5.1: Dimensionless groups

Because dimensionless quantities are the only meaningful quantities, we can understand the world better by describing it in terms of dimensionless quantities. But we need to find them.

5.1.1 Finding dimensionless groups

To illustrate finding these dimensionless quantities, let’s try an example that uses familiar physics: When learning a new idea, it is helpful to try it on a familiar example. We’ll find a train’s inward acceleration as it moves on a curved track. The larger the acceleration, the more the track or the train needs to tilt so that the passengers do not feel uncomfortable and (if the track is not tilted enough) the train does not tip over

Our goal is the relation between the train’s acceleration a, its speed v, and the track’s radius of curvature r. In our state of knowledge now, the relation could be almost anything. Here are a few possibilities:

\[\frac{a+ v^{2}}{r} = \frac{v^{3}}{a}; \: \frac{r+a^{2}}{v}= \frac{a^{2}}{v + r}; \: \frac{v}{ra + v^{3}}= \frac{a+v}{r^{2}}.\]

Although those possibilities are bogus, among the vast sea of possible relations, one relation is correct.

To find the constraint that will shrink the sea to a few drops, first defocus our eyes and see all the choices as examples of the general form

Even though the blobs might be complicated functions, they must have identical dimensions. By dividing both sides by blob B, we get a simpler form:

Now each side is dimensionless. Therefore, whatever the relation between a, v, and r, we can write it in dimensionless form.

The preceding process is not limited to this problem. In any valid equation, all the terms have identical dimensions. For example, here is the total energy in a spring–mass system:

\[E_{total} = \frac{1}{2} mv^{2} + \frac{1}{2} k x^{2}.\]

The kinetic-energy term (mv²/2) and the potential-energy term (kx²/2) have the same dimensions (of energy). The same process of dividing by one of the terms turns any equation into a dimensionless equation. As a result, any equation can be written in dimensionless form. Because the dimensionless forms are a small fraction of all possible forms, this restriction offers us a huge reduction in complexity.

To benefit from this reduction, we must compactly describe all such forms: Any dimensionless form can be built from dimensionless groups. A dimensionless group is the product of the quantities, each raised to a power, such that the product has no dimensions. In our example, where the quantities are a, v, and r, any dimensionless group G has the form

\[G \equiv a^{x}n^{y}r^{z},\]

where G has no dimensions and where the exponents x, y, and z are real numbers (possibly negative or zero).

Because any equation describing the world can be written in dimensionless form, and because any dimensionless form can be written using dimensionless groups, any equation describing the world can be written using dimensionless groups.

That news is welcome, but how do we find these groups?

a	LT^-2	acceleration
v	LT^-1	speed
r	L	radius

The first step is to tabulate the quantities with their description and dimensions. By convention, capital letters are used to represent dimensions. For now, the possible dimensions are length (L), mass (M), and time (T). Then, for example, the dimensions of v are length per time or more compactly, LT^-1 .

Then, by staring at the table, we find all possible dimensionless groups. A dimensionless group contains no length, mass, or time dimensions. For our example, let’s start by getting rid of the time dimension. It occurs in a as T⁻² and in v as T⁻¹. Therefore, any dimensionless group must contain a/v². This quotient has dimensions of L⁻¹. To make it dimensionless, multiply it by the only quantity that is purely a length, which is the radius r. The result, ar/v², is a dimensionless group. In the a^xv^yr^z form, it is a¹v^-2r¹.

Are there other dimensionless groups?

To get rid of time, we started with a/v² and then ended, inevitably, with the group ar/v². To make another dimensionless group, we would have to choose another starting point. However, the only starting points that get rid of time are powers of a/v²—for example, v²/a or a²/v⁴—and those choices lead to the corresponding power of ar/v². Therefore, any dimensionless group can be formed from ar/v². Our three quantities a, r, and v produce exactly one independent dimensionless group.

As a result, any statement about the circular acceleration can be written using only ar/v². All dimensionless statements using only ar/v² have the general form

\[\frac{ar}{v^{2}}=\textrm{dimensionless constant,}\]

because there are no other independent dimensionless groups to use on the right side.

Why can’t we use ar/v² on the right side?

We can, but it doesn’t create new possibilities. As an example, let’s try the following dimensionless form:

\[\frac{ar}{v^{2}} =3(\frac{ar}{v^{2}})-1.\]

Its solution is ar/v² =1/2, which is another example of our general form

\[\frac{ar}{v^{2}}=\textrm{dimensionless constant.}\]

But what if we use a more complicated function?

Let’s try one:

\[\frac{ar}{v^{2}}=(\frac{ar}{v^{2}})^{2}-1.\]

Its solutions are

\[\frac{ar}{v^{2}}=\left \{ \begin{array}{ll} \phi \\-1/\phi, \\ \end{array} \right. \]

where \(\phi\) is the golden ratio (1.618…). Deciding between these solutions would require additional information or requirements, such as the sign convention for the acceleration. Even so, each solution is another example of the general form

\[\frac{ar}{v^{2}}=\textrm{dimensionless constant.}\]

Using that form, which we cannot escape, the acceleration of the train is

\[a \sim \frac{v^{2}}{r}\]

where the ~ contains the (unknown) dimensionless constant. In this case, the dimensionless constant is 1. However, dimensional analysis, as this procedure is called, does not tell us its value—which would come from a calculus analysis or, approximately, from a lumping analysis (Section 6.3.4).

Using a ~ v²/r, we can now estimate the inward acceleration of the train going around a curve. Imagine a moderately high-speed train traveling at v ≈ 60 meters per second (approximately 220 kilometers or 135 miles per hour). At such speeds, railway engineers specify that the track’s radius of curvature be at least 2 or 3 kilometers. Using the smaller radius of curvature, the inward acceleration becomes

\[a \sim \frac{(60 m s^{-1})^{2}}{2 \times 10^{3}m} = 1.8 m s^{-2}.\]

Because no quantity with dimensions is large or small on its own, this acceleration by itself is meaningless. It acquires meaning in comparison with a relevant acceleration: the gravitational acceleration g. For this train, the dimensionless ratio a/g is approximately 0.18. This ratio is also tan θ, where θ is the train’s tilt that would make the passengers feel a net force perpendicular to the floor. With a/g ≈ 0.18, the comfortable tilt angle is approximately 10^∘. Indeed, tilting trains can tilt up to 8^∘. (This range is usually sufficient, as a full tilt is disconcerting: One would see a tilted ground but would still feel gravity acting as it normally does, along one’s body axis.)

Using our formula for circular acceleration, we can also estimate the maximum walking speed. In walking, one foot is always in contact with the ground. As a rough model of walking, the whole body, represented as a point mass at its center of mass (CM), pivots around the foot in contact with the ground--as if the body were an inverted pendulum. If you walk at speed v and have leg length l, then the resulting circular acceleration (the acceleration toward the foot) is v²/l. When you walk fast enough such that this acceleration is more than g, gravity cannot supply enough acceleration.

This change happens when \(v \sim \sqrt{gl}\) Then your foot leaves the ground, and the walk turns into a run. Therefore, gait is determined by the dimensionless ratio v²/gl. This ratio, which also determines the speed of water waves (Problem 5.15) and of ships (Problem 5.64), is called the Froude number and abbreviated Fr.

Here are three ways to check whether our pendulum model of walking is reasonable. First, the resulting formula,

\[v_{max} \sim \sqrt{gl}\]

explains why tall people, with a longer leg length, generally walk faster than short people.

Second, it predicts a reasonable maximum walking speed. With a leg length of l ~ 1 meter, the limit is 3 meters per second or about 7 miles per hours:

\[v_{max} \sim \sqrt{10 m s^{-2} \times 1m} \approx 3 m s^{-1}.\]

This prediction is consistent with world-record racewalking speeds, where the back toe may not leave the ground until the front heel has touched the ground. The world records for 20-kilometer racewalking are 1h:24m:50s for women and 1h:17m:16s for men. The corresponding speeds are 3.9 and 4.3 meters per second.

The third test is based on gait. In a fascinating experiment, Rodger Kram and colleagues [31] reduced the effective gravity g. This reduction changed the speed of the walking–running transition, but the speed still satisfied v²/gl ~ 0.5. The universe cares only about dimensionless quantities!

Another moral of this introduction to dimensional analysis is that every dimensionless group is an abstraction. Here, the group ar/v² tells us that the universe cares about a, r, or v through the combination ar/v² . This relation is described by the tree diagram (the edge label of -2 is the exponent of v in v^-2 ). Because there is only one independent dimensionless group, the universe cares only about ar/v². Therefore, the universe cares about a, r, and v only through the combination ar/v². This freedom not to worry about individual quantities simplifies our picture of the world.

5.1.2 Counting dimensionless groups

Finding the circular acceleration required finding all possible dimensionless groups and showing that all groups could be constructed from one group—say, ar/v². That reasoning followed a chain of constraints: To get rid of the time, the dimensionless group had to contain the quotient a/v²; to get rid of the length, the quotient needed to be multiplied by the correct power of r.

For each problem, do we have to construct a similar chain of reasoning in order to count the dimensionless groups?

Following the constraints is useful for finding dimensionless groups, but there is a shortcut for counting independent dimensionless groups. The number of independent groups is, roughly, the number of quantities minus the number of dimensions. A more precise statement will come later, but this version is enough to get us started.

Let’s test it using the acceleration example. There are three quantities: a, v, and r. There are two dimensions: length (L) and time (T). There should be, and there is, one independent dimensionless group.

Let's also test the shortcut on a familiar physics formula: W=mg, where W is an object's weight, m is its mass, and g is the gravitational acceleration. There are three quantities (W, m, and g) made from three dimensions (M, L, and T). Our shortcut predicts no dimensionless groups at all. However, W/mg is dimensionless, which refutes the prediction!

W	MLT^-2	weight
m	M	mass
g	LT^-2	gravity

What went wrong?

Although the three quantities seem to contain three dimensions, the three dimensional combinations actually used—force (MLT⁻²), mass (M), and acceleration (LT⁻²)—can be constructed from just two dimensions. For example, we can construct them from mass and acceleration.

But acceleration is not a fundamental dimension, so how can we use it?

The notion of fundamental dimensions is a human convention, part of our system of measurement. Dimensional analysis, however, is a mathematical process. It cares neither about the universe nor about our conventions for describing the universe. That lack of care may appear heartless and seem like a disadvantage. However, it means that dimensional analysis is independent of our arbitrary choices, giving it power and generality.

As far as dimensional analysis is concerned, we can choose any set of dimensions as our fundamental dimensions. The only requirement is that they suffice to describe the dimensions of our quantities. Here, mass and acceleration suffice, as the rewritten dimensions table shows: Using the notation [a] for "the dimensions of a," the dimensions of W are M x [a] and the dimensions of g are just [a].

W	M x [a]	weight
m	M	mass
g	[a]	gravity

In summary, the three quantities contain only two independent dimensions. Three quantities minus two independent dimensions produce one independent dimensionless group. Accordingly, the revised counting shortcut is

number of quantities

- number of independent dimensions

_____________________________________

number of independent dimensionless groups

This shortcut, known as the Buckingham Pi theorem [4], is named after Edgar Buckingham and his uppercase Pi (Π) notation for dimensionless groups. (It is also the rank–nullity theorem of linear algebra [42].)

Exercise \(\PageIndex{1}\): Bounding the number of independent dimensionelss groups

Why is the number of independent dimensionless groups never more than the number of quantities?

Exercise \(\PageIndex{2}\): Counting dimensionless groups

How many independent dimensionless groups do the following sets of quantities produce?

a. period T of a spring–mass system in a gravitational field: T, k (spring constant), m, x₀ (amplitude), and g.

b. impact speed v of a free-falling object: v, g, and h (initial drop height).

c. impact speed v of a downward-thrown free-falling object: v, g, h, and v₀ (launch velocity).

Exercise \(\PageIndex{3}\): Using angular frequency instead of speed

Redo the dimensional-analysis derivation of circular acceleration using the radius r and the angular frequency \(\omega\) as independent variables. Using a = v²/r, find the dimensionless constant in the general dimensionless form

\[\textrm{dimensionless group proportional to } a= \textrm{dimensionless constant}\]

Exercise \(\PageIndex{4}\): Impact speed of a dropped object

Use dimensional analysis to estimate the impact speed of a freely falling rock that is dropped from a height ℎ

Exercise \(\PageIndex{5}\): Speed of gravity waves on deep water

In this problem, you use dimensional analysis to find the speed of waves on the open ocean. These waves are the ones you would see from an airplane and are driven by gravity. Their speed could depend on gravity g, their angular frequency \(\omega\), and the density of water \(\rho\). Do the analysis as follows.

v	LT^-1	wave speed
g	LT^-2	gravity
\(\omega\)	T^-1	angular freq.
\(rho\)	ML^-3	water density

a. Explain why these quantities produce one independent dimensionless group.

b. What is the group proportional to v?

c. With the further information that the dimensionless constant is 1, predict the speed of waves with a period of 17 seconds (you can measure the period by timing the interval between each wave’s arrival at the shore). This speed is also the speed of the winds that produced the waves. Is it a reasonable wind speed?

d. What would the dimensionless constant be if, in the table of quantities, angular frequency \(\omega\) is replaced by the period T?

Exercise \(\PageIndex{6}\): Using period instead of speed

In finding a formula for circular acceleration (Section 5.1.1), our independent variables were the radius r and the speed v. Redo the dimensional-analysis derivation using the radius r and the period T as the independent variables.

a. Explain why there is still only one independent dimensionless group.

b. What is the independent dimensionless group proportional to a?

c. What dimensionless constant is this group equal to?