University Physics IV · unit UPIV-03
Special Relativity II
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.
Extra challenging questions
A selection from this pathway. Open a topic below for its complete practice and challenge set.
Proper time and time dilation: infer an input with uncertainty
Proper time is the interval between two events measured by a single clock that is present at both; time dilation is the statement that every other inertial frame assigns a larger interval, Δt = γΔτ. Use the supplied topic model Y_out = 1/sqrt(max(1-b*b,0.0001)), where Y_out is LORENTZ FACTOR γ (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; max(a,b) selects the larger value. b is Speed β = v/c, dimensionless. Keep t0 (Proper time Δτ) = 2.2 μs fixed. For this exercise use only 0.279687 ≤ b ≤ 0.870313 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 = 1.3014039, with a bounded reading uncertainty ±0.038975. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown b by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0000590625, 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.
LORENTZ FACTOR γ against b
Predictions from the supplied Proper time and time dilation 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): LORENTZ FACTOR γ
Horizontal axis (x): b: Speed β = v/c
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| b: Speed β = v/c | LORENTZ FACTOR γ |
|---|---|
| 0.2796875 | 1.0415678363642908 |
| 0.42734375 | 1.106085524403928 |
| 0.575 | 1.2222646627042817 |
| 0.72265625 | 1.4467433547506021 |
| 0.8703125 | 2.030457261535957 |
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(b) = 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(0.279687) = 1.04157 and Y_out(0.870313) = 2.03046. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives b_line = 0.279687 + (1.3014 − 1.04157)(0.870313 − 0.279687)/(2.03046 − 1.04157) = 0.434877. 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 b = 0.639969; substituting back gives Y_out = 1.3014039.
- The interpolation error in the input is |b_line − b| = 0.205091. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1.2624289 and 1.3403789. Sort the roots even if the response decreases: 0.61036 ≤ b ≤ 0.665881.
- The lower and upper input deviations are 0.0296083 and 0.0259126. 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 b ≈ 0.639969; compatible interval [0.61036, 0.665881]. Endpoint interpolation gives 0.434877.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Proper time along a worldline: infer an input with uncertainty
The proper time along a worldline is the reading of a clock carried along it, accumulated as dt/γ(t) instant by instant, so it depends on the whole speed history rather than on any single γ. Use the supplied topic model Y_out = 1/sqrt(max(1-b1^2,0.0001)), where Y_out is OUTBOUND γ₁ (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; max(a,b) selects the larger value. b1 is Outbound speed β₁, measured in c. Keep b2 (Return speed β₂) = 0.8 c; w (Fraction parked at the station) = 0 fixed. For this exercise use only 0.190312 c ≤ b1 ≤ 0.859687 c 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 = 1.2484009, with a bounded reading uncertainty ±0.034468. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown b1 by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0000669375 c, 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.
OUTBOUND γ₁ against b1
Predictions from the supplied Proper time along a worldline 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): OUTBOUND γ₁
Horizontal axis (x): b1: Outbound speed β₁ / c
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| b1: Outbound speed β₁ / c | OUTBOUND γ₁ |
|---|---|
| 0.1903125 | 1.0186166846000015 |
| 0.35765625 | 1.0708319904299257 |
| 0.5249999999999999 | 1.1749469908920933 |
| 0.6923437499999999 | 1.3858710987673009 |
| 0.8596874999999999 | 1.9576355032536286 |
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(b1) = 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(0.190312) = 1.01862 and Y_out(0.859687) = 1.95764. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives b1_line = 0.190312 + (1.2484 − 1.01862)(0.859687 − 0.190312)/(1.95764 − 1.01862) = 0.354113 c. 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 b1 = 0.598631 c; substituting back gives Y_out = 1.2484009.
- The interpolation error in the input is |b1_line − b1| = 0.244518 c. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1.2139329 and 1.2828689. Sort the roots even if the response decreases: 0.566926 c ≤ b1 ≤ 0.626399 c.
- The lower and upper input deviations are 0.0317053 c and 0.0277674 c. 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 b1 ≈ 0.598631 c; compatible interval [0.566926, 0.626399] c. Endpoint interpolation gives 0.354113 c.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
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.
Length contraction and proper length: infer an input with uncertainty
The length of a moving object is the distance between the positions of its two ends taken at the same instant in the measuring frame, and it comes out shorter than the proper length by the factor 1/γ. Use the supplied topic model Y_out = 1/sqrt(max(1-b*b,0.0001)), where Y_out is LORENTZ FACTOR γ (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; max(a,b) selects the larger value. b is Speed β = v/c, dimensionless. Keep l0 (Proper length L₀) = 100 m fixed. For this exercise use only 0.190312 ≤ b ≤ 0.859687 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 = 1.2484009, with a bounded reading uncertainty ±0.034468. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown b by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0000669375, 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.
LORENTZ FACTOR γ against b
Predictions from the supplied Length contraction and proper length 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): LORENTZ FACTOR γ
Horizontal axis (x): b: Speed β = v/c
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| b: Speed β = v/c | LORENTZ FACTOR γ |
|---|---|
| 0.1903125 | 1.0186166846000015 |
| 0.35765625 | 1.0708319904299257 |
| 0.5249999999999999 | 1.1749469908920933 |
| 0.6923437499999999 | 1.3858710987673009 |
| 0.8596874999999999 | 1.9576355032536286 |
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(b) = 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(0.190312) = 1.01862 and Y_out(0.859687) = 1.95764. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives b_line = 0.190312 + (1.2484 − 1.01862)(0.859687 − 0.190312)/(1.95764 − 1.01862) = 0.354113. 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 b = 0.598631; substituting back gives Y_out = 1.2484009.
- The interpolation error in the input is |b_line − b| = 0.244518. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1.2139329 and 1.2828689. Sort the roots even if the response decreases: 0.566926 ≤ b ≤ 0.626399.
- The lower and upper input deviations are 0.0317053 and 0.0277674. 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 b ≈ 0.598631; compatible interval [0.566926, 0.626399]. Endpoint interpolation gives 0.354113.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.