# Frequency and energy of EM radiation

Am I correct that the direct proportionality between the energy and frequency of an EM wave can be obtained from classical electromagnetism? Of course there's the Planck–Einstein relation ##E = h \nu## for a photon, but that entails QM. I'm wondering about the relation ##E \propto \nu## for an EM wave—is that knowable without QM?

Ibix
In quantum mechanical terms, a classical EM wave is ##N## photons, where ##N>>1##. The energy is therefore ##Nh\nu##. Given the frequency of the wave, then, you can't know the total energy without knowing ##N## - which means counting photons, which implies using QM.

I think, anyway.

Delta2
phyzguy
In classical electromagnetism, an electromagnetic wave can have any arbitrary amount of energy, and an arbitrarily large or small amount of energy can be added to or extracted from the wave. It is only in QM that the energy of the wave is quantized in units of hν.

Delta2
Thank you both for your responses.

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?

Chandra Prayaga
In Classical EM, the energy of the wave is proportional to the square of the amplitude of the electric field, and is independent of the frequency. You can have any energy of the wave by changing the amplitude, at any frequency.

Klystron, davenn and phyzguy
In classical electromagnetism, an electromagnetic wave can have any arbitrary amount of energy, and an arbitrarily large or small amount of energy can be added to or extracted from the wave. It is only in QM that the energy of the wave is quantized in units of hν.
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.
In Classical EM, the energy of the wave is proportional to the square of the amplitude of the electric field, and is independent of the frequency. You can have any energy of the wave by changing the amplitude, at any frequency.

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

In Classical EM, the energy of the wave is proportional to the square of the amplitude of the electric field, and is independent of the frequency. You can have any energy of the wave by changing the amplitude, at any frequency.

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.

Chandra Prayaga
Does 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?
In classical EM, there IS no relation between Energy and frequency. The energy is proportional to the square of the electric field.
before QM, was it known that increasing the frequency of an emitted signal increased its energy in direct proportion?
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.

Klystron and davenn
Ibix
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.
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-5974014

Thanks for the replies. Cheers!

davenn
Gold Member
2021 Award
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

This has no direct bearing of the original question, but I am trying to clarify my minor epiphany of post #6.

Please correct me if I am wrong: if we have two waves, one of frequency ν1 and one of a greater frequency ν2, the one with the greater amplitude will be of greater energy assuming we measure an equal number of wavelengths of each wave. If the amplitude of the frequency ν2 is the smaller of the two by a small amount and we measured the two waves for equal times, we would get more wavelengths at frequency ν2 and might have a greater energy incident on the detector that we get for frequency ν1.

mjc123
Homework Helper
No, the 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 cm2 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).

great responses

Hopefully @SiennaTheGr8 got the understanding

Dave

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 wavelength transforms, and then invoke the inverse-proportionality of wavelength and frequency to prove that frequency transforms in the "opposite" way (i.e., swap the "primed" and "unprimed" quantities: ##\lambda^\prime / \lambda = \nu / \nu^\prime##).

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?

No, the 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 cm2 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).
Of course. I am embarrassed that did not occur to me.

Thank you.

ZapperZ
Staff Emeritus
Actually, contrary to a few of the responses in this thread, there IS a connection between energy of a "wave" and its frequency. This connection is in addition to its dependence of the amplitude of the wave.

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.

Chandra Prayaga
I would like to look at further details of post #16 of ZapperZ which raises very important questions.
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.

ZapperZ
Staff Emeritus
I would like to look at further details of post #16 of ZapperZ which raises very important questions.
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.

Chandra Prayaga
Let us take a detector for which the response is as fast as we want. The current (or any response function) is proportional to the square of the electric field:
i(t ) = k E02 cos2(ωt ) = ½ k E02 (1 + cos(2ωt ))
If we accumulate the signal over a time T, we are measuring the total charge accumulated:
q(T ) = ½ k E02 ∫(1 + cos(2ωt ))dt, where the integration is from 0 to T

q
(T ) = ½ k E02T + ½ k E02T sin(2ωT ) / (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.

ZapperZ
Staff Emeritus
Let us take a detector for which the response is as fast as we want. The current (or any response function) is proportional to the square of the electric field:
i(t ) = k E02 cos2(ωt ) = ½ k E02 (1 + cos(2ωt ))
If we accumulate the signal over a time T, we are measuring the total charge accumulated:
q(T ) = ½ k E02 ∫(1 + cos(2ωt ))dt, where the integration is from 0 to T

q
(T ) = ½ k E02T + ½ k E02T sin(2ωT ) / (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.

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

ZapperZ
Staff Emeritus
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.

