Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP Posting-Version: version B 2.10.1 6/24/83; site ucf-cs.UUCP Path: utzoo!watmath!clyde!floyd!whuxle!mit-eddie!genrad!decvax!mcnc!duke!ucf-cs!giles From: giles@ucf-cs.UUCP (Bruce Giles) Newsgroups: net.physics Subject: Re: Re: eV Revisited Message-ID: <1189@ucf-cs.UUCP> Date: Wed, 15-Feb-84 23:05:14 EST Article-I.D.: ucf-cs.1189 Posted: Wed Feb 15 23:05:14 1984 Date-Received: Sat, 18-Feb-84 01:43:33 EST References: <2513@fortune.UUCP> Organization: University of Central Florida Lines: 79 [munch munch ... gobble gobble] You could approach the problem as we did in general relativity; it's certainly a cleaner approach: let M, T, and L represent the "dimensions" under traditional approches (i.e. speed = distance/time, force = mass * distance/time ^ 2); and P, Q, and R represent non-collinear "dimensions" under your new approach (i.e. speed = dimensionless; mass = length...). Express P, Q, and R in terms of M, T, and L: P = M^pm * T^pt * L^pl Q = M^qm * T^qt * L^ql R = M^rm * T^rt * L^rl Now invert the above set of equations to find M, T, and L in terms of P, Q, and R: M = P^mp * Q^mq * R^mr T = P^tq * Q^tq * R^tr L = P^lr * Q^lq * R^lr M = (M^pm * T^pt * L^pl)^mp * (M^qm * T^qt * L^ql)^mq * (M^rm * T^rt * L^rl)^mr T = (M^pm * T^pt * L^pl)^tp * (M^qm * T^qt * L^ql)^tq * (M^rm * T^rt * L^rl)^tr L = (M^pm * T^pt * L^pl)^lp * (M^qm * T^qt * L^ql)^lq * (M^rm * T^rt * L^rl)^lr [I think this step is valid, but when I write it down it looks funny....] M = M^(pm*mp+qm*mq+rm*mr) * T^(pt*mp+qt*mq+rt*mr) * L^(pl*mp+ ql*mq+rl*mr) T = M^(pm*tp+qm*tq+rm*tr) * T^(pt*tp+qt*tq+rt*tr) * L^(pl*tp+ ql*tq+rl*tr) L = M^(pm*lp+qm*lq+rm*lr) * T^(pt*lp+qt*lq+rt*lr) * L^(pl*lp+ ql*lq+rl*lr) It follows directly that: pm*mp + qm*mq + rm*mr = 1 pt*mp + qt*mq + rt*mr = 0 pl*mp + ql*mq + rl*mr = 0 pm*tp + qm*tq + rm*tr = 0 pt*tp + qt*tq + rt*tr = 1 pl*tp + ql*tq + rl*tr = 0 pm*lp + qm*lq + rm*lr = 0 pt*lp + qt*lq + rt*lr = 0 pl*lp + ql*lq + rl*lr = 1 or, in a slightly more human-readable form, (pm qm rm) (mp) (1) (pt qt rt) (mq) = (0) (pl ql rl) (mr) (0) (pm qm rm) (tp) (0) (pt qt rt) (tq) = (1) (pl ql rl) (tr) (0) (pm qm rm) (lp) (0) (pt qt rt) (lq) = (0) (pl ql rl) (lr) (1) The square matrix is known (look at the definitions of the components..); hence the problem has been reduced to solving three matrix equations. An example of what I mean by P, Q, and R is shown below; just replace 'c' with what ever you want to normalize to.... P = c = m^0 * t^-1 * l^1 [traditional dimensional analysis] --> pm = 0; pt = -1; pl = 1. ave discordia going bump in the night ... bruce giles decvax!ucf-cs!giles university of central florida giles.ucf-cs@Rand-Relay orlando, florida 32816