Dear Richard,

The photon has always mass zero. Relativistically

m=m0 gamma

and if m0=0 you always have zero. The mass always is a inertial concept and this tell us that never is possible to produce an acceleration of a photon. They move at the velocity of light independently of the state of motion of the physical system which produces it or which observe it.

Dear Richard,

The photon has always mass zero. Relativistically

m=m0 gamma

and if m0=0 you always have zero. The mass always is a inertial concept and this tell us that never is possible to produce an acceleration of a photon. They move at the velocity of light independently of the state of motion of the physical system which produces it or which observe it.

Both are correct. In units where c=1, the invariant mass is:

m2 = E2 - p2

The evidence supports theoretical prediction that m=0

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

This is a grandiose question.

John Archibald Wheeler once dealt with it at length in his big book on Einstein that he edited jointly with Zurek, "Quantum Theory and Measurement" of 1983.

This is an important question. In addition to the helpful observations already made, there is a bit more to add.

It has been observed that when sterile neutrinos decay, they give off a photon with exactly half their mass in energy. See Section 3.5, p. 8 in

https://arxiv.org/pdf/astro-ph/0702164.pdf

More to the point, consider the planar optical cavity (c) represented in the attached image from page 2 in

https://arxiv.org/pdf/1512.01130.pdf

Then consider the wave aspect of the effective mass. The general dynamics of the scalar electric field in the cavity is governed by the Helmholtz equation given in the attached notes. M. Richard, 2015, p. 2,

observes the following.

The photon mass given by the derived identity (see the attached notes)

is m||. Observe that m|| is not only the effective mass but also the exact

mass for the cavity photon from an inertial point of view.

E=mc2

if m=0, E=0

Is there a place/thing in the universe where there is no energy and matter?

May be, photon has 0.00000..........1 invariant mass

Hello Daniel,

The relativistic mass formula m=gamma mo is for a particle with a non-zero invariant mass mo so it does not apply to the photon with zero invariant mass and whose energy is E=hf. A photon experiences a vector acceleration when it is reflected from a mirror or when it is scattered from an electron as in Compton scattering, so its invariant mass can be calculated here. Please see my article Article A Photon Has Inertial Mass in Mirror Reflection and Compton Scattering

HI George,

How do you get the photon's inertial mass M=E/c^2 from the relativistic energy-momentum formula m^2 = E^2 - p^2 ? The experimental checks on the photon's mass m are for its invariant mass m (which equals zero to a very high degree of experimental accuracy) and not for its claimed inertial mass M=E/c^2 = hf/c^2 which can be far from zero.

Hello Otto,

What was Wheeler’s conclusion about the inertial mass of a photon in his book on Einstein "Quantum Theory and Measurement" that you mentioned? Einstein mentioned photons carrying "inertia" in his 1905 article where he first derived the mass-energy relationship, in "Does the Inertia of a Body Depend upon its Energy-Content?" at http://www.fourmilab.ch/etexts/einstein/E_mc2/www/, in the last line of his article, where he wrote "If the theory corresponds to the facts, radiation conveys inertia between the emitting and absorbing bodies." at the end of the article.

Hello James,

Could you please briefly summarize your main point about the inertial mass of a photon? Thanks.

Hello Mesut,

E=mc^2 is usually the formula for the rest energy of a resting particle (or other object) having invariant mass m such as an electron. The formula for the total energy E of a moving invariant-mass particle with speed v is E = gamma mc^2 where gamma =1/sqrt( 1-v^2/c^2). A photon has invariant mass m=0 but still has energy E=hf where f is the photon's frequency, so E is not zero for a photon having invariant mass m=0. If we write E=Mc^2 , it may be that for a photon, M is the inertial mass (not the invariant mass) of a photon, so the photon's inertial mass could be M=E/c^2. This is the question we are discussing here. It is possible that a photon has an extremely small but non-zero rest mass, but this has never been detected.

