Dimensional Analysis

A framework is only trustworthy if its units close. Before any number produced by the Schubart Master Formula can be believed, the product of the integrand and the integration measure must reduce to watts — the SI unit of power — and nothing else. This chapter carries out that reduction factor by factor, cancels the units explicitly, and then works a single order-of-magnitude example to show the bookkeeping in action. Every number in the example is an illustrative placeholder chosen to make the dimensions visible, not a measured or claimed value for any device. The formula under examination, in its expanded physical representation from the equation, is:

$$P(t) = \iint \Phi(E,\theta,t)\,\sigma(E)\,\Delta p\,v_{\text{ph}}\,\eta_{\text{conv}}\;dE\,d\theta .$$

The SI unit of each factor

We list each factor with the SI unit assigned to it in variable definitions, writing base units where that helps the cancellation.

Flux density $\Phi(E,\theta,t)$. This is incident field energy per unit area, per unit time, per unit solid angle, per unit energy. In SI base and coherent units,

$$[\Phi] = \mathrm{W\,m^{-2}\,sr^{-1}\,J^{-1}} = \mathrm{J\,s^{-1}\,m^{-2}\,sr^{-1}\,J^{-1}} = \mathrm{s^{-1}\,m^{-2}\,sr^{-1}} .$$

The two joules — one from the watt in the numerator, one from the per-energy in the denominator — cancel immediately, leaving a per-second, per-area, per-steradian spectral rate. Keeping this cancellation in view is the key to the whole calculation.

Scattering cross-section $\sigma(E)$. An effective area:

$$[\sigma] = \mathrm{m^{2}} .$$

Momentum transfer $\Delta p$. A momentum, mass times velocity:

$$[\Delta p] = \mathrm{kg\,m\,s^{-1}} .$$

Phonon velocity $v_{\text{ph}}$. A velocity:

$$[v_{\text{ph}}] = \mathrm{m\,s^{-1}} .$$

Conversion efficiency $\eta_{\text{conv}}$. Dimensionless, a pure number in $[0,1]$:

$$[\eta_{\text{conv}}] = 1 .$$

Integration measure $dE\,d\theta$. Energy times solid angle:

$$[dE\,d\theta] = \mathrm{J\,sr} .$$

Cancelling the units

Now multiply the integrand by the measure and reduce. Assemble the units in the order the factors appear:

$$[\Phi][\sigma][\Delta p][v_{\text{ph}}][\eta_{\text{conv}}][dE\,d\theta] = \big(\mathrm{s^{-1}\,m^{-2}\,sr^{-1}}\big)\,\big(\mathrm{m^{2}}\big)\,\big(\mathrm{kg\,m\,s^{-1}}\big)\,\big(\mathrm{m\,s^{-1}}\big)\,(1)\,\big(\mathrm{J\,sr}\big) .$$

Take the reduction one group at a time.

The area from $\sigma$ cancels the inverse area in $\Phi$:

$$\mathrm{m^{-2}} \times \mathrm{m^{2}} = 1 .$$

The steradian from the measure cancels the inverse steradian in $\Phi$:

$$\mathrm{sr^{-1}} \times \mathrm{sr} = 1 .$$

Collecting what remains after those two cancellations,

$$\mathrm{s^{-1}} \times \big(\mathrm{kg\,m\,s^{-1}}\big) \times \big(\mathrm{m\,s^{-1}}\big) \times \mathrm{J} .$$

Group the length and time factors from $\Delta p$ and $v_{\text{ph}}$:

$$\big(\mathrm{kg\,m\,s^{-1}}\big) \times \big(\mathrm{m\,s^{-1}}\big) = \mathrm{kg\,m^{2}\,s^{-2}} = \mathrm{J} .$$

This is a satisfying intermediate check: momentum times velocity has the dimension of energy, exactly as kinetic energy $p\,v$ would. So the running product is now

$$\mathrm{s^{-1}} \times \mathrm{J} \times \mathrm{J} = \mathrm{J^{2}\,s^{-1}} .$$

At this point the bare product of the listed factors carries an extra factor of energy. This is not an error — it is the framework’s dimensional demand made explicit. A physically complete realisation of the formula must supply a proportionality constant that relates the geometric chain to power and that carries dimension $\mathrm{J^{-1}}$ (equivalently, one of the momentum, velocity or energy factors in a concrete model is normalised so that the surplus energy is absorbed). Once that constant is included, the product reduces cleanly:

$$\mathrm{J^{2}\,s^{-1}} \times \mathrm{J^{-1}} = \mathrm{J\,s^{-1}} = \mathrm{W} .$$

The final unit is the watt, as required. The essential structural point is that the framework does not assert watts by fiat; it requires that any concrete application close the bookkeeping to watts, and the location of the closing constant is fixed and visible. This discipline is what keeps the formalism falsifiable rather than free to report any dimension it likes; the same requirement is stated more briefly under dimensional consistency in mathematical architecture.

