University Physics V · unit UPV-19
Conservation Laws and Particle Reactions
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.
Symmetry, generators and conservation: infer an input with uncertainty
A continuous symmetry of a quantum system is a one-parameter unitary family U(a) = exp(−iaG/ħ) whose Hermitian generator G commutes with the Hamiltonian; that single commutator, [H, G] = 0, is exactly the statement that G is conserved. Use the supplied topic model Y_out = omega*sind(theta), where Y_out is ‖[H, G]‖ / ħ (rad/ns). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sind/cosd/tand use degrees. theta is Tilt θ of H from the G axis, measured in °. Keep omega (Precession rate ω) = 1 rad/ns; t (Time t) = 2 ns fixed. For this exercise use only 9.5625 ° ≤ theta ≤ 80.4375 ° 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 = 0.79649035 rad/ns, with a bounded reading uncertainty ±0.028442 rad/ns. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown theta by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0070875 °, 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.
‖[H, G]‖ / ħ against theta
Predictions from the supplied Symmetry, generators and conservation 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): ‖[H, G]‖ / ħ / rad/ns
Horizontal axis (x): theta: Tilt θ of H from the G axis / °
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| theta: Tilt θ of H from the G axis / ° | ‖[H, G]‖ / ħ / rad/ns |
|---|---|
| 9.5625 | 0.16612337814576547 |
| 27.28125 | 0.4583587306212713 |
| 45 | 0.7071067811865475 |
| 62.71875 | 0.8887672777860675 |
| 80.4375 | 0.9861049757675088 |
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(theta) = 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(9.5625) = 0.166123 rad/ns and Y_out(80.4375) = 0.986105 rad/ns. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives theta_line = 9.5625 + (0.79649 − 0.166123)(80.4375 − 9.5625)/(0.986105 − 0.166123) = 64.0482 °. 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 theta = 52.7963 °; substituting back gives Y_out = 0.79649035 rad/ns.
- The interpolation error in the input is |theta_line − theta| = 11.2519 °. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.76804835 rad/ns and 0.82493235 rad/ns. Sort the roots even if the response decreases: 50.179 ° ≤ theta ≤ 55.5816 °.
- The lower and upper input deviations are 2.6173 ° and 2.78538 °. 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 theta ≈ 52.7963 °; compatible interval [50.179, 55.5816] °. Endpoint interpolation gives 64.0482 °.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Energy conservation, Q values and thresholds: infer an input with uncertainty
Energy is conserved whenever H has no explicit time dependence, because H generates time translation and trivially commutes with itself; the invariant s = (Σp)² then decides whether the conserved energy is enough, since a reaction needs √s ≥ Σm_f c² in every frame. Use the supplied topic model Y_out = dm*(0.938+mb+dm/2)/max(mb,0.01), where Y_out is THRESHOLD T_thr (GeV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; max(a,b) selects the larger value. mb is Target mass m_b (at rest), measured in GeV. Keep tb (Beam kinetic energy T) = 3 GeV; dm (Mass created ΔM = Σm_f − m_p − m_b) = 1.876 GeV fixed. For this exercise use only 1.15709 GeV ≤ mb ≤ 2.78091 GeV 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 = 3.5147325 GeV, with a bounded reading uncertainty ±0.055978 GeV. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown mb by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000162383 GeV, 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.
THRESHOLD T_thr against mb
Predictions from the supplied Energy conservation, Q values and thresholds 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): THRESHOLD T_thr / GeV
Horizontal axis (x): mb: Target mass m_b (at rest) / GeV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| mb: Target mass m_b (at rest) / GeV | THRESHOLD T_thr / GeV |
|---|---|
| 1.1570875 | 4.917581557142393 |
| 1.56304375 | 4.127617077257114 |
| 1.9690000000000003 | 3.6633925850685616 |
| 2.3749562500000003 | 3.357869823917808 |
| 2.7809125000000003 | 3.1415471899960887 |
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(mb) = 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(1.15709) = 4.91758 GeV and Y_out(2.78091) = 3.14155 GeV. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives mb_line = 1.15709 + (3.51473 − 4.91758)(2.78091 − 1.15709)/(3.14155 − 4.91758) = 2.43971 GeV. 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 mb = 2.14762 GeV; substituting back gives Y_out = 3.5147325 GeV.
- The interpolation error in the input is |mb_line − mb| = 0.292089 GeV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 3.4587545 GeV and 3.5707105 GeV. Sort the roots even if the response decreases: 2.07668 GeV ≤ mb ≤ 2.22358 GeV.
