What is more correct and what is the difference? Polarised waves are waves with vibrations in one direction perpendicular to energy propagation

  1. "vibrations in one plane"

  2. "vibrations in one direction in a plane"


I would say it depends on the kind of polarisation. For linear polarisation, you can say that the vibrations are in one direction in a plane perpendicular to the direction of propagation. For circularly or elliptically polarised light, the most you can say is that the vibrations are in the plane perpendicular to the direction of propagation, since the actual direction will vary with position / time.

  • $\begingroup$ whats the difference though between in one direction and in one direction in a plane $\endgroup$ – user143525 Nov 6 '14 at 9:18

The word "polarisation" simply means the relative geometry of the electric, magnetic and Poynting vectors of an electromagnetic field.

A pure polarisation state is any linear superposition of polarisation base states - which I define below. Since Maxwell's equations are linear, any linear superposition of these basic solutions is also a solution.

For any solution to Maxwell's equations in a homogeneous dielectric medium (or free-space), you can get another valid solution by swapping the roles of the electric and magnetic field vectors, and applying the scale factors as follows:

$$\vec{E} \mapsto \sqrt{\frac{\mu}{\epsilon}} \vec{H}$$ $$\vec{H} \mapsto -\sqrt{\frac{\epsilon}{\mu}} \vec{E}$$

and the new solution will have the same Poynting vectors (Intensity vectors). Therefore we have two orthogonal solutions:

\begin{align}F_1 &= (\vec{E}_1(\vec{r},\,t),\,\vec{H}_1(\vec{r},\,t))\\ F_2 &= \left(\sqrt{\frac{\mu}{\epsilon}}\,\vec{H}_1(\vec{r},\,t),\,-\sqrt{\frac{\epsilon}{\mu}}\,\vec{E}_1(\vec{r},\,t)\right)\\ & = (\vec{E}_2(\vec{r},\,t),\,\vec{H}_2(\vec{r},\,t))\end{align}

If the fields are time-harmonic, we can use the phasor notation, and then any solution of Maxwell's equations that has the same Poynting vector as the two solutions above can be represented by two complex numbers $\left(\begin{array}{c}\alpha_1\\\alpha_2\end{array}\right)$ that represent the superposition $\alpha_1\,F_1 + \alpha_2 F_2$ and where $\alpha_1|^2+|\alpha_2|^2=1$. Thus the two complex numbers fully define the geometry of the fields at any time and place.

You can see how a simple plane wave in the $\hat{Z}$ direction fits in with the above. The basic solutions, if you choose linear polarisations to be your base states, are:

$$\begin{array}{lclcclcl}E_1(x,\,y,\,z,\,t) &=& E_0\,e^{i\,(k\,z - \omega\,t)}\,\hat{X}&&&H_1(x,\,y,\,z,\,t) &=& \sqrt{\frac{\epsilon}{\mu}}\,E_0\,e^{i\,(k\,z - \omega\,t)}\,\hat{Y}\\E_2(x,\,y,\,z,\,t) &=& E_0\,e^{i\,(k\,z - \omega\,t)}\,\hat{Y}&&&H_2(x,\,y,\,z,\,t) &=& -\sqrt{\frac{\epsilon}{\mu}}\,E_0\,e^{i\,(k\,z - \omega\,t)}\,\hat{X}\end{array}$$

but you can equally take the left and right circular polarisations as your base states:

$$\begin{array}{lclcclcl}E_1(x,\,y,\,z,\,t) &=& \frac{E_0}{\sqrt{2}}\,e^{i\,(k\,z - \omega\,t)}\,(\hat{X}+i\,\hat{Y})&&&H_1(x,\,y,\,z,\,t) &=& \sqrt{\frac{\epsilon}{\mu}}\,\frac{E_0}{\sqrt{2}}\,e^{i\,(k\,z - \omega\,t)}\,(\hat{Y}-i\,\hat{X})\\E_2(x,\,y,\,z,\,t) &=& \frac{E_0}{\sqrt{2}}\,e^{i\,(k\,z - \omega\,t)}\,(\hat{Y}-i\,\hat{X})&&&H_2(x,\,y,\,z,\,t) &=& -\sqrt{\frac{\epsilon}{\mu}}\,\frac{E_0}{\sqrt{2}}\,e^{i\,(k\,z - \omega\,t)}\,(\hat{X}+i\,\hat{Y})\end{array}$$

these represent plane waves where the electric and magnetic field vectors are orthogonal but which rotate uniformly about the $\hat{Z}$ axis at angular speed $\omega$. Although a little harder to visualise, these are often taken to be THE polarisation base states because they represent the angular momentum eigenstates of the electromagnetic field. I say more about this, and the special Riemann-Silberstein notation for describing the circular polarisation decomposition of the electromagnetic field, in my answer to the question "What does the complex electric field show?".

Any of the above are equally well "pure polarisation states" and can be transformed into one another by unitary (power conserving) linear transformations. The description of the transformations of pure polarisations by optical components leads to the Jones Calculus (see Wikipedia page of this name). The optical component's action is described by a $2\times 2$ unitary matrix $U$ that multiplies the $2\times1$ vector $\vec{\alpha}=\left(\begin{array}{c}\alpha_1\\\alpha_2\end{array}\right)$. Another neat description of the polarisation state, if you only care about relative phases between the polarisation base states, is that by the Stokes parameters:

$$s_j = \vec{\alpha}^\dagger\,\sigma_j\,\vec{\alpha}$$

where the $\sigma_j$ are the $2\times 2$ Pauli spin matrices. Indeed we can describe the state of the light by the $2\times 2$ matrix $S = s_1\,\sigma_1 + s_2\,\sigma_2+s_3\,\sigma_3$. When a lossless optical component with Jones transformer $U$ acts on the Jones vector $\vec{\alpha}=\left(\begin{array}{c}\alpha_1\\\alpha_2\end{array}\right)$, the Stokes matrix transforms by $S\mapsto U^\dagger \,S\,U=U^{-1}\,S\,U$ and indeed the $3\times 1$ Stokes vector $\vec{s}=\left(\begin{array}{c}s_1\\s_2\\s_3\end{array}\right)$ keeps its same length and the transformation wrought on the space of $3\times 1$ Stokes vectors is rigidly rotated. The Stokes vector's head lives on the surface of the unit sphere - the Poincaré sphere.

Now we get to partially polarised light: this is quite a subtle concept especially if you ask for a classical description. The quantum description is simple and elegant: it is a classical statistical mixture of pure polarisation states. I discuss its mathematical description in my answer here to the question "Partially polarized light with jones vectors?", wherein I describe the density matrix formalism to fully describe a partially polarised light state.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.