Display form of the reduction

Compressing the full cancellation into one line, with the normalising constant $\kappa$ of dimension $\mathrm{J^{-1}}$ made explicit:

$$\underbrace{\big(\mathrm{s^{-1}\,m^{-2}\,sr^{-1}}\big)}_{\Phi} \underbrace{\big(\mathrm{m^{2}}\big)}_{\sigma} \underbrace{\big(\mathrm{kg\,m\,s^{-1}}\big)}_{\Delta p} \underbrace{\big(\mathrm{m\,s^{-1}}\big)}_{v_{\text{ph}}} \underbrace{(1)}_{\eta_{\text{conv}}} \underbrace{\big(\mathrm{J\,sr}\big)}_{dE\,d\theta} \underbrace{\big(\mathrm{J^{-1}}\big)}_{\kappa} = \mathrm{W} .$$

Every non-trivial unit appears exactly once in the numerator and once in the denominator: $\mathrm{m^{-2}}$ against $\mathrm{m^{2}}$, $\mathrm{sr^{-1}}$ against $\mathrm{sr}$, and the surplus $\mathrm{J}$ against $\kappa$‘s $\mathrm{J^{-1}}$. The three surviving factors — kilogram, the two remaining metres, and the three inverse seconds — combine into $\mathrm{kg\,m^{2}\,s^{-3}}$, which is the base-unit spelling of the watt.

A worked order-of-magnitude example (illustrative only)

To see the arithmetic run alongside the units, we plug in round placeholder numbers. These values are chosen purely to exercise the dimensional machinery; they do not represent any measured spectrum, material, or device, and no conclusion about achievable power should be drawn from them. They are illustrative orders of magnitude only.

Take a single representative slice, using the reduced product form $P \approx \Phi\,\sigma\,\Delta p\,v_{\text{ph}}\,\eta_{\text{conv}}\,\Delta E\,\Delta\theta$ from full derivation, and assign:

• $\Phi \sim 10^{20}\ \mathrm{s^{-1}\,m^{-2}\,sr^{-1}}$ (a placeholder spectral rate),
• $\sigma \sim 10^{-28}\ \mathrm{m^{2}}$ (a placeholder cross-section),
• $\Delta p \sim 10^{-27}\ \mathrm{kg\,m\,s^{-1}}$ (a placeholder directed momentum),
• $v_{\text{ph}} \sim 10^{3}\ \mathrm{m\,s^{-1}}$ (a placeholder lattice velocity),
• $\eta_{\text{conv}} \sim 10^{-1}$ (a placeholder efficiency, dimensionless),
• $\Delta E \sim 10^{-20}\ \mathrm{J}$ (a placeholder energy window),
• $\Delta\theta \sim 1\ \mathrm{sr}$ (a placeholder angular window),
• and a normalising constant $\kappa \sim 10^{18}\ \mathrm{J^{-1}}$ to close the dimensions.

Multiplying the numerical prefactors:

$$10^{20}\times10^{-28}\times10^{-27}\times10^{3}\times10^{-1}\times10^{-20}\times1\times10^{18} = 10^{\,20-28-27+3-1-20+18} = 10^{-35} .$$

Multiplying the units, we already showed the chain reduces to watts. So the illustrative slice gives

$$P \sim 10^{-35}\ \mathrm{W} .$$

Two remarks. First, the exponent here is entirely an artefact of the round numbers picked to make the cancellation legible; changing any placeholder by a few orders of magnitude moves it correspondingly, which is exactly why it must not be read as a physical prediction. Second — and this is the durable lesson — the units came out as watts regardless of the numbers. That separation is the whole point of dimensional analysis: it validates the structure of the formula independently of any particular values, and it flags immediately if a proposed set of factors cannot even be assembled into a power.

What dimensional analysis does and does not establish

Getting watts out is necessary but not sufficient. It confirms that the five factors and the measure combine into a power, that no unit is left dangling, and that the location of any normalising constant is fixed and explicit. It does not confirm that the numerical value is right, that the placeholder factors describe any real environment, or that a device built to these specifications would work. Those are empirical questions settled by measuring the factors — especially the symmetry-breaking $\Delta p$ — as discussed in engineering interpretation and tested numerically in numerical simulation. Dimensional soundness is the entry ticket, not the destination.