Ziang said:
May you tell me how can the direction of the field vector be changed when the observer is moving as same direction as the vector?
I'll do this explicitly. If you can't follow this then you have some learning to do - maybe read Dale's links.
The electromagnetic field tensor is, in general,$$F^{\mu\nu}=\pmatrix{
0&-E_x/c&-E_y/c&-E_z/c\cr
E_x/c&0&-B_z&B_y\cr
E_y/c&B_z&0&-B_x\cr
E_z/c&-B_y&B_x&0\cr}$$It's easy to see how you can split that up into a sum of two tensors, one that describes the electric field and one that describes the magnetic field:
$$\pmatrix{
0&-E_x/c&-E_y/c&-E_z/c\cr
E_x/c&0&0&0\cr
E_y/c&0&0&0\cr
E_z/c&0&0&0\cr}
+\pmatrix{
0&0&0&0\cr
0&0&-B_z&B_y\cr
0&B_z&0&-B_x\cr
0&-B_y&B_x&0\cr}$$
For the case of an electromagnetic wave heading in the z-direction with its E-field in the x direction, the electromagnetic field tensor is $$F^{\mu\nu}=\pmatrix{
0&-B\sin \left(kz-\omega t\right)&0&0\cr
B\sin \left(kz-\omega t\right)&0&0&B\sin \left(kz-\omega t\right)\cr
0&0&0&0\cr
0&-B\sin \left(kz-\omega t\right)&0&0\cr }$$where I have used that the amplitude ##E## of the E-field is c times the amplitude ##B## of the B field. Transforming it into another frame is straightforward - ##F^{\mu'\nu'}=\Lambda^{\mu'}{}_{\mu} F^{\mu\nu}\Lambda^{\nu'}{}_\nu##, where $$\Lambda^{\mu'}{}_{\mu}=
\pmatrix{
\gamma&-{{v}\over{c}}\gamma&0&0\cr
-{{v}\over{c}}\gamma&\gamma&0&0\cr
0&0&1&0\cr
0&0&0&1\cr }$$is the Lorentz transform (or, in matrix notation, ##F'=\Lambda F \Lambda^T##). Decomposing our tensor into the electric and magnetic components before transforming, we find that the electric component becomes$$\pmatrix{0&-B\sin \left(k z-\omega t\right)&0&0\cr B\sin \left(k
z-\omega t\right) &0&0&0\cr 0&0&0&0\cr 0&0&0&0\cr }$$and the magnetic component becomes
$$\pmatrix{0&0&0&-B{{v \sin \left(k z-\omega t\right) }\over{
\sqrt{c^2-v^2}}}\cr 0&0&0&B{{c \sin \left(k z-\omega t\right)
}\over{\sqrt{c^2-v^2}}}\cr 0&0&0&0\cr B{{v \sin \left(k z-\omega t
\right) }\over{\sqrt{c^2-v^2}}}&-B{{c \sin \left(k z-\omega t
\right) }\over{\sqrt{c^2-v^2}}}&0&0\cr }$$That is, what the first frame calls an electric field, the other frame sees the same (a special case because the E field is parallel to the velocity - generally a B field component will appear). But what the first frame calls a magnetic field, the other frame sees as both a magnetic field and an additional electric field. Adding them together, we see that the electric field in the primed frame is parallel to ##(1,0,\gamma v/c)##.
It's straightforward to transform the direction of travel of the light to get ##(-\gamma v/c,0,1)## and hence see that the electric field is perpendicular to the direction of the light since the dot product is zero. The magnetic field is also perpendicular, trivially so since it only includes a y component in either frame. It's also straightforward to show that the magnitudes of both the electric and magnetic components have changed by a factor of ##\gamma##.
So, in short, the transformed electromagnetic wave is an electromagnetic wave.
Edit: Note that I was lazy and didn't bother transforming z and t into z' and t', which is why there doesn't appear to have been any Doppler effect. If you do carry out that substitution the Doppler effect (concealed behind my laziness) will become apparent. It has no effect on the argument above, since all it does is change ##kz-\omega t## into ##\vec{k'}\cdot\vec{x'}-\omega't'## and then promptly cancels out of all the direction vectors I was interested in.