Richard

Dear Richard,

The formula E=hf is not related with mass of the photon. And the formula E=m0 c2 is only valid at rest which is not the case for a photon which never can be at rest. But the formula m=m0 gamma is general and tell you that some particle, case of the photon, with rest mass zero must has the same value for other system. The proper energy of a photon is E=pc without taking into account the mass (p=h/lambda).

The idea is that you never could accelerate a photon

Hello Daniel,

When a photon is reflected from a mirror, it undergoes a change of vector velocity because it changes its direction, even though its speed remains constant. This reflection is an acceleration, in vector math. So it is incorrect to say that a photon cannot be accelerated. Please see the article that I mentioned. Yes, a photon has zero invariant mass in all reference frames, but its inertial mass M=E/c^2 would be different and non-zero in different reference frames, since its E would be different in different frames.

Dear Richard Gauthier

According to the usual (and confusing) terminology a body at rest has a "proper mass" or "rest mass" m_0, whereas a body moving with velocity v has a "relativistic mass" or "mass" m, given by m = E/c^2. Since for the photon m_0 = 0, we can associate to the photon only the relativistic mass, m. But, we cannot apply to the photon the relation involving m and m_0, ie, m = m_0/sqrt(1 - v^2/c^2). Because, in this case we would have a mathematical indetermination (using for the photon v = c, we get zero/zero). There is only one mass here, m, that does not depend on the reference frame.

Dear Vasconcellos,

The things are worse, this expression is never true in the case of the photon which has a non rest system of coordinates associated. But the expression m = m_0/sqrt(1 - v^2/c^2) would be possible to write in the hypothetical case that photon would take a dynamic mass. For instance, m=h\nu/c^2. And even in such a case you would be obtain crazzy things as \gamma= infinite.

The photon has zero rest mass and also dynamic too. It never can get to be massive in Maxwell's electrodynamics!

Dear Richard,

The famous Einstein's formula E=mc2 is only a special case Pμ Pμ=(mc)2 (square of 4- linear momentum), which gives( plus/minus) E=mc2 when p=o, definition of rest mass. This is the usual definition and it is not necessary to complicate more the things. M= E/c2 is only true for particles with rest mass different than zero. Thus no for the photon and Richard your formula initial is not right..

Well Daniel

I do not agree with the first part of your comment. But with the second part yes. For the first part M= E/c^2 = sqrt(p^2c^2 +m_0^2 c^4)/c^2 is a parametrization. If m_0=0 => M= E/c^2 = p/c. This is the relativistic mass of the photon.

The concept of "relativistic mass" M=gamma m=E/c^2 is not universally recognized since it is hardly different from total energy E, in the case of a particle with invariant mass m not equal to zero like an electron. For a photon, "relativistic mass" M=E/c^2 is also hardly recognized since it is also hardly different from total photon energy E. The photon's possible inertial mass has been problematic since a photon seems to have inertia yet it has invariant mass equal to zero. Yet "inertial mass", even for a particle with m greater than zero, has never really been explained. It is just an unexplained proportionality constant in F=ma. Newton failed to explain inertial mass. Even today there is no agreed explanation of the inertial mass of an object. My double-helix photon model has calculated inertial mass M=E/c^2 , derived from its two rotating internal vector momenta. You can see it at Conference Paper Entangled Double-Helix Superluminal Photon Model Defined by ...

Dear Daniel and Richard

I do not have to go long to show why I do not agree with your statements. Even beginning students in relativity are accustomed to solving problems involving the relations between mass, energy and linear momentum in relativity. One type of problem they have to face is, for example, the one I collected for you: "Compute the effective mass of a 5000 Angström photon (Schaum's outline series - Modern Physics - Gautreau and Savin)". The answer is simple m_eff = E_photon / c^2 = h nu / c^2 = hc / (lambda c^2) = 4.42 x 10-36 kg ... (pg. 43). This is a classical problem they have to solve!

Daniel said: M= E/c^2 is only true for particles with rest mass different than zero

But this is not the true. In the equation m_eff= E/c^2 = sqrt(p^2c^2 +m_0^2 c^4)/c^2, since the rest mss of the photon m_0=0 => m_eff= E_photon/c^2 = p/c. This is the relativistic mass of the photon, or its effective mass. So, does the photon has mass? After all, since the photon has energy, and energy is equivalent to mass, yes. What the photon does not have is rest mass. The parametrization M= E/c^2 corresponds to the so called mass-energy equivalence relation.