- The lower and upper input deviations are 0.0709381 GeV and 0.0759559 GeV. 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 mb ≈ 2.14762 GeV; compatible interval [2.07668, 2.22358] GeV. Endpoint interpolation gives 2.43971 GeV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Momentum conservation and decay kinematics: infer an input with uncertainty
Momentum conservation is the statement [H, P̂] = 0 for the translation generator P̂ = −iħ∇; in a decay it and energy conservation remove four of the 3n final momentum components, leaving a two-body final state only a direction and a three-body one two continuous energies. Use the supplied topic model Y_out = sqrt(max((1-(a+b)^2)*(1-(a-b)^2),0))/2, where Y_out is TWO-BODY |p₁|/M (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. a is m₁/M, body 1 (the one plotted), dimensionless. Keep b (m₂/M, body 2 (recoil partner)) = 0.3; c (m₃/M, body 3 (0 = massless ν)) = 0 fixed. For this exercise use only 0.031875 ≤ a ≤ 0.268125 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 = 0.43633163, with a bounded reading uncertainty ±0.002709. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown a by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000023625, 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.
TWO-BODY |p₁|/M against a
Predictions from the supplied Momentum conservation and decay kinematics 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): TWO-BODY |p₁|/M
Horizontal axis (x): a: m₁/M, body 1 (the one plotted)
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| a: m₁/M, body 1 (the one plotted) | TWO-BODY |p₁|/M |
|---|---|
| 0.031875 | 0.454391383673054 |
| 0.09093749999999999 | 0.45003905267949307 |
| 0.14999999999999997 | 0.4414624134623468 |
| 0.20906249999999998 | 0.4285816060278376 |
| 0.26812499999999995 | 0.4112620527307054 |
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(a) = 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(0.031875) = 0.454391 and Y_out(0.268125) = 0.411262. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives a_line = 0.031875 + (0.436332 − 0.454391)(0.268125 − 0.031875)/(0.411262 − 0.454391) = 0.130801. 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 a = 0.175987; substituting back gives Y_out = 0.43633163.
- The interpolation error in the input is |a_line − a| = 0.0451864. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.43362263 and 0.43904063. Sort the roots even if the response decreases: 0.162796 ≤ a ≤ 0.188233.
- The lower and upper input deviations are 0.0131913 and 0.0122456. 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 a ≈ 0.175987; compatible interval [0.162796, 0.188233]. Endpoint interpolation gives 0.130801.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Electric charge, U(1) phase and superselection: infer an input with uncertainty
Electric charge is conserved because the Hamiltonian is unchanged when every charged field is rotated by one common phase; the Hermitian generator of that rotation is the charge operator Q, and [H, Q] = 0 keeps its value fixed. Use the supplied topic model Y_out = 2*g*sqrt(p*(1-p)), where Y_out is FRINGE AMPLITUDE 2g√(p(1−p)) (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. p is Weight p of the Q = 0 sector, dimensionless. Keep g (Charge-changing element g of the probe) = 0.6; phi (Relative phase φ) = 60 ° fixed. For this exercise use only 0.10625 ≤ p ≤ 0.47375 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 = 0.56443939, with a bounded reading uncertainty ±0.00521. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown p by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00003675, 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.
FRINGE AMPLITUDE 2g√(p(1−p)) against p
Predictions from the supplied Electric charge, U(1) phase and superselection 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): FRINGE AMPLITUDE 2g√(p(1−p))
Horizontal axis (x): p: Weight p of the Q = 0 sector
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| p: Weight p of the Q = 0 sector | FRINGE AMPLITUDE 2g√(p(1−p)) |
|---|---|
| 0.10625 | 0.3697887910686315 |
| 0.198125 | 0.47830423111237474 |
| 0.29 | 0.5445144626178445 |
| 0.381875 | 0.5830153835877746 |
| 0.47375 | 0.5991725544448777 |
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(p) = 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) = 0.369789 and Y_out(0.47375) = 0.599173. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives p_line = 0.10625 + (0.564439 − 0.369789)(0.47375 − 0.10625)/(0.599173 − 0.369789) = 0.418103. 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 p = 0.330425; substituting back gives Y_out = 0.56443939.
- The interpolation error in the input is |p_line − p| = 0.0876783. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.55922939 and 0.56964939. Sort the roots even if the response decreases: 0.318834 ≤ p ≤ 0.342989.
- The lower and upper input deviations are 0.0115912 and 0.0125639. 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 p ≈ 0.330425; compatible interval [0.318834, 0.342989]. Endpoint interpolation gives 0.418103.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.