One of my tasks this week, before heading off to the Caribbean for a relaxing vacation, is going to be to find a new pair of polarized sunglasses that aren't ridiculously ugly. This seems like a decent hook of a physics post, explaining why "polarized" is a selling point for sunglasses, but first, I probably ought to explain what we means when we talk about the polartization of light.
As you know, even if your name isn't Bob, light can be thought of as an electromagnetic wave. You have an electric field oscillating in space and time, and a magnetic field with it, oscillating at the same frequency. The changing electric field creates a changing magnetic field, which creates a changing electric field, which creates a changing magnetic field, and so on. The fields are perpendicular to each other and to the direction of motion, and support each other as they move through space at the speed of light. The usual graphical representation is shown in the right-hand figure below.
(Image taken from Hyperphysics.)
Now, the electric and magnetic fields are vectors, and as such point in a particular direction in space. We can use the direction of one of the two (by convention, we pick the electric field) to define the polarization of the light field. The actual electric field will oscillate up and down, but at any given instant, it will be pointing in some direction (unless you're really unlucky, and it's zero at that instant, in which case wait a bit, and it'll come back), and we use that axis to define the polraization. If the field is pointing either up or down, we call it vertical polarization, and if it's pointing either right or left, we call it horizontal polarization.
Of course, electric fields aren't restricted to pointing exactly along vertical or horizontal axes, but can be at any arbitrary angle to those axes. That's ok, though, because we can think of any arbitrary angle of polarization as being a combination of horizontally and vertically polarized light, with appropriate amplitude.
This is sort of the reverse process from the bad treasure maps you would make as a kid-- if you want to hide something a bit west of north, you can define the position as, say, four steps north followed by three steps west, starting from some landmark. That works out to be five steps away from the landmark in a north-by-northwest sort of direction, but it's easier to specify in terms of steps along the north-south and east-west axes.
We can do the same thing with electric fields. A field with a magnitude of 5 units at an angle of 37 degrees left of the vertical axis can be broken up into a vertical field with a magnitude of 4 units, and a horizontal field with a magnitude of 3 units. Or, in terms of polarization, we can think of a light field with a polarization angle 37 degrees from the vertical as being 4 units of vertically polarized light, plus 3 units of horizontally polarized light.
Ordinary light from the sun, or a light bulb is unpolarized, a term which is a little deceptive. Individual light waves will, of course, be polarized, but each wave has its own polarization, and they're not correlated in any way. The polarization of the waves measured at any particular spot will flucuate randomly, and very rapidly, with no preferred direction.
We can make polarized light using special materials in which the transmission of light through the material is different for different orientations of the polarization. A naturally occurring example is calcite, which has a different index of refraction for different polarizations. A propery cut calcite crystal will bend horizontally and vertically polarized light by different amounts, and can be used to separate the two. If you block one beam, you're left with either horizontally or vertically polarized light.
These days, polarized light is more likely to be made from a Polaroid type material (invented by the same guy responsible for the concept of shaking things like instant-film pictures, who thought up one good name and ran with it). Polaroid sheets are made of polymers that allow one polarization of light to pass through, while strongly absorbing the other.
You can create light of an arbitrary polarization by shining unpolarized light at a polarizer of whatever type you like, arranged to transmit light with its polarization along a particular direction. If you take two polarizers and put one after the other, you find that the amount of light making it through the second varies depending on the angle between the two. If the polarizers are aligned in the same direction, all of the light makes it through the second polarizer. If they're perpendicular, none of the light makes it through (you get, say, vertical polarization from the first one, which is absorbed by a horizontal polarizer).
For angles in betwee, you pass only the component along the direction of the polarizer. So, for example, if your first polarizer is set up to transmit 5 units of light at 37 degrees to the vertical, a vertical polarizer would pass only the 4 units of vertically polarized light on, and a horizontal polarizer would pass only the 3 units of horizontally polarized light.
