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I know that a sinusoidal plane wave can be represented by the wave equation $$ \psi (x,t)=A\, \cos(kx-\omega t) $$ I have also seen that a plane wave can be represented in complex exponential form as $$ \psi (x,t)=A\, e^{i(kx-\omega t)} $$ I know that the Euler Formula is $$ e^{i \theta}=\cos(\theta)+i\sin(\theta)$$ My question is this: is the reason $e^{i (kx-\omega t)}$ can replace $\cos(kx-\omega t)$ without the $i\sin(\theta)$ term from Euler Formula being present because only the real part of the Euler Formula is what we are interested in?

Note: I do not know the derivation of the complex form from the sinusoidal form. If the real reason has to do with the derivation, could someone please explain the derivation?

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  • $\begingroup$ Basically, you are right. We just ignore the imaginary part of the solution, and we prefer this form because using $\sin$ and $\cos$ requires us to remember trig identities; using $e^{i}$ instead just requires basic exponential manipulations, which are easier. However, it is also true that the complex form comes in useful in many other cases where you care about imaginary parts. $\endgroup$ – Mike Feb 24 '17 at 17:21
  • $\begingroup$ @Mike i was tempted to answer this but it can be confusing since if we are dealing with non linear equations the replacement although introduces huge simplification isn't correct. $\endgroup$ – lakehal Feb 24 '17 at 17:28
  • $\begingroup$ @lakehal But the wave equation is linear, by construction. If you're talking about different equations, the solutions will be different, and there's generally no reason to expect this replacement to give the solution. I don't understand what would be confusing here. $\endgroup$ – Mike Feb 24 '17 at 19:25
  • $\begingroup$ @Mike I agree with your comments. Why don't you make it an answer? $\endgroup$ – LedHead Feb 24 '17 at 20:08
  • $\begingroup$ @led23head Oh, fine... $\endgroup$ – Mike Feb 25 '17 at 0:12
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I will explain to you the derivation of Euler's formula simply.

define

$$ f(x)=\cos(x)+i\sin(x)\\ \partial_xf(x)=-\sin(x)+i\cos(x)=i(\cos(x)+i\sin(x))=if(x) $$

from this you see that : $f(x)=e^{ix}$.

The reason we keep only the cosine term has nothing to do with the derivation.

We are interested in $\psi(x,t)$. With $\psi(x,t)$ some physical real observable, the idea is then to solve the equation for a complex $\psi$ which is easier and at the end of the calculations impose on your function to be real.

As an exemple consider the harmonic oscillator $\partial^2_x \psi +w^2\psi=0$ , the solution for a complex $\psi$ is $\psi=Ae^{iwx}+Be^{-iwx}$. Asking for a real $\psi$ gives $\psi = A \cos(wx+\phi)$ or equivalently $\psi = A \sin(wx+\phi)$.

So the answer is that, you need to solve your equation for a complex function ( which is simpler) and at the end of your calculations remember that your function mus be real.Which in many cases means taking the real part but not always.

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Basically, you are right: the reason we do that is that we only care about the real part of the solution (for basic physics problems), and that gives it to us.

Now, you might ask why you can do this. One really important fact about the wave equation is that it is "linear". This means that you can add two solutions to each other and get another solution, and you can multiply a solution by a constant and get another solution. In particular, if \begin{equation} \psi(x, t) = A\, e^{i(k x - \omega t)} \tag{1} \end{equation} is a solution and its complex conjugate $\bar{\psi}$ is too, then you also know that $(\psi + \bar{\psi})/2$ is a solution. That is, the real part alone is a solution.

So that's why you can do it, but there are also pedagogical reasons you should do it. By introducing these exponential solutions now in a familiar setting, you can start to get good at the mathematics, and that'll get even handier later. Here are just a few reasons off the top of my head:

1) It's easier to manipulate complex exponentials than to manipulate trig functions. High school hotshots may care about trig identities, but nobody else does. They're annoying and dumb, and you should forget all about them. What you should get good at is exponentials. Just a few simple rules, and you can rederive any old trig identity you want, and then some.

2) It's a good introduction to the ideas of Fourier methods. It's true that you can use Fourier methods with sines and cosines, but they're just much prettier with complex exponentials. And they're extremely powerful. For example, you can show that (subject to some basic assumptions), any solution to the wave equation can be expressed as a combination of solutions like equation (1), with various values for $A$ and $\omega$ (hence also $k$). More generally, you'll frequently see Fourier methods as nice ways of understanding other differential equations.

3) There are other physical systems where you actually do care about both the real and imaginary parts of the solution. For example, when you talk about gravitational waves, you find that they have two components. If you add the first component to $i$ times the second component, they satisfy the wave equation, and the solution is proportional to $e^{i(kx-\omega t)}$ — not just its real part. And of course, you'll also find many examples in quantum physics. Quantum fields are inherently complex, so again you can find solutions like $e^{i(kx-\omega t)}$. If you want to get really fancy, geometric algebra is full of examples (or here for a free copy) of systems where the solution looks like this, except that $i$ is replaced by geometric objects that have the familiar algebraic property that they square to $-1$.

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Thanks to Mike for his comment, which I include in case it gets deleted.

There are several things wrong with this answer. First, the OP was evidently asking about real-valued fields, and K was correct in thinking that we just ignore the imaginary part, so starting off your answer with "No" is just wrong. Now, it is true that another solution in the complex domain does not ignore the imaginary part, but that's a different problem. Also, what you give is not the general solution to the 1-D wave equation; it is the general form of a particular eigenfunction of that equation. The general solution is a series of such solutions with different values of ω

No, we can use either the cos or the sin version, but in many cases in quantum mechanics, there are two solutions, and using the Euler notation condenses them into one compact expression.

by

From Wikipedia Wave Solution

$${\displaystyle f(x)=Ae^{\pm ikx}}$$

$ f(x ) = A e^{\pm ikx} $, with wave number $ {\displaystyle k=\omega /c} $ .

The total wave function for this eigenmode is then the linear combination

$${\displaystyle u_{\omega }(x,t)=e^{-i\omega t}\left(Ae^{-ikx}+Be^{ikx}\right)=Ae^{-i(kx+\omega t)}+Be^{i(kx-\omega t)},} $$

The plus and minus signs indicate the direction of travel of the wave.

It's the General Solution to a 1 D differential equation

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  • $\begingroup$ There are several things wrong with this answer. First, the OP was evidently asking about real-valued fields, and K was correct in thinking that we just ignore the imaginary part, so starting off your answer with "No" is just wrong. Now, it is true that another solution in the complex domain does not ignore the imaginary part, but that's a different problem. Also, what you give is not the general solution to the 1-D wave equation; it is the general form of a particular eigenfunction of that equation. The general solution is a series of such solutions with different values of $\omega$. $\endgroup$ – Mike Feb 24 '17 at 17:19
  • $\begingroup$ @Mike thank you, I will put your comment in the post, as they often get deleted, and then leave the answer as a reminder to myself not to rush to an assumption. $\endgroup$ – user146020 Feb 24 '17 at 17:28

protected by Qmechanic Feb 25 '17 at 0:27

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