PDA

View Full Version : Photon model

Fazor
2007-Sep-12, 07:10 PM
I've been following the Tired Light thread in ATM, but don't want to side-track that discussion.

Just some questions about photons, in regards to frequency and energy. For starters, how accurate is this visualization:
http://www.monos.leidenuniv.nl/smo/basics/images/wave.gif
in that:

1) Is the magnetic field always at a right angle to the electric field?
2) Do the fields rotate along the axis of travel, or are the arrows just representing oscillations in strength of these fields?

Obviously my knowlege is limited, so if the arrows are actually representing the orientation of fields; Aren't these fields a "figure 8" shape? So is there an implied field in the opposite dirrection of the arrows as well?

I know I get myself in trouble trying to physically visualize these things (photons), just trying to grasp it.

Fazor
2007-Sep-12, 07:50 PM
Alright, also a second question.
Looking at the included diagram (sorry was bored and love MSPaint):

If we measure the speed of a photon emitted from an object that we (the observer) are moving away from, is it correct that we would measure the speed of said photon to be slower than c, even if the photon is traveling through space at c?

I've always read it as a photon (barring any medium) will always measure at c, but that doesn't make sense to me.

edit: the attached bmp is a bit unnecessary, as words can depict question well enough. But I included it anyway.

01101001
2007-Sep-12, 08:13 PM
I've always read it as a photon (barring any medium) will always measure at c, but that doesn't make sense to me.

It's not exactly an intuitive concept. It collides with our experience with slow-paced Newtonian physics.

Wikipedia: Postulates of special relativity (http://en.wikipedia.org/wiki/Postulates_of_special_relativity)

Second postulate (invariance of c)

Light is always propagated in empty space with a definite velocity c that is independent of the state of motion of the emitting body.

korjik
2007-Sep-12, 08:27 PM
I've been following the Tired Light thread in ATM, but don't want to side-track that discussion.

Just some questions about photons, in regards to frequency and energy. For starters, how accurate is this visualization:
http://www.monos.leidenuniv.nl/smo/basics/images/wave.gif
in that:

1) Is the magnetic field always at a right angle to the electric field?
2) Do the fields rotate along the axis of travel, or are the arrows just representing oscillations in strength of these fields?

Obviously my knowlege is limited, so if the arrows are actually representing the orientation of fields; Aren't these fields a "figure 8" shape? So is there an implied field in the opposite dirrection of the arrows as well?

I know I get myself in trouble trying to physically visualize these things (photons), just trying to grasp it.

Unless you are need the quantum effects, there really is no reason to try to visualize photons as photons. Classical wave mecanics works just as well. Speaking of which, I believe your ray is the classical view, which means that you are looking at many photons making up a EM wave.

If I remember correctly, in any isotropic medium, the B and E fields are perpendicular to the propagation vector k and to each other. In other situations, the vectors may not be all perpendicular.

The fields can rotate about the direction of travel, but dont have to. This is known as circular polarization. You can have either right hand or left hand polarization depending on which way the rotation is. If you stick your thumb in the direction of motion, and curl your fingers, that gives you the direction of rotation for each type of circular polarization.

Visualizing properly, the fields should be s-curves, not figure 8 curves. A figure-8 curve would imply two opposing vectors for each component (say B in the +z and -z directions) which would sum to zero.

01101001
2007-Sep-12, 08:39 PM
Just some questions about photons, in regards to frequency and energy. For starters, how accurate is this visualization:

Accurate on what level? Obviously a photon is not black line through space with red and blue arrows growing on it. The diagram is a schematic, so it represents things symbolically.

Maybe just to get started, those field arrows might be contrasted with the concept of a vector field (http://en.wikipedia.org/wiki/Vector_field) (nice pictures there; more arrows, but it sometimes seems arrows are all we've got).

In mathematics a vector field is a construction in vector calculus which associates a vector to every point in a (locally) Euclidean space.

Vector fields are often used in physics to model, for example, the speed and direction of a moving fluid throughout space, or the strength and direction of some force, such as the magnetic or gravitational force, as it changes from point to point.

So when I try to visualize the "mechanical" operation of a photon, I tend to focus on all the 3D points around the path, and what the magnetic and charge forces are like everywhere, over time. That's pretty hard to draw, though.

1. The magnetic and electrical force fields are at right angles.
2. I just dabble, and I rarely need to think of light this way, but I think that you could say the fields are able to rotate, considering the polarization of the fields. Wikipedia: Polarization (http://en.wikipedia.org/wiki/Polarization) has some neat diagrams of various electric field vectors in 3-space, mapped onto two more familiar looking planes.

Oh, and if you go on to consider the normal incoherence of most light, composed of various frequencies, the accurate depiction quickly becomes arduous, and symbolic representation begins to be most attractive.

Fazor
2007-Sep-12, 08:49 PM
It's not exactly an intuitive concept. It collides with our experience with slow-paced Newtonian physics.

Wikipedia: Postulates of special relativity (http://en.wikipedia.org/wiki/Postulates_of_special_relativity)

Okay well the reason I asked that question is this:
If light is always propigated with a velocity of C, then why does it red/blue-shift? How does the relative movement of two bodies change a photon's frequency?

Ken G
2007-Sep-13, 01:48 AM
There are really two questions there: why does light move always at c, and why does it Doppler shift. Note that it is not necessary for light to always move at c to get a Doppler shift-- the same thing happens with sound, yet sound moves at a different speed if the air is moving. Thus the speed of light always being c really just means that the Doppler shift for light is slightly different than that for sound. But if no one is moving anywhere close to c, the difference is not terribly important, and it suffices to understand the Doppler shift for sound.

The way that works is, if the source is moving, it changes the wavelength. The source sends out one wave, and then "catches up" slightly in one direction, and sends out the next wave. Thus along the direction of motion, the wavelength is sqeezed, and along the opposite direction, the wavelength is stretched. Since the speed of the wave "forgets" the motion of the source as soon as the wave is emitted, the shorter wavelength in front causes a higher frequency to be observed (blueshift) and the longer wavelength in back causes a lower frequency to be observed (redshift).

If you are dealing with sound, and you want to move the receiver instead of the source, then you also get a shift but it is not quite the same-- it happens due to the apparent change in the speed of sound rather than any change in wavelength. If you are dealing with light, there is no "ponderable" medium, so you are not allowed to see the difference depending on whether the source is moving or the receiver-- relativity says you can't tell which is moving. So a correction is needed which is called 'time dilation' and you are off to the relativistic races, but you don't necessarily have to delve that deeply just to expect an observed frequency shift of some kind.

As for why light always moves at c, it's "because" light is a particle with no rest mass, so you cannot "catch up to it"-- it would then be at rest and could not exist!

Fazor
2007-Sep-13, 02:37 PM
Ah there's that damnedable (hope that doesn't count as profanity, i just like it 'caus its old sounding) relativity again. But basically, the speed of the emitted photon is constant and independant of the source's motion, which I understand, but the frequency is effected. And this is what is known as "time dialation", which gives me a starting point.

Fazor
2007-Sep-13, 03:02 PM
Okay, I'm going to probably use some terms wrong here, and wax theoretical that might not make sense, so hopefully this will be inteligable:

Would it be safe to say that a photon always acts and exists in the reference frame it was created in? In other words, the time of travel from creation to detection will always be the time it takes a photon at c to transverse the distance between origin and detection at the time of emission?

Por ejemplo; If we could take a detector that's far enough away from an emitter, and have it moving at a good percentage of c, we could fire a photon and measure the time between emission and detection. According to my understanding of light and relativity, the time it takes to reach the detector should be the distance between where the emitter and the detector were upon emission, divided by c, regardles of the two object's motion.

Ken G
2007-Sep-13, 03:13 PM
Would it be safe to say that a photon always acts and exists in the reference frame it was created in? No, it would be far more accurate to say that a photon loses all allegience to the frame it was created in the instant it leaves its source-- and it becomes a "citizen of the universe" that may be addressed in any reference frame at all if you simply know the transformation rules.

In other words, the time of travel from creation to detection will always be the time it takes a photon at c to transverse the distance between origin and detection at the time of emission?
That statement is true in any reference frame, but they might mean different things by "the time of emission".

According to my understanding of light and relativity, the time it takes to reach the detector should be the distance between where the emitter and the detector were upon emission, divided by c, regardles of the two object's motion.
No, you are treating that distance as though it was absolute, but it depends on reference frame, as does the time elapsed. The only thing that doesn't is the inferred speed.