1
$\begingroup$

In A.P. French's Special relativity the author said,

We suppose that an amount $E$ of radiant energy (a burst of photons) is emitted from one end of a box of mass $M$ and length $L$ that is isolated from its surroundings and is initially stationary. The radiation carries momentum $E$/c.

The mass and length of the box are irrelevant here.

He said the momentum of the radiation is $E_{radiation}/c$. We know that the momentum of a photon with energy $E_{photon}$ is $p_{photon}=E_{photon}/c$.

So is $E_{radiation}$ the sum of the energy of each photon, $E_{photon}$?

$\endgroup$
1
  • $\begingroup$ Yes, and if we know the mass of the box we can calculate its velocity .... $\endgroup$ – PhysicsDave Jan 18 '20 at 17:35
1
$\begingroup$

Your question is one of the most fundamental aspects of QM, that is, the relation between the classical EM wave of frequency ν and how it is built up by a herd of photons of E=hν.

It is not simple to follow if one does not have the mathematical background. Conceptually watching the build up of interference fringes from single photons in a two slit experiment might give you an intuition of how even though light is composed of individual elementary particles, photons, the classical wave pattern emerges when the ensemble becomes large.

What is the relation between electromagnetic wave and photon?

It is a very complicated mathematical theory, but it is beautifully describing how individual photons build up the classical EM wave.

Now based on that you could think that the energy of the classical EM wave is just the sum of the individual photons. Though, it is a little bit different.

Light is an electromagnetic wave, and think of the wave being made up of a large number of photons $N$ (even though technically it's a little more complicated than that). The wave has a frequency $f$. The energy of a single photon is $hf$, which might lead you to believe that the total energy is $E_{tot}=Nhf$, but in fact it's slightly different: $E_{tot}=(N+1/2)hf$ The extra $(1/2)hf$ is sometimes called "zero-point energy", and it's interesting because it means that even when you have zero photons there is still some electromagnetic energy there.

How is light related to photons?

$\endgroup$
0
$\begingroup$

The answer is yes. The total energy is the sum of the individual photon energies.

$\endgroup$
1
  • 1
    $\begingroup$ Please elaborate what leads you to this conclusion. What models did you use, what assumptions did you make? $\endgroup$ – electronpusher Jan 4 '20 at 1:12
-3
$\begingroup$

Is the purchasing power of my checking account equal to the sum of the purchasing power of each dollar in my checking account? Well, in a sense yes. But in another sense, not really, because there is no such thing as an individual dollar in my checking account. If I have a five dollar balance, there is no meaningful way to point to individual dollars that make up that balance.

Likewise with photons.

$\endgroup$
2
  • 1
    $\begingroup$ Not the downvoter, but your response is more of a comment than an answer. Answers should have a physics justification for the conclusion rather than an economic example. $\endgroup$ – Bill N Jan 2 '20 at 23:57
  • $\begingroup$ You can't back that up. $\endgroup$ – my2cts Jan 3 '20 at 20:45

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.