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Using Wave Equation to Prove that EM Waves are Light

  1. Jul 17, 2014 #1
    1. The problem statement, all variables and given/known data
    I'm working on using the wave equation to prove that EM waves are light.


    2. Relevant equations
    Here's what I'm working with:

    E = Em sin(kx-wt)
    B = Bm sin(kx-wt)

    ∂E/∂x = -∂B/∂t
    -∂B/∂x = μ0ε0 ∂E/∂t

    and the wave equation: ∂2y/∂x2 = 1/v^2(∂2y/∂t2)


    3. The attempt at a solution

    I've differentiated the two equations with respect to x and t (after substituting in the equations for E and B) to get something resembling the wave equation, where y=E and y=B.

    ∂2E/∂x2 = 1/v^2(∂2E/∂t2) --> (w^2)Bm sin(kx-wt) = (1/v^2)(k^2)Bm sin(kx-wt) (1/μ0ε0)

    which simplifies to: w^2 = (k^2)/(v^2)(1/μ0ε0)

    Now I'm stuck, because I can't figure out a way to prove this last relation.
     

    Attached Files:

  2. jcsd
  3. Jul 18, 2014 #2

    Simon Bridge

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    Here - let me help, you just wrote:$$\frac{\partial^2E}{\partial x^2} = \frac{1}{v^2}\frac{\partial^2E}{\partial t^2} \implies \omega^2 B_m\sin(kx-\omega t) = \frac{1}{v^2}k^2 B_m\sin(kx-\omega t)\frac{1}{\mu_0\epsilon_0}$$ ... which simplifies to: $$\omega^2 = \frac{k^2}{v^2}\frac{1}{\mu_0\epsilon_0} $$ ... is this correct?
    I'm guessing you wanted that first equation to be a "B" equation?
    It is unclear were the permitivity and permiability came from in the RHS of the second equation.
    I suspect you have got a bit mixed up between the different equations.

    Note: if you have a stationary periodic function ##y(x)=\sin kx##, then the same function as a wave moving in the ##+x## direction with speed ##v## is ##y(x-vt) = \sin k(x-vt) = sin(kx-\omega t)## so ##\omega=kv## ... does that help?

    Also: $$c^2=\frac{1}{\mu_0\epsilon_0}$$

    It is not clear how you expect to "prove" that EM waves are light by this approach.
     
    Last edited: Jul 18, 2014
  4. Jul 18, 2014 #3
    Thanks for the help!

    So, if w = kv, then (w^2) = (k^2)(v^2) and:


    w^2 = (k^2)/(v^2)(1/μ0ε0) simplifies to (v^2) = (1/v^2)(c^2) which goes to (v^4) = (c^2)

    ...now?
     
  5. Jul 18, 2014 #4

    mfb

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    Where did the (1/μ0ε0) go?
    Edit: Ah, you used Simon Bridge's equation. Looks like there is an error in the derivation of your equation, not sure where.

    This is impossible. You can show that electromagnetic waves travel at the same speed of light, you can show their energy/momentum relation agrees with light and so on - but that does not prove light is an electromagnetic wave, it could be a wave of something different.
     
  6. Jul 18, 2014 #5
    I've looked over my derivation (attached in original post) and can't seem to find any errors.
     
  7. Jul 18, 2014 #6

    Simon Bridge

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    You seem to have got mixed up between the different equations.
    Go back through the derivation one step at a time, and document your reasoning.
    Note: you cannot use this approach to prove that EM waves are light.

    You are starting from:$$\frac{\partial}{\partial x}E= -\frac{\partial}{\partial t}B\\ \frac{\partial}{\partial x}B= -\frac{1}{c^2}\frac{\partial}{\partial t}E$$
    Differentiating the first equation wrt x, and the second equation wrt t, gets you:$$\frac{\partial^2}{\partial x^2}E= -\frac{\partial}{\partial x}\frac{\partial}{\partial t}B\\
    \frac{\partial}{\partial t}\frac{\partial}{\partial x}B= -\frac{1}{c^2}\frac{\partial^2}{\partial t^2}E$$
    IF$$\frac{\partial}{\partial x}\frac{\partial}{\partial t}B=\frac{\partial}{\partial t}\frac{\partial}{\partial x}B$$

    THEN... you should be able to take it from there.
     
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