| HPS 0628 | Paradox | |
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Back to Additive Measures
John D. Norton
Department of History and Philosophy of Science
University of Pittsburgh
http://www.pitt.edu/~jdnorton
The idea of an additive measure is so familiar that one might wonder whether there can be other measures. It is a simple matter to contrive very many other measures. Here is a simple example.
The volume of spherical drop of liquid is an additive measure in this sense. If it is produced by combining many smaller drops, then the volume of the large drop is just the sum of the volumes of the smaller drops:
volume large drop = volume small drop-1 + volume small drop-2 + ...
Now consider the areas of these drops. These areas are a different sort of measure of the drops. We might not normally think of measuring drops in terms of their area. Things are otherwise if they are hard spheres and our job is to paint them. Then we want to know how much paint is needed to cover their surfaces. We might wonder how the amount of paint needed to cover all the individual spheres relates to the amount needed to paint them when all the matter of the spheres is combined into a single sphere. We are asking how the area measures of the individual spheres relates to the area measure of the combined sphere.
The area of these spheres can be important in other
contexts. If they are droplets of water, their surface area determines how
quickly they evaporate. How much faster does water in a mist
evaporate than the same amount of water in single drop? The answer depends
on comparing their surface areas.
Since volume increases with the cube of radius and area with the square of radius, we can conclude that the volume of the drop is proportional to the area raised to the power 3/2. Using this fact we can now rewrite the rule for volumes as:
(area large drop)3/2 = (area small drop-1)3/2 + (area small drop-2)3/2 + ...
That is, areas do not conform with a simple additive rule, but a more complicated rule of "three-halves power." Another way to write this rule is as
(area large drop)
= [ (area small drop-1)3/2
+ (area small drop-2)3/2 + ... ]2/3
Here is how the rule works
in practice. Imagine that we have 8 drops, each with area 4. The normal
rule of arithmetic addition would sum the areas as:
4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 = 32
The "three-halves" rule, however, gives a different result. To recover it,
first we calculate:
43/2 = (43)1/2 = (64)1/2
= 8
We then add up eight of these to recover:
43/2 + 43/2 + 43/2 + 43/2 +43/2
+ 43/2 +43/2 + 43/2
= 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 = 64
We then take the 2/3 power to find the final area:
642/3 = (642)1/3 = 40961/3 =
16.
Another example will be familiar to anyone who has played around with electrical circuits. When a voltage V is placed over a resistor, a current i flows in it. Ohm's law tells us that the resistance R of the resistor is given by R = V/i. There are two ways we combine resistors. In one case, we wire two resistors in series:
Their combined resistance is just given by the
arithmetic sum of the two component resistances:
Rseries = R1 + R2
We can also wire them in parallel:
Then their combined resistance is given by a different formula:
Rparallel = 1/[ 1/R1 + 1/R2 ]
Alternative rules such as these are used much less frequently. The reason is more a matter of practicality. The theory of additive measures is simple and well developed and, with ingenuity, can be made to apply to many systems. If our attachment to additive measure is strong enough, we might only use the additive volume measure in our deliberations and introduce considerations of area as a derivative quantity sparingly.
June 28, October 4, 2021. February 23, 2023.
Copyright, John D. Norton