University Advanced Quantum Mechanics · unit UAQM-05
Lorentz Covariance of the Dirac Equation
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.
Lorentz generators: rotations J and boosts K: infer an input with uncertainty
A Lorentz transformation is a real 4 × 4 matrix Λ that leaves c²t² − x² − y² − z² unchanged, Λᵀ g Λ = g, and its generators are the three rotation matrices J and three boost matrices K whose exponentials, multiplied together, give every Λ reachable continuously from the identity. Use the supplied topic model Y_out = (1+b1*b2*cosd(alpha))/sqrt((1-b1^2)*(1-b2^2)), where Y_out is γ OF THE PRODUCT (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; sind/cosd/tand use degrees. b1 is First boost v₁/c, along x, dimensionless. Keep b2 (Second boost v₂/c) = 0.8; alpha (Angle α of v₂ from x) = 90 ° fixed. For this exercise use only 0.145625 ≤ b1 ≤ 0.854375 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 = 2.0423252, with a bounded reading uncertainty ±0.053655. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown b1 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.
γ OF THE PRODUCT against b1
Predictions from the supplied Lorentz generators: rotations J and boosts K 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): γ OF THE PRODUCT
Horizontal axis (x): b1: First boost v₁/c, along x
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| b1: First boost v₁/c, along x | γ OF THE PRODUCT |
|---|---|
| 0.145625 | 1.6846250043881952 |
| 0.32281249999999995 | 1.7609423493452323 |
| 0.49999999999999994 | 1.9245008972987527 |
| 0.6771874999999999 | 2.2650759534518237 |
| 0.8543749999999999 | 3.2072439797353334 |
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(b1) = 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(0.145625) = 1.68463 and Y_out(0.854375) = 3.20724. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives b1_line = 0.145625 + (2.04233 − 1.68463)(0.854375 − 0.145625)/(3.20724 − 1.68463) = 0.312128. 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 b1 = 0.577963; substituting back gives Y_out = 2.0423252.
- The interpolation error in the input is |b1_line − b1| = 0.265835. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1.9886702 and 2.0959802. Sort the roots even if the response decreases: 0.545546 ≤ b1 ≤ 0.606383.
- The lower and upper input deviations are 0.0324167 and 0.0284202. 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 b1 ≈ 0.577963; compatible interval [0.545546, 0.606383]. Endpoint interpolation gives 0.312128.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
The covariance condition S⁻¹ γ^μ S = Λ^μ_ν γ^ν: infer an input with uncertainty
The covariance condition is the matrix equation S⁻¹ γ^μ S = Λ^μ_ν γ^ν that the 4 × 4 matrix S(Λ) in ψ′(x′) = S(Λ)ψ(x) must satisfy for the Dirac equation, with unchanged γ matrices, to hold in every inertial frame. Use the supplied topic model Y_out = kappa^4, where Y_out is det S = κ⁴ (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. kappa is Overall factor κ in S, dimensionless. Keep beta (Frame speed β = v/c) = 0.6; phi (Rapidity φ written into S) = 0.69 fixed. For this exercise use only 0.60625 ≤ kappa ≤ 1.39375 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 = 1.3941798, with a bounded reading uncertainty ±0.18886. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown kappa by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00007875, 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.
det S = κ⁴ against kappa
Predictions from the supplied The covariance condition S⁻¹ γ^μ S = Λ^μ_ν γ^ν 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): det S = κ⁴
Horizontal axis (x): kappa: Overall factor κ in S
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| kappa: Overall factor κ in S | det S = κ⁴ |
|---|---|
| 0.60625 | 0.13508496246337887 |
| 0.8031249999999999 | 0.41603759775161714 |
| 0.9999999999999999 | 0.9999999999999996 |
| 1.196875 | 2.0520842286109917 |
| 1.3937499999999998 | 3.773458009338377 |
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(kappa) = 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(0.60625) = 0.135085 and Y_out(1.39375) = 3.77346. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives kappa_line = 0.60625 + (1.39418 − 0.135085)(1.39375 − 0.60625)/(3.77346 − 0.135085) = 0.878772. 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 kappa = 1.08663; substituting back gives Y_out = 1.3941798.
- The interpolation error in the input is |kappa_line − kappa| = 0.207853. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1.2053198 and 1.5830398. Sort the roots even if the response decreases: 1.04779 ≤ kappa ≤ 1.12169.
