University Physics III · unit UPIII-08
Special Relativity
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.
Galilean relativity and the invariance problem: infer an input with uncertainty
Galilean relativity says the laws of mechanics take the same form in every inertial frame, the frames being related by x′ = x − vt and t′ = t, so that velocities subtract while accelerations, forces and Newton's laws are identical in all of them. Use the supplied topic model Y_out = 0.2998*t*(1-b), where Y_out is FORWARD REACH (c − v)t′ (m). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. b is Frame speed v, measured in c. Keep t (Time since emission t′) = 2 ns fixed. For this exercise use only 0.053125 c ≤ b ≤ 0.446875 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 = 0.42372983 m, with a bounded reading uncertainty ±0.013811 m. 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 0.000039375 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.
FORWARD REACH (c − v)t′ against b
Predictions from the supplied Galilean relativity and the invariance problem 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): FORWARD REACH (c − v)t′ / m
Horizontal axis (x): b: Frame speed v / c
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| b: Frame speed v / c | FORWARD REACH (c − v)t′ / m |
|---|---|
| 0.053125 | 0.56774625 |
| 0.1515625 | 0.5087231249999999 |
| 0.25 | 0.4497 |
| 0.34843749999999996 | 0.39067687500000003 |
| 0.44687499999999997 | 0.3316537500000001 |
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(0.053125) = 0.567746 m and Y_out(0.446875) = 0.331654 m. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives b_line = 0.053125 + (0.42373 − 0.567746)(0.446875 − 0.053125)/(0.331654 − 0.567746) = 0.293312 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 b = 0.293312 c; substituting back gives Y_out = 0.42372983 m.
- The interpolation error in the input is |b_line − b| = 0 c. This is within the requested numerical tolerance.
- Invert both limits Y_out = 0.40991883 m and 0.43754083 m. Sort the roots even if the response decreases: 0.270279 c ≤ b ≤ 0.316346 c.
- The lower and upper input deviations are 0.0230337 c and 0.0230337 c. 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 b ≈ 0.293312 c; compatible interval [0.270279, 0.316346] c. Endpoint interpolation gives 0.293312 c.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
The postulates and the relativity of simultaneity: infer an input with uncertainty
Two events are simultaneous in a frame when that frame's light-synchronised clocks, one at each event, read the same time; a frame moving at v along the line joining them assigns the pair a time difference Δt′ = −γvΔx/c², so simultaneity belongs to the frame, not to the events. Use the supplied topic model Y_out = 1/sqrt(1-b^2), 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. b is Train speed β = v/c, dimensionless. Keep len (Train length L in S) = 100 m; delay (Front-strike delay in S) = 0 ns fixed. For this exercise use only 0.06375 ≤ b ≤ 0.53625 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 = 1.0683653, with a bounded reading uncertainty ±0.0099491. 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 0.00004725, 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 The postulates and the relativity of simultaneity 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: Train 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: Train speed β = v/c | LORENTZ FACTOR γ |
|---|---|
| 0.06375 | 1.0020382460277781 |
| 0.18187499999999998 | 1.0169612264723313 |
| 0.29999999999999993 | 1.0482848367219182 |
| 0.41812499999999997 | 1.1008493598805413 |
| 0.5362499999999999 | 1.1847510185485324 |
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(0.06375) = 1.00204 and Y_out(0.53625) = 1.18475. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives b_line = 0.06375 + (1.06837 − 1.00204)(0.53625 − 0.06375)/(1.18475 − 1.00204) = 0.235273. 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 = 0.351975; substituting back gives Y_out = 1.0683653.
- The interpolation error in the input is |b_line − b| = 0.116701. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1.0584162 and 1.0783144. Sort the roots even if the response decreases: 0.327625 ≤ b ≤ 0.374137.
- The lower and upper input deviations are 0.0243501 and 0.0221624. 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 ≈ 0.351975; compatible interval [0.327625, 0.374137]. Endpoint interpolation gives 0.235273.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Time dilation, proper time & length contraction: infer an input with uncertainty
Proper time is the interval between two events read by one clock present at both; any inertial frame in which the events occur at different places measures the longer interval Δt = γΔτ, and an object moving at v in that frame is shorter along v by the same factor, L = L₀/γ. Use the supplied topic model Y_out = 1/sqrt(1-b*b), 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. b is Speed of the clock β = v/c, dimensionless. Keep no other adjustable input fixed. For this exercise use only 0.095625 ≤ b ≤ 0.804375 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 = 1.177485, with a bounded reading uncertainty ±0.025932. 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 0.000070875, 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 Time dilation, proper time & length contraction 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 of the clock 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 of the clock β = v/c | LORENTZ FACTOR γ |
|---|---|
| 0.095625 | 1.0046036669149399 |
| 0.2728125 | 1.0394284068994142 |
| 0.45 | 1.1197850219117087 |
| 0.6271874999999999 | 1.2839140979959294 |
| 0.804375 | 1.683156202973745 |
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(0.095625) = 1.0046 and Y_out(0.804375) = 1.68316. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives b_line = 0.095625 + (1.17748 − 1.0046)(0.804375 − 0.095625)/(1.68316 − 1.0046) = 0.2762. 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 = 0.527962; substituting back gives Y_out = 1.177485.
- The interpolation error in the input is |b_line − b| = 0.251762. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1.151553 and 1.203417. Sort the roots even if the response decreases: 0.495877 ≤ b ≤ 0.556321.
- The lower and upper input deviations are 0.032085 and 0.028359. 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 ≈ 0.527962; compatible interval [0.495877, 0.556321]. Endpoint interpolation gives 0.2762.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
The Lorentz transformation: infer an input with uncertainty
The Lorentz transformation is the linear rule x′ = γ(x − vt), t′ = γ(t − vx/c²), y′ = y, z′ = z that converts an event's coordinates in one inertial frame into those of a frame moving at v along x, keeping the speed of light c in both. Use the supplied topic model Y_out = 1/sqrt(1-b*b), 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. b is Boost speed β = v/c, dimensionless. Keep xe (Event position x in S) = 1.5 m; te (Event time ct in S) = 1 m fixed. For this exercise use only 0.06375 ≤ b ≤ 0.53625 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 = 1.0683653, with a bounded reading uncertainty ±0.0099491. 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 0.00004725, 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 The Lorentz transformation 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: Boost 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: Boost speed β = v/c | LORENTZ FACTOR γ |
|---|---|
| 0.06375 | 1.0020382460277781 |
| 0.18187499999999998 | 1.0169612264723313 |
| 0.29999999999999993 | 1.0482848367219182 |
| 0.41812499999999997 | 1.1008493598805413 |
| 0.5362499999999999 | 1.1847510185485324 |
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(0.06375) = 1.00204 and Y_out(0.53625) = 1.18475. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives b_line = 0.06375 + (1.06837 − 1.00204)(0.53625 − 0.06375)/(1.18475 − 1.00204) = 0.235273. 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 = 0.351975; substituting back gives Y_out = 1.0683653.
- The interpolation error in the input is |b_line − b| = 0.116701. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1.0584162 and 1.0783144. Sort the roots even if the response decreases: 0.327625 ≤ b ≤ 0.374137.
- The lower and upper input deviations are 0.0243501 and 0.0221624. 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 ≈ 0.351975; compatible interval [0.327625, 0.374137]. Endpoint interpolation gives 0.235273.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.