Linear Algebra

Topic: Dimensional Analysis 151any complete relationship among quantities with dimensional formulas can bealgebraically manipulated into a form where there is some function f such thatf(Π 1 , . . . , Π n ) = 0for a complete set {Π 1 , . . . , Π n } of dimensionless products. (The first examplebelow describes what makes a set of dimensionless products ‘complete’.) Weusually want to express one of the quantities m 1 for instance, in terms of theothers, and for that we will assume that the above equality can be rewrittenm 1 = m −p 22· · · m −p kk· ˆf(Π 2 , . . . , Π n )where Π 1 = m 1 m p 22 · · · mp kkis dimensionless and the products Π 2, . . . , Π n don’tinvolve m 1 (as with f, here ˆf is just some function, this time of n − 1 arguments).Thus, to do dimensional analysis we should find which dimensionless productsare possible.For example, consider again the formula for a pendulum’s period.dimensionalquantity formulaperiod p L 0 M 0 T 1length of string l L 1 M 0 T 0mass of bob m L 0 M 1 T 0acceleration due to gravity g L 1 M 0 T −2arc of swing θ L 0 M 0 T 0By the first fact cited above, we expect the formula to have (possibly sums ofterms of) the form p p 1l p 2m p 3g p 4θ p 5. To use the second fact, to find whichcombinations of the powers p 1 , . . . , p 5 yield dimensionless products, considerthis 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 1 M 0 T −2 ) p 4(L 0 M 0 T 0 ) p 5= L 0 M 0 T 0It gives three conditions on the powers.p 2 + p 4 = 0p 3 = 0p 1 − 2p 4 = 0Note that p 3 = 0 so the mass of the bob does not affect the period. Gaussianreduction and parametrization of that system gives this⎛ ⎞ ⎛ ⎞ ⎛ ⎞p 1 1 0p 2−1/20∣{p⎜ 3=0p⎟ ⎜ ⎟ 1 +0p⎜ ⎟ 5 p1 , p 5 ∈ R}⎝p 4 ⎠ ⎝ 1/2⎠⎝0⎠p 5 0 1(we’ve taken p 1 as one of the parameters in order to express the period in termsof the other quantities).