# Plane Waves vs. Waves on a String

1. Oct 4, 2008

### WarPhalange

I already took 3 quarters of EM and I'm ashamed to say I didn't learn anything the final quarter, where we covered the most interesting topics. Bah.

One thing that I'm still confused about are plane waves. I understand the description of a regular sine wave on string. If you wiggle it once, you'll get a single wave traveling along the string and I can tell you the amplitude and how fast it's going and where it is.

I've also seen pictures like this depicting a photon:

Where E and B are perpendicular to one another and the photon is traveling in the direction of propagation of both. But what I don't get are the E and B fields actually. A more energetic photon will have higher frequencies for the E and B fields, correct?

But where are these fields? Let me explain. I'll take the Yellow arrows as being the E field, and I am standing at a point where E = 0. Then, as a photon zooms by me, will I gradually feel the E-field increase to a maximum and then decrease back to 0, then go negative, and finally back to 0 and then it will stay at 0 forever? Where the time it takes for this to happen is the 1/frequency of the photon (so one wavelength).

Because that picture makes it seem like the E and B fields extend infinitely in the x direction, which is where the photon is traveling.

2. Oct 5, 2008

### Staff: Mentor

Do the MIT pages really claim that this represents a single photon? Can you give a link to a page that states that? That diagram looks like the ones that you find in many textbooks describing classical electromagnetic waves.

3. Oct 5, 2008

### WarPhalange

Well then that's where my confusion stems from I suppose. That's not a single photon, but a ray of light, then?

And the EM wave will always stay the same because that's just the phase that the light is in? Or will the EM fields still oscillate?

4. Oct 5, 2008

### crazy_photon

You're correct in your original post: the more energetic photons would have E and B fields oscillating faster. Afterall, energy of a photon is defined by the frequency of oscillations of E/B fields.
MIT page is showing you a continuous monochromatic (single frequency) electromagnetic wave. You can also have pulsed electromagnetic waves (like the ones i'm sending and receiving right now through my wireless connection on my laptop). In a way you can think of these pulses of EM field as photons, then many of these pulses (with proper phase relationship maintained) would form the wave shown on MIT page.

As for plane wave - its a mathematical construct only. Albeit being useful, it does not correspond to reality, because it extends to infinity - i.e. it comes from infinite source. Here's helpful physical model (which works for this demonstration): the disturbance of water surface due to dropped pebble is circular (such wave is called a spherical wave originating from a point-source). Now, if we drop a stick into the water, you'll notice that disturbance propagates perpendicular to stick (close to the middle of the stick) but there is more circular-looking disturbance closer to the edges. Of course, if you drop an 'infinite' stick you'll get rid of these edge-effects and hence obtain your mathematical equivalent of the (unrealistic) plane wave. Close to the middle of the stick though, plane wave looks like a good approximation of the disturbance however, so that's why we use it.

Cheers

5. Oct 5, 2008

### Staff: Mentor

A plane EM wave is an idealization, but it approximately represents the field from a monochromatic point source, far away from the source. Or more practically, inside a laser beam, not too close to the edge of the beam.

At each point along the wave, the E and B fields oscillate back and forth, perpendicular to the propagation direction of the wave, in such a way that the envelope moves forward with speed c.

6. Oct 9, 2008

### Antenna Guy

At each point along the wave's propagation, the E and B fields have a value. The E and B fields oscillate with distance as the wave propagates.

I'm sure that's what you meant.

Regards,

Bill