Verdicts

Floor confirmed, formula refuted means: the physical statement (there is a floor, more bandwidth lowers energy per bit) holds, while the specific formula and the “optimum” prediction do not.

Why these three

Of the twenty, these are the three whose answer is a closed form rather than a model output. The others either depend on supplied data — a soil-moisture reanalysis, a cache trace, an earthquake catalogue — or on a numerical search over an assumed spectrum. For those, a “proof” is not available; for these three it is, and in two of them the notebook itself labels the entry exact under assumptions.

The tests here are cheap on purpose. Every number below comes from a script that runs in under a minute on a laptop, and each script checks the theorem by a route that does not use the theorem: capacity is evaluated directly, the reaction-diffusion equation is integrated in time, the separation integral is done by quadrature.

1 · The per-bit energy floor

The claim

Transmission has a per-bit floor. The energy per bit of a communication link is bounded below by the ratio of noise to bandwidth, so widening bandwidth lowers the floor but raises total power.

Mechanism. Shannon capacity sets the bits per second per hertz; the energy per bit is power divided by capacity, which has a minimum.

Prediction. Energy per bit should bottom out at an optimal bandwidth.

Refutation. Energy per bit falling indefinitely with bandwidth would remove the floor.

The model

An ideal band-limited Gaussian channel with average power P, one-sided noise spectral density N_0, bandwidth B. Capacity and energy per bit at capacity are

C(B)=B\log_2\left(1+\frac{P}{N_0B}\right),\qquad E_b=\frac{P}{C(B)} .

Write s=P/(N_0B) for the signal-to-noise ratio. Then

\frac{E_b}{N_0}=\frac{P/N_0}{B\log_2(1+s)}=\ln 2\;\frac{s}{\ln(1+s)} \qquad (1.1)

The rate of approach is also exact. Expanding \ln(1+s)=s-s^2/2+O(s^3),

\frac{E_b}{N_0}-\ln 2=\frac{\ln 2}{2}\,s+O(s^2)=\frac{\ln2}{2}\frac{P}{N_0 B}+O(B^{-2}), \qquad (1.2)

so the gap halves with every doubling of bandwidth. The numerical check gives \big[(E_b/N_0-\ln2)\big/\big(\tfrac{\ln2}{2}s\big)\big]\to1.0001 at B=10^{12}, i.e. the asymptotic law holds to four digits.

What the mechanism implies for total power

The claim also says that widening bandwidth “raises total power”. That is a statement about a fixed information rate R, and it is false in the same model.

So at fixed rate, bandwidth lowers the required power, monotonically, toward the same \ln2 N_0R floor. The third clause of the claim — “raises total power” — describes what real transceivers do (circuit power rises with bandwidth), not what the ideal model does. It is an empirical statement wearing the model’s clothes.

Where the claim’s formula breaks

The phrase “bounded below by the ratio of noise to bandwidth” has two readings, and both fail.

• If “noise” means the noise power N=N_0B, then N/B=N_0, which is an energy (J) and numerically the right order: the true floor is \ln2\,N_0=0.693\,N_0. The claim overstates the floor by 44%, and — more importantly — asserts that it moves with B, which it does not.
• If “noise” means the spectral density N_0, then N_0/B has units of \mathrm{W\,s\,Hz^{-1}}, not energy. A quantity with the wrong dimension cannot bound an energy per bit. As B\to\infty it would go to zero, which would “prove” that arbitrarily small energy per bit is possible — the opposite of what the claim intends.

Either way the shape of the claim is wrong: the floor is a constant, and the approach to it is monotone rather than minimised.

How close is close

From (1.1), a given closeness \epsilon to the floor requires s/\ln(1+s)\le 1+\epsilon, i.e. s\lesssim 2\epsilon:

With P/N_0=1 MHz these are 50 and 500 MHz. The gain is real but logarithmic in cost: each further factor of ten in bandwidth buys a factor of ten less excess, never zero.

Numbers

Table 1 gives the exact value from (1.1) at the bandwidth where the SNR is 20,10,\ldots,-120 dB, and Figure 1 shows the two monotone curves.

2 · The invasion front

The claim

Invasion speed is growth against dispersal. The spread rate of an invader is set by the ratio of population growth rate to dispersal distance, so a fast-growing invader with short dispersal can outrun a slow one with long dispersal.

Mechanism. Spread combines reproduction and movement; whichever limits produces the slower front.

Prediction. Spread should track the ratio.

Refutation. Spread tracking dispersal alone would remove growth.

