Finite coordinate changes move coefficients while preserving predictions
Transform a beta function and an observable consistently, and identify which perturbative invariants require assumptions.
On this page
The question for today
What you’ll understand
- Transform a single-coupling beta function by the chain rule.
- Derive the invariance of its first two coefficients under specified redefinitions.
- Translate observable coefficients and interpret residual higher-order differences.
A renormalization scheme assigns coordinates to a theory rather than directly labeling an observable. A finite analytic change of coupling reshuffles both the beta function and the perturbative coefficients of predictions. Certain leading beta coefficients survive such a change, but that statement has specific single-coupling and mass-independent assumptions. We will derive the transformation and use an observable conversion to explain why holding the same numerical coupling in two schemes does not constitute a physical comparison.
A beta function transforms as a vector field
Take a dimensionless four-dimensional marginal coupling g and define a new coordinate g′=g+c₁g²+c₂g³ plus higher powers, with no explicit scale dependence. Applying the chain rule gives β′(g′)=(dg′/dg)β(g). Express the right-hand side back in terms of g′ rather than leaving a mixture of variables. The inverse series begins g=g′−c₁g′²+(2c₁²−c₂)g′³. Without this inversion, even the apparent coefficient of the next power can be misidentified.
For β=b₁g²+b₂g³+b₃g⁴, the first two coefficients remain b₁ and b₂, while the third changes as shown below. This invariance assumes a single marginal coupling, an analytic near-identity redefinition, and compatible mass-independent conventions without an explicit μ-dependent finite transformation. Several couplings, threshold-dependent definitions, dimensionful parameters, or a nonstandard leading normalization require a more general analysis. It is therefore safer to quote the transformation assumptions than to say that two beta coefficients are universal without qualification.
Physical series coefficients transform together with the coupling
Suppose a dimensionless physical quantity has expansion A=g+r₁g² through second order. Substituting the inverse transformation changes its next coefficient to r₁−c₁ in the primed scheme. This cancellation is the simplest example of scheme independence: the value assigned to a coupling and the coefficient multiplying its square are not separately observable. The same physical input must be used to determine g and g′. Comparing both formulas at the same numerical input generally violates that requirement.
Even after a consistent conversion, evaluating two truncated formulas as ordinary numbers can leave a difference of higher order. Such a difference is expected because the substitution itself generates unretained powers. It does not mean exact observables depend on arbitrary schemes. Conversely, the numerical size of a scheme variation is not a rigorous bound on all omitted terms. A pathological transformation can manufacture poor-looking coefficients or a misleading apparent convergence pattern without teaching anything about the underlying physical accuracy.
Fixed-point coordinates can move while critical slopes remain invariant
For an exact flow, an invertible smooth transformation maps a zero of β to a zero of β′. Its coordinate usually changes. At the fixed point, the derivative of β with respect to its coupling is invariant because the extra derivative of the coordinate map multiplies β itself and vanishes. This statement concerns an exact regular transformation near the fixed point. A zero produced by truncating a weak-coupling series at order-one coupling may move or disappear under allowed higher-order redefinitions and is not automatically a controlled physical fixed point.
The function lab illustrates the finite quadratic part of a coordinate change. Reset and set both controls to a=0.5 and x=0.20, predicting y=0.020. Reset and set a=1 and x=0.20, predicting y=0.040. Interpret y as δg=c₁g²; adding g=0.20 manually gives primed coordinates 0.220 and 0.240. This does not calculate an observable or beta coefficients. Its coefficient changes compare two independent coordinate definitions, not an actual scale trajectory or a change of the underlying interaction.
Work it through
Transform the first scheme-dependent beta coefficient
Let β(g)=−g²+0.5g³+2g⁴+O(g⁵). Define g′=g+g² with c₂=0. Find the beta function through g′⁴ and identify the first coefficient that changes.
Read b₁=−1, b₂=0.5, b₃=2 and c₁=1, c₂=0 in the stated single-coupling convention.
The first two coefficients remain −1 and 0.5. The transformed third coefficient is 2−1×0.5+(0−1)×(−1)=2.5.
Therefore β′=−g′²+0.5g′³+2.5g′⁴+O(g′⁵). This is a coordinate transformation of the same flow, provided the coupling conversion is also used.
The transformed beta function is −g′²+0.5g′³+2.5g′⁴ through the stated order; the g⁴ coefficient is the first one changed here.
The living notebook
Explore the model
Interactive lab · function
Display the finite coordinate correction
Reset and set BOTH controls to a=0.5 and x=0.20, predicting y=0.020. Reset and set a=1 and x=0.20, predicting y=0.040. Assign y=δg=c₁g², then add g manually. This is a coordinate illustration without RG integration.
- y at chosen x
- 0.608
- Tangent slope · 2ax
- 1.35
- Signed area · ax³/3
- 0.182
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 forChanging a scheme coordinate does not by itself change a physical prediction when its coefficients are converted too.
Pause & reason
Check your understanding
Your turn
Try it yourself
A physical quantity has A=g+3g²+O(g³). Define g′=g+2g². Find the second-order primed expression. For g=0.050 use g′=0.055 and compare the two unexpanded second-order values. Explain their difference.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Use the inverse series g=g′−2g′²+O(g′³); the numerical substitution introduces terms beyond the retained order.
Compare with the solution
The primed series is A=g′+(3−2)g′²+O(g′³)=g′+g′²+O(g′³). The old polynomial gives 0.050+3×0.0025=0.0575; the new gives 0.055+0.003025=0.058025. Their difference 0.000525 starts at higher order when g′=g+2g² is expanded. It is a truncation effect, not physical scheme dependence.
Keep the idea
Take it with you
- A beta function transforms by the coupling-coordinate chain rule.
- Leading-coefficient invariance has explicit assumptions.
- Observable coefficients must be translated along with their running parameters.
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