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Quantum theory, show variation of S zero, integrate by parts

  1. Jan 11, 2017 #1
    1. The problem statement, all variables and given/known data

    Hi,

    Please see attached.

    I am trying to show the second equality , expressing all as a total derivative (I can then show that ##\delta S = ##)

    2. Relevant equations

    See above

    3. The attempt at a solution

    So the ## m ## term is pretty obvious, simply using the chain rule.

    It is the first term I am stuck on. So looking by the sign, it looks like we have done integration by parts twice.

    My working so far is to go by parts initially as:

    ##w=\partial^{u}\phi ##
    ##\partial w = \partial_{v}\partial^{u} \phi ##
    ## \partial z = \partial_{u}\partial_{v} \phi ##
    ## z= \partial_{u} \phi ##

    to get, since we are allowed to assume vanishing of the field ##\phi ## at inifnity:

    ## - \int \partial_{u} \phi ( \partial_{v} \partial^{u} \phi ) ##

    I am now stuck of what to do, I can't see a move that will get the desired expression for a choice of integration by parts, which makes me question whether this was the correct first move to make.

    Many thanks in advance.
     

    Attached Files:

  2. jcsd
  3. Jan 12, 2017 #2

    Orodruin

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    Is ##a## constant? It is just the same type of relation as the mass term. Just switch the order of the derivatives.
     
  4. Jan 12, 2017 #3
    The ##\partial_{u} ## is not hitting the ## \partial^{u} \phi ## though?

    I have the first line as ## a^{v}\partial_{u}\partial_{v}\phi \partial^{u} \phi = a^{v}(\partial^2 _{uv} \phi)(\partial^{u} \phi )##,
    whereas I have the bottom line as ## (a^{v}\partial^{2}_{uv}\phi)(\partial_{v}\partial^{u} \phi) ##..

    thanks
     
  5. Jan 12, 2017 #4

    Orodruin

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    No this is not the bottom line - it does not contain four derivatives ... Just use the same argument as for the mass term, or more generally ##f\, df/dx = 0.5 df^2/dx##.
     
  6. Jan 12, 2017 #5
    erm, so I should have used the product rule on the bottom line?
    Which would give arise to two terms, whereas the top line has one term..?
     
  7. Jan 12, 2017 #6

    Orodruin

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    I don't get your meaning. Both the bottom and top lines have two terms. One from the mass and one from the kinetic term.
     
  8. Jan 12, 2017 #7
    apologies, corresponding to the kinetic term, I am only referring to here.
     
  9. Jan 12, 2017 #8

    Orodruin

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    Is your problem that you do not see that
    $$
    (\partial_a \partial_b \phi) (\partial^a \phi) = \frac{1}{2} \partial_b [(\partial_a \phi)(\partial^a\phi)] ?
    $$
     
  10. Jan 12, 2017 #9

    Orodruin

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    The kinetic term has one term only in both lines in your OP.
     
  11. Jan 12, 2017 #10
    oh right got it !
    by two terms I was reffering to the product rule, but then you sea-saw giving the factor of ##1/2##, thank you !
     
  12. Jan 12, 2017 #11
    [QUOTE="binbagsss, post: 5663557, member: 252335"

    My working so far is to go by parts initially as:

    ##w=\partial^{u}\phi ##
    ##\partial w = \partial_{v}\partial^{u} \phi ##
    ## \partial z = \partial_{u}\partial_{v} \phi ##
    ## z= \partial_{u} \phi ##

    [/QUOTE]

    May I ask, in general how should you approach integration by parts when four -derivaitves are involved.
    Should I choose a different index like ##\partial_a## than u and v which are already used in the problem
    Am I okay to use ##\partial_v ##, but then I must be consistent throughout? i.e. not change to ##\partial_u##
    I'm a bit confused..
    Many thanks !
     
  13. Jan 12, 2017 #12

    Orodruin

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    There is nothing different from the usual integration by parts, you just do it in the direction of the given index. In other words, ##\partial_\mu## is really a lot of different terms. You can do partial integration with respect to each direction separately. In reality, it really is just the divergence theorem.
     
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