# Why photon is mass-less?

Discussion in 'Physics & Math' started by Saint, Feb 9, 2020.

1. ### HalcRegistered Senior Member

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And then the wall must move. If I accelerate, then so must the wall, per Newton's 3rd law. My body countering the push back from the wall is what normally prevents that.

3. ### SaintValued Senior Member

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text book's theory is for idealised cases,
actual world is more complicated.

5. ### SaintValued Senior Member

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Why photon can be wave AND particle at the same time?

7. ### James RJust this guy, you know?Staff Member

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Didn't you study any physics to gain your engineering qualification, Saint? Or just no quantum physics (not even at an introductory level)?

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I'm willing to be corrected, but I believe the duel nature of light/photons, that is being observed as particle and wave depending on experiment, is actually unknown. A mystery of quantum theory?

9. ### exchemistValued Senior Member

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It is what we observe and QM is the mathematical model that predicts when particle-like or wavelike behaviour can be expected. So at one level the answer to your question is to learn QM.

But as to WHY QM gives the right answers, all we can say is this is how nature seems to behave. Just as we cannot answer WHY relativity seems to be a good model. The job of science is to model the physical world so that we can predict what to expect it to do. These WHY questions eventually become metaphysical and beyond the scope of science.

10. ### SaintValued Senior Member

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The job of science is to model the physical world so that we can predict what to expect it to do

Agree.
This include the big bang theory.

11. ### exchemistValued Senior Member

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Yes. So things like what the rate of expansion should be, for example. Which we find, to our surprise, seems to be accelerating, something NOT predicted by the model.

Hence all the work now on "dark energy", an ad-hoc hypothesis to explain the discrepancy.

But, as a lecturer in mechanics, I'm sure you know all this.

12. ### exchemistValued Senior Member

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[duplicate post deleted]

13. ### QuarkHeadRemedial Math StudentValued Senior Member

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agreed. But a more pertinent question might perhaps be "WHY the photon has momentum, since it is massless?" This bothered me for a while because my physics education was only elementary. I figured it, and here it is......

Classically momentum is given by $p=mv$ ($m$ is mass, $v$ is velocity). Enter the quantum world - Nils Bohr, the founding father of Quantum Mechanics postulates that, for an electron "orbiting" an atomic nucleus with orbital radius $r$ the only allowable angular momenta are integer multiples of $\frac{h}{2\pi}\equiv \hbar$.

Enter Louis de Broglie, who states that whatever can be said about one fundamental particle must be true of any other. Einstein agrees, hence the photon. But the photon orbits nothing (or almost nothing), so we can drop the requirement for a radius, and write $p=\hbar$ for its momentum

No mass term!

14. ### exchemistValued Senior Member

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Yes but hold on. An electron does not have to "orbit" anything. It still has mass when it is free.

The "orbiting" imposes a periodicity on its motion and it is this periodicity that results in quantisation of its allowed energy and momentum (pictorially, the waves overlap each other in successive passes, so they can only interfere constructively in whole wave multiples, i.e. a standing wave pattern.) This is why the harmonic oscillator, molecular vibration and and molecular rotation are also quantised systems.

Free motion, on the other hand, is not quantised.

Last edited: Feb 28, 2020
15. ### QuarkHeadRemedial Math StudentValued Senior Member

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I did not mean to imply that the electron's orbit is what "gave" it mass.
If that is how my post read, I apologise - only a madman would make such a claim.

16. ### exchemistValued Senior Member

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I thought it was a bit odd, coming from you.

But I've clearly misconstrued what you were saying. Do you want to have another go?

P.S. If one uses the relativistic formula for momentum, p=γmv, in which γ = 1/(1-v²/c²), it ceases to be a mystery that a photon can have momentum.

17. ### Neddy BateValued Senior Member

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I think you are missing a square root in your gamma equation. But for a photon, v=c and so gamma becomes undefined (division by zero), so I don't see how that equation helps explain a photon's momentum. What did you have in mind?

The link below says you have to use the general equation, and apply the Plank relationship, but I don't really understand what they are doing there either:
http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/relmom.html#c2

EDIT: Oh now I get it, the v in that webpage's equation is frequency, and then they put it in terms of wavelength (lamda).

Last edited: Feb 29, 2020
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18. ### Q-reeusValued Senior Member

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Interesting physics - photon momentum has one unique magnitude!
Perhaps you really meant |p| = νℏ/c, where v is the photon frequency! And btw ℏ is the full Planck's constant, not the reduced one:
https://en.wikipedia.org/wiki/Photon#Physical_properties (last equation line in sub-section)

While it's true a free photon has no rest mass, radiation confined inside a container (e.g. cavity resonator), adds to net rest mass.

19. ### CptBorkRobbing the Shalebridge CradleValued Senior Member

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This reasoning wouldn't make sense since Planck's constant has units of angular momentum, not regular. The real primary reasons the photon must be massless (i.e. no rest mass) are because

a) Photons always travel at the speed of light as measured by any observer, hence according to Relativity they can't have a rest mass. The general energy-momentum relation for Relativistic particles is $E^2=p^2c^2+m^2c^4$, where $m$ is the rest mass, and photons obey the classical electromagnetic relation $E=pc$, hence they must have zero rest mass.

b) Assigning a rest mass to the photon would break the gauge invariance postulate of quantum electrodynamics and also lead to non-conservation of electric charge. Even when renormalization is taken into account, the Ward-Takahashi identities ensure that it doesn't add any mass to the photon. Within the Standard Model, the W and Z gauge bosons acquire masses through interaction with the Higgs boson, but the gauge symmetries of the photon remain intact and it remains massless even after accounting for the Higgs.

20. ### Write4UValued Senior Member

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I can identify with that, I think.
It acquires only kinetic force (virtual mass)?
Does this mean that the Higgs field actually prevents the particle from traveling at greater than "c"?

21. ### exchemistValued Senior Member

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Yes thanks for pointing out the typo. And sure, one gets an undefined value, but one can see that if m->0 and v->c, momentum does not automatically vanish, in the way it does without the γ.

22. ### CptBorkRobbing the Shalebridge CradleValued Senior Member

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I don't think you understand how the Higgs mechanism works to add mass to the picture. Basically when you set up the fundamental equation of the Standard Model, which is called the "Lagrangian", there are certain symmetries that need to be included in the equation (i.e. changes of variables that don't alter the general form of the equation), and these in turn require that all the particles in the model have zero rest mass, including quarks and electrons. Then by adding the Higgs boson into the picture and allowing it to have a fixed non-zero vacuum expectation value, it allows you to mathematically assign masses to the various particles in the theory without breaking the underlying symmetries used to set it all up in the first place.

What my comment was referring to is that the section of the Standard Model describing electroweak forces includes four force-carrying gauge bosons: neutrally charged photons and Z bosons, and both positive and negatively-charged W bosons. When you add the Higgs boson to the picture, it "spontaneously" breaks the Standard Model's electroweak symmetry and mathematically assigns masses to the Z and W bosons, but there is an underlying sub-symmetry which still remains intact during this process and thus leaves the photon massless and always travelling at $c$ relative to any observer.

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23. ### Write4UValued Senior Member

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Welcome CrystalGauge, looking forward to your posts and hopefully learning a thing or two....