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Imaginary Exponents

Mathematics, of course, provides a simple solution to this problem: just have  $\kappa$  be an imaginary number, say

\begin{displaymath}\kappa \; \equiv \; i \, \omega
\qquad \hbox{\rm where} \qquad
i \; \equiv \; \sqrt{-1}
\end{displaymath}

and  $\omega$  is a positive real constant. Let's see if this trial solution ``works'' (i.e. take its second derivative and see if we get back our equation of motion):
x(t) = $\displaystyle x_0 \, e^{i \, \omega \, t}$ (13.14)
$\displaystyle \dot{x}$ = $\displaystyle i \, \omega \, x_0 \, e^{i \, \omega \, t}$ (13.15)
$\displaystyle \ddot{x}$ = $\displaystyle - \omega^2 \, x_0 \, e^{i \, \omega \, t}$ (13.16)
$\displaystyle \hbox{\rm or} \qquad
\ddot{x}$ = $\displaystyle - \omega^2 \, x$ (13.17)
$\displaystyle \hbox{\rm so} \qquad \omega$ $\textstyle \equiv$ $\displaystyle \sqrt{k \over m}$ (13.18)

OK, it works. But what does it describe? For this we go back to our series expansions for the exponential, sine and cosine functions and note that if we let   $z \equiv i \theta$,  the following mathematical identity holds:13.5

\begin{displaymath}e^{i \, \theta} \; = \; \cos(\theta) \; + \; i \; \sin(\theta)
\end{displaymath} (13.19)

Thus, for the case at hand, if   $\theta \equiv \omega \, t$  [you probably knew this was coming] then

\begin{displaymath}x_0 \, e^{i \, \omega \, t} \; = \;
x_0 \; \cos(\omega t) \; + \; i \; x_0 \; \sin(\omega t)
\end{displaymath}

-- i.e. the formula for the projection of uniform circular motion, with an imaginary part ``tacked on.'' What does this mean?

I don't know.

What! How can I say, ``I don't know,'' about the premiere paradigm of Mechanics? We're supposed to know everything about Mechanics! Let me put it this way: we have happened upon a nice tidy mathematical representation that works - i.e. if we use certain rules to manipulate the mathematics, it will faithfully give correct answers to our questions about how this thing will behave. The rules, by the way, are as follows:

Keep the imaginary components through all your calculations until the final ``answer,'' and then throw away any remaining imaginary parts of any actual measurable quantity.
The point is, there is a difference between understanding how something works and knowing what it means. Meaning is something we put into our world by act of will, though not always conscious will. How it works is there before us and after we are gone. No one asks the ``meaning'' of a screwdriver or a carburetor or a copy machine; some of the conceptual tools of Physics are in this class, though of course there is nothing to prevent anyone from putting meaning into them.13.6


next up previous
Next: Damped Harmonic Motion Up: The Spring Pendulum Previous: The Spring Pendulum
Jess H. Brewer
1998-10-09