Chandra Prayaga
In fact it does not make ay sense. But a turbine is not a simple harminic oscillator. The SHO is a system in which the spring plus the block attached to it form an isolated system with fixed total energy which never changes throughout the oscillation. A turbine is a lot more complicated system which is receiving energy from somewhere, and is depositing energy somewhere else.

ZapperZ
Staff Emeritus
In fact it does not make ay sense. But a turbine is not a simple harminic oscillator. The SHO is a system in which the spring plus the block attached to it form an isolated system with fixed total energy which never changes throughout the oscillation. A turbine is a lot more complicated system which is receiving energy from somewhere, and is depositing energy somewhere else.

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.

Chandra Prayaga
Ah. I am afraid I did not realize that you were suggesting a driven oscillator. I agree in fact, that is a very good model for a detector of EM waves. A radio receiver. So, if we look at our detector as modelled by a driven oscillator (LCR circuit of the receiver), then a driven oscillator always has a resonant behavior, with maximum energy transfer from the driving agency (the EM wave) occuring at the natural frequency of the oscillator. The energy transfer does not increase with increasing frequency. On either side of the resonance frequency, the response falls off.

ZapperZ
Staff Emeritus
Ah. I am afraid I did not realize that you were suggesting a driven oscillator. I agree in fact, that is a very good model for a detector of EM waves. A radio receiver. So, if we look at our detector as modelled by a driven oscillator (LCR circuit of the receiver), then a driven oscillator always has a resonant behavior, with maximum energy transfer from the driving agency (the EM wave) occuring at the natural frequency of the oscillator. The energy transfer does not increase with increasing frequency. On either side of the resonance frequency, the response falls off.

I only mentioned a driven oscillator so that there is no amplitude drop-off. Not sure why "resonance" now comes into this.

If you look at a signal generator, or even an RF source, you have to put in a lot more power into the generator to create EM signals with the same amplitude but at a higher frequency. If I use that RF signal in a particle accelerator, I get a lot more energy output at a higher frequency than at a lower frequency (higher average current of the accelerated particles), etc. And all of this within the wave picture of EM radiation, not the quantum picture. There are just MORE of "things done" per second at a higher frequency!

I feel as if this is going nowhere because I keep repeating the SAME thing, and you are taking this off on a tangent somewhere. So I'm done with this one.

Zz.

Chandra Prayaga
OK. We were discussing the detection of EM radiation, because you made a very important point about accumulation of the energy by the detector. I pointed out that whatever method you use, the energy detected does not increase with frequency. Also, please note that any mechanical or electrical oscillating system has a resonance. You cannot avoid it. Now, in this last post, you are talking of generation of EM radiation, and not detection.
But, if you don't want to carry out the discussion further, that is up to you.

If you look at a signal generator, or even an RF source, you have to put in a lot more power into the generator to create EM signals with the same amplitude but at a higher frequency. If I use that RF signal in a particle accelerator, I get a lot more energy output at a higher frequency than at a lower frequency (higher average current of the accelerated particles), etc. And all of this within the wave picture of EM radiation, not the quantum picture. There are just MORE of "things done" per second at a higher frequency!

Dear ZapperZ, what you say can intuitively make sense, but it surprises me as well as Chandra Prayaga, maybe because we are lacking some mechanical waves concepts and using a different perspective.
When for example Electro-magnetic waves are introduced in a classical approach, the Electric field is assumed to be ##E_0 \cos(\omega t - kx)## and the energy carried the Electro-magnetic wave per unit volume is
$$w = \displaystyle \frac{1}{2} \epsilon_0 E_0^2$$
as stated for example here. It does not depend on ##\omega## at all. Not only: in a power line, for example a home socket, the product of current and voltage is something like
$$V_0 \cos^2 (\omega t) = V_0 \displaystyle \frac{1 + \cos(2 \omega t)}{2}$$
but only the ##V_0 / 2## term, which again does not depend on ##\omega##, carries active power and is able to definitively deliver energy from the generator to the load (a light bulb, an oven, etc.). So, maybe our misunderstanding is a matter of definitions.

In fact, in another thread almost about this same topic, a link is provided: http://spiff.rit.edu/classes/phys207/lectures/waves/wave_energy.html. In that page, the average kinetic energy transmitted to the particles of a cylinder crossed by a mechanical wave ##A \cos (kx - \omega t)## is
$$KE_{\mathrm{avg}} = \displaystyle \frac{1}{4} M \omega^2 A^2$$
where ##M## is the mass of the cylinder. It depends on ##\omega## and completely agrees with you. It is obtained, I guess,
• stating that ##A \cos (kx - \omega t) = x(t)## is the position of the particles of the cylinder,
• then deriving this expression to obtain their velocity,
• and finally squaring the derivative to obtain the kinetic energy.
What is different here from the Electro-magnetic waves / home socket perspective? Both are about the transferred energy, they should give coherent results, but they apparently don't. Please, consider joining again this thread, we are only trying to better understand.
Thank you,

Emily

Klystron
If you can, please, give a quantitative description with some examples, like the cosinusoidal functions in my post. Also the introduction of the exact name for each quantity (transferred energy, available power, reactive power, etc.) surely would be another help to understand.

