Date: Mon, 4 Jan 1999
Math in One Lesson
Thomas M. Miovas, Jr.
I finally had a chance to read "Mathematics in One Lesson," and
I don't recall anyone answering these questions from Ed Matthews
in the interim, so I'll have a go at it.
Ed Matthews wrote:
>
>I have two questions based on the recent TIA article.
>
>(1) How does multiplication by units of different types fit into this
>theory? For instance, momentum is the product of mass and velocity. Mass
>is some multiple of a uniform unit, while velocity is some multiple of
>_another_ uniform unit. Does their product then become a multiple of yet
>another type of uniform unit?
For the metric system of measurements, at least, there is no
unit of measurement for either velocity or momentum. The "units"
of velocity are meters per second (m/s) in a specific direction.
The "units" of momentum are kilogram meters per second (kgm/s)
[Where "k" means "1,000" and "g" means grams. The actual unit is
the gram, but that is inconveniently small, so the kilogram is
used instead]. The primary reason there are no such units for
velocity and momentum is due to the unit economy purpose of
having units of measurement in the first place -- i.e. if every
valid combination of units of measurement each had its own
designated unit of measurement, one wouldn't be able to keep
track of them all.
I think momentum could have been designated its own unit of
measurement, since Newton called it a "quantity of motion,"
except for a certain controversy over the concept "mass." In
Newton's conception, "mass" meant "resistance to change of
motion" for inertial motion (locomotion). That is, momentum
incorporated the object's current motion (v) with its resistance
to change of that motion (m), potentially as one measurement.
However, "mass" also involved the resultant weight of something
in a gravitational field. Since Newton (and scientists long
after him) couldn't integrate the two meanings of "mass," I
think a separate unit of measurement for momentum was not
incorporated into the system of measurement simply to avoid
confusion.
Let's take a unit of measurement that does involve other units
of measurement that are seemingly multiplied or divided by each
other. The "Newton" (N) is a unit of measurement of force. This
unit can be broken down into kilograms (kg), meters (m), and
seconds (s) as in: 1N = (1kg)(1m)/(1s^2). But it's confusing to
write it this way. A better way is: 1N = (1kg)(1m/1s)/1s. Which
means that a force of one newton acting on a body of one
kilogram will get that body up to a speed of one meter per
second in one second -- *that's* the unit of measurement. Notice
that this involves a specific relationship of standard units of
measurement to each other. One is not actually multiplying and
dividing different units of measurements. There is no such thing
as a kilogram number of meter groups, for instance. Instead, the
ratio of the (subsumed) units of measurement (arrived at by
observation) *is* the unit of measurement. This process doesn't
have to involve one unit of measurement to one unit of
measurement ratios to be valid, but generally speaking, the
metric system does it this way for its convenience.
>
>(2) How does this theory explain differential coefficients in calculus?
>(These are the "dx" and "dy" in the equations.) These things aren't
>infinitely small (maybe arbitrarily small?). Are they units?
The symbols in calculus are notations of mathematical
operations, rather than being units of measurements in and of
themselves. As multiplication is re-iterative addition and
division is re-iterative subtraction, so calculus involves
re-iterative multiplication and re-iterative division.
