# In Mechanical Wave v = w/k. EM wave w/k = c. How Equated ?

1. May 13, 2015

### morrobay

With ω/k = 2π/T / 2π/λ = velocity for both transverse mechanical waves and EM waves.
I can understand velocity as distance over time in mechanical wave. But how is the ratio Em/Bm = ω/k = c.
That is the maximum amplitudes of the E and B fields in the y and z planes corresponding to c in x direction ?
Is it correct to say that the " leading edges" of the E and B fields in y and z planes are at c ?

http://www.santarosa.edu/~lwillia2/42/WaveEquationDerivation.pdf

Last edited: May 13, 2015
2. May 14, 2015

### morrobay

Too late edit : With dE/dx and t constant, dB/dt and x constant --> ∂E/∂x = - ∂B/∂t .... then to ω/k = Em/Bm = c. Does answer the change in x/change in t question. Again still not sure how this is related to amplitudes ?

3. May 14, 2015

### SteamKing

Staff Emeritus
The velocity of propagation for a transverse wave is not related to the amplitude of the wave at all.

4. May 14, 2015

### morrobay

Well thats my question , with ω/k = Emax/Bmax = c. (from Halliday-Resnick) The propagation velocity is related to amplitude. Unless I am misinterpreting Em/Bm

5. May 14, 2015

### morrobay

This is from a page in the text : kEm cos (kx-ωt) = ωBm cos (kx-ωt).
ω/k = Em/Bm = c
Thus the speed of wave c is the ratio of the amplitude of the electric and magnetic components of the wave.
So can someone show how Em/Bm = ω/k ?

6. May 15, 2015

### Staff: Mentor

Consider the following plane wave as an example: $$\vec E = \hat x E_m \cos (kz - \omega t) \\ \vec B = \hat y B_m \cos (kz - \omega t)$$ where $\hat x$ and $\hat y$ are unit vectors in the x and y directions. That is, $\vec E$ and $\vec B$ are in the x and y directions respectively, and the wave propagates in the z direction. Substitute these into the third Maxwell equation in free space: $$\vec \nabla \times \vec E = - \frac {\partial \vec B}{\partial t}$$ and you will get the desired result.

[added: I originally had $E_m$ instead of $B_m$ in my equation for $\vec B$ above. I've fixed this.]

Last edited: May 17, 2015