# Sean Carroll's claim on renormalisation?

So I'm reading Sean Carroll's old blog post:

But to a modern physicist, this seems like a misguided quest. First, because renormalization theory teaches us that $$\alpha$$ isn’t really a number at all; it’s a function. In particular, it’s a function of the total amount of momentum involved in the interaction you are considering. Essentially, the strength of electromagnetism is slightly different for processes happening at different energies. Atiyah isn’t even trying to derive a function, just a number.

This is basically the objection given by Sabine Hossenfelder. But to be as charitable as possible, I don’t think it’s absolutely a knock-down objection. There is a limit we can take as the momentum goes to zero, at which point $$\alpha$$ is a single number.

I'm kind of rusty in QFT but can someone give me a heuristic reasoning to the lines:

it’s a function of the total amount of momentum involved in the interaction you are considering

And

There is a limit we can take as the momentum goes to zero, at which point $$\alpha$$ is a single number.

• Wikipedia? Oct 25, 2020 at 18:15
• For your second question, are you asking what a limit is? Oct 25, 2020 at 19:29
• More basic WP. Oct 26, 2020 at 13:44
• @G. Smith I'm asking what makes the limit when momentum goes to $0$ special? Oct 27, 2020 at 2:50
• That’s the limit that applies to the world around us, such as the value of $\alpha$ relevant to atomic structure. Oct 27, 2020 at 3:42

In QFT (quantum field theory) the interactions between particles are described perturbatively as successive contributions of terms at different orders in the coupling between the underlying fields. Higher order corrections to the tree level means loops, which are infinite. To overcome this difficulty, in QED (quantum electrodynamics) the bare coupling $$e$$ is renormalized to an observed value $$e_R$$ at some momentum $$Q$$ of the interaction. If we express the effective electric charge vs. the momentum, we have
$$e^2_{eff} (Q) = \frac{e^2_R}{1 - \frac{e^2_R}{12 \pi^2} ln \frac{Q^2}{m^2}}$$
Here the renormalized coupling is defined as $$e_R = e_{eff} (m)$$, i.e. with $$Q = m$$.
Note: $$\alpha = \frac{e^2}{4 \pi}$$
If the renormalization is defined at $$Q \to 0$$, the renormalized coupling $$e_R$$ measures the electric charge of the Coulomb interaction at long distance, in principle at infinity.