research29 min studyLesson 682 in Physics

Finite subtraction conditions define different useful coordinates

Compare stable-particle pole conditions with MS and modified MS without treating scheme parameters as identical numbers.

On this page

What you’ll understand

  • Impose pole-position and residue conditions for a stable scalar.
  • Distinguish MS from modified MS in an explicit dimensional convention.
  • Convert parameter meanings before comparing physical predictions.

Removing a divergent pole does not fix every finite local term. An on-shell prescription chooses mass and field normalization through the propagator near a stable-particle pole. Minimal subtraction instead removes regulator poles, leaving finite parts in amplitudes and running parameters. These are alternative descriptions of the same predictions when their parameters are translated consistently. We will compare the conditions and avoid identifying a scale-dependent MS mass numerically with a pole mass without a conversion.

A stable pole supplies an on-shell renormalization condition

Use a scalar propagator with inverse denominator s−m²−ΣR(s), where s is Minkowski momentum squared and ΣR is the renormalized self-energy. In an on-shell scheme choose m=M, the stable-particle pole mass, and require both ΣR(M²)=0 and its first derivative at the pole to vanish. The first condition places the pole correctly; the second gives unit residue in the chosen renormalized field. The residue is a normalization convention for that field rather than an additional directly measurable scattering probability.

These elementary conditions assume a real isolated stable pole and sufficiently regular self-energy near it. Unstable particles require a complex-pole treatment and care with gauge invariance. Massless gauge theories can have infrared singularities in on-shell quantities, while colored quark pole masses have a nonperturbative limitation discussed later. Thus an on-shell prescription is useful in an appropriate setting, but it is not a universal assertion that every field corresponds to a stable asymptotic particle with a precisely observable real pole.

DR−1(s)=s−m2−ΣR(s),m=M,ΣR(M2)=0,ΣR′(M2)=0D_R^{-1}(s)=s-m^2-\Sigma_R(s),\qquad m=M,\qquad \Sigma_R(M^2)=0,\qquad \Sigma_R'(M^2)=0
s, m², M², and ΣR have energy-squared units; the derivative is with respect to s and is dimensionless. Overall factors of i in the propagator are suppressed. The displayed pole conditions concern a stable scalar with a simple pole.↔ Scroll wide equations horizontally.

MS removes poles while modified MS also removes universal constants

With d=4−ε, a typical one-loop scalar integral contains the combination 2/ε−γE+ln4π plus finite logarithms and constants. MS subtracts only the regulator pole. Modified MS subtracts that pole together with the universal Euler-constant and logarithm terms. Equivalently, these prescriptions can be related by a reference-scale convention. The precise factor of two changes when a reference instead defines d=4−2ε, so a quoted subtraction formula is incomplete unless the dimensional convention is stated.

Neither minimal scheme sets all finite self-energy values or derivatives to zero at a particle pole. A finite conversion therefore relates its mass to an on-shell mass. Likewise its field normalization need not give unit residue without the appropriate external-leg factors. Keeping the same numerical mass and coupling in two schemes generally compares two differently specified theories at that accuracy. A valid comparison first matches the same physical inputs, then transforms all parameters and perturbative coefficients through a consistent order.

ΔMS=2ϵ,ΔMS‾=2ϵ−γE+ln⁡(4π),d=4−ϵ\Delta_{\mathrm{MS}}=\frac2\epsilon,\qquad \Delta_{\overline{\mathrm{MS}}}=\frac2\epsilon-\gamma_E+\ln(4\pi),\qquad d=4-\epsilon
These dimensionless combinations describe a common one-loop integral normalization, with γE Euler’s constant. They specify what is subtracted from that integral, not a universal coefficient for every interaction or diagram. Finite momentum-dependent parts are not arbitrary local counterterms.↔ Scroll wide equations horizontally.

A subtraction point controls the organization of an approximation

Minimal schemes often simplify high-energy calculations because their renormalization constants do not depend on masses. On-shell inputs can be convenient when measured stable masses and low-energy charges are natural reference quantities. Both choices leave the exact physical answer unchanged, but their finite-order series can distribute corrections differently. Very large logarithms of momentum over mass can make one organization inconvenient. Choosing a subtraction scale near a relevant hard scale reduces those logarithms without declaring that the scale itself is observable.

The function lab illustrates why a local value subtraction and a local slope subtraction are separate conditions. Reset and set both controls to a=1 and x=0, predicting y=0 and slope zero. Reset and set a=1 and x=1, predicting y=1 and slope two. A tangent subtraction at the chosen point removes a value and a linear variation, analogous only algebraically to pole and residue conditions. The plotted parabola is not a self-energy, and its position control is not a physical momentum-shell prescription.

Find the finite terms selected by pole conditions

Near a specified stable pole s=M²=9 GeV², suppose a regulated self-energy is 2 GeV²/ε+3 GeV²+0.10(s−M²). Define ΣR=Σloop+c₀+c₁(s−M²). Find c₀ and c₁ for on-shell conditions, and compare with subtracting the pole alone.

  1. Evaluate at the pole: Σloop(M²)=2 GeV²/ε+3 GeV². The on-shell value condition therefore requires c₀=−2 GeV²/ε−3 GeV².

  2. The loop slope is 0.10. The unit-residue condition requires c₁=−0.10, so the renormalized slope also vanishes.

  3. Pole-only subtraction would choose c₀=−2 GeV²/ε and c₁=0 in this example, leaving the finite value three and slope 0.10. Matching the same physics then requires finite mass and field conversions.

The result

The on-shell counterterms include c₀=−2 GeV²/ε−3 GeV² and c₁=−0.10; a pole-only prescription leaves the displayed finite terms in the self-energy.

Explore the model

Interactive lab

Interactive lab · function

Distinguish value and derivative conditions

Reset and set BOTH controls to a=1 and x=0, predicting y=0 and slope 0. Reset and set a=1 and x=1, predicting y=1 and slope 2. The tangent illustrates two local Taylor data; it is not a loop self-energy or a subtraction-scheme calculator.

Distinguish value and derivative conditionsThis lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Current readings: y at chosen x: 0.608; Tangent slope · 2ax: 1.35; Signed area · ax³/3: 0.182. All parameters are adjustable with the labeled controls below.−2.4−1.201.22.4−9−4.504.59xy(x, ax²)
y = ax²TangentSigned integral
y at chosen x
0.608
Tangent slope · 2ax
1.35
Signed area · ax³/3
0.182
t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forRemoving a local value and fixing a local slope are distinct requirements.

Check your understanding

What must be done before comparing an on-shell calculation with a modified-MS calculation for the same physics?

Try it yourself

At μ=m, suppose a model gives M²=m²(1+κA)+O(κ²), with modified-MS m=10 GeV, κ=0.02, and A=3. Find M² through first order and M expanded through first order. Explain why simply setting M=m misses a finite conversion.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Expand the square root consistently: √(1+x)=1+x/2+O(x²).

Compare with the solution

The first-order squared mass is M²=100×1.06=106 GeV². Expanding the mass itself gives M=10×(1+0.06/2)=10.3 GeV+O(κ²). Taking the unexpanded square root gives about 10.296 GeV, whose difference is beyond this order. The finite six-percent squared-mass shift remains even at μ=m, where the explicit logarithm vanishes.

Take it with you

  • Pole position and residue define distinct stable-particle conditions.
  • MS and modified MS differ in their specified universal finite subtractions.
  • Scheme comparisons require matched parameters and consistent truncation.

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