4
$\begingroup$

My question is somehow related to: The relation between critical surface and the (renormalization) fixed point but there is another problem:

The problem is that if we accept that all points on the critical surface are critical in the manner that their corresponding correlation length is infinite, then according to the scaling hypothesis a system whose parameters lie on the critical surface should be scale invariant. And therefore its parameters shouldn't change under the RG transformation. So each point on the critical surface should be a fixed point and hence there is not any RG flow over the critical surface. It implies that the RG transformation doesn't push any point on the critical surface to our first fixed point.

From the above statements one can deduce that the RG method is inconsistent. Surely there should be something which I'm missing but what's that?

$\endgroup$
9
  • 3
    $\begingroup$ There are always higher order corrections. So, you have a leading scaling behaviour, this is the same for all the models described by points on the critical surface, but they differ from each other due to so-called "irrelevant operators". These are terms that flow to zero under the RG action, they give rise to corrections to the leading scaling behavior. $\endgroup$ Aug 3, 2016 at 20:25
  • $\begingroup$ @CountIblis Regardless of the order of correction, if there are some irrelevant directions, then there exist corresponding paths in the parameter space that which cross the fixed point, so the problem should not be related to approximations. And I don't understand what do you mean in rest or your statements. $\endgroup$
    – Hossein
    Aug 3, 2016 at 20:37
  • 1
    $\begingroup$ Note that you need an infinite number of RG steps before you arrive at the exact critical point. So, the scale invariant behavior you see in a model that is described by a parameter on the critical surface that is not a critical point, only refers to phenomena that exist when you have zoomed out by an infinite amount. If you are not on the critical surface, then zooming out by an infinite amount only leads to a trivial scale invariant behavior (correlation lenght = 0). Obviously, the correlation length has to be infinite for an infinite rescaling to still yield an infinite correlation length. $\endgroup$ Aug 3, 2016 at 20:44
  • $\begingroup$ @CountIblis So you are saying that there is a singularity in the correlation length as a function of parameters. Because for a point on the critical surface the effect of RG is to push it toward the fixed point, and we know that RG reduces the correlation length by the scaling factor. So the more a point on the critical surface is close to the fixed point, the shorter correlation length it has, so for points arbitrary close to the critical point the correlation length goes to zero and then suddenly it jumps to infinite exactly at the fixed point. It seems really strange. $\endgroup$
    – Hossein
    Aug 3, 2016 at 20:58
  • $\begingroup$ Continue: But in Kardar's book he claims that the correlation length on the basin of attraction is infinite not zero. $\endgroup$
    – Hossein
    Aug 3, 2016 at 21:00

1 Answer 1

1
$\begingroup$

I'll attempt to give an answer to this old post.

I think the statement:

[...] a system whose parameters lie on the critical surface should be scale-invariant. And therefore its parameters shouldn't change under the RG transformation.

Is wrong. Infinite correlation length does not imply being on a fixed point of the RG transformation, it just means that we are on a critical manifold of some critical fixed point. If a system has a divergent correlation length then yes, we expect it to be scale-invariant, but that does not mean that the parameters don't change. It means rather that, by coarse-graining and rescaling, the symmetries of the system are preserved and the free energy doesn't change. Only the fixed points have, by definition, the property of being invariant under the RG transformation.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.