Wave Theory of Light: Energy Independent of Frequency?

In summary, the energy of light in classical wave theory is dependent on its brightness and independent of its frequency. However, when considering energy per unit time, frequency does play a role in determining the overall energy of the light. This is due to the fact that higher frequencies result in more cycles in a given unit of time, resulting in a higher energy. However, the energy of a photon is directly proportional to its frequency, not the square of its frequency. This can be seen in the equation E=hf, where h is Planck's constant. Other types of waves, such as sound and water waves, also exhibit a similar relationship between frequency and energy.
  • #1
kkmans
7
0
For wave theory of light, it said that energy of light is depend only on its brightness and independent of its frequency.
I would like to ask which equation shows that wave energy is independent of frequency..??

thank you.
 
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  • #2
kkmans said:
For wave theory of light, it said that energy of light is depend only on its brightness and independent of its frequency.

You should cite this source. That is the one thing we try to make people do when they talk about something they read or heard.

I would like to ask which equation shows that wave energy is independent of frequency..??

Actually, it DOES! Maybe the energy PER CYCLE or per period isn't dependent on frequency in classical wave theory of light, but energy in a unit time does! Think about it. The higher the frequency, the more cycles in that unit time that you measure, the more energy you will get. So yes, even in classical wave theory of light, frequency does matter IF you are looking at it per unit time.

Zz.
 
  • #3
kkmans said:
For wave theory of light, it said that energy of light is depend only on its brightness and independent of its frequency.
I would like to ask which equation shows that wave energy is independent of frequency..??

thank you.

Then energy of light actually is dependant on frequency by the relation [tex]E=hf[/tex] where h is Planck's constant.
 
  • #4
americanforest said:
Then energy of light actually is dependant on frequency by the relation [tex]E=hf[/tex] where h is Planck's constant.

That can be restated:

The energy of a photon is dependant on the frequency. I think its incorrect to say the energy of light with respect to that equation.
 
  • #5
ranger said:
That can be restated:

The energy of a photon is dependant on the frequency. I think its incorrect to say the energy of light with respect to that equation.


But the light is made up of photons. The only way this dependancy couldn't be extended to light frequencies in general is if there were less photons in high frequency light than in low frequency light. Is this true?
 
  • #6
the equation of light intensity

the equation of average intensity
I = .5 nceεo sqr(E)
n refractive index of medium
c velocity of light
εo permativity of medium
E max value of Electric filed
as we know that light is an electromagnetic wave has 2 perpendicular fields electric and magnetic
 
  • #7
americanforest said:
But the light is made up of photons. The only way this dependancy couldn't be extended to light frequencies in general is if there were less photons in high frequency light than in low frequency light. Is this true?

It doesn't mean less or more photons, it means photons with lower/higher energy levels.
 
  • #8
ranger, that wasn't what he said. He's saying that the only way kkmans' scenario would work is if the numbers of photons were intrinsically related to the frequency of the light 'beam'/ray/whatever. This is quite obviously not true.
 
  • #9
Sojourner01, I thought that's what I said in my previous reply.
 
  • #10
You said exactly opposite that:

ranger said:
It doesn't mean less or more photons
 
  • #11
Does one not measure brightness/intensity in units of power per area?

The intensity is proportional to the square of the amplitude, not dependent on the wave frequency.

When one later discovers light quantised in packets of energy proportional to frequency, one must infer (to avoid contradiction) that for a red and blue light source of equal measured brightness there will be a higher rate of photons received from the lower frequency source.
 
  • #12
cesiumfrog said:
When one later discovers light quantised in packets of energy proportional to frequency, one must infer (to avoid contradiction) that for a red and blue light source of equal measured brightness there will be a higher rate of photons received from the lower frequency source.


This is what I was trying to say in my earlier post, I'm sorry if I wasn't clear.
 
  • #13
>You should cite this source. That is the one thing we try to make people do when they talk about something they read or heard.
oops, I read this from wikipedia.

let's forget about the light, what about other waves like sound and water ?
If we consider wave as the particles performing SHM, their energy is depends on the frequency.
Is there anything wrong with my concept..??

thank you.
 
  • #14
kkmans said:
>You should cite this source. That is the one thing we try to make people do when they talk about something they read or heard.
oops, I read this from wikipedia.

let's forget about the light, what about other waves like sound and water ?
If we consider wave as the particles performing SHM, their energy is depends on the frequency.
Is there anything wrong with my concept..??

thank you.

