kk9w@pur-ee.UUCP (Andersen) (02/08/84)
I think some people missed the point of what I said in my last note on this subject of mass vs. energy. I think what I wrote was clear, but here is another attempt. Mass and energy are related to each other by a constant. This constant is c^2. If you know something's mass and want the energy associated with that mass, you use the well known equation E=mc^2. Likewise, to obtain mass from energy, divide the energy by c^2. The point of the whole thing is this. You cannot specify mass by using units of energy. You have to use units of mass, and the unit of mass associated with eV is eV/c^2. Not simply eV. Not a big point, but one I thought I should make. Dave pur-ee!kk9w
guy@rlgvax.UUCP (Guy Harris) (02/09/84)
> Mass and energy are related to each other by a constant. This > constant is c^2. If you know something's mass and want the energy > associated with that mass, you use the well known equation E=mc^2. > Likewise, to obtain mass from energy, divide the energy by c^2. > The point of the whole thing is this. You cannot specify mass > by using units of energy. You have to use units of mass, and the > unit of mass associated with eV is eV/c^2. Not simply eV. However, given that c^2 is a constant, you can specify mass by giving the energy equivalent of that mass; saying that the electron has a mass of .511 MeV is more convenient than saying it has a mass of .511 MeV/c^2, so the e-/e+ mass is conventionally given as .511 MeV. Guy Harris {seismo,ihnp4,allegra}!rlgvax!guy
crl@pur-phy.UUCP (Charles LaBrec) (02/10/84)
One of the first things you get used to in physics is differing unit systems. It is entirely possible (and in fact done) to devise a system of units where the unit of mass = the unit of energy. In relativity, you often see c = 1, and perhaps time in units of distance. In quantum electrodynamics, a frequent practice is to define hbar, c, and the mass of the electron = 1. One of the things that irks me sometimes is the variety of unit systems in electromagnetics. There are the MKS (SI), CGS(esu), CGS(emu), and Gaussian systems. Looking at more common systems, there are two English systems, one that uses "pound-mass" and the other "slugs" to measure mass. (I believe something that weighs 1 lb.-force has a mass of 1 lb-mass in the former.) Of course, a kilowatt-hr is a unit of energy (even though people commonly say, "How much power did we use?"). So, the bottom line is, don't hold unit systems sacred. Charles LaBrec UUCP: pur-ee!Physics:crl, purdue!Physics:crl INTERNET: crl @ pur-phy.UUCP
barryg@sdcrdcf.UUCP (Barry Gold) (02/10/84)
[fuel to feed the mailer]
Given the trend of physics since Einstein, it is probably a mistake to try
to distinguish between mass and energy. Measuring mass in eV makes it easy
to figure out how high an EMF you need to accelerate it. It also helps
one remember that mass and energy are the same thing.
Changing matter into energy is no different than transforming energy from
one form to another (the function performed by a transducer). Thus, there
is in principle no difference between the functions of:
a nuclear reactor
a hydro-electric plant
a loudspeaker
an LED.
Barry Gold
usenet: {decvax!allegra|ihnp4}!sdcrdcf!barryg
Arpanet: barry@BNL
--
Barry Gold
usenet: {decvax!allegra|ihnp4}!sdcrdcf!barryg
Arpanet: barry@BNLdarrelj@sdcrdcf.UUCP (Darrel VanBuer) (02/11/84)
The natural unit of velocity under relativity IS c (i.e. c=1) since it's the
omly priveledged unit. thus E=m with the right units for velocity.
--
Darrel J. Van Buer, PhD
System Development Corp.
2500 Colorado Ave
Santa Monica, CA 90406
(213)820-4111 x5449
...{allegra,burdvax,cbosgd,hplabs,ihnp4,sdccsu3,trw-unix}!sdcrdcf!darrelj
VANBUER@USC-ECL.ARPArpw3@fortune.UUCP (02/11/84)
#R:rlgvax:-168600:fortune:8600013:000:2555
fortune!rpw3 Feb 11 05:04:00 1984
Guy is right. It's just that jargon is confusing to some. Physicists
have gotten so used to talking about mass in eV that they don't realize
that people don't understand that they don't mean the mass of an electron
IS 511KeV, they mean the mass of the electron is EQUIVALENT to 511KeV
when said mass is converted to energy (via the c^2).
