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- Thread starter SiennaTheGr8
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I think, anyway.

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phyzguy

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I don't think my question has quite been answered, though (or something is going over my head), so I'll rephrase:

Does the proportionality ##E \propto \nu## for a monochromatic EM wave come out of classical electromagnetism? Put differently: before QM, was it known that increasing the frequency of an emitted signal increased its energy in direct proportion?

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Chandra Prayaga

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As I understand it, that is the case becasue you can select an arbitrary portion of the wave...the longer the waves is let to land on a surface the more energy the surface receives.

Hmmm...that is also part of my upstanding of the wave description. I had not noticed the conflict before.

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If there is no relation between the energy and frequency of light in classical EM, and if classical EM is inherently compatible with special relativity, then I'm suddenly puzzled by the fact that the relativistic Doppler formulas for frequency and energy look identical (which suggests that they are indeed directly proportional). I'm missing something, I fear.

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Chandra Prayaga

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In classical EM, there IS no relation between Energy and frequencyDoes the proportionality E∝νE∝νE \propto \nu for a monochromatic EM wave come out of classical electromagnetism? Put differently: before QM, was it known that increasing the frequency of an emitted signal increased its energy in direct proportion?

before QM, was it known that increasing the frequency of an emitted signal increased its energy in direct proportion?

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The amplitudes change when you do a Lorentz transform. Factors of ##\gamma## are involved, and the B field in one frame can add or subtract a bit to the E field in the other, and likewise the E field can contribute to the transformed B field. I did the maths for a particular special case here (please disregard the slightly combative air of the post - it's not directed at you!): https://www.physicsforums.com/threads/constancy-of-the-speed-of-light.943215/page-3#post-5974014If there is no relation between the energy and frequency of light in classical EM, and if classical EM is inherently compatible with special relativity, then I'm suddenly puzzled by the fact that the relativistic Doppler formulas for frequency and energy look identical (which suggests that they are indeed directly proportional). I'm missing something, I fear.

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Thanks for the replies. Cheers!

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davenn

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In classical EM, there IS no relation between Energy and frequency. The energy is proportional to the square of the electric field.

We must clearly state what we are talking about. Even in QM, it is not true that the energy of a "signal" increases with increasing frequency. It is the energy of an individual photon that is proportional to the frequency. In a signal, you can emit a lot of photons at a lower frequency, or a few photons at a higher frequency, and the energy of the signal would be the same in both cases.

great responses

Hopefully @SiennaTheGr8 got the understanding

Dave

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Please correct me if I am wrong: if we have two waves, one of frequency ν

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mjc123

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Well, let's test that.

My confusion was (is?) essentially that the energy carried by an EM wave transforms between frames in precisely the same way that the wave's frequency does, and this was giving me the (wrong) idea that these quantities must be directly proportional in classical EM theory (as they are for a photon à la Planck/Einstein). Part of this confusion stems from the fact that many derivations of the relativistic Doppler formula begin by demonstrating how

But now I think I was drawing an unwarranted conclusion: although a direct proportionality might be enough to conclude that light-wave properties transform the same way, the converse isn't true: transforming the same way doesn't imply direct proportionality. And, in fact, the identical transformation of energy and frequency is "happenstance" from the point of view of classical EM (the Planck–Einstein relation is a red herring). To get the energy transformation, you've got to transform amplitude (or intensity?) and volume(?), and after some algebra the stars align and you get something that's identical in form to the frequency transformation.

Is that the gist of it?

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Of course. I am embarrassed that did not occur to me.intensity(energy per unit area per unit time) is proportional to the square of the amplitude. It is more meaningful to talk of the intensity than the energy of a classical EM wave. If you measure the energy incident on 1 cm^{2}for 1 second, the wave with greater amplitude will deliver more energy, irrespective of the frequency. If you were to measure the same number of wavelengths, you would measure the higher frequency wave for a shorter time, and measure less energy (assuming equal amplitudes).

