# Does a rotating electric field vector (in circular light) moves slower than the speed of light?

We can imagine circular light as a combination of two rotating vectors moving through space: the electric field vector that´s rotating together in sync with the magnetic field vector. As the vector rotates doesn´t it take some velocity away from the propagating speed? In other words doesn't circular light propagate a very bit less than the speed of light?

• What exactly, in your opinion, is moving in a spiral (in real space) for circular-polarized light? Because the illustrations commonly shown to students are mixed-dimensional 3d plots where one axis is the real space axis (let it be x), and the other two are the magnitudes of the electric field ($E_y$ and $E_z$). Think about how you could(n't) possibly add together speed of motion along $x$ with speed of change of electric field. – LLlAMnYP May 17 '16 at 11:57
• @LLIAMnYP The other two dimensions represent space as well. And in this (y,z) space the electric field vector is making its turns. The two other are not representing the electric field (hence, Ey and Ez). So the electric field is rotating in real space, happy spiraling forwards. :-) – descheleschilder May 17 '16 at 23:50
• There is no real (y-z) space in which an arrow rotates. This is simply a misconception. The direction of the field is a property of space, not something material which is "spinning". – LLlAMnYP May 18 '16 at 6:13

Vector plots like this

should not be shown to students. Some falsely begin to believe, that the blue line actually represents something in real space. That is not true.

So what is shown here? This is, in fact, a snapshot of a single moment in time. I have arbitrarily chosen a convention (let me stress arbitrarily again), where I represent the electric field at a point by an arrow, which starts at that point and extends in the direction of the electric field to a length, proportional to the magnitude of the field. The proportionality between the length of the arrow and the magnitude of the field is again arbitrarily chosen by me. There is no physical significance behind it, thus there is no physical significance behind the blue line. The only motion of a photon is along the red line.

Still not convinced? I can freely change my convention and have the mid-point of every arrow placed at the point where I want to show the field. Then the above plot would transform to something like this:

Now the length of the blue line has changed, although nothing has changed about the distribution of the E-field. It's still the same wave. Same speed.

I can just as well adopt a convention, where the field at a given point is shown by an arrow, whose arrowhead is at that point, like so:

There is no longer a helix traced out. No spiral motion to take away from the speed of a photon.

The last plot is in no way better or worse than the first two. There exists no special rule, that the tail of the arrow absolutely must start at the point, at which we wish to represent the field. The 2nd and 3rd plots are just as valid.

Finally, here's how a propagating circular-polarized wave looks like:

It is, essentially, moving as a whole, so again, there isn't some spinning motion to take away from the speed of the wavefront.

• Nice diagrams but the problem is drawing your last diagram in a book. So the best that can be done is the wave profile which is your first diagram. If you sit at a point in space what do you "see"? A rotating electric field. And if you move away from the source? Again a rotating electric field but slightly delayed in time. I am afraid that I just cannot visualise what a photon is doing. – Farcher May 18 '16 at 14:57
• @Farcher in an infinite free space, the wavefunctions of a particle are of the form of $e^{ikz-i\omega t}$. Keeping things as simple as possible, a photon is its electric field. It's not "doing" anything, it's infinitely extended in space (in order to have a specific energy/momentum) and the train of crests of the wave is forever propagating, as in my last diagram. If you desire a more localized phonon, take a wavepacket that has a less defined energy, but somewhat defined location. But that waveform will, as a whole, fly forever until it hits something. It's not "spinning". – LLlAMnYP May 18 '16 at 15:16
• I have never thought of the photon as spinning, indeed as a pointed out before I have no idea what a photon does. All I observe is the effect of a rotating electric field. – Farcher May 18 '16 at 15:23
• @Farcher Well, what I'm trying to say, is that a photon is merely a disturbance (deviation from zero) of the electric field, somewhere in space, and all that's happening, is that it's flying along at $c$. If you stand on one spot, $E$ appears to be rotating, but if you are (hypothetically, since SR would kick in otherwise) co-moving with it, everything is perfectly static. I'm not sure, what would be a satisfactory analogy of a photon "doing something", if this doesn't seem satisfactory enough. – LLlAMnYP May 18 '16 at 15:30
• @WetSavannaAnimalakaRodVance no, but I'm doing a PhD with ellipsometry as the primary experimental method. I've basically just spent 4.5 years studying polarized light. – LLlAMnYP Jun 7 '16 at 10:18

Circularly polarised light can be thought of as the sum of two linearly polarised beams which are at right angles to one another and ninety degrees out of phase.
The linearly polarised beams components travel as the same speed as each other and their sum.

• Your answer would beg the question "but doesn't the motion along a sinusoid take away from the speed of light?" :-) – LLlAMnYP May 17 '16 at 11:58
• @LLlAMnYP The original question is interesting as is your comment because I think that is a asking whether or not waves actually interfere (not superpose but change wavelength/frequency/speed) with one another. I do not think they do in this case. – Farcher May 17 '16 at 13:45
• IMO, this is a deeply rooted misconception. People begin to think of circular polarized light as actual rotating arrows, flying through space, as if the photon is sitting on the end of the arrow and tracing out a helix. – LLlAMnYP May 17 '16 at 14:07
• @LLIAMnYP "but doesn't the motion along a sinusoid take away from the speed of light?" The electric field in a sinusoid wave doesn´t have a speed other than the forward one. It´s only the amplitude of the field that changes, wich gives the electric field not an extra velocity. – descheleschilder May 17 '16 at 22:53
• @descheleschilder Uhh, how do you reason, that a sinusoid is "only a change of amplitude" but a circular polarization is not? – LLlAMnYP May 18 '16 at 6:24