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Chapter 1
Dimensional Analysis
The first step in modeling any physical phenomena is the identification of the relevant vari- ables, and then relating these variables via known physical laws. For sufficiently simple phenomena we can usually construct a quantitative relationship among these variables from first principles; however, for many complex phenomena (which often occur in engineering applications) such an ab initio theory is often difficult, if not impossible. In these situations modeling methods are indispensable, and one of the most powerful modeling methods is dimensional analysis. You have probably encountered dimensional analysis in your previous physics courses when you were admonished to “check your units” to ensure that the left and right hand sides of an equation had the same units (so that your calculation of a force had the units of kg m/s^2 ). In a sense, this is all there is to dimensional analysis, although “checking units” is certainly the most trivial example of dimensional analysis (incidentally, if you aren’t in the habit of checking units, do it!). Here we will use dimensional analysis to actually solve problems, or at least infer some information about the solution. Much of this material is taken from Refs. [1] and [2]; Ref. [3] provides many interesting applications of dimensional analysis and scaling to biological systems (the science of allometry). The basic idea is the following: physical laws do not depend upon arbitrariness in the choice of the basic units of measurement. In other words, Newton’s second law, F = ma, is true whether we choose to measure mass in kilograms, acceleration in meters per second squared, and force in newtons, or whether we measure mass in slugs, acceleration in feet per second squared, and force in pounds. As a concrete example, consider the angular frequency of small oscillations of a point pendulum of length l and mass m:
ω =
√ g l
where g is the acceleration due to gravity, which is 9.8 m/s^2 on earth (in the SI system of units; see below). To derive Eq. (1.1), one usually needs to solve the differential equation which results from applying Newton’s second law to the pendulum (do it!). Let’s instead deduce (1.1) from dimensional considerations alone. What can ω depend upon? It is reasonable to assume that the relevant variables are m, l, and g (it is hard to imagine others, at least for a point pendulum). Now suppose that we change the system of units so that the unit of
mass is changed by a factor of M , the unit of length is changed by a factor of L, and the unit of time is changed by a factor of T. With this change of units, the units of frequency will change by a factor of T −^1 , the units of velocity will change by a factor of LT −^1 , and the units of acceleration by a factor of LT −^2. Therefore, the units of the quantity g/l will change by T −^2 , and those of (g/l)^1 /^2 will change by T −^1. Consequently, the ratio
ω √ g/l
is invariant under a change of units; Π is called a dimensionless number. Since it doesn’t depend upon the variables (m, g, l), it is in fact a constant. Therefore, from dimensional considerations alone we find that
ω = const. ×
√ g l
A few comments are in order: (1) the frequency is independent of the mass of the pen- dulum bob, a somewhat surprising conclusion to the uninitiated; (2) the constant cannot be determined from dimensional analysis alone. These results are typical of dimensional analysis—uncovering often unexpected relations among the variables, while at the same time failing to pin down numerical constants. Indeed, to fix the numerical constants we need a real theory of the phenomena in question, which goes beyond dimensional considerations.
1.1 Units
Before proceeding further with dimensional considerations we first need to discuss units of measurement.
1.1.1 The SI system of units
In this course we will adopt the SI system of units,^1 which is described in some detail in the Physicist’s Desk Reference [4] (which I will abbreviate as PDR from now on), pp. 4–10. In the SI system the base, or defined, units, are the meter (m), the kilogram (kg), the second (s), the kelvin (K), and the ampere (A).^2 The definitions of these units in terms of fundamental physical processes are given in the PDR. All other units are derived. For instance, the SI unit of energy, the joule (J), is equal to 1 kg m^2 /s^2. The derived units are also listed in the PDR. The SI system is referred to as a LM T -class, since the defined units are length L, mass M , and time T (if we add thermal and electrical phenomena, then we have a LM T θI-class in the SI system).
(^1) In some older texts this is referred to as the MKS system. (^2) We should also add the mole (mol) and the candela (cd), but these will seldom enter into our models.
Table 1.1: Dimensions of some commonly encountered physical quantities in the LM T θI- class. [L] Length L [M ] Mass M [T ] Time T [v] Velocity LT −^1 [a] Acceleration LT −^2 [F ] Force M LT −^2 [ρ] Mass density M L−^3 [p] Pressure M L−^1 T −^2 [α] Angle 1 [E] Energy M L^2 T −^2 [θ] Temperature θ [S] Entropy M L^2 T −^2 θ−^1 [I] Electric current I [Q] Electric charge IT [E] Electric field M LT −^3 I−^1 [B] Magnetic field M T −^2 I−^1
physical quantity φ by [φ]; thus, [v] = LT −^1. A dimensionless quantity would have [φ] = 1; i.e., its numerical value is the same in all systems of units within a given class. What about more complicated quantities such as force? From Newton’s second law, F = ma, so that [F ] = [m][a] = M LT −^2. Proceeding in this way, we can easily construct the dimensions of any physical quantity; some of the more commonly encountered quantities are included in Table 1.1. We see that all of the dimensions in the Table are power law monomials, of the form (in the LM T -class) [φ] = CLaM bT c, (1.7)
where C and (a, b, c) are constants. In fact, this is a general result which can be proven mathematically; see Sec. 1.4 of Barenblatt’s book, Ref. [1]. This property is often called dimensional homogeneity, and is really the key to dimensional analysis. To see why this is useful, consider again the determination of the period of a point pendulum, in a more abstract form. We have for the dimensions [ω] = T −^1 , [g] = LT −^2 , [l] = L, and [m] = M. If ω is a function of (g, l, m), then its dimensions must be a power-law monomial of the dimensions of these quantities. We then have
[ω] = T −^1 = [g]a[l]b[m]c = (LT −^2 )aLbM c = La+bT −^2 aM c, (1.8)
with a, b, and c constants which are determined by comparing the dimensions on both sides of the equation. We see that
a + b = 0, − 2 a = − 1 , c = 0. (1.9)
The solution is then a = 1/2, b = − 1 /2, c = 0, and we recover Eq. (1.2). A set of quantities (a 1 ,... , ak) is said to have independent dimensions if none of these quantities have dimensions which can be represented as a product of powers of the dimensions of the remaining quantities. As an example, the density ([ρ] = M L−^3 ), the velocity ([v] = LT −^1 ), and the force ([F ] = M LT −^2 ) have independent dimensions, so that there is no product of powers of these quantities which is dimensionless.^3 On the other hand, the density, velocity, and pressure ([p] = M L−^1 T −^2 ) are not independent, for we can write [p] = [ρ][v]^2 ; i.e., p/(ρv^2 ) is a dimensionless quantity. Now suppose we have a relationship between a quantity a which is being determined in some experiment (which we will refer to as the governed parameter), and a set of quantities (a 1 ,... , an) which are under experimental control (the governing parameters), which is of the form a = f (a 1 ,... , ak, ak+1,... an), (1.10)
where (a 1 ,... , ak) have independent dimensions. For example, this would mean that the dimensions of the governed parameter a is determined by the dimensions of (a 1 ,... , ak), while all of the as’s with s > k can be written as products of powers of of the dimensions of (a 1 ,... , ak); e.g., ak+1/ap 1 · · · ark, would be dimensionless, with (p,... , r) an appropriately chosen set of constants. With this set of definitions, it is possible to prove that Eq. (1.10) can be written as
a = ap 1 · · · arkΦ
( ak+ a pk+ 1 · · ·^ a
rk+ k
an ap 1 n · · · ar kn
) , (1.11)
with Φ some function of dimensionless quantities only. The great simplification is that while the function f in Eq. (1.10) was a function of n variables, the function Φ in Eq. (1.11) is only a function of n − k variables. Eq. (1.11) is a mathematical statement of Buckingham’s Π- Theorem, which is the central result of dimensional analysis. The formal proof can again be found in Barenblatt’s book [1]. Dimensional analysis cannot supply us with the dimensionless function Φ—we need a real theory for that. As a simple example of how this works, let’s return to the pendulum, but this time we’ll assume that the mass can be distributed, so that we relax the condition of the mass being concentrated at a point. The governed parameter is the frequency ω; the governing parameters are g, l (which we can interpret as the distance between the pivot point and the center of mass), m, and the moment of inertia about the pivot point, I. Since [I] = M L^2 , the set (g, m, l, I) is not independent; we can choose as our independent parameters (g, m, l) as before, with I/ml^2 a dimensionless parameter. In the notation developed above, n = 4
(^3) Prove this formally by writing [F ]a[v]b[ρ]c (^) = 1, and then show that the only solution is a = b = c = 0.
Alternatively, show that it is impossible to write [F ] = [ρ]a[v]b^ for any a, b.
1.3.2 Gravity waves on water
Next consider waves on the surface of water, which are called gravity waves (or sometimes capillary waves). How does the frequency ω depend upon the wavenumber^4 k of the wave? The relationship ω = ω(k) is known as the dispersion relation for the wave. The relevant variables would appear to be (ρ, g, k), which have dimensions [ρ] = M L−^3 , [g] = LT −^2 and [k] = L−^1 ; these quantities have independent dimensions, so n = 3, k = 3. Now we can determine the exponents:
[ω] = T −^1 = [ρ]a[g]b[k]c = M aL−^3 a+b−cT −^2 b, (1.16)
so that a = 0, − 3 a + b − c = 0, − 2 b = − 1 , (1.17)
with the solution a = 0, b = c = 1/2. Therefore,
ω = C
√ gk, (1.18)
with C another undetermined constant. We see that the frequency of water waves is propor- tional to the square root of the wavenumber, in contrast to sound or light waves, for which the frequency is proportional to the wavenumber. This has the interesting consequence that
the group velocity of these waves is vg = ∂ω/∂k = (C/2)
√ g/k, while the phase velocity is
vφ = ω/k = C
√ g/k, so that vg = vφ/2. Recall that the group velocity describes the large scale “lumps” which would occur when we superimpose two waves, while the phase velocity describes the short scale “wavelets” inside the lumps. For water waves these wavelets travel twice as fast as the lumps. You might worry about the effects of surface tension σ on the dispersion relationship. We can include these in our dimensional analysis by recalling that the surface tension is the energy per unit area of the surface of the water, so it has dimensions [σ] = M T −^2. The dimensions of the surface tension are not independent of the dimensions of (ρ, g, k); in fact, it is easy to show that [σ] = [ρ][g][k]−^2 , so that σk^2 /ρg is dimensionless. Then using the same arguments as before, we have
ω =
√ gkΦ
( σk^2 ρg
) , (1.19)
with Φ some undetermined function. A calculation of the dispersion relation for gravity waves starting from the fundamental equations of fluid mechanics [5] gives
ω =
√ gk
√ 1 + σk^2 /ρg, (1.20) (^4) Recall that k = 2π/λ, with λ the wavelength of the wave.
so that our function Φ(x) is Φ(x) =
1 + x. (1.21)
Dimensional analysis enabled us to deduce the correct form of the solution, i.e., the possible combinations of the variables. Of course, only a complete theory could provide us with the function Φ(x). What have we gained? We originally started with ω being a function of the four variables (ρ, g, k, σ); what dimensional analysis tells us is that it is really only a function of the combination σk^2 /ρg, even though we don’t know the function. Notice that this is an important fact if you are trying to measure the dependence of ω on the physical parameters (ρ, g, k, σ). If you needed to make (say) 10 separate measurements on each variable while holding the others fixed, then without dimensional analysis you would naively need to make 10^4 separate measurements. Dimensional analysis tells you that you only really need to measure the combinations gk and σk^2 /ρg, so only need to make 10^2 measurements to characterize ω. Dimensional analysis can be a labor-saving device!
1.3.3 Energy in a nuclear explosion
We next turn to a famous example worked out by the eminent British fluid dynamicist G. I. Taylor.^5 In a nuclear explosion there is an essentially instantaneous release of energy E in a small region of space. This produces a spherical shock wave, with the pressure inside the shock wave thousands of times greater that the initial air pressure, which may be neglected. How does the radius R of this shock wave grow with time t? The relevant governing variables are E, t, and the initial air density ρ 0 , with dimensions [E] = M L^2 T −^2 , [t] = T , and [ρ 0 ] = M L−^3. This set of variables has independent dimensions, so n = 3, k = 3. We next determine the exponents:
[R] = L = [E]a[ρ 0 ]b[t]c = M a+bL^2 a−^3 bT −^2 a+c^ (1.22)
so that a + b = 0, 2 a − 3 b = 1, − 2 a + c = 0, (1.23)
with the solution a = 1/5, b = − 1 /5, c = 2/5. Therefore
R = CE^1 /^5 ρ− 0 1 /^5 t^2 /^5 , (1.24)
with C an undetermined constant. If we could plot the radius R of the shock as a function of time t on a log-log plot, the slope of the line should be 2/5. The intercept of the graph would provide information about the energy E released in the explosion, if the constant C could be determined. By solving a model shock-wave problem Taylor estimated C to be
(^5) Taylor’s name is associated with many phenomena in fluid mechanics: the Rayleigh-Taylor instability,
Saffman-Taylor fingering, Taylor cells, Taylor columns, etc.
