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dhd40
2008-Jan-05, 03:54 PM
Electromagnetic waves are transverse waves that travel at the speed of light

Question 1: If the EM wave travels parallel to the x-direction, its transverse components lie in the y-z-plane. Right or wrong?

Question 2: If the distance travelled in x-direction is measured in meters, what is the amplitude of the transverse components, measured in meters?

Iīm not interested in the energy of the electric or magnetic field components. Iīm interested in the spacial components in the y-z-plane. If there are no such spacial components, why do we call the EM waves “transverse” waves? If there are no such spacial components, are the spacial amplitudes in y- and z-direction zero? Is this allowed (singularity)? How far (in meters) do the electric and magnetic field components stretch out in y-z-direction?

OMG, Iīm awaiting a lot of :lol: :silenced: :cry: :( :mad: :doh: etc

Jeff Root
2008-Jan-05, 10:43 PM
I've asked this question before. Maybe here on BAUT.

The 'standard' coordinates assign the Z axis to the direction of travel,
with X and Y axes in the planes of the electric and magnetic fields.

-- Jeff, in Richfield MN

KaiYeves
2008-Jan-05, 11:53 PM
(Title)
Do they grow hair?

mr obvious
2008-Jan-06, 03:12 AM
Electromagnetic waves are transverse waves that travel at the speed of light

Question 1: If the EM wave travels parallel to the x-direction, its transverse components lie in the y-z-plane. Right or wrong?

Question 2: If the distance travelled in x-direction is measured in meters, what is the amplitude of the transverse components, measured in meters?

Iīm not interested in the energy of the electric or magnetic field components. Iīm interested in the spacial components in the y-z-plane. If there are no such spacial components, why do we call the EM waves “transverse” waves? If there are no such spacial components, are the spacial amplitudes in y- and z-direction zero? Is this allowed (singularity)? How far (in meters) do the electric and magnetic field components stretch out in y-z-direction?

OMG, Iīm awaiting a lot of :lol: :silenced: :cry: :( :mad: :doh: etc

Clarification requested: You mention that you are not interested in the enery of the EM field components. However, the EM field components are the components that are perpendicular to the direction of the EM wave. Thus, your question appears to be asking for something that is itself being excluded.

Basically the EM wave travels in x (by your coordinate choice). This means the E field is in y and the B field is in z (keeping in mind the designation of y and z are arbitrary; here they are chosen to keep things simple, I could randomly rotate the y and z axes about x and then would have to introduce sine and cosine terms into the E/B fields). Because the fields' fluctuations are perpendicular to the wave direction, they are considered transverse waves. The amplitude of the fluctuation for each field is the strength of the field and are thus not measured in meters. Interestingly, in a vacuum, the (peak) amplitude of the E field divided by the amplitude of the B field is c, the speed of light.

I have taken some of this material from Halliday/Resnick/Walker 6th Ed.

Ken G
2008-Jan-06, 07:15 AM
Question 1: If the EM wave travels parallel to the x-direction, its transverse components lie in the y-z-plane. Right or wrong?Right-- but be careful, "lying in the plane" for a vector just specifies its direction-- the vector itself does not occupy any space, it is not a spatial vector (it is the direction and magnitude of a field that exists independently at each point, classically).


Question 2: If the distance travelled in x-direction is measured in meters, what is the amplitude of the transverse components, measured in meters?
You cannot measure the amplitude of an EM wave in meters, it is not a displacement like a water wave. It is an EM field, so it's like an arrow at every point whose length has nothing to do with spatial dimensions at all, only its direction does.

dhd40
2008-Jan-06, 04:11 PM
I've asked this question before. Maybe here on BAUT.

The 'standard' coordinates assign the Z axis to the direction of travel,
with X and Y axes in the planes of the electric and magnetic fields.

-- Jeff, in Richfield MN

Which answers did you get? Where can I find them?

dhd40
2008-Jan-06, 04:12 PM
(Title)
Do they grow hair?

?sorry?

dhd40
2008-Jan-06, 04:28 PM
Clarification requested: You mention that you are not interested in the enery of the EM field components. However, the EM field components are the components that are perpendicular to the direction of the EM wave.
Yes, the energy of the EM field components can be calculated. My thinking was: How far (in nm, cm, m, etc) do they stretch out in the YZ-plane?


(snip)keeping in mind the designation of y and z are arbitrary
true

Because the fields' fluctuations are perpendicular to the wave direction, they are considered transverse waves.
Yes. But how FAR (nm, mm, etc) do they fluctuate?


