Does light intensity oscillate really fast since it is a wave?












7












$begingroup$


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?










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  • 3




    $begingroup$
    Related Have we directly observed the electric component to EM waves?
    $endgroup$
    – Farcher
    6 hours ago










  • $begingroup$
    Of course, if you slowed "time" down enough, you'd eventually get to a point where you'd just "see" individual photons reflected from the wall. Interestingly enough, you can try this even without slowing time - a dark room is quite enough :) Your eyes and brain compensate a bit to make something of a picture, but you'll see randomly colored blobs in your vision. Needless to say, their intensity is the same everywhere - at the level of "individual photons" (observed individual photon-eye interactions, let's say), there is no "intensity" - either you detect the photon, or you don't.
    $endgroup$
    – Luaan
    5 hours ago






  • 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
    3 hours ago










  • $begingroup$
    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
    45 mins ago
















7












$begingroup$


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?










share|cite|improve this question









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.







$endgroup$








  • 3




    $begingroup$
    Related Have we directly observed the electric component to EM waves?
    $endgroup$
    – Farcher
    6 hours ago










  • $begingroup$
    Of course, if you slowed "time" down enough, you'd eventually get to a point where you'd just "see" individual photons reflected from the wall. Interestingly enough, you can try this even without slowing time - a dark room is quite enough :) Your eyes and brain compensate a bit to make something of a picture, but you'll see randomly colored blobs in your vision. Needless to say, their intensity is the same everywhere - at the level of "individual photons" (observed individual photon-eye interactions, let's say), there is no "intensity" - either you detect the photon, or you don't.
    $endgroup$
    – Luaan
    5 hours ago






  • 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
    3 hours ago










  • $begingroup$
    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
    45 mins ago














7












7








7


1



$begingroup$


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?










share|cite|improve this question









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






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share|cite|improve this question




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edited 6 hours ago









Ruslan

9,81843173




9,81843173






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asked 7 hours ago









AdgornAdgorn

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394




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  • 3




    $begingroup$
    Related Have we directly observed the electric component to EM waves?
    $endgroup$
    – Farcher
    6 hours ago










  • $begingroup$
    Of course, if you slowed "time" down enough, you'd eventually get to a point where you'd just "see" individual photons reflected from the wall. Interestingly enough, you can try this even without slowing time - a dark room is quite enough :) Your eyes and brain compensate a bit to make something of a picture, but you'll see randomly colored blobs in your vision. Needless to say, their intensity is the same everywhere - at the level of "individual photons" (observed individual photon-eye interactions, let's say), there is no "intensity" - either you detect the photon, or you don't.
    $endgroup$
    – Luaan
    5 hours ago






  • 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
    3 hours ago










  • $begingroup$
    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
    45 mins ago














  • 3




    $begingroup$
    Related Have we directly observed the electric component to EM waves?
    $endgroup$
    – Farcher
    6 hours ago










  • $begingroup$
    Of course, if you slowed "time" down enough, you'd eventually get to a point where you'd just "see" individual photons reflected from the wall. Interestingly enough, you can try this even without slowing time - a dark room is quite enough :) Your eyes and brain compensate a bit to make something of a picture, but you'll see randomly colored blobs in your vision. Needless to say, their intensity is the same everywhere - at the level of "individual photons" (observed individual photon-eye interactions, let's say), there is no "intensity" - either you detect the photon, or you don't.
    $endgroup$
    – Luaan
    5 hours ago






  • 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
    3 hours ago










  • $begingroup$
    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
    45 mins ago








3




3




$begingroup$
Related Have we directly observed the electric component to EM waves?
$endgroup$
– Farcher
6 hours ago




$begingroup$
Related Have we directly observed the electric component to EM waves?
$endgroup$
– Farcher
6 hours ago












$begingroup$
Of course, if you slowed "time" down enough, you'd eventually get to a point where you'd just "see" individual photons reflected from the wall. Interestingly enough, you can try this even without slowing time - a dark room is quite enough :) Your eyes and brain compensate a bit to make something of a picture, but you'll see randomly colored blobs in your vision. Needless to say, their intensity is the same everywhere - at the level of "individual photons" (observed individual photon-eye interactions, let's say), there is no "intensity" - either you detect the photon, or you don't.
$endgroup$
– Luaan
5 hours ago




