University Physics III · topic UPIII-08.07
Spacetime diagrams, the invariant interval, and causality
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.
Spacetime diagrams, the invariant interval, and causality: predict a finite change
The spacetime interval between two events, Δs² = (cΔt)² − (Δx)², takes the same value in every inertial frame, and its sign says whether the events could be causally connected. Use the supplied topic model Y_out = (10*b1/(b1+b2))^2*(1-b2^2), where Y_out is LEG 2 INTERVAL Δs² (ly²). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. b2 is Return speed β₂ = v/c, dimensionless. Keep b1 (Outbound speed β₁ = v/c) = 0.6 fixed. For this exercise use only 0.185 ≤ b2 ≤ 0.815 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 b2 = 0.3236 and 0.4118. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δb2 across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
LEG 2 INTERVAL Δs² against b2
Predictions from the supplied Spacetime diagrams, the invariant interval, and causality 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): LEG 2 INTERVAL Δs² / ly²
Horizontal axis (x): b2: Return speed β₂ = v/c
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| b2: Return speed β₂ = v/c | LEG 2 INTERVAL Δs² / ly² |
|---|---|
| 0.185 | 56.42078786157655 |
| 0.34249999999999997 | 35.7725446601327 |
| 0.49999999999999994 | 22.31404958677686 |
| 0.6575 | 12.92410941903253 |
| 0.815 | 6.03723357764487 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of b2 separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At b2 = 0.3236, substitute into Y_out = (10*b1/(b1+b2))^2*(1-b2^2): Y₀ = 37.7829 ly².
- At b2 = 0.4118, the same fixed-parameter model gives Y₁ = 29.2019 ly².
- Signed change: ΔY_out = Y₁ − Y₀ = -8.58096 ly²; the input change is Δb2 = 0.0882.
- Average sensitivity is ΔY_out/Δb2 = -97.2898 ly² per unit input. 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 = 37.7829 ly², Y₁ = 29.2019 ly²; ΔY_out = -8.58096 ly²; average sensitivity = -97.2898 ly² 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.
Spacetime diagrams, the invariant interval, and causality: infer an input with uncertainty
The spacetime interval between two events, Δs² = (cΔt)² − (Δx)², takes the same value in every inertial frame, and its sign says whether the events could be causally connected. Use the supplied topic model Y_out = (10*b1/(b1+b2))^2*(1-b2^2), where Y_out is LEG 2 INTERVAL Δs² (ly²). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. b2 is Return speed β₂ = v/c, dimensionless. Keep b1 (Outbound speed β₁ = v/c) = 0.6 fixed. For this exercise use only 0.185 ≤ b2 ≤ 0.815 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 = 17.796371 ly², with a bounded reading uncertainty ±1.7639 ly². Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown b2 by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000063, 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.
LEG 2 INTERVAL Δs² against b2
Predictions from the supplied Spacetime diagrams, the invariant interval, and causality 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): LEG 2 INTERVAL Δs² / ly²
Horizontal axis (x): b2: Return speed β₂ = v/c
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| b2: Return speed β₂ = v/c | LEG 2 INTERVAL Δs² / ly² |
|---|---|
| 0.185 | 56.42078786157655 |
| 0.34249999999999997 | 35.7725446601327 |
| 0.49999999999999994 | 22.31404958677686 |
| 0.6575 | 12.92410941903253 |
| 0.815 | 6.03723357764487 |
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(b2) = 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(0.185) = 56.4208 ly² and Y_out(0.815) = 6.03723 ly². The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives b2_line = 0.185 + (17.7964 − 56.4208)(0.815 − 0.185)/(6.03723 − 56.4208) = 0.667963. 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 b2 = 0.5693; substituting back gives Y_out = 17.796371 ly².
- The interpolation error in the input is |b2_line − b2| = 0.0986628. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 16.032471 ly² and 19.560271 ly². Sort the roots even if the response decreases: 0.54098 ≤ b2 ≤ 0.599427.
- The lower and upper input deviations are 0.0283199 and 0.0301273. 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 b2 ≈ 0.5693; compatible interval [0.54098, 0.599427]. Endpoint interpolation gives 0.667963.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.