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I have this equation

$$\dfrac{\partial E_z}{\partial y^2} - \mu_0 \varepsilon_0 \dfrac{\partial E_z}{\partial t^2} = 0$$

Why is $v^2 = 1 / (\mu_0 \varepsilon_0)$ ?

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Why do I know that this is the speed? –  kame Sep 12 '11 at 13:13
    
Because it has the velocity squared units. –  Vladimir Kalitvianski Sep 12 '11 at 14:19

2 Answers 2

Take the function $E_z(x,t) = f([x-vt])$, where $f$ is an arbitrary function of the single variable $x-vt$, such as $e^{-(x-vt)^2}$. You can verify, by taking partial derivatives and using the chain rule, that any such $f$ solves the PDE if and only if $v^2 = \frac{1}{\mu_0\epsilon_0}$. It also turns out that all solutions can be written as linear combinations of two such solutions, one $v$ positive and one with $v$ negative.

If you set $u = x-vt$, so that $E_z$ is constant when viewed from the point of view of $u$, we see that that $u$ frame is moving relative to the $x$ frame as speed $v$. Thus, all solutions to the PDE are waves traveling right at $v$, left at $v$, or a combination of those two.

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If we consider any wave traveling in a single spatial dimension, then the function, $f(y,t)$, that describes that wave satisfies the PDE $$ \frac{\partial^2 f}{\partial y^2} - \frac{1}{v^2}\frac{\partial^2 f}{\partial t^2} = 0. $$

Empirically we identify the constant, $v$, with the velocity of that wave. It is fairly easy to see from D'Alembert, why, from a theoretical point of view, this is the case.

Now, comparing the above equation to $$ \frac{\partial^2 E_z}{\partial y^2} - \mu_0\epsilon_0\frac{\partial^2 E_z}{\partial t^2} = 0, $$ we would identify $$ v^2 = \frac{1}{\mu_0\epsilon_0}. $$

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