I often see $I(t)\propto |E(t)|^2$. What is the exact form ? Which constants are missing to make this an equality and why are they so often omitted ?


Intensity is related to the Poynting vector (https://en.wikipedia.org/wiki/Poynting_vector) by simply taking the magnitude. Note that $\vec{S}=\vec{E}\times \vec{H}$, and $|\vec{H}| = \sqrt{\frac{\epsilon_o}{\mu_o}}|\vec{E}|$ for an electromagnetic wave in vacuum. Thus,

$$|\vec{S}|=|\vec{E}\times\vec{H}|=|\vec{E}|\cdot |\vec{H}|=\sqrt{\frac{\epsilon_o}{\mu_o}}|\vec{E}|^2$$

where the second equality follows from using $|\vec{A}\times\vec{B}|=|\vec{A}||\vec{B}|\sin(\theta)$ and $\theta =90^\circ$ since magnetic and electric fields are perpendicular to the direction of propagation for waves.

Another way to write this would be

$$I=|\vec{S}|=\frac{|\vec{E}|^2}{Z_0} $$

Where $Z_0$ is the impedance of free space, with a value of about 377 ohms.

The constants are frequently omitted if we are doing a theoretical derivation since constants typically factor out of the entire problem and are not an interesting consideration. In experiment, we can typically perform some calibration for the constants. Constants are, however, useful for dimensional considerations and can be useful for checking that your final units are correct.

  • 1
    $\begingroup$ Are you sure about $\sqrt{\frac{\nu_0}{\epsilon_0}}$ $\endgroup$ – Hans Wurst Aug 3 '17 at 21:43
  • 2
    $\begingroup$ Thank you for pointing that out, I had it upside down. $\sqrt{\epsilon/\mu}$ has units of 1/ohms, which is correct, and the opposite of what I had before. Kind of ironic given my last sentence about dimensional analysis in my answer. $\endgroup$ – QtizedQ Aug 3 '17 at 21:51
  • 1
    $\begingroup$ I know that constants are often set to one in calculations and reintroduced later on as in atomic units for example, but it makes no sense to do this in a simple equality where no caclulations are involved; yet it is done extremely often when talking about the intensity of light. I never understood why this inaccuracy is so ubiquitous in the equation for intensity of light. It is very confusing for someone with little experience in EM theory. $\endgroup$ – Hans Wurst Aug 4 '17 at 6:47
  • 1
    $\begingroup$ For intensity calculations that you've seen, perhaps the relative intensities are more important than the absolute intensities. In this way, the constants would drop out when you take ratios of two numbers. An example would be calculating intensity peaks from diffraction. $\endgroup$ – QtizedQ Aug 4 '17 at 10:18
  • $\begingroup$ That is a good point. Quantities of interest are often transmittance or absorbance which are ratio's of intensities where the units drop out. I didn't think of that. $\endgroup$ – Hans Wurst Aug 4 '17 at 10:33

To answer the question why are the constants so often omitted, I would like to supplement QtizedQ's point (that the constants are of less interest because often relative intensity ratios are used). Focusing on proportionality rather than a strict equation with constants also leaves "amplitude" ambiguous, where it could refer to either the electric field or the magnetic field. Because (in a vacuum) the amplitudes of both perpendicular fields of the electromagnetic wave are proportional, we can say the intensity is proportional to both the electric field and the magnetic field (by different constants). The consequence of the proportionality between the two amplitudes is that when discussing electromagnetic radiation qualitatively, it doesn't matter which amplitude we are talking about - we can simply say "intensity is proportional to the amplitude". In fact, this statement is true of waves in general (both transverse like light and longitudinal like sound). Different physical wave systems will have different constants, and there are a variety of quadratic equations between intensity and amplitude, but if we ignore the constants, we can make a broad statement that is true for waves in general: $I \propto A^2$.

The proportionality form is more general, more convenient, and perhaps even pedagogically more tractable.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.