See, in an accelerator near the speed of light it all gets blurred anyway,
since nobody really CARES what the rest "mass" is when the relativistic
"mass/energy" is 100-1000 times rest mass. So the language gets
sloppy. People naturally contract "rest mass equivalent energy" ==>
rest mass energy => rest "mass".
<<fun ON>>
Note that other units can be converted for amusement and amazement...
Anybody remember Cpt. Grace Hopper's standard guest lecture, with
the punch-line when she reaches into the bag and pulls out some
"nanoseconds"? Little pieces of telephone wire, 11.9 inches long!
Somebody once showed me the corresponding equivalence of mass and length,
if I can just remember it. Let's see (this is a bit weird)...
1. F = m * a force due to acceleration
2. F = G * (m1 * m2) / r^2 force due to gravity (r = distance)
1,2=> 3. a = G * m / r^2 acceleration (of a point mass)
due to gravity.
4. a = S / t^2 Def'n of acceleration (S = distance)
3,4=> 5. S / t^2 = G * m / r^2
or, m = 1/G * S * (r/t)^2 (rearranging)
6. c = c0 * (r/t) (let c0 be just the number ~2.99e8
with no units)
5,6=> 7. m = 1/(G*c0^2) * c^2 * S
and voila, pulling the same trick of scaling by c^2, we get
8. m = K * S where constant K = c^2 / (G * c0^2)
or, S = K' * m where K' = 1 / K
Anyway, the dimensional analysis holds, if you want to play the game.
The way I heard it, by the time you have all the scaling constants
worked out, the mass of the Sun is ~ 1.1 kilometers, which led a wag to
suggest that that's how far the Sun has ejected itself from our space-time
due to its mass. Rubber-sheet geometry lives! :-)
Rob Warnock
UUCP: {sri-unix,amd70,hpda,harpo,ihnp4,allegra}!fortune!rpw3
DDD: (415)595-8444
USPS: Fortune Systems Corp, 101 Twin Dolphins Drive, Redwood City, CA 94065
p.s. You can also get length in terms of mass by starting with
the following two equations and playing the same games (hint:
expand everything out to mass, distance, and time, and start
cancelling terms 'til you're left with m = S), but the constant
of proportionality comes out different. Any explanations?
1. E = m * c^2 Einstein
2. E = F * S force * distance
...
N. S = k" * m (what's K"?)giles@ucf-cs.UUCP (Bruce Giles) (02/16/84)
[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
gwyn%brl-vld@sri-unix.UUCP (02/19/84)
From: Doug Gwyn (VLD/VMB) <gwyn@brl-vld> The reason you don't get the same conversion factor for mass<->distance working it out two ways is that in one of them you have tried to mix a relativistic fundamental equivalence (E=mc^2) with a Newtonian formula (E=Fs) that is not appropriate. One should not just toss a bunch of formulas together at random; your equations should be describing some single THING or SITUATION in order for them all to apply at once. It is true that if sufficient care is taken one can reduce the number of "fundamental" physical units (standards of measurement). Instead of separate time and space units, one can use the fundamentals of special relativity to relate the units to each other via the speed of light. (A careful treatment of this would require discussion of canonical forms of the metric tensor and their physical interpretation.) Similarly, other types of physical quantity have fundamental relations that let one interconvert what appear to be different aspects of an object or situation. The interesting question is, how FEW independent standards of physical measurement are there? From the standpoint of geometrical theories of physics such as relativity, there would appear to be only one basic unit (could measure any non-pure-number attribute, e.g. unit of length, or of mass, or of time, ...). Amazingly, according to Einstein-Schr"odinger unified field theory, there are NO inpendendent measurement standards. This is a consequence of the closure of the theory; it is sometimes phrased "the universe is self- gauging". A more usual way of looking at this would be to allow one unit (say length) in terms of which the natural curvature of the universe would be expressed (the reciprocal of a very large number, using "laboratory size" units), instead of the theoretically preferable approach of using the inherent curvature as a "natural" unit. (In this theory, the local metric is intimately related to the large-scale structure of space-time via a non-zero CONSTANT that spontaneously occurs in the development of the theory [think of it as a constant of integration if you wish].)