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My questions are in italics. In the article [1] a dimensional regularization is presented on an electrostatic example of an infinite wire with constant linear charge density $\lambda$. It is shown that the direct computation of the scalar potential gives infinity: $$ \phi({\bf x}) = {\lambda\over 4\pi\epsilon_0}\int_{-\infty}^\infty { d l \over |{\bf x} - {\bf l}| } = {\lambda\over 4\pi\epsilon_0}\int_{-\infty}^\infty { d l \over (x^2 + y^2 + (z-l)^2)^{1\over 2} } = $$ $$ = {\lambda\over 4\pi\epsilon_0}\int_{-\infty}^\infty { d u \over \sqrt{x^2 + y^2 + u^2} } = \infty $$ But with dimensional regularization in the modified minimal subtraction scheme we get eventually: $$ \phi_{\overline{\rm MS}}({\bf x}) = {\lambda\over 4\pi\epsilon_0} \log{\Lambda^2\over x^2 + y^2} $$ where $\Lambda$ is the auxiliary scale parameter. One can then calculate the electric field (let's set $y=0$ from now on) as follows: $$ E_x = -{\partial \over \partial x} \phi_{\overline{\mathrm{MS}}}(x) = -{\partial \over \partial x} {\lambda\over 4\pi\epsilon_0} \log{\Lambda^2\over x^2}= $$ $$ = - {\lambda\over 4\pi\epsilon_0} {x^2\over\Lambda^2} \Lambda^2 \left(-{2\over x^3}\right) = {\lambda\over 2\pi\epsilon_0} {1\over x} $$

The article claims that the original scalar potential is scale invariant: $\phi(kx) = \phi(x)$. But since both $\phi(kx)$ and $\phi(x)$ are infinite, I don't understand the argument.

The article claims that the dimensional regularization preservers translational symmetry. However, the only way to make $\phi_{\overline{\mathrm{MS}}}(kx)=\phi_{\overline{\mathrm{MS}}}(x)$ is to choose different $\Lambda$ for each side. Are we allowed to do that?

I thought that we have to set $\Lambda$ once and for all and then just keep calculating with it and it must cancel at the end.

Update: based on Michael's comment below I realized that the article claims translational invariance of the original problem, i.e. $\phi_{\overline{\rm MS}}(x, y, z+h)=\phi_{\overline{\rm MS}}(x, y, z)$ and that is obviously true, because $\phi_{\overline{\rm MS}}({\bf x})$ does not depend on $z$. So I think that answers this particular question. Still a clarification from an expert would be nice.

[1] Olness, F., & Scalise, R. (2011). Regularization, renormalization, and dimensional analysis: Dimensional regularization meets freshman E&M. American Journal of Physics, 79(3), 306. doi:10.1119/1.3535586, available online here.

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First of all the symmetry you mention x -> kx is scale invariance, not translation invariance. Dim. reg. explicitly breaks scale invariance by introducing the scale $\mu$. Second, unlike cutoff reg., dim. reg. preserves translation invariance along the y-axis because the entire y-axis is integrated over, not just a segment of it. Third, I think you're right about having to choose one scale and stick with it (at least till the end of the calculation, and you can talk about renormalisation group transforms), though I'll wait for someone with more expertise than me to comment on that. –  Michael Brown Jan 1 '13 at 6:41
Michael, you are right! I've just updated the question with your remarks. –  Ondřej Čertík Jan 1 '13 at 6:59
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2 Answers

up vote 3 down vote accepted

First of all, the procedure is called dimensional regularization, not dimensional renormalization. Regularization is the process by which we make sums and integrals non-singular so that their results aren't infinite – the result "infinity" carries no meaningful physical information because the results of measurements in physics are always particular finite numbers.

After the regularization, the results of integrals are manageable finite expressions although they may still diverge in the limits we call "physical".

Renormalization is another step in which we carefully distinguish bare values of parameters (in the action) and the observed values, making sure that the theory with the appropriate values of the parameters agrees with the observations. Renormalization is something we would have to do even if the underlying integrals were convergent. It usually follows a regularization procedure but is independent of it.

Dimensional regularization is just a methodology to evaluate particular integrals – not mentioning what physical quantities or parameters are expressed by these integrals – so it's clearly a regularization technique, not renormalization technique. A broader technique to see these integrals in the loop diagrams and give them the right interpretations for the amplitudes may lead to $\overline{MS}$, em-es-bar, which is a renormalization scheme (a renormalization scheme is given by the choice of the renormalization scale as well as the exact definition of physically measurable quantities, i.e. scattering amplitudes of particles with certain energies, that play the role of the coupling constants for Taylor expansions etc.). But dim. reg. itself is just a regularization technique.

Now, the new parameter $\Lambda$ in dim. reg. is auxiliary, newly added, so we finally expect or want it to drop out of the physical expressions. Indeed, $\Lambda$ of dim. reg. also drops out of the final physical expressions although it's only after the full calculation of the physics quantities including the renormalization. In particular, various quantities linked to certain scales may depend on $\Lambda$, or the ratio $\mu/\Lambda$ involving a new auxiliary scale, but the observed/predicted cross sections are linked to other observed cross sections etc. by formulae that contain neither $\Lambda$ nor $\mu$.