Hello C.A.Z.

Thanks you for that example. To calculate E/c^2 for a photon and to call this its "effective mass" is easy. But what does "effective mass" or "relativistic mass" mean for a photon? One point of my photon article is that it not energy per se that gives a particle such as a photon its inertial mass, relativistic mass or effective mass, but the changing momentum vector "dp/dt" within the particle in relation to the internal acceleration "a" of the motion within the particle. This is a new approach to explaining the nature of inertial mass of a particle.

Richard

Hi Richard,

RG: How do you get the photon's inertial mass M=E/c^2 from the relativistic energy-momentum formula m^2 = E^2 - p^2 ?

The simplest approach I know is to consider a pair of photons of the same frequency but moving in opposite directions. The total energy is 2E because energy is a scalar but the momentum is a vector and the sum is zero. The total mass is therefore m=2E. To make that a 'practical' thought experiment, you can consider the photons inside a perfectly reflecting box, or such a box containing a large amount of light energy.

RG: The experimental checks on the photon's mass m are for its invariant mass m (which equals zero to a very high degree of experimental accuracy) and not for its claimed inertial mass M=E/c^2 = hf/c^2 which can be far from zero.

Yes, that's why I started by saying "both are correct". I am of the opinion that the term "relativistic mass" is misleading but the inertia of light is meaningful as we can see from radiation pressure. The problem with a common definition of relativistic mass as

mrel = gamma * m0 where gamma =1/sqrt( 1-v^2/c^2)

is that for m0=0 and v=c you get mrel = 0/0 which is indeterminate.

The definitions I prefer are that mass is the norm (or magnitude) of the energy-momentum 4-vector, hence invariant, while "relativistic mass" is really another name for the total energy in a specific frame.

Dear Vasconcellos,

The mass of the photon is zero (look at every table of particles that you want) as I have tried to explain but what is so important it must be zero for keeping alive the nowadays physics.

1. If you could introduce a mass mp just with any kind of consideration, then this must would against special relativity and Lorentz transformations. This mp would no follow Lorentz transformations and even it would introduce a singularity in the spacetime.

2. Electrodynamics (Proca) would be no gauge invariant and QED would't apply the renormalization recipes.

3. Electrodynamics in cosmology would be put in doubt as giving us the real information of distances and so, because the horizon would be more nearby that we usually assume.

Dear Richard,

My advice is that don't waste too much time looking for defining different kinds of mass. The mass is a well defined concept in physics that would be very dangerous to have several ones for obtaining different values. Trying to define pure electrodynamic concepts by the second the Newton's second law is not too serious due to have the electromagnetic field always moving at the velocity c and which was the source of relativity at the beginning of the past century.

Dear George,

Your sentence is not right: not for its claimed inertial mass M=E/c^2 = hf/c^2 which can be far from zero. A wave is not at rest by definition.

It is easy to see that your definition: The definitions I prefer are that mass is the norm (or magnitude) of the energy-momentum 4-vector. Is not enough for defining the mass because you would have always two terms into de square root.

Hi George,

Yes, I know that trick of looking at two oppositely moving, equal energy photons which taken together as a single object act as if they have invariant mass (and also inertial mass) M=2E. But this doesn't say anything about the inertial mass of an individually moving photon. Plus, the decision to treat two oppositely moving photons as a single "object" is a subjective one--they could be light years away from each other. You're right that the formula for relativistic mass M=gamma m doesn't work for a photon that has invariant mass m equal to zero. That's why I developed an independent way of calculating the inertial mass for a single particle, based on its hypothesized circulating internal momentum. Please see my article Working Paper Origin of the Electron's Inertia and Relativistic Energy‐Mom...

