Imaginary numbers aren't so complex

There's a further way of seeing the solution for a harmonic oscillator. We can use complex numbers. This involves the use $i \equiv \sqrt{-1}$. An imaginary number is a multiple of $i$, like $3.1 i$, $-2i$, etc. A complex number is the sum of a regular (real) number and an imaginary number. But actually these complex numbers aren't so complex and have a lot of really nifty properties.

The use of complex numbers is extremely important in physics and is used all the time in a wide variety of fields, so this should give you an introduction as to how to use them and what they mean.

Let's guess another functional form and see if it solves eqn. 1.1 This time guess

x(t) ~=~ A~e^{\lambda t}
\end{displaymath} (1.23)

This doesn't seem right. An exponential form like this grows, or decays. It doesn't oscillate!. OK, but suppose that we're too stupid to see this and think that it might be a good solution all the same. Sometimes trying stupid things leads to new and interesting discoveries, and this is one of those times!

So let's calculate the velocity

v ~=~ {dx\over dt} ~=~ \lambda A e^{\lambda t}
\end{displaymath} (1.24)


a ~=~ {dv\over dt} ~=~ \lambda^2 A e^{\lambda t}
\end{displaymath} (1.25)


a ~=~ \lambda^2 x
\end{displaymath} (1.26)

Comparing this with eqn. 1.1, we have that
\lambda^2 ~=~ -{k\over m}
\end{displaymath} (1.27)

Notice the negative sign! This says that

\lambda ~=~ \pm \sqrt{-{k\over m}} ~=~ \pm \sqrt{-1} \sqrt{k\over m} ~=~ \pm i \sqrt{k\over m}
\end{displaymath} (1.28)

So we get that $\lambda$ is imaginary.

We therefore know that one solution to the harmonic oscillator problem is

x(t) ~=~ A e^{i\sqrt{k/m} t} ~=~ A e^{i\omega t}
\end{displaymath} (1.29)

where $A$ can be any number, and another solution is
x(t) ~=~ B e^{-i\sqrt{k/m} t} ~=~ B e^{-i\omega t}
\end{displaymath} (1.30)

where here $B$ is any number you like, and I have purposely made it different from $A$ as the two numbers don't have to be linked.

It turns out that another solution to this problem would be to sum these two solutions together. So the general solution to this problem can be written as the sum of the two above solutions. If you don't want to keep writing all these pesky constants, you could say that the solution to this equation was

x(t) ~=~ \{ e^{i\omega t},e^{-i\omega t} \}, ~~~ \omega ~=~ \sqrt{k\over m}
\end{displaymath} (1.31)

which means you take any linear combination of these two functions and it'll be a solution to the harmonic oscillator equation, eqn. 1.1.

Similarly, we learned previously that the general solution can also be written as

x(t) ~=~ \{ \cos{\omega t}, \sin{\omega t}\}
\end{displaymath} (1.32)

They both solve the same equation 1.1 which can be rewritten

{d^2 x\over dt^2} ~=~ -{k\over m} x
\end{displaymath} (1.33)

So there must be some relation between them! You don't need to know what it is to follow what comes later, but you might be curious. It's
e^{ix} ~=~ \cos x+i\sin x
\end{displaymath} (1.34)

That's called Euler's equation. Let's take $x ~=~ \pi$. This says the
e^{\i\pi} ~=~ -1
\end{displaymath} (1.35)

Pretty amazing!

josh 2010-01-05