Nondimensionalization

Big Idea

In Doing Some Modeling: Why Inertia Is Irrelevant inertia looked negligible because $\tau_i/T \approx 10^{-9}$ , a comparison of two numbers made outside the equation. To delete the inertial term and ask how much error that commits, the ratio has to be inside the equation. Measure every variable in a unit taken from the problem instead of in seconds and metres. If the units are well chosen, every rescaled variable is of order one, and all the information about how big a term is moves into the constant in front of it. Four constants collapse into one number, $\ve = \tau_i/T$ , and it stands in front of the inertial term.

The question¶

Doing Some Modeling: Why Inertia Is Irrelevant ended with an observation: except for a few microseconds after each change in force, a cell’s velocity is $v \approx F/\gamma$ . Turning that into an equation of motion means deleting the term $m\ddot{\mathbf x}$ from Newton’s second law, which leaves the overdamped force balance $\gamma\dot{\mathbf x} = \mathbf F$ . The reduced equation cannot be exactly right: a cell that loses its traction while moving glides to a stop under Newton’s law, but stops dead under the force balance. So the question of this part is:

How much error do we make by using the overdamped equation in place of Newton’s law?

The first step is to put the equation in a form where “small” means something.

The equation¶

In one dimension, with $F$ the net force the cell generates for itself, Newton’s second law with Stokes drag (3) is the following.

Definition 1 (Equation of motion)

m\,\ddot x = F(t) - \gamma\,\dot x \qquad\Longleftrightarrow\qquad m\,\dot v = F(t) - \gamma\,v, \quad v = \dot x .

(1)

Second order in $x$ , so it needs $x(0)$ and $\dot x(0)$ . Since $x$ does not appear on the right it is first order in $v$ , and $x$ follows by one more integration.

We work with the equation for the velocity,

m\,\frac{\dd v}{\dd t} = F(t) - \gamma\,v , \qquad v(0) = v_0 .

(2)

No value of $F$ is needed anywhere in this part: the whole argument is about time scales.

Units¶

Symbol	Meaning	SI units	At cell scale
$t$	time	$\mathrm{s}$	s, min
$x$	position	$\mathrm{m}$	$\mu$ m
$v = \dot x$	velocity	$\mathrm{m\,s^{-1}}$	$\mu$ m/min
$a = \ddot x$	acceleration	$\mathrm{m\,s^{-2}}$	never measured
$m$	mass	$\mathrm{kg}$	pg
$F$	force	$\mathrm{N} = \mathrm{kg\,m\,s^{-2}}$	pN
$\gamma$	drag coefficient	$\mathrm{N\,s\,m^{-1}} = \mathrm{kg\,s^{-1}}$	pN s $\mu\mathrm{m}^{-1}$
$\eta$	viscosity	$\mathrm{Pa\,s} = \mathrm{kg\,m^{-1}s^{-1}}$	water: 10^-3

Every term of (1) is a force:

\underbrace{\mathrm{kg}\cdot\mathrm{m\,s^{-2}}}_{m\ddot x} \;=\; \underbrace{\mathrm{N}}_{F} \;=\; \underbrace{\mathrm{kg\,s^{-1}}\cdot\mathrm{m\,s^{-1}}}_{\gamma\dot x} \;=\; \mathrm{kg\,m\,s^{-2}} . \qquad\checkmark

(3)

Both spellings of the units of $\gamma$ are useful. As $\mathrm{N\,s\,m^{-1}}$ it is a force per unit velocity, which is what $\gamma v = F$ says; as $\mathrm{kg\,s^{-1}}$ the kilograms cancel against $m$ and leave seconds.

Nothing in the equation is a time except $t$ itself. The units say which combinations of the constants, and of the cell’s own length $L$ and speed $v_0$ , are times.

Example 1 (Which combinations are times)

Reference units¶

Definition 2 (Reference unit, dimensionless variable)

A reference unit (or scale) for a variable is a fixed constant carrying that variable’s units, chosen from the problem. Writing

x = X\,\tilde x \qquad\text{equivalently}\qquad \tilde x = \frac{x}{X}

(6)

replaces $x$ by the dimensionless $\tilde x$ , which counts how many $X$ ’s. Dimensionless variables carry a tilde.

Definition 3 (Order one)

A dimensionless quantity is of order one, written $O(1)$ , if it is neither large nor small compared with 1: as a working rule, between about 0.1 and 10. A scale is well chosen when the corresponding dimensionless variable, and its derivatives, are $O(1)$ over the range we care about.

There are three reasons to scale.

