# Why do we know all photons have same speed as the speed of light?

• B
• fxdung
In summary, we know the velocity of all photons are the same as the velocity of light because that is what is predicted by theory.f

#### fxdung

Why do we know the velocity of all photons are the same as the velocity of light?Can we deduce this or we must have experiment test?What is the experiment test?

Why do we know the velocity of all photons are the same as the velocity of light?Can we deduce this or we must have experiment test?What is the experiment test?
Photons are light. They have the speed of their speed. Is that what you really meant to ask?

vanhees71 and davenn
The answer to your question is related to the question of whether the photon has zero invariant-mass.
(Based on my reply in https://physics.stackexchange.com/questions/457472/effect-of-non-zero-photon-rest-mass-in-optics ...)

Quoting from Feynman's Lectures on Gravitation (1962)
(Lecture 2, 2.2 Difficulties of Speculative Theories):

He asked, "Tell me, Professor Feynman, how sure are you that the photon has no rest mass?"
...[snip]...
My answer was that, if we agreed that the mass of the photon was related to the frequency as $\omega=\sqrt{k^2+m^2}$
photons of different wavelengths would travel with different velocities. Then in observing an eclipsing double star, which was sufficiently far away, we would observe the eclipse in blue light and red light at different times. Since nothing like this is observed, we can put an upper limit on the mass, which, if you do the numbers, turns out to be of the order of $10^{-9}$ electron masses.

More effects are described in the following article.

"The mass of the photon"
Liang-Cheng Tu, Jun Luo and George T Gillies
Rep. Prog. Phys. 68 (2005) 77–130 ( doi:10.1088/0034-4885/68/1/R02 )
https://iopscience.iop.org/article/10.1088/0034-4885/68/1/R02/

Ch 3 Implications of a photon mass 83
3.1. The dispersion of light 83
3.2. The Yukawa potential in static fields 84
3.3. The longitudinal photon 84
3.4. Special relativity with nonzero photon mass 85
3.5. AB and AC effects with finite photon mass 85
3.6. Monopoles and the photon mass 87
3.7. The Casimir effect for massive photons 88
3.8. Photon mass and blackbody radiation 89
3.9. Other implications

https://en.wikipedia.org/wiki/Photon#Experimental_checks_on_photon_mass

This post gives more of the quote by Feynman
https://physics.stackexchange.com/q...-that-photons-move-at-the-universal-speed-lim

sysprog, Klystron, atyy and 1 other person
I am starting read Special Relativity,then I like to know why photon has same speed as of EM wave speed of light, without relying on Special Relativity theory.

I am starting read Special Relativity,then I like to know why photon has same speed as of EM wave speed of light, without relying on Special Relativity theory.
So again: photons are light [and are em waves]. They have the speed of their speed.

However, if you are asking how that speed can be determined; the speed of EM waves/light can be calculated from Maxwell's equations or determined experimentally, such as via Michelson's method:
https://www.saburchill.com/physics/chapters3/0007.html

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DennisN, davenn and jim mcnamara
I am starting read Special Relativity,then I like to know why photon has same speed as of EM wave speed of light, without relying on Special Relativity theory.
I am not sure that this makes any sense. A photon is a concept of QED so it doesn’t make sense to ask about a photon outside of that context. QED has relativity built into it. So I don’t think there is a way to ask about a photon without relying on relativity.

bhobba, vanhees71 and Delta2
So to build Special Relativity we only need base on EM wave light speed c,but need not base on speed of photon c?(But it seems to me that to build eq E=mc2 we must rely on eq E=pc of photon,then we also need notion photon to build SR theory?)

sophiecentaur
Relativity is a theory of spacetime geometry. It doesn’t need photons. It doesn’t even actually need light, although historically that is how it came about.

bhobba, sophiecentaur, vanhees71 and 1 other person
Does it only depend on invariant of spacetime interval?

atyy
Does it only depend on invariant of spacetime interval?
Yes, everything follows from that. You can also derive it from the basic symmetry principles like homogeneity, isotropy, etc.

atyy, sophiecentaur and vanhees71
There's another subtle reason for why "particles" of zero mass do not make sense in Newtonian physics: If you try to formulate a quantum theory of massless particles within a Galilei invariant theory you don't get anything that makes sense as a physical theory. It's a pretty fascinating property of the Galilei group underlying the Galilei-Newton spacetime model in comparison to the Poincare group underlying the Einstein-Minkowski spacetime model underlying special relativity.

sophiecentaur and Dale
" Then in observing an eclipsing double star, which was sufficiently far away, we would observe the eclipse in blue light and red light at different times."
What a brilliant example of a free experiment with fantastic potential resolution. A time difference (say a second) in possibly millions of years. Such good value.

Nugatory and vanhees71
Photon(particle) has statistical manifestation, then we know average speed is c,but why we know each photon(particle) has speed c?

Delta2
Photon(particle) has statistical manifestation

Not in the sense you mean. Any measurement of the speed of a photon will give exactly ##c##. There is no statistical distribution in such a measurement.

bhobba, vanhees71 and Delta2
Then in what sense does photon have statistical manifestation?

in what sense does photon have statistical manifestation?

