Work out B0', E0', k' and omega', and show that the phase (kr-omega t) is

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Homework Statement

An electromagnetic planewave with frequency ω is propagating in the x direction in a reference frame S; it is polarised in the y direction with amplitude E$_{0}$. The k-$\omega$ 4-vector can be written (k$_{x}$, k$_{y}$, k$_{z}$, i$\omega$/c).

a) Write down and define the fields of the plane wave E(x, y, z, t) and B(x, y, z, t), being sure to define all terms you use.
b) The plane wave is observed from a reference frame S' moving in the x direction with speed v relative to S. Work out B'$_{0}$, E'$_{0}$, k' and $\omega$', and show that phase (kr - $\omega$t) is Lorentz invariant.
c) Find the speed of the wave in S' and check that the product of the intensities E'$_{0}$B'$_{0}$ is unchanged.

The Attempt at a Solution

a) E(x, y, z, t)=E$_{0}$e$^{i(k_{E}.r - \omega_{E}t+\phi_{E}}$)
B(x, y, z, t)=B$_{0}$e$^{i(k_{B}.r - \omega_{B}t+\phi_{B}}$)

But there seem to be so many equations for E and B around that I don't really know which one they are asking for in this case.

b) Does one have to use matrices? If so, what are the matrix representations of B and E? I lookled it up and couldn't find it anywhere. If not, how is the question meant to be done?

So

x'=$\frac{x-vt}{(1-\frac{v^{2}}{c^{2}})^{1/2}}$
y'=y
z'=z
t'=$\frac{t-v\frac{x^{2}}{c^{2}}}{(1-\frac{v^{2}}{c^{2}})^{1/2}}$

But I am not quite sure how that is supposed to help me to the question. Do I have to find the x, y, z and t components of the E and B fields? How is that meant to be done?

E$_{x0}$=$\frac{1}{4\pi\epsilon_{0}}$$\frac{qx_{0}}{(x^{2}_{0}+y^{2}_{0}+z^{2}_{0})^{3/2}}$

E$_{y0}$=$\frac{1}{4\pi\epsilon_{0}}$$\frac{qy_{0}}{(x^{2}_{0}+y^{2}_{0}+z^{2}_{0})^{3/2}}$

E$_{z0}$=$\frac{1}{4\pi\epsilon_{0}}$$\frac{qz_{0}}{(x^{2}_{0}+y^{2}_{0}+z^{2}_{0})^{3/2}}$

E'$_{x0}$=$\frac{\frac{qx}{4\pi\epsilon(x^{2}_{0}+y^{2}_{0}+z^{2}_{0})^{3/2}}-vt}{(1-\frac{v^{2}}{c^{2}})^{1/2}}$

E'$_{y0}$=$\frac{1}{4\pi\epsilon_{0}}$$\frac{qy_{0}}{(x^{2}_{0}+y^{2}_{0}+z^{2}_{0})^{3/2}}$

E'$_{z0}$=$\frac{1}{4\pi\epsilon_{0}}$$\frac{qz_{0}}{(x^{2}_{0}+y^{2}_{0}+z^{2}_{0})^{3/2}}$

Is this correct? What am I supposed to do about the B0', k' and omega'?

Please help.

 

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From some webpage on cramster.com that had a similar question:

E(x, y, z, t)=E$_{0}$cos (kx-$\omega$t)y-hat [equation 1]

E'$_{x}$=E$_{x}$=0
E'$_{z}$=E$_{z}$=0

E'$_{y}$=$\gamma$(E$_{y}$ - vB$_{z}$)
E$_{y}$=E$_{0}$cos(kx-$\omega$t)
B$_{z}$=$\frac{E_{0}}{c}$cos(kx-$\omega$t)


E'$_{y}$=$\gamma$(E$_{0}$cos(kx-$\omega$t)-$\frac{E_{0}v}{c}$cos(kx-$\omega$t)
E'=$\gamma$(E$_{0}$cos(kx-$\omega$t)-$\frac{E_{0}v}{c}$cos(kx-$\omega$t)

=$\alpha$E$_{0}$cos(kx-$\omega$t)

where
$\alpha$=$\gamma$(1-v/c)

so I figured that:

E'(x, y, z, t)=E'$_{0}$[cos (kx-$\omega$t)]y-hat

E'$_{0}$=$\frac{E'}{[cos (kx-ωt)]y-hat}$

=$\frac{\alpha E_{0}cos(kx-ωt)}{[cos (kx-ωt)]y-hat}$
=$\frac{\alpha E_{0}}{y-hat}$

(substituting in [equation 1]):

=$\alpha$$\frac{E(x, y, z, t)}{[cos (kx-ωt)](y-hat) (y-hat)}$

Stupid question, but what do I do with the y-hat y-hat on the denominator? I have never seen unit vectors in the denominator before. Do I just assume that the dot product is taken? If so, why? Or should the y-hats have not been there in the first place?