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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