3.4 Dimentional Homogeneity

Should we reduce to its fundamental dimensions each of the quantities occurring in a valid physical equation, all terms of the equation must then consist of like fundamental dimensions raised to like powers. The equation must be dimensionally homogeneous. This is a truism for definitional equations, and for those equations of empirical basis it expresses a condition of physical necessity. The principle of dimensional homogeneity, or its equivalent, is the basic axiom for the operational rules in the general method of dimensional analysis.

3.4.1 The Dimentional Constraints on Definitional and Empirical Equations

The principle of homogeneity applies to equations of purely dimensional content, and it applies to equations that contain scale magnitudes and quantities of measurement. It excludes any equation that might be mathematically complete but physically meaningless. For example, the equation

$$x + y = z$$

where $x + y$ sums to a quantity $z$, is mathematically correct for numbers. But let $x$ and $y$ have dimensional content. Let $x$, say, be the displacement of a uniformly accelerated body starting from rest, and let $y$ be its velocity over time. Accordingly, $$y=at\;\mbox{, }\;x=\frac{1}{2}at^2$$ $a$ being the uniform acceleration and $t$ the time variable. Our mathematically complete equation becomes $$z=\frac{1}{2}at^2+at$$ which is physically meaningless since we have attempted to sum incompatible natural quantities (displacement and velocity). By reducing each term on the RHS to its fundamental dimensions, the violation of the homogeneity principle becomes obvious. Since $[a] = LT^{-2}$, then $$[\frac{1}{2}at^2]=LT^{-2}T^2=L$$ $$[at]=LT^{-2}T=LT^{-1}$$ and obviously, the terms do not reduce to like dimensions.

Now let us suppose $x$ and $y$ to be some measured magnitudes of a body, say its length $x$ = 1 meter and its mass $y$ = 3 kilograms. By the definitions 1 m = 100 cm and 3 kg = 3000 g, we can write, in good mathematical faith, $$1m + 3kg = 100cm + 3000g$$ or even $$1m - 3000g = 100cm - 3kg$$ Terms on opposing sides are nicely balanced in the algebraic sense, but what does it mean to subtract 3000 grams of mass from 1 meter of length? The dimensions of the terms in the equation are $$[1m]=[100cm]=L$$ $$[3kg]=[3000g] =M$$ and the homogeneity principle is again violated; we must reject the equation (and others like it) as having no physical significance. Common sense, of course, would lead us to such distinctions in the first place.

The assignment of dimensions to the elements of integrals, differential equations, difference equations, and integro-differential equations is just as straightforward (but perhaps not so obvious) as that of simple algebraic structures. For expositional purposes, let the symbol “$X$” in the following examples be the dimension of the quantity $x$:
Differentials:
$$\begin{align*} &[dx]=X \\ &[d^2x] = X \\ &[dx^2] = X^2 \\ \end{align*}$$

Differential operators:
$$\begin{align*} &\Big[\frac{d}{dx}\Big]=X^{-1} \\ &\Big[\frac{dM}{dx}\Big]=[M]X^{-1} \\ &[\dot{x}]=\Big[\frac{dx}{dt}\Big]=XT^{-1} \\ &\Big[\frac{d^n}{dx^n}\Big]=X^{-n}\; \mbox{ (n a positive integer)} \\ &\Big[\frac{d^nM}{dx^n}\Big]=[M]X^{-n} \\ \end{align*}$$

$$\begin{align*} &[\ddot{x}]=\Big[\frac{d^2x}{dt^2}\Big]=XT^{-2} \\ &\Big[\frac{\partial}{\partial x}\Big]=X^{-1} \\ &\Big[\frac{\partial M}{\partial x}\Big]=[M]X^{-1} \\ &\Big[\frac{\partial^n}{\partial x^n}\Big]=X^{-n} \\ &\Big[\frac{\partial^n M}{\partial x^n}\Big]=[M]X^{-n} \\ \end{align*}$$

Laplacian:
$$\begin{align*} &[\nabla\phi]=\bigg[\frac{\partial^2\phi}{\partial x^2}+\frac{\partial^2\phi}{\partial y^2}+\frac{\partial^2\phi}{\partial z^2}\bigg]=[\phi]L^{-2}\;\;\mbox{each term} \end{align*}$$

