Linear Algebra

154 Chapter 2. Vector Spaces
The classic example is a pendulum. An investigator trying to determine
the formula for its period might conjecture that these are the relevant physical
quantities.
quantity
dimensional
formula
period p L 0 M 0 T 1
length of string ℓ L 1 M 0 T 0
mass of bob m L 0 M 1 T 0
acceleration due to gravity g L 1 M 0 T −2
arc of swing θ L 0 M 0 T 0
To find which combinations of the powers in p p1 ℓ p2 m p3 g p4 θ p5 yield dimensionless
products, consider this equation.
(L 0 M 0 T 1 ) p1 (L 1 M 0 T 0 ) p2 (L 0 M 1 T 0 ) p3 (L 1 M 0 T −2 ) p4 (L 0 M 0 T 0 ) p5 = L 0 M 0 T 0
It gives three conditions on the powers.
p2 + p4 =0
p3 =0
p1 − 2p4 =0
Note that p3 is 0—the mass of the bob does not affect the period. The system’s
solution space can be described in this way (p1 is taken as one of the parameters
in order to express the period in terms of the other quantities).
⎛
⎜
{ ⎜
⎝
p1
p2
p3
p4
p5
⎞
⎟
⎠ =
⎛ ⎞
1
⎜
⎜−1/2
⎟
⎜ 0 ⎟
⎝ 1/2 ⎠
0
p1
⎛ ⎞
0
⎜
⎜0
⎟
+ ⎜
⎜0
⎟
⎝0⎠
1
p5
�
� p1,p5 ∈ R}
Here is the linear algebra. The set of dimensionless products is the set of
products p p1 ℓ p2 m p3 a p4 θ p5 subject to the conditions in the above linear system.
This forms a vector space under the ‘+’ addition operation of multiplying two
such products and the ‘·’ scalar multiplication operation of raising such a product
to the power of the scalar (see Exercise 5). The term ‘complete set of
dimensionless products’ in Buckingham’s Theorem means a basis for this vector
space.
We can get a basis by first taking p1 =1andp5 = 0, and then taking p1 =0
and p5 = 1. The associated dimensionless products are Π1 = pℓ −1/2 g 1/2 and
Π2 = θ. The set {Π1, Π2} is complete, so we have
p = ℓ 1/2 g −1/2 · ˆ f(θ)
= � ℓ/g · ˆ f(θ)