The amplitude of the fluctuation for each field is the strength of the field and are thus not measured in meters.(snip)
Thatīs why I said that Iīm not interested in the energy of the EM components

Neverfly
2008-Jan-06, 04:43 PM
(Title)
Do they grow hair?

?sorry?

It would be the silliest question...

dhd40
2008-Jan-06, 04:49 PM
(snip)
You cannot measure the amplitude of an EM wave in meters, it is not a displacement like a water wave. It is an EM field, so it's like an arrow at every point whose length has nothing to do with spatial dimensions at all, only its direction does.

I think I understand what you mean, and I agree with you in this. But AMPLITUDE in your sense is energy (or something like this), wheras Iīm not thinking of an energy amplitude but a spatial (is that wording correct at all?) amplitude.
Perhaps this example can clarify: If you drop a stone into your pool the resulting water waves will have some energy, depending on masses (number of water molecules) moving up and down (z-direction), how FAR they move up and down, how fast they move up and down (thereīs actually nothing moving in the xy-direction, if I understand water waves correctly, at least not if your pool is deep and wide).

Yet, there is a spatial amplitude, in addition to the energy amplitude.
This doesnīt hold true for EM waves?

Ken G
2008-Jan-06, 05:03 PM
I think I understand what you mean, and I agree with you in this. But AMPLITUDE in your sense is energy (or something like this), wheras Iīm not thinking of an energy amplitude but a spatial (is that wording correct at all?) amplitude.But that's just my point-- an EM wave does not have spatial amplitude, it's just not the kind of wave whose amplitude is spatial (material waves have spatial amplitudes, but those are just material analogs to what a wave really is. I suppose gravity waves and magnetic Alfven waves have a spatial amplitude too, and they are not really material waves, so amplitude is a subtle issue, and I think that's what underlies your question. But light waves don't conserve the density of anything, and don't represent anything with a spatial amplitude).


Perhaps this example can clarify: If you drop a stone into your pool the resulting water waves will have some energy, depending on masses (number of water molecules) moving up and down (z-direction), how FAR they move up and down, how fast they move up and down (thereīs actually nothing moving in the xy-direction, if I understand water waves correctly, at least not if your pool is deep and wide).Again, light waves are different from water waves in this respect.


Yet, there is a spatial amplitude, in addition to the energy amplitude.
This doesnīt hold true for EM waves?Right, nor for the "wave functions" of quantum mechanics.

dhd40
2008-Jan-06, 05:12 PM
But that's just my point-- an EM wave does not have spatial amplitude, it's just not the kind of wave whose amplitude is spatial


Thatīs how I learned it, and I still believe in it. BUT, if so, why/how does an EM wave (a single photon!) "feel" the width of a slit, let alone a double-slit?

mr obvious
2008-Jan-06, 05:56 PM
Thatīs why I said that Iīm not interested in the energy of the EM components

But the EM components are those that are fluctuating. There is no physical (as far as I know) object moving back and forth in the transverse direction; illustrations in texts are those of fields. So I am afraid I must reiterate that you are asking for something and then excluding the correct response from the answer.

For your comment on slits, you may be confusing the transverse "amplitude" of the EM components with wavelength, which is a measure of energy of the EM wave. If you are thinking about interference patterns with two slits, those are caused by the pathlengths of photons from individual slits resulting in the photons being out of phase at the location of a dark band.

Also I might note that your water wave analogy is not appropriate for EM since EM does not require a medium for propagation.

KaiYeves
2008-Jan-06, 06:20 PM
It would be the silliest question...
Correct, that is what I meant.

Ken G
2008-Jan-06, 07:56 PM
Thatīs how I learned it, and I still believe in it. BUT, if so, why/how does an EM wave (a single photon!) "feel" the width of a slit, let alone a double-slit?
The vector field exists in space, but the amplitude vectors do not. It's all about the difference between one vector and a field of vectors. That's what mr obvious is saying also.

dhd40
2008-Jan-06, 09:02 PM
But the EM components are those that are fluctuating. There is no physical (as far as I know) object moving back and forth in the transverse direction
True, no Aether


For your comment on slits, you may be confusing the transverse "amplitude" of the EM components with wavelength, which is a measure of energy of the EM wave.
Well, Iīm aware of E=h*f.
Let me try again: If an EM wave (travelling in x-direction) has no spatial components in y,z-direction, why does its behaviour (e.g. interference) depend on the width and distance of slits? If it (or the fields) has (have) zero extension in y,z-direction, where does it know from that slit-walls exist (unless it hits them)

mr obvious
2008-Jan-06, 10:33 PM
True, no Aether
Well, Iīm aware of E=h*f.
Let me try again: If an EM wave (travelling in x-direction) has no spatial components in y,z-direction, why does its behaviour (e.g. interference) depend on the width and distance of slits? If it (or the fields) has (have) zero extension in y,z-direction, where does it know from that slit-walls exist (unless it hits them)

The behavior of the interference patterns depend on the wavelength, and not on any spatial components in the y/z direction. If you are willing to accept that I can generate a (uniform, say) E field in the y direction and cancel it out by generating a second, uniform E field in the -y direction then you can see that when a EM wave (whose E field fluctuates in the z-direction) meets another there is the possibility that at a given location, the E fields will cancel. Repeat for the B field (which fluctuates in phase with the E field in a given coherent EM wave). When the E/B fields in one wave cancel another, you get a dark band (destructive interference); when they add, you get a bright band (constructive).

When you pass a beam through two slits, you are effectively splitting a beam into two different EM waves.
......... A
.___________________________________
............o
..............o...o
................o.....o
...................o.......o
._____________ _____ __________________
.....................1........2
(Ignore the dots, the beam is represented by the o's.)
As you can see, I hope, the distance from slit 1 to point A is different from the distance from slit 2 to point A. The difference in that distance will be some multiple of the wavelength of the light used (I'm assuming monochromatic illumination). If the multiple is in phase, you get a bright band. If the multiple is out of phase, you get a dark band. As you move point A from left to right, you can see that the difference in distance will also change continuously, so you get the interference pattern. At a point where A is midway between points 1 and 2 (horizontally speaking - it's still on the board in back) you get a bright pattern because the difference in distances is zero.

So, the interference pattern depends on the wavelength, and wavelength depends on the z-directional variation in the E/B field strength. A photon does not have to "sense" how wide (laterally) the slit is. It just passes through (if it doesn't hit a non-slit region) and then interferes, or not, with any other fields in the area.

However, this is not to say that transverse field amplitudes are not relevant anywhere. Polarization depends on the orientation of the E/B fields, but that is not a physical limitation (e.g., that is not a matter of not having 'space' to fluctuate but rather of selective absorption of fields in a given orientation). Before we jump into that (assuming you want to), do you understand what I've written here?

Ken G
2008-Jan-07, 02:43 AM
Let me try again: If an EM wave (travelling in x-direction) has no spatial components in y,z-direction, why does its behaviour (e.g. interference) depend on the width and distance of slits? If it (or the fields) has (have) zero extension in y,z-direction, where does it know from that slit-walls exist (unless it hits them)

As I said, the EM wave you are describing does occupy space, but its amplitude has no spatial component. You can even use a sound wave to understand that-- the wave amplitude "lives" in the longitudinal direction, yet you get a double-slit diffraction pattern when sound waves encounter side-by-side slits. The spatial interference is there because the wave expands to fill the space. In other words, it is the wave itself that fills the space and encounters the slits, each wave amplitude does not need to fill space, that's irrelevant.

dhd40
2008-Jan-07, 02:50 PM
The behavior of the interference patterns depend on the wavelength, and not on any spatial components in the y/z direction. If you are willing to accept that I can generate a (uniform, say) E field in the y direction and cancel it out by generating a second, uniform E field in the -y direction then you can see that when a EM wave (whose E field fluctuates in the z-direction) meets another there is the possibility that at a given location, the E fields will cancel. Repeat for the B field (which fluctuates in phase with the E field in a given coherent EM wave). When the E/B fields in one wave cancel another, you get a dark band (destructive interference); when they add, you get a bright band (constructive).

When you pass a beam through two slits, you are effectively splitting a beam into two different EM waves.
......... A
.___________________________________
............o
..............o...o
................o.....o
...................o.......o
._____________ _____ __________________
.....................1........2
(Ignore the dots, the beam is represented by the o's.)
As you can see, I hope, the distance from slit 1 to point A is different from the distance from slit 2 to point A. The difference in that distance will be some multiple of the wavelength of the light used (I'm assuming monochromatic illumination). If the multiple is in phase, you get a bright band. If the multiple is out of phase, you get a dark band. As you move point A from left to right, you can see that the difference in distance will also change continuously, so you get the interference pattern. At a point where A is midway between points 1 and 2 (horizontally speaking - it's still on the board in back) you get a bright pattern because the difference in distances is zero.

Up to here I understand everything (or so I think)


A photon does not have to "sense" how wide (laterally) the slit is. It just passes through (if it doesn't hit a non-slit region) and then interferes, or not, with any other fields in the area.