Thank you.

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The issue here is what is meant by "energy". When I measure the energy of something, I open a device, measure the amount of energy being captured by the device for a window of time, and then I stop. If I open the window for 10 seconds, then I know that this is the total amount of energy that I captured in 10 seconds. If I wish to know how much energy is there in 1 second, I simply divide the total amount by 10.

Now this appears to be elementary, but for a cw source, such as a wave, this is not pedantic. Your photodiode that measures the amount of EM energy it has received needs to have a finite-length time window to acquire this energy. So the time factor here plays a role.

So how does that connects to the frequency of the wave. Think about it. If you have a spring of spring constant k1, and then you have another spring of spring constant k2, where k2>k1, then for the SAME AMPLITUDE of oscillation, spring k2 would have made more oscillation per second than spring k1! So a device that measure the amount of energy entering it per unit time would have measured MORE energy for something that oscillates MORE per second than something that oscillates less per second. Thus, the device measure more energy for a higher frequency wave than a lower frequency wave, even if they both have the same amplitude!

So yes, even classical physics at that time agreed with the notion that for a wave, the MEASURED energy per second gets higher with higher frequency. It is just that this energy also depends on amplitude (intensity) of the oscillation, something that QM picture does not agree with for light.

Zz.

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Chandra Prayaga

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Let us use a photodetector to measure the energy. The result of the measurement depends on the response time of the detector.

Consider the case where the detector response time is much longer than the time period of the wave. The detector output is then proportional to the average power, which takes out any dependence on frequency. In this case therefore, the detector output depends only on the square of the amplitude, and not on the frequency.

So for visible EM waves, the energy does not depend on the frequency.

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Let us use a photodetector to measure the energy. The result of the measurement depends on the response time of the detector.

Consider the case where the detector response time is much longer than the time period of the wave. The detector output is then proportional to the average power, which takes out any dependence on frequency. In this case therefore, the detector output depends only on the square of the amplitude, and not on the frequency.

So for visible EM waves, the energy does not depend on the frequency.

If you use a detector with such a response time, then you are not accurately measuring the energy of the wave.

The scenario that I presented assumes that the response time of a detector is not a factor. The detector is only that as a means to measure an amount of energy over a time window. It can be ideal (after all, we also do not care about the quantum efficiency of the detector).

Zz.

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Chandra Prayaga

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If we accumulate the signal over a time

q

The first term is the same one that you would get with a slow detector. The second term oscillates with the choice of the accumulation time, with no guarantee that it is automatically larger for larger frequencies.

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i(t) =k Ecos_{0}^{2}^{2}(ωt) = ½k E(1 + cos(2_{0}^{2}ωt))

If we accumulate the signal over a timeT, we are measuring the total charge accumulated:

q(T) = ½k E∫(1 + cos(2_{0}^{2}ωt))dt, where the integration is from 0 toT(

qT) = ½k E½_{0}^{2}T +k Esin(2_{0}^{2}TωT ) /(2ωT)

The first term is the same one that you would get with a slow detector. The second term oscillates with the choice of the accumulation time, with no guarantee that it is automatically larger for larger frequencies.

Why not?

Consider a spring system. The more it oscillates per second, the more WORK it can do in that time. An engine with a faster rep rate produces more energy than something with a lower rep rate. We are not talking about duty cycle here.

Zz.

- #21

Chandra Prayaga

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The work done by a harmonic oscillator in each cycle is zero.

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The work done by a harmonic oscillator in each cycle is zero.

So you are saying that a turbine does no work when it is running. Does this makes sense to you?

Zz.

- #23

Chandra Prayaga

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It’s the same as an oscillating system. Look at the oscillating current you get out of such a turbine.

A driven spring system is the exact same thing. The faster it more it moves per second, the more work it does, the more energy it generates in the unit time.

Zz.

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Chandra Prayaga

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