Nature of electromagnetic radiation.

  • #1
PrincePhoenix
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According to wikipedia,

"Electromagnetic radiation (often abbreviated E-M radiation or EMR) is a phenomenon that takes the form of self-propagating waves in a vacuum or in matter. It consists of electric and magnetic field components which oscillate in phase perpendicular to each other and perpendicular to the direction of energy propagation."

What is meant by "electric and magnetic field components"? And what oscillates in electromagnetic waves? In mechanical waves the particles of the medium oscillates like air, water molecules in water etc
 

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  • #2
Born2bwire
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An electromagnetic wave is made up of electric and magnetic fields oriented and related to each other according to a specific fashion. The amplitude of the fields oscillate with time and space.
 
  • #3
PrincePhoenix
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Thank you. That cleared it up a bit.
 
  • #4
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How does this oscillation of electric and magnetic field strengths relate to the idea of photons?
 
  • #5
Born2bwire
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How does this oscillation of electric and magnetic field strengths relate to the idea of photons?
Fields were originally borne out of classical electrodynamics. Photons are a product of quantum electrodynamics. We assume that the electromagnetic fields are propagated in quanta of energy called photons. Each photon has an energy of 0.5\hbar\omega. So the photons are related to the frequency and magnitude of the resulting fields. Classically, the energy of the fields are dependent solely upon their amplitude. Thus, if I have a wave that has an energy of 1 J and a frequency of 1 MHz, then it would have fewer photons than a 1 J wave with a frequency of 1 GHz since each photon of the higher frequency wave would have more energy associated with it. This is represented by the rate that photons are observed. This is perfectly observable and is called granularity if I recall correctly. In addition, shot noise is a kind of noise in low energy signals that correlates to the fact that there are low numbers of photons being received.

So in short, given waves with the same amplitude, the higher the frequency, the lower the rate of photons. Likewise, given a constant rate of photons, the higher the frequency of the photons the lower the amplitude of the observed fields in the wave.
 
  • #6
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The amplitude of the fields oscillate with time and space.
When you say the "amplitude oscillates", do you mean that the field strengths oscillate (the amplitude of the wave being proportional to the maximum field strengths)?

Each photon has an energy of 0.5\hbar\omega.
I thought it was simply hbar times omega (angular frequency) = h times nu (frequency) = hc divided by lambda (wavelength). Are there different conventions for defining these symbols? What do your two forward slashes represent? If division, how would you bracket this?
 
  • #7
Born2bwire
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When you say the "amplitude oscillates", do you mean that the field strengths oscillate (the amplitude of the wave being proportional to the maximum field strengths)?



I thought it was simply hbar times omega (angular frequency) = h times nu (frequency) = hc divided by lambda (wavelength). Are there different conventions for defining these symbols? What do your two forward slashes represent? If division, how would you bracket this?
The value of the electric and magnetic field components of the wave vary both spatially and temporally.

Sorry, I was confusing the total energy with the photon energy. The total energy is equal to (n+0.5)*\hbar\omega where n is the number of photons. The ground state energy is 0.5*\hbar\omega, each photon represents an excitation of \hbar\omega above the ground state.
 
  • #8
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Thus, if I have a wave that has an energy of 1 J and a frequency of 1 MHz, then it would have fewer photons than a 1 J wave with a frequency of 1 GHz since each photon of the higher frequency wave would have more energy associated with it.
So [itex]E_w \propto \nu n[/itex], energy of wave = frequency times number of photons.

So in short, given waves with the same amplitude, the higher the frequency, the lower the rate of photons.
[itex]A \propto \nu n[/itex], assuming the "number of photons" is the same thing as the "rate of photons" (neither having been defined exactly)?

Likewise, given a constant rate of photons, the higher the frequency of the photons the lower the amplitude of the observed fields in the wave.
But doesn't [itex]E_w \propto \nu n \propto A[/itex] imply a higher amplitude? So maybe by "number of photons" you meant something different from the "rate of photons".
 
  • #9
Born2bwire
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Classically, the energy density of a wave is related to:

[tex]\mathbf{E}\cdot\mathbf{D}+\mathbf{H}\cdot\mathbf{B}[/tex]

The power density is related to the Poynting vector which is related to:

[tex]\mathbf{E}\times\mathbf{H}[/tex]

Both the power and energy is related purely to the amplitude of the fields.