The model and the theorem

For density u(x,t) in one space dimension, with growth rate r and diffusivity D,

u_t = D\,u_{xx} + r\,u(1-u),\qquad u_0\ge 0 \text{ compactly supported},\ u_0\not\equiv 0. \qquad (2.1)

Three steps carry the proof, and each is worth keeping separate because each is where a claim usually goes wrong.

(i) No wave can be slower. Substituting U(x-ct) gives DU''+cU'+rU(1-U)=0. Linearising at the leading edge U\to0 where U(1-U)\approx U gives DU''+cU'+rU=0, with characteristic equation D\lambda^2+c\lambda+r=0. Real roots require c^2\ge4Dr. For c<2\sqrt{Dr} the roots are complex and U oscillates through zero, which a density cannot do. Hence c\ge 2\sqrt{Dr}.

(ii) The minimal speed is the marginal one. Writing the characteristic roots as \lambda_\pm=\big(-c\pm\sqrt{c^2-4Dr}\big)/(2D), a wave of speed c has a leading tail e^{-\lambda_- x} with \lambda_-<\sqrt{r/D}; at c=2\sqrt{Dr} the roots coincide at \lambda_*=\sqrt{r/D}. Equivalently: look for solutions of the linearised equation growing like e^{\lambda(x-vt)}; this requires v=D\lambda+r/\lambda, whose minimum over \lambda>0 is 2\sqrt{Dr} at \lambda=\lambda_*. The speed is a minimum over decay rates, which is why it is a product: D and r enter symmetrically.

(iii) The minimum is the one selected. Compact initial data have a sharp enough tail that the front is “pulled” by its leading edge, and comparison with a super-solution (Aronson–Weinberger) gives both bounds. The classical statement, and the correction to the position, are in the classical references [3], [4], [5] and [6].

This matters for measurement: any finite-time estimate of the speed from a step-like start is biased low, by an amount that shrinks only logarithmically. The bias is not a numerical artefact, and it is large when D/r is large. The fit below handles it, and the fitted coefficient is a free check on the whole picture.

Why the ratio cannot be the answer

The same argument in one line: the equation has the scaling symmetry x\mapsto\sqrt{D/r}\,y, t\mapsto\tau/r, under which (2.1) becomes the equation with D=r=1. Every solution is a rescaling of one solution, so the speed is \sqrt{Dr} times a universal constant. The ratio r/D is not invariant under that rescaling and therefore cannot determine the speed.

Two numerical consequences, both measured below:

• Same ratio, different speed. (D,r)=(1,1) and (4,4) both have r/D=1. Measured speeds: 1.988 and 7.942 — a factor of four.
• Reversed ranking. (D,r)=(9,\,0.25) has r/D=0.028; (D,r)=(0.111,\,9) has r/D=81, nearly 3000 times larger. Measured speeds: 2.873 and 2.020. The ratio ranks the second invader three orders of magnitude faster; the invader it ranks slower is in fact faster.

The qualitative sentence in the claim — “a fast-growing invader with short dispersal can outrun a slow one with long dispersal” — is true under the exact rule, but for a different reason: growth and dispersal are multiplicatively interchangeable. A factor 4 in growth and a factor 4 in dispersal each multiply the speed by 2. The claim’s ratio makes them linearly interchangeable, which is a different (and false) statement.

Where the constant 2 comes from, and when it does not apply

The constant is not universal. Take the Allee-type growth g(u)=r\,u(1-u)(u-a) with 0<a<\tfrac12, where low density reproduces poorly.

At a=0.1 this predicts c=0.566\sqrt{Dr} against the KPP value 2\sqrt{Dr}; the simulation measures 0.565 for D=r=1. The \sqrt{Dr} scaling survives, the constant does not. A claim of the form “speed is set by a combination of growth and dispersal” is therefore safe; a claim of the form “speed is 2\sqrt{Dr}” is only true inside the KPP hypotheses — concave growth, homogeneous habitat, no advection.

Numbers

Equation (2.1) was integrated with a semi-implicit scheme (implicit diffusion, explicit reaction, zero-flux boundaries, \Delta x=0.08, \Delta t=0.02). The front position is interpolated at u=0.5 and fitted two ways: a straight line, and the Bramson form x=at+b\ln t+c. The second slope a is the estimate used in the last column.

3 · Osmotic minimum work

The claim

Desalination has a floor. The energy needed to separate salt from water is bounded below by the ratio of osmotic pressure to recovery, so higher recovery costs disproportionately more.

Mechanism. The minimum work is the free energy of mixing, which diverges as the concentrate approaches saturation.

