I ask because I have a real problem with wave/particle duality (one of the biggest cop outs in the whole of physics, in my opinion)

A radar pulse is definitely a wave.

It is spatially bounded. (at least in one dimension!) It has energy. It is a 'quanta'.

Is this not particle-like? Could a pulse of sufficiently high energy have mass in accordance with E=MC^2 and E=hf?

Apologies for the ignorance of a simple engineer.

  • $\begingroup$ A particle, by definition of classical mechanics, is the approximation of the motion of an extended body by the motion of its center of mass, i.e. we are neglecting any and all other properties, including rotation, vibration and composition. By that definition an electromagnetic wavelet is not a particle. It's also not a quantum in the sense of QM because there are no restrictions on its angular momentum, charge etc.. A wavelet is a wavelet, which is a category of physical object in its own right. Would you ask in EE if a capacitor is like a strangely behaving inductor? So why ask in physics? $\endgroup$
    – CuriousOne
    Jul 30, 2016 at 8:24
  • $\begingroup$ Could we not consider then that a radar pulse, if bounded in all three dimensions, would be a true particle? Is this the essence of string theory? $\endgroup$
    – N.G. near
    Jul 30, 2016 at 8:36
  • $\begingroup$ A particle is an approximation of the dynamics of an extended physical body, i.e. matter. That's how the word "particle" is defined in physics. That's the only definition of "particle" and it is linked to matter, not radiation. That's how it should be taught in high school. If they are not teaching it that way, then they are performing a very poor public service in science education. What you are describing is called a "wavelet". I don't know why you are having a problem with wave-particle duality. We haven't been using that in physics since the 1930s. $\endgroup$
    – CuriousOne
    Jul 30, 2016 at 8:47
  • $\begingroup$ Ok, so I'll stick to the script then. $\endgroup$
    – N.G. near
    Jul 30, 2016 at 8:57
  • 1
    $\begingroup$ This question and my answer to it is relevant at least in that it helps clear up what "particle" means in quantum mechanics. $\endgroup$
    – DanielSank
    Jul 30, 2016 at 19:47

3 Answers 3


In quantum mechanics the electromagnetic field interacts as quantized particles, photons. We ask if you can neglect the discrete theory for the classical continuous theory.

I'll use typical numbers found in wikipedia.

A typical radar pulse lasts $t=1 \mu s $ at a frequency of $f=3$ GHz.

I would estimate the maximum power of the pulse arriving at a $1 m^2$ cross section at less than $1 MW$. The energy of the pulse is therefore $E=P\times t=1 J$.

The energy of a quantum of light is given by

$E_{quantum} = h f$, in this case $E_{quantum} \approx 10^{-24}J$

This means the radar pulse carries about $10^{24}$ discrete quanta, which means we can safely approximate it as a continuous wave.


$E=mc^2$ is true only for a particle at rest, the photon is never at rest. The full equation is $$E^2=p^2c^2+m^2c^4.$$

A photon has no mass, but two photons can meet to form an electron and a positron, see pair production.


Quanta of a electromagnetic wave or photon is the minimum amount of energy that can be transferred from one place to another. Increasing the energy of a pulse means you are increasing number of photons in the pulse and we can write


Where equivalent mass can be written as


$$p=\frac {h\nu}{c}$$



where $v=c$ is the velocity of light, n are number of photons, hence you can see that by increasing the energy you can not increase the equivalent mass of the photon as the frequency of the photon remains same, you are increasing just the numbers of the photons.

Radars emit electromagnetic radiation in microwave range hence their photons has very small energy, which makes really hard to observe particle like behavior from them.

I hope this will help

EDIT: Regorusly one may argue that mass of photon is not physical entity. However in several situations one may define the effective momentum of the photon as

$p=\frac{E}{c}$ and from here one may deduce $m=\frac{p}{c}$, This analogy is not correct regorusly but it works in several situations. It is like scalar diffraction theory which is not regorusly correct but works fine in several occasions

  • 3
    $\begingroup$ $h\nu \neq mc^2$, so the rest of your explanation doesn't work. $\endgroup$
    – garyp
    Jul 30, 2016 at 10:41
  • $\begingroup$ @garyp This is a common and well accepted method to define the effective mass of a photon. $\endgroup$
    – hsinghal
    Jul 30, 2016 at 10:45
  • 1
    $\begingroup$ I've not seen that. Can you provide a reference showing its use? $\endgroup$
    – garyp
    Jul 30, 2016 at 10:49
  • 2
    $\begingroup$ So what is $m_0$ for a photon? $E=mc^2$ is only true for rest mass. $E=\gamma m_0 c^2$ is a different equation! More importantly, $mc^2 = \gamma m_oc^2$ is a point of view that has been abandoned for decades precisely because it causes confusion like this. Einstein himself warn against the concept of relativistic mass. For inexplicable reasons, it lives on in some books and web sites. $\endgroup$
    – garyp
    Jul 30, 2016 at 11:05
  • 1
    $\begingroup$ In a way you are right. since $m_0=0$ for photon hence the above experssion do not hold but in this case you may take $E=pc$ (since $m_0=0$) and $p=mv$ and may get the expression for non rest mass of the photon. $\endgroup$
    – hsinghal
    Jul 30, 2016 at 11:08

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