University Physics IV · unit UPIV-14
Particle Physics
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.
Relativistic kinematics and invariant mass: infer an input with uncertainty
The invariant mass of a system is M = √(E_total² − |p_total c|²)/c², assembled from the total energy and the total momentum, and every inertial observer computes the same value for it. Use the supplied topic model Y_out = (4*eb*eb-2*mt*mt)/(2*max(mt,0.05)), where Y_out is TARGET BEAM TO MATCH (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. mt is Target rest mass m, measured in GeV/c². Keep eb (Beam energy per beam E) = 46 GeV fixed. For this exercise use only 0.301875 GeV/c² ≤ mt ≤ 1.79812 GeV/c² 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 = 3483.0959 GeV, with a bounded reading uncertainty ±169.7 GeV. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown mt by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000149625 GeV/c², 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.
TARGET BEAM TO MATCH against mt
Predictions from the supplied Relativistic kinematics and invariant mass 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): TARGET BEAM TO MATCH / GeV
Horizontal axis (x): mt: Target rest mass m / GeV/c²
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| mt: Target rest mass m / GeV/c² | TARGET BEAM TO MATCH / GeV |
|---|---|
| 0.301875 | 14018.745744047617 |
| 0.6759375 | 6260.257950618355 |
| 1.0499999999999998 | 4029.426190476191 |
| 1.4240624999999998 | 2970.355617113781 |
| 1.7981249999999998 | 2351.7646139641993 |
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(mt) = 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(0.301875) = 14018.7 GeV and Y_out(1.79812) = 2351.76 GeV. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives mt_line = 0.301875 + (3483.1 − 14018.7)(1.79812 − 0.301875)/(2351.76 − 14018.7) = 1.65304 GeV/c². 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 mt = 1.21459 GeV/c²; substituting back gives Y_out = 3483.0959 GeV.
- The interpolation error in the input is |mt_line − mt| = 0.438448 GeV/c². This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 3313.3959 GeV and 3652.7959 GeV. Sort the roots even if the response decreases: 1.1582 GeV/c² ≤ mt ≤ 1.27675 GeV/c².
- The lower and upper input deviations are 0.0563901 GeV/c² and 0.06216 GeV/c². 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 mt ≈ 1.21459 GeV/c²; compatible interval [1.1582, 1.27675] GeV/c². Endpoint interpolation gives 1.65304 GeV/c².
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Quarks, leptons and generations: infer an input with uncertainty
The fermion inventory is the Standard Model's list of matter fields, sorted not by mass but by the SU(3), SU(2) and U(1) charges each one carries: six quarks in colour triplets, six colourless leptons, arranged in three identical-looking generations. Use the supplied topic model Y_out = 258.66/max(4*(rs-91.19)^2+(1.9964+n*gv/1000)^2,0.05), where Y_out is σ AT THIS ENERGY (nb). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; max(a,b) selects the larger value. rs is Beam energy √s, measured in GeV. Keep n (Light neutrino species N) = 3; gv (Width per species Γ(νν)) = 167 MeV fixed. For this exercise use only 88.6375 GeV ≤ rs ≤ 91 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 = 23.140874 nb, with a bounded reading uncertainty ±2.2699 nb. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown rs by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00023625 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.
σ AT THIS ENERGY against rs
Predictions from the supplied Quarks, leptons and generations 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): σ AT THIS ENERGY / nb
Horizontal axis (x): rs: Beam energy √s / GeV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| rs: Beam energy √s / GeV | σ AT THIS ENERGY / nb |
|---|---|
| 88.6375 | 8.008537545632807 |
| 89.228125 | 11.956831803043125 |
| 89.81875 | 18.80027004851515 |
| 90.409375 | 29.818404730684946 |
| 91 | 40.53338232900863 |
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(rs) = 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(88.6375) = 8.00854 nb and Y_out(91) = 40.5334 nb. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives rs_line = 88.6375 + (23.1409 − 8.00854)(91 − 88.6375)/(40.5334 − 8.00854) = 89.7367 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 rs = 90.0786 GeV; substituting back gives Y_out = 23.140874 nb.
- The interpolation error in the input is |rs_line − rs| = 0.341961 GeV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 20.870974 nb and 25.410774 nb. Sort the roots even if the response decreases: 89.9494 GeV ≤ rs ≤ 90.1973 GeV.