ZapperZ
Staff Emeritus
Dear ZapperZ, what you say can intuitively make sense, but it surprises me as well as Chandra Prayaga, maybe because we are lacking some mechanical waves concepts and using a different perspective.
When for example Electro-magnetic waves are introduced in a classical approach, the Electric field is assumed to be ##E_0 \cos(\omega t - kx)## and the energy carried the Electro-magnetic wave per unit volume is
$$w = \displaystyle \frac{1}{2} \epsilon_0 E_0^2$$
as stated for example here. It does not depend on ##\omega## at all. Not only: in a power line, for example a home socket, the product of current and voltage is something like
$$V_0 \cos^2 (\omega t) = V_0 \displaystyle \frac{1 + \cos(2 \omega t)}{2}$$
but only the ##V_0 / 2## term, which again does not depend on ##\omega##, carries active power and is able to definitively deliver energy from the generator to the load (a light bulb, an oven, etc.). So, maybe our misunderstanding is a matter of definitions.

In fact, in another thread almost about this same topic, a link is provided: http://spiff.rit.edu/classes/phys207/lectures/waves/wave_energy.html. In that page, the average kinetic energy transmitted to the particles of a cylinder crossed by a mechanical wave ##A \cos (kx - \omega t)## is
$$KE_{\mathrm{avg}} = \displaystyle \frac{1}{4} M \omega^2 A^2$$
where ##M## is the mass of the cylinder. It depends on ##\omega## and completely agrees with you. It is obtained, I guess,
• stating that ##A \cos (kx - \omega t) = x(t)## is the position of the particles of the cylinder,
• then deriving this expression to obtain their velocity,
• and finally squaring the derivative to obtain the kinetic energy.
What is different here from the Electro-magnetic waves / home socket perspective? Both are about the transferred energy, they should give coherent results, but they apparently don't. Please, consider joining again this thread, we are only trying to better understand.
Thank you,

Emily

What exactly does it mean by "average" here? You will notice that for sinusoidal waves, both time average and spatial average are done over one complete cycle! After all, you can't carry the integration to infinity.

So the expression you have here is the average value relevant to one cycle. It is why there is no time dependence (or spatial dependence) on the energy of such a wave. If one were to measure power, which is what I started with (i.e. the amount of energy measured in a unit time), then which wave will produce more energy for a fixed amplitude (i.e. they have the SAME average energy per cycle): the one that makes more complete cycle per unit time, or the one that makes less complete cycle per unit time?

Zz.

If one were to measure power, which is what I started with (i.e. the amount of energy measured in a unit time), then which wave will produce more energy for a fixed amplitude (i.e. they have the SAME average energy per cycle): the one that makes more complete cycle per unit time, or the one that makes less complete cycle per unit time?

Case 1: Let's say we have a 12 V battery, from witch we draw 1 Ampere for one second, zero Amperes for one second, 1 Ampere for one second ... and so on.

Now the RMS voltage of that current is ## \frac{12}{ \sqrt 2} ##

And the average power is 6 W.

Case 2: Let's say we have a 12 V battery, from witch we draw 1 Ampere for two second, zero Amperes for two seconds, 1 Ampere for two seconds ... and so on.

The RMS voltage of that current is still ## \frac{12}{ \sqrt 2} ##

And the average power is still 6 W.

(The voltage varies between 0 and 12, that is the idea, for some reason I talked about amperage)

Last edited:
ZapperZ
Staff Emeritus
Case 1: Let's say we have a 12 V battery, from witch we draw 1 Ampere for one second, zero Amperes for one second, 1 Ampere for one second ... and so on.

Now the RMS voltage of that current is ## \frac{12}{ \sqrt 2} ##

And the average power is 6 W.

Case 2: Let's say we have a 12 V battery, from witch we draw 1 Ampere for two second, zero Amperes for two seconds, 1 Ampere for two seconds ... and so on.

The RMS voltage of that current is still ## \frac{12}{ \sqrt 2} ##

And the average power is still 6 W.

(The voltage varies between 0 and 12, that is the idea, for some reason I talked about amperage)

I don't think this is what we're talking about. The average value doesn't change, because this is the same value averaged over one cycle. But there is a difference if your measurement window measures 10 complete oscillations, or 20 complete oscillations. The device will measure the the total energy within that window. This is what I tried to emphasize in my first post here.

Zz.