University Quantum Field Theory I · topic UQFTI-13.09
Muonic hydrogen: where vacuum polarisation dominates
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.
Muonic hydrogen: where vacuum polarisation dominates: predict a finite change
Muonic hydrogen is a muon bound to a proton; its orbit is small enough to sit inside the electron's vacuum-polarisation cloud, which makes the electron loop the largest radiative correction to its energy levels. Use the supplied topic model Y_out = (0.149264+1.325503/m), where Y_out is ORBIT ÷ CLOUD κ = 2mₑa (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. m is Orbiting particle mass, measured in m_μ. Keep no other adjustable input fixed. For this exercise use only 0.408125 m_μ ≤ m ≤ 2.69188 m_μ 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 m = 0.91055 m_μ and 1.23027 m_μ. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δm across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
ORBIT ÷ CLOUD κ = 2mₑa against m
Predictions from the supplied Muonic hydrogen: where vacuum polarisation dominates 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): ORBIT ÷ CLOUD κ = 2mₑa
Horizontal axis (x): m: Orbiting particle mass / m_μ
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| m: Orbiting particle mass / m_μ | ORBIT ÷ CLOUD κ = 2mₑa |
|---|---|
| 0.40812499999999996 | 3.3970508300153144 |
| 0.9790625 | 1.5031132180019153 |
| 1.5499999999999998 | 1.0044272258064517 |
| 2.1209375 | 0.7742248958302638 |
| 2.691875 | 0.6416728228465289 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of m separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At m = 0.91055 m_μ, substitute into Y_out = (0.149264+1.325503/m): Y₀ = 1.60498.
- At m = 1.23027 m_μ, the same fixed-parameter model gives Y₁ = 1.22667.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.378313; the input change is Δm = 0.319725 m_μ.
- Average sensitivity is ΔY_out/Δm = -1.18325 1 per m_μ. Its negative sign means that this response decreases 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 = 1.60498, Y₁ = 1.22667; ΔY_out = -0.378313; average sensitivity = -1.18325 1 per m_μ.
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.
Muonic hydrogen: where vacuum polarisation dominates: infer an input with uncertainty
Muonic hydrogen is a muon bound to a proton; its orbit is small enough to sit inside the electron's vacuum-polarisation cloud, which makes the electron loop the largest radiative correction to its energy levels. Use the supplied topic model Y_out = (0.149264+1.325503/m), where Y_out is ORBIT ÷ CLOUD κ = 2mₑa (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. m is Orbiting particle mass, measured in m_μ. Keep no other adjustable input fixed. For this exercise use only 0.408125 m_μ ≤ m ≤ 2.69188 m_μ 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 = 0.88515885, with a bounded reading uncertainty ±0.036523. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown m by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000228375 m_μ, 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.
ORBIT ÷ CLOUD κ = 2mₑa against m
Predictions from the supplied Muonic hydrogen: where vacuum polarisation dominates 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): ORBIT ÷ CLOUD κ = 2mₑa
Horizontal axis (x): m: Orbiting particle mass / m_μ
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| m: Orbiting particle mass / m_μ | ORBIT ÷ CLOUD κ = 2mₑa |
|---|---|
| 0.40812499999999996 | 3.3970508300153144 |
| 0.9790625 | 1.5031132180019153 |
| 1.5499999999999998 | 1.0044272258064517 |
| 2.1209375 | 0.7742248958302638 |
| 2.691875 | 0.6416728228465289 |
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(m) = 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(0.408125) = 3.39705 and Y_out(2.69188) = 0.641673. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives m_line = 0.408125 + (0.885159 − 3.39705)(2.69188 − 0.408125)/(0.641673 − 3.39705) = 2.49007 m_μ. 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 m = 1.80121 m_μ; substituting back gives Y_out = 0.88515885.
- The interpolation error in the input is |m_line − m| = 0.688853 m_μ. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.84863585 and 0.92168185. Sort the roots even if the response decreases: 1.71604 m_μ ≤ m ≤ 1.89528 m_μ.
- The lower and upper input deviations are 0.0851685 m_μ and 0.094064 m_μ. 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 m ≈ 1.80121 m_μ; compatible interval [1.71604, 1.89528] m_μ. Endpoint interpolation gives 2.49007 m_μ.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.