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I'm still in high school, and while I can't complain about the quality of my teachers (all of them have done at least a bachelor, some a masters) I usually am cautious to believe what they say straight away. Since I'm interested quite a bit in physics, I know more about it than other subjects and I spot things I disagree with more often, and this is the most recent thing:

While discussing photons, my teacher made a couple of statements which might be true but sound foreign to me:

  • He said that under certain conditions, photons have mass. I didn't think this was true at all. I think he said this to avoid confusion regarding $E=mc^2$, however, in my opinion it only adds to the confusion since objects with mass can't travel with the speed of light, and light does have a tendency to travel with the speed of light.. I myself understand how photons can have a momentum while having no mass because I lurk this site, but my classmates don't.

  • He said photons don't actually exist, but are handy to envision. This dazzled my mind. Even more so since he followed this statement by explaining the photo-electric effect, which to me seems like a proof of the existence of photons as the quantum of light. He might have done this to avoid confusion regarding the wave-particle duality.

This all seems very odd to me and I hope some of you can clarify.

Ylyk Coitus
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  • This question may be useful http://physics.stackexchange.com/q/34067/ even though is closed. – J L Feb 07 '13 at 19:28
  • Photons do have a mass inside a superconductor. Which is why, inside a superconductor, the electromagnetic force becomes short-range. Perhaps that's what your teacher meant. – Dmitry Brant Feb 07 '13 at 20:07
  • @DmitryBrant I know this is to some extent just a matter of semantics, but I personally feel it's somewhat misleading to call the effective mass of a photon inside of a superconductor its mass. – joshphysics Feb 07 '13 at 21:10
  • Also Ylyk, you might be interested to read the section in http://en.wikipedia.org/wiki/Photon#Experimental_checks_on_photon_mass that talks about experimental checks on photon mass. – joshphysics Feb 07 '13 at 21:12
  • I would definitely ignore anything that your teacher or classmates have to say about physics that you can not find in a good text book. Until you reach topics of quantum gravity and quantum information theory, these things are well understood within the physics community (although not by all its members). I would also ignore most media releases until you can sift the good from the bad. A good list of freely available books can be found athttp://physics.stackexchange.com/questions/6157/list-of-freely-available-physics-books – Freedom Feb 08 '13 at 11:45
  • Perhaps he meant "only photons in vacuum lack mass," because any light traveling through transparent materials is actually composed of mass-carrying particles: polaritons, an EM/matter coupling where the EM fields propagate at group-velocity far slower than c. He could also have said that light in glass is no longer photons? The simplified high-school version is that the light-particles traveling through glass have mass, that's why they don't propagate at c. (Far easier to just ignore QM when explaining FO and lenses and prisms! Stick to EM waves only.) – wbeaty Apr 15 '23 at 20:44

5 Answers5

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  1. Photons are massless. This should not cause confusion with $E=mc^2$ because the expression for the relativistic energy of the photon is $E = h \nu$, there $\nu$ is the photon's frequency and $h$ is Planck's constant. You can also understand the relativistic energy of the photon by noting that $E = pc$ where $p=\hbar k$ is the magnitude of its momentum, and photons possess momentum, as you point out. $k$ is the photon's wavevector and $\hbar = h/2\pi$ is the reduced Plank's constant. To tie this all together, we have the formula $E=\sqrt{m^{2}c^{4}+p^{2}c^{2}}$.

  2. Photons exist! As you point out, the existence of photons has physical consequences that can be measured. Perhaps, as you also mention, your teacher is trying to insist that you resist thinking of photons purely as particles (which is probably not such a bad idea), but the statement "photons don't exist," in my opinion, should be considered just as fallacious as "the keyboard on which I'm typing doesn't exist.

You're being prudent by taking everything your teacher says with a large grain of salt. In my experience in physics (heck in life as a whole), it's good to take everything that anyone ever says with a grain of salt (including my response for that matter).

