University Advanced Quantum Mechanics · topic UAQM-15.09
Landau's criterion and frictionless flow of a Bose gas
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.
Landau's criterion and frictionless flow of a Bose gas: predict a finite change
Landau's criterion: a weak obstacle moving at speed v through a uniform fluid at zero temperature can create elementary excitations, and so feel drag, only if v exceeds v_c, the minimum over all k of the energy per unit momentum Eₖ/(ħk) of those excitations. Use the supplied topic model Y_out = 0.1*n*as^3, where Y_out is DILUTENESS n aₛ³ (× 10⁻⁶). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. as is Scattering length aₛ, measured in nm. Keep n (Density n) = 1 10²⁰ m⁻³; v (Obstacle speed v) = 7.5 mm s⁻¹ fixed. For this exercise use only 0.53125 nm ≤ as ≤ 4.46875 nm 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 as = 1.3975 nm and 1.94875 nm. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δas across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
DILUTENESS n aₛ³ against as
Predictions from the supplied Landau's criterion and frictionless flow of a Bose gas 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): DILUTENESS n aₛ³ / × 10⁻⁶
Horizontal axis (x): as: Scattering length aₛ / nm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| as: Scattering length aₛ / nm | DILUTENESS n aₛ³ / × 10⁻⁶ |
|---|---|
| 0.53125 | 0.0149932861328125 |
| 1.515625 | 0.3481571197509766 |
| 2.5 | 1.5625 |
| 3.484375 | 4.230334091186523 |
| 4.46875 | 8.923971557617188 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of as separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At as = 1.3975 nm, substitute into Y_out = 0.1*n*as^3: Y₀ = 0.272933 × 10⁻⁶.
- At as = 1.94875 nm, the same fixed-parameter model gives Y₁ = 0.740062 × 10⁻⁶.
- Signed change: ΔY_out = Y₁ − Y₀ = 0.46713 × 10⁻⁶; the input change is Δas = 0.55125 nm.
- Average sensitivity is ΔY_out/Δas = 0.847401 × 10⁻⁶ per nm. 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 = 0.272933 × 10⁻⁶, Y₁ = 0.740062 × 10⁻⁶; ΔY_out = 0.46713 × 10⁻⁶; average sensitivity = 0.847401 × 10⁻⁶ per nm.
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.
Landau's criterion and frictionless flow of a Bose gas: infer an input with uncertainty
Landau's criterion: a weak obstacle moving at speed v through a uniform fluid at zero temperature can create elementary excitations, and so feel drag, only if v exceeds v_c, the minimum over all k of the energy per unit momentum Eₖ/(ħk) of those excitations. Use the supplied topic model Y_out = 0.1*n*as^3, where Y_out is DILUTENESS n aₛ³ (× 10⁻⁶). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. as is Scattering length aₛ, measured in nm. Keep n (Density n) = 1 10²⁰ m⁻³; v (Obstacle speed v) = 7.5 mm s⁻¹ fixed. For this exercise use only 0.53125 nm ≤ as ≤ 4.46875 nm 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 = 2.5234326 × 10⁻⁶, with a bounded reading uncertainty ±0.37627 × 10⁻⁶. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown as by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00039375 nm, 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.
DILUTENESS n aₛ³ against as
Predictions from the supplied Landau's criterion and frictionless flow of a Bose gas 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): DILUTENESS n aₛ³ / × 10⁻⁶
Horizontal axis (x): as: Scattering length aₛ / nm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| as: Scattering length aₛ / nm | DILUTENESS n aₛ³ / × 10⁻⁶ |
|---|---|
| 0.53125 | 0.0149932861328125 |
| 1.515625 | 0.3481571197509766 |
| 2.5 | 1.5625 |
| 3.484375 | 4.230334091186523 |
| 4.46875 | 8.923971557617188 |
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(as) = 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(0.53125) = 0.0149933 × 10⁻⁶ and Y_out(4.46875) = 8.92397 × 10⁻⁶. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives as_line = 0.53125 + (2.52343 − 0.0149933)(4.46875 − 0.53125)/(8.92397 − 0.0149933) = 1.6399 nm. 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 as = 2.93312 nm; substituting back gives Y_out = 2.5234326 × 10⁻⁶.
- The interpolation error in the input is |as_line − as| = 1.29322 nm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 2.1471626 × 10⁻⁶ and 2.8997026 × 10⁻⁶. Sort the roots even if the response decreases: 2.77943 nm ≤ as ≤ 3.07221 nm.
- The lower and upper input deviations are 0.1537 nm and 0.139087 nm. 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 as ≈ 2.93312 nm; compatible interval [2.77943, 3.07221] nm. Endpoint interpolation gives 1.6399 nm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.