We usually think of this in terms of the intensity of the light passed through the polarizer, as a fraction of the initial intensity. Intensity goes like electric field squared, so the vertical polarizer passes 4/5 of the initial electric field, which means 16/25 of the initial intensity. The horizontal polarizer would pass 9/25 of the original intensity.
In general, the amount of light transmitted through a polarizer at an arbitrary angle is given by Malus's Law, which says that the transmitted intensity is the initial intensity multiplied by the cosine squared of the angle between the polarization of the light and the polarizer. In our simple example, cos(37)=4/5, so everything works out.
This behavior is why polarization is such a useful tool for quantum optics. We can treat the polarization of light as a combination of two polarization states (H and V), and easily prepare any arbitrary combination of those two that we want (there are also devices that allow you to rotate the polarization of light without decreasing the intensity, but I won't talk about them here). We can then analyze that light using polarizers, and get an excellent realization of the ideal quantum measurement-- light either passes through a polarizer or it doesn't, and the light that makes it through a verical polarizer is perfectly vertically polarized afterwards. This is why so many of the classic experiments on the fundamentals of quantum mechanics have been done with polarized photons.
The discussion above has assumed that we have linearly polarized light, that is, light where the electric field oscillates back and forth along the same axis all the time. This doesn't have to be the case, though-- the horizontal and vertical components of the field are independent, and we can delay one of them relative to the other. If we delay one by one-quarter of an oscillation relative to the other, we get circular polarization, which looks like this:
At some instant, the horizontally polarized wave will be at a maximum, and the vertically polarized wave will be at zero. At that instant, the polarization looks horizontal. A little bit later, though, the horizontally polarized wave has decreased a little, while the vertically polarized wave has started to increase. Now, the light looks like it's polarized at a slight angle-- mostly horizontal, but with a small vertical component. A little while later, the horizontal component has decreased some more, and the vertical component has increased some more, and the angle is bigger. And so on. Eventually, you hit a point where the vertically polarized wave is at a maximum, and the horizontally polarized wave is zero, and you have pure vertical polarization.
If you sit at one point in space, and look at the direction of the electric field, you'll see it spinning around in a circle like the hand on a clock. It rotates very fast-- one complete circle per oscillation of the light wave-- and is always changing. We can classify this as either right-hand circular or left-hand circular polarization, depending on the direction of rotation.
Circular polarization can be generated using materials like calcite in which one polarization travels faster than the other (called "birefringent" materials), and cutting them to just the right thickness to produce the necessary delay. A piece of birefringent material of the right thickness to make circular polarization is called a "quarter wave plate," because it delays one wave by a quarter of an oscillation. Exactly what thickness you need depends on the material you're using and the wavelength of the light you're working with. Quarter wave plates tend to be very expensive.
Circular polarization might seem like just a curiousity, but in fact, it's very important for AMO physics. Cicrularly polarized light carries angular momentum, and thus can be used to selectively produce atoms in particular states. Atoms absorbing right-hand-circular light increase their angular momentum, while atoms absorbing left-hand-circular light decrease their angular momentum, and this allows us to exert fine control over the exact state of an electron inside an atom, which in turn lets us do some amazing experiments.
What does this have to do with ugly sunglasses? Come back tomorrow, and I'll explain.
but this is a description of light as a 100% wave phenomenon. How does one translate this to photons ? Is there a one-to-one correspondence between "one photon" and "one lightwave". And does "circularly polarized" light mean that "each individual photon" continually changes its state (i.e. the direction of 'its' associated electric field)?
In terms of photons, each photon in a polarized beam has a probability of being transmitted through the polarizer that is equal to the fraction of the light passed in the wave picture. So, sticking with that 37-degree example, each photon would have a 16/25=64% probability of going through a vertical polarizer, and a 9/25=36% probability of going through a horizontal polarizer.
As for the field, yes, to the degree that it makes sense to talk about the electric field of a single photon, that field would be changing direction in a circular manner, all the time.
Have I mentioned how much I love these "Basic Concept" posts?
Thank you, sir.
