Linear Algebra

Topic: Dimensional Analysis 155
where ˆ f is a function that we cannot determine from this analysis (by other
means we know that for small angles it is approximately the constant function
ˆf(θ) =2π).
Thus, analysis of the relationships that are possible between the quantities
with the given dimensional formulas has given us a fair amount of information: a
pendulum’s period does not depend on the mass of the bob, and it rises with
the square root of the length of the string.
For the next example we try to determine the period of revolution of two
bodies in space orbiting each other under mutual gravitational attraction. An
experienced investigator could expect that these are the relevant quantities.
quantity
dimensional
formula
period of revolution p L 0 M 0 T 1
mean radius of separation r L 1 M 0 T 0
mass of the first m1 L 0 M 1 T 0
mass of the second m2 L 0 M 1 T 0
gravitational constant G L 3 M −1 T −2
To get the complete set of dimensionless products we consider the equation
(L 0 M 0 T 1 ) p1 (L 1 M 0 T 0 ) p2 (L 0 M 1 T 0 ) p3 (L 0 M 1 T 0 ) p4 (L 3 M −1 T −2 ) p5 = L 0 M 0 T 0
which gives rise to these relationships among the powers
p1
p2
+3p5 =0
p3 + p4 − p5 =0
− 2p5 =0
with the solution space
⎛ ⎞
1
⎜
⎜−3/2
⎟
{ ⎜ 1/2 ⎟
⎝ 0 ⎠
1/2
p1
⎛ ⎞
0
⎜ 0 ⎟
+ ⎜
⎜−1
⎟
⎝ 1 ⎠
0
p4
�
� p1,p4 ∈ R}
(p1 is taken as a parameter so that we can state the period as a function of the
other quantities). As with the pendulum example, the linear algebra here is
that the set of dimensionless products of these quantities forms a vector space,
and we want to produce a basis for that space, a ‘complete’ set of dimensionless
products. One such set, gotten from setting p1 = 1 and p4 = 0, and also
setting p1 = 0 and p4 = 1 is {Π1 = pr−3/2m 1/2
1 G1/2 , Π2 = m −1
1 m2}. With
that, Buckingham’s Theorem says that any complete relationship among these
quantities must be stateable this form.
p = r 3/2 m −1/2
1 G−1/2 · ˆ f(m −1
1 m2)
= r3/2
√ ·
Gm1
ˆ f(m2/m1)