Dilatations and action on derivatives of fields

Question

How do derivatives of fields transform under dilatations?

Specifically I am interested on what I misunderstand with the example:

Consider a theory that has a field $A_\mu$ that transforms under dilatation as $x\to \lambda x$ and $A_\mu(x)\to \lambda^\Delta A_\mu(\lambda x)$. The infinitesimal form of this is $$A_\mu\to A_\mu +\epsilon (x_\sigma\partial^\sigma+\Delta)A_\mu.$$ Now this I find by just setting $\lambda=1+\epsilon$ and expanding to lowest order in $\epsilon$.

Now in https://arxiv.org/abs/1101.4886 equation (20) it is claimed that then also infinitesimally $$F_{\mu\nu}\to F_{\mu \nu}+\epsilon (x_\mu \partial^\mu+\Delta+1)F_{\mu \nu}, \qquad \Delta=(D-2)/2.\tag{20}$$

However I get a different result.

Using $\partial_\mu\to \frac{1}{\lambda}\partial_\mu$ and then expanding $\frac{1}{\lambda}\partial_\nu \lambda^\Delta A_\mu(\lambda x)$ with $\lambda =1+\epsilon$ to lowest order in $\epsilon$ I find $$F_{\mu\nu}\to F_{\mu \nu}+\epsilon (x_\mu \partial^\mu+\Delta)F_{\mu \nu}$$
the $1/\lambda$ cancels the additional one they get.

What am I doing wrong?

Answer 1

By using: $$\frac{1}{1+\epsilon}\simeq 1-\epsilon$$ and $$(1+\epsilon)^\Delta\simeq 1+\Delta\epsilon$$ and $$A_\mu(x+\epsilon x)\simeq A_\mu(x)+\epsilon x^\sigma\partial_\sigma A_\mu(x)$$ We find: $$\frac{1}{\lambda}\partial_\nu\lambda^\Delta A_\mu(\lambda x)\simeq(1-\epsilon)(1+\Delta\epsilon)\partial_\nu(A_\mu+\epsilon x^\sigma\partial_\sigma A_\mu)\simeq\partial_\nu A_\mu-\epsilon\partial_\nu A_\mu+\Delta\epsilon\partial_\nu A_\mu+\epsilon\partial_\nu(x^\sigma\partial_\sigma A_\mu)=\partial_\nu A_\mu-\epsilon\partial_\nu A_\mu+\Delta\epsilon\partial_\nu A_\mu+\epsilon\delta_\nu^\sigma\partial_\sigma A_\mu+\epsilon x^\sigma\partial_\sigma\partial_\nu A_\mu=\partial_\nu A_\mu-\epsilon\partial_\nu A_\mu+\Delta\epsilon\partial_\nu A_\mu+\epsilon\partial_\nu A_\mu+\epsilon x^\sigma\partial_\sigma\partial_\nu A_\mu=\partial_\nu A_\mu+\Delta\epsilon\partial_\nu A_\mu+\epsilon x^\sigma\partial_\sigma\partial_\nu A_\mu$$ Now by considering: $$F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu$$ We find: $$F_{\mu\nu}\to\partial_\mu A_\nu+\Delta\epsilon\partial_\mu A_\nu+\epsilon x^\sigma\partial_\sigma\partial_\mu A_\nu-(\partial_\nu A_\mu+\Delta\epsilon\partial_\nu A_\mu+\epsilon x^\sigma\partial_\sigma\partial_\nu A_\mu)=F_{\mu\nu}+\epsilon(\Delta+x^\sigma\partial_\sigma)F_{\mu\nu}$$

I find precisely what you find and I think they have corrected the paper since now it says it right!