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

Redefinition of a scalar field

  1. Jan 12, 2007 #1
    Hello everyone. I was hoping that someone could possibly help me with a problem I've got.

    If you have an action for two independent scalar fields, say A and B (arbitrary functions of (x_mu), both without any zeros), then can I redefine the action in terms of two new scalar fields A and C=AB? Are these two independent variables? I hope they are, but suspect they aren't. Sorry if its a stupid question, I dont know the criteria for two fields to be independent. Any help would be great.

    Thank you,

  2. jcsd
  3. Jan 13, 2007 #2


    User Avatar
    Homework Helper

    redefining scalar fields as linear combinations of each other is definitely ok, but as products? doubtful... why do I say that? well, I am not so concerned about whether they are independent at this point (however, they may well be indpt??) but more immediate problem is how you gonna fix the issue of dimensionality. As you may know the action must be dimensionless, multiplying two fields together changes the dimension and you'll probably need to introduce a dimensionful constant in the definition to counter that which could then lead to renormalisation problem of your theory...(well, I guess QFT is not what you are looking at, r u? ignore this last comment if you are not thinking about QFT) Anyway, all I can say is that regardless whether they are indpt, the first and foremost thing to consider is to make sure that your resultant action is dimensionally correct.
  4. Jan 13, 2007 #3
    Thanks for the reply. I hadnt considered the dimensionality of the problem to be honest, Ive been setting constants to unity in my work, and I was going to restore them at the end. But yeah good point (its definately a classical calculation so no worries about renormalization.)

    I cant make my mind up on the independence of the fields, and whether Ive lost a degree of freedom somewhere. I guess I should calculate the propagators, but I really dont want to unless I have to. I've never seen anyone redefine fields in a non linear way before, so I dont know. I cant find any relevant papers/books on the matter.

    Thank you,

  5. Jan 13, 2007 #4


    User Avatar
    Homework Helper

    I must admit that I haven't thought through this carefully, but at first glance, a product of fields look a lot like an interaction term to me (?)
  6. Jan 13, 2007 #5
    Well what I've got is an action with kinetic terms that look like

    S ~ d (AB) d(AB) + B d(A) d(A)

    d = four-derivative. So Ive got cross derivative terms in the original variables A and B, i.e. d(A)d(B), which I dont really want. So if I can redefine the fields, in the way I have, then Ill get standard kinetic terms (sort of)

    S ~ d(C)d(C) + C/A d(A) d(A),

    C = AB.

    So that is my problem. Although, thinking about it, I could redefine the cross derivative terms as something like

    d(A)d(B) = d(A d(B)) - A dd(B)

    and get rid of the divergence term (as you do). Something like that. But then I've got second derivatives in my action, which I dont really want either. Hhhm.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook