Speed of Light Question.......VERY Difficult.

Factual Questions
1

The speed of light is (roughly) 186,000 MPS from point to point. This is a straight line measurement.

But if light is a waveform - and waves contain varying degrees of VERTICAL movement within them - then does not this vertical component have to travel FASTER than 186,000 MPS? A wave, after all, is NOT a straight-line phenomenon.

And since different wavelengths of light have higher or lower vertical components to them, how can ALL of them be said to travel the same distance in the same period of time?

2

I think you are confusing electromagnetic waves with waves in the water. There is no vertical movement. That is just a convention in how they are visualized on paper and in electronic instruments.

3

I think you are confusing the envelope speed with the propagation speed. Since no information resides in the envelope it doesn’t have to be bound to c.

4

Ditto mks57. Perhaps you picture the wave as a wiggly line. That wiggly line is a graph of a variable on the vertical axis versus distance on the horizontal axis. But, the variable is field strength perpendicular to the direction of travel. The variable is NOT a distance.

You could plot your checkbook balance over time, with dollars as the vertical axis and months as the horizontal axis. If one month was an inch, you wouldn’t measure one inch on the dollars axis and say it represented a month.

The point is, those wiggly lines are graphs of field strength versus distance, or field strength versus time. They aren’t two dimensional shapes in space.

5

I think your picture is a bit off, as has been noted. An electromagnetic wave in free space has an associated mafgnetic field and electric field with amplitudes at right angles to the direction of proagation, but the wave proagates at the speed of light along that direction. A wavefront has a 3-D figure to it, and, in fact, that’s an interesting part of wave propagation, but the speed of the wavefront will be the speed of light. See Wikipedia on this, or any book on Electromagnetic waves:

Different wavelengths have different frequencies as well, so that the product of wavelength and frequency, which gives the velocity, is always c. In short, the peaks and troughs of the different colors don’t line up – they can’t

Things get more interesting in materials. In amorphous solids and liquids there is dispersion, and the different wavelengths will have different velocities. In crystals things can get really interesting – where the symmetry is broken, you can get different velocities along different directions for different polarizations of the same wavelength, so that projections along different directions will have different velocities. But they’re all less than the free-space velocity c.

6

In the context of talking about light waves - I thought there was a spatial dimension to the amplitude variation of light waves, and that’s the way I was led to understand polarization, like in polarized sunglasses was I wrong?
I also remember from dusty physics corners of my mind (physics obviously isn’t my strong point) that certain speeds (such as angular or rotational speeds) could exceed the speed of light without throwing all of physics and Einstein’s work into a tizzy.

Help me understand…
ETA: Cheers, Mangetout, I’m glad I’m not the only one…

7

This is a really great explanation. Could you please expand it though, to explain how polarisation works? - because to the layman, that would seem to be support for the notion of light being a two-dimensional wave that will fit through a narrow gap in one orientation, but not another.

8

I’m confused too because if electromagnetic “waves” don’t have a 2D space how do they create interferences patterns?

9

Because the “waving” portion of the E-M wave is a field strength acting in a particular direction, not the actual movement of some artifact in that direction.

I thin the problem is a lot of physics books draw pictures of E-M waves using a series of “vector arrows” perpendicular to the presumed direction of motion. Subconsciously, students think of this as depicting a real thing swinging back and forth like the presumed position of, say, the surface of a pond with a wave travelling across it. The E-M vectors are really just describing the strength of the field at certain points (and a fixed time) according to the Maxwell equations. A vertically polarized lens doesn’t “block the slanted arrow” from getting thru, it diminishes the strength of an electric field in directions other than the preferred vertical (by dissipating some of that energy in the physical structure, e.g. by moving its charged particles around).

Napier’s checkbook analogy is a good one; to expand on it, we often talk about our checking account as “high”, “low”, “climbing”, or “falling” based on the behavior of the line on a graph.We realize, however, that these words are metaphors and do not literally describe the behavior; a “falling stock market” does not mean Wall Street has just gone over a cliff. Similarly, the pictures of vectors in textbooks are just descriptive; if we press them too hard, we are led to misconceptions.

10

This looks like as good a place for this question as any:

Let’s say I am a photon, having been freed from my static energy state I am now hurling through space at a speed of precisely c. During my trip I decide to take a little spin, say, on the Y axis relative to an observer following me also at the speed of light (the observer is also a photon). Would this make my upper surface (again relative to the parallel observer) actually exceed the speed of light? Would this motion affect my ability to travel c and slow me down? Would the upper survace traveling c+1 and the lower surface traveling c-1 mathematically average out to an overall mass moving at exactly c in the first place?

The reason I ask and think it’s relative is that, If my uneducated guesses to these questions are correct, the spin of a particle (especially one with no mass) is irrelavent to it’s speed as expressed by momentum (but definatly relavent to speed as expressed in relation to the vector, but that is a different story). Therefore a particle of mass 0 traveling at the speed of light and spinning could spin as fast as c while still traveling c and have no effects whatsoever.

This means that the perceived verticle motion of light (which, as I understand it, is really more of a simultanious lateral motion rather than a verticle one, but that would be much more of a bitch trying to express on graph paper and mentally visualize) can have any speed up to c without affecting the forward momentum of the object moving at the speed of c.

Or not, I’m kind of just guessing here.

What do you think?

