# A Spectral intensities from E&M fields

1. Jan 5, 2017

### hilbert2

A question about the relation of light intensities to the actual time-dependent EM field:

Suppose I have a system where the functions $\mathbb{E}(\mathbf{r},t)$ and $\mathbb{B}(\mathbf{r},t)$ (time and position dependent electric and magnetic fields) are known. I would like to have an explicit formula for calculating the position, wavelength and direction dependent spectral intensity (as defined here: https://en.wikipedia.org/wiki/Radiant_intensity#Spectral_intensity ) from these functions... I suppose I should write the fields as some kind of a Fourier integral, convert the plane wave components of different wavelengths and directions to Poynting vectors, and then take the time averages of the Poynting vector magnitudes over one period of oscillation. Am I going to the right direction here?

I'm currently doing work on the modelling of radiative heat transfer, and there one obviously doesn't use the Maxwell equations to calculate radiative quantities, as the wavelengths of IR radiation are much smaller than the dimensions of the physical systems considered (you would need a huge discrete lattice to simulate that kind of a system using discretized Maxwell equations). However, in my earlier studies I have mainly approached these kinds of problems using the actual microscopic fields, and I'd like to see the explicit connection of these macroscopic intensity functions to them.

2. Jan 5, 2017

For infrared systems, many of them can be treated as blackbodies, and/or graybodies with an emissivity $\epsilon<1$. In this case the Planck spectral density function (either in wavelength or frequency), gives the necessary spectral energy distribution and it is not necessary to compute electromagnetic field strengths. You can convert to electromagnetic field strengths if need be, but it is a lot of extra computation. $\\$ One thing that is helpful in such a computation, if you ever treat a system that is monochromatic (or nearly monochromatic), is the individual photons in the same mode can be considered to have random phases, so that the resultant amplitude of the electric field $E$ of a single mode will be proportional to $\sqrt{N}$. Thereby the energy in that mode, which is proportional to $E^2$, will thereby be proportional to the number of photons $N$. $\\$ Note also, in using the Planck Spectral density function, that it integrates from $0$ to $+ \infty$ (in wavelength or frequency) to give the Stefan-Boltzmann result $M=\sigma T^4$.

Last edited: Jan 5, 2017
3. Jan 5, 2017

### hilbert2

Thanks for the answer. The problem of calculating the intensities in a cavity with black walls is obviously quite easy... I was thinking more about a case where we are measuring the intensity of radiation at some distance from a finite size black body in vacuum. Doesn't a point source of EM radiation produce some kind of a Bessel function shaped waves around it?

4. Jan 5, 2017

The Planck blackbody derivation counts photons and assigns an energy $E_p=\hbar \omega$ to each one. In addition, the Planck derivation for the radiation emerging out of an aperture of area A from the blackbody cavity uses the effusion rate formula (number of particles per unit area per unit time) $R= \frac{n v_{mean}}{4}$ where $n$ is the density of particles/photons and $v_{mean}=c$ is their average speed. $\\$ The conversion from irradiance (watts/cm^2) which also uses the letter $E (watts/cm^2)$ in radiometry (sometimes H, but H is a somewhat obsolete letter) to electric field $E$ can be equated directly to the Poynting vector. Note: the watts/cm^2 mixes M.K.S. and cg.s. but otherwise, the conversions are straightforward to get the $E$ of the electric field. In c.g.s. units, I believe Poynting vector $S=c \frac{E^2}{4 \pi}$. (I might be off by a factor of 2 here=I don't have an E&M textbook handy at the moment.) $\\$ (And note: we have 3 E's floating around here=photon energy $E_p$ , irradiance $E$ (watts/cm^2), and $E$ of electric field, but hopefully you have them sorted out.)

Last edited: Jan 5, 2017
5. Jan 5, 2017

@hilbert2 Be sure and see the latest edited version of post #4.

6. Jan 5, 2017

### hilbert2

Yes, I saw it, thanks. I'm currently reading a textbook about radiative heat transfer, and it's probably not too difficult material to grasp. It's just written from an engineering viewpoint, and I was interested to see how it relates to the more abstract concepts that I have studied before.

