Linear Algebra

152 Chapter Two. Vector SpacesThe set of dimensionless products contains all terms p p 1l p 2m p 3a p 4θ p 5subject to the conditions above. This set forms a vector space under the ‘+’operation of multiplying two such products and the ‘·’ operation of raising sucha product to the power of the scalar (see Exercise 5). The term ‘complete set ofdimensionless products’ in Buckingham’s Theorem means a basis for this vectorspace.We can get a basis by first taking p 1 = 1, p 5 = 0, and then taking p 1 = 0,p 5 = 1. The associated dimensionless products are Π 1 = pl −1/2 g 1/2 and Π 2 = θ.Because the set {Π 1 , Π 2 } is complete, Buckingham’s Theorem says thatp = l 1/2 g −1/2 · ˆf(θ) = √ l/g · ˆf(θ)where ˆf is a function that we cannot determine from this analysis (a first yearphysics text will show by other means that for small angles it is approximatelythe constant function ˆf(θ) = 2π).Thus, analysis of the relationships that are possible between the quantitieswith the given dimensional formulas has given us a fair amount of information: apendulum’s period does not depend on the mass of the bob, and it rises withthe square root of the length of the string.For the next example we try to determine the period of revolution of twobodies in space orbiting each other under mutual gravitational attraction. Anexperienced investigator could expect that these are the relevant quantities.dimensionalquantity formulaperiod p L 0 M 0 T 1mean separation r L 1 M 0 T 0first mass m 1 L 0 M 1 T 0second mass m 2 L 0 M 1 T 0gravitational constant G L 3 M −1 T −2To get the complete set of dimensionless products we consider the equation(L 0 M 0 T 1 ) p 1(L 1 M 0 T 0 ) p 2(L 0 M 1 T 0 ) p 3(L 0 M 1 T 0 ) p 4(L 3 M −1 T −2 ) p 5= L 0 M 0 T 0which results in a systemp 2 + 3p 5 = 0p 3 + p 4 − p 5 = 0p 1 − 2p 5 = 0with this solution.⎛ ⎞ ⎛ ⎞1 0−3/20∣{1/2p⎜ ⎟ 1 +−1p⎜ ⎟ 4 p1 , p 4 ∈ R}⎝ 0⎠⎝ 1⎠1/2 0