University Physics V · topic UPV-17.02
Antiparticles, charge conjugation, and CPT
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.
Antiparticles, charge conjugation and CPT: predict a finite change
An antiparticle is what a relativistic wave equation's negative-energy solution becomes when it is read forward in time: the particle's mass and spin, every additive charge reversed, and, by the CPT theorem, the same total lifetime. Use the supplied topic model Y_out = 197.327/max(m, 0.001), where Y_out is COMPTON LENGTH ħ/mc (fm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; max(a,b) selects the larger value. m is Rest energy mc², measured in MeV. Keep p (Momentum pc) = 1.2 MeV fixed. For this exercise use only 0.2125 MeV ≤ m ≤ 1.7875 MeV 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 = 0.559 MeV and 0.7795 MeV. 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.
COMPTON LENGTH ħ/mc against m
Predictions from the supplied Antiparticles, charge conjugation and CPT 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): COMPTON LENGTH ħ/mc / fm
Horizontal axis (x): m: Rest energy mc² / MeV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| m: Rest energy mc² / MeV | COMPTON LENGTH ħ/mc / fm |
|---|---|
| 0.2125 | 928.5976470588236 |
| 0.60625 | 325.4878350515464 |
| 1 | 197.327 |
| 1.3937499999999998 | 141.57991031390137 |
| 1.7874999999999999 | 110.39272727272729 |
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 = 0.559 MeV, substitute into Y_out = 197.327/max(m, 0.001): Y₀ = 353 fm.
- At m = 0.7795 MeV, the same fixed-parameter model gives Y₁ = 253.146 fm.
- Signed change: ΔY_out = Y₁ − Y₀ = -99.8544 fm; the input change is Δm = 0.2205 MeV.
- Average sensitivity is ΔY_out/Δm = -452.854 fm per MeV. 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 = 353 fm, Y₁ = 253.146 fm; ΔY_out = -99.8544 fm; average sensitivity = -452.854 fm per MeV.
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.
Antiparticles, charge conjugation and CPT: infer an input with uncertainty
An antiparticle is what a relativistic wave equation's negative-energy solution becomes when it is read forward in time: the particle's mass and spin, every additive charge reversed, and, by the CPT theorem, the same total lifetime. Use the supplied topic model Y_out = 197.327/max(m, 0.001), where Y_out is COMPTON LENGTH ħ/mc (fm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; max(a,b) selects the larger value. m is Rest energy mc², measured in MeV. Keep p (Momentum pc) = 1.2 MeV fixed. For this exercise use only 0.2125 MeV ≤ m ≤ 1.7875 MeV 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 = 168.18837 fm, with a bounded reading uncertainty ±8.6693 fm. 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 0.0001575 MeV, 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.
COMPTON LENGTH ħ/mc against m
Predictions from the supplied Antiparticles, charge conjugation and CPT 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): COMPTON LENGTH ħ/mc / fm
Horizontal axis (x): m: Rest energy mc² / MeV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| m: Rest energy mc² / MeV | COMPTON LENGTH ħ/mc / fm |
|---|---|
| 0.2125 | 928.5976470588236 |
| 0.60625 | 325.4878350515464 |
| 1 | 197.327 |
| 1.3937499999999998 | 141.57991031390137 |
| 1.7874999999999999 | 110.39272727272729 |
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(0.2125) = 928.598 fm and Y_out(1.7875) = 110.393 fm. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives m_line = 0.2125 + (168.188 − 928.598)(1.7875 − 0.2125)/(110.393 − 928.598) = 1.67625 MeV. 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 = 1.17325 MeV; substituting back gives Y_out = 168.18837 fm.
- The interpolation error in the input is |m_line − m| = 0.502997 MeV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 159.51907 fm and 176.85767 fm. Sort the roots even if the response decreases: 1.11574 MeV ≤ m ≤ 1.23701 MeV.
- The lower and upper input deviations are 0.057511 MeV and 0.063762 MeV. 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 ≈ 1.17325 MeV; compatible interval [1.11574, 1.23701] MeV. Endpoint interpolation gives 1.67625 MeV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.