# Is AC any different from light?

I’m kinda bored of writing electricity & electric circuits, because that’s what I’ve been doing the whole damn week..!!! For now, I’m hibernating my writing of stuff under those two tags with this being the last post. Now, that doesn’t mean I don’t like (or Bah… I’ve quit?) writing about those, just that I’m not satisfied with my “research-before-writing-posts”. Okay, this post is something I always wanted to write as it reveals the true nature of alternating current and the inverse-square laws which are always interesting…

## AC by definition, is oscillation…

Well, in my first post in the “electricity series”, I explained all about the motion of free electrons in DC circuits and finally did a dirty skim through AC. I did that because, AC wouldn’t be a big trouble to grasp once you’ve understood DC since, again AC is no different from flipping the voltage source a few times which is a crude analogy and is presumably around 50 times (Hz) for domestic consumptions.

Alrighty then… Now, I presume you’re clear with AC that though the charges oscillate back & forth (unlike DC where they’re moving), the electric power keeps on transmitting from one end to the other end of the circuit. We’ve also seen that when current passes through a wire, magnetic field is produced around it, which can be visualized with the help of fine metal fillings. The free electrons are oscillated back & forth inside the wire, causing electric and magnetic fields to curl around the conductor. This is how the electric power is transmitted. And, it should also be emphasized that whenever charges accelerate, electromagnetic waves are radiated, depending on the frequency of oscillation. So, how is this propagation of disturbance different from an EM wave?

What did I mean by disturbance propagation? First of all, it’s worth noting that the speed of electric current is neither instantaneous, nor very negligible. It’s a misconception that electricity is instantaneous, in a sense that “a light bulb turns on almost immediately” once you switch it on. Our brain’s inability to grasp sufficient information from “what’s-happening-around”, work out and render the output with superior quality, doesn’t really mean that it’s instantaneous. Now, to the negligible part. The drift speed of free electrons (only a few mm/s) isn’t the speed of electricity itself. Though the electrons’ contribution to the current is given by $v_d$, the influence, the signal propagation, the disturbance (whatever you wanna call it), is always at a significant fraction of speed of light.

These are the electromagnetic interactions, the same mystical thing that came out of Maxwell equations, that whenever you shake a charge, it emits electromagnetic radiation. This is what I meant by the phrase, “disturbance that propagates close to speed of light”.

## In what way does AC signal differ?

The signal transmission in AC is just another electromagnetic radiation like Light, X-rays, or radio waves. The only difference lies in the fact that the frequency of oscillation of charges in AC is very low. In case of radio waves, you shake the charge to around a million times, whereas for light, it goes around a trillion times a second. Crudely speaking, for gamma rays, you shake mercilessly to a maximum of about an octillion times a second (which is what happens in cosmic rays). Meh… Don’t ask me how..!!! But, when we mean we shake the charges, we don’t go around and shake ’em with some diabolical supervillain device. We just energize the atoms to emit such harmful radiation. For instance, take Hiroshima (nuclear fission).

Well, the wavelength of gamma rays are about $10^{-15}\ \mathrm {m}$, so the requirement is satisfied only within the nuclei where the energy levels are essentially in the Fermi scale. That’s the reason why gamma rays always come out of the nuclei…

Feynman has listed AC within the different species of EM waves, naming it as “Electrical disturbance” in one of his lectures (don’t forget to read his short poetry below the table). You may also notice that as the frequency increases, the particle aspect of light starts showing up, which is not necessary for us now.

Okay, metals are extremely efficient by huge orders of magnitude for such low frequency transmission (i.e) the signal can propagate without any appreciable loss of energy. Moreover, the thing is constrained to travel within a small volume. In case, say you’re shaking the charge in air (or in vacuum, this disturbance needs no medium to propagate), the signal is so weak that it can never be perceived, even for smaller distances, let alone be used for transmitting information, which is a good reason why you can’t charge your cell phone with a “live” plug far away from you.

As told previously, it’s all based on efficiency. But, as you shake it further and reach the radio frequency portion of the spectrum, outer space is a way too efficient. Because, as the frequency increases, metals start heating up, leading to wastage of energy in the form of heat. Moreover, as the metal heats up (especially at the resonant frequency, when it’s heated tremendously), it starts emitting it’s own blackbody radiation (which is also the same reason why fire glows) which is dependent only on the temperature of the metal. So, you fly into the air.

