4
$\begingroup$

If an electric field $E$ oscillates as $E_0\sin(ωt)$ then the average value of $E^2$ over one period of oscillation will be

$$E_0^2\left< \sin^2(ωt)\right>=E_0^2/2$$ since the average value of $\sin^2(ωt)$ is well known to be $1/2$.

However if we write $E$ using complex numbers as $E_0e^{iωt}$ and then take real parts, as is often the case, then we have

$$\left< E^2 \right> = E_0^2\left< (e^{iωt})^2 \right> = E_0^2\left< e^{2iωt} \right> $$

$$=E_0^2 \left< \cos(2ωt) + i\sin(2ωt)\right>=0$$ as the average value of an unsquared $\sin$ or $\cos$ will be equal to $0$ over a single period of oscillation, and the real and imaginary parts (should be) independent of one another.

What's gone wrong here? Is there a fault in my above assumption?

$\endgroup$
1
  • $\begingroup$ Let me know if this belongs better on maths SE, but I figured taking time averages of quantities lies in physics often enough that it should be ok here. $\endgroup$
    – Poo2uhaha
    Nov 12, 2021 at 19:01

3 Answers 3

6
$\begingroup$

When you use the complex representation of the electric field, you have to modify the way you calculate the intensity. Instead of calculating $\langle E^2 \rangle$ you need to replace it with $1/2 \langle E E^* \rangle$, where $E^*$ is the complex conjugate.

$\endgroup$
5
$\begingroup$

So the complex numbers that we use in Classical Electromagnetism are more of a mathematical trick. The electromagnetic waves are in fact "real" quantities, i.e. they are represented by sines and cosines. However, it is often easier to work with complex exponentials, and so we cheat a little: since any linear combination of the sines and cosines is mathematically allowed, we choose the combination with a complex number $$\cos(\omega t) + i \sin(\omega t) = e^{i\omega t}.$$

The understanding is then that we will work with such "complex" waves, but at the end we will always return the "real" part. The imaginary part is just along for the ride, to make calculations simpler. (The definition of what is the "real" part is also not set in stone, you could work with either $\text{Re}(e^{i\omega t})$ or $\text{Im}(e^{i\omega t})$, and as long as you're consistent, that's perfectly acceptable.)

However, this argument only works so long as the operations you perform are linear operations. Happily for us, Classical EM is linear, since Maxwell's Equations are linear. Thus, this "complex" wave satisfies the same equations that a "real" wave would.

Thus, you cannot use the complex representation of the wave when you are working with physical quantities that are non-linear. In particular, when calculating physical quantities that depend non-linearly on the wave, like the power, intensity, or even the Poynting vector, one should always take the real part. Thus, your first approach (using $\sin^2(\omega t)$) is the right one.

It turns out that you can use the complex representation if you redefine what you mean by the average, using a complex conjugate representation. (See these notes for a review of phasor notation.) Thus, the average of the product of two waves $\mathbf{A}$ and $\mathbf{B}$ is given by: $$\text{average of $\mathbf{A}\mathbf{B}$} = \frac{1}{2}\text{Re}(\mathbf{A}\mathbf{B}^*)$$ So, for example, you can see why the average of $\mathbf{E}^2$ is: $$\text{average of $\mathbf{E}^2$} = \frac{1}{2}\langle \mathbf{E} \mathbf{E}^* \rangle = \frac{1}{2}.$$ Similarly, it is possible to define the "Poynting vector" for a complex wave as (ignoring some constants): $$\mathbf{S} = \mathbf{E} \times \mathbf{H}^*,$$ and so on. But when in doubt, always go back to the real notation when performing non-linear operations on a complex wave.

$\endgroup$
2
  • 1
    $\begingroup$ Great answer! Though, the first equation (Euler's formula) is incorrect. Instead it should be $e^{i\omega t} = \cos(\omega t) + i \sin(\omega t)$, and with the convention of the question one takes the imaginary part to be the actual wave. You could also mention that some author's use cosines to represent waves, and hence take the real part of the complex expression to be the actual wave. $\endgroup$
    – ummg
    Nov 12, 2021 at 20:02
  • 1
    $\begingroup$ Good gods, did I actually make a mistake with Euler's formula!? :( Thanks for the suggestion will make some changes :) $\endgroup$
    – Philip
    Nov 12, 2021 at 20:04
3
$\begingroup$

The real part of the product of two complexes is not the product of the real parts. So it is mandatory to go back to real notation before calculating the average.

Another solution is to use the trick : $$\langle x(t)y(t)\rangle=\frac{1}{2}\Re\{\underline{x}{\underline{y}}^*\},$$ with the exponent * which represents the conjugate complex.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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