Hello, I was hoping someone would be able to clarify a problem I've got. A lagrange multiplier can be introduced into an action to impose a constraint right?
I was wondering what relation lagrange multipliers have to gauge conditions, which are imposed by hand. Am I correct in saying that...
hhhmm,
ok, thanks for the responses. It turns out that my particular function is not invertible for an entirely different reason. However thank you all.
I was confused because I started with the domain (-infty,+infty), and 'removed' (-infty,0), i.e. an open interval at 0. So I was thinking...
yes sorry. By my horrible notation I meant (for example) (-2,2) -> 4. So if we take (0,+infty) then we have only (2) -> 4 (a bijection) from the domain.
But if we keep x=0, we still have (0,0) -> (0) don't we? Its still a double zero, so there is no bijective map from x=0 to its image y=0. I...
Hi, just a quick question concerning the invertibility of multivalued functions. Specifically, I am looking at
y=x^2
as a simple example. So for an inverse to exist we have to restrict the domain to (0,+infty) right (doesnt matter which branch I am taking, so Ill take this one.)
Now my...
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...
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...
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...