When talking about a plane wave of the form $$\vec E=\vec E_0\cos(\vec k \cdot \vec r-\omega t)$$ We can replace it by $$\vec E=\vec E_0\exp[i(\vec k \cdot \vec r-\omega t)]$$ so that it is easier for calculation and then only taking the real part as the physical quantity.

How can we be sure that the complex part of the wave equation will never become real and contribute to our calculations and let us arrive at an incorrect answer?


marked as duplicate by Emilio Pisanty, AccidentalFourierTransform, Qmechanic Apr 2 '18 at 17:18

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  • $\begingroup$ Related: physics.stackexchange.com/q/77156/2451 and links therein. $\endgroup$ – Qmechanic Apr 2 '18 at 12:25
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    $\begingroup$ The answer might be in the link above. In short, we know we must take only the real part, so we must be careful about not to mix them. Sums, derivatives and integrals are linear operations, so they don't mix real and imaginary part. However, a cross product, for example, will do, so you cannot do a product, unless you "correct it" either taking real parts only or conjugating the first one. $\endgroup$ – FGSUZ Apr 2 '18 at 13:08
  • $\begingroup$ Please clarify you question. What do you mean by "complex part" of the wave equation? The solution of the wave equation can be a complex function which has a real part and an imaginary part. $\endgroup$ – freecharly Apr 2 '18 at 14:52
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    $\begingroup$ Possible duplicate of What is the physical significance of the imaginary part when plane waves are represented as $e^{i(kx-\omega t)}$? as well as links therein. $\endgroup$ – Emilio Pisanty Apr 2 '18 at 15:04

We have

\begin{align} E(r,t) &= E_0 \cos(kr - \omega t - \phi)\\ &=E_0\left(e^{i(kr-\omega t -\phi)} + e^{-i(kr - \omega t + \phi)} \right)\\ &= \tilde{E}_0 e^{i(kr-\omega t)} + \tilde{E}^* e^{-i(kr-\omega t)}\\ &= E^{(+)} + E^{(-)} \end{align}

Note I've set $\tilde{E}_0 = E_0 e^{-i\phi}$.

By (a particular choice of) convention, the first term is called the positive frequency term and the second term is called the negative frequency term.

As has been mentioned, if you are adding waves or performing other linear manipulations you can drop the negative frequency term and just work with the positive one, adding in the corresponding negative frequency term at the end of the calculation to recover a real final answer.

If you are confused, I recommend performing the manipulations I have shown above so that you have an expression in terms of complex exponentials rather than sines and cosines. However, instead of dropping the negative rotating part as is often (somewhat mysteriously) recommended in my courses and textbooks, just go ahead and keep it. You now have two terms to drag around instead of one but you will see that it is easier to perform manipulations on the complex expressions instead of sinusoidal expressions. You will also see that whenever you do something to the positive frequency term you basically do the complex conjugate thing to the negative frequency term. Then, if you insist, you can take the intuition you have built to understand that in some circumstances you can drop the negative frequency term so that you have less things to write down.

As a practice example try to answer the following question:

What is the amplitude and phase of the wave which is the sum of the two following waves:

\begin{align} E_1(r,t) &= E_1 \cos(kr-\omega t -\phi_1)\\ E_2(r,t) &= E_2 \sin(kr - \omega t + \phi_2) \end{align}

This can be solved in sinusoidal form using trig identities for sums and differences inside sines and cosines. It can also be solved using complex exponentials as described above. I recommend doing it both ways to see the differences.

Finally, to really put the nail in the coffin of this sinusoidal vs. exponential representation I recommend using the exponential formulas to derive the trigonometric identities needed to solve the above exercise. These trig identities can alternatively be derived from geometric considerations and drawing funny triangles and labeling the sides but I have a very hard time doing it that way. Once you get used to it, it is quite simple to prove them using the exponential representation as I hope you'll discover.

edit: Let me add a bit more to directly address your question: You ask how we know the complex part won't become real and influence the answer. What you should realize is that when you are working in the complex representation (after having ignored the negative frequency part) the complexity of the expression is actually critical to capture the phase of the wave. What you should in fact be concerned about is "how do we know that the negative frequency part will not develop a positive frequency part and affect the answer?" The answer is that if all of the operations are linear then it is only the coefficients of the exponentials which are affected, never the arguments of the exponentials which contain the phase and frequency terms. However, if you begin multiply waves (say you're mixing or homo/heterodyning signals or looking at other non-linear processes) you will see that terms pop up with different factors up in the exponential than you had originally. In this case I would not recommend dropping negative frequency terms as you could easily miss something.

To summarize: In sinusoidal representation the relevant information is contained in the amplitudes and phases of the sines and cosines and in the complex representation the relevant information is contained in the (complex) amplitude of the coefficients of the positive and negative frequency terms, bearing in mind that the information in the coefficient of the positive frequency term is redundant with the information in the negative frequency coefficient as those two terms are complex conjugates of each other.


Actually the imaginary part of $e^{i\omega t}$, i.e. the $\sin(\omega t)$ part, may eventually produce a real component since there is no guarantee that $\vec E_0$ is real. This happens for instance in E&M waves where, given an electric field in the form $$ \vec E_c=\hat x E_0 e^{-\alpha z} e^{-i \beta z} e^{i\omega t} $$ with the physical electric field $$ \vec E=\hat x E_0 e^{-\alpha z} \cos(\omega t-\beta z)\, , $$ (I'm using the subscript $c$ to indicate the complex nature of the quantity) the magnetic field is of the form $$ \vec H_c=\frac{\hat y}{\eta}e^{-\alpha z} e^{-i \beta z}e^{i\omega t} =\frac{\hat y}{\vert\eta\vert }e^{-\alpha z} e^{-i \beta z}e^{i\omega t}e^{-i\theta_\eta} $$ where $\eta=\vert \eta\vert e^{i\theta_\eta} $ is the effective complex impedence of the medium.

You can write then the real part as \begin{align} \vec H&=\frac{\hat y}{\vert \eta\vert}e^{-\alpha z} \cos(\omega t-\beta z-\theta_\eta)\, ,\\&= \frac{\hat y}{\vert \eta\vert}e^{-\alpha z} \left(\cos(\omega t-\beta z)\cos(\theta_\eta)+\sin(\omega t-\beta z)\sin(\theta_\eta)\right)\, . \end{align} The key point, however, is that the appearance of a sine and cosine simply indicates there is a phase shift, given by $\theta_\eta$ between the electric and magnetic fields.

The same general type of situation occurs in LRC circuits, as there can be a phase shift between the voltage and the current.

  • $\begingroup$ What do you mean by "complex part"? A complex function can have a real part and an imaginary part but no complex part. $\endgroup$ – freecharly Apr 2 '18 at 14:54
  • $\begingroup$ @freecharly Good observation. changed "complex part" to "imaginary part". Thanks. $\endgroup$ – ZeroTheHero Apr 2 '18 at 14:59

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