Can Scalar Field Redefinition Ensure Independence?

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Homework Help Overview

The discussion revolves around the redefinition of an action involving two independent scalar fields, A and B, and whether a new scalar field C defined as the product AB maintains independence from A. Participants are exploring the implications of this redefinition on dimensionality and the structure of the action.

Discussion Character

  • Conceptual clarification, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants are considering the independence of the fields A and C, questioning whether redefining fields as products affects their independence. There is also discussion about the dimensionality of the action and the potential need for dimensionful constants.

Discussion Status

Some participants have raised concerns about the dimensionality of the action when redefining fields, while others are uncertain about the independence of the fields and the implications of introducing non-linear combinations. There is an acknowledgment of the complexity of the problem, with various interpretations being explored.

Contextual Notes

One participant notes that they are working under the assumption of setting constants to unity and that the calculation is classical, which may influence the discussion regarding renormalization and dimensionality. There is also mention of cross derivative terms in the original action that the original poster wishes to avoid.

Joey_Joe_Jojo
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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 don't know the criteria for two fields to be independent. Any help would be great.

Thank you,

Ste
 
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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 going to 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.
 
Thanks for the reply. I hadnt considered the dimensionality of the problem to be honest, I've been setting constants to unity in my work, and I was going to restore them at the end. But yeah good point (its definitely a classical calculation so no worries about renormalization.)

I can't make my mind up on the independence of the fields, and whether I've lost a degree of freedom somewhere. I guess I should calculate the propagators, but I really don't want to unless I have to. I've never seen anyone redefine fields in a non linear way before, so I don't know. I can't find any relevant papers/books on the matter.

Thank you,

Ste
 
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 (?)
 
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 I've got cross derivative terms in the original variables A and B, i.e. d(A)d(B), which I don't 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 don't really want either. Hhhm.
 

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