University Quantum Field Theory I · topic UQFTI-08.08
A massive vector field for contrast: Proca's propagator and three polarisations
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.
Proca field: propagator and three polarisations: predict a finite change
The Proca field is a real vector field with Lagrangian −¼F_μνF^μν + ½M_V²A_μA^μ. It describes a spin-one particle of mass M_V with three polarisation states, all of positive norm, and it has no gauge freedom to fix. Use the supplied topic model Y_out = mv/sqrt(k^2+mv^2), where Y_out is CONSERVED M_V/E (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. k is Momentum |k|, measured in GeV. Keep mv (Vector mass M_V) = 20 GeV; d (Non-conservation δ) = 0.02 fixed. For this exercise use only 107.144 GeV ≤ k ≤ 893.856 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. Calculate Y_out at k = 280.221 GeV and 390.36 GeV. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δk across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
CONSERVED M_V/E against k
Predictions from the supplied Proca field: propagator and three polarisations 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): CONSERVED M_V/E
Horizontal axis (x): k: Momentum |k| / GeV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| k: Momentum |k| / GeV | CONSERVED M_V/E |
|---|---|
| 107.14375 | 0.18349563546002057 |
| 303.821875 | 0.06568587930590002 |
| 500.5 | 0.03992817392616205 |
| 697.1781249999999 | 0.028675276629026726 |
| 893.8562499999999 | 0.02236936363042669 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of k separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At k = 280.221 GeV, substitute into Y_out = mv/sqrt(k^2+mv^2): Y₀ = 0.0711913.
- At k = 390.36 GeV, the same fixed-parameter model gives Y₁ = 0.0511676.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.0200237; the input change is Δk = 110.14 GeV.
- Average sensitivity is ΔY_out/Δk = -0.000181802 1 per GeV. 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 = 0.0711913, Y₁ = 0.0511676; ΔY_out = -0.0200237; average sensitivity = -0.000181802 1 per GeV.
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.
Proca field: propagator and three polarisations: infer an input with uncertainty
The Proca field is a real vector field with Lagrangian −¼F_μνF^μν + ½M_V²A_μA^μ. It describes a spin-one particle of mass M_V with three polarisation states, all of positive norm, and it has no gauge freedom to fix. Use the supplied topic model Y_out = mv/sqrt(k^2+mv^2), where Y_out is CONSERVED M_V/E (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. k is Momentum |k|, measured in GeV. Keep mv (Vector mass M_V) = 20 GeV; d (Non-conservation δ) = 0.02 fixed. For this exercise use only 107.144 GeV ≤ k ≤ 893.856 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 = 0.034049568, with a bounded reading uncertainty ±0.001752. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown k by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0786712 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.
CONSERVED M_V/E against k
Predictions from the supplied Proca field: propagator and three polarisations 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): CONSERVED M_V/E
Horizontal axis (x): k: Momentum |k| / GeV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| k: Momentum |k| / GeV | CONSERVED M_V/E |
|---|---|
| 107.14375 | 0.18349563546002057 |
| 303.821875 | 0.06568587930590002 |
| 500.5 | 0.03992817392616205 |
| 697.1781249999999 | 0.028675276629026726 |
| 893.8562499999999 | 0.02236936363042669 |
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(k) = 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(107.144) = 0.183496 and Y_out(893.856) = 0.0223694. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives k_line = 107.144 + (0.0340496 − 0.183496)(893.856 − 107.144)/(0.0223694 − 0.183496) = 836.827 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 k = 587.038 GeV; substituting back gives Y_out = 0.034049568.
- The interpolation error in the input is |k_line − k| = 249.788 GeV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.032297568 and 0.035801568. Sort the roots even if the response decreases: 558.277 GeV ≤ k ≤ 618.919 GeV.
- The lower and upper input deviations are 28.7618 GeV and 31.8802 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 k ≈ 587.038 GeV; compatible interval [558.277, 618.919] GeV. Endpoint interpolation gives 836.827 GeV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.