Why did you quoted one part of my response while ignore the rest that would have answered your question here?

ZapperZ said:
Actually, it DOES! Maybe the energy PER CYCLE or per period isn't dependent on frequency in classical wave theory of light, but energy in a unit time does! Think about it. The higher the frequency, the more cycles in that unit time that you measure, the more energy you will get. So yes, even in classical wave theory of light, frequency does matter IF you are looking at it per unit time.

Doesn't THAT address what you just asked?

Zz.
 
  • #15
hmm..I understand if I consider it per unit time, the intensity is directly proportional to the energy, not the frequecy.
My problem is that is the expression of "energy" include frequency ?
Like other waves, the particles are performing SHM, and their energy is proportional to the square of the frequceny.
 
  • #16
Like other waves, the particles are performing SHM, and their energy is proportional to the square of the frequceny.

No. No no no. The energy of a photon is directly proportional to its frequency. Not the square. Linear dependence. No question about it.

Review your ideas of what 'intensity' is. I sense some confusion of concepts creeping in here.
 
  • #17
ZapperZ said:
The higher the frequency, the more cycles in that unit time that you measure, the more energy you will get. So yes, even in classical wave theory of light, frequency does matter IF you are looking at it per unit time.
Do you have a reference to support that?
 
  • #18
>energy is proportional to the square of the frequceny
If we consider waves as particle performing SHM, then their energy
= 1/2m ω^2 A^2, where ω = 2πf

Can I use this concept in wave theory of light ?
 
  • #19
cesiumfrog said:
Do you have a reference to support that?

No I don't. Rather, look at a mass-spring system. Find the energy per second of the system at a particular amplitude. Now replace the spring so that you have a system that doubles the natural frequency. Now measure again the amount of energy of the system per second with the SAME amplitude. Are you telling me you need a "reference" to figure out that the amount of energy in that unit time has increased?

Zz.
 
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  • #20
The thing I consider confusing in this thread is the way frequency is being thought of. It simply is a number of an event in an amount of time. In this case we are talking about waves. Therefore, the number of oscillations per unit time, be it light waves or mechanical waves the same principals apply. The relation to energy in the quantum mechanical formula E=hf i think is being miss interpreted. Here the frequency is just a multiplier where as Planck's constant is a number related to the most discrete quantisation of energy. Remember where this formula was derived from.
 
  • #21
I think people are forgetting the OP. The original question is based on the wave theory of light. We are not bringing in any quantum effect of light here.

Zz.
 
  • #22
ZapperZ said:
No I don't. Rather, look at a mass-spring system. Find the energy per second of the system at a particular amplitude. Now replace the spring so that you have a system that doubles the natural frequency. Now measure again the amount of energy of the system per second with the SAME amplitude. Are you telling me you need a "reference" to figure out that the amount of energy in that unit time has increased?

According to both Griffiths Electrodynamics (p.381) and Saleh & Teich Photonics (p.44), the intensity of a monochromatic light wave is proportional to the square of the amplitude and independent of frequency (just like the average of [itex]cos^2\ \omega t[/itex] over [itex]t[/itex]). Energy per second is simply intensity, integrated over an area.
 
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  • #23
Do you have a definitive piece of reasoning that makes you certain that the power of a beam is equivalent to the energy residing in the photons that comprise it? It sounds obvious but isn't necessarily, so be careful.
 
  • #24
cesiumfrog said:
According to both Griffiths Electrodynamics (p.381) and Saleh & Teich Photonics (p.44), the intensity of a monochromatic light wave is proportional to the square of the amplitude and independent of frequency (just like the average of [itex]cos^2\ \omega t[/itex] over [itex]t[/itex]). Energy per second is simply intensity, integrated over an area.

I have the 2nd Edition of Griffiths, and pg. 381 has nothing of that sort.

If you look in the section titled "Energy and Momentum of Electromagnetic Waves" (section 8.2.2 in Griffiths 2nd Edition), you would have seen that the energy of a monochromatic plane wave is

[tex]U = \epsilon_0 E^2_0 cos^2(\kappa x - \omega t + \delta)[/tex]

This clearly has a frequency term. The confusion comes in when you are looking at the total energy that has been integrated over the volume of space that is relevant to the problem. THEN you have integrated out the effects of frequency or the spatial extent of the oscillation. You then no longer have either the time dependent, or the volume dependent. Now all you care about is how much of that goes through an area that you are looking at.