Prediction. Specific energy should rise steeply with recovery.

Refutation. Energy independent of recovery would remove the thermodynamic floor.

The mechanism sentence is the correct one — the minimum work is a free energy of mixing. The formula in the first sentence is not, and the two can be compared directly.

Setup

Feed of volume V_f at salt concentration c_f, ideal solution with van ’t Hoff osmotic pressure \pi(c)=\nu RTc (\nu=2 for NaCl). Recover a permeate volume V_p=rV_f of pure water at recovery r\in(0,1). Salt is conserved, so while permeate volume V has been drawn the retentate concentration is

c(V)=\frac{c_f}{1-V/V_f},\qquad c_r=c(V_p)=\frac{c_f}{1-r}. \qquad (3.1)

Shape of the curve

The claimed floor is not a floor

The claim’s bound is \pi_f/r. Comparing with (3.2),

w(r)-\frac{\pi_f}{r}=\frac{\pi_f}{r}\Big[-\ln(1-r)-1\Big],

so the sign is that of -\ln(1-r)-1, which changes at e^{-1}=1-r:

r^{*}=1-e^{-1}=0.63212\ldots

So “bounded below by the ratio of osmotic pressure to recovery” is true only above 63% recovery, and even there the bound is not tight — it is a second, unrelated curve that happens to cross the first. At r=0.3 the claim overstates the minimum by a factor 2.8; at r=0.05 by a factor 19; at r=0.95 it understates it by a factor 3.

What the claim gets right is the direction and the convexity: higher recovery does cost disproportionately more, and the marginal cost diverges. What it gets wrong is the size, and the functional form — the exact curve is logarithmic in the brine concentration factor, not hyperbolic in recovery.

Numbers

35 g/L NaCl, \nu=2, T=298.15 K gives c_f=0.5989 mol/L and \pi_f=\nu RTc_f=2.97 MPa =0.825 kWh m^{-3}.

The practical reading: at 50% recovery the ideal minimum is 1.14 kWh per cubic metre of product. Reported plant consumptions for seawater reverse osmosis cluster around 2–4 kWh m^{-3} [9], i.e. two to three times the ideal value, which is what a second-law efficiency of 30–55% looks like. Two things push the real minimum down relative to Table 3 and two push the real consumption up: activity coefficients make seawater’s osmotic pressure lower than the van ’t Hoff value (≈2.4–2.6 MPa at 25 °C, so the ideal curve is perhaps 15% high), while concentration polarisation, pump and turbine inefficiencies, pre-treatment and the finite membrane permeability all add. The bound is a bound on the process, not a forecast of the device.

4 · One shape, three times

The three claims fail in the same way, which is the reason for reporting them together.

Two of the three claims are internally inconsistent in the same way: the stated prediction is contradicted by the theorem that supplies the stated mechanism. In #16 the mechanism is Shannon capacity; the capacity theorem gives a monotone curve, not an optimum. In #17 the mechanism is free energy of mixing; that free energy is -\ln(1-r), not 1/r. In #10 the mechanism sentence (“whichever limits produces the slower front”) is actually a correct description of a minimum-speed selection principle — and the minimum is a geometric mean, which is exactly what a ratio cannot express.

5 · What this does and does not show

Does. It gives a proof, a numerical replication and a verdict for three questions, with the exact constants. It fixes three statements that can be quoted directly: the floor E_b/N_0>\ln2 with no optimum, the speed v=2\sqrt{Dr}, and the work w=\pi_f[-\ln(1-r)]/r with a crossing at 1-1/e.

Does not. It does not show that the notebook’s questions are wrong — all three are good questions, and two of the three mechanism sentences are correct. It does not test the theorems against nature: no link, organism or plant is measured here. It does not claim novelty for any of the three results, which are classical or published; the citations are in the reference list and predate this report by decades.

The three idealisations are worth naming, because each one is where a real test would begin. The channel model assumes Gaussian noise, infinite coding blocklength and no circuit power. The invasion model assumes a homogeneous habitat, no advection, and growth that is concave near zero. The separation model assumes an ideal solution, no concentration polarisation, and reversible operation. Everything above is exact inside those assumptions and says nothing outside them.

6 · Reproducing this

cd code
./run_all.sh          # runs all three checks, writes ../data and ../figures

Each script is standalone and prints its checks to stdout. The three are:

Machine checks, all passing, are collected in verification/VERIFICATION.md and reproduced in data/*-checks.json. Versions: Python 3.14, NumPy 2.5.4, SciPy 1.18.1, SymPy 1.14.0, Matplotlib 3.11.2.

References