$begingroup$
Of course, if you slowed "time" down enough, you'd eventually get to a point where you'd just "see" individual photons reflected from the wall. Interestingly enough, you can try this even without slowing time - a dark room is quite enough :) Your eyes and brain compensate a bit to make something of a picture, but you'll see randomly colored blobs in your vision. Needless to say, their intensity is the same everywhere - at the level of "individual photons" (observed individual photon-eye interactions, let's say), there is no "intensity" - either you detect the photon, or you don't.
$endgroup$
– Luaan
5 hours ago




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
3 hours ago




$begingroup$
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
3 hours ago












$begingroup$
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
45 mins ago




$begingroup$
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
45 mins ago










4 Answers
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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, the sinusoidal pattern of impinging intensity . After all, mathematics allows us to materialize thought experiments as this one.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    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
    6 hours ago






  • 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
    6 hours ago










  • $begingroup$
    @EL_DON Calculating the variance in the magnitude caused by the noise in the case of shining a flashlight on the wall is implicitly left as an exercise for the reader.
    $endgroup$
    – JiK
    1 hour ago










  • $begingroup$
    I'm struggling with this answer--a sinusoidal pattern of impinging intensity? What exact part in the video do you mean?
    $endgroup$
    – smcs
    40 mins ago










  • $begingroup$
    @JiK what matters is that the answer is arguing that intensity doesn't change in time unless the light source is coherent, but this is wrong. The justification for no change uses the word average and also the $E$ field must change in order for there to be a wave at all. The only wave with constant intensity vs time is a coherent and circularly polarized one. If you could detect variation in intensity for coherent light vs. time, you could also detect variation for incoherent light, unless we're talking about a spatial average over the area of the detector.
    $endgroup$
    – EL_DON
    9 mins ago



















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$begingroup$

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 vec{E}timesvec{B}$ 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.






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$endgroup$









  • 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
    7 hours ago








  • 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
    4 hours ago










  • $begingroup$
    @ahemmetter: The Poynting vector is a vector, not an intensity $I$.
    $endgroup$
    – Vladimir Kalitvianski
    4 hours ago






  • 1




    $begingroup$
    The intensity is directly proportional to the magnitude of the Poynting vector.
    $endgroup$
    – ahemmetter
    2 hours ago



















2












$begingroup$

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.






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    "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.






    share|cite|improve this answer









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      4 Answers
      4






      active

      oldest

      votes








      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      8












      $begingroup$

      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, the sinusoidal pattern of impinging intensity . After all, mathematics allows us to materialize thought experiments as this one.






      share|cite|improve this answer









      $endgroup$









      • 1




        $begingroup$
        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
        6 hours ago






      • 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
        6 hours ago










      • $begingroup$
        @EL_DON Calculating the variance in the magnitude caused by the noise in the case of shining a flashlight on the wall is implicitly left as an exercise for the reader.
        $endgroup$
        – JiK
        1 hour ago










      • $begingroup$
        I'm struggling with this answer--a sinusoidal pattern of impinging intensity? What exact part in the video do you mean?
        $endgroup$
        – smcs
        40 mins ago










      • $begingroup$
        @JiK what matters is that the answer is arguing that intensity doesn't change in time unless the light source is coherent, but this is wrong. The justification for no change uses the word average and also the $E$ field must change in order for there to be a wave at all. The only wave with constant intensity vs time is a coherent and circularly polarized one. If you could detect variation in intensity for coherent light vs. time, you could also detect variation for incoherent light, unless we're talking about a spatial average over the area of the detector.
        $endgroup$
        – EL_DON
        9 mins ago
















      8












      $begingroup$

      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, the sinusoidal pattern of impinging intensity . After all, mathematics allows us to materialize thought experiments as this one.






      share|cite|improve this answer









      $endgroup$









      • 1




        $begingroup$
        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
        6 hours ago






      • 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
        6 hours ago










      • $begingroup$
        @EL_DON Calculating the variance in the magnitude caused by the noise in the case of shining a flashlight on the wall is implicitly left as an exercise for the reader.
        $endgroup$
        – JiK
        1 hour ago










      • $begingroup$
        I'm struggling with this answer--a sinusoidal pattern of impinging intensity? What exact part in the video do you mean?
        $endgroup$
        – smcs
        40 mins ago










      • $begingroup$
        @JiK what matters is that the answer is arguing that intensity doesn't change in time unless the light source is coherent, but this is wrong. The justification for no change uses the word average and also the $E$ field must change in order for there to be a wave at all. The only wave with constant intensity vs time is a coherent and circularly polarized one. If you could detect variation in intensity for coherent light vs. time, you could also detect variation for incoherent light, unless we're talking about a spatial average over the area of the detector.
        $endgroup$
        – EL_DON
        9 mins ago














      8












      8








      8





      $begingroup$

      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, the sinusoidal pattern of impinging intensity . After all, mathematics allows us to materialize thought experiments as this one.






      share|cite|improve this answer









      $endgroup$



      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, the sinusoidal pattern of impinging intensity . After all, mathematics allows us to materialize thought experiments as this one.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered 7 hours ago









      anna vanna v

      161k8153454




      161k8153454








      • 1




        $begingroup$
        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
        6 hours ago






      • 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
        6 hours ago










      • $begingroup$
        @EL_DON Calculating the variance in the magnitude caused by the noise in the case of shining a flashlight on the wall is implicitly left as an exercise for the reader.
        $endgroup$
        – JiK
        1 hour ago










      • $begingroup$
        I'm struggling with this answer--a sinusoidal pattern of impinging intensity? What exact part in the video do you mean?
        $endgroup$
        – smcs
        40 mins ago










      • $begingroup$
        @JiK what matters is that the answer is arguing that intensity doesn't change in time unless the light source is coherent, but this is wrong. The justification for no change uses the word average and also the $E$ field must change in order for there to be a wave at all. The only wave with constant intensity vs time is a coherent and circularly polarized one. If you could detect variation in intensity for coherent light vs. time, you could also detect variation for incoherent light, unless we're talking about a spatial average over the area of the detector.
        $endgroup$
        – EL_DON
        9 mins ago














      • 1




        $begingroup$
        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
        6 hours ago






      • 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
        6 hours ago










      • $begingroup$
        @EL_DON Calculating the variance in the magnitude caused by the noise in the case of shining a flashlight on the wall is implicitly left as an exercise for the reader.
        $endgroup$
        – JiK
        1 hour ago










      • $begingroup$
        I'm struggling with this answer--a sinusoidal pattern of impinging intensity? What exact part in the video do you mean?
        $endgroup$
        – smcs
        40 mins ago










      • $begingroup$
        @JiK what matters is that the answer is arguing that intensity doesn't change in time unless the light source is coherent, but this is wrong. The justification for no change uses the word average and also the $E$ field must change in order for there to be a wave at all. The only wave with constant intensity vs time is a coherent and circularly polarized one. If you could detect variation in intensity for coherent light vs. time, you could also detect variation for incoherent light, unless we're talking about a spatial average over the area of the detector.
        $endgroup$
        – EL_DON
        9 mins ago








      1




      1




      $begingroup$
      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
      6 hours ago




      $begingroup$
      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
      6 hours ago




      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
      6 hours ago




      $begingroup$
      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
      6 hours ago












      $begingroup$
      @EL_DON Calculating the variance in the magnitude caused by the noise in the case of shining a flashlight on the wall is implicitly left as an exercise for the reader.
      $endgroup$
      – JiK
      1 hour ago




      $begingroup$
      @EL_DON Calculating the variance in the magnitude caused by the noise in the case of shining a flashlight on the wall is implicitly left as an exercise for the reader.
      $endgroup$
      – JiK
      1 hour ago












      $begingroup$
      I'm struggling with this answer--a sinusoidal pattern of impinging intensity? What exact part in the video do you mean?
      $endgroup$
      – smcs
      40 mins ago




      $begingroup$
      I'm struggling with this answer--a sinusoidal pattern of impinging intensity? What exact part in the video do you mean?
      $endgroup$
      – smcs
      40 mins ago












      $begingroup$
      @JiK what matters is that the answer is arguing that intensity doesn't change in time unless the light source is coherent, but this is wrong. The justification for no change uses the word average and also the $E$ field must change in order for there to be a wave at all. The only wave with constant intensity vs time is a coherent and circularly polarized one. If you could detect variation in intensity for coherent light vs. time, you could also detect variation for incoherent light, unless we're talking about a spatial average over the area of the detector.
      $endgroup$
      – EL_DON
      9 mins ago