Difference operators:
$$\begin{align*} &[\Delta x]=X \\ &[\Delta^2x]=X \\ &[\Delta^n x]=X \\ \end{align*}$$

Indexed differences:
$$\begin{align*} &[x_{i+1}-x_i]=X \end{align*}$$

Indefinite integrals:
$$\begin{align*} &\bigg[\int f(x)dx\bigg]-[f(x)]X \\ &\bigg[\int\!\!\!\int f(x,y)dxdy\bigg]=[f(x,y)][dy][dx]=[f(x,y)][y]X \\ \end{align*}$$

Definite integrals:
$$\begin{align*} &\bigg[\int_xf(t)dt\bigg]=[f(x)]X \\ &\bigg[\int\!\!\!\int_Rf(x,y)dxdy\bigg]=\bigg[\int_{y_1}dy\int_{x_1}[f(x,y)dx\bigg]=[f(x_1,y_1)][dy_1][dx_1] \end{align*}$$

Let us use the following integro-differential equation to illustrate the homogeneity of analytical equations that have physical meaning. The equation arises from the principle of momentum conservation as applied to the two-dimensional steady flow of a viscous fluid at a boundary. The quantities $u$, $U$ are velocities, while $x$, $y$ are the two-dimensional space variables of displacement, and $v$ is the kinematic viscosity of the fluid: $$u\frac{\partial u}{\partial x}-\frac{\partial u}{\partial y}\bigg[\frac{\partial}{\partial y}\int_0^yudy\bigg]=U\frac{dU}{dx}+x\frac{\partial^2u}{\partial y^2}$$ The terms of the equation separately reduce to fundamental dimensions as follows. Note that the dimensional set {$L,V$} accommodates this equation equally as well as the set {$L,T$}.

$$\begin{align*} &\Big[u\frac{\partial u}{\partial x}\Big]=V\frac{V}{L}=LT^{-1}LT^{-1}L^{-1}=LT^{-2} \\ &\Big[\frac{\partial u}{\partial y}\Big(\frac{\partial}{\partial x}\int_0^yudy\Big)\Big]=\Big[\frac{\partial u}{\partial y}\Big]\Big[\frac{\partial}{\partial x}\Big]\Big[u(y)\Big]\Big[dy\Big] \\ &= \frac{V}{L}\frac{1}{L}VL=LT^{-1}L^{-1}LT^{-1}L=LT^{-2} \\ &\Big[U\frac{dU}{dx}\Big]=LT^{-1}LT^{-1}L^{-1}=LT^{-2} \\ &\bigg[v\frac{\partial^2u}{\partial y^2}\bigg] = L^2T^{-1}\frac{LT^{-1}}{L^2} LT^{-2} \\ \end{align*}$$

Hence, our equation is dimensionally homogeneous; its dimensions are those of acceleration. We can also write the dimensions of any term as $$LT^{-2}=\frac{MLT^{-2}}{M}=\frac{F}{M}$$ which are the dimensions of force per unit mass (of fluid).

3.4.2 Intensive and Extensive Properties

Properties that depend on the total quantity of matter (or total effect) being measured are called extensive properties. If we take twice as much matter of a given substance, for example, it will contain twice as much volume and twice as much mass, implying that mass and volume are to be regarded as extensive properties of a substance. Properties independent of the quantity of matter or total effect being measured are called intensive properties. The density of a substance and its temperature are examples of the intensive properties of the substance. The density of water, say, is the same–under similar conditions–whether we measure one cupful or one cubic kilometer. Intensive properties are properties of the substances themselves, and, like the density of water, can be used in the identification of a substance, since, generally speaking, unlike substances differ in their intensive properties. The name of an intensive property will often bear the prefix “specific”. Examples are specific volume (the volume occupied by a unit mass of a substance) and specific luminous intensity (the luminous intensity per unit area of source).

