University Physics IV · topic UPIV-03.08
The energy-momentum four-vector and its invariant
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.
Energy–momentum four-vector: predict a finite change
The energy-momentum four-vector is the single object (E/c, p) whose components mix under a Lorentz boost exactly as (ct, x) do, and whose invariant length E² − (pc)² equals (mc²)² in every inertial frame. Use the supplied topic model Y_out = hypot(p,m), where Y_out is LAB E (MeV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; hypot(a,b) means √(a²+b²). m is Rest energy mc², measured in MeV. Keep p (Lab momentum pc) = 320 MeV; v (Speed of the new frame) = 0.4 c fixed. For this exercise use only 42.5 MeV ≤ m ≤ 357.5 MeV 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 m = 111.8 MeV and 155.9 MeV. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δm across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
LAB E against m
Predictions from the supplied Energy–momentum four-vector 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): LAB E / MeV
Horizontal axis (x): m: Rest energy mc² / MeV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| m: Rest energy mc² / MeV | LAB E / MeV |
|---|---|
| 42.5 | 322.80992859576054 |
| 121.25 | 342.201055667571 |
| 200 | 377.3592452822641 |
| 278.75 | 424.38374438708183 |
| 357.5 | 479.79813463580706 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of m separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At m = 111.8 MeV, substitute into Y_out = hypot(p,m): Y₀ = 338.968 MeV.
- At m = 155.9 MeV, the same fixed-parameter model gives Y₁ = 355.956 MeV.
- Signed change: ΔY_out = Y₁ − Y₀ = 16.9883 MeV; the input change is Δm = 44.1 MeV.
- Average sensitivity is ΔY_out/Δm = 0.385222 MeV per MeV. 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 = 338.968 MeV, Y₁ = 355.956 MeV; ΔY_out = 16.9883 MeV; average sensitivity = 0.385222 MeV per MeV.
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.
Energy–momentum four-vector: infer an input with uncertainty
The energy-momentum four-vector is the single object (E/c, p) whose components mix under a Lorentz boost exactly as (ct, x) do, and whose invariant length E² − (pc)² equals (mc²)² in every inertial frame. Use the supplied topic model Y_out = hypot(p,m), where Y_out is LAB E (MeV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; hypot(a,b) means √(a²+b²). m is Rest energy mc², measured in MeV. Keep p (Lab momentum pc) = 320 MeV; v (Speed of the new frame) = 0.4 c fixed. For this exercise use only 42.5 MeV ≤ m ≤ 357.5 MeV 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 = 396.81308 MeV, with a bounded reading uncertainty ±11.1 MeV. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown m by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0315 MeV, 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.
LAB E against m
Predictions from the supplied Energy–momentum four-vector 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): LAB E / MeV
Horizontal axis (x): m: Rest energy mc² / MeV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| m: Rest energy mc² / MeV | LAB E / MeV |
|---|---|
| 42.5 | 322.80992859576054 |
| 121.25 | 342.201055667571 |
| 200 | 377.3592452822641 |
| 278.75 | 424.38374438708183 |
| 357.5 | 479.79813463580706 |
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(m) = 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(42.5) = 322.81 MeV and Y_out(357.5) = 479.798 MeV. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives m_line = 42.5 + (396.813 − 322.81)(357.5 − 42.5)/(479.798 − 322.81) = 190.989 MeV. 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 m = 234.65 MeV; substituting back gives Y_out = 396.81308 MeV.
- The interpolation error in the input is |m_line − m| = 43.6612 MeV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 385.71308 MeV and 407.91308 MeV. Sort the roots even if the response decreases: 215.348 MeV ≤ m ≤ 252.969 MeV.
- The lower and upper input deviations are 19.3024 MeV and 18.3185 MeV. 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 m ≈ 234.65 MeV; compatible interval [215.348, 252.969] MeV. Endpoint interpolation gives 190.989 MeV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.