University Physics IV · topic UPIV-14.07
Colour charge, confinement, and asymptotic freedom
Connect the ideas, make your working visible, and test the result. Use the hints when you need a starting point; open the full solution after your attempt.
Further practice
Work through a fresh case before tackling the extension.
Colour, confinement and asymptotic freedom: predict a finite change
Colour is the three-valued charge of the strong interaction, and its coupling α_s shrinks as the probing energy rises — asymptotic freedom — while growing too large for perturbation theory at hadronic distances, where the energy between separating quarks rises linearly, which is why quarks are only ever found inside colour-neutral hadrons. Use the supplied topic model Y_out = 10^lq, where Y_out is PROBE SCALE Q (GeV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. lq is Probe scale log₁₀(Q/GeV), dimensionless. Keep nc (Colour charges N_c) = 3; nf (Active quark flavours n_f) = 5 fixed. For this exercise use only 0.31875 ≤ lq ≤ 2.68125 on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. Calculate Y_out at lq = 0.8385 and 1.16925. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δlq across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
PROBE SCALE Q against lq
Predictions from the supplied Colour, confinement and asymptotic freedom model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): PROBE SCALE Q / GeV
Horizontal axis (x): lq: Probe scale log₁₀(Q/GeV)
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| lq: Probe scale log₁₀(Q/GeV) | PROBE SCALE Q / GeV |
|---|---|
| 0.31875 | 2.083291298989999 |
| 0.9093749999999999 | 8.116616003248668 |
| 1.5 | 31.622776601683793 |
| 2.0906249999999997 | 123.20405444827627 |
| 2.68125 | 480.00968490811147 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of lq separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At lq = 0.8385, substitute into Y_out = 10^lq: Y₀ = 6.89446 GeV.
- At lq = 1.16925, the same fixed-parameter model gives Y₁ = 14.7656 GeV.
- Signed change: ΔY_out = Y₁ − Y₀ = 7.87111 GeV; the input change is Δlq = 0.33075.
- Average sensitivity is ΔY_out/Δlq = 23.7978 GeV per unit input. Its positive sign means that this response increases over the specified interval.
- This is a secant slope between two settings. It equals a point derivative only for a locally linear relation or in a justified small-interval limit.
Answer Y₀ = 6.89446 GeV, Y₁ = 14.7656 GeV; ΔY_out = 7.87111 GeV; average sensitivity = 23.7978 GeV per unit input.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Extra challenging questions
Work through the linked stages and explain what each result means.
Colour, confinement and asymptotic freedom: infer an input with uncertainty
Colour is the three-valued charge of the strong interaction, and its coupling α_s shrinks as the probing energy rises — asymptotic freedom — while growing too large for perturbation theory at hadronic distances, where the energy between separating quarks rises linearly, which is why quarks are only ever found inside colour-neutral hadrons. Use the supplied topic model Y_out = 10^lq, where Y_out is PROBE SCALE Q (GeV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. lq is Probe scale log₁₀(Q/GeV), dimensionless. Keep nc (Colour charges N_c) = 3; nf (Active quark flavours n_f) = 5 fixed. For this exercise use only 0.31875 ≤ lq ≤ 2.68125 on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. A reading gives Y_out = 57.527434 GeV, with a bounded reading uncertainty ±8.3166 GeV. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown lq by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00023625, using a numerical bracket if required. Find the full input interval consistent with the bounded reading uncertainty and compare it with a symmetric input-error assumption. Do not assume that the supplied calibration points prove monotonicity outside this branch.
PROBE SCALE Q against lq
Predictions from the supplied Colour, confinement and asymptotic freedom model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): PROBE SCALE Q / GeV
Horizontal axis (x): lq: Probe scale log₁₀(Q/GeV)
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| lq: Probe scale log₁₀(Q/GeV) | PROBE SCALE Q / GeV |
|---|---|
| 0.31875 | 2.083291298989999 |
| 0.9093749999999999 | 8.116616003248668 |
| 1.5 | 31.622776601683793 |
| 2.0906249999999997 | 123.20405444827627 |
| 2.68125 | 480.00968490811147 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
The model's endpoint predictions bracket the reading. Solve Y_out(lq) = Y_measured and then solve again at Y_measured − u and Y_measured + u; a nonlinear inverse can give unequal input deviations.
Reveal the worked solution
- Endpoint predictions are Y_out(0.31875) = 2.08329 GeV and Y_out(2.68125) = 480.01 GeV. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives lq_line = 0.31875 + (57.5274 − 2.08329)(2.68125 − 0.31875)/(480.01 − 2.08329) = 0.592823. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
- Use the actual equation, not the chord, and bracket its root. Evaluating the midpoint and retaining the sign-changing half-interval gives lq = 1.75988; substituting back gives Y_out = 57.527434 GeV.
- The interpolation error in the input is |lq_line − lq| = 1.16705. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 49.210834 GeV and 65.844034 GeV. Sort the roots even if the response decreases: 1.69206 ≤ lq ≤ 1.81852.
- The lower and upper input deviations are 0.0678143 and 0.0586414. They differ because the inverse response is nonlinear; a symmetric error bar would only be a local approximation. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer lq ≈ 1.75988; compatible interval [1.69206, 1.81852]. Endpoint interpolation gives 0.592823.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.