Again, look at the simplest wave equation, of which a mass-spring system is the simplest visual example. The amount of energy in a particular time that the system makes clearly depends on the frequency if you fix the amplitude. This is the most transparent example I can give.

Zz.
 
  • #25
ZapperZ said:
I have the 2nd Edition of Griffiths, and pg. 381 has nothing of that sort. If you look in the section titled "Energy and Momentum of Electromagnetic Waves" [..]
In the 3rd edition, a section with that name derives that for a monochromatic wave with electric field amplitude [itex]E_0[/itex] the intensity is [itex]\frac 1 2 c \epsilon_0 E^2_0[/itex] (eq. 9.63).

ZapperZ said:
[tex]U = \epsilon_0 E^2_0 cos^2(\kappa x - \omega t + \delta)[/tex]This clearly has a frequency term.
And just as clearly, if you average this over unit time (as you https://www.physicsforums.com/showthread.php?p=1224951#post1224951" specify, rather than integrating over just one cycle), that frequency term will dissappear.

ZapperZ said:
Again,[..] a mass-spring system is the simplest visual example.
Regardless of the behaviour of such a system, the fact it contains non-zero rest-mass is enough for me to doubt the equivalence. It seems more analogous to a free electron in an electric field.
 
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  • #26
cesiumfrog said:
In the 3rd edition, a section with that name derives that for a monochromatic wave with electric field amplitude [itex]E_0[/itex] the intensity is [itex]\frac 1 2 c \epsilon_0 E^2_0[/itex] (eq. 9.63).


And just as clearly, if you average this over unit time (as you https://www.physicsforums.com/showthread.php?p=1224951#post1224951"specify, rather than integrating over just one cycle), that frequency term will dissappear.

But "averaging" isn't the issue here, is it? When you average over time, you integrate out the time dependence. That's why you don't have any frequency dependence on the energy, because you sum up over the time period that you want.

All you need to do is sum up the total energy that impinges on a surface for a particular time. Then do the same for another EM wave with a different frequency. Is the energy "received" the same?

You haven't addressed my simple example with the mass-spring system.

Zz.
 
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  • #27
ZapperZ said:
All you need to do is sum up the total energy that impinges on a surface for a particular time. Then do the same for another EM wave with a different frequency. Is the energy "received" the same?

You haven't addressed my simple example with the mass-spring system.

(I haven't addressed your mass-spring example because it is fundamentally unlike the situation in question.)

Yes, the total energy "received" impinging on a surface for a particular time will be equal and the same if repeated with EM waves of different frequency (but equal amplitude). This is what the textbooks are saying, with the caveat that the "particular time" must not be small compared to the period of oscillation, and it completely contradicts your initial answer to the OP.
 
  • #28
cesiumfrog said:
(I haven't addressed your mass-spring example because it is fundamentally unlike the situation in question.)

Yes, the total energy "received" impinging on a surface for a particular time will be equal and the same if repeated with EM waves of different frequency (but equal amplitude). This is what the textbooks are saying, with the caveat that the "particular time" must not be small compared to the period of oscillation, and it completely contradicts your initial answer to the OP.

Why is it not the same?

The energy of the mass-spring system is also characterized by the amplitude of the oscillation. You can still write it in the same form as the monochromatic light.

And if we're talking about the OP, I will also point out that there's no mention of any "average" of any kind here. What you are quoting the "textbook" is not the relevant case here.

Zz.
 
  • #29
kkmans said:
I would like to ask which equation shows that wave energy is independent of frequency..?
Look in any first-year general physics textbook for the energy transmitted by any classical wave (string, sound, EM, etc).

ZapperZ said:
(section 8.2.2 in Griffiths 2nd Edition), you would have seen that the energy of a monochromatic plane wave is
[tex]U = \epsilon_0 E^2_0 cos^2(\kappa x - \omega t + \delta)[/tex]
This clearly has a frequency term. The confusion comes in when you are looking at the total energy [..] THEN you have integrated out the effects of frequency
Clearly this is the authoritive source (albeit a less developed version) so you should be able to find in the following paragraphs an expression for the intensity (likely represented as "I" rather than "U").

The first thing you should notice about the dependence you're giving is that it does not cause the "integrated" energy (over any period of time exceeding about a [itex]10^{-15}[/itex]th of a second) to increase with frequency, and this really should contradict what you have said. Moreover, if you cling to this definition, then you can never define a practical "energy per unit time" on any measurable scale.

ZapperZ said:
Why is it not the same?
You're asking why, given a particular amplitude, the energy transmitted through a particular point by possible electromagnetic waves (with different frequencies) is not proportional to the internal energies of different stationary isolated mass-springs (in which the position of the mass changes, but also with a particular amplitude, and the frequencies are equivalent with those of the EM waves)?

I really think I need you to explain what these two examples have in common. Then, why do you choose increasing the spring stiffness to change frequency (which also increases energy) instead of just decreasing the mass to change frequency (which does not change the energy)? Thirdly, how do you then derive from this the equivalent equation to in Griffiths?
 
  • #30
cesiumfrog said:
Look in any first-year general physics textbook for the energy transmitted by any classical wave (string, sound, EM, etc).Clearly this is the authoritive source (albeit a less developed version) so you should be able to find in the following paragraphs an expression for the intensity (likely represented as "I" rather than "U").

The first thing you should notice about the dependence you're giving is that it does not cause the "integrated" energy (over any period of time exceeding about a [itex]10^{-15}[/itex]th of a second) to increase with frequency, and this really should contradict what you have said. Moreover, if you cling to this definition, then you can never define a practical "energy per unit time" on any measurable scale.

I can't? Why not?

You asked why my mass-spring system is similar to this, yet you asked me to look up ANY classical wave. Isn't the SHO done by the mass-spring system A CLASSICAL WAVE description?

If you are telling me that the frequency isn't a part of the total energy, then it means that there is no time dependence at all in the instantaneous energy. For example, if I have a point (or a surface) in a particular location, and I have a plane monochromatic EM wave passing through that point and I look at the number of oscillation that crosses through that point, I see the time dependence in the magnitude of the E-field. If I have a charge at that point, that oscillating field will be doing work onto the charge. If I have the same EM field, with the same amplitude, but with a larger frequency, I see MORE oscillation of that charge, meaning more work was done onto the charge by the field in the same time period.

In fact, I can stop time, and at any part of the oscillation, I can figure out what is the energy of that system, similar to the energy of the mass at any point of the oscillation. Different part of the oscillation will produce different energy that the charge has acquired. There IS a time dependence! In fact, I make use of this effect everytime I run our accelerator! I shoot a 10 fs laser pulse at our photocathode at different RF phase, and each phase produces different energy of the emitted photoelectrons. If I have a cw laser, I will have a spread in energy being emitted due to the spread in the energy given to the electron from the RF field! The energy from the field imparted to the charge has a time dependence and thus, a frequency dependence! It is only when you integrate out over time would you remove the implicit time dependence.

Zz.
 
  • #31
ZapperZ said:
Isn't the SHO done by the mass-spring system A CLASSICAL WAVE description?
No. Simple harmonic motion is not a propagating wave.

If you'd used a chain of coupled mass-springs, that would be different.
ZapperZ said:
If you are telling me that the frequency isn't a part of the total energy, then it means that there is no time dependence at all in the instantaneous energy.
It is incorrect to claim that the total energy increases with frequency. Nonetheless, there is a time dependence in the instantaneous energy: it oscillates from one femtosecond to the next, which is why the conventional measure of intensity is technically the average per unit time.

ZapperZ said:
For example, if I have a point (or a surface) in a particular location, and I have a plane monochromatic EM wave passing through that point and I look at the number of oscillation that crosses through that point, I see the time dependence in the magnitude of the E-field. If I have a charge at that point, that oscillating field will be doing work onto the charge. If I have the same EM field, with the same amplitude, but with a larger frequency, I see MORE oscillation of that charge, meaning more work was done onto the charge by the field in the same time period.

This is where I think you are becoming confused; the energy transmitted in an electromagnetic wave is a different concept to the energy transferred to an electron (indeed, you are aware that if the light wave is incident on a metal, much of the energy is merely reflected and so forth).

Now, a driven mass-spring oscillator is analogous to an antenna in the path of an EM wave. The electrons behave like the mass whilst the field they produce in the antenna acts as the restoring spring. For a constant driving amplitude, the energy transferred to the electrons will increase as the frequency approaches a resonance (although it will be a maximum, since higher frequency does not give the current enough time to gather).

ZapperZ said:
[..]There IS a time dependence! In fact, I make use of this effect everytime I run our accelerator! I shoot a 10 fs laser pulse at our photocathode at different RF phase[..]
We both agree the the higher frequency wave rises faster (and then falls faster, while the lower frequency is sustaining it's maximum longer). For a very short pulse envelope, I've no doubt controlling the phase will affect the total energy of the pulse itself.

