Dimensional Analysis

Derived quantities versus fundamental quantities

To measure physical quantities we use two types of units: fundamental units and
derived units. There are innumerous derived units, but the number of fundamental
units is kept to minimum. In Mechanics there are three fundamental units: unit
of length [L], unit of time [T], and unit of mass [M]. To measure the
fundamental units we use scales called standards. Standards may be different:
meter or foot, second or year, kilogram or slag. Expression of a derived unit
versus fundamental units is called dimensionality.

The scales can be different but the relationships between the quantities must
remain the same: the ratio of two numerical values of the same quantity must not
depend on the scale used in their measurement. For instance, the ratio of two
areas, e.g. room floor and parking lot, does not depend on whether we measure
them in square meters or squatter feet. Hence, dimension of a derived physical
quantity must be a power term of the fundamental units:

[D]=[L]^m [T]^n [M]^k

Indeed, let’s consider a quantity D=Lm, e.g. area D of a square with a side L.
Then, the ratio of two squares is D1/D2=(L1/L2)m. Now let us increase the scale
of L by times (going from meters to feet 3); then D1()/D2()=(L1/L2)m
=D1/D2. That would not be the case if, e.g. D=e^L. For instance, force is a
derived unit; its dimensionality is determined as: [F]=[M][L][T]2.

Equation verification

All equations must be consistent with respect to dimensions. That is, the
dimensions of the physical quantities on the left hand side of an equation must
be equal to the dimensions of the physical quantities on the right hand side.
For instance, the dimension of physical work is [W]=[M][L]2[T]2 . Let’s verify
if the laymen’s definition of physical work as “force times distance” is proper.

[F][L]=[M][L][T]2[L] =[M][L]2[T]2= [W]

Relationship between physical quantities

Dimensional analysis can also be used to obtain relationships between physical
quantities without a thorough analysis of the phenomenon. Let us consider as an
example period of a simple pendulum. A simple pendulum has its mass concentrated
at a point (called the center of mass). Although the definition of a point is a
position without dimension, the length of the string supporting the mass (the
bob) must be large in comparison with the diameter of the bob. The question
before us is: How does its period P depend on the physical parameters of the
pendulum? We may speculate that the period of a simple pendulum depends upon the
length of the pendulum L, the mass of the pendulum bob m, and the magnitude of
the bob’s displacement from the position of equilibrium h.

Let us try the formula:

P=L^m m^n h^k

There are three problems with this formula:

[P]=[L]/[T] but there is no [M] in the left hand side

There is no [T] in the right hand side

[h]=[L], then how do we combine h with L?

Problem 1 means that P is independent of m; this is an important conclusion of
the Dimensional Analysis. Problem 2 means that we have to be looking for another
quantity that affects motion of a pendulum. Such quantity is gravity and we
should include acceleration due to gravity g into the list. Now the formula
looks like that:

P=L^m g^n h^k

Actually, Problem 3 is the hardest. It is not possible to resolve the issue of
h. This requires experimental study, which shows that the bob’s displacement
from the equilibrium does not affect the period unless the angular displacement
is too large. Finally we try the formula:

P=L^m g^n

Taking into account that [g]=[L] [T]2 the only possibility to satisfy this is
the formula

P∝√(L/g)

This is a great formula, which tells us a lot about motion of a pendulum. The
exact formula has a coefficient 2, which can be found experimentally.

Thus, a combination of Dimensional Analysis and Experiment is a powerful
scientific tool.