From a circular loop to a defect observable

A circular Wilson loop gives one number for each coupling. How can its derivative measure the response of a moving heavy probe? The missing link is a local operator on the line: the displacement operator, which records how the defect responds when its contour moves.

Use the finite-color circular loop, what protected means, and observable normalization. The theory is massless four-dimensional $\mathcal N=4$ SYM in its conformal vacuum, with sixteen real Poincaré and sixteen real conformal supercharges. The straight half-BPS line preserves half of each family. Its circular conformal image preserves the corresponding combinations; a generic bent contour need not preserve them.

Our physical target is the fundamental line of $SU(N)$, $N\geq2$, with gauge rank $N-1$. Keep $\lambda=g_{\rm YM}^2N$ and the normalized trace $\operatorname{Tr}/N$. We will explicitly label the auxiliary $U(N)$ Gaussian answer whenever it is used.

A local operator on a line

The scalar-coupled Euclidean line has the inherited insertion

$$W[C,n]=\frac1N\operatorname{Tr}\mathcal P \exp\!\left(i\int_C a_\mu dx^\mu+ \int_C |dx|\,n^I\Phi_I\right),\qquad n^In^I=1.$$

The scalar direction $n$ is constant for the straight half-BPS line and its ordinary circular image. The connection and scalar normalization are those used to obtain the Gaussian integral; changing the scalar coupling changes the observable.

Parametrize a straight line by its length coordinate $s$. A small perpendicular displacement $\xi^i(s)$, $i=1,2,3$, defines an insertion $\mathbb D_i(s)$ through

$$\delta W=\int ds\,\xi^i(s)\,W[\mathbb D_i(s)]. \tag{N20.1}$$

The brackets mean an insertion inside the path-ordered line, with its parallel transport included. They do not denote a separate gauge-invariant bulk operator. Define normalized defect correlations by $\langle\!\langle\mathcal O_1\mathcal O_2\rangle\!\rangle =\langle W[\mathcal O_1\mathcal O_2]\rangle/\langle W\rangle$.

Since $ds$ and $\xi$ each have length dimension one, (N20.1) fixes $\mathbb D$ to have scaling dimension two. Translation along the straight line, its conformal symmetry, and transverse rotations then fix the separated-point two-point function to

$$\langle\!\langle\mathbb D_i(s)\mathbb D_j(0)\rangle\!\rangle =\frac{C_D(\lambda,N)\delta_{ij}}{s^4},\qquad s\ne0. \tag{N20.2}$$

Reflection positivity gives $C_D\geq0$, with a positive coefficient for a nontrivial displacement operator. The normalization of $\mathbb D$ is fixed by a physical displacement, not by an arbitrary rescaling. Its dimension is protected; its coefficient can vary with $\lambda$. This is precisely the distinction discussed in CHMS, section 4.1.

Fix the cusp convention

Let $\varphi$ be the small angle through which the tangent turns at a Euclidean cusp; $\varphi=0$ is a straight line. Keep the same scalar direction on both segments initially. After the straight-line subtractions, define the logarithmic cusp response by

$$\langle W_{\rm cusp}\rangle\sim \exp[-\Gamma_{\rm cusp}(\varphi,0)\log(L/\varepsilon)], \qquad \Gamma_{\rm cusp}(\varphi,0)=-B\varphi^2+O(\varphi^4). \tag{N20.3}$$

$L$ and $\varepsilon$ are infrared and ultraviolet length cutoffs. $B$ is dimensionless and is called the Bremsstrahlung function. The negative Euclidean sign is part of this convention. It is not the large-boost, lightlike cusp coefficient. Continuing $\varphi=i\chi$ reverses the quadratic sign, but does not turn a small-angle expansion into a large-$\chi$ result.

One can also rotate the scalar direction across the cusp by an internal angle $\vartheta$, with $n_{\rm left}\cdot n_{\rm right}=\cos\vartheta$. The generalized cusp is BPS at $\vartheta=\pm\varphi$. Near the straight configuration, reflection symmetries and this BPS condition give

$$\Gamma_{\rm cusp}(\varphi,\vartheta) =B(\vartheta^2-\varphi^2) +O(\varphi^4,\varphi^2\vartheta^2,\vartheta^4). \tag{N20.4}$$

This specifies the conventions of CHMS (1), (6), (9) and (62).