Physical equations often contain a mixture of intensive and extensive quantities, but the principle of dimensional homogeneity still holds. When multiplied by density $\rho$, for example, our last sample equation becomes $$\rho u\frac{\partial u}{\partial x}-\rho \frac{\partial u}{\partial y}\bigg[\frac{\partial}{\partial x}\int_0^yudy\bigg]=\rho U\frac{dU}{dx}+\mu\frac{\partial^2u}{\partial y^2}$$ where $\mu$ is now the dynamic viscosity of the fluid [$v \equiv \mu/\rho$] being the definition of kinematic viscosity]. By dimensional reduction, the first term becomes $$\Big[\rho u\frac{\partial u}{\partial x}\Big]=ML^{-3}LT^{-1}LT^{-1}L^{-1}=ML^{-2}T^{-2}$$ which, of course, is typical of all the terms. We can also write the term dimensions of each term as $$ML^{-2}T^{-2}=MLL^{-3}T^{-2}=\frac{MLT^{-2}}{L^3}=\frac{F}{L^3}$$ which are the dimensions of force per unit volume (of fluid).

3.4.3 Conversion Factors

Very often the magnitude of a quantity must be transformed from one set of scale units to a different set of scale units (pounds to grams, miles per hour to meters per second, and so on). Such a transformation is commonly called a conversion of units, and conversions between two units can be made whenever the two units have the same dimensions (pounds and grams both have dimension $M$, miles per hour and meters per second both have dimensions $LT^{-1}$, and so on), and whenever we know the equation (the identity) that relates the two units. To convert, say, the magnitude of the mass of some object from units of pounds to units of kilograms, we make use of the identity $$1 lb=0.45359237kg$$ or the identity $$1kg=2.2046226lb$$ (where the number of significant figures we actually choose to employ depends, of course, on the accuracy appropriate to the application). From the first identity we can write the conversion factor $$\frac{0.45359237kg}{1lb}=1$$ and from the second we can write the conversion factor $$\frac{2.2046226lb}{1kg}=1$$ In either case, the factor is dimensionless and equal to unity. Any conversion factor is a dimensionless ratio of scale units identically equal to unity.

For the sake of illustration, let us transform 4.2 pounds scale magnitude of mass to the equivalent magnitude in units of the kilogram. We start by making the obvious statement $$4.2lb=4.2lb$$ and then operate on the RHS with the appropriate conversion factor. Since we want “lb” to cancel on the RHS (leaving “kg”), we choose the factor that has “lb” in the denominator: $$4.2lb=4.2\rlap{---}lb\bigg(\frac{0.454kg}{1\rlap{---} lb}\bigg)$$ Perform the necessary multiplication and get $$4.2lb=1.907kg$$ We follow much the same procedure for conversions that require a combination of conversion factors. Suppose we desire to have the scale velocity of 30 miles per hour in units of the centimeter and the second. Accordingly, we are concerned here with essentially two chains of conversions, one for dimension L and one for dimension T.

L: mile $\rightarrow$ foot $\rightarrow$ inch $\rightarrow$ centimeter.
T: hour $\rightarrow$ minute $\rightarrow$ second.

We write the required identities and apply them one by one.

For dimension L:
    Identity: 1 mi = 5280 ft
    Conversion factors: $\frac{5280 ft}{1mi}=1$ or $\frac{1mi}{5280ft}=1$
    Identity: 1 ft = 12 in
    Conversion factors: $\frac{12in}{1ft}$ or $\frac{1ft}{12in}=1$
    Identity: 1 in = 2.54 cm
    Conversion factors: $\frac{2.54cm}{1in}=1$ or $\frac{1in}{2.54cm}=1$

