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Speed of light

  1. May 6, 2004 #1
    How exactly did Maxwell prove speed of light is [tex]\frac{1}{\sqrt{\mu\epsilon}}[/tex]
  2. jcsd
  3. May 7, 2004 #2
    Well, that is mathematically rather sophisticated. To prove it, the following two equations are needed (2 of the 4 famous equations that explain classical electrodynamics completely:)

    [tex]\oint\vec{B}d\vec{s}=\mu_0 I+\mu_0\varepsilon_0\frac{d}{dt}\oint E_n dA[/tex]

    [tex]\oint\vec{E}d\vec{s}=-\frac{d}{dt}\int B_n dA[/tex]

    Now, there are two possibilties to get the speed of light:

    1) The equations can be transformed into a wave equation (difficult) and you can get the speed of the wave.

    2) A special and simple case can be constructed where it is rather easy to get the speed. Though, this way isn't really sufficient.
  4. May 7, 2004 #3


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    The speed of light was already closely approximated at the time (I believe). What Maxwell really did is to suggest that light was a form of electromagnetic radiation through the close numerical agreement of c ~ 1/√με.
  5. May 7, 2004 #4
    Hehe. I like the first treatment better... Its more "convincing" as of now. :)
  6. May 11, 2004 #5
    In a vacuum, Maxwell's Equation (differential form) reduce to

    [Tex] \nabla * E = 0 [/Tex]

    \nabla * B = 0[/Tex]

    \nabla \times B = \epsilon_0 \mu_0 \frac{\delta E} {\delta t}

    \nabla \times E = -\frac{\delta B}{\delta t}

    These are obtainable from the integral forms Sitewinder mentioned, which themselves can be experimentally verified with batteries and wire loops and compasses and such.

    Take the curl of both sides of the bottom two equations. For example, the last one is

    \nabla \times \nabla \times E = -\frac{\delta \nabla \times B}{\delta t}

    Use a vector calculus identity (think of it as a special case of the chain rule, if you want):
    \nabla \times \nabla \times A = \nabla(\nabla * A) - \nabla^2 A
    (I fudged the first term; I know that it's zero in a vacuum because div E is 0.)

    Substitute in -mu0 epsilon0 dE/dt for curl of B, and you've got
    \nabla^2 E = \mu_0 * \epsilon_0 * -\frac{\delta^2 E}{ \delta t^2}
    This is the 3D version of the wave equation; you can show that any wave (spherical, cylindrical, plane) satisfies this equation, with a propagation speed of
    [tex] \frac{1}{(\mu_0 \epsilon_0 )^(0.5)}[/tex]

    Last edited: May 11, 2004
  7. May 13, 2004 #6
    Just a short note: Rocketcity showed a valid method of determining the speed of an electromagnetic waves equations using the vector operators.However, vector operators were introduced by Oliver Heaviside several years later after Maxwell published his original findings. So, the method that Rocket city posted is not the "exact" method that James Maxwell would have used to prove what the speed of electromagnetic waves was. He would have started usinfg the integral-differential forms of his equations instead.
    Last edited: May 13, 2004
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