University Quantum Field Theory I · topic UQFTI-08.01
The Maxwell Lagrangian and the missing momentum of A^0
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.
Further practice
Work through a fresh case before tackling the extension.
The Maxwell Lagrangian and the missing momentum of A⁰: predict a finite change
The momentum conjugate to A₀ in the Maxwell Lagrangian is identically zero, because ∂₀A₀ appears only in F₀₀ = 0; A⁰ is therefore not a dynamical coordinate but a Lagrange multiplier enforcing Gauss's law, a constraint on the fields at one instant. Use the supplied topic model Y_out = lam*pi/(2*w), where Y_out is PEAK δA⁰ = ∂ₜΛ (ρ₀/k²). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. w is Ramp duration kΔt, dimensionless. Keep lam (Final gauge amplitude ℓ) = 0.6 ρ₀/k³; t0 (Switch-on time kt₀) = 2 fixed. For this exercise use only 1.31875 ≤ w ≤ 3.68125 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. Calculate Y_out at w = 1.8385 and 2.16925. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δw across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
PEAK δA⁰ = ∂ₜΛ against w
Predictions from the supplied The Maxwell Lagrangian and the missing momentum of A⁰ 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): PEAK δA⁰ = ∂ₜΛ / ρ₀/k²
Horizontal axis (x): w: Ramp duration kΔt
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| w: Ramp duration kΔt | PEAK δA⁰ = ∂ₜΛ / ρ₀/k² |
|---|---|
| 1.31875 | 0.7146751060299055 |
| 1.909375 | 0.4936053923807203 |
| 2.5 | 0.37699111843077515 |
| 3.090625 | 0.30494731521195156 |
| 3.68125 | 0.25602113305994917 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of w separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At w = 1.8385, substitute into Y_out = lam*pi/(2*w): Y₀ = 0.512634 ρ₀/k².
- At w = 2.16925, the same fixed-parameter model gives Y₁ = 0.434472 ρ₀/k².
- Signed change: ΔY_out = Y₁ − Y₀ = -0.0781624 ρ₀/k²; the input change is Δw = 0.33075.
- Average sensitivity is ΔY_out/Δw = -0.236319 ρ₀/k² per unit input. Its negative sign means that this response decreases over the specified interval.
- This is a secant slope between two settings. It equals a point derivative only for a locally linear relation or in a justified small-interval limit.
Answer Y₀ = 0.512634 ρ₀/k², Y₁ = 0.434472 ρ₀/k²; ΔY_out = -0.0781624 ρ₀/k²; average sensitivity = -0.236319 ρ₀/k² per unit input.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Extra challenging questions
Work through the linked stages and explain what each result means.
The Maxwell Lagrangian and the missing momentum of A⁰: infer an input with uncertainty
The momentum conjugate to A₀ in the Maxwell Lagrangian is identically zero, because ∂₀A₀ appears only in F₀₀ = 0; A⁰ is therefore not a dynamical coordinate but a Lagrange multiplier enforcing Gauss's law, a constraint on the fields at one instant. Use the supplied topic model Y_out = lam*pi/(2*w), where Y_out is PEAK δA⁰ = ∂ₜΛ (ρ₀/k²). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. w is Ramp duration kΔt, dimensionless. Keep lam (Final gauge amplitude ℓ) = 0.6 ρ₀/k³; t0 (Switch-on time kt₀) = 2 fixed. For this exercise use only 1.31875 ≤ w ≤ 3.68125 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 = 0.34149293 ρ₀/k², with a bounded reading uncertainty ±0.012821 ρ₀/k². Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown w 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, 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.
PEAK δA⁰ = ∂ₜΛ against w
Predictions from the supplied The Maxwell Lagrangian and the missing momentum of A⁰ 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): PEAK δA⁰ = ∂ₜΛ / ρ₀/k²
Horizontal axis (x): w: Ramp duration kΔt
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| w: Ramp duration kΔt | PEAK δA⁰ = ∂ₜΛ / ρ₀/k² |
|---|---|
| 1.31875 | 0.7146751060299055 |
| 1.909375 | 0.4936053923807203 |
| 2.5 | 0.37699111843077515 |
| 3.090625 | 0.30494731521195156 |
| 3.68125 | 0.25602113305994917 |
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(w) = 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(1.31875) = 0.714675 ρ₀/k² and Y_out(3.68125) = 0.256021 ρ₀/k². The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives w_line = 1.31875 + (0.341493 − 0.714675)(3.68125 − 1.31875)/(0.256021 − 0.714675) = 3.24099. 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 w = 2.75987; substituting back gives Y_out = 0.34149293 ρ₀/k².
- The interpolation error in the input is |w_line − w| = 0.481115. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.32867193 ρ₀/k² and 0.35431393 ρ₀/k². Sort the roots even if the response decreases: 2.66001 ≤ w ≤ 2.86753.
- The lower and upper input deviations are 0.0998672 and 0.107659. 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 w ≈ 2.75987; compatible interval [2.66001, 2.86753]. Endpoint interpolation gives 3.24099.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.