Re: FW: Dinosaur Weights

[Dinogeorge@aol.com]

In a message dated 97-08-23 20:06:10 EDT, EDELS@classic.msn.com (Allan )
writes:

<< I guess it is a matter of approximation.  We say the density of water is 1

 gram to 1 cubic centimeter.  I suspect that it is true until you reach the 
 cubic meter size and above.  I don't know in which direction (without
looking 
 it up somewhere) the discrepancy lies.>>

A long time ago, I learned that a kilogram was defined to be the mass of one
liter of water = 1000 cubic centimeters. Later I found out that one liter of
water is not quite 1000 cubic centimeters (or vice versa: I don't remember
either). The difference is minute--a few thousandths of a percent. Which
suggests a temperature effect of some kind--measuring the liter of water at
one temperature but a cubic centimeter of water at another temperature,
thereby slightly altering the density or whatever. But I sure don't know
whether this is correct.

<<  This is sort of like "E=mc2"  -  which 
 is true, but not precise in all instances.  The more complete version of the

 formula is "E=mc2 + m2c4" [in case this doesn't pass through the email 
 correctly - the first formula is E = m c (squared) and the second formula is
E 
 = m c (squared) plus m (squared) c (to the fourth)].  The lesser terms tend
to 
 fall out of the formulas because they are usually unimportant to the task at

 hand.  On occasion, it does matter.  Some scientists like to speak as 
 precisely as possible, so they will remind you that this number is not
exact.  
 Just as I wish some paleontologists would say that "This is what I think the

 dinosaurs were like" vs. "This is what the dinosaurs were like"!!! >>

I think the higher-order terms in the equation depend on higher powers of
v/c, not just c. So they would not become important except at velocities
>very< close to c. Otherwise the higher order terms totally overwhelm mc^2.
At velocity zero, of course, e=mc^2 exactly: the rest mass of the object.