Does light intensity oscillate really fast since it is a wave? The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Have we directly observed the electric component to EM waves?How does light oscillate?Light Intensity vs Visibility/Brightness?How does distance affect light intensity?Maximum light intensity?What does “intensity of light” mean?Why does the intensity of light decrease as you move away from a particular point (described in question)?Wave optics, intensity distributionRelationship between intensity and amplitude of light waveFeynman’s Treatment of an Opaque Wallwhy does the intensity of light does not vary with time in youngs double slit experiment
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Does light intensity oscillate really fast since it is a wave?
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Have we directly observed the electric component to EM waves?How does light oscillate?Light Intensity vs Visibility/Brightness?How does distance affect light intensity?Maximum light intensity?What does “intensity of light” mean?Why does the intensity of light decrease as you move away from a particular point (described in question)?Wave optics, intensity distributionRelationship between intensity and amplitude of light waveFeynman’s Treatment of an Opaque Wallwhy does the intensity of light does not vary with time in youngs double slit experiment
$begingroup$
If you shine light on a wall, what will be seen is a "patch" with constant intensity. However, if light is viewed as a wave, then it is oscillations of the electromagnetic field changing from 0 to the amplitude and back really fast. So my question is, if I were able to look at the world at extreme slow motion, a quadrillion times slower or so, and I shined a beam of light at a wall, will I see a "patch" with oscillating intensity, with maximum brightness at the peaks of the wave and minimum when the field is 0? If so, is the constant brightness seen normally just our puny mortal eyes capturing only the average of this oscillating brightness?
visible-light waves electromagnetic-radiation intensity
edited Apr 9 at 7:31
Ruslan
9,85243173
asked Apr 9 at 6:17
AdgornAdgorn
6415
New contributor
Adgorn is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
If you shine light on a wall, what will be seen is a "patch" with constant intensity. However, if light is viewed as a wave, then it is oscillations of the electromagnetic field changing from 0 to the amplitude and back really fast. So my question is, if I were able to look at the world at extreme slow motion, a quadrillion times slower or so, and I shined a beam of light at a wall, will I see a "patch" with oscillating intensity, with maximum brightness at the peaks of the wave and minimum when the field is 0? If so, is the constant brightness seen normally just our puny mortal eyes capturing only the average of this oscillating brightness?
visible-light waves electromagnetic-radiation intensity
edited Apr 9 at 7:31
Ruslan
9,85243173
asked Apr 9 at 6:17
AdgornAdgorn
6415
New contributor
Adgorn is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
4
$begingroup$
Related Have we directly observed the electric component to EM waves?
$endgroup$
– Farcher
Apr 9 at 7:03
3
$begingroup$
What you are talking about is the amplitude. Intensity usually means the average power and that would be constant. And you can't talk about seeing patches in those timescales but observing electric field change.
$endgroup$
– Džuris
Apr 9 at 10:23
2
$begingroup$
If you were able to look at the world at extreme slow motion, a quadrillion times slower or so, then you would have to be using some kind of "vision" that would not, in any way, resemble how your vision actually works.
$endgroup$
– Solomon Slow
Apr 9 at 13:14
$begingroup$
Possible duplicate of Have we directly observed the electric component to EM waves?
$endgroup$
– ahemmetter
Apr 9 at 14:07
add a comment |
$begingroup$
If you shine light on a wall, what will be seen is a "patch" with constant intensity. However, if light is viewed as a wave, then it is oscillations of the electromagnetic field changing from 0 to the amplitude and back really fast. So my question is, if I were able to look at the world at extreme slow motion, a quadrillion times slower or so, and I shined a beam of light at a wall, will I see a "patch" with oscillating intensity, with maximum brightness at the peaks of the wave and minimum when the field is 0? If so, is the constant brightness seen normally just our puny mortal eyes capturing only the average of this oscillating brightness?
visible-light waves electromagnetic-radiation intensity
edited Apr 9 at 7:31
Ruslan
9,85243173
asked Apr 9 at 6:17
AdgornAdgorn
6415
New contributor
Adgorn is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
If you shine light on a wall, what will be seen is a "patch" with constant intensity. However, if light is viewed as a wave, then it is oscillations of the electromagnetic field changing from 0 to the amplitude and back really fast. So my question is, if I were able to look at the world at extreme slow motion, a quadrillion times slower or so, and I shined a beam of light at a wall, will I see a "patch" with oscillating intensity, with maximum brightness at the peaks of the wave and minimum when the field is 0? If so, is the constant brightness seen normally just our puny mortal eyes capturing only the average of this oscillating brightness?
visible-light waves electromagnetic-radiation intensity
visible-light waves electromagnetic-radiation intensity
edited Apr 9 at 7:31
Ruslan
9,85243173
asked Apr 9 at 6:17
AdgornAdgorn
6415
New contributor
Adgorn is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Apr 9 at 7:31
Ruslan
9,85243173
asked Apr 9 at 6:17
AdgornAdgorn
6415
New contributor
Adgorn is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Apr 9 at 7:31
Ruslan
9,85243173
edited Apr 9 at 7:31
Ruslan
9,85243173
edited Apr 9 at 7:31
Ruslan
9,85243173
9,85243173
asked Apr 9 at 6:17
AdgornAdgorn
6415
New contributor
Adgorn is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Apr 9 at 6:17
AdgornAdgorn
6415
asked Apr 9 at 6:17
AdgornAdgorn
6415
6415
New contributor
Adgorn is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Adgorn is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Adgorn is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
4
$begingroup$
Related Have we directly observed the electric component to EM waves?
$endgroup$
– Farcher
Apr 9 at 7:03
3
$begingroup$
What you are talking about is the amplitude. Intensity usually means the average power and that would be constant. And you can't talk about seeing patches in those timescales but observing electric field change.