Optical rotation is commonly indicative of resolved chiral configuration - certainly in a randomly ordered medium like dilute low viscosity solution or gas phase. This is an electronic effect and is dangerous as a diagnostic. Circular birefringence across the spectrum must integrate to zero. Look at a CD or ORD output - optically active media routinely zero at selected wavelengths (e.g., within optical transitions).
Individual atoms are chiral, 100% left-handed. Geometric handedness and optical handedness are defined in opposite observing directions. A geometrically right-handed helix default rotates light left-handed even though the senses of handedness are superposable.
Solid state is complicated. Silver thiogallate, AgGaS2 with non-polar achiral tetragonal space group I-42d (#122), has immense optical rotatory power: 522 degrees/millimeter along  at 497.4 nm.
Twistane (C10H16) has no chromophores at all, not even a double bond. Its lowest energy sigma->sigma* transition is well below 250 nm. It has [alpha]D = 440 degrees (589 nm). Unlike helices that can add small perturbations in-phase to large numbers, twistane is compact and nearly equiaxial without a dipole moment or polarizability. What does the intense differential chiral grabbing?
Depending on your definition of "ridiculously ugly," as well as your definition of "ridiculously expensive," you might take a look at Maui Jim sunglasses. They make some nice ones.
Atoms absorbing right-hand-circular light increase their angular momentum,
Not to put too fine a spin on things, but their relative momentum I hope, allowing for negative quantum values in some states.
Btw, it was quite some time since I had to measure polarization of light. But wouldn't another deceptive thing with unpolarized light be that it can be measured as 'circular polarized' in a (too) simple polarimeter that may have to take an undirected time average? Or am I misremembering?
Question: What is the transmission efficiency of the polarizers you use in your laser experiments?
Comment: Be sure your polarizing sunglasses absorb plenty of UV as well as doing their glare reduction thing.
One detail: In your reply about photons, you forgot to mention that the photon in a pure 37deg polarization state becomes a photon in a pure vertical state if it is one of the lucky ones to get transmitted through the vertical polarizer, ditto for the horizontal one. That, along with my fav demo (H,V,45 = dark; H,45,V = gray), is a great way to introduce the idea of projection operators in quantum mechanics.
On your trip, be sure to look for the darkish band at 90 deg to the sun when you are wearing your polarizing sunglasses down in the Caribbean. This effect of Raleigh scattering is really striking with a clear sky. And be careful using your laptop, since LCD displays are polarized.
Finally, some of us like to think that photons are fundamentally circularly polarized, so linear states are a superposition in that basis. I always found that preference for one basis space over the other to be one of the curious differences between the classical and quantum description of light: waves are planar and particles are chiral.
I may be about to ask a spectacularly dumb question, but:
I guess I still don't get where the photon fits in. Photons are described as carrying "electromagnetic" radiation, right? But here we have two fields, the electric and the magnetic fields. Is the photon in the electric field or the magnetic field? Or is the photon's position described by some other field, and it just creates distortions in the electric and magnetic fields as it goes? Or is the idea that when we talk about the "electric and magnetic" fields we're really talking about one complex field, and the electric and magnetic are the real and imaginary parts of that field's values? Or what?
I have a great deal of respect for Hyperphysics [GSU].
However, I suspect that David Hestenes [ASU] has a better conception that EM waves are rotating helices rather than perpendiclar sine waves?
D Hestenes, 'The Kinematic Origin of Complex Wave Functions'
[Zitterbewegung interactions generating resonances]
Catenoid to helicoid transformation [after equation 14]
Isosurface tutorial [Â© Mike Williams 2003, 2004, 2005]
1 - Alphabetical Index
a - Polar Transformation
b - Mod()
c - Soap Bubble Surfaces
2 - Isosurface Index, Figures associated with:
a - "By transforming from cartesian to cylindrical polar coordinates, it's possible to bend any isosurface into a circle around the origin."
b - 'It's possible to use a single isosurface to produce multiple copies of a shape by using the mod operator in a substitution."
c - Mathematical Zoo "This is a collection of surfaces that mathematicians have found interesting for various reasons."
