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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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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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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