However, in terms of quantum electrodynamics, we still consider that the electromagnetic wave to be a field, using quantum field theory. However, the fields that make up the electromagnetic wave are not the electric and magnetic fields. They are instead the fields related to the scalar and vector potentials, \Phi and \mathbf{A}, that are the analogues to the same classical potentials that make up the classical fields. The electric and magnetic fields are now the observables but are no longer the primitives so to speak. These fields have discrete energy levels. The difference between these energy levels are the quanta that we call photons. So each excitation above the ground state for the fields is a single photon.

Thus, the total energy of the fields is represented by the number of photons. However, we cannot measure the total energy of the fields because we cannot place detectors everywhere in space (of course the same argument applies to the classical case, we cannot permeate a volume with field detectors either). Whenever the fields interact, they do so as a point-like interaction with a quanta of energy. That is to say, all interactions are done via the photons. This is how the electromagnetic field is both a wave yet interacts like a particle. I should say that what this means is that we do not assume that there are a bunch of photons running around in space. All we can say is that the interactions with the fields are done via photons but that cannot be construed to how the fields behave outside of these interactions. This is also why I feel it is not really meaningful to talk about the number of photons and energy when it comes to measurement since measurement is conined locally meaning that we are really talking about energy density, not total energy. This relates to how I do not think we can say that space is uniformily permeated with photons. If I said that we have a wave with total energy of 1 W, then we can say that it has X number of photons. However, I do not think we can say that if this same wave has an energy density of Y W/m^3 that we can say that the volume has a photon density.

If we place a detector to measure the field, it does so through photon interactions. We can only measure the number of photons that appear at the detector over time, the photon rate. So, we can measure the electric and magnetic fields (like with a probe or antenna), use this to calculate both the energy (by taking into account the area of the detector) and the power of the field. If we measure the photons (like with a simple CCD), it is generally more meaningful to talk about the power since we are basically just counting up the photons that hit our detector. We could of course integrate in time over the counts to get an average of the energy but I do not think that it is too meaningful to be able to find an instantaneous energy due to the granularity of the photon count, it is more useful by looking at it over a statistical sample. This is exasperated when we have high-frequency low power signals because now we will have a very low rate of photons. We can even attenuate the fields so much that our detectors only measure single photon events. At this point, we could measure, say on average, 1 photon every 30 seconds. Under these conditions, we cannot say that the energy of the wave is zero in between detection events, we have to take into account the statistical sample of measurements over a long period of time to get a meaningful number to the energy of the wave. However, I would argue that we can more readily get a measurement of the power by looking at the rate of detection.

Anyway, since the number of photons is related to the energy, then the rate of photons that interact with a detector over time is related to the power. Both the energy and power are related classically to the amplitude of the fields. Since the energy of each photon is related to the frequency of the signal, then we have the relationships I mentioned above. The additional dependence of energy and power on the number/rate of the photons and their frequency has noticeable consequences, like the existence of shot noise as I mentioned previously.
 
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  • #10
jtbell
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[itex]A \propto \nu n[/itex], assuming the "number of photons" is the same thing as the "rate of photons" (neither having been defined exactly)?
That should be A^2 not A. To be more precise:

The energy density (joules per m^3) at each point in a classical electromagnetic field is

[tex]u = u_E + u_B =\frac {\epsilon_0}{2} E^2 + \frac{1}{\mu_0} B^2[/tex]

In an electromagnetic wave, E = cB everywhere, and [itex]c^2 = 1 / \epsilon_0 \mu_0[/itex], so we can collect this together to

[tex]u = \epsilon_0 E^2 = \frac {1}{\mu_0} B^2[/tex]

Again, this is for each point in the wave. The average energy density is half of the maximum, which is easiest to prove for a simple sinusoidal wave (left as an exercise for the student :wink:):

[tex]\langle u \rangle = \frac {\epsilon_0 }{2} E_0^2 = \frac {1}{2 \mu_0} B_0^2[/tex]

where we now use the amplitudes (maximum values) of the E and B fields.

The energy density is also related to the average number density of photons (photons per m^3):

[tex]\langle u \rangle = \langle n \rangle hf[/tex]

Equating the two <u>'s we get

[tex]\langle n \rangle = \frac {\epsilon_0}{2hf} E_0^2 = \frac {1}{2 \mu_0 hf} B_0^2[/tex]
 

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