Hey Chad, have a nice vacation (I myself am blogging from an island in the mediterranean).
One detail: the pic of electrical and magnetic field seems incorrect even for didactical purposes, where is the phase shift ?
A field with a magnitude of 5 units at an angle of 37 degrees left of the vertical axis can be broken up into a vertical field with a magnitude of 4 units, and a horizontal field with a magnitude of 3 units.
Dumb question, but how are you getting 37 degrees to the vertical instead of 30 degrees for this? 3, 4, and 5 units would make this a 30-60-90 triangle unless I've missed something horribly obvious (which, admittedly, is always, a possibility).
Gotta love polarizers.
BTW, linearly-polarized light separates naturally into the left-handed and right-handed circular polarizations in a magnetized plasma. This is because
a) the plasma causes a certain dispersion of the light's wavelength
b) the magnetic field causes an increased dispersion for the right-hand polarized mode (called the extraordinary mode in analogy to crystal optics, such as the mentioned calcite example in the post), but not for the left-hand polarized mode.
As an added bonus, not all frequencies can propagate for all values of the plasma's number density or magnetic field strength. This makes for really interesting work interpreting data from wave instruments and radio soundings on satellites or on the ground.
Dumb question, but how are you getting 37 degrees to the vertical instead of 30 degrees for this? 3, 4, and 5 units would make this a 30-60-90 triangle unless I've missed something horribly obvious (which, admittedly, is always, a possibility).
The small angle in a 3-4-5 triangle is about 37 degrees-- you can check it by taking the inverse sine of 3/5. A 30-60-90 triangle has sides of 1,2, and sqrt(3).
Ok, so it was something horribly obvious. Thanks :-).
their relative momentum
Duh! I both misread the post and misconstrued this, it is absolute momentum we discuss.
Is the photon in the electric field or the magnetic field?
I wish the QFT literati would have chimed in on virtual photons and the rest, but FWIW:
A non-virtual photon is emitted when the electric and magnetic field couples and radiates energy in some direction from accelerated charges. (Photons are energy quanta, as Planck and Einstein taught us.)
When you have a macroscopic emitter, such as a radio antenna for radio waves, you can rather easily observe the near field where the coupling happens, and the far field of the radio waves/emitted photons. When you have a microscopic emitter, such as an atom for light waves, you usually only think of the far field effects described in the post.
the electric and magnetic are the real and imaginary parts of that field's values
In the classical picture of Maxwell's equations, the electric and magnetic field are both vector fields.
But when you are looking at radiation fields it can be more convenient to rewrite them as a scalar and vector potential instead.
And when you are looking at relativistic radiation it can be convenient to combine them into a four-vector potential. So the magnetic field can be interpreted as a low-velocity relativistic effect if you wish.
Answering comment #8:
Photons are responsible for all electric and magnetic effects, not just radiation. The lesson of Maxwell, who invented the first unified theory in physics, was that electric and magnetic effects are not two different things. The totality of this unification was not fully appreciated until QED came along, although it might not have fully sunk in until electro-weak unification (by Weinberg and Salam) was confirmed experimentally. The electric (Coulomb) force is due to the exchange of virtual photons. Interestingly, these photons can have longitudinal polarization, the third possibility one would normally expect for a spin-1 object that has mass. [Mutterings about such photons being "off shell" and thus having mass would appear here.]
Answering comment #10 for Chad:
There is no phase difference between the E and M parts of the wave. They are in lock step because the spatial variation of one is responsible for the temporal variation of the other, and (the important bit) vice versa with a minus sign.
However, your question made me look at that first graph and notice that it is in error! The magnetic field is wrong by 180 deg. Why? Because E x B (x is the vector cross product) poynts (bad pun) in the direction of the wave velocity. [E x B is, up to factors that depend on the units being used, equal to S, the Poynting vector, whose average value is the wave intensity and whose direction is the same as that of the wave.]
