# About effective (relativistic) mass of photon

Discussion in 'Physics & Math' started by Ultron, Jun 14, 2016.

1. ### UltronRegistered Senior Member

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This is non trivial issue, which is unfortunately not very well described in good sources, but I will try.

So my understanding is that the mainstream idea is that photon is massless, meaning that photon has zero invariant mass (rest mass). This was confirmed by experiments which set the lower possible value of photon at incredible low value. That means there was no experiment which would find any invariant mass of photon.

But there is also the energy/momentum of photon, which can be also be called effective mass (relativistic mass).

From wikipedia:
https://en.wikipedia.org/wiki/Mass–energy_equivalence#Massless_particles

Massless particles have zero rest mass. Their relativistic mass is simply their relativistic energy, divided by c2, or mrel = E/c2.[27][28] The energy for photons is E = hf, where h is Planck's constant and f is the photon frequency. This frequency and thus the relativistic energy are frame-dependent.
If an observer runs away from a photon in the direction the photon travels from a source, and it catches up with the observer—when the photon catches up, the observer sees it as having less energy than it had at the source. The faster the observer is traveling with regard to the source when the photon catches up, the less energy the photon has. As an observer approaches the speed of light with regard to the source, the photon looks redder and redder, by relativistic Doppler effect (the Doppler shift is the relativistic formula), and the energy of a very long-wavelength photon approaches zero. This is why a photon is massless—this means that the rest mass of a photon is zero.
.....
Relation to gravity
In physics, there are two distinct concepts of mass: the gravitational mass and the inertial mass. The gravitational mass is the quantity that determines the strength of the gravitational field generated by an object, as well as the gravitational force acting on the object when it is immersed in a gravitational field produced by other bodies. The inertial mass, on the other hand, quantifies how much an object accelerates if a given force is applied to it. The mass–energy equivalence in special relativity refers to the inertial mass. However, already in the context of Newton gravity, the Weak Equivalence Principle is postulated: the gravitational and the inertial mass of every object are the same. Thus, the mass–energy equivalence, combined with the Weak Equivalence Principle, results in the prediction that all forms of energy contribute to the gravitational field generated by an object. This observation is one of the pillars of the general theory of relativity.
The above prediction, that all forms of energy interact gravitationally, has been subject to experimental tests. The first observation testing this prediction was made in 1919.[29] During a solar eclipse, Arthur Eddington observed that the light from stars passing close to the Sun was bent. The effect is due to the gravitational attraction of light by the Sun. The observation confirmed that the energy carried by light indeed is equivalent to a gravitational mass. Another seminal experiment, the Pound–Rebka experiment, was performed in 1960.[30] In this test a beam of light was emitted from the top of a tower and detected at the bottom. The frequency of the light detected was higher than the light emitted. This result confirms that the energy of photons increases when they fall in the gravitational field of the Earth. The energy, and therefore the gravitational mass, of photons is proportional to their frequency as stated by the Planck's relation.

Another source:
http://physics.stackexchange.com/qu...-have-no-mass-how-can-they-have-momentum?rq=1

There are two important concepts here that explain the influence of gravity on light (photons).
1. The theory of Special Relativity, proved in 1905 (or rather the 2nd paper of that year on the subject) gives an equation for the relativistic energy of a particle;
$$E^2 = (m_0 c^2)^2 + p^2 c^2$$
where $m_0$ is the rest mass of the particle (0 in the case of a photon). Hence this reduces to $E = pc$. Einstein also introduced the concept of relativistic mass (and the related mass-energy equivalence) in the same paper; we can then write
$$m c^2 = pc$$
where $m$ is the relativistic mass here, hence
$$m = p/c$$
In other words, a photon does have relativistic mass proportional to its momentum.
2. De Broglie's relation, an early result of quantum theory (specifically wave-particle duality), states that
$$\lambda = h / p$$
where $h$ is simply Planck's constant. This gives
$$p = h / \lambda$$
Hence combining the two results, we get
$$m = E / c^2 = h / \lambda c$$
again, paying attention to the fact that $m$ is relativistic mass.
And here we have it: photons have 'mass' inversely proportional to their wavelength! Then simply by Newton's theory of gravity, they have gravitational influence. (To dispel a potential source of confusion, Einstein specifically proved that relativistic mass is an extension/generalisation of Newtonian mass, so we should conceptually be able to treat the two the same.)

My comment and important disclaimer:
Effective mass of photon means that you can calculate bending of light based on Newton theory. But this calculation gives only the half value compared to value calculated by GR which is also value confirmed by observations. So Im not trying to imply, that effective mass of photon is the only cause of bending of light.
I have tried to discuss the components contributing to this bending in previous thread:

But unfortunately it seems that some people dont understand what Im trying to say with this example.

Last edited: Jun 14, 2016

3. ### LaymanTotally Internally ReflectedValued Senior Member

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If your theory doesn't give you anything close to the correct answer, then it should give you some indication that isn't correct...

5. ### UltronRegistered Senior Member

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182
What theory are you talking about? Did you even read the original post?

7. ### LaymanTotally Internally ReflectedValued Senior Member

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Yes, I read it. You started with E = mc^2, and then you ended up with E = mc^2 in the form m = E/c^2. Then there was some word salad all mixed in the middle of it.

It is wrong, because the Higgs Field doesn't give mass to photons, only energy... It was not known if the Higgs Boson actually existed before, but now since the Higgs Boson has been discovered, I think that it is safe to say that the properties of the Higgs Field can invalidate this theory of the relativistic mass increase of light.