where D = κ/cp is the thermal diffusivity of the metal bar; it has dimensions [D] = [κ]/[cp] = L^2 T −^1 , as it should. Eq. (1.27) is the diffusion equation for heat. The diffusion equation will appear in many other contexts during this course. It usually results from combining a continuity equation with an empirical law which expresses a current or flux in terms of some local gradient. Suppose that the bar is very long, so that we can consider the idealized case of an infinite bar. At an initial time t = 0, we add an amount of heat Q 0 (with dimensions [Q 0 ] = M L^2 T −^2 ) at some point of the bar, which we will arbitrarily call x = 0. We could do this, for instance, by briefly holding a match to the bar. The heat is conserved at all times, so that cpA
∫ (^) ∞
−∞
τ (x, t)dx = Q 0. (1.28)
How does this heat diffuse away from x = 0 as a function of time t; i.e., what is τ (x, t)? We first identify the important parameters. The temperature τ certainly depends upon x, t, and the diffusivity D; we see from Eq. (1.28) that it also depends upon the initial conditions through the combination Q ≡ Q 0 /cpA. What are the dimensions? We have [x] = L, [t] = T , [D] = L^2 T −^1 , and [Q] = Lθ, so that n = 4. These dimensions are not independent, for the quantity x/(Dt)^1 /^2 is dimensionless, so that k = 3. We will choose as our independent quantities (t, D, Q). Now express τ in terms of these variables:
[τ ] = θ = [t]a[D]b[Q]c = L^2 b+cT a−bθc. (1.29)
We find 2 b + c = 0, a − b = 0, c = 1, (1.30)
which has the solution a = − 1 /2, b = − 1 /2, c = 1. Therefore, dimensional analysis tells us that the solution of the diffusion equation is of the form
τ (x, t) =
Q
(Dt)^1 /^2
( x (Dt)^1 /^2
) , (1.31)
with Φ a function which we still need to determine. The important point is that Φ is only a function of the combination x/(Dt)^1 /^2 , and not x and t separately. To determine Φ, let’s introduce the dimensionless variable z = x/(Dt)^1 /^2. Now use the chain rule to calculate various derivatives of τ :
∂τ ∂x
Q
(Dt)^1 /^2
∂z ∂x
dΦ(z) dz
=
Q
Dt
dΦ(z) dz
∂^2 τ ∂x^2
Q
(Dt)^3 /^2
d^2 Φ(z) dz^2
∂τ ∂t
Q
D^1 /^2 t^3 /^2
Φ(z) +
Q
(Dt)^1 /^2
∂z ∂t
dΦ(z) dz
= −
Q
D^1 /^2 t^3 /^2
[ Φ(z) + z
dΦ(z) dz
]
. (1.34)
Substituting Eqs. (1.33) and (1.34) into the diffusion equation (1.27), and canceling various factors, we obtain a differential equation for Φ,
d^2 Φ(z) dz^2
z 2
dΦ(z) dz
Φ(z) = 0. (1.35)
Dimensional analysis has reduced the problem from the solution of a partial differential equa- tion in two variables to the solution of an ordinary differential equation in one variable! The normalization condition, Eq. (1.28), becomes in these variables ∫ (^) ∞
−∞
Φ(z) dz = 1. (1.36)
You might think that Eq. (1.35) is hard to solve; however, it turns out that it is an exact differential, d dz
[ dΦ dz
z 2
] = 0, (1.37)
which we can integrate once to obtain
dΦ dz
z 2
Φ = const. (1.38)
However, since any physically reasonable solution would have both Φ → 0 and dΦ/dx → 0 as x → ∞, the integration constant must be zero. We now need to solve a first order differential equation, which we do by dividing Eq. (1.38) by Φ, multiplying by dz, and integrating, with the result that ln Φ = −z^2 /4 + const., or
Φ(z) = Ce−z (^2) / 4 , (1.39)
with C a constant. To determine C, we use the normalization condition, Eq, (1.36):
C
∫ (^) ∞
−∞
e−z
(^2) / 4 dz = C(4π)^1 /^2 = 1, (1.40)
where the integral (known as a Gaussian integral) can be found in integral tables. Therefore C = 1/(4π)^1 /^2. Returning to our original variables, we have
τ (x, t) =
Q
(4πDt)^1 /^2
e−x (^2) / 4 Dt
. (1.41)
This is the complete solution for the temperature distribution in a one-dimensional bar due to a point source of heat.^7
(^7) For the mathematically sophisticated, I’ll mention that the same solution can be obtained using the
method of Fourier transforms applied to the diffusion equation.
Bibliography
[1] G. I. Barenblatt, Dimensional Analysis (New York: Gordon and Breach Science Pub- lishers, 1987)
[2] H. L. Langhaar, Dimensional Analysis and the Theory of Models (New York: Wiley, 1951).
[3] T. A. McMahon and J. T. Bonner, On Size and Life (New York: Scientific American Library, 1983).
[4] H. L. Anderson (ed.), A Physicist’s Desk Reference (New York: American Institute of Physics, 1989)
[5] L. D. Landau and E. M. Lifshitz, Fluid Mechanics (New York: Pergamon Press, 1959), §61.
[6] G. I. Taylor, “The formation of a blast wave by a very intense explosion. II. The atomic explosion of 1945,” Proc. Roy. Soc. London A201, 159 (1950).