$endgroup$
– Džuris
Apr 9 at 10:23
2
$begingroup$
If you were able to look at the world at extreme slow motion, a quadrillion times slower or so, then you would have to be using some kind of "vision" that would not, in any way, resemble how your vision actually works.
$endgroup$
– Solomon Slow
Apr 9 at 13:14
$begingroup$
Possible duplicate of Have we directly observed the electric component to EM waves?
$endgroup$
– ahemmetter
Apr 9 at 14:07
add a comment |
4
$begingroup$
Related Have we directly observed the electric component to EM waves?
$endgroup$
– Farcher
Apr 9 at 7:03
3
$begingroup$
What you are talking about is the amplitude. Intensity usually means the average power and that would be constant. And you can't talk about seeing patches in those timescales but observing electric field change.
$endgroup$
– Džuris
Apr 9 at 10:23
2
$begingroup$
If you were able to look at the world at extreme slow motion, a quadrillion times slower or so, then you would have to be using some kind of "vision" that would not, in any way, resemble how your vision actually works.
$endgroup$
– Solomon Slow
Apr 9 at 13:14
$begingroup$
Possible duplicate of Have we directly observed the electric component to EM waves?
$endgroup$
– ahemmetter
Apr 9 at 14:07
4
4
$begingroup$
Related Have we directly observed the electric component to EM waves?
$endgroup$
– Farcher
Apr 9 at 7:03
$begingroup$
Related Have we directly observed the electric component to EM waves?
$endgroup$
– Farcher
Apr 9 at 7:03
3
3
$begingroup$
What you are talking about is the amplitude. Intensity usually means the average power and that would be constant. And you can't talk about seeing patches in those timescales but observing electric field change.
$endgroup$
– Džuris
Apr 9 at 10:23
$begingroup$
What you are talking about is the amplitude. Intensity usually means the average power and that would be constant. And you can't talk about seeing patches in those timescales but observing electric field change.
$endgroup$
– Džuris
Apr 9 at 10:23
2
2
$begingroup$
If you were able to look at the world at extreme slow motion, a quadrillion times slower or so, then you would have to be using some kind of "vision" that would not, in any way, resemble how your vision actually works.
$endgroup$
– Solomon Slow
Apr 9 at 13:14
$begingroup$
If you were able to look at the world at extreme slow motion, a quadrillion times slower or so, then you would have to be using some kind of "vision" that would not, in any way, resemble how your vision actually works.
$endgroup$
– Solomon Slow
Apr 9 at 13:14
$begingroup$
Possible duplicate of Have we directly observed the electric component to EM waves?
$endgroup$
– ahemmetter
Apr 9 at 14:07
$begingroup$
Possible duplicate of Have we directly observed the electric component to EM waves?
$endgroup$
– ahemmetter
Apr 9 at 14:07
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
You would need a coherent beam, because in waves it is not only intensity but also phase that makes a difference. In an incoherent beam, as sunlight, you would not get any changes in this thought experiment, because the average intensity would hold even at wavelength distances.
In a coherent laser beam you should see in your thought experiment what is shown towards the end of this youtube video,(at 2' 09") the sinusoidal in time pattern of impinging intensity . The video draws the oscillating electric field E, and the intensity left on the screen will be $=E^2$, which will also be oscillating in time.
After all, mathematics allows us to materialize thought experiments as this one.
answered Apr 9 at 6:28
anna vanna v
162k8153456
$endgroup$
4
$begingroup$
An incoherent beam would mean noise in the electric field magnitude, not constant magnitude. You said yourself "average". If you average a coherent beam over a wavelength, it is also constant intensity. I do agree that the coherent beam makes for the better thought experiment.
$endgroup$
– EL_DON
Apr 9 at 7:09
1
$begingroup$
I would just add that visible light oscillates $10^14$ times per second, so the oscillations certainly cannot be seen by eye alone. There have been some ingenious experiments capturing the oscillation though, see science.sciencemag.org/content/305/5688/1267
$endgroup$
– Void
Apr 9 at 7:26
1
$begingroup$
@smcs i have edited
$endgroup$
– anna v
Apr 9 at 14:32
3
$begingroup$
Good job not getting sidetracked by the biological impossibility of the question. However, the sum of a set of randomly phase shifted sinusoids of the same frequency still oscillates. Adding waves of different frequencies still makes an oscillating wave, albiet a less regular one. This is because average amplitude will remain zero, while average power will be the sum of the individual waves. Watching such a wave in very slow motion should seem to flash in a very noisy pattern.
$endgroup$
– Vaelus
Apr 9 at 15:05
1
$begingroup$
@annav but sometimes the E-field amplitudes of the individual photons would add up to a positive number, and (unless there's a large DC-bias for some reason) sometimes they'd add up to a negative number, and in between they cross zero. So the instantaneous amplitude $I(t)sim E^2(t)$ also goes to zero. That the beam is incoherent just means the zero-crossings wouldn't be very correlated, and so an incoherent beam would flicker very irregularly, but it would still flicker.
$endgroup$
– aquirdturtle
Apr 10 at 0:10
|
show 12 more comments
$begingroup$
All depends how you "look" at the light.
For example, in a linearly polarized EMW the electric and the magnetic fields oscillate like $sin(omega t)$ or $cos(omega t)$ at a given point of space. Then the question is: what device do you use for "detecting" the EMF?
For EMF there are two notions expressed via fields: it is the energy density $ propto E^2+B^2$ and the energy-momentum flux $ propto vecEtimesvecB$ which may oscillate or not at a given point, depending on their phase shifts.
Some devices deal with a light "spot", where there are many points of the space involved, so you must average (sum up, integrate) local things. Some devices have inertial response and effectively average over time the incident wave too.
