University Advanced Quantum Mechanics · topic UAQM-06.01
Large and small components: eliminating chi to order v/c
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.
Large and small components: eliminating χ to order v/c: predict a finite change
Eliminating the small component means solving the lower two rows of the Dirac equation for χ in terms of the upper pair φ, χ ≈ (σ⋅π/2mc)φ for a slow electron, so that φ obeys a closed two-component equation of its own. Use the supplied topic model Y_out = sqrt(t*(t+1021.998))/(t+1021.998), where Y_out is FREE EXACT |χ|/|φ| (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. t is Free electron's kinetic energy T, measured in keV. Keep z (Nuclear charge Z of the 1s ion) = 92 fixed. For this exercise use only 25.7188 keV ≤ t ≤ 179.281 keV 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 t = 59.5025 keV and 81.0012 keV. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δt across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
FREE EXACT |χ|/|φ| against t
Predictions from the supplied Large and small components: eliminating χ to order v/c 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): FREE EXACT |χ|/|φ|
Horizontal axis (x): t: Free electron's kinetic energy T / keV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| t: Free electron's kinetic energy T / keV | FREE EXACT |χ|/|φ| |
|---|---|
| 25.71875 | 0.1566761838398359 |
| 64.109375 | 0.2429541882182584 |
| 102.5 | 0.3019135390524935 |
| 140.890625 | 0.3480743226194173 |
| 179.28125 | 0.38631844835115303 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of t separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At t = 59.5025 keV, substitute into Y_out = sqrt(t*(t+1021.998))/(t+1021.998): Y₀ = 0.23456.
- At t = 81.0012 keV, the same fixed-parameter model gives Y₁ = 0.270993.
- Signed change: ΔY_out = Y₁ − Y₀ = 0.036433; the input change is Δt = 21.4988 keV.
- Average sensitivity is ΔY_out/Δt = 0.00169465 1 per keV. 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 = 0.23456, Y₁ = 0.270993; ΔY_out = 0.036433; average sensitivity = 0.00169465 1 per keV.
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.
Large and small components: eliminating χ to order v/c: infer an input with uncertainty
Eliminating the small component means solving the lower two rows of the Dirac equation for χ in terms of the upper pair φ, χ ≈ (σ⋅π/2mc)φ for a slow electron, so that φ obeys a closed two-component equation of its own. Use the supplied topic model Y_out = sqrt(t*(t+1021.998))/(t+1021.998), where Y_out is FREE EXACT |χ|/|φ| (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. t is Free electron's kinetic energy T, measured in keV. Keep z (Nuclear charge Z of the 1s ion) = 92 fixed. For this exercise use only 25.7188 keV ≤ t ≤ 179.281 keV 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 = 0.32342261, with a bounded reading uncertainty ±0.0094344. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown t by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0153563 keV, 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.
FREE EXACT |χ|/|φ| against t
Predictions from the supplied Large and small components: eliminating χ to order v/c 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): FREE EXACT |χ|/|φ|
Horizontal axis (x): t: Free electron's kinetic energy T / keV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| t: Free electron's kinetic energy T / keV | FREE EXACT |χ|/|φ| |
|---|---|
| 25.71875 | 0.1566761838398359 |
| 64.109375 | 0.2429541882182584 |
| 102.5 | 0.3019135390524935 |
| 140.890625 | 0.3480743226194173 |
| 179.28125 | 0.38631844835115303 |
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(t) = 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(25.7188) = 0.156676 and Y_out(179.281) = 0.386318. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives t_line = 25.7188 + (0.323423 − 0.156676)(179.281 − 25.7188)/(0.386318 − 0.156676) = 137.223 keV. 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 t = 119.392 keV; substituting back gives Y_out = 0.32342261.
- The interpolation error in the input is |t_line − t| = 17.8307 keV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.31398821 and 0.33285701. Sort the roots even if the response decreases: 111.777 keV ≤ t ≤ 127.339 keV.
- The lower and upper input deviations are 7.61456 keV and 7.94758 keV. 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 t ≈ 119.392 keV; compatible interval [111.777, 127.339] keV. Endpoint interpolation gives 137.223 keV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.