Confession: First time I wrote "Poynting vector" in my notes it was spelled "Pointing". Poynting is a persons name, pointing is what it does. Lesson 1: as a student, always read the book before class. Lesson 2: as a lectturer, always write names of things on the board.
Observation about comments #11 and #14:
Before the advent of calculators, problems could only contain angles whose trig functions had simple values. The 37-53-90 triangle (it is actually 36.9 deg to 3 sig figs) gets a lot of use as a result, particularly because hand arithmetic is really easy with 0.6 (sin), 0.8 (cos), and 0.75 (tan). You need to know sqrt(3) = 1.732 (where 1732 is the year Geo Washington was born) to work with the 30-60-90 triangle beyond slide-rule precision.
Feynman's explanation in his book QED, Penguin, 1990, p.120:
'Photons, it turns out, come in four different varieties, called polarizations, that are related geometrically to the directions of space and time. Thus there are photons polarized in the [spatial] X, Y, Z, and [time] T directions. (Perhaps you have heard somewhere that light comes in only two states of polarization - for example, a photon going in the Z direction can be polarized at right angles, either in the X or Y direction. Well, you guessed it: in situations where the photon goes a long distance and appears to go at the speed of light, the amplitudes for the Z and T terms exactly cancel out. But for virtual photons going between a proton and an electron
in an atom, it is the T component that is the most important.)'
Interestingly, these photons can have longitudinal polarization, the third possibility one would normally expect for a spin-1 object that has mass. [Mutterings about such photons being "off shell" and thus having mass would appear here.]
CCPhysicist, thanks, that was exactly what I wanted to see (learn more about). Since I haven't ventured into QFT land I can at most reproduce your first sentence. ("Quantization" and/or "force carriers" is obviously the name of the game here. :-)
So EM fields are massive (outside the relativistic mass contribution)?
The magnetic field is wrong by 180 deg.
I missed that! Fortunate I wasn't the only one...
in situations where the photon goes a long distance and appears to go at the speed of light, the amplitudes for the Z and T terms exactly cancel out. But for virtual photons going between a proton and an electron in an atom, it is the T component that is the most important.
Sounds like we are looking at near field effects here. I didn't know it affected the quantizations so much. Now I must look into QFT, this is too exciting to wait!
There is no such thing as relativistic mass. There is (at least for practitioners who prefer to call energy "energy" rather than "relativistic mass") only the invariant mass m (which is zero for a real photon), along with the total energy E (which is equal to mc^2 + K, the kinetic energy) and the 3-momentum p. These are related by E^2 = (pc)^2 + (mc^2)^2. You can trivially obtain Maxwell's result, p = E/c, when m = 0. Those are real photons, in the far field, which only have two possible polarizations.
Aside: Note that E = mc^2 when p=0, which is where the idea of rest energy comes from. Even though a photon is never at rest, you can determine experimentally (and from Maxwell's equations) that E=0 in the p=0 limit so its mass is zero. You can also see that K = p^2/2m, hence 1/2 mv^2, in the non-relativistic case where K << mc^2 by just substituting E = K + mc^2 into the main formula, expanding, and then dropping the K^2 term.
The hand-waving version of QFT is that Heisenberg lets you borrow some energy, but only for a short time (which automatically puts you in the near field). That virtual photon no longer has E = pc, hence it acts like it has a non-zero mass. It takes a book to say this more coherently (note the page that Jonathon is quoting), and no one says it better than Feynman in QED.
There is no such thing as relativistic mass.
Sorry, first textbook was old and non-english, introducing a terminology I still struggle to get rid off now that the usage has changed. I meant the kinetic energy contribution, obviously.
Yes, the einsteinian derivation of the energy-mass equivalence is particularly cute. Lorentz invariance gives a Taylor expansion for the kinetic energy where one can identify the invariant mass. [Heh! I see when I check now that my textbook from 1970 does this entirely in "relativistic mass", not momentum, as I remembered it.]
Um, okay, the off shell photon just "acts like it has a non-zero mass". So we are not discussing invariant mass, and the described contribution of the uncertainty principle is of course the zero-point energy of the field ground state.