University Quantum Field Theory I · topic UQFTI-12.09
The photon self-energy is transverse, so the photon stays massless
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.
A transverse self-energy keeps the photon massless: predict a finite change
The photon self-energy Π^μν(q) obeys q_μΠ^μν = 0, so it equals (q²g^μν − q^μq^ν)Π(q²); the factor q² lets it rescale the photon propagator by Z₃ = 1/(1 − Π(0)) but never move its pole from q² = 0. Use the supplied topic model Y_out = sqrt(2*a/((1-p0)+sqrt((1-p0)^2-4*s*a))), where Y_out is PHOTON MASS (mₑ). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. a is Non-transverse piece a, measured in mₑ². Keep p0 (Π(0), the self-energy at q² = 0) = -0.08; s (Slope s, in Π(q²) = Π(0) + s q²/mₑ²) = -0.1 fixed. For this exercise use only 0.053125 mₑ² ≤ a ≤ 0.446875 mₑ² 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 a = 0.13975 mₑ² and 0.194875 mₑ². Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δa across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
PHOTON MASS against a
Predictions from the supplied A transverse self-energy keeps the photon massless 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): PHOTON MASS / mₑ
Horizontal axis (x): a: Non-transverse piece a / mₑ²
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| a: Non-transverse piece a / mₑ² | PHOTON MASS / mₑ |
|---|---|
| 0.053125 | 0.22128667415882272 |
| 0.1515625 | 0.37223383152899087 |
| 0.25 | 0.47615329323032035 |
| 0.34843749999999996 | 0.5599335845349833 |
| 0.44687499999999997 | 0.6316882865124185 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of a separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At a = 0.13975 mₑ², substitute into Y_out = sqrt(2*a/((1-p0)+sqrt((1-p0)^2-4*s*a))): Y₀ = 0.357609 mₑ.
- At a = 0.194875 mₑ², the same fixed-parameter model gives Y₁ = 0.421333 mₑ.
- Signed change: ΔY_out = Y₁ − Y₀ = 0.0637249 mₑ; the input change is Δa = 0.055125 mₑ².
- Average sensitivity is ΔY_out/Δa = 1.15601 mₑ per mₑ². Its positive sign means that this response increases 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 = 0.357609 mₑ, Y₁ = 0.421333 mₑ; ΔY_out = 0.0637249 mₑ; average sensitivity = 1.15601 mₑ per mₑ².
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.
A transverse self-energy keeps the photon massless: infer an input with uncertainty
The photon self-energy Π^μν(q) obeys q_μΠ^μν = 0, so it equals (q²g^μν − q^μq^ν)Π(q²); the factor q² lets it rescale the photon propagator by Z₃ = 1/(1 − Π(0)) but never move its pole from q² = 0. Use the supplied topic model Y_out = sqrt(2*a/((1-p0)+sqrt((1-p0)^2-4*s*a))), where Y_out is PHOTON MASS (mₑ). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. a is Non-transverse piece a, measured in mₑ². Keep p0 (Π(0), the self-energy at q² = 0) = -0.08; s (Slope s, in Π(q²) = Π(0) + s q²/mₑ²) = -0.1 fixed. For this exercise use only 0.053125 mₑ² ≤ a ≤ 0.446875 mₑ² 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 = 0.51485865 mₑ, with a bounded reading uncertainty ±0.017524 mₑ. 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 0.000039375 mₑ², 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.
PHOTON MASS against a
Predictions from the supplied A transverse self-energy keeps the photon massless 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): PHOTON MASS / mₑ
Horizontal axis (x): a: Non-transverse piece a / mₑ²
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| a: Non-transverse piece a / mₑ² | PHOTON MASS / mₑ |
|---|---|
| 0.053125 | 0.22128667415882272 |
| 0.1515625 | 0.37223383152899087 |
| 0.25 | 0.47615329323032035 |
| 0.34843749999999996 | 0.5599335845349833 |
| 0.44687499999999997 | 0.6316882865124185 |
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
- Endpoint predictions are Y_out(0.053125) = 0.221287 mₑ and Y_out(0.446875) = 0.631688 mₑ. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives a_line = 0.053125 + (0.514859 − 0.221287)(0.446875 − 0.053125)/(0.631688 − 0.221287) = 0.334786 mₑ². 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 a = 0.293312 mₑ²; substituting back gives Y_out = 0.51485865 mₑ.
- The interpolation error in the input is |a_line − a| = 0.0414731 mₑ². This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.49733465 mₑ and 0.53238265 mₑ. Sort the roots even if the response decreases: 0.273247 mₑ² ≤ a ≤ 0.314139 mₑ².
- The lower and upper input deviations are 0.0200656 mₑ² and 0.0208266 mₑ². 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 ≈ 0.293312 mₑ²; compatible interval [0.273247, 0.314139] mₑ². Endpoint interpolation gives 0.334786 mₑ².
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.