8. ### UltronRegistered Senior Member

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182
Please, just stop. What I have posted is not my theory, it is just standard relativity theory copied from wikipedia. And dont try to mix it with Higgs field, which is something different.

Last edited: Jun 14, 2016
9. ### LaymanTotally Internally ReflectedValued Senior Member

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Anybody can make a post on Wikipedia. Wikipedia isn't a reliable scientific source, and it cannot be used as references in scientific papers.

I think the first step is wrong, because momentum includes mass. Then the momentum would be equal to zero. Then the equation you start with there is just 0=0.

This type of operation was never accepted by the physics community. If anyone ever tried to publish something like this or that used this, the theory wouldn't be accepted. If you don't want to believe me, that is your prerogative. Then you have been warned.

10. ### LaymanTotally Internally ReflectedValued Senior Member

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Another thing is that E=mc^2 describes the conversion from mass to energy. When mass converts to energy, that energy is photons. Then you would be saying that massive photons would convert to massless photons...

The side of the equation that describes photons is the energy side, not the side with mass.

Last edited: Jun 14, 2016
11. ### exchemistValued Senior Member

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Careful! A photon most certainly has momentum, so you need to recognise that momentum does not imply rest mass in this case.

What Ultron is quoting seems to me to be standard stuff, not rubbish. The only issue is how far it is useful to push the concept of relativistic mass, given that is merely a ratio of momentum to velocity. I see he acknowledges that one cannot use relativistic mass plus Newtonian gravity to get the correct bending of light. So far, so good, it seems to me.

12. ### LaymanTotally Internally ReflectedValued Senior Member

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I think it does in this case, because the right side of the equation of E = mc^2 + pc is intended to represent massive particles. A photon has momentum due to energy density. Energy density is not going to be converted to energy in this equation...
In the steps taken in the equation he says m = p/c from substituting the part of the equation he just take out from setting it equal to zero... Then he gets E=mc^2 even though he set mc^2 to zero... Looks to be wrong on so many different levels... Times like this I wish Sixstrings was still here to rant on about how dumb it is when people try to use E = mc^2 in derivations, because they always end up getting nonsense out of it...
Have you ever heard of the correspondence principle?
https://en.wikipedia.org/wiki/Correspondence_principle

13. ### exchemistValued Senior Member

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The first line of your response is definitely wrong. The equation does apply to photons as well as to massive particles. The first terms vanishes because the photon has zero rest mass and you are left with E=pc. Applying Planck's relation E=hν, you get p= hν/c and hence p = h/λ ....or, rearranging, λ = h/p, which is de Broglie's relation, one of the most fundamental relations in quantum theory. These are all absolutely standard, well-known, results.

And yes of course I have heard of the correspondence principle. But at the moment I do not follow why you think it is relevant.

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14. ### UltronRegistered Senior Member

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It seems to me, that you dont realize that the calculation in OP was not done by me, it is quote from post on physics.stackexchange, and the original poster is graduate in mathematics and theoretical physics and is rated as top contributor for relativity in stackexchange:

http://physics.stackexchange.com/users/13/noldorin

Im by far not such expert on physics as he is, but it seems to me, that by raising questions about his calculation you just display your lack of knowledge about the topic.

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15. ### LaymanTotally Internally ReflectedValued Senior Member

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For a photon, the relativistic momentum expression

approaches zero over zero, so it can't be used directly to determine the momentum of a zero rest mass particle.

16. ### LaymanTotally Internally ReflectedValued Senior Member

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I am actually completely astonished that I am the only person here saying that they see something wrong with all of this... Maybe I should just give it time, until someone else has to scratch their itch to say how wrong something is about it...

17. ### exchemistValued Senior Member

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Layman, you have cut and pasted this from the link I gave, without bothering to read the very next section, in which it specifically addresses this issue, as follows:

QUOTE

Momentum of Photon
For a photon, the relativistic momentum expression

approaches zero over zero, so it can't be used directly to determine the momentum of a zero rest mass particle. But the general energy expression can be put in the form

and by setting rest mass equal to zero and applying the Planck relationship, we get the momentum expression:

UNQUOTE

OK? All clear now?

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18. ### exchemistValued Senior Member

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That is because, hard thought it may be to appreciate it, it is you who are wrong!

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19. ### LaymanTotally Internally ReflectedValued Senior Member

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No, I still don't understand why I have read books written by theoretical physicist that say that they never were able to discover a way to accurately predict the relativistic mass of a photon, but then you guys are claiming that it is a general known fact...

20. ### LaymanTotally Internally ReflectedValued Senior Member

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I would have to ask who was the original discoverer of this theory, and what experiment proved it? What paper was it originally published in?

21. ### exchemistValued Senior Member

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We are not just claiming it, we have both given you links to reputable sources saying exactly the same thing.

I think you need to cite some references for this idea that you can't "predict the relativistic mass of a photon". I rather suspect you may have misunderstood or misinterpreted something a bit more nuanced. The relations we have been talking about are all pretty standard stuff, you know. Even to a chemist like me!

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22. ### exchemistValued Senior Member

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You mean de Broglie's relation? Or Planck's relation?

For you to ask such a question reveals that you have no background whatever in quantum theory. I am happy to take you on that journey if you like, but not on this thread, as its purpose is not the education of Layman.

Shall we start another one, in which I can go through the history of the development of the fundamental concepts of quantum theory? It's in many ways a wonderful story, I think.