7. Jan 6, 2017

### hilbert2

Just one more question about the simplest possible special case. Suppose I have a monochromatic radiation field of wavelength $\lambda$ where the EM wave is moving to the direction of the unit vector $\mathbf{u}_x$. Now at time t the electric and magnetic fields are

$\mathbb{E} (x,y,z,t) = E_0 \sin \left( \frac{2\pi x}{\lambda_0} - \omega t \right) \mathbf{u}_y$
$\mathbb{B} (x,y,z,t) = B_0 \sin \left( \frac{2\pi x}{\lambda_0} - \omega t \right) \mathbf{u}_z$

Now the Poynting vector is $\mathbf{S}(x,y,z,t)=\frac{E_0 B_0}{\mu_0}\cos^2 \left( \frac{2\pi x}{\lambda_0} - \omega t \right)\mathbf{u}_x$. The intensity of the wave in the direction of the x-axis is the average of the magnitude of the Poynting vector over one period:

$I=\frac{E_0 B_0}{2\mu_0}$.

So what's the wavelength- and direction-dependent spectral intensity $I(\Omega, \lambda)$ as defined in the Wikipedia article ? Is it proportional to the Dirac delta function $\delta ( \lambda - \lambda_0 )$, where $\lambda_0$ is the wavelength of our plane electromagnetic wave, or it is simply zero when $\lambda \neq \lambda_0$ and has a finite value when $\lambda = \lambda_0$ ? And if we measure the intensity in some direction that is not parallel with the x-axis but is in angle $\alpha$ relative to it, is the intensity zero, or is it proportional to something like $\cos \alpha$ ?

8. Jan 6, 2017

One thing to mention first is E and B are related. In cgs units they are equal. For the MKS units, I think the E and H are equal, but I would need to look it up. For monochromatic case, the spectral irradiance (in power per unit area per unit time) per spectral wavelength is given by (in c.g.s. units) $S(\lambda)=c\frac{E^2}{4 \pi}\delta(\lambda-\lambda_o)$. $\\$ Now to apply to a practical case: If you start with a typical commercial $P= 2 E-3 \, watts$ (2 mwatt) HeNe laser, it typicaly will have a beam divergence of 1 milliradian , making a solid angle divergence $\Omega = 1.0 E-6 \, steradians$ (approximately). The intensity $I$ in radiiometric units is given by $I=\frac{P}{\Omega}$ (watts/steradian). The irradiance $E(watts/cm^2)$, but only in the direction of the beam, is given by $E=\frac{I}{s^2}$ where $s$ is the distance to the observation point. (Typically the beam will not have uniform intensity but will fall off in intensity as you go off center, but I'm giving you the simplest case here.) If your observation plane is at some angle to the direction of the beam, you will pick up a $cos(\theta)$ factor. (And notice in the above, the steradian is a dimensionless parameter, so it doesn't appear in the irradiance units (watts/cm^2). This irradiance $E$ and the $S(\lambda)$ above are equivalent. Here $\lambda_o=632.8 \, nm$. (Don't confuse the irradiance $E$ with the electromagnetic field $E$.) $\\$ To relate to the Planck function in heat transfer, etc. $L(\lambda,T)=\frac{2hc^2}{\lambda^5} \frac{1}{e^{\frac{hc}{\lambda k T}}-1}$, and heat radiated per unit area $M(\lambda, T)=L(\lambda,T) \pi$. For a blackbody of area $A$, the spectral irradiance $E(\lambda)$ at a distance $s$ is given by $E(\lambda)=\frac{L(\lambda,T)A}{s^2}$.

Last edited: Jan 6, 2017
9. Jan 6, 2017

### hilbert2

In SI units, the amplitudes are related as $|E_0 |=c|B_0 |$. I think the Poynting vector in my last post should have been

$\mathbf{S}(x,y,z,t)=\frac{E_0 B_0}{\mu_0}\sin^2 \left( \frac{2\pi x}{\lambda_0} - \omega t \right)\mathbf{u}_x$

I hope this didn't cause any confusion.

10. Jan 6, 2017

I should mention, in the case of a flat blackbody of area A, the irradiance $E$ falls of as one goes off-axis with a $cos(\theta)$ intensity pattern (Lambert's law). This is basically why $M=\pi L$ rather than a $2 \pi$ factor that you might expect of what gets radiated over a hemisphere.