      $begingroup$
      @JiK what matters is that the answer is arguing that intensity doesn't change in time unless the light source is coherent, but this is wrong. The justification for no change uses the word average and also the $E$ field must change in order for there to be a wave at all. The only wave with constant intensity vs time is a coherent and circularly polarized one. If you could detect variation in intensity for coherent light vs. time, you could also detect variation for incoherent light, unless we're talking about a spatial average over the area of the detector.
      $endgroup$
      – EL_DON
      9 mins ago











      6












      $begingroup$

      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 vec{E}timesvec{B}$ 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.






      share|cite|improve this answer











      $endgroup$









      • 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
        7 hours ago








      • 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
        4 hours ago










      • $begingroup$
        @ahemmetter: The Poynting vector is a vector, not an intensity $I$.
        $endgroup$
        – Vladimir Kalitvianski
        4 hours ago






      • 1




        $begingroup$
        The intensity is directly proportional to the magnitude of the Poynting vector.
        $endgroup$
        – ahemmetter
        2 hours ago
















      6












      $begingroup$

      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 vec{E}timesvec{B}$ 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.






      share|cite|improve this answer











      $endgroup$









      • 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
        7 hours ago








      • 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
        4 hours ago










      • $begingroup$
        @ahemmetter: The Poynting vector is a vector, not an intensity $I$.
        $endgroup$
        – Vladimir Kalitvianski
        4 hours ago






      • 1




        $begingroup$
        The intensity is directly proportional to the magnitude of the Poynting vector.
        $endgroup$
        – ahemmetter
        2 hours ago














      6












      6








      6





      $begingroup$

      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 vec{E}timesvec{B}$ 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.






      share|cite|improve this answer











      $endgroup$



      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 vec{E}timesvec{B}$ 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.







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited 19 mins ago

























      answered 7 hours ago









      Vladimir KalitvianskiVladimir Kalitvianski

      11.2k11334




      11.2k11334








      • 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
        7 hours ago








      • 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
        4 hours ago










      • $begingroup$
        @ahemmetter: The Poynting vector is a vector, not an intensity $I$.
        $endgroup$
        – Vladimir Kalitvianski
        4 hours ago






      • 1




        $begingroup$
        The intensity is directly proportional to the magnitude of the Poynting vector.
        $endgroup$
        – ahemmetter
        2 hours ago














      • 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
        7 hours ago








      • 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
        4 hours ago










      • $begingroup$
        @ahemmetter: The Poynting vector is a vector, not an intensity $I$.
        $endgroup$
        – Vladimir Kalitvianski
        4 hours ago






      • 1




        $begingroup$
        The intensity is directly proportional to the magnitude of the Poynting vector.
        $endgroup$
        – ahemmetter
        2 hours ago








      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
      7 hours ago






      $begingroup$
      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
      7 hours ago






      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
      4 hours ago




      $begingroup$
      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
      4 hours ago












      $begingroup$
      @ahemmetter: The Poynting vector is a vector, not an intensity $I$.
      $endgroup$
      – Vladimir Kalitvianski
      4 hours ago




      $begingroup$
      @ahemmetter: The Poynting vector is a vector, not an intensity $I$.
      $endgroup$
      – Vladimir Kalitvianski
      4 hours ago




      1




      1




      $begingroup$
      The intensity is directly proportional to the magnitude of the Poynting vector.
      $endgroup$
      – ahemmetter
      2 hours ago




      $begingroup$
      The intensity is directly proportional to the magnitude of the Poynting vector.
      $endgroup$
      – ahemmetter
      2 hours ago











      2












      $begingroup$

      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.






      share|cite|improve this answer











      $endgroup$


















        2












        $begingroup$

        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.






        share|cite|improve this answer











        $endgroup$
















          2












          2








          2





          $begingroup$

          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.






          share|cite|improve this answer











          $endgroup$



          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.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 6 hours ago

























          answered 7 hours ago









          EL_DONEL_DON

          2,3332726




          2,3332726























              1












              $begingroup$

              "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.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                "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.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  "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.






                  share|cite|improve this answer









                  $endgroup$



                  "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.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 5 hours ago









                  my2ctsmy2cts

                  5,7892719




                  5,7892719






















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