However, some devices are much more sensitive to the electric field than to the magnetic one (photo effect, for example), so they clearly "feel" oscillations.
There are also some devices (magnetic antennas, for example) that are more sensitive to the magnetic field (some radio-receivers).
In other words, the incident fields get into the equation of motion of the detector charges and currents, and the detector features determine what you get in reality.
answered Apr 9 at 6:29
Vladimir KalitvianskiVladimir Kalitvianski
11.3k11334
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2
$begingroup$
I've heard this before (from an E&M professor, I think), but I don't see how you satisfy Maxwell's equations with it. $nablatimes E=-partial B/partial t$ doesn't work if $Epropto cos(omega t-kcdot x)$ unless $B$ is also a cosine because they each take one derivative. Also, the poynting vector goes like $Etimes B$, so you'd get periodic variation in energy flux, anyway.
$endgroup$
– EL_DON
Apr 9 at 6:48
1
$begingroup$
The fields in a linearly polarized wave are in phase - that is a direct consequence of Maxwell's equations. As such also $|vec E times vec B|$ is oscillating, not a constant. Only when you average the Poynting vector over one period $langle vec S rangle$, the intensity becomes constant.
$endgroup$
– ahemmetter
Apr 9 at 9:17
$begingroup$
@ahemmetter: The Poynting vector is a vector, not an intensity $I$.
$endgroup$
– Vladimir Kalitvianski
Apr 9 at 9:30
1
$begingroup$
The intensity is directly proportional to the magnitude of the Poynting vector.
$endgroup$
– ahemmetter
Apr 9 at 11:46
add a comment |
$begingroup$
The EM field strength in a linearly polarized, coherent light wave does indeed cross through zero in between + and - peaks, just like the surface of a pond goes through its natural rest height in between going up and down as ripples go by. If you slowed the wave down, you'd change its frequency, which is the same as changing its color. You could change from blue, to red, to infrared, and all the way through radio waves and other invisible colors. So no, you could not see the EM field changing as a flashing light for a slow wave (the slow changes couldn't simulate your vision receptors) , but you could set up an electric field meter and measure the change in field as the (no longer visible) wave went by. What does it really mean for the field to cross through zero? Not much; just like having the surface of a pond cross through its equilibrium height doesn't mean the ripples are gone, neither does a moment of 0 electric field mean the light wave is gone.
answered Apr 9 at 6:37
EL_DONEL_DON
2,3522726
$endgroup$
add a comment |
$begingroup$
"Light intensity" in my opinion means the number of optical photons absorbed by a light detector. Such detectors resonate with the electric field and absorb the photons with a characteristic time of many periods of oscillation. So you can say that the oscillation is observed by the detector, but is not translated into a rapid oscillation of the detected intensity. The latter varies with the number of photons absorbed instead, which is at a generally much longer timescale. However, extremely short laser pulses can approach the timescale of the light oscillation period.
answered Apr 9 at 8:32
my2ctsmy2cts
5,7872719
$endgroup$
add a comment |
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4 Answers
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active
oldest
votes
4 Answers
4
active
oldest
votes
active
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votes
active
oldest
votes
$begingroup$
You would need a coherent beam, because in waves it is not only intensity but also phase that makes a difference. In an incoherent beam, as sunlight, you would not get any changes in this thought experiment, because the average intensity would hold even at wavelength distances.
In a coherent laser beam you should see in your thought experiment what is shown towards the end of this youtube video,(at 2' 09") the sinusoidal in time pattern of impinging intensity . The video draws the oscillating electric field E, and the intensity left on the screen will be $=E^2$, which will also be oscillating in time.
After all, mathematics allows us to materialize thought experiments as this one.
answered Apr 9 at 6:28
anna vanna v
162k8153456
$endgroup$
4
$begingroup$
An incoherent beam would mean noise in the electric field magnitude, not constant magnitude. You said yourself "average". If you average a coherent beam over a wavelength, it is also constant intensity. I do agree that the coherent beam makes for the better thought experiment.
$endgroup$
– EL_DON
Apr 9 at 7:09
1
$begingroup$
I would just add that visible light oscillates $10^14$ times per second, so the oscillations certainly cannot be seen by eye alone. There have been some ingenious experiments capturing the oscillation though, see science.sciencemag.org/content/305/5688/1267
$endgroup$
– Void
Apr 9 at 7:26
1
$begingroup$
@smcs i have edited
$endgroup$
– anna v
Apr 9 at 14:32
3
$begingroup$
Good job not getting sidetracked by the biological impossibility of the question. However, the sum of a set of randomly phase shifted sinusoids of the same frequency still oscillates. Adding waves of different frequencies still makes an oscillating wave, albiet a less regular one. This is because average amplitude will remain zero, while average power will be the sum of the individual waves. Watching such a wave in very slow motion should seem to flash in a very noisy pattern.
$endgroup$
– Vaelus
Apr 9 at 15:05
1
$begingroup$
@annav but sometimes the E-field amplitudes of the individual photons would add up to a positive number, and (unless there's a large DC-bias for some reason) sometimes they'd add up to a negative number, and in between they cross zero. So the instantaneous amplitude $I(t)sim E^2(t)$ also goes to zero. That the beam is incoherent just means the zero-crossings wouldn't be very correlated, and so an incoherent beam would flicker very irregularly, but it would still flicker.
$endgroup$
– aquirdturtle
Apr 10 at 0:10
|
show 12 more comments
$begingroup$
You would need a coherent beam, because in waves it is not only intensity but also phase that makes a difference. In an incoherent beam, as sunlight, you would not get any changes in this thought experiment, because the average intensity would hold even at wavelength distances.
