda^{-1}x))}={(\Lambda^{-1})^{\nu}}_{\mu}{(\Lambda)_{\alpha}}^{\beta}\frac{\partial((\partial_{\nu}\phi)(\Lambda^{-1}x))}{\partial((\partial_{\beta}\phi)(\Lambda^{-1}x))}##

    Hmm. I guess that's very important. I was just trying to practice my skills in tensor manipulations since I'm still new to this kind of math.
     
  15. May 4, 2016 #14
    Wait! My order of indices on ##{(\Lambda)_{\alpha}}^{\beta}## in the final result are not the same as your order of indices in ##{\Lambda^{\alpha}}_{\beta}##.

    Did I make a mistake in the second step of my calculation in the previous post?
     
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