A question about electromagnetic wave function

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The discussion centers on the use of complex exponentials, specifically e^(iωt), in wave functions within physics. The primary reasons for this choice include the periodic nature of the imaginary exponential function and its mathematical convenience over trigonometric functions. The participants highlight that complex exponentials are solutions to governing partial differential equations, such as the 1D wave equation, which admits both periodic and non-periodic solutions. The method of variable separation is noted for its role in selecting physically acceptable solutions, often leading to the use of sine and cosine functions. The conversation also touches on Fourier analysis as a critical mathematical tool that connects these concepts, although some participants express a desire for deeper mathematical insights into this relationship. Overall, the use of e^(iωt) is justified by its simplicity, periodicity, and compatibility with the underlying physics of wave phenomena.
somy
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Hi there.
We always put the time dependent part of the wave functions as e^(iwt).
Of course there is a reason! but I don't know it.
Can you help me?
Thanks in advance.
Somy :smile:
 
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This is due two reasons:

1) The imaginary exponential is a periodical function

e^{iz} = \cos(z) + i \sin(z)

and

2) It is a very good function to work with it. So physicists, who are very intelligent people, try to use easy functions, and it is more easy to work with and exponential rather than trigonometric ones.

Maths are the easy part of physics, that's why we try to use the more simplest things to work with.

Happy new year !
 
Last edited:
somy said:
Hi there.
We always put the time dependent part of the wave functions as e^(iwt).
Of course there is a reason! but I don't know it.
Can you help me?
Thanks in advance.
Somy :smile:

To add to what Migui has said,i'm sure you know that the coordinate part of the wavefunctions is put under the form of complex exponentials as well:
\exp(i\vec{k}\cdot\vec{r})

Daniel.

PS.Migui has given u a simple nontechnical explanation.There's much more behind an explanantion for this fact.To give you a hint:distributions and Fourier transformations.
 
Of course the main reason that is used, is because it is the solution to the governing partial differential equation.
 
Yap,Integral,but think about the 1D wave equation:

\frac{\partial^{2}u(x,t)}{\partial x^{2}}=\frac{1}{v^{2}}\frac{\partial^{2}u(x,t)}{\partial t^{2}}

Solve it and tell me whether the result contains complex exponentials.

Daniel.
 
Looks to me like u(t) = e^{ivt} is a solution to the time equation. What am I missing?
 
The point?

First of all,your solution is wrong wrt dimensions.Exponential of a meter is kinda hard to swallow.Even complex.

That equation (1D wave eq.posted above) is a typical example of the fact that the method of variable separation reduces the number of solutions,and even affects the physically acceptable solutions.In this specific example,it selects the periodic (sine/cosine) waves propagating along the "x" axis.Both regressive and progressive.
I'm sure you still remember (and u still have time to show it in 2004 :-p ) 2 facts:
1)The 1D wave equation admits nonperiodical solutions.
2)The method of variable separation works best in the case that the solution of the equation is unique,working hand in hand with initial conditions,boundary conditions and Fourier representations.Here,uniqueness suffers from inexistance.

Daniel.

PS.Do you remember the solution and how to get it?
 
In an unbounded domain, the general solution is (let's stick to 1D and use Kurt's notation)
u(x,t)=f_{\rm f}(x-vt)+f_{\rm b}(x+vt)
where the f's are arbitrary functions. A fun demonstration is to hit a clothesline sharply with your fist and watch the wave impulse (the f) race away without changing shape, rebound off the end and come back. So why do we always end up using e^{\pm i\omega t}? Or to put it terms of physical objects: why are these arbitrary functions appropriate for the clothesline but not for say a guitar string? This is indeed a very good question; one that bothered me a lot when I learned it. As Kurt said, it is partly to do with the great mathematical tool called Fourier analysis. But the convenience or not of using harmonic functions also has to do with whether the length of the wave shape f is short or long with respect to the length of the 1D domain. A third reason is that if you depart slightly from the wave equation (say v is a function of the wavelength), then the general solution given above is not very helpful.
 
So are you trying to say that if it isn't physical it isn't a solution?

Humm...
 
  • #10
Thanks all,
But I still didn't get the answer!
Well, I know that they tell us we use this function because of simplicity, but I want to know some deeper mathematical points. dextercioby told me about the Fourier transfer, but I really coudn't find the relation between these.
Can you give me something more?
Thanks in advance.
Somy
 
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