How can $e^{i\theta}$ ever be a real quantity?

Question

From Euler's formula, $e^{i\theta} = \cos(\theta) + i\sin(\theta)$, which seems to be a complex quantity involving real and imaginary parts. Yet, D.J. Griffiths in his book on electrodynamics mentions this:

in the formula, we are dealing with \begin{align*} A_R &= \text{Amplitude of reflected wave}\\ A_I &= \text{Amplitude of incident wave}\\ A_T &= \text{Amplitude of transmitted wave}\\ \delta &= \text{Phase angle}\\ t &= \text{time} \end{align*} and the $\tilde{A}$ variants related to their uninflected roots by $\tilde{A} = A e^{i \theta}$.

So, why does he say that the "real" amplitudes and phases have that form when the left hand side of the equation is complex?

You have complex quantities of the form $e^{i\delta}$ multiplied by a real quantity $A$. When the phases are said to be equal then the complex quantities $e^{i\delta}$ cancel out leaving the real quantities "A". — Farcher, May 01 '17 at 19:30
I've typeset a bunch of your math for you using the MathJax engine that is active on the site. You should check that I have transcribed it correctly. You should also do the text from the book, and the image is not serachable. — dmckee --- ex-moderator kitten, May 01 '17 at 19:31
Possible duplicate of What is the physical significance of the imaginary part when plane waves are represented as $e^{i(kx-\omega t)}$? — Emilio Pisanty, May 01 '17 at 20:04

score 3 · Answer 1 · answered May 01 '17 at 19:37

The phrase is "amplitudes and phases". The equation he quotes has the real amplitude, $A_T$ etc, and the phase, which is where the complex part of the expression arises.

The point is that we define the amplitude of a wave as the biggest deflection it reaches, while the phase tells us "are we there yet". By including the phase we can see that at the same moment in time, the reflected wave can have a phase inversion compared to the incident wave.

score 3 · Answer 2 · answered May 01 '17 at 20:01

So, why does he say that the "real" amplitudes and phases have that form when the left hand side of the equation is complex?

Consider the complex number

$$z = \sigma + i \omega$$

Both $\sigma$ and $\omega$ are real numbers; $\sigma$ is the real part of $z$ while $\omega$ is the imaginary part of $z$.

But we can also write

$$z = |z|e^{i\phi},\qquad |z| = \sqrt{\sigma^2 + \omega^2},\qquad \tan\phi = \frac{\omega}{\sigma}$$

Both $|z|$ and $\phi$ are real numbers; $|z|$ is the modulus of $z$ while $\phi$ is the argument of $z$.

So, even though $A_Re^{i\delta_R}$ is a complex number (a phasor in fact), it is also true that $A_R$ and $\delta_R$, the amplitude and phase of the phasor, are real and Griffiths seems to be emphasizing this.

score 2 · Answer 3 · edited May 01 '17 at 22:37

"So, why does he say that the "real" amplitudes and phases have that form when the left hand side of the equation is complex?"

Both sides of the equation are complex. $ A_R $ and $and A_I$ are real but they are both multiplied by the complex exponential factor. The exponential factor is complex but the angles in exponent ($ \delta_R, $ etc ) are real. So there is no problem with the equation, neither physical nor mathematical. The fact that $ A_R $ and $ A_I$ are real is not in contradiction with the equation.

score 1 · Answer 4 · answered May 01 '17 at 20:16

In that passage, Griffiths is not claiming that any quantity of the form $e^{i\theta}$ is real. On the contrary, Griffiths derives some expression for the complex-valued phasor amplitude $\tilde A$, and then goes on to say

the real amplitude $A$ and phase $\delta$ are then related to the phasor amplitude as $$Ae^{i\delta} = \tilde A.$$

Here $A$ and $\delta$ are both real numbers and they are the amplitude and the phase of the wave in the sense that you have waveforms that look like $$f(x,t) = A\cos(kx -\omega t+\delta),$$ i.e. they are real-valued and measurable.

@Sakazuki What about it? It's correct, if that's what you were wondering. — Emilio Pisanty, May 02 '17 at 16:04

score 0 · Answer 5 · answered May 01 '17 at 19:58

Imagine you had a complex amplitude

$$ (a+\hat{i} b) {\rm e}^{\hat{i} \theta} = a {\rm e}^{\hat{i} \theta} + b {\rm e}^{\hat{i} \left( \theta + \frac{\pi}{2} \right)} $$

So a complex amplitude results in an additional real amplitude with a $\frac{\pi}{2}$ phase angle.

How can $e^{i\theta}$ ever be a real quantity?

5 Answers5