Linear Algebra

Topic: Dimensional Analysis 153As earlier, the set of dimensionless products of these quantities forms avector space and we want to produce a basis for that space, a ‘complete’ set ofdimensionless products. One such set, gotten from setting p 1 = 1 and p 4 = 0and also setting p 1 = 0 and p 4 = 1 is {Π 1 = pr −3/2 m 1/21G 1/2 , Π 2 = m −11 m 2 }.With that, Buckingham’s Theorem says that any complete relationship amongthese quantities is stateable this form.p = r 3/2 m −1/21G −1/2 · ˆf(m −11 m 2) = r3/2√ Gm1· ˆf(m 2 /m 1 )Remark. An important application of the prior formula is when m 1 is themass of the sun and m 2 is the mass of a planet. Because m 1 is very much greaterthan m 2 , the argument to ˆf is approximately 0, and we can wonder whetherthis part of the formula remains approximately constant as m 2 varies. One wayto see that it does is this. The sun is so much larger than the planet that themutual rotation is approximately about the sun’s center. If we vary the planet’smass m 2 by a factor of x (e.g., Venus’s mass is x = 0.815 times Earth’s mass),then the force of attraction is multiplied by x, and x times the force acting onx times the mass gives, since F = ma, the same acceleration, about the samecenter (approximately). Hence, the orbit will be the same and so its periodwill be the same, and thus the right side of the above equation also remainsunchanged (approximately). Therefore, ˆf(m 2 /m 1 ) is approximately constant asm 2 varies. This is Kepler’s Third Law: the square of the period of a planet isproportional to the cube of the mean radius of its orbit about the sun.The final example was one of the first explicit applications of dimensionalanalysis. Lord Raleigh considered the speed of a wave in deep water andsuggested these as the relevant quantities.The equationdimensionalquantity formulavelocity of the wave v L 1 M 0 T −1density of the water d L −3 M 1 T 0acceleration due to gravity g L 1 M 0 T −2wavelength λ L 1 M 0 T 0(L 1 M 0 T −1 ) p 1(L −3 M 1 T 0 ) p 2(L 1 M 0 T −2 ) p 3(L 1 M 0 T 0 ) p 4= L 0 M 0 T 0gives this systemwith this solution space.p 1 − 3p 2 + p 3 + p 4 = 0p 2 = 0−p 1 − 2p 3 = 0⎛ ⎞10{ ⎜ ⎟⎝−1/2⎠ p ∣1 p1 ∈ R}−1/2