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Why do you get electric field of a light wave in following form: $E(x,t)=A cos(kx-\omega t- \theta)$?( look at: https://public.me.com/ricktrebino -> OpticsI-02-Waves-Fields.ppt, p. 18)

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Far from the source, the propagating wave has a spherical form and propagates along the radius. So its local amplitude is a harmonic function of time. As to the coordinate dependence, it is something like $E=\frac{E_0}{R}\cdot cos(\vec{k}\vec{R}-\omega t - \theta)$. If R changes within $\delta R \approx 1/k<<R$, the R-dependence of the amplitude is weak (nearly constant), so locally you have a plane wave in direction of $\vec{R}$. –  Vladimir Kalitvianski May 2 '12 at 16:41

2 Answers 2

If the light wave has frequency $\nu$, wavelength $\lambda$, then it becomes:

$$E=E_0\sin(\frac{2\pi}{\lambda}x-2\pi\nu t+\phi)$$

$\phi$ is arbitrary.

I'm not sure of this, but you can only calculate $E_0$ if you have the amplitude of magnetic field and the energy density of the wave($U$):

$$U=\frac12\epsilon_0E_0^2+\frac1{2\mu_0}E_0^2$$

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shouldn't $nu$ be $T$ (period) ? –  Nic May 2 '12 at 17:17
    
@Nic yep, thanks :) –  Manishearth May 2 '12 at 17:19

As it happens I've just been reading "17 Equations That Changed the World" by Ian Stewart and he gives the derivation. I strongly recommend the book, but if you just want the derivation you can find it on Wikipedia. Since we're not supposed to just give links I'll copy the stuff from Wikipedia here:

Start with Maxwell's equations:

$$ \nabla \cdot \boldsymbol{E} = 0$$ $$ \nabla \times \boldsymbol{E} = -\frac{\partial \boldsymbol{B}}{\partial t}$$ $$ \nabla \cdot \boldsymbol{B} = 0$$ $$ \nabla \times \boldsymbol{B} = \mu_0\epsilon_0\frac{\partial \boldsymbol{E}}{\partial t}$$

Taking the curl of the curl equation for $\boldsymbol{E}$ gives:

$$ \nabla \times (\nabla \times \boldsymbol{E}) = -\frac{\partial}{\partial t} \nabla \times \boldsymbol{E} = -\mu_0\epsilon_0\frac{\partial^2 \boldsymbol{E}}{\partial t^2}$$

But for any vector space $\boldsymbol{V}$ there is an identity:

$$\nabla \times (\nabla \times \boldsymbol{V}) = \nabla(\nabla \cdot \boldsymbol{V}) - \nabla^2 \boldsymbol{V}$$

so $$\nabla \times (\nabla \times \boldsymbol{E}) = \nabla(\nabla \cdot \boldsymbol{E}) - \nabla^2 \boldsymbol{E} = - \nabla^2 \boldsymbol{E}$$

because $\nabla \cdot \boldsymbol{E} = 0$ so:

$$\nabla^2 \boldsymbol{E} = \mu_0\epsilon_0\frac{\partial^2 \boldsymbol{E}}{\partial t^2}$$

and this is just the wave equation:

$$ c^2\nabla^2 \boldsymbol{E} = \frac{\partial^2 \boldsymbol{E}}{\partial t^2}$$

where $\mu_0\epsilon_0$ is equal to $c^{-2}$, $c$ is the speed of light.

You wanted to know why

$$\boldsymbol{E}(x,t)=\boldsymbol{A} \cos(kx-\omega t- \theta)$$

is one possible equation for a light wave, well substitute it into the equation above and you get:

$$ -c^2 \boldsymbol{A} k^2 \cos(kx-\omega t- \theta) = -\boldsymbol{A} \omega^2 \cos(kx-\omega t- \theta)$$

and obviously this satifies the wave equation if $c = \omega/k$. This is one solution of the wave equation, but of course there are lots of others.

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Hmm, but doesn't the amplitude factor out and get cancelled here? –  Manishearth May 2 '12 at 15:46
    
@Manishearth: I spotted an error and was editing my post while you commented. Does your comment still apply? –  John Rennie May 2 '12 at 15:54
    
unfortunately, yes, since it still doesn't determine $A$. Unless I've misinterpreted the question--which I now think I have :/ . I thought the OP wanted the relation between a given light wave and the equation. +1 then :) –  Manishearth May 2 '12 at 15:58
    
$A$ is just an arbitrary constant isn't it? It's the intensity of the light wave, but the wave can be any intensity. –  John Rennie May 2 '12 at 16:25
    
The wave amplitude $A$ is determined either with the boundary conditions or with the source in the field equation. In your case (no source) it is the boundary conditions that "supply" the wave amplitude. –  Vladimir Kalitvianski May 2 '12 at 16:26

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