Now itīs getting tricky. IIRC (some 45 years ago), you have to adjust width and distance of the slits in order to get interference. But I may be completely wrong
And how about the one-at-a-time-photon double-slit experiment? It (the photon) interferes with its own EM fields (as I understand it). If you would erect a thin wall between the two slits, perpendicular to the slit-plane, at which height of the wall would interference disappear? (remember: No two photons at the same time!). Would it make a difference, if the wall was transparent or opaque, or if the wall would be erected on the detector-oriented side of the slit-plane?


However, this is not to say that transverse field amplitudes are not relevant anywhere. Polarization depends on the orientation of the E/B fields, but that is not a physical limitation (e.g., that is not a matter of not having 'space' to fluctuate but rather of selective absorption of fields in a given orientation).

Actually, "polarization" would have been my next question. But your description (absorption of fields...) seems to explain this.

dhd40
2008-Jan-07, 03:16 PM
As I said, the EM wave you are describing does occupy space, but its amplitude has no spatial component. You can even use a sound wave to understand that-- the wave amplitude "lives" in the longitudinal direction, yet you get a double-slit diffraction pattern when sound waves encounter side-by-side slits. The spatial interference is there because the wave expands to fill the space. In other words, it is the wave itself that fills the space and encounters the slits, each wave amplitude does not need to fill space, that's irrelevant.

I understand this in connection with sound waves. But how about a very narrow laser beam. If you have a sensitive light detector and move it towards the beam in a direction perpendicular to the beamīs direction, at which distance from the laser beam will you get a signal?

Ken G
2008-Jan-07, 03:22 PM
I understand this in connection with sound waves. But how about a very narrow laser beam. If you have a sensitive light detector and move it towards the beam in a direction perpendicular to the beamīs direction, at which distance from the laser beam will you get a signal?

That depends on the width of the beam-- it's no different from if the beam is passing through dusty air so you can actually see it as a cylinder of a certain width. Note that none of that has any connection to the amplitude of the EM wave-- you could turn the laser brightness up and down, varying the amplitude of the EM field oscillations, without changing the width of that visible "cylinder". In fact, your question is a good way to help understand the difference between the spatial extent of a wave, and the amplitude of a wave, which are very different things but can get a bit confused, especially for transverse waves.

dhd40
2008-Jan-07, 08:09 PM
That depends on the width of the beam-- it's no different from if the beam is passing through dusty air so you can actually see it as a cylinder of a certain width.
Actually, I was thinking of a laser beam in 100% "clean" vacuum: No scattering from dust or gas.
Assuming a diameter of the cylinder of 0,7297353 mm (:) ), how close to the wall of the cylinder do you have to move the detector?


Note that none of that has any connection to the amplitude of the EM wave-- you could turn the laser brightness up and down, varying the amplitude of the EM field oscillations, without changing the width of that visible "cylinder".

Hmmm, yes. Have to think about this for a while

mr obvious
2008-Jan-08, 12:01 AM
Now itīs getting tricky. IIRC (some 45 years ago), you have to adjust width and distance of the slits in order to get interference. But I may be completely wrong
And how about the one-at-a-time-photon double-slit experiment? It (the photon) interferes with its own EM fields (as I understand it). If you would erect a thin wall between the two slits, perpendicular to the slit-plane, at which height of the wall would interference disappear? (remember: No two photons at the same time!). Would it make a difference, if the wall was transparent or opaque, or if the wall would be erected on the detector-oriented side of the slit-plane?


You need the slits to be sufficiently narrow that you are not unleashing a ton of photons on the back plate (the place where you do the readout). If the slits are too wide, it becomes difficult to see interference patterns because of all the excess photons bouncing around. The distance between slits can be adjusted to conform to light sources of various sizes. Obviously if the slit is too wide then you'll only get light through one slit.

A single-photon two-slit experiment departs a bit from classical EM into the realm of quantum optics. The best 'model' I've heard is that while you are not observing a photon, it's wave properties dominate. Thus, if a photon is capable of (e.g., not forbidden from) passing through either of two slits, it actually passes through both. When the photon hits the back plate, the waveform collapses and the photon is 'forced' to show up as a single hit. However, because the waveform of the photon was able to interfere with itself, there are some locations that the photon will not show up on the back plate. Someone more versed in QO/QM can correct me or explain better, perhaps.