- The lower and upper input deviations are 0.0388318 and 0.0350653. 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 kappa ≈ 1.08663; compatible interval [1.04779, 1.12169]. Endpoint interpolation gives 0.878772.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Infinitesimal transformations and the generators σ^μν: infer an input with uncertainty
The generators σ^μν = (i/2)[γ^μ, γ^ν] are six 4 × 4 matrices that say how a Dirac spinor responds to an infinitesimal Lorentz transformation Λ = 1 + ω: the spinor is multiplied by S = 1 − (i/4)σ_μνω^μν. Use the supplied topic model Y_out = (exp(eta)+exp(-eta))/2*(exp(s)+exp(-s))/2+(exp(eta)-exp(-eta))/2*(exp(s)-exp(-s))/2*cosd(phi), where Y_out is EXACT AT δη (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; exp(x) means eˣ; sind/cosd/tand use degrees. s is Boost step δη, dimensionless. Keep eta (Electron rapidity η₀) = 0.7; phi (Boost angle to momentum φ) = 0 ° fixed. For this exercise use only -0.63 ≤ s ≤ 0.7875 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 = 1.4695496, with a bounded reading uncertainty ±0.070065. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown s by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00014175, 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.
EXACT AT δη against s
Predictions from the supplied Infinitesimal transformations and the generators σ^μν 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): EXACT AT δη
Horizontal axis (x): s: Boost step δη
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| s: Boost step δη | EXACT AT δη |
|---|---|
| -0.63 | 1.002451000580082 |
| -0.275625 | 1.0914066216492795 |
| 0.07874999999999999 | 1.318863252914981 |
| 0.43312499999999987 | 1.7136854555602468 |
| 0.7875 | 2.3259767133185223 |
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(s) = 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(-0.63) = 1.00245 and Y_out(0.7875) = 2.32598. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives s_line = -0.63 + (1.46955 − 1.00245)(0.7875 − -0.63)/(2.32598 − 1.00245) = -0.129736. 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 s = 0.234675; substituting back gives Y_out = 1.4695496.
- The interpolation error in the input is |s_line − s| = 0.364411. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1.3994846 and 1.5396146. Sort the roots even if the response decreases: 0.166489 ≤ s ≤ 0.297045.
- The lower and upper input deviations are 0.0681865 and 0.06237. 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 s ≈ 0.234675; compatible interval [0.166489, 0.297045]. Endpoint interpolation gives -0.129736.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Finite spinor rotations and boosts: infer an input with uncertainty
A finite spinor transformation is the 4 × 4 matrix S(Λ) found by exponentiating topic 5.03's generators: a rotation through φ about n̂ multiplies a Dirac spinor by cos(φ/2) − iΣ⋅n̂ sin(φ/2), and a boost of rapidity η along n̂ by cosh(η/2) + α⋅n̂ sinh(η/2). Use the supplied topic model Y_out = (exp(eta)+exp(-eta))/2, where Y_out is γ = u†u ÷ ūu (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; exp(x) means eˣ. eta is Boost rapidity η, dimensionless. Keep phi (Rotation angle φ) = 120 ° fixed. For this exercise use only 0.265625 ≤ eta ≤ 2.23438 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 = 2.2825138, with a bounded reading uncertainty ±0.18705. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown eta by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000196875, 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.
γ = u†u ÷ ūu against eta
Predictions from the supplied Finite spinor rotations and boosts 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): γ = u†u ÷ ūu
Horizontal axis (x): eta: Boost rapidity η
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| eta: Boost rapidity η | γ = u†u ÷ ūu |
|---|---|
| 0.265625 | 1.035486235419229 |
| 0.7578125 | 1.3011472102964041 |
| 1.25 | 1.8884238771610158 |
| 1.7421875 | 2.9424784360370806 |
| 2.234375 | 4.7238505734895115 |
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(eta) = 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(0.265625) = 1.03549 and Y_out(2.23438) = 4.72385. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives eta_line = 0.265625 + (2.28251 − 1.03549)(2.23438 − 0.265625)/(4.72385 − 1.03549) = 0.931255. 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 eta = 1.46656; substituting back gives Y_out = 2.2825138.
- The interpolation error in the input is |eta_line − eta| = 0.535308. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 2.0954638 and 2.4695638. Sort the roots even if the response decreases: 1.3704 ≤ eta ≤ 1.55342.
- The lower and upper input deviations are 0.0961633 and 0.086856. 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 eta ≈ 1.46656; compatible interval [1.3704, 1.55342]. Endpoint interpolation gives 0.931255.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.