For dimension T:
    Identity: 1 hr = 60 min
    Conversion factors: $\frac{60min}{1hr}=1$ or $\frac{1hr}{60min}=1$
    Identity: 1 min = 60 sec
    Conversion factors: $\frac{60sec}{1min}=1$ or $\frac{1min}{60sec}=1$
To proceed with the conversion we make the obvious statement $$\frac{30mi}{1hr}=\frac{30mi}{1hr}$$ and convert the units for each dimension. Starting with L (as a matter of choice; we could start with T just as legitimately), we follow the chain mile $\rightarrow$ foot $\rightarrow$ …, and employ the conversion factor that cancels the unit “mi”: $$\frac{30mi}{1hr}=\frac{30 \rlap{-----}mi}{1hr}\Big(\frac{5280ft}{1\rlap{-----}mi}\Big)$$ We continue the process and arrive at the unit “cm”: $$\frac{30mi}{1hr}=\frac{30 \rlap{-----}mi} {1hr}\Big(\frac{5280\rlap{----}ft} {1\rlap{-----}mi}\Big)\Big(\frac {12\rlap{---}in}{1\rlap{----}ft}\Big)\Big(\frac{2.54cm}{1\rlap{---}in}\Big)$$ Now we complete the process for units of T. Again we follow the, appropriate conversion chain (hour $\rightarrow$ minute $\rightarrow$ …). Since “hr” appears in a denominator, we choose the hr $\rightarrow$ min conversion that will cancel “hr”, and so on: $$\frac{30mi}{1hr}=\frac{30}{1\rlap{----}hr}\Big(\frac{5280}{1}\Big)\Big(\frac{12}{1}\Big)\Big(\frac{2.54}{1}\Big)\Big(\frac{1\rlap{----}hr}{60\rlap{------}min}\Big)\Big(\frac{1\rlap{------}min}{60sec}\Big)$$ And finally, we perform the necessary multiplication and get the desired conversion $$30mi/hr=1340cm/sec$$ The rational operations of algebra extend with great convenience to the operations (and conversions) on scale units.

Example.
Convert the area magnitude of 1 square inch to units of square centimeters:
The identity between the inch and the centimeter is $$1in=2.54cm$$ and by squaring both sides of the identity we get $$\begin{align*} (1in)^2&=(2.54cm)^2 \\ &=(2.54)^2cm^2 \\ 1in^2&=6.4516cm^2 \\ \end{align*}$$

Example. Convert 20 in² to units of cm²:
Make the obvious statement $$20in^2=20in^2$$ From the previous example we can write the conversion factor $$\frac{6.4516cm^2}{1in^2}=1$$ and thus $$20in^2=\frac{20\rlap{---}in^2}{1}\Big(\frac{6.4516cm^2}{1\rlap{---}in^2}\Big)$$ $$20in^2=129.032cm^2$$

Example. The density of CCl₄ at 0$^{\circ}$C is 1.600 grams per cubic centimeter. Convert this intensive property to units of the ounce and the cubic inch:
The units to be converted have dimensions M and L³. The chains of conversions are

M: gram $\rightarrow$ pound $\rightarrow$ ounce,
L³: cm³ $\rightarrow$ in³

Identities and conversion factors are
    gram $\rightarrow$ pound:
    1 lb = 453.6 g   or   2.205 lb = 100 g (since 1 kg = 100 g),
    $\frac{453.6g}{1lb}=1$   or   $\frac{1lb}{453.6g} =1$   or   $\frac{2.205lb}{100g}=1$   or   $\frac{100g}{2.205lb}$.
    pound $\rightarrow$ ounce:
    16 oz = 1 lb
    $\frac{16oz}{1lb}=1$   or   $\frac{1lb}{16oz}=1$.
    cm³ $\rightarrow$ in³:
    1 in = 2.54 cm
    1 in³ = 16.387 cm³,
    $\frac{16.387cm^3}{1in^3}=1$   or   $\frac{1in^3}{16.387cm^3}=1$.

Make the obvious statement $$\frac{1.600g}{1cm^3}=\frac{1.600g}{1cm^3}$$ Apply the chain of conversion factors, choosing those that cancel in succession: $$\frac{1.600g}{1cm^3}=\frac{1.600\rlap{--}g}{1\rlap{-----}cm^3}\Big(\frac{1\rlap{---}lb}{453.6\rlap{--}g}\Big)\Big(\frac{16oz}{1\rlap{---}lb}\Big)\Big(\frac{16.387\rlap{-----}cm^3}{1in^3}\Big)$$ $$1.600g\cdot cm^{-3}=0.9248oz\cdot in^{-3}$$ which is the desired conversion.