I am taking Analytical Mechanics, and in Goldstein's book, chapter 6 (page 241) about linear oscillations, he says the following:

"... $\eta_i=Ca_ie^{-i\omega t}$ (6.11) ... It is understood of course that it is the real part of (6.11) that is to correspond to the actual motion."

Equation 6.11 is the solution for the equation $T_i\ddot{\eta_i}+V_i\eta_i=0$ (1).

I understand why we chose the solution to be of the form 6.11.

It is also clear to me that if 6.11 is a solution to (1), then its real and imaginary parts will also be solutions, and also that the solution of (1) must be real, as it is a real homogenous linear differential equation.

What I don't understand is why we say that (only) the real part corresponds to the actual motion of the system.

In addition, wouldn't it be better if we denoted the solution to the equation (1) as $\eta_{real}$,and the solution in eq. 6.11 as $\eta_{complex}$?

(I read this and this before posting, but I don't get it yet...)

  • $\begingroup$ Take a look at this series of videos by 3blue1brown - Differential Equations. Especially this one - But what is a Fourier series? From heat flow to circle drawings $\endgroup$
    – mmesser314
    Commented May 5 at 1:10
  • 1
    $\begingroup$ The best explanation of this notation on this stack is probably Emilio Pisanty's answer about complex wave notation. $\endgroup$ Commented May 5 at 16:07
  • $\begingroup$ Try a simple example, eg where does the circle $x^2+(y-2)^2=1$ cross the X axis? $\endgroup$
    – PM 2Ring
    Commented May 5 at 17:54
  • $\begingroup$ @PM2Ring That example is not so simple. Maybe that was a little joke with us. The answer is that that equation never intersects the x-axis. It does however intersect the y=0 Argand plane defined by the x and i(x) axes, at i(x) = z = $\pm i \sqrt{3}$ but all the complex solutions to that equation bypass the real $-\infty <x < \infty$, y=0, z=0 axis. Have a look at this plot of all the complex solutions on a 3D plot where the z axis represents the imaginary axis. geogebra.org/classic/xjpr6xxk . I think you will find it interesting. $\endgroup$
    – KDP
    Commented May 6 at 19:42
  • $\begingroup$ Actually, I have generalised the solution a bit further and plotted the solution as a curved surface in 3D space. Only the black circle is a purely real. The blue curved surfaces are the complex solutions, with the z-component representing the imaginary part. geogebra.org/classic/tcpmd8s9 $\endgroup$
    – KDP
    Commented May 7 at 4:45

3 Answers 3


I think the reason for the sentence

"It is understood of course that it is the real part of (6.11) that is to correspond to the actual motion"

Is that $\eta_i$ represents (likely) the displacement of a particle from a certain position. Something for example measured in meters and therefore, obviously, a real quantity. This is the reason for the "of course" in the sentence as a complex value wouldn't make sense.

This however is a bit cheating. We have a mathematical description of reality. Said description allowed to be augmented to complex values. But if the description is correct it must give back real values for proper (real) data.

And this is indeed what happens. First you should notice that there are actually two solutions

$$ \eta_i = C_\pm e^{\pm i \omega t} $$

Any linear combination of the above is still a solution. The point is that, if we did things correctly, for proper initial conditions (i.e. initial conditions that are compatible with meaningful data) the solution should automatically turns out real.

To summarize, the solution of the equation, being a second order ODE, depends on two initial conditions (for example initial position and velocity). If these values correspond to a meaningful motion, the end solution turns out automatically real.


You are correct, the imaginary part is also a solution. And you could even go further: any combination of the real part and the imaginary part of a complex solution can be used. If the complex solution is some function $f(t)$, we could therefore write any allowed real solution $g(t)$ as: $$ g(t) = a \text{Re} f(t) + b \text{Im} f(t) $$ with real coefficients $a$ and $b$. But it is often easier to use a complex number $\alpha$: $$ g(t) = \text{Re}\ ( \alpha f(t)) = \text{Re}\, \alpha\ \text{Re} f(t) - \text{Im}\,\alpha\ \text{Im} f(t) $$ which does of course just use the same two degrees of freedom, just combining $\text{Re} f(t)$ and $\text{Im} f(t)$. But it leads to the somewhat strange situation that to describe a real quantity we use a complex amplitude! (This is done all the time in electrical signal analysis, for instance.)


to solve this ode

$$\ddot\eta+\omega^2\eta=0\quad , \omega^2=\frac VT$$

you make this ansatz

$$\eta(t)=(a+b\,i)\,e^{i\,\omega\,t}+(a-b\,i)\,e^{-i\,\omega\,t}\quad,a,b\in Re\tag 1$$

from the Initial conditions $~\eta(0)=\eta_0~,\dot\eta(0)=\dot\eta_0~$

you obtain

$$a=\frac 12 \eta(0)~,b=-\frac 12 \frac{\dot\eta(0)}{\omega}\tag 2$$

thus the solution is

$$\eta(t)=\eta(0)\,\cos(\omega\,t)+\frac{\dot\eta(0)}{\omega}\,\sin(\omega\,t)\tag 3$$

the solution is real not complex !

take the real part of equation (1) you obtain


this is also solution of the ode , again the initial conditions $~a~,b~$ are equal to equation (2) thus you obtain the solution of equation (3), this is way you can take the real part of the ansatz , equation (1)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.