Hello Daniel and all,

I was not really searching for a new kind of mass. I was more interested in modeling particles as composed of circulating superluminal energy-- see Data Transluminal-Energy Quantum Models of the Photon and the Electron

Article Relativity Simplified by Modified Minkowski Metric Spacetime...

Hi Daniel,

DB: Your sentence is not right: not for its claimed inertial mass M=E/c^2 = hf/c^2 which can be far from zero. A wave is not at rest by definition.

That was Richard's sentence, not mine, I only quoted it.

DB: It is easy to see that your definition: The definitions I prefer are that mass is the norm (or magnitude) of the energy-momentum 4-vector. Is not enough for defining the mass because you would have always two terms into de square root.

You have four terms to be pedantic. If the energy is E and the momentum vector is (px py pz) with scalar components then the mass is given by:

m = sqrt(E2 - px2 - py2 - pz2)

The result is of course a scalar too, this is a well-known result.

Hi Richard,

RG: Yes, I know that trick of looking at two oppositely moving, equal energy photons which taken together as a single object act as if they have invariant mass (and also inertial mass) M=2E. But this doesn't say anything about the inertial mass of an individually moving photon.

Well if they have the same frequency as I specified, each contributes half, but it is a trick in a way.

The problem is that if you analyse a single photon on a box, it has to bounce of the inner ends which imparts a small change of momentum to the box at each reflection. If the box starts at rest and the photon has momentum +p then the box gains 2p while the photon reverses direction to have momentum -p on the first reflection, then it hits the other end and the box is stopped while the photon momentum reverts to +p. In other words the box is moving with a sawtooth distance characteristic. Similarly for a box containing a large amount of incoherent light, there is an effect on the box similar to Brownian motion, essentially the shot noise component of the internal radiation pressure.

In the though experiment we can assume two photons of equal frequency but with opposite directions which impact the ends simultaneously in the rest frame of the box hence the box never moves.

Using a single photon, we can average over twice the propagation time, L/c where L is the length of the box (assuming the light path is normal to the box ends). If you then apply a brief force to the box while the photon is in transit, the box acquires a speed. The photon will reflect from one end with a reduced frequency due to Doppler (assuming it was heading in the same direction as you accelerated the box) but get a blue shift when it returns to the other end. You still get the sawtooth motion but now it is asymmetric and if you again average over a complete trip (two reflections), you can show that the addition of the light in the box reduces the mean speed, hence the apparent acceleration for a given force applied to the box is also less than it would be for the box alone.

Basically, because over time the photon will move with the box, it has gained a little momentum from the applied external force, the box therefore gains slightly less and the total inertial mass is higher than that of the box alone.

The two-photon "trick" makes the analysis simpler but doesn't change the outcome.

Dear George,

The mass m never is a Lorentz scalar, in fact it is tensor of second order. On the other hand your mathematical expression is not is not necessarily real and less univalued. This tell you that your definition is not coherent with the concept of mass.

P.S. On thing is terms (energy and momentum) and other is their components.

Dear George

You are completely right. The mass is a Lorentz scalar. This is a well known concept.

Dear Vasconcellos,

Thus the mass has the same value for every observer that you have watching it. Good boy, yes!

DB: Thus the mass has the same value for every observer that you have watching it.

Yes, that is correct, mass in this case is a Lorentz invariant. Bear in mind that we are talking here about special relativity, I just found this entry in Wikipedia which summarises the point:

https://en.wikipedia.org/wiki/Mass_in_general_relativity#Review_of_mass_in_special_relativity

The situation is more complex in GR but from the same page above, we have these definitions of Komar and Bondi mass:

https://en.wikipedia.org/wiki/Mass_in_general_relativity#Types_of_mass_in_general_relativity

In particular the last sentence of the latter section states "Note that mass is computed as the length of the energy–momentum four vector, which can be thought of as the energy and momentum of the system 'at infinity'." which provides the link to the SR case.

@Baldomir: "The photon has always mass zero. Relativistically

m=m0 gamma

and if m0=0 you always have zero. "

But you have gamma=∞, so this is certainly not a good explanation. The "inertial mass" of the photon is indeed finite and equal to h υ/c2. But inertial mass is an outdated concept, because energy is the carrier of inertia, not mass. The mass of a particle is nowadays considered an invariant (and numerically equal to its "rest mass").