- The lower and upper input deviations are 0.129218 GeV and 0.118634 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 rs ≈ 90.0786 GeV; compatible interval [89.9494, 90.1973] GeV. Endpoint interpolation gives 89.7367 GeV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Interactions and their gauge bosons: infer an input with uncertainty
A gauge boson is the spin-1 field that a local phase symmetry forces into existence, and the mass it carries sets how far the interaction it mediates can reach — unless, as for the massless gluon, confinement ends that reach first. Use the supplied topic model Y_out = 10^u, where Y_out is MEDIATOR MASS Mc² (MeV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. u is Mediator mass log₁₀(Mc²/MeV), dimensionless. Keep v (Separation log₁₀(r/fm)) = 0 fixed. For this exercise use only 1.80375 ≤ u ≤ 4.79625 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 = 4257.6994 MeV, with a bounded reading uncertainty ±629.11 MeV. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown u by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00029925, 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.
MEDIATOR MASS Mc² against u
Predictions from the supplied Interactions and their gauge bosons 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): MEDIATOR MASS Mc² / MeV
Horizontal axis (x): u: Mediator mass log₁₀(Mc²/MeV)
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| u: Mediator mass log₁₀(Mc²/MeV) | MEDIATOR MASS Mc² / MeV |
|---|---|
| 1.80375 | 63.64290574262721 |
| 2.551875 | 356.34855330614226 |
| 3.3 | 1995.2623149688789 |
| 4.048125 | 11171.847531298365 |
| 4.79625 | 62553.267470760664 |
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(u) = 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(1.80375) = 63.6429 MeV and Y_out(4.79625) = 62553.3 MeV. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives u_line = 1.80375 + (4257.7 − 63.6429)(4.79625 − 1.80375)/(62553.3 − 63.6429) = 2.00459. 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 u = 3.62917; substituting back gives Y_out = 4257.6994 MeV.
- The interpolation error in the input is |u_line − u| = 1.62458. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 3628.5894 MeV and 4886.8094 MeV. Sort the roots even if the response decreases: 3.55974 ≤ u ≤ 3.68903.
- The lower and upper input deviations are 0.0694372 and 0.0598504. 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 u ≈ 3.62917; compatible interval [3.55974, 3.68903]. Endpoint interpolation gives 2.00459.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Feynman diagrams and perturbative amplitudes: infer an input with uncertainty
A Feynman diagram is shorthand for one term in the perturbative expansion of a transition amplitude: each vertex supplies a factor of the coupling, each internal line a propagator, and only the sum over all diagrams of a given order means anything physically. Use the supplied topic model Y_out = (floor(1/max(c,0.03))*ln(floor(1/max(c,0.03)))-floor(1/max(c,0.03))+0.5*ln(2*pi*floor(1/max(c,0.03)))+1/(12*floor(1/max(c,0.03)))+floor(1/max(c,0.03))*ln(max(c,0.03)))/2.302585, where Y_out is log10 SMALLEST TERM (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; ln(x) means the natural logarithm; max(a,b) selects the larger value. c is Expansion parameter λ, dimensionless. Keep n (Truncate after order n) = 8 fixed. For this exercise use only 0.0905625 ≤ c ≤ 0.539437 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 = -0.57572834, with a bounded reading uncertainty ±0.046298. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown c by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0000448875, 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.
log10 SMALLEST TERM against c
Predictions from the supplied Feynman diagrams and perturbative amplitudes 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): log10 SMALLEST TERM
Horizontal axis (x): c: Expansion parameter λ
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| c: Expansion parameter λ | log10 SMALLEST TERM |
|---|---|
| 0.09056249999999999 | -3.8724111031522868 |
| 0.20278124999999997 | -1.391659103601069 |
| 0.31499999999999995 | -0.7268737523907601 |
| 0.4272187499999999 | -0.4375278447607972 |
| 0.5394374999999999 | -0.26707221773105216 |
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(c) = 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(0.0905625) = -3.87241 and Y_out(0.539437) = -0.267072. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives c_line = 0.0905625 + (-0.575728 − -3.87241)(0.539437 − 0.0905625)/(-0.267072 − -3.87241) = 0.501009. 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 c = 0.364376; substituting back gives Y_out = -0.57572834.
- The interpolation error in the input is |c_line − c| = 0.136633. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = -0.62202634 and -0.52943034. Sort the roots even if the response decreases: 0.345463 ≤ c ≤ 0.384325.
- The lower and upper input deviations are 0.0189136 and 0.0199491. 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 c ≈ 0.364376; compatible interval [0.345463, 0.384325]. Endpoint interpolation gives 0.501009.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.