Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: EoM from action, indices confused, (QFT)

  1. Dec 25, 2017 #1
    1. The problem statement, all variables and given/known data

    Using the E-L equations to get the EoM from the action.


    2. Relevant equations


    I am using E-L equations in the form:

    ## \frac{\partial}{\partial_u} \frac{\partial L}{\partial_u \phi}-\frac{\partial L}{\partial \phi} ##

    where ##L ## is the Lagrangian

    3. The attempt at a solution

    I am a bit confused with the term ## \frac{1}{2} \partial_u \phi \partial^u \phi ## of alot of actions.

    The correct answer is ## \frac{1}{2} \partial_u \partial^u \phi ##

    As it is for the first term of the EoM from [1] I get:

    ## \frac{\partial}{\partial_u} \frac{1}{2} \partial^u \phi = \frac{1}{2}\partial_u \partial^u \phi ##

    However if I write it as ##\frac{1}{2}(\partial_u \phi)^2 ## , the factor of half cancels, but my indices don't work out :

    ## \frac{\partial}{\partial_u} \partial_u \phi = \partial_u \partial_u \phi ##


    Many thanks
     
  2. jcsd
  3. Dec 25, 2017 #2

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member
    2017 Award

    Writing it as ##(\partial_u \phi)^2/2## is a bastard notation often used in physics. At face value you might think it means what you are taking it to mean, but what it really means is ##g^{ab}(\partial_a\phi)(\partial_b \phi)/2##.
     
  4. Dec 25, 2017 #3
    i dont understand the differnce, the metric in your expression raises the index to get the same expression, no?
     
  5. Dec 25, 2017 #4

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member
    2017 Award

    Yes, the point being that the expression ##(\partial_a \phi)^2## while seemingly containing two covariant indices really contains a contraction between a covariant and a contravariant index and needs to be treated as such.
     
  6. Dec 25, 2017 #5
    oh so as in when i differentiate what i have done is treated them as seperate indices but i need to apply some sort of chain rule instead?
     
  7. Dec 25, 2017 #6

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member
    2017 Award

    Right. ##\partial^a\phi## is not indepentent from ##\partial_a\phi##.
     
  8. Dec 25, 2017 #7

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member
    2017 Award

    You mean when you take the derivative ##\partial_u## of ##\partial L/\partial(\partial_u\phi)##? In general yes, but in most cases QFT is done on a Minkowski space background meaning that that term vanishes. In a curved background (or in curvilinear coordinates) you will generally not obtain ##\partial_a\partial^a\phi =0## but ##\nabla_a\partial^a\phi =0##, which is fortunate since the first expression is not invariant whereas the second is.
     
  9. Dec 26, 2017 #8
    so looking at the OP, what is the term coming from the chain rule I am missing that you have hinted at.
    From the above discussion all I have is:

    ##\frac{1}{2} \partial_u (g^{ua} \partial_{a} \phi )
    =\frac{1}{2} ( g^{ua} \partial_u \partial_{a} \phi + \partial_{a} \phi \partial_u g^{ua} ) ##

    this is equal to what I had previously , sincce the 2nd term is zero...

    Please can I have more guidance,

    thanks
     
  10. Dec 26, 2017 #9

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member
    2017 Award

    Why do you still have the 1/2?
     
  11. Dec 26, 2017 #10

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member
    2017 Award

    Also I am not sure I understand what you think the problem is ...
     
  12. Dec 26, 2017 #11
    becuse in genreal the action will include other terms such as ## m^2 \phi ^2 + \lambda \frac{\phi}{4!} ## i.e ##\frac{\partial L}{\partial \phi } \neq 0 ##
     
  13. Dec 26, 2017 #12

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member
    2017 Award

    Are you dealing with a complex scalar field or a real scalar field? The 1/2 should disappear in the case of a real scalar field because you have two derivative terms. In the case of a complex field you do not have the factor of 1/2 from the beginning. In the case of a real scalar field, the mass term also comes with a 1/2.
     
  14. Dec 26, 2017 #13
    yep the m^2 term comes with a half.
    no i mean im stil struggling with this derivative term.

    ok so I want ## \frac{\partial }{\partial_u} (\partial_u \phi \partial^u \phi) = \frac{\partial }{\partial_u} (\partial_u \phi g^{au} \partial_a \phi)= g^{au}\partial_a \phi (1) + g^{au}\partial_u \phi \frac{\partial}{\partial_u \phi} (\partial_a \phi) + 0 ##

    using the product rule

    for the 2nd term I get ## \frac{\partial}{\partial_u \phi} (\partial_a \phi) =\delta^{au} ##

    this must be wrong since this then gives me a factor of ##4## via the trace of the metric and so the index is not raised as required also.
     