Why a coupling derivative appears

The required extra input is a supersymmetric family of loops, not dimensional analysis. On a unit spatial circle let its scalar profile be

$$n_\alpha(\tau)=(\sin\alpha\cos\tau,\sin\alpha\sin\tau, \cos\alpha,0,0,0).$$

The latitude relation used in CHMS (10)–(12) states

$$\langle W_\alpha\rangle(\lambda) =W_\circ(\lambda\cos^2\alpha). \tag{N20.5}$$

We take this supersymmetric localization input in its fixed loop prescription; we are not proving a new localization theorem. In particular, the source’s footnote 2 records a qualification about the fluctuation-determinant argument supporting this input. The response calculation below shows how the stated relation leads to $B$.

Expanding its right side gives

$$\frac{\langle W_\alpha\rangle-W_\circ}{W_\circ} =-\alpha^2\lambda\partial_\lambda\log W_\circ+O(\alpha^4). \tag{N20.6}$$

The logarithm comes from dividing the variation by the original loop. Omitting it would compute a different, unnormalized response.

The same variation inserts the two transverse scalars in the profile. Write their straight-line correlation as $\gamma\delta_{ab}/s^2$, $a,b=1,2$. Mapping the dimension-one insertions to the unit circle gives $\gamma\delta_{ab}/[2(1-\cos(\tau-\tau'))]$. The term from varying $n_3$ is a one-point insertion and vanishes in the inherited conformal prescription. The quadratic coefficient is therefore

$$I_\varepsilon= \frac{\pi\gamma}{2}\int_\varepsilon^{2\pi-\varepsilon} \frac{\cos u}{1-\cos u}\,du =\pi\gamma\left[\cot\frac\varepsilon2-\pi+\varepsilon\right]. \tag{N20.7}$$

Here the factor $1/2$ comes from the second-order expansion, the integral over one circle position gives $2\pi$, and the scalar propagator supplies another $1/2$. Subtracting the power divergence $2\pi\gamma/\varepsilon$ leaves $I_{\rm finite}=-\pi^2\gamma$. Comparison with (N20.6) yields $\gamma=\lambda\partial_\lambda\log W_\circ/\pi^2$.

To connect this response to the internal cusp angle, map the straight line to cylinder time $u$. The dimension-one correlation becomes $\gamma/[2(\cosh u-1)]$. The energy response is minus one half its integral, because the cylinder amplitude has the form $e^{-\Gamma_{\rm cusp}T}$. Thus

$$I_c=-\frac\gamma4\int_{|u|>\varepsilon} \frac{du}{\cosh u-1} =-\frac\gamma2\left[\coth\frac\varepsilon2-1\right] =-\frac\gamma\varepsilon+\frac\gamma2+O(\varepsilon).$$

Keeping the finite response gives $\Gamma_{\rm cusp}(0,\vartheta)=\gamma\vartheta^2/2+O(\vartheta^4)$. Consequently

$$\boxed{B(\lambda,N)=\frac{1}{2\pi^2} \lambda\partial_\lambda\log W_\circ(\lambda,N).} \tag{N20.8}$$

This restores the intermediate steps of CHMS (13)–(24). The finite parts belong to the specified conformal defect prescription; they do not authorize an arbitrary finite redefinition of the line. Use the logarithm in the source’s (2) and (20), and the cusp sign in (1) and (62): the displayed (25) omits that logarithm, and (54) has the opposite sign to those definitions.

Extract the answer at finite color number

The known normalized circle is

$$W_U=\frac1N e^{\lambda/(8N)}P_N(\lambda),\qquad P_N(\lambda)=L_{N-1}^{(1)}\!\left(-\frac{\lambda}{4N}\right), \qquad W_{SU}=e^{-\lambda/(8N^2)}W_U.$$

Here $P_N$ is the positive-coefficient polynomial derived in the prerequisite. Differentiating at fixed $N$ gives

$$B_U=\frac{\lambda}{2\pi^2} \left(\frac1{8N}+\frac{P_N'}{P_N}\right), \qquad \boxed{B_{SU}=B_U-\frac{\lambda}{16\pi^2N^2}.} \tag{N20.9}$$

The subtraction is the central $U(1)$ contribution with its inherited fundamental charge. It does not change $\lambda$ or assert a factorization of every global sector of $U(N)$. At $N=1$, the Abelian circle gives $B_{U(1)}=\lambda/(16\pi^2)$; the formal traceless answer vanishes.

For a visible weak-coupling check, use the finite-color circle expansion

$$W_U=1+\frac\lambda8+ \frac{2+N^{-2}}{384}\lambda^2+ \frac{1+2N^{-2}}{9216}\lambda^3+O(\lambda^4).$$

For $W=1+a\lambda+b\lambda^2+c\lambda^3+\cdots$, $\log W=a\lambda+(b-a^2/2)\lambda^2+(c-ab+a^3/3)\lambda^3+\cdots$. Applying this before differentiating gives

$$\begin{aligned} B_U&=\frac\lambda{16\pi^2} -(1-N^{-2})\frac{\lambda^2}{384\pi^2} +(1-N^{-2})\frac{\lambda^3}{6144\pi^2}+O(\lambda^4),\\ B_{SU}&=(1-N^{-2})\left[ \frac\lambda{16\pi^2}-\frac{\lambda^2}{384\pi^2} +\frac{\lambda^3}{6144\pi^2}\right]+O(\lambda^4). \end{aligned} \tag{N20.10}$$

The common factor displayed through cubic order is not an all-orders assertion. Setting $N\to\infty$ reproduces the first three terms in CHMS (26).

For a complete finite-color example, $W_{SU(2)}=(1+\lambda/16)e^{\lambda/32}$ gives

$$B_{SU(2)}=\frac{\lambda}{2\pi^2} \left(\frac1{32}+\frac1{16+\lambda}\right)>0 \quad(\lambda>0). \tag{N20.11}$$

At $\lambda=4$ this is $13/(80\pi^2)$. Positivity and the zero-coupling limit are immediate. At fixed $N=2$, its large-$\lambda$ growth is linear. This differs from the planar result $B_0=\sqrt\lambda\,I_2(\sqrt\lambda)/[4\pi^2I_1(\sqrt\lambda)]$, obtained by differentiating $W_0=2I_1(\sqrt\lambda)/\sqrt\lambda$. Its leading strong-coupling behavior is $\sqrt\lambda/(4\pi^2)$. The planar lesson explains the ordered limits; neither planar nor strong coupling was used in (N20.9)–(N20.11).

The displacement and radiation normalizations

The displacement correlation on the cylinder is $C_D/[4(\cosh u-1)^2]$. Repeating the energy-response argument for the geometric angle gives

$$\Gamma_{\rm cusp}(\varphi,0) =-\frac{\varphi^2 C_D}{2}\, \operatorname{FP}\int_{-\infty}^{\infty} \frac{du}{4(\cosh u-1)^2}+O(\varphi^4).$$

$\operatorname{FP}$ means the finite part after the specified power subtractions. With $c=\coth(\varepsilon/2)$, the cut integral is

$$\int_{|u|>\varepsilon}\frac{du}{4(\cosh u-1)^2} =\frac{c^3}{12}-\frac c4+\frac16 =\frac{2}{3\varepsilon^3}-\frac{1}{3\varepsilon} +\frac16+O(\varepsilon).$$

The finite $1/6$ fixes the factor in CHMS (61)–(62):

$$\boxed{C_D=12B.} \tag{N20.12}$$

There is an independent weak-coupling normalization check. With scalar and gauge propagators normalized as $g^2/(4\pi^2x^2)$, the line’s gauge-force contribution at separated points is $g^2C_F\delta_{ij}/(\pi^2s^4)$, and its scalar-gradient contribution is $g^2C_F\delta_{ij}/(2\pi^2s^4)$. Here $C_F=(N^2-1)/(2N)$ for $SU(N)$ and $g=g_{\rm YM}$. Their sum gives $C_D=3g^2C_F/(2\pi^2)$, precisely twelve times the leading term of (N20.10). The derivative and Euclidean gauge-field factors are both needed for that positive result.

In Lorentzian signature, the ordered displacement correlation has Fourier transform

$$\int dt\,e^{i\omega t}\frac{C_D}{(t-i0)^4} =\frac{\pi C_D}{3}\omega^3\Theta(\omega). \tag{N20.13}$$

For example, closing the contour in the upper half-plane for $\omega>0$ picks the third derivative of $e^{i\omega t}$ at the pole; for $\omega<0$ no pole contributes. This nonnegative spectral weight describes energy absorbed from a small prescribed displacement. For $\xi(t)=\eta e^{-i\omega t}+\bar\eta e^{i\omega t}$ acting for a long time $T$, absorption probability per time is $\pi C_D|\eta|^2\omega^3/3$. Multiplying by $\omega$ gives energy, while $T^{-1}\int dt\,\ddot\xi^2=2|\eta|^2\omega^4$. Hence

$$\boxed{\Delta E=\frac{\pi C_D}{6}\int dt\,|\dot{\boldsymbol v}|^2 =2\pi B\int dt\,|\dot{\boldsymbol v}|^2.} \tag{N20.14}$$

This is the small-velocity, prescribed infinitely heavy probe result of CHMS (63)–(67). It is not a dynamical quark recoil calculation or an exact formula at arbitrary acceleration and velocity without qualification. The acceleration-squared integral has energy dimension one; $B$ is dimensionless.