In a coherent laser beam you should see in your thought experiment what is shown towards the end of this youtube video,(at 2' 09") the sinusoidal in time pattern of impinging intensity . The video draws the oscillating electric field E, and the intensity left on the screen will be $=E^2$, which will also be oscillating in time.
After all, mathematics allows us to materialize thought experiments as this one.
answered Apr 9 at 6:28
anna vanna v
162k8153456
$endgroup$
4
$begingroup$
An incoherent beam would mean noise in the electric field magnitude, not constant magnitude. You said yourself "average". If you average a coherent beam over a wavelength, it is also constant intensity. I do agree that the coherent beam makes for the better thought experiment.
$endgroup$
– EL_DON
Apr 9 at 7:09
1
$begingroup$
I would just add that visible light oscillates $10^14$ times per second, so the oscillations certainly cannot be seen by eye alone. There have been some ingenious experiments capturing the oscillation though, see science.sciencemag.org/content/305/5688/1267
$endgroup$
– Void
Apr 9 at 7:26
1
$begingroup$
@smcs i have edited
$endgroup$
– anna v
Apr 9 at 14:32
3
$begingroup$
Good job not getting sidetracked by the biological impossibility of the question. However, the sum of a set of randomly phase shifted sinusoids of the same frequency still oscillates. Adding waves of different frequencies still makes an oscillating wave, albiet a less regular one. This is because average amplitude will remain zero, while average power will be the sum of the individual waves. Watching such a wave in very slow motion should seem to flash in a very noisy pattern.
$endgroup$
– Vaelus
Apr 9 at 15:05
1
$begingroup$
@annav but sometimes the E-field amplitudes of the individual photons would add up to a positive number, and (unless there's a large DC-bias for some reason) sometimes they'd add up to a negative number, and in between they cross zero. So the instantaneous amplitude $I(t)sim E^2(t)$ also goes to zero. That the beam is incoherent just means the zero-crossings wouldn't be very correlated, and so an incoherent beam would flicker very irregularly, but it would still flicker.
$endgroup$
– aquirdturtle
Apr 10 at 0:10
|
show 12 more comments
$begingroup$
You would need a coherent beam, because in waves it is not only intensity but also phase that makes a difference. In an incoherent beam, as sunlight, you would not get any changes in this thought experiment, because the average intensity would hold even at wavelength distances.
In a coherent laser beam you should see in your thought experiment what is shown towards the end of this youtube video,(at 2' 09") the sinusoidal in time pattern of impinging intensity . The video draws the oscillating electric field E, and the intensity left on the screen will be $=E^2$, which will also be oscillating in time.
After all, mathematics allows us to materialize thought experiments as this one.
answered Apr 9 at 6:28
anna vanna v
162k8153456
$endgroup$
You would need a coherent beam, because in waves it is not only intensity but also phase that makes a difference. In an incoherent beam, as sunlight, you would not get any changes in this thought experiment, because the average intensity would hold even at wavelength distances.
In a coherent laser beam you should see in your thought experiment what is shown towards the end of this youtube video,(at 2' 09") the sinusoidal in time pattern of impinging intensity . The video draws the oscillating electric field E, and the intensity left on the screen will be $=E^2$, which will also be oscillating in time.
After all, mathematics allows us to materialize thought experiments as this one.
answered Apr 9 at 6:28
anna vanna v
162k8153456
edited Apr 9 at 14:11
answered Apr 9 at 6:28
anna vanna v
162k8153456
answered Apr 9 at 6:28
anna vanna v
162k8153456
answered Apr 9 at 6:28
anna vanna v
162k8153456
162k8153456
4
$begingroup$
An incoherent beam would mean noise in the electric field magnitude, not constant magnitude. You said yourself "average". If you average a coherent beam over a wavelength, it is also constant intensity. I do agree that the coherent beam makes for the better thought experiment.
$endgroup$
– EL_DON
Apr 9 at 7:09
1
$begingroup$
I would just add that visible light oscillates $10^14$ times per second, so the oscillations certainly cannot be seen by eye alone. There have been some ingenious experiments capturing the oscillation though, see science.sciencemag.org/content/305/5688/1267
$endgroup$
– Void
Apr 9 at 7:26
1
$begingroup$
@smcs i have edited
$endgroup$
– anna v
Apr 9 at 14:32
3
$begingroup$
Good job not getting sidetracked by the biological impossibility of the question. However, the sum of a set of randomly phase shifted sinusoids of the same frequency still oscillates. Adding waves of different frequencies still makes an oscillating wave, albiet a less regular one. This is because average amplitude will remain zero, while average power will be the sum of the individual waves. Watching such a wave in very slow motion should seem to flash in a very noisy pattern.
$endgroup$
– Vaelus
Apr 9 at 15:05
1
$begingroup$
@annav but sometimes the E-field amplitudes of the individual photons would add up to a positive number, and (unless there's a large DC-bias for some reason) sometimes they'd add up to a negative number, and in between they cross zero. So the instantaneous amplitude $I(t)sim E^2(t)$ also goes to zero. That the beam is incoherent just means the zero-crossings wouldn't be very correlated, and so an incoherent beam would flicker very irregularly, but it would still flicker.
$endgroup$
– aquirdturtle
Apr 10 at 0:10
|
show 12 more comments
4
$begingroup$
An incoherent beam would mean noise in the electric field magnitude, not constant magnitude. You said yourself "average". If you average a coherent beam over a wavelength, it is also constant intensity. I do agree that the coherent beam makes for the better thought experiment.
$endgroup$
– EL_DON
Apr 9 at 7:09
1
$begingroup$
I would just add that visible light oscillates $10^14$ times per second, so the oscillations certainly cannot be seen by eye alone. There have been some ingenious experiments capturing the oscillation though, see science.sciencemag.org/content/305/5688/1267
$endgroup$
– Void
Apr 9 at 7:26
1
$begingroup$
@smcs i have edited
$endgroup$
– anna v
Apr 9 at 14:32
3
$begingroup$
Good job not getting sidetracked by the biological impossibility of the question. However, the sum of a set of randomly phase shifted sinusoids of the same frequency still oscillates. Adding waves of different frequencies still makes an oscillating wave, albiet a less regular one. This is because average amplitude will remain zero, while average power will be the sum of the individual waves. Watching such a wave in very slow motion should seem to flash in a very noisy pattern.
$endgroup$
– Vaelus
Apr 9 at 15:05
1
$begingroup$
@annav but sometimes the E-field amplitudes of the individual photons would add up to a positive number, and (unless there's a large DC-bias for some reason) sometimes they'd add up to a negative number, and in between they cross zero. So the instantaneous amplitude $I(t)sim E^2(t)$ also goes to zero. That the beam is incoherent just means the zero-crossings wouldn't be very correlated, and so an incoherent beam would flicker very irregularly, but it would still flicker.
$endgroup$
– aquirdturtle
Apr 10 at 0:10
4
4
$begingroup$
An incoherent beam would mean noise in the electric field magnitude, not constant magnitude. You said yourself "average". If you average a coherent beam over a wavelength, it is also constant intensity. I do agree that the coherent beam makes for the better thought experiment.
$endgroup$
– EL_DON
Apr 9 at 7:09
$begingroup$
An incoherent beam would mean noise in the electric field magnitude, not constant magnitude. You said yourself "average". If you average a coherent beam over a wavelength, it is also constant intensity. I do agree that the coherent beam makes for the better thought experiment.
$endgroup$
– EL_DON
Apr 9 at 7:09
1
1
$begingroup$
I would just add that visible light oscillates $10^14$ times per second, so the oscillations certainly cannot be seen by eye alone. There have been some ingenious experiments capturing the oscillation though, see science.sciencemag.org/content/305/5688/1267
$endgroup$
– Void
Apr 9 at 7:26
$begingroup$
I would just add that visible light oscillates $10^14$ times per second, so the oscillations certainly cannot be seen by eye alone. There have been some ingenious experiments capturing the oscillation though, see science.sciencemag.org/content/305/5688/1267
$endgroup$
– Void
Apr 9 at 7:26
1
1
$begingroup$
@smcs i have edited
$endgroup$
– anna v
Apr 9 at 14:32
$begingroup$
@smcs i have edited
$endgroup$
– anna v
Apr 9 at 14:32
3
3
$begingroup$
Good job not getting sidetracked by the biological impossibility of the question. However, the sum of a set of randomly phase shifted sinusoids of the same frequency still oscillates. Adding waves of different frequencies still makes an oscillating wave, albiet a less regular one. This is because average amplitude will remain zero, while average power will be the sum of the individual waves. Watching such a wave in very slow motion should seem to flash in a very noisy pattern.
$endgroup$
– Vaelus
Apr 9 at 15:05
$begingroup$
Good job not getting sidetracked by the biological impossibility of the question. However, the sum of a set of randomly phase shifted sinusoids of the same frequency still oscillates. Adding waves of different frequencies still makes an oscillating wave, albiet a less regular one. This is because average amplitude will remain zero, while average power will be the sum of the individual waves. Watching such a wave in very slow motion should seem to flash in a very noisy pattern.
$endgroup$
– Vaelus
Apr 9 at 15:05
1
1
$begingroup$
@annav but sometimes the E-field amplitudes of the individual photons would add up to a positive number, and (unless there's a large DC-bias for some reason) sometimes they'd add up to a negative number, and in between they cross zero. So the instantaneous amplitude $I(t)sim E^2(t)$ also goes to zero. That the beam is incoherent just means the zero-crossings wouldn't be very correlated, and so an incoherent beam would flicker very irregularly, but it would still flicker.
$endgroup$
– aquirdturtle
Apr 10 at 0:10
$begingroup$
@annav but sometimes the E-field amplitudes of the individual photons would add up to a positive number, and (unless there's a large DC-bias for some reason) sometimes they'd add up to a negative number, and in between they cross zero. So the instantaneous amplitude $I(t)sim E^2(t)$ also goes to zero. That the beam is incoherent just means the zero-crossings wouldn't be very correlated, and so an incoherent beam would flicker very irregularly, but it would still flicker.
$endgroup$
– aquirdturtle
Apr 10 at 0:10
|
show 12 more comments
$begingroup$
All depends how you "look" at the light.
For example, in a linearly polarized EMW the electric and the magnetic fields oscillate like $sin(omega t)$ or $cos(omega t)$ at a given point of space. Then the question is: what device do you use for "detecting" the EMF?
For EMF there are two notions expressed via fields: it is the energy density $ propto E^2+B^2$ and the energy-momentum flux $ propto vecEtimesvecB$ which may oscillate or not at a given point, depending on their phase shifts.
Some devices deal with a light "spot", where there are many points of the space involved, so you must average (sum up, integrate) local things. Some devices have inertial response and effectively average over time the incident wave too.
However, some devices are much more sensitive to the electric field than to the magnetic one (photo effect, for example), so they clearly "feel" oscillations.
There are also some devices (magnetic antennas, for example) that are more sensitive to the magnetic field (some radio-receivers).