  15. Dec 28, 2017 #14

    please can somebody help with my working here, is the delta term wrong? thanks
     
  16. Dec 28, 2017 #15

    haushofer

    User Avatar
    Science Advisor

    You're using very confusing notation, which I think is the reason people are not responding. I fix your notation and get

    ## \frac{\partial }{\partial_c} (\partial_b \phi \partial^b \phi) = \partial_c \Bigl(\eta^{ab} \partial_a \phi \partial_b \phi \Bigr) = \eta^{ab} \Bigl( (\partial_c \partial_a) \phi \partial_b \phi + \partial_a \phi ( \partial_c \partial _b \phi ) \Bigr) = \Bigl( (\partial_c \partial_a) \phi \partial^a \phi + \partial^a \phi ( \partial_c \partial_a \phi ) \Bigr)##

    This equals

    ## 2 \Bigl( (\partial_c \partial_a) \phi \partial^a \phi \Bigr)##
     
  17. Dec 28, 2017 #16

    stevendaryl

    User Avatar
    Staff Emeritus
    Science Advisor

    Why do you want that? The Euler-Lagrange equations of motion are:

    [itex]\partial_u \frac{\partial \mathcal{L}}{\partial (\partial_u \phi)} = \frac{\partial \mathcal{L}}{\partial \phi}[/itex]

    But [itex]\frac{\partial \mathcal{L}}{\partial (\partial_u \phi)} = \partial^u \phi[/itex] (assuming [itex]\mathcal{L} = \frac{1}{2} \partial_u \phi \partial^u \phi [/itex] + terms involving powers of [itex]\phi[/itex])

    So unless you have a strange Lagrangian, you shouldn't have to compute [itex]\partial_u (\partial_v \phi \partial^v \phi)[/itex]
     
  18. Dec 29, 2017 #17

    okay well I was trying to incrorporate Oroduin's hints since, as said in the OP, i can't get rid of the factor of 1/2 in the derivative term.

    To give a specfic example if I consider: ##-1/2 \partial_u \phi \partial^u \phi - 1/2 m^2 \phi^2 ##
    then, my previous method, before making this thread was:
    ##\frac{\partial L}{\partial \phi}= m^2 \phi##
    ##\frac{\partial L}{\partial_u \phi}= \partial^u \phi (-1/2) ##

    therefore yielding ## \partial_u \frac{\partial L}{\partial_u \phi} - \frac{\partial L}{\partial \phi} = -1/2 \partial^u\partial_u \phi + m^2 \phi ##
    whereas the EoM is : ## \partial^u \partial_u \phi -m^2 \phi^2 ##

    But now, trying to use Oroduin's hint I rewrite as ## L=-1/2 \partial_u \phi g^{au} \partial_a \phi +m^2term... ##
    Now looking at the derivative term I get ## \frac{\partial L}{\partial_u \phi} = =-1/2g^{au} ( \partial_a \phi \frac{\partial}{\partial_u \phi}(\partial_u \phi) + \partial_u \phi \frac{\partial}{\partial_u \phi}(\partial_a \phi))=-1/2 g^{au} ( \partial_a \phi (1) + \partial_u \phi \delta_{au}) ##
    where I have used the product rule, and taken out the metric since it is the minkoski...

    ##= -1/2 (\partial^u \phi + g^{uu}\partial_u \phi) ##...
     
    Last edited: Dec 29, 2017
  19. Dec 29, 2017 #18

    stevendaryl

    User Avatar
    Staff Emeritus
    Science Advisor

    You have to be careful about indices. If you look at just the term [itex]\frac{1}{2} g^{au} (\partial_a \phi) (\partial_u \phi)[/itex] and you take a derivative with respect to [itex]\partial_u \phi[/itex], you have to realize that in [itex]\frac{1}{2} g^{au} (\partial_a \phi) (\partial_u \phi)[/itex], [itex]u[/itex] is a dummy index. You can replace it by [itex]v[/itex], say. So you have:

    [itex]\frac{\partial}{\partial(\partial_u \phi)} \frac{1}{2} g^{av} (\partial_a \phi) (\partial_v \phi)[/itex]


    Now, [itex]\frac{\partial}{\partial(\partial_u \phi)} (\partial_a \phi) = 0[/itex] unless [itex]a=u[/itex]. So we can summarize it as:

    [itex]\frac{\partial}{\partial(\partial_u \phi)} (\partial_a \phi) = \delta^u_a[/itex]

    Similarly,

    [itex]\frac{\partial}{\partial(\partial_u \phi)} (\partial_v \phi) = \delta^u_v[/itex]

    So using the product rule gives you:

    [itex]\frac{\partial}{\partial(\partial_u \phi)} \frac{1}{2} g^{av} (\partial_a \phi) (\partial_v \phi) = \frac{1}{2} g^{av} (\delta^u_a \partial_v \phi + \delta^u_v \partial_a \phi)[/itex]

    Then we can use [itex]g^{av} \delta^u_a = g^{uv} \delta^u_a[/itex]. Summing over [itex]a[/itex] gives [itex]g^{uv}[/itex]. Similarly, [itex]g^{av} \delta^u_v = g^{au} \delta^u_v[/itex]. Summing over [itex]v[/itex] gives [itex]g^{au}[/itex]. So this simplifies to:


    [itex]\frac{1}{2} (g^{uv} \partial_v \phi + g^{au} \partial_a \phi)[/itex]

    Those two terms are the same, since [itex]a[/itex] and [itex]v[/itex] are dummies and [itex]g^{au} = g^{ua}[/itex]. So we have:
    [itex]g^{uv} \partial_v \phi[/itex]

    The EoM are just:

    [itex]\partial_u (g^{uv} \partial_v \phi) = \pm m^2 \phi[/itex]

    (I don't remember whether the right side is a + or a -)
     
  20. Dec 29, 2017 #19

    stevendaryl

    User Avatar
    Staff Emeritus
    Science Advisor

    In flat spacetime, you can take the [itex]g^{uv}[/itex] outside the derivative.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted