Tuesday, 14 November 2017

homework and exercises - Derivation of the speed of light using the integral forms of Maxwell's Equations


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.



Answer



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: $$\varepsilon_0 \mu_0 \frac{\partial^2 E_x}{\partial t^2} = \frac{\partial^2 E_x}{\partial x^2}$$


No comments:

Post a Comment

Understanding Stagnation point in pitot fluid

What is stagnation point in fluid mechanics. At the open end of the pitot tube the velocity of the fluid becomes zero.But that should result...