EM wave direction after Rayleigh scattering I have a problem in figuring out where and how a beam of light is scattered by a molecule (Rayleigh scattering), in the two pictures I posted two different situations seem to be illustrated: in the first picture the incident wave after scattering is spread out in all the space with different intensity depending on the angle while in the second image the incident beam is scattered in just one direction. Now my questions:
1. What is the right image? What happens once the incident beam is scattered? Does it become a spherical wave that spread all over the space? Does it have a definite propagation direction?
2. Since the scattered light is due to the induced electric dipole, does an oscillating electric dipole always radiate spherical waves? If yes, is it possible to have a spherical wave with different inesities in different points of its wave-front like in the first image?
3. Should I think of the first image as the possible scenario of a Rayleigh scattering i.e.: illustrates the various intensities the scattered light will have depending on the angle at which it is emitted?


 A: A molecule excited by light will oscillate in the direction of the electric field. It becomes an oscillating dipole, which has a donut-shaped emission pattern (perhaps hinted by your final image). Moreover, the emitted light will be polarized along the direction of oscillation.
Have you ever noticed that the sky is polarized? Try looking at a blue sky with polarized glasses, turning your head side to side. This is a consequence of the dipolar emission of atmospheric molecules, and there are critters in the animal kingdom which are adapted to see this and use it to their advantage.
A: *

*The second image is a reasonable picture. The first could be possible but would need qualification and explanation.

Rayleigh scattering occurs when the electric dipole moment of the molecule is made to oscillate in the direction of the electric field of the incoming wave.
The oscillating electric dipole then emits "dipole radiation", which has a dipole antenna pattern $\propto \sin^2 \theta$; the intensity of dipole radiation is not spherically symmetric and there is no radiation along the axis of the oscillating dipole.
However, all depends on the polarisation state of the incoming light. If it is plane polarised, then the intensity of the scattered light has a dipolar anisotropy (but the wavefronts would still be spherical) as described above. If it is unpolarized, then there is no preferred axis for the molecular dipole oscillations, but no dipole oscillation in the same direction as the incident light -  the scattered light would then have an intensity $\propto 1 + \cos^2 \theta$.


*Yes, see above. The "spherical" waves will be anisotropic in intensity, especially if the incoming light has any linear polarisation.


*Yes. The arrows could represent the Poynting vector of the scattered light, which can vary with the scattering angle.
A: Your first diagram is a graph using polar coordinates of the intensity of scattered light $r$ as a function of the scattering angle $\theta$ as defined in the second diagram.

So the intensity os the scattered light is maximum in the incident light direction and the reverse direction.
Your second diagram as well as being useful when defining the scattering angle can be used to illustrate why it is that the scattered light is partially polarized.

Note that the scattering occurs in three dimensions.
A: The question boils down to:
What are the EM fields associated with an osscilating dipole?
$$\rho = -\vec{P}\cdot \nabla \delta^3(r)$$
$$\vec{J} = \frac{d\vec{P}}{dt}\delta^3(r) $$
If your uncomfortable with that, simply just imagine an oscillating current density at a single point in space
$$\vec{J} = I_{0}\cos(\omega t) \delta^3(r) \hat k$$
[use charge conservation to prove it fits the same form as a dipole]
The  radiation fields from this distribution are:
$$\vec{E} =\frac{1}{r} \frac{\mu_0 I_{0}\omega}{4\pi} \sin(\frac{\omega}{c}r-\omega t)\sin(\theta)\hat \theta$$
$$\vec{B} =\frac{1}{r} \frac{\mu_0 I_{0}\omega}{4\pi c} \sin(\frac{\omega}{c}r-\omega t)\sin(\theta)\hat \phi$$
The radiation has symmetry along the variable $\phi$, it does not have spherical symmetry
When $\theta = 0, \pi$ [the oscillation axis] the EM field is zero, meaning no light is produced.
From these equations it is clear that the propagation direction is in the direction of $\hat r$
This can also be seen the the poynting vector
$$\vec{S} \propto (\hat \theta × \hat \phi) \propto \hat r$$
Edit:
The full three dimensional plane wave solution is in the form $$\vec{E} =\vec{E}_{0}\cos(\vec{k} \cdot \vec{r} - \omega)$$
$\vec{k}$ is the wave vector, it specifies the direction the wave is travelling, or the measured direction of the phase velocity
For all planes perpendicular to the wave vector, $\vec{k} \cdot \vec{r}$ is a constant. This means that the field will have the same value, for all planes in  perpendicular to the direction of travel,which by definition constitutes, a wave front. Whose whose phase velocity in the direction of the wave vector is "c"
Visualising this makes it clear to see why there is a direction of travel in the direction of k
The simplification of $\vec{k} \cdot \vec{r}$ for the case where the wave vector is in the direction of the x unit vector, is:
$$\vec{k} \cdot \vec{r} = kx$$
Which reduces to the standard 1 dimensional equation. Please note in the 1d equations, this one dimensional variation of E with x,  does not mean the field exists only on the x axis. It exists as that same values for all planes perpendicular to the wave vector [x,y,z]
So where is the wave vector in my equation for the dipole field? Well even though it is not a plane wave, it can still be expressed as a dot product with a wave vector.
$$\vec{E} =\frac{1}{r} \frac{\mu_0 I_{0}\omega}{4\pi} \sin(\vec{k}\cdot \vec{r}-\omega t)\sin(\theta)\hat \theta$$
Here the wave vector i am defining is
$$\vec{k}= \frac{\omega}{c}\hat r$$
It is easy to see that for all surfaces perpendicular to $\vec{r}$ the field has the same value [atleast for the cos part of of E field, which is responsible for the wave l
part]
This constitutes a spherical  surface where the cos part has the same value.
Phase velocity is the  speed at which a  part of the wave with the same value, moves in the direction  of $\vec{k}$ with respect to time. The speed at which e.g a peak moves.
Doing this for the wave of a dipole
Set the cos to a constant value:
$$\sin(\vec{k}\cdot \vec{r}-\omega t) = c$$
We want to know how the location of field value c changes with respect to time,-  $\vec{r}(t)$
$$|\vec{r}| [\hat k \cdot \hat r] = \sin^1(c) + \frac{\omega}{|\vec{k}|} t$$
For this very specific case $\hat k = \hat r$ and this expression reduces to
$$|\vec{r}|  = \sin^1(c) + \frac{\omega}{|\vec{k}|} t$$
$$\frac{d|\vec{r}|}{dt}  = \frac{\omega}{|\vec{k}|} $$
Which substituting in is
$\frac{d|\vec{r}|}{dt} = c$$
The fact this expression has a $|\vec{r}| $ in it and r can be any value, instead of a specific vector direction like
x  \hat i, means the wave velocity has a velocity of C, for all directions with spherical symetry.
This does not mean the field is soherically symmetric, just that the wave vector is, whose phase velocity pointing out of that sphere, is c