Cheers!

user12345
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joshphysics
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  • It depends on the definition of mass, obviously. Photons have zero rest mass. But physicists working in relativity also use "mass" as a synonim of "energy". – Martino Feb 07 '13 at 22:48
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    @Bzazz Nuclear and particle physicists use the work "mass" to mean only the rest mass. Every time. While there is no difficulty in defining the "relativistic mass" the concept obscures and confuses rather than clarifying. – dmckee --- ex-moderator kitten Feb 07 '13 at 22:57
  • Yes, that's why I specified physicists working on relativity. In all the courses of GR I had, you put c=1 and talk of mass and energy equivalently. Of course, one seldom talks about the mass of a photon, because, we are more familiar with speaking of photon energy. But if we want to look for "the mass of the photon" as in the question, this is what comes to my mind first. Remember that relativistic mass is effectively mass, in the sense that also experiences gravity. – Martino Feb 07 '13 at 23:05
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    @Bzazz I think this sort of statement is the type of corruption of physics that we should be concerned about. Although any object with energy can create curvature, Photons are bosonic and do not directly interact with the higgs boson, and are therefore massless. See some good answer at http://physics.stackexchange.com/questions/23161/why-dont-photons-interact-with-the-higgs-field – Freedom Feb 08 '13 at 10:57
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    I definitely agree with @HalSwyers on this one. – joshphysics Feb 08 '13 at 16:22
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He is doing a good job trying to communicate the weirdness of Photons, but a poor one at being consistant. It's difficult to communicate how strange they are, without using the relationships you described.

First of all photons have no mass. At rest. The concept of describing a photon at rest is a bit weird even as there is no such thing.

Once your talking about photons in motion it is tempting to say they have mass, but most physicists these days don't speak of it this way. Rather its best to say it has ENERGY. While people like to say things can have relativistic mass, this is incorrect, mass is a constant: the mass a thing has never changes no matter how fast you travel: rather the energy to accelerate it increases (inertia).

So a photon has no mass: but it does have momentum. This is the fascinating thing about light. And leads to the fact that momentum is not always related to mass as you mentioned.

Examine $E=mc^2$ in more detail and the correct equation actually is $E^2 = m^2 c^4 + p^2 c^2$. As photons have momentum related to its energy/frequency this is fine.

The best way to think of light is as photons. Light is a photon, that is the fact, BUT light waves are only a model. The appearance of wave phenomenon is because quantum mechanically light interacts probabilistically, and this nature allows it to display wave like properties.

The science of Quantum Electro-Dynamics examines this strange behavior.

For now think of light as a photon and a wave, it allows every day behavior to be modeled well and is perfectly fine.

But in reality light is a photon particle (this is why the photo-electric effect works) that when described using Schrodinger's Equations (a description of probabilites) can be transformed into a description of a wave of Electro-Magnetism.

dmckee --- ex-moderator kitten
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Eric_
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    There are two reasons why a photon can't be described by the Schrödinger equation: A single particle theory for a photon is non-sense and Schrödinger's equation is non-relativistic. – J L Feb 07 '13 at 20:08
  • For modeling interactions between photons to show that they display wave characteristics, Shrodinger's equation works as a good approximation to demonstrate his equations and Maxwells are equivalent. Nontheless like I mentioned QED manages it better: see Feynman's books for good explanations without equations. – Eric_ Feb 07 '13 at 20:41
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    Eric, Schrödinger's equation is explicitly built on the Newtonian relationship between kinetic energy, momentum and mass (the rest mass for those that insist). It really, really doesn't do to use it with photons. If you must do non-field-theoretical QM with light, use the Klein-Gordon equation. However, you've got the right relationship in your post: $E^2 = m^2c^4 + p^2c^2$ is the correct answer. Don't detract from it by using the wrong wave equation. – dmckee --- ex-moderator kitten Feb 07 '13 at 22:45
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    In any case, welcome to Physics.SE. WE have the MathJax rendering engine active on the site which lets you write latex-alike math inside pairs of $'s (for inline) or $$ (for block typesetting). I've done this post for you. – dmckee --- ex-moderator kitten Feb 07 '13 at 22:47
  • I have sat in a college class and watched step by step as a Schrodinger equation is transformed into a wave equation with a form identical to maxwells. The point is not that it explains light, but that probabilistic quantum mechanics explains how a wave nature can appear for ANY and ALL particle phenomenon. There is wave particle duality in everything, not just light. – Eric_ Feb 08 '13 at 00:00
  • Yes. However, Schrödinger's equation is an approximate version of physics just as Newtonian mechanics is approximate and it does not apply to light. The easiest way to see that is to notice that it has a $1/m$ in it and the mass of a photon is zero. If you want to do QM on particles moving at a significant fraction of $c$ you use the Dirac equation for fermions and the Klein-Gordon equation for bosons. You really do. You can even get KG from $E^2 = m^2c^4 + p^2c^2$ the same way you get the Schrödinger equation from $E = \frac{p^2}{2m}$, it just works over a larger velocity range. – dmckee --- ex-moderator kitten Feb 08 '13 at 02:13
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Regarding your two points:

He said that under certain conditions, photons have mass.

Massless particles move at the speed of light $c$ in vacuum. By his statement your teacher may have been alluding to the fact that photons travel at speeds slower than $c$ when they travel through media, like glass for example. However, I would rather phrase this as something like "in the transmission of a photon through a medium, the photon's transit time through that medium is such that it travels as if it had a mass." The travel of photons through media is a rather complex affair, which I don't fully understand, involving interactions with charged particles in the medium and quasiparticle states, and I'm not sure to what extent the incident photon even retains its identity whilst in the medium (maybe that's another question).

He said photons don't actually exist, but are handy to envision.

There is no question that photons exist. Particle physicists deal with "hard" (i.e high energy) photons, which behave like particles - scatter off other charged particles etc. Good evidence that "soft" (low energy) photons exist comes from antibunching experiments.

twistor59
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1) I would refer your high school teacher to some of the good answers found in this physics stack exchange question. The answer with the highest votes is really good and is referring to the electroweak theory which governs the electroweak section of the Standard Model.

General Relativity respects special relativity, and therefore respects that the invariant mass of the photon is zero, since the photon is described by a null-like vector in its rest frame.

In particle physics, as discussed in the links above, the current understanding is that mass is a measure of the relative interaction strength of a particle with the Higgs field as mediated by the Higgs boson. The photon does not directly interact with the Higgs boson, and therefore has no mass.

2) As far as visualizing the photon, I would venture the easiest way is to think in terms of classical EM theory (which is a gauge theory btw) where we consider the orthogonal oscillating electric and magnetic fields as representing the photon.

Community
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Freedom
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$E = mc^2$ is a popular formula but only valid in a special case when "the total additive momentum is zero for the system under consideration" $^1$

The more general "energy-momentum relation" is: $E^2 = (mc^2)^2 + (pc)^2$

(Also, here's a neat little video for your classmates ;) -> http://www.youtube.com/watch?v=NnMIhxWRGNw )

You can read more about this and the correct four-vector notation under:

$^1$ http://en.wikipedia.org/wiki/Mass%E2%80%93energy_equivalence

Even today in the time of quantum optics, there are still quite a lot of papers, that try to grasp what a "photon" is, e.g.:

  1. What is a photon - David Finkelstein
  2. Light reconsidered - Arthur Zajonc
  3. The concept of the photon - revisited - Ashok Muthukrishan, Marlan O. Scully, M.Suhail Zubairy

and many more all together in this nice review: http://www.sheffield.ac.uk/polopoly_fs/1.14183!/file/photon.pdf

Or in the words of Roy Glauber: "A photon is what a photodetector detects"

Mike
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  • That sheffield link went stale. Instead see the author's website, http://gerdbreitenbach.de/gallery/photon_what_is_a.pdf , 2003 OPN Trends article collection "What is a Photon?" – wbeaty Apr 15 '23 at 20:48