For your detector question, your precision is rather impressive given that the last '3' is in the Angstrom range. But I'm not sure what you're getting at - for the detector to detect the photon, the photon needs to be absorbed by something that's acting as a detector (alternatively, the photon needs to affect something that then affects something else acting as a detector, and so forth). Additionally, how the detector absorbs the photon will depend on the arrangement of the surface of the detector elements. This causes all sorts of problems because of uncertainty - the 'detecting' electron say) is not fixed in space but is itself a probability cloud. Thus, whether the photon interacts with, and is absorbed by, the electron is a matter of chance and is not a yes/no a priori determination.

I think some confusion may stem from the use of vectors to illustrate fields. There is no physical manifestation of length. For example, consider a point charge q, represented by a + :
q...............A
+..............->

At a location A from the point charge,there is a electric field 'pointing' right. However, you do not actually see anything like that in real life (a tiny EM arrow) - it's just a convention we use. Your original question is analogous to asking how long such an arrow is, if I tell you the exact magnitude of the charge q. Your detector question is analogous to, how close do I have to have a detector to point A to detect some contribution from the field at A. With no spatial component, the field at A is entirely localized at A, so the 'length' of the arrow is not a measurable length in meters, and you need to be on point A to measure the field at A. If you move 10e-200 meters away from A, you are no longer measuring the field at A (theoretically speaking).

Ken G
2008-Jan-08, 03:20 AM
Actually, I was thinking of a laser beam in 100% "clean" vacuum: No scattering from dust or gas. The dust I had in mind has no effect on the beam other than letting you see it.

Assuming a diameter of the cylinder of 0,7297353 mm (:) ), how close to the wall of the cylinder do you have to move the detector?
0,7297353 mm, though the last few decimals are meaningless because you can't resolve it better than its wavelength. Regardless of amplitude (laser brightness).

dhd40
2008-Jan-08, 03:30 PM
You need the slits to be sufficiently narrow that you are not unleashing a ton of photons on the back plate (the place where you do the readout). If the slits are too wide, it becomes difficult to see interference patterns because of all the excess photons bouncing around.

Interesting, I wasnīt aware of this


For your detector question, your precision is rather impressive given that the last '3' is in the Angstrom range.

QEDīs precision is impressive, indeed (fine-structure constant0.00729735253 = 1/137,035977..)

Thanks for all the other detailed explanations, I really appreciate this

dhd40
2008-Jan-08, 03:51 PM
0,7297353 mm, though the last few decimals are meaningless because you can't resolve it better than its wavelength. Regardless of amplitude (laser brightness).

Yes, of course. (see my response to mr obvious)
By now, I think we should close this thread, because any further details from your (and mr obviousī) expertise will not be very helpful for a layman in this field.
For example when you say "the wave expands to fill the space", or when mr obvious says "the waveform collapses and the photon is 'forced' to show up as a single hit" I feel that there must be some truth behind it, and I heard and read this before. Still, itīs so counterintuitive. Itīs difficult for someone who is not trained in this to grasp it, but thatīs not much different from curved space-time, entanglement, etc, etc

Thanks to all for your patience with a silly question(er)

Jeff Root
2008-Jan-09, 12:39 AM
The question I asked was "Do photons have extension in space
perpendicular to the direction of motion, and if so, how much?"

I have not seen an answer.

-- Jeff, in Richfield, MN

Ken G
2008-Jan-09, 02:33 PM
Here's your answer: the wave function of the photon tells you this. So you have to answer your own question, when you specify what you mean by a "photon". One meaning is a plane wave solution to the wave equation, which would have infinite extent, so is merely an idealization. If you take that meaning for "photon", your question takes your own idealization too seriously.

mr obvious
2008-Jan-09, 11:33 PM
The question I asked was "Do photons have extension in space
perpendicular to the direction of motion, and if so, how much?"

I have not seen an answer.

-- Jeff, in Richfield, MN

If I may add more uncertainty to Ken G's comment, many (all?) very small particles do not have simple sizes (e.g., a rest radius). For electrons, for example, see
http://en.wikipedia.org/wiki/Classical_electron_radius
You'll note that only under classical assumptions can a size be calculated. Indeed, the wiki article postulates that electrons are generally considered point particles. I'd wager that photons are similar.

This is because once you know something about the velocity of the particle (which for light isn't much of a mystery) its position becomes uncertain. Therefore, you are asking, what is the size of a particle whose location cannot be determined (e.g., a particle whose existence is a probability cloud)? You see the difficulty in answering that question.

Another complication (by no means unrelated) can be seen by rephrasing the question: how small can you make a hole such that a photon can still pass through? Given that quantum tunneling (assuming it holds for photons) suggests that even with no hole, photons can appear on the other side, one can see why no one is particularly enthusiastic about performing this experiment.

Again, apologies if my quantum explanation is weird.