Is it possible to actually measure/deduce the amplitude of a wave? We know that the length of am infrared wave is a fraction of a millimetre, do you have any idea what is the range of the physical length of a less/more intense wave?

Or it makes no sense to talk about the width of an em wave as we can do about a water wave?


2 Answers 2


It makes no sense. The light is an electromagnetic wave, which means that it is a combination of two fields: electric $\vec E$ and magnetic $\vec B$ at each point in space. The two fields are perpendicular to each other.

The energy flux is given by the Poynting vector: $$\vec {\mathscr{P}}=\frac{1}{\mu_0}\vec E\times\vec B$$ $$\mathscr{P}=\frac{1}{\mu_0}EB$$ using $\mu_0=4\pi\cdot 10^{-7} \rm\,\frac{kg\cdot m}{s^2\cdot A^2}$.

Also, there is an identity regarding electric and magnetic field: $$\vec E=\vec B\times\vec c$$ $$E=Bc$$ where $c=299792458\rm\,\frac{m}{s}$. Therefore, using the two scalar equations, we get $$B=\sqrt{\frac{\mathscr{P}\mu_0}{c}}$$ and $$E=\sqrt{\mathscr{P}\mu_0c}$$

So, if you know the energy flux in $\frac{W}{m^2}$, you can get the amplitude of the magnetic and electric component of an electromagnetic wave. However, as said before, you cannot ascribe the width of light, since the two fields are just some vectors, ascribed to points in space, they are not arrows in space with some definite length.


We have fallen for an inaccurate formulation in physics. What has been called the amplitude of EM radiation is in fact the intensity of the radiation. The amplitude of EM radiation per definition has to do with the intensity of the light.
The intensity of a monochromatic radiation - all photons of the same wavelength - has to do with the number of photons through a given area in a given time.

Hans Bethe wrote a paper in 1944, "Theory of Diffraction by Small Holes," Phys. Rev. 66, 163. He describes how he envisages the interaction between radiation and hole:
"The method is based on the use of fictitious magnetic charges and currents in the diffracting hole which has the advantage of automatically satisfying the boundary conditions on the conducting screen. The charges and currents are adjusted so as to give the correct tangential magnetic, and normal electric, field in the hole."

Since EM radiation consists of photons (and always does, there is no other method of generating EM radiation than by emitting photons), we can talk about an effective cross-section. If the hole is too small, the electric and magnetic fields of the electrons at the surface of the hole absorb or reflect the photons. If the aperture is too large, only the photons that come close to the edge are deflected, while the photons in the middle of the beam leave the aperture unhindered.

  • $\begingroup$ so, amplitude just regards the number of photons? it does NOT depend on the acceleration/displcement of the electrons in the antenna? $\endgroup$
    – user157860
    Commented May 30, 2022 at 8:56
  • $\begingroup$ For emission (from an antenna rod) the current is responsible for the number of accelerated electrons aka photons and the voltage is responsible for value of the acceleration aka the wavelength of the emitted photons. This holds also for the reversible process of incomming photons to the acceleration of electrons in the receiving conductor. $\endgroup$ Commented May 30, 2022 at 9:18
  • $\begingroup$ I understand that more electrons means more photons, but how does greater acceleration/displacement affect amplitude? thanks. If you elaborate your answer to explain this I'll accept your answer $\endgroup$
    – user157860
    Commented May 30, 2022 at 9:24
  • $\begingroup$ With a higher acceleration of the electrons, photons with a higher frequency are emitted. BTW, there is a third parameter. If the frequency of the antenna generator is increased, more photons (per time unit) are also emitted. $\endgroup$ Commented May 31, 2022 at 3:37

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