In the sense that there are many measurements you can make on a photon that will not always give exactly the same result, but will give a statistical distribution of results. A measurement of the photon's speed just isn't one of them.

russ_watters, vanhees71 and Delta2
Then because speed of photon is definite, then the position of photon is very uncertainly?

Then because speed of photon is definite, then the position of photon is very uncertainly?
Yes. You cannot say 'where it is' ever. Hence the fact that people say a photon has a presence at either of the two Young's Slits; it could be anywhere in space during the, (poorly described as) flight. There is no problem with the fact that it has to end up somewhere definite because it will experience no effect of time (SR) and so the initial worry about that information getting to all parts of the photon at the instant it's detected, is groundless.
The uncertainty principle tells you that any imaging process (telescope / camera etc) will limit the possible paths between source and detector (truncated by the aperture) and blurs any image. We explain this in terms of diffraction (waves) because it's much easier.

We have a lot of experiments, which clearly show that photons do NOT all travel at the same speed. That phenomenon is called dispersion. There would be no rainbows if all photons traveled at the speed of light!
However, in all experiments that indicate cause of dispersion, it seems related to interaction of photons with matter. Not to any intrinsic rest mass of photons.
Indeed, if photons had rest mass, this would show as index of refraction increasing without bound for low frequency radio waves, independent of the environment. No such phenomenon has shown up.

weirdoguy and Delta2
Although position of photon is undefinite,but I think the position of photon is somewhere in "wave packet" of ME fields?(And is it the number of photons in a ME wave packet is constant when it still flight?)

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We have a lot of experiments, which clearly show that photons do NOT all travel at the same speed.

That's not correct. Refracted light consists of the (quantum) interference of incoming photons with photons
re-emitted by atoms in ay medium. The fundamental speed of light is unaffected. On Bruce Sherwood’s homepage (https://brucesherwood.net/) you find an article “Refraction and the speed of light” dealing with this question.

lomidrevo
Photons travel at the same speed(that of light) only in vacuum.

If we suppose position of photon is somewhere in EM wave packet, then how can we explain Vavilov experiment?But it is reasonable to suppose the position of photon is limited in wave packet of EM field.

If we suppose position of photon is somewhere in EM wave packet, then how can we explain Vavilov experiment?But it is reasonable to suppose the position of photon is limited in wave packet of EM field.
Is that according to NMQFT (Newtonian-Maxwellian-Quantum Field Theory)?

Photons travel at the same speed(that of light) only in vacuum.

This is a fairy tale. When light travels through a medium like, for example, a glass plate, it appears to slow down. The apparent "slower speed" is the result of the superposition of two radiative electric fields:
The incoming light, traveling at speed c, and the light re-radiated by the atoms in the medium (oscillating charges driven by the incoming light) in the forward direction, traveling at speed c, too.
The superposition shifts the phase of the radiation in the air downstream of the glass plate in the same way that would occur if the light were to go slower than c in the glass plate. If one wants to understand the essential aspects of the phenomena, I recommend to read chapter 31 “The Origin of the Refractive Index” in “The Feynman Lectures on Physics, Volume I". (http://www.feynmanlectures.caltech.edu/I_31.html).

russ_watters, lomidrevo and DrChinese
What do you mean when saying NMQFT?Vavilov exp means when we weaken the beam of light to threshold,some time we see the light(enough photons go to eye), sometime we do not see the light( not enough photons go to the eye)

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We have a lot of experiments, which clearly show that photons do NOT all travel at the same speed.
Can you name just one that measures c in a vacuum and gets a spread of values? Dispersion and observed gravitational bending of light do not disprove that c is constant - you just need to describe the phenomena in the right terms.

What do you mean when saying NMQFT?
If we suppose position of photon is somewhere in EM wave packet, then how can we explain Vavilov experiment?But it is reasonable to suppose the position of photon is limited in wave packet of EM field.
Well-defined position of a particle (without measurement) is a concept from Newtonian physics.

Generally, your questions mix concepts from Maxwellian EM, QM and QFT, so the whole thread is build on shifting sands.

Dale
Can the number of photons is constant while they still flight?

Then because speed of photon is definite, then the position of photon is very uncertainly?
It's unfortunate that you start to learn about quantum mechanics on the example of photons. The problem is that photons were the first discovery of the fact that classical physics needs to be substituted by quantum mechanics due to Planck's discovery of the theoretical foundation of the black-body radiation spectrum. Now many textbook writers think they have to talk about "photons" in the introductory sections of quantum-mechanics textbooks and also in the popular-science literature.

The photon is a rather complicated notion, and unfortunately what's written in the said introductory sections of textbooks or in the popular-science literature gives a picture about photons from the period we now call the "old quantum theory". It's roughly from original research of the years from 1900-1924, before the "modern quantum theory" was discovered by Born and Jordan using an idea by Heisenberg ("matrix mechanics") as well as Schrödinger ("wave mechanics"), and Dirac ("transformation theory") in 1925/26.