What I'm disagreeing with is the content of your initial post:
ZapperZ said:
The higher the frequency, the more cycles in that unit time that you measure, the more energy you will get.
Your assumption that each cycle has equal energy is incorrect; the wattage is constant with amplitude and divided equally among whatever number of cycles occur in the second. Unless you are switching a pulse so quickly for it to matter which part of the cycle you cut, increasing the number of cycles per second does not increase the energy.

Note that the topic of this thread is the energy of classical light waves, not the photelectric effect. Classically, you would not expect ten minutes of blue light to have a different effect to ten minutes of red light (of equal intensity), because both transmit the same amount of energy to your apparatus. Hence it was a surprise that a different effect is seen, more so once shown (by then going to UV) it isn't just a result of one colour happening to be on a resonance.
 
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  • #32
ZapperZ said:
If you are telling me that the frequency isn't a part of the total energy, then it means that there is no time dependence at all in the instantaneous energy. For example, if I have a point (or a surface) in a particular location, and I have a plane monochromatic EM wave passing through that point and I look at the number of oscillation that crosses through that point, I see the time dependence in the magnitude of the E-field. If I have a charge at that point, that oscillating field will be doing work onto the charge. If I have the same EM field, with the same amplitude, but with a larger frequency, I see MORE oscillation of that charge, meaning more work was done onto the charge by the field in the same time period.
Sorry if I get into your discussion. What you say is infact very interesting. It seems to me you have found a classical reason to introduce the quantum of electromagnetic field!
As cesiumfrog says, from Poynting theorem it comes that energy per unit time is only given by |E |^2 and not by frequency. Now we know that, with a higher frequency of a field with given amplitude, the number of photons per unit time is lower and this, together with the fact the energy of every photon is higher, mantain the equality of energy.

But you, rightly, says that a *continuous* EM field of the same amplitude but higher frequency makes an higher work on an electron, e.g., and this gives a paradox, in my opinion.

So, the solution is that the field cannot be continuous!

I'm probably making some mistake.
 
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  • #33
lightarrow said:
Sorry if I get into your discussion. What you say is infact very interesting. It seems to me you have found a classical reason to introduce the quantum of electromagnetic field!
As cesiumfrog says, from Poynting theorem it comes that energy per unit time is only given by |E |^2 and not by frequency. Now we know that, with a higher frequency of a field with given amplitude, the number of photons per unit time is lower and this, together with the fact the energy of every photon is higher, mantain the equality of energy.

But you, rightly, says that a *continuous* EM field of the same amplitude but higher frequency makes an higher work on an electron, e.g., and this gives a paradox, in my opinion.

So, the solution is that the field must not be continuous!

I'm probably making some mistake.

No, cesiumfrog is correct. In my case, I WAS applying it to the excitation of of a charged particle, and the example would be the line antenna case or even the photoelectric effect. When we do such energy transfer, then yes, the frequency does come in for such a process, even in classical terms. It is only when we deal with the different intensity versus a fixed frequency is there now an issue with classical wave.

Zz.
 

1. What is the wave theory of light?

The wave theory of light is a scientific explanation of the nature of light that describes it as a form of electromagnetic radiation. According to this theory, light travels in the form of waves and can exhibit properties such as diffraction, interference, and polarization.

2. How does the wave theory of light differ from the particle theory of light?

The wave theory of light differs from the particle theory of light in that it describes light as a wave rather than a stream of particles. This theory also explains the behavior of light in terms of its frequency and wavelength, while the particle theory focuses on the energy and momentum of light particles.

3. What is meant by "energy independent of frequency" in the wave theory of light?

In the wave theory of light, the energy of a light wave is directly proportional to its frequency. This means that as the frequency of a light wave increases, so does its energy. However, the wave theory also states that the energy of a light wave is independent of its frequency, meaning that the energy of a light wave is not affected by changes in its frequency.

4. How does the wave theory of light explain the phenomenon of color?

The wave theory of light explains color as a result of the different frequencies of light waves. When white light passes through a prism, it is separated into its component colors because each color has a different frequency. This theory also explains how different colors can be created by combining light waves of different frequencies.

5. What are some practical applications of the wave theory of light?

The wave theory of light has many practical applications in fields such as optics, telecommunications, and photography. It is used to design and improve optical instruments such as microscopes and telescopes, and is also the basis for technologies such as fiber optics and lasers. Additionally, the wave theory of light is essential in understanding and manipulating light in photography and film production.

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