University Physics IV · topic UPIV-03.03
The twin paradox and the asymmetry that resolves it
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.
The twin paradox: predict a finite change
The twin paradox is the apparent contradiction that two twins each measure the other's clock as slow, resolved by the fact that only the twin who stays inertial has a straight worldline, and the straight worldline between two meetings carries the most proper time. Use the supplied topic model Y_out = (8/max(b,0.05))*(f-b*b*(f-clamp(floor(2*f),0,1))), where Y_out is BOB'S CLOCK ON HER SLICE (yr). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; max(a,b) selects the larger value; clamp(x,a,b) confines x to [a,b]. f is Where Alice is in her trip, dimensionless. Keep b (Cruise speed β = v/c) = 0.8 c fixed. For this exercise use only 0.10625 ≤ f ≤ 0.89375 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 f = 0.2795 and 0.38975. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δf across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
BOB'S CLOCK ON HER SLICE against f
Predictions from the supplied The twin paradox 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): BOB'S CLOCK ON HER SLICE / yr
Horizontal axis (x): f: Where Alice is in her trip
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| f: Where Alice is in her trip | BOB'S CLOCK ON HER SLICE / yr |
|---|---|
| 0.10625 | 0.38249999999999995 |
| 0.303125 | 1.0912499999999994 |
| 0.5 | 8.200000000000001 |
| 0.6968749999999999 | 8.90875 |
| 0.8937499999999999 | 9.6175 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of f separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At f = 0.2795, substitute into Y_out = (8/max(b,0.05))*(f-b*b*(f-clamp(floor(2*f),0,1))): Y₀ = 1.0062 yr.
- At f = 0.38975, the same fixed-parameter model gives Y₁ = 1.4031 yr.
- Signed change: ΔY_out = Y₁ − Y₀ = 0.3969 yr; the input change is Δf = 0.11025.
- Average sensitivity is ΔY_out/Δf = 3.6 yr 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 = 1.0062 yr, Y₁ = 1.4031 yr; ΔY_out = 0.3969 yr; average sensitivity = 3.6 yr 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.
The twin paradox: infer an input with uncertainty
The twin paradox is the apparent contradiction that two twins each measure the other's clock as slow, resolved by the fact that only the twin who stays inertial has a straight worldline, and the straight worldline between two meetings carries the most proper time. Use the supplied topic model Y_out = (8/max(b,0.05))*(f-b*b*(f-clamp(floor(2*f),0,1))), where Y_out is BOB'S CLOCK ON HER SLICE (yr). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; max(a,b) selects the larger value; clamp(x,a,b) confines x to [a,b]. f is Where Alice is in her trip, dimensionless. Keep b (Cruise speed β = v/c) = 0.8 c fixed. For this exercise use only 0.10625 ≤ f ≤ 0.89375 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 = 8.51185 yr, with a bounded reading uncertainty ±0.16585 yr. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown f by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00007875, 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.
BOB'S CLOCK ON HER SLICE against f
Predictions from the supplied The twin paradox 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): BOB'S CLOCK ON HER SLICE / yr
Horizontal axis (x): f: Where Alice is in her trip
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| f: Where Alice is in her trip | BOB'S CLOCK ON HER SLICE / yr |
|---|---|
| 0.10625 | 0.38249999999999995 |
| 0.303125 | 1.0912499999999994 |
| 0.5 | 8.200000000000001 |
| 0.6968749999999999 | 8.90875 |
| 0.8937499999999999 | 9.6175 |
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(f) = 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(0.10625) = 0.3825 yr and Y_out(0.89375) = 9.6175 yr. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives f_line = 0.10625 + (8.51185 − 0.3825)(0.89375 − 0.10625)/(9.6175 − 0.3825) = 0.799467. 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 f = 0.586625; substituting back gives Y_out = 8.51185 yr.
- The interpolation error in the input is |f_line − f| = 0.212842. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 8.346 yr and 8.6777 yr. Sort the roots even if the response decreases: 0.540556 ≤ f ≤ 0.632694.
- The lower and upper input deviations are 0.0460694 and 0.0460694. They agree on this branch; this does not establish linearity for changes in other inputs or outside the interval. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer f ≈ 0.586625; compatible interval [0.540556, 0.632694]. Endpoint interpolation gives 0.799467.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.