Exercises

1. A harmless-looking normalization change

Replace the circular answer by $W'_\circ=e^{c\lambda}W_\circ$, with constant real $c$. Both satisfy $W(0)=1$. Calculate the apparent shift in $B$ and $C_D$ if one blindly applies (N20.8). Does a source-only bulk counterterm have the same effect?

Hint 1

Take the logarithm before the coupling derivative.

Hint 2

Distinguish redefining the line itself from multiplying both the numerator and denominator of its normalized expectation by the same factor.

Solution

The naive results shift by $\delta B=c\lambda/(2\pi^2)$ and $\delta C_D=6c\lambda/\pi^2$. They would even change the leading weak-coupling coefficient. The identity $W(0)=1$ does not exclude this redefinition.

However, the sourced latitude identity and the physical displacement normalization were established for the inherited line prescription. A coupling-dependent finite multiplier is not permission to change its measured separated-point coefficient; applying the same formula to a redefined circular number without checking the full defect prescription is invalid. A source-only bulk counterterm cancels between the normalized numerator and denominator and changes neither $W_\circ$ nor this derivative. A coupling-independent trace multiplier likewise drops out of $\partial_\lambda\log W$.

2. Change the probe history

Let a heavy probe make the small transverse excursion $\xi(t)=A e^{-t^2/(2T^2)}$, with $A/T\ll1$. Find its leading emitted energy in terms of $B$, and then for the $SU(2)$ theory at $\lambda=4$. What happens if the same excursion is performed twice as slowly?

Hint 1

$\ddot\xi=A(t^2/T^4-1/T^2)e^{-t^2/(2T^2)}$. Rescale $u=t/T$ before integrating.

Hint 2

Use the Gaussian moments of $1,u^2,u^4$ to integrate $(u^2-1)^2e^{-u^2}$.

Solution

The integral is

$$\int dt\,\ddot\xi^2 =\frac{A^2}{T^3}\int du\,(u^2-1)^2e^{-u^2} =\frac{3\sqrt\pi A^2}{4T^3}.$$

Thus $\Delta E=3\pi^{3/2}BA^2/(2T^3)$ to leading order in the small velocity. At the stated coupling and color number, $B=13/(80\pi^2)$, so $\Delta E=39A^2/(160\sqrt\pi T^3)$. The result is positive and has dimension inverse length. Doubling $T$ at fixed $A$ reduces the leading energy by eight and further improves the small-velocity condition. The approximation concerns the trajectory; the value of $B$ used here is the exact finite-color coupling function.

3. Extract a coefficient with the correct sign

For the same $SU(2)$ theory at $\lambda=4$, find the quadratic cusp response for (a) an internal angle $\vartheta$ on a straight contour, (b) a geometric angle $\varphi$ with no internal turn, and (c) equal small angles. Evaluate $C_D$. Explain why none of these quadratic results determines the lightlike cusp coefficient.

Hint 1

Use $\Gamma_{\rm cusp}=B(\vartheta^2-\varphi^2)$, rather than assigning a positive sign to both Euclidean responses.

Hint 2

The displacement norm is positive even though the geometric Euclidean cusp coefficient is negative. The lightlike quantity involves a different, large Lorentzian angle limit.

Solution

The leading terms are respectively $13\vartheta^2/(80\pi^2)$, $-13\varphi^2/(80\pi^2)$, and zero. The equal-angle configuration lies on the BPS locus; the full generalized cusp vanishes there, not merely its quadratic expansion. Meanwhile $C_D=12B=39/(20\pi^2)>0$.

Continuing the second result to $\varphi=i\chi$ gives $+B\chi^2$ near $\chi=0$. The lightlike coefficient instead controls the asymptotic growth as $\chi\to\infty$. Analytic continuation does not extend the domain of a truncated Taylor series. The small-angle response, the separated displacement norm and the slow-probe radiation coefficient agree through the derived normalizations, while the large-angle function requires additional information.

The circular matrix integral now determines a local, normalized piece of defect data. The protected dimension fixes the power $s^{-4}$; the coupling derivative supplies its nontrivial coefficient and, with the stated kinematics, the radiation response.

For a finite-angle continuation, extract the cusp slope with a specified source profile. The calculation keeps the actual deformation weight and tests when a smoothing bias can imitate new defect data.