Only dimensionless numbers can be called small. The cell’s mass is 10^-12 kg, which sounds tiny, but that is a statement about the kilogram: the same mass is 10⁹ femtograms. “The mass is negligible” is not a meaningful sentence; “ $\ve = 10^{-9}$ ” is.
Parameters collapse. Each scaled variable removes one parameter. Four mechanical constants become one below.
Size moves into the prefactors. If every scale is well chosen, every tilde variable and derivative is $O(1)$ , so terms can be ranked by their prefactors. A term with prefactor 10^-9 multiplying something $O(1)$ contributes 10^-9.

The catch is the “if”

A badly chosen scale leaves a tilde variable that is not $O(1)$ , and reading the prefactors then gives the wrong answer. The model usually signals this: a term expected to be $O(1)$ comes out enormous.

Choosing the scales¶

Two scales are measured, one follows from them, and one is a decision.

Variable	Scale	Value	Where it comes from
length	$L$	10^-5 m	measured: the cell’s diameter
velocity	$V = v_0$	10^-8 m/s	measured: the speed cells crawl at
time	$T = L/V$	10³ s	derived: Definition 2
force	$\gamma V$	10^-14 N	chosen: the drag on a cell moving at $V$

The time scale is the question’s, not the equation’s; the equation’s own time, $\tau_i$ , will appear as the coefficient left over. The force scale is a genuine choice, and a bet: it makes the drag on a cell moving at its observed speed of order one, which amounts to betting that drag is the force that matters. If the bet were wrong, a term expected to be $O(1)$ would come out large.

The substitution¶

Three replacements:

t = T\,\tilde t, \qquad v(t) = V\,\tilde v(\tilde t), \qquad F(t) = \gamma V\,\tilde F(\tilde t) .

(7)

Each says: old quantity $=$ fixed unit $\times$ pure number. The units live in $T$ , $V$ , $\gamma V$ ; the tildes are numbers. By the chain rule, with $\tilde t(t) = t/T$ ,

\frac{\dd v}{\dd t} = V\,\frac{\dd\tilde v}{\dd\tilde t}\,\frac{\dd\tilde t}{\dd t} = \frac{V}{T}\,\frac{\dd\tilde v}{\dd\tilde t} .

(8)

Substitute in the denominator too

The derivative contains $t$ as well as $v$ , so it needs both replacements. Replacing $v$ and forgetting $\dd t$ gives $V\,\dd\tilde v/\dd\tilde t$ , which has the units of a velocity; the correct expression has units $[V/T] = \mathrm{m/s^2}$ , an acceleration. Each derivative with respect to $t$ brings one factor of $1/T$ .

Replace each term of (2):

\underbrace{m\cdot\frac{V}{T}\,\frac{\dd\tilde v}{\dd\tilde t}}_{\text{inertia}} \;=\; \underbrace{\gamma V\,\tilde F(\tilde t)}_{\text{applied force}} \;-\; \underbrace{\gamma V\,\tilde v(\tilde t)}_{\text{drag}} .

(9)

Every term is a force, so divide by the force scale $\gamma V$ :

\underbrace{\frac{m}{\gamma T}}_{\text{dimensionless}}\, \frac{\dd\tilde v}{\dd\tilde t} = \tilde F - \tilde v .

(10)

The two force terms now have coefficient 1, $V$ has dropped out, and the only prefactor that can be large or small is $m/(\gamma T)$ , in front of the inertial term.

The initial condition must be scaled with the equation. The instant $t = 0$ is the instant $\tilde t = 0$ , so $v(0) = v_0$ becomes $\tilde v(0) = v_0/V =: \tilde v_0$ , a pure number. With $V = v_0$ it equals 1, but we keep the symbol, since the reduction is about what happens to it.

Definition 4 (The dimensionless equation of motion)

With $t = T\tilde t$ , $v = v_0\tilde v$ , $F = \gamma v_0\tilde F$ , the model (2) together with its initial condition reads

\ve\,\frac{\dd\tilde v}{\dd\tilde t} = \tilde F(\tilde t) - \tilde v, \qquad \tilde v(0) = \tilde v_0, \qquad \ve = \frac{m}{\gamma T} = \frac{\tau_i}{T} = \frac{m v_0}{\gamma L} .

(11)

The single parameter $\ve$ is the ratio of the inertial time of Definition 1, which belongs to the cell, to the migration time of Definition 2, which belongs to the question. For a crawling cell

\ve = \frac{\tau_i}{T} \approx \frac{10^{-6}\ \mathrm{s}}{10^{3}\ \mathrm{s}} = 10^{-9} .

(12)