## How can it be efficient now?

The same properties which were inefficient to transmit low frequency electrical disturbances will help us now. Outer space is three dimensional. In such a circumstance, the radiation from a point source (say, antenna) starts spreading out everywhere (unlike wires, where the thing is constrained to only one direction). The power spreads outwards in the form of concentric spherical wavefronts with the point source as origin. We can calculate the power distribution as a function of distance, which we’ll discuss soon. And, the power you get at the receiver end (radio for radio waves, eyes for visible light and so on) is only a fraction of the power emitted by the source.

### Let’s have a look at flux & power…

In order to brief about the “fraction”, I have to go through a new thing (But, you can always skip to the end). This figure you see above, has a great significance for gravitation, electromagnetic radiation, etc. You should’ve noticed that the laws of Newton, Coulomb, etc. are based on inverse-squares which is the case for three dimensions, that the intensity of such phenomena varies as the square of distance. More generally, for $N$ dimensions, the intensity varies with distance to the power of $N-1$

$\mathrm{Intensity_{(N\ dimensions)}} \propto$ $\mathrm{\frac{1}{(Distance)^{N-1}}}$

How shall we comprehend the above? Before that, we need to see what “flux” means. Intensity is a rather dull word to use here, because it can mean a lot of things. For our convenience, let’s take a point source emitting electromagnetic waves (it can be a star, an antenna, a molten rod, etc.). In this case, we substitute “intensity” with a new fancy name, it’s called the luminosity (though officially, it should be called the “irradiance”). It’s just a measure of the total power output (we’re talking about light energy here) from the source itself. Roughly, it determines how bright the source appears to the detector. For sun, it’s about $3.846\times 10^{26}\ \mathrm{Watts}$. But, we’ve just now learned that the intensity (No… our new candidate) luminosity, varies with distance (squared).

What if you require the amount of power incident on a certain region (over a certain area), which is quite far away from the source? Say, you’re interested in calculating the power incident on your terrace, or balcony. How would you do it? Well, we’ve got a name for the “power-incident-over-an-area-at-a-specific-distance-from-source”. It’s called the flux. So, our general expression would be…

$\mathrm{Flux} =$ $\mathrm{\frac{Luminosity}{4\pi r^2}}$

What’s that $4\pi r^2$? It’s the area of the power carrying spherical wavefront (with $r$ as its radius, which is also the distance from the source). If you’ve somehow measured the flux incident on your instrument (from a star) and you know the distance to reach the star, then you can calculate its total power output. Two guys Stefan & Boltzmann tapped this thing gently, by which they showed the purpose of this entire mystery. They used Planck’s law and mathematically showed that the flux is proportional to the fourth power of temperature. Their law modified the above expression as…

$\mathrm{L=4\pi r^2\sigma T^4}$

where $L$ is the luminosity (I’ve shortened it) and $\sigma$ represents their constant which also has numerous purposes. This law has a great significance in astronomy & astrophysics. Just by using the sufficient parameters, you can find the temperature of the source (which is how the sun’s surface temperature was determined, and later verified with blackbodies).

Anyways, that luminosity-flux thing gives the fraction of the total power you get in your radio from a broadcasting station far away. The antennae in the station emit a bulk amount of radiation. But, you get only a millionth part or something. It doesn’t mean that the signal has lost its energy by traveling. No, that’s a rude statement to make. Energy always comes in lumps – Photons, discrete packets of stuff. In this case, you obtain less number of stuff (flux is low) while some friend of you, relatively closer to the station receives more number of stuff (huge flux). The distribution of power is different. The energy still remains the same. That’s enough for today.

But, there’s more to it, which has its roots into signal processing, that they modify a thing called “carrier wave” according to the signal that carries information, etc., etc. I don’t wanna get my hands further dirty, and destroy my reputation (if I had any) by explaining that…

Now, let’s comprehend as promised before. The expression says that if there are three dimensions, the wavefronts emerge out as concentric spheres, the surfaces of the spheres represent the flux, which explains the $4\pi r^2$. For 2D, the stuff comes out as concentric circles with the circumference representing the flux, where the $4\pi r^2$ will be replaced by $2\pi r$.

I took the explanation so briefly because, it’s a wonderful phenomena and as I said, it’s very much worth it. But, this isn’t really necessary for our discussion today.

### Back to our topic…

So, we’ve come to the conclusion that the space around is helpful in distributing power, unlike materials where it’s wasted away as heat. And that too, is applicable only for high frequency EM waves, whereas it’s the other way around for low frequency waves due to high efficiency of galvanic conduction.

It should also be noted that when these electromagnetic waves incident on a metal surface, there’s still a small amount of current flow because the energy and momentum of the photon is transferred to the free electron, causing it to displace slightly. If the surface is insulator, it’s gone as heat. Do you still believe that AC signal is different from light?