In other words, the incident fields get into the equation of motion of the detector charges and currents, and the detector features determine what you get in reality.
answered Apr 9 at 6:29
Vladimir KalitvianskiVladimir Kalitvianski
11.3k11334
$endgroup$
2
$begingroup$
I've heard this before (from an E&M professor, I think), but I don't see how you satisfy Maxwell's equations with it. $nablatimes E=-partial B/partial t$ doesn't work if $Epropto cos(omega t-kcdot x)$ unless $B$ is also a cosine because they each take one derivative. Also, the poynting vector goes like $Etimes B$, so you'd get periodic variation in energy flux, anyway.
$endgroup$
– EL_DON
Apr 9 at 6:48
1
$begingroup$
The fields in a linearly polarized wave are in phase - that is a direct consequence of Maxwell's equations. As such also $|vec E times vec B|$ is oscillating, not a constant. Only when you average the Poynting vector over one period $langle vec S rangle$, the intensity becomes constant.
$endgroup$
– ahemmetter
Apr 9 at 9:17
$begingroup$
@ahemmetter: The Poynting vector is a vector, not an intensity $I$.
$endgroup$
– Vladimir Kalitvianski
Apr 9 at 9:30
1
$begingroup$
The intensity is directly proportional to the magnitude of the Poynting vector.
$endgroup$
– ahemmetter
Apr 9 at 11:46
add a comment |
$begingroup$
All depends how you "look" at the light.
For example, in a linearly polarized EMW the electric and the magnetic fields oscillate like $sin(omega t)$ or $cos(omega t)$ at a given point of space. Then the question is: what device do you use for "detecting" the EMF?
For EMF there are two notions expressed via fields: it is the energy density $ propto E^2+B^2$ and the energy-momentum flux $ propto vecEtimesvecB$ which may oscillate or not at a given point, depending on their phase shifts.
Some devices deal with a light "spot", where there are many points of the space involved, so you must average (sum up, integrate) local things. Some devices have inertial response and effectively average over time the incident wave too.
However, some devices are much more sensitive to the electric field than to the magnetic one (photo effect, for example), so they clearly "feel" oscillations.
There are also some devices (magnetic antennas, for example) that are more sensitive to the magnetic field (some radio-receivers).
In other words, the incident fields get into the equation of motion of the detector charges and currents, and the detector features determine what you get in reality.
answered Apr 9 at 6:29
Vladimir KalitvianskiVladimir Kalitvianski
11.3k11334
$endgroup$
2
$begingroup$
I've heard this before (from an E&M professor, I think), but I don't see how you satisfy Maxwell's equations with it. $nablatimes E=-partial B/partial t$ doesn't work if $Epropto cos(omega t-kcdot x)$ unless $B$ is also a cosine because they each take one derivative. Also, the poynting vector goes like $Etimes B$, so you'd get periodic variation in energy flux, anyway.
$endgroup$
– EL_DON
Apr 9 at 6:48
1
$begingroup$
The fields in a linearly polarized wave are in phase - that is a direct consequence of Maxwell's equations. As such also $|vec E times vec B|$ is oscillating, not a constant. Only when you average the Poynting vector over one period $langle vec S rangle$, the intensity becomes constant.
$endgroup$
– ahemmetter
Apr 9 at 9:17
$begingroup$
@ahemmetter: The Poynting vector is a vector, not an intensity $I$.
$endgroup$
– Vladimir Kalitvianski
Apr 9 at 9:30
1
$begingroup$
The intensity is directly proportional to the magnitude of the Poynting vector.
$endgroup$
– ahemmetter
Apr 9 at 11:46
add a comment |
$begingroup$
All depends how you "look" at the light.
For example, in a linearly polarized EMW the electric and the magnetic fields oscillate like $sin(omega t)$ or $cos(omega t)$ at a given point of space. Then the question is: what device do you use for "detecting" the EMF?
For EMF there are two notions expressed via fields: it is the energy density $ propto E^2+B^2$ and the energy-momentum flux $ propto vecEtimesvecB$ which may oscillate or not at a given point, depending on their phase shifts.
Some devices deal with a light "spot", where there are many points of the space involved, so you must average (sum up, integrate) local things. Some devices have inertial response and effectively average over time the incident wave too.
However, some devices are much more sensitive to the electric field than to the magnetic one (photo effect, for example), so they clearly "feel" oscillations.
There are also some devices (magnetic antennas, for example) that are more sensitive to the magnetic field (some radio-receivers).
In other words, the incident fields get into the equation of motion of the detector charges and currents, and the detector features determine what you get in reality.
answered Apr 9 at 6:29
Vladimir KalitvianskiVladimir Kalitvianski
11.3k11334
$endgroup$
All depends how you "look" at the light.
For example, in a linearly polarized EMW the electric and the magnetic fields oscillate like $sin(omega t)$ or $cos(omega t)$ at a given point of space. Then the question is: what device do you use for "detecting" the EMF?
For EMF there are two notions expressed via fields: it is the energy density $ propto E^2+B^2$ and the energy-momentum flux $ propto vecEtimesvecB$ which may oscillate or not at a given point, depending on their phase shifts.
Some devices deal with a light "spot", where there are many points of the space involved, so you must average (sum up, integrate) local things. Some devices have inertial response and effectively average over time the incident wave too.
However, some devices are much more sensitive to the electric field than to the magnetic one (photo effect, for example), so they clearly "feel" oscillations.
There are also some devices (magnetic antennas, for example) that are more sensitive to the magnetic field (some radio-receivers).
In other words, the incident fields get into the equation of motion of the detector charges and currents, and the detector features determine what you get in reality.
answered Apr 9 at 6:29
Vladimir KalitvianskiVladimir Kalitvianski
11.3k11334
edited Apr 9 at 13:40
answered Apr 9 at 6:29
Vladimir KalitvianskiVladimir Kalitvianski
11.3k11334
answered Apr 9 at 6:29
Vladimir KalitvianskiVladimir Kalitvianski
11.3k11334
answered Apr 9 at 6:29
Vladimir KalitvianskiVladimir Kalitvianski
11.3k11334
11.3k11334
2
$begingroup$
I've heard this before (from an E&M professor, I think), but I don't see how you satisfy Maxwell's equations with it. $nablatimes E=-partial B/partial t$ doesn't work if $Epropto cos(omega t-kcdot x)$ unless $B$ is also a cosine because they each take one derivative. Also, the poynting vector goes like $Etimes B$, so you'd get periodic variation in energy flux, anyway.
$endgroup$
– EL_DON
Apr 9 at 6:48
1
$begingroup$
The fields in a linearly polarized wave are in phase - that is a direct consequence of Maxwell's equations. As such also $|vec E times vec B|$ is oscillating, not a constant. Only when you average the Poynting vector over one period $langle vec S rangle$, the intensity becomes constant.
$endgroup$
– ahemmetter
Apr 9 at 9:17
$begingroup$
@ahemmetter: The Poynting vector is a vector, not an intensity $I$.
$endgroup$
– Vladimir Kalitvianski
Apr 9 at 9:30
1
$begingroup$
The intensity is directly proportional to the magnitude of the Poynting vector.
$endgroup$
– ahemmetter
Apr 9 at 11:46
add a comment |
2
$begingroup$
I've heard this before (from an E&M professor, I think), but I don't see how you satisfy Maxwell's equations with it. $nablatimes E=-partial B/partial t$ doesn't work if $Epropto cos(omega t-kcdot x)$ unless $B$ is also a cosine because they each take one derivative. Also, the poynting vector goes like $Etimes B$, so you'd get periodic variation in energy flux, anyway.
$endgroup$
– EL_DON
Apr 9 at 6:48
1
$begingroup$
The fields in a linearly polarized wave are in phase - that is a direct consequence of Maxwell's equations. As such also $|vec E times vec B|$ is oscillating, not a constant. Only when you average the Poynting vector over one period $langle vec S rangle$, the intensity becomes constant.
$endgroup$
– ahemmetter
Apr 9 at 9:17
$begingroup$
@ahemmetter: The Poynting vector is a vector, not an intensity $I$.
$endgroup$
– Vladimir Kalitvianski
Apr 9 at 9:30
1
$begingroup$
The intensity is directly proportional to the magnitude of the Poynting vector.
$endgroup$
– ahemmetter
Apr 9 at 11:46
2
2
$begingroup$
I've heard this before (from an E&M professor, I think), but I don't see how you satisfy Maxwell's equations with it. $nablatimes E=-partial B/partial t$ doesn't work if $Epropto cos(omega t-kcdot x)$ unless $B$ is also a cosine because they each take one derivative. Also, the poynting vector goes like $Etimes B$, so you'd get periodic variation in energy flux, anyway.
$endgroup$
– EL_DON
Apr 9 at 6:48
$begingroup$
I've heard this before (from an E&M professor, I think), but I don't see how you satisfy Maxwell's equations with it. $nablatimes E=-partial B/partial t$ doesn't work if $Epropto cos(omega t-kcdot x)$ unless $B$ is also a cosine because they each take one derivative. Also, the poynting vector goes like $Etimes B$, so you'd get periodic variation in energy flux, anyway.
$endgroup$
– EL_DON
Apr 9 at 6:48
1
1
$begingroup$
The fields in a linearly polarized wave are in phase - that is a direct consequence of Maxwell's equations. As such also $|vec E times vec B|$ is oscillating, not a constant. Only when you average the Poynting vector over one period $langle vec S rangle$, the intensity becomes constant.
$endgroup$
– ahemmetter
Apr 9 at 9:17
$begingroup$
The fields in a linearly polarized wave are in phase - that is a direct consequence of Maxwell's equations. As such also $|vec E times vec B|$ is oscillating, not a constant. Only when you average the Poynting vector over one period $langle vec S rangle$, the intensity becomes constant.
$endgroup$
– ahemmetter
Apr 9 at 9:17
$begingroup$
@ahemmetter: The Poynting vector is a vector, not an intensity $I$.
$endgroup$
– Vladimir Kalitvianski
Apr 9 at 9:30
$begingroup$
@ahemmetter: The Poynting vector is a vector, not an intensity $I$.
$endgroup$
– Vladimir Kalitvianski
Apr 9 at 9:30
1
1
$begingroup$
The intensity is directly proportional to the magnitude of the Poynting vector.
$endgroup$
– ahemmetter
Apr 9 at 11:46
$begingroup$
The intensity is directly proportional to the magnitude of the Poynting vector.
$endgroup$
– ahemmetter
Apr 9 at 11:46
add a comment |
$begingroup$
The EM field strength in a linearly polarized, coherent light wave does indeed cross through zero in between + and - peaks, just like the surface of a pond goes through its natural rest height in between going up and down as ripples go by. If you slowed the wave down, you'd change its frequency, which is the same as changing its color. You could change from blue, to red, to infrared, and all the way through radio waves and other invisible colors. So no, you could not see the EM field changing as a flashing light for a slow wave (the slow changes couldn't simulate your vision receptors) , but you could set up an electric field meter and measure the change in field as the (no longer visible) wave went by. What does it really mean for the field to cross through zero? Not much; just like having the surface of a pond cross through its equilibrium height doesn't mean the ripples are gone, neither does a moment of 0 electric field mean the light wave is gone.
answered Apr 9 at 6:37
EL_DONEL_DON
2,3522726
$endgroup$
add a comment |
$begingroup$
The EM field strength in a linearly polarized, coherent light wave does indeed cross through zero in between + and - peaks, just like the surface of a pond goes through its natural rest height in between going up and down as ripples go by. If you slowed the wave down, you'd change its frequency, which is the same as changing its color. You could change from blue, to red, to infrared, and all the way through radio waves and other invisible colors. So no, you could not see the EM field changing as a flashing light for a slow wave (the slow changes couldn't simulate your vision receptors) , but you could set up an electric field meter and measure the change in field as the (no longer visible) wave went by. What does it really mean for the field to cross through zero? Not much; just like having the surface of a pond cross through its equilibrium height doesn't mean the ripples are gone, neither does a moment of 0 electric field mean the light wave is gone.
answered Apr 9 at 6:37
EL_DONEL_DON
2,3522726
$endgroup$
add a comment |
$begingroup$
The EM field strength in a linearly polarized, coherent light wave does indeed cross through zero in between + and - peaks, just like the surface of a pond goes through its natural rest height in between going up and down as ripples go by. If you slowed the wave down, you'd change its frequency, which is the same as changing its color. You could change from blue, to red, to infrared, and all the way through radio waves and other invisible colors. So no, you could not see the EM field changing as a flashing light for a slow wave (the slow changes couldn't simulate your vision receptors) , but you could set up an electric field meter and measure the change in field as the (no longer visible) wave went by. What does it really mean for the field to cross through zero? Not much; just like having the surface of a pond cross through its equilibrium height doesn't mean the ripples are gone, neither does a moment of 0 electric field mean the light wave is gone.
answered Apr 9 at 6:37
EL_DONEL_DON
2,3522726
$endgroup$
The EM field strength in a linearly polarized, coherent light wave does indeed cross through zero in between + and - peaks, just like the surface of a pond goes through its natural rest height in between going up and down as ripples go by. If you slowed the wave down, you'd change its frequency, which is the same as changing its color. You could change from blue, to red, to infrared, and all the way through radio waves and other invisible colors. So no, you could not see the EM field changing as a flashing light for a slow wave (the slow changes couldn't simulate your vision receptors) , but you could set up an electric field meter and measure the change in field as the (no longer visible) wave went by. What does it really mean for the field to cross through zero? Not much; just like having the surface of a pond cross through its equilibrium height doesn't mean the ripples are gone, neither does a moment of 0 electric field mean the light wave is gone.
answered Apr 9 at 6:37
EL_DONEL_DON
2,3522726
edited Apr 9 at 7:11
answered Apr 9 at 6:37
EL_DONEL_DON
2,3522726
answered Apr 9 at 6:37
EL_DONEL_DON
2,3522726
answered Apr 9 at 6:37
EL_DONEL_DON
2,3522726
2,3522726
add a comment |
add a comment |
$begingroup$
"Light intensity" in my opinion means the number of optical photons absorbed by a light detector. Such detectors resonate with the electric field and absorb the photons with a characteristic time of many periods of oscillation. So you can say that the oscillation is observed by the detector, but is not translated into a rapid oscillation of the detected intensity. The latter varies with the number of photons absorbed instead, which is at a generally much longer timescale. However, extremely short laser pulses can approach the timescale of the light oscillation period.
answered Apr 9 at 8:32
my2ctsmy2cts
5,7872719
$endgroup$
add a comment |
$begingroup$
"Light intensity" in my opinion means the number of optical photons absorbed by a light detector. Such detectors resonate with the electric field and absorb the photons with a characteristic time of many periods of oscillation. So you can say that the oscillation is observed by the detector, but is not translated into a rapid oscillation of the detected intensity. The latter varies with the number of photons absorbed instead, which is at a generally much longer timescale. However, extremely short laser pulses can approach the timescale of the light oscillation period.
answered Apr 9 at 8:32
my2ctsmy2cts
5,7872719
$endgroup$
add a comment |
$begingroup$
"Light intensity" in my opinion means the number of optical photons absorbed by a light detector. Such detectors resonate with the electric field and absorb the photons with a characteristic time of many periods of oscillation. So you can say that the oscillation is observed by the detector, but is not translated into a rapid oscillation of the detected intensity. The latter varies with the number of photons absorbed instead, which is at a generally much longer timescale. However, extremely short laser pulses can approach the timescale of the light oscillation period.
answered Apr 9 at 8:32
my2ctsmy2cts
5,7872719
$endgroup$
"Light intensity" in my opinion means the number of optical photons absorbed by a light detector. Such detectors resonate with the electric field and absorb the photons with a characteristic time of many periods of oscillation. So you can say that the oscillation is observed by the detector, but is not translated into a rapid oscillation of the detected intensity. The latter varies with the number of photons absorbed instead, which is at a generally much longer timescale. However, extremely short laser pulses can approach the timescale of the light oscillation period.
answered Apr 9 at 8:32
my2ctsmy2cts
5,7872719
answered Apr 9 at 8:32
my2ctsmy2cts
5,7872719
answered Apr 9 at 8:32
my2ctsmy2cts
5,7872719
answered Apr 9 at 8:32
my2ctsmy2cts
5,7872719
5,7872719
add a comment |
add a comment |
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$begingroup$
Related Have we directly observed the electric component to EM waves?
$endgroup$
– Farcher
Apr 9 at 7:03
3
$begingroup$
What you are talking about is the amplitude. Intensity usually means the average power and that would be constant. And you can't talk about seeing patches in those timescales but observing electric field change.
$endgroup$
– Džuris
Apr 9 at 10:23
2
$begingroup$
If you were able to look at the world at extreme slow motion, a quadrillion times slower or so, then you would have to be using some kind of "vision" that would not, in any way, resemble how your vision actually works.
$endgroup$
– Solomon Slow
Apr 9 at 13:14
$begingroup$
Possible duplicate of Have we directly observed the electric component to EM waves?
$endgroup$
– ahemmetter
Apr 9 at 14:07