University Advanced Quantum Mechanics · topic UAQM-11.06
The Casimir force between conducting plates
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 Casimir force between conducting plates: predict a finite change
The Casimir force is the attraction between two neutral, parallel conducting plates in vacuum, π²ħc/(240d⁴) per unit area for ideal plates a distance d apart, which arises because the plates change the zero-point energies of the field modes between them. Use the supplied topic model Y_out = 1.30013*(1000/d)^4, where Y_out is IDEAL PRESSURE π²ħc/240d⁴ (mPa). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. d is Plate separation d, measured in nm. Keep lam (Cutoff length λ) = 50 nm fixed. For this exercise use only 391.25 nm ≤ d ≤ 1808.75 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 d = 703.1 nm and 901.55 nm. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δd across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
IDEAL PRESSURE π²ħc/240d⁴ against d
Predictions from the supplied The Casimir force between conducting plates 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): IDEAL PRESSURE π²ħc/240d⁴ / mPa
Horizontal axis (x): d: Plate separation d / nm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| d: Plate separation d / nm | IDEAL PRESSURE π²ħc/240d⁴ / mPa |
|---|---|
| 391.25 | 55.48420623873223 |
| 745.625 | 4.206345439114935 |
| 1100 | 0.8880062837237893 |
| 1454.3749999999998 | 0.29059034265535155 |
| 1808.7499999999998 | 0.12147099895924326 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of d separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At d = 703.1 nm, substitute into Y_out = 1.30013*(1000/d)^4: Y₀ = 5.32008 mPa.
- At d = 901.55 nm, the same fixed-parameter model gives Y₁ = 1.96801 mPa.
- Signed change: ΔY_out = Y₁ − Y₀ = -3.35207 mPa; the input change is Δd = 198.45 nm.
- Average sensitivity is ΔY_out/Δd = -0.0168913 mPa per nm. 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 = 5.32008 mPa, Y₁ = 1.96801 mPa; ΔY_out = -3.35207 mPa; average sensitivity = -0.0168913 mPa 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.
The Casimir force between conducting plates: infer an input with uncertainty
The Casimir force is the attraction between two neutral, parallel conducting plates in vacuum, π²ħc/(240d⁴) per unit area for ideal plates a distance d apart, which arises because the plates change the zero-point energies of the field modes between them. Use the supplied topic model Y_out = 1.30013*(1000/d)^4, where Y_out is IDEAL PRESSURE π²ħc/240d⁴ (mPa). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. d is Plate separation d, measured in nm. Keep lam (Cutoff length λ) = 50 nm fixed. For this exercise use only 391.25 nm ≤ d ≤ 1808.75 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 = 0.52255494 mPa, with a bounded reading uncertainty ±0.060163 mPa. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown d by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.14175 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.
IDEAL PRESSURE π²ħc/240d⁴ against d
Predictions from the supplied The Casimir force between conducting plates 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): IDEAL PRESSURE π²ħc/240d⁴ / mPa
Horizontal axis (x): d: Plate separation d / nm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| d: Plate separation d / nm | IDEAL PRESSURE π²ħc/240d⁴ / mPa |
|---|---|
| 391.25 | 55.48420623873223 |
| 745.625 | 4.206345439114935 |
| 1100 | 0.8880062837237893 |
| 1454.3749999999998 | 0.29059034265535155 |
| 1808.7499999999998 | 0.12147099895924326 |
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(d) = 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(391.25) = 55.4842 mPa and Y_out(1808.75) = 0.121471 mPa. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives d_line = 391.25 + (0.522555 − 55.4842)(1808.75 − 391.25)/(0.121471 − 55.4842) = 1798.48 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 d = 1255.93 nm; substituting back gives Y_out = 0.52255494 mPa.
- The interpolation error in the input is |d_line − d| = 542.556 nm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.46239194 mPa and 0.58271794 mPa. Sort the roots even if the response decreases: 1222.17 nm ≤ d ≤ 1294.92 nm.
- The lower and upper input deviations are 33.7537 nm and 38.9986 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 d ≈ 1255.93 nm; compatible interval [1222.17, 1294.92] nm. Endpoint interpolation gives 1798.48 nm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.