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Having just finished physics 2, I've been (slightly) exposed to showing that light is a wave with speed $1/\sqrt{\mu _0 \epsilon _0 }$ using the differential forms of Maxwell's equations, though this is the only derivation I've come across. Can you show the same thing using the integral forms? My first thought is that it may be more difficult since the wave equation is often given as a differential equation.

Note: I have not taken vector calculus (or even multivariable), and do not have sufficient mathematical (or even physical) background to explicitly do the derivation. I'm merely asking for a hopefully understandable solution or a source to the solution.

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The hand-wavy way to do it is to consider a wave solution like the one below, and apply Faraday's law to loop 1, and Ampere's law to loop 2:

EM Wave

If you make the loops narrow enough, i.e., their widths are $dx$, then $$\oint_1\!\vec{E}\cdot \vec{ds} = -\frac{d\Phi_B}{dt} \to \frac{\partial E_y}{\partial x} = -\frac{\partial B_z}{dt}$$ $$\oint_2\!\vec{B}\cdot \vec{ds} = \varepsilon_0\mu_0\frac{d\Phi_E}{dt} \to \frac{\partial B_z}{\partial x} = -\varepsilon_0\mu_0\frac{\partial E_x}{dt}$$ Now differentiate the first equation wrt $t$ and the second wrt $x$, and combine to obtain the wave equation: $$\varepsilon_0 \mu_0 \frac{\partial^2 E_x}{\partial t^2} = \frac{\partial^2 E_x}{\partial x^2}$$

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The differential and integral forms of Maxwell's equations are truly equivalent; they are essentially the same set of equations.

One can convert between the two using two mathematical theorems:

Divergence Theorem (Wikipeda - Divergence Theorem) Stokes' Theorem (Wikipedia - Stokes Theorem)

The divergence theorem states that the flux over a closed surface of a vector field is equal to integrating the divergence of that vector field over the volume enclosed by the surface. Stokes' theorem states that the line integral of a vector field over a closed path is equal to integrating the curl of that vector field over any surface bounded by the closed path.

Now, what you are seeking to show is a wave equation, which as you have stated is a differential form; which is as follows:

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

Now, to get there from the integral form, what you do is the following: Take the third equation, for instance, which states:

$\oint \limits_C \vec{E} \cdot \vec{dl} = -\frac{\partial}{\partial t} \iint\limits_S \vec{B}\cdot \vec{dS} $

Now Stokes Theorem states that:

$\oint \limits_C \vec{E} \cdot \vec{dl} = \iint\limits_S \vec{\nabla} \times \vec{E} \cdot \vec{dS}$

Plug it in, and you get:

$\iint\limits_S \vec{\nabla} \times \vec{E} \cdot \vec{dS} = -\frac{\partial}{\partial t} \iint\limits_S \vec{B}\cdot \vec{dS}$

Change order of time derivative and integral (yes, you can do that) and collect under same integral sign:

$\iint\limits_S (\vec{\nabla} \times \vec{E} + \frac{\partial \vec{B}}{\partial t} )\cdot \vec{dS}$

Now, since this must hold for any surface S, the integrand itself must be zero. So we get:

$\vec{\nabla} \times \vec{E} + \frac{\partial \vec{B}}{\partial t} = 0$

Now, we are on our way to getting the wave equation, but what we have done as the first step is simply converting the integral form of the equation to its differential form! To continue, you would need to do the same to the first two equations, then apply $(\vec{\nabla}\times)$ operator to the equation we just obtained above... Use a vector identity, and get there...

You may say, this is not really using the integral forms to come up with the wave equation! Well, it is starting from the integral forms of the equations, but what choice do we have if what we are trying to reach a second-order differential equation anyways??

(The above procedure will give you the wave equation for $\vec{E}$. To get the one for $\vec{B}$, you will need to start with equation four...)

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  • $\begingroup$ This is simply starting with the integral form and then moving to the differential form again. $\endgroup$ – Rob Jeffries Dec 17 '14 at 7:44
  • $\begingroup$ I am fully aware of that, and stated it in the answer. The problem with the integral forms is that, the equations are scalars. (A number on the left, a number on the right.) They are very useful when you are trying to calculate specific answers (hence they are taught first). However, the differential forms are vector field equations themselves, much useful for formal derivations. Any attempt to reach the wave equation will have to convert to the differential form at some point... $\endgroup$ – safkan Dec 17 '14 at 11:41
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I think what I do is similar to what lionelbrits has posted. I calculate the voltage around a square loop anenna two ways: first, as the rate of change of magnetic flux from the peak of the magnetic wave passing through; and second, as the maximum difference of the voltages along the two vertical legs as calculated by integrating the e-vector.

If you assume that the E-vector and the H-vector are sinusoidal, and that the ratio between them is 377 ohms, then the only way those two antenna calculations give the same answer is if the speed of light is 3 x 10^8.

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