11

I think the problem you are having is that you are trying to reason out the behavior of a photon based upon everyday experience. This is always an error with fundamental particles, which are not made of stuff and therefore cannot just be considered really little tiny balls. Photons are bosonic fundamental force carriers; that is to say, they have integer spin and do not obey the Pauli exclusion principle (i.e. two or more can occupy the same quantum state in the same locus simultaneously, which isn’t strictly relevant but is interesting). Photons don’t have a “surface” to measure a spin from. They do have a property that physicists call “spin” but it is not related to normal kinetic angular momentum (although it is, in a sense, mathematically similar). Even more confusingly, although photons interact with other things only in given energy quanta, like a particle, they also interact in such a way to create wavelike interference, hence the so-called “wave/particle duality.” In fact, there is no conflict or paradox; it’s just that photons aren’t really particles like we see under a microscope; they’re fundamentally different?

Does this answer your question, or at least make it muddier than a campaign finance reform platform?

Stranger

12

I have a question for the OP–why do you think this question is “VERY Difficult”? Don’t you know the members of the SDMB live for these types of questions? They eat these questions for lunch! In fact, some posters can answer as many as six questions of this type before breakfast! :slight_smile:

13

I don’t disbelieve you but this raises more questions. If I’m interpreting you correctly then the photon can be thought to be moving in a straight line (with the caveat that it’s always dicey to be thinking of quantum objects as macroscopic objects) and the wave field is emanating from the photon. How is the field strength propagated? (Maybe a better question is ‘how is it measured?’)

14

I love the smell of photons in the morning…

…Smells like…

…duality.

15

mmmm, wavicles.

16

This is actually a bit of a tensies snack; lunch would be more like, “How is the cat simultaneously alive and dead?”, “Why won’t my hand go through my desk?”, or “If the Red Sox can win the World Series why can’t we build a car that runs on Barack Obama’s good intentions?”

This is a bit easier; basically unteaching what one is presented by superficial pop science and replacing it with a somewhat more accurate notion of what is really occurring.

Stranger

17

Field strength is measured in terms of its interaction with charged particles (most typically an electron). This, of course, results in destruction, or at least alteration, of the photon.

As far as propagation…back in the 19th Century it was assumed to be due to waves in a medium called the luminiferous aether. Unfortunately, attempts to measure the properties and flow of this media were anomolous–it seemed to be moving along with the observer, giving the same properties at all speeds and orientations–and a Swiss patent clerk decided it was all bugger and came up with a totally new theory that didn’t require any kind of luminiferous aether. (Despite popular notions, he wasn’t the only one–Henri Poincairé reached the same conclusion independently, and perhaps even before–but Einstein was the first to publish widely on the topic.) The photon is self-propagating; the field is itself the photon, which provides its own medium, and interacts with itself (demonstratively in quantum electrodynamics).

That begs the question, of course, of “What is a photon?” which we can only answer in terms of its observed properties. It is possible that a more fundamental field–the Higgs field–and its associated force carrier (the Higgs boson) may govern all interactions, including self-interactions. It’s also possible that the fundamental order of the universe is governed by a game of hyperdimensional croquet with flamingoes as the mallets and hedgehogs as the balls. Wrap your prefrontal association complex around that.

Stranger

18

Ok, dictionary.com has enlightened me that “hurling” really is a synonym for “hurtling”. But I gotta say, when I first read that I got a mental image of a photon, barfing out chunks of quarks . . .

19

Lotta great questions and I want to interfere constructively with many of them.

Three big statements that will send some of the questions into new directions all at once: Photons aren’t balls and they don’t have surfaces with surface velocities with spin. And, observers can’t move at c. And, there is no real wave-particle duality. Well, let me reword that last one: when we refer to a wave-particle duality, though we may say radiation sometimes behaves as waves and sometimes behaves as particles, it would be more correct to say that radiation always does exactly the same thing, and we have two ideas, “waves” and “particles”, such that sometimes one seems more correct and sometimes the other does, but the problem is with the ideas.

Now, then, polarization. The drawings that illustrate how a wavy shaped sturdy rigid bent wire would fit through a wire grating are very irritating. They share several features with the real situation but also with the incorrect picture of two dimensional wave shapes being involved in polarization. This is most misleading. In polarized light there is some kind of order or organization to the direction in which the electromagnetic field is changing. Most simply, linearly polarized light, like what you see through Polaroid sunglasses, has all the field fluctuation happening in the same direction.

In radio, charge moving back and forth in an antenna creates the electromagnetic field radiating away from it. If the charge moves up and down, like it does in an AM radio station whose entire tower is a conductor, then the radio emission is vertically polarized.

When light goes through a transparent solid, it generally actually behaves as vibrations in the electrons and to a much lesser extend the nuclei of the atoms of the structure. This is why the nature of the solid influences the transmission of light. The electrons are very light and their connection to the rest of the solid gives them a high stiffness to mass ratio so the vibrations pass very rapidly. When light shines perpendicularly on a glass face, the vibrations it creates also create their own light headed back the other way, which is why you see a bit of a reflection. If you start angling the light further and further from perpendicular, at some point the vibrations in the surface caused by field strength variation in the plane of the incoming and reflecting beams will be aligned with the direction of the reflected light beam; that is, they will be trying to push and pull on the reflected light beam, if you will, and won’t have any transverse motion. At further angles their transverse motion is in the opposite transverse motion for the reflected beam. At the angle where we go through this minimum, none of the light waves whose field direction are in the plane of the beams have a transverse component in the reflected beam, and they aren’t reflected at all, and so only the perpendicularly oriented field waves are part of the reflected beam, which makes it polarized.

Hmm. This is hard to say well, and even harder when I should be going and am worried about the SDMB timing out. Maybe I’ll try again when there’s more time.

20

You’ve definitely amplified the intensity of my curiosity. I’ll have to ponder your explanation of polarization, but I’m glad to have been dissuaded of the notion that those diagrams were literally correct.