First of all you must forget to think about photons as something like point-like particles. Photons are in no sense in any way completely localizable. It's not even possible to define a position observable for them. This is due to the fact that they are described by quantized massless vector fields, the electromagnetic field. For such fields there's no way to define a position operator. All that can be said about a photon, which is a specific state of the quantized electromagnetic field (a single-photon Fock state) is with which probability it is observed at a certain place defined by the detector placed at this position to detect it.

Other observables are well defined, among them energy, momentum, and angular momentum, and these observables can be measured. The masslessness of the field implies that a photon with a quite well determined momentum ##\vec{p}## has also a pretty well determined energy ##E=c|\vec{p}|##, and this implies that the "speed" ##\vec{v}=c^2 \vec{p}/E=c \vec{p}/\|\vec{p}|##, i.e., ##|\vec{v}|=c##. That's what's meant when you say a "photon is massless" or a "photon moves with the speed of light".

sysprog, lomidrevo, Klystron and 4 others
With Fock state, how can we define probability the divice find out photon at a position?

That's not so easy to describe. One way to detect photons is to use the photoelectric effect. In this case you can use the usual dipole approximation for the interaction of a photon with an atom (within the detector), assuming that the wavelength of the photon is much larger than the typical size of the atom. E.g., to typical atomic size is of the order of an Angstrom, i.e., ##10^{-10} \; \text{m}##. If you have light or photons in the visible range of the spectrum the wavelength is around ##500 \; \text{nm}=5 \cdot 10^{-7} \; \text{m}##, such that for this application you can use the dipole approximation.

The mathematical analysis then gives for the detection probability that it is proportional to the energy density of the em. field at the position of the detector, i.e., it's proportional to the expectation value ##\langle \vec{E}^2(t,\vec{x}) \rangle##.

It's very often a good approximation to think of the photons simply as an electromagnetic wave and the probability to detect a photon as given by the corresponding intensity of the electromagnetic wave.

In this rough picture the difference between a single-photon state and a classical em. wave (which in terms of quantum electrodynamics is represented by a socalled coherent state of high intensity) is that you don't get a continuous distribution of light but just one single point on a detector screen (a photoplate or in our times a CCD cam). You never know, where an individual photon will be detected but only the probability to detect it at a given position on the screen. A single photon is either completely absorbed and detected or it's not absorbed and stays undetected. There's no way that a single photon can be detected at two or more spots at once, and that's all that's left from what was considered "particle like" in the (short) era of "old quantum mechanics". Only if you repeat an experiment (like shooting single photons through a double slit) you build up a distribution of photon-detection points which then follows the probability distribution predicted by QED. What's "wavelike" in this pattern is the fact that you get an interference pattern as with classical electromagnetic waves when looking at this pattern build up from many photons.

What was hand-wavingly and pretty desperately known as "wave-particle dualism", and Einstein himself, who promoted the photon picture as "a heuristic point of view" only, was very puzzled by it, is today resolved by modern quantum mechanics and the assumption that all that can be known about some object (here a photon or the electromagnetic field) is in what quantum state they are "prepared", and this implies only statistical knowledge about the outcome of measurements (here the position of the detection of em. fields/photons on a screen).

hutchphd and sysprog
Can you name just one that measures c in a vacuum and gets a spread of values?
No, and that´ s my point. We know what it looks like when photons have different speeds, and it always appears related to the effects of medium on photon, not intrinsic properties of photon. Whereas rest mass of photon would show up as low frequency radio waves having index of refraction that diverges at low frequencies independent of medium.

weirdoguy
We know what it looks like when photons have different speeds
Do we know? Are you sure? Because I have no idea.
Whereas rest mass of photon would show up as low frequency radio waves having index of refraction that diverges at low frequencies independent of medium.
That doesn't make any sense to me. How would that suggest a rest mass of photons?

weirdoguy
It's correct that in-medium photons have a different spectral function than free photons in the vacuum, but that doesn't make the massive.

The only place where photons get really massive is within superconductors, where the electromagnetic gauge symmetry gets "Higgsed" through the formation of a "Cooper-pair condensate".

You can see this from the point of view of the BCS model (Bardeen, Cooper, Shriever): Start from a model of a superconductor first neglecting the electromagnetic field. Formally you have a gas of electrons with an effective attractive interaction (through the electron-phonon interaction). Whenever you have fermions with an effective attractive interaction there's a phase transition leading to spontaneous symmetry breaking and the formation of Cooper pairs and a non-vanishing ground-state expectation value for the corresponding (quasibosonic) field, leading to condensation.

Now you gauge this model in the usual way to get the electromagnetic interaction in. The so far global spontaneously broken symmetry is now a local gauge symmetry, which cannot be spontaneously broken. What happens instead is that the gauge field turns out to be massive by "eating up" what was the Nambu-Goldstone mode of the spontaneously broken global symmetry, which means that the photon gets massive and the Nambu-Goldstone mode is gone.