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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.

Extra challenging · 01

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.

Explore the question data

‖[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

‖[H, G]‖ / ħ against thetaPredictions 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. Horizontal axis: theta: Tilt θ of H from the G axis / °. Vertical axis: ‖[H, G]‖ / ħ / rad/ns. Exact values are available in the data table below.0.13330.35470.57610.79751.0196.72825.864564.1483.27xy

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.

Reading 1theta: Tilt θ of H from the G axis / °: 9.563‖[H, G]‖ / ħ / rad/ns: 0.1661
Read the exact data as a table
‖[H, G]‖ / ħ against theta — supplied values
theta: Tilt θ of H from the G axis / °‖[H, G]‖ / ħ / rad/ns
9.56250.16612337814576547
27.281250.4583587306212713
450.7071067811865475
62.718750.8887672777860675
80.43750.9861049757675088

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
  1. 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.
  2. 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.
  3. 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.
  4. The interpolation error in the input is |theta_line − theta| = 11.2519 °. This is larger than the requested numerical tolerance.
  5. 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 °.
  6. 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.

Extra challenging · 02

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.

Explore the question data

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

THRESHOLD T_thr against mbPredictions 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. Horizontal axis: mb: Target mass m_b (at rest) / GeV. Vertical axis: THRESHOLD T_thr / GeV. Exact values are available in the data table below.3.0713.554.034.5094.9891.0921.5311.9692.4072.846xy

Horizontal axis (x): mb: Target mass m_b (at rest) / GeV

Axes show the supplied calibration interval; the origin may lie outside the plot.

Reading 1mb: Target mass m_b (at rest) / GeV: 1.157THRESHOLD T_thr / GeV: 4.918
Read the exact data as a table
THRESHOLD T_thr against mb — supplied values
mb: Target mass m_b (at rest) / GeVTHRESHOLD T_thr / GeV
1.15708754.917581557142393
1.563043754.127617077257114
1.96900000000000033.6633925850685616
2.37495625000000033.357869823917808
2.78091250000000033.1415471899960887

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
  1. 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.
  2. 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.
  3. 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.
  4. The interpolation error in the input is |mb_line − mb| = 0.292089 GeV. This is larger than the requested numerical tolerance.
  5. 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.
  6. 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.

Extra challenging · 03

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.

Explore the question data

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

TWO-BODY |p₁|/M against aPredictions 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. Horizontal axis: a: m₁/M, body 1 (the one plotted). Vertical axis: TWO-BODY |p₁|/M. Exact values are available in the data table below.0.40950.42120.43280.44450.45610.022430.086210.150.21380.2776xy

Horizontal axis (x): a: m₁/M, body 1 (the one plotted)

Axes show the supplied calibration interval; the origin may lie outside the plot.

Reading 1a: m₁/M, body 1 (the one plotted): 0.03188TWO-BODY |p₁|/M: 0.4544
Read the exact data as a table
TWO-BODY |p₁|/M against a — supplied values
a: m₁/M, body 1 (the one plotted)TWO-BODY |p₁|/M
0.0318750.454391383673054
0.090937499999999990.45003905267949307
0.149999999999999970.4414624134623468
0.209062499999999980.4285816060278376
0.268124999999999950.4112620527307054

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
  1. 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.
  2. 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.
  3. 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.
  4. The interpolation error in the input is |a_line − a| = 0.0451864. This is larger than the requested numerical tolerance.
  5. Invert both limits Y_out = 0.43362263 and 0.43904063. Sort the roots even if the response decreases: 0.162796 ≤ a ≤ 0.188233.
  6. 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.

Extra challenging · 04

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.

Explore the question data

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))

FRINGE AMPLITUDE 2g√(p(1−p)) against pPredictions 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. Horizontal axis: p: Weight p of the Q = 0 sector. Vertical axis: FRINGE AMPLITUDE 2g√(p(1−p)). Exact values are available in the data table below.0.36060.42250.48450.54640.60830.091550.19080.290.38920.4885xy

Horizontal axis (x): p: Weight p of the Q = 0 sector

Axes show the supplied calibration interval; the origin may lie outside the plot.

Reading 1p: Weight p of the Q = 0 sector: 0.1062FRINGE AMPLITUDE 2g√(p(1−p)): 0.3698
Read the exact data as a table
FRINGE AMPLITUDE 2g√(p(1−p)) against p — supplied values
p: Weight p of the Q = 0 sectorFRINGE AMPLITUDE 2g√(p(1−p))
0.106250.3697887910686315
0.1981250.47830423111237474
0.290.5445144626178445
0.3818750.5830153835877746
0.473750.5991725544448777

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
  1. 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.
  2. 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.
  3. 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.
  4. The interpolation error in the input is |p_line − p| = 0.0876783. This is larger than the requested numerical tolerance.
  5. Invert both limits Y_out = 0.55922939 and 0.56964939. Sort the roots even if the response decreases: 0.318834 ≤ p ≤ 0.342989.
  6. 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.

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