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Re: [Phys-l] use of lgarithms



On 10/28/09 09:31, John Mallinckrodt wrote:
Here are two possible ways of thinking about it.

1. First of all it is important to note that one simply cannot take
the log of something like "10 meters."

Why not? People do it all the time.

If you think you can, then
tell me if the result is the same as the log of "1000 cm."

It is.

So you
are already implicitly dealing with the units in the following manner:

Rewrite the equation x = (1/2)gt^2 as

x/ul = (g/2)/(ul/ut^2) * (t/ut)^2 [1]

That's one way of doing it. But as my friend Tim Toady
likes to say, TIMTOWTDI.

The fact that equation [1] works for *any* ul and ut that
you can think of should be a tip-off.

Here's the deal: In ordinary introductory-level unit
analysis, the units are factors on each side, and all
the units multiplied together should be the same on
both sides. If at any point you divide meters by
centimeters, you get a dimensionless factor of 100.

Well, if you take the logarithm of both sides, the
language changes slightly and the ideas change hardly
at all. The units become loonits, i.e. the logarithms
of the previous units. The loonits are _terms_ on
each side, and all the loonits added together should
be the same on both sides. If at any point you subtract
log(cm) from log(m), you get an additive contribution
of log(100).

Additive units are not particularly uncommon. Familiar
examples include
-- Earthquakes, measured in Richter numbers
-- Acoustic or electromagnetic power, measured in dB
-- Proton concentration, measured in pH.