Scale invariance

Below equation 7 of the paper you mentioned, they make it rather clear what they mean by the scale invariance. They mean that the expression, the integral in equation 7, is formally scale-invariant under $k\to kx$. If we just rescale $y\to ky$ as well, the integration variable, the factors of $k$ cancel between $dy$ and $1/\sqrt{x^2+y^2}$.

The integral itself is divergent but if we were satisfied with this unphysical answer $\infty$, it would be OK for the scale invariance because $\phi(x)=\phi(kx)=\infty$.

As they make it clear between equations 7 and 12, this scale invariance is subtle for divergent integrals because while we may say that $\infty=\infty$, it's still true that $\infty-\infty$ which appears in physically important quantities (work that is done) is an indeterminate form whose value may be any finite (or infinite) number. In particular, if you rescale $x,y$ by $k$, the $\overline{MS}$ renormalized expression changes additively by $(\lambda/4\pi\epsilon_0)\ln k$ with some sign. It's a purely additive shift that is independent of $x,y,z$ and such an additive shift may be undone by a simple $U(1)$ gauge transformation whose gauge parameter is something like $(\lambda/4\pi\epsilon_0)\ln(k)\cdot t$ i.e. linear in time (because we want to eliminate the temporal $A_0$ component).

So the renormalized value of the potential isn't quite scale-invariant because, as you correctly said – and it is easy to verify it by looking at the actual logarithmic expression for the potential – the value of $\Lambda$ would have to be rescaled as well. But if $\Lambda$ isn't rescaled, the change of the potential under $\vec x\to k\vec x$ is just a simple constant additive shift that is equivalent to a gauge transformation so it's still true that all gauge-invariant quantities that may be calculated out of such a potential (and physical, measurable, observable quantities have to be gauge-invariant) are scale-invariant. More precisely, they are "covariant" and get rescaled by the right power $k^\Delta$ where $\Delta$ refers to their dimension. That's why the electric field goes like $1/x$ etc.

For this reason, one may use a bit sloppy language and say that the potential itself is scale-invariant. But the extent to which this statement is true for the potential in various forms – formal expression for the integral, the naive result of the integral, or the result in a renormalization scheme – depends on the details in the way sketched above.

On the other hand, as you wrote, the translational invariance in the direction along the wire is uncontroversial, manifest, and protected by all forms of the potential or the field strength.

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Hi Lubos Motl, I'm editing dimensional renormalization -> dimensional regularization in the question formulation to increase usefulness for future readers. –  Qmechanic Jan 1 '13 at 15:40
Thank you Luboš for your time and your reply. I will think about it for some time to fully understand it. Thanks Qmechanic for the update. –  Ondřej Čertík Jan 2 '13 at 1:34
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I hardly count as an expert, but I can probably shed some light on this.

When the authors claim that the original potential is scale invariant, they do so on the basis of the formula being scale invariant, not just the result. They show their proof on page 307:

$$\begin{align}V(kx) &= \frac{\lambda}{4\pi\epsilon_0}\int_{-\infty}^{\infty}\mathrm{d}y\frac{1}{\sqrt{(kx)^2 + y^2}} \\ &= \frac{\lambda}{4\pi\epsilon_0}\int_{-\infty}^{\infty}\mathrm{d}\biggl(\frac{y}{k}\biggr)\frac{1}{\sqrt{x + \bigl(\frac{y}{k}\bigr)^2}} \\ &= \frac{\lambda}{4\pi\epsilon_0}\int_{-\infty}^{\infty}\mathrm{d}z\frac{1}{\sqrt{x^2 + z^2}} = V(x)\end{align}$$

This works because all lengths in the problem, in particular the limits of integration, are unchanged under the rescaling: when they change variables from $\frac{y}{k}$ to $z$, the limits change as $\pm\infty\to\frac{\pm\infty}{k}$. However, if you attempt to fix the infiniteness of the potential using cutoff regularization, by chopping off the integral at some finite values $\pm L$, that is no longer true. The rescaling converts the limits $\pm L\to\frac{\pm L}{k}$, which means you don't get the original formula back.

In the case of the dimensional regularization, you're right that you have to be consistent with your choice of $\Lambda$ (their $\mu$), and if you choose the same value, you will find that $\phi_{\overline{\text{MS}}}(kx) \neq \phi_{\overline{\text{MS}}}(x)$. The point they're making is that it doesn't matter. All that matters, since $\phi$ is a potential, is that the difference between potentials at different locations is scale invariant:

$$\phi_{\overline{\text{MS}}}(kx) - \phi_{\overline{\text{MS}}}(ky) = \phi_{\overline{\text{MS}}}(x) - \phi_{\overline{\text{MS}}}(y)$$

And it is. In other words, you can explain away the difference between $\phi_{\overline{\text{MS}}}(kx)$ and $\phi_{\overline{\text{MS}}}(x)$ as the addition of a constant value to the potential, which has no physical significance.

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Your answer is very clear, thanks David! –  Ondřej Čertík Jan 2 '13 at 1:35
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