"The mass always is a inertial concept and this tell us that never is possible to produce an acceleration of a photon."

In fact, photons can be accelerated easily. Light bending is a sideways acceleration of photons. The speed of photons in flat spacetime cannot be changed, but their direction of flight can, and this is acceleration, too.

Moreover, in a gravitational field (i.e. curved spacetime), the following is also wrong:

"They move at the velocity of light independently of the state of motion of the physical system which produces it or which observe it."

The speed of light of a photon deflected by the sun is smaller near the sun and is larger far away from it, so the photon even gets accelerated along the direction of its motion. All of these speeds are lower than c, as Gerardus 't Hooft has remarked in another thread. Whether this has any physical significance, remained open in that discussion.

I think that I have tried to explain this question from all the points of views, even some of them quite dangerous. In mathematics there are two very important objects

∞ and 0.∞

The first exists and would be the mass of a photon if it had at rest a very small value (enough to be non zero). It is in this case the mass of the photon would be infinite and therefore without associated motion.

The second is just a non valid mathematical operation. Thus in the case of having rest mass zero for the photon we cannot obtain any relativistic transformation that change it. This is exactly the main principle of special relativity: velocity of light is independent of the state of motion, which was not more than a confirmation of experiments as the Michelson-Morley.

In any case, although wikipedia is not the most specialiced reference, I add it and I wish to everybody a glad weekend

https://en.wikipedia.org/wiki/Massless_particle

The point I was making Daniel is that using the energy-momentum relation allows you to deal with the physics while avoiding having any infinities in the maths.

Hello George,

Thank you for your detailed reply and about the thought experiment to show that the photon has inertial mass. I think I have read that explanation before. However, I don't think that this thought experiment shows WHY a photon has inertial mass M, as defined by F=Ma. I have no problem with the photon having momentum and transferring it. This is not the same as having inertial mass or having energy. I also have no problem with the average energy E of a photon or two corresponding to a mass M=E/c^2. That is what Einstein showed (or tried to show, as there is some controversy about his derivation) in his 1905 article "Does the Inertia of a Body Depend upon its Energy-Content?" But why does energy have inertia (or inertial mass) in the first place? Einstein and others came back to this proof over the years to try to show better proofs of E=mc^2 , as described in Hans C. Ohanian's article "Einstein's E=mc^2 Mistakes" at https://arxiv.org/pdf/0805.1400v2.pdf (a very interesting read). But none of these proofs or purported proofs of E=mc^2 really shows that the photon itself (or an electron in itself) carries inertial mass, which is the M in Newton's second law F=Ma. The origin of inertial mass of objects, except as a constant in F=Ma for a particular material object such an electron or a photon, remains a mystery. My double-helix photon model contains rotating internal momenta which change as a result of the rotating dipole of electric charge in the model, in a way that yields the derivation of M=E/c^2 from the way the internal momentum of the photon model changes in relation to the internal Coulomb forces within the photon model, in accordance with F=Ma applied internally within the photon model. You may say that F=Ma has been replaced by 4-momentum vectors and energy tensors, but I think this is not a complete answer. The fact of inertia remains. I think the fundamental question is "Why does energy have inertia in the first place?" Maybe inertia mass depends on the hypothetical (for now) internal momentum of a particle (like a photon or an electron), which itself depends on the particle's energy.

RG: I think I have read that explanation before.

I certainly don't claim originality, even Einstein analysed light in a box, although my version is distinctly not rigorous.

RG: However, I don't think that this thought experiment shows WHY a photon has inertial mass M, as defined by F=Ma.

It is important to note that F=ma was Newtonian and has to be used with care in SR, there is more than one acceleration.

RG: I have no problem with the photon having momentum and transferring it. This is not the same as having inertial mass or having energy.

In that case I think you need to be more precise in stating the question.

In the equation F=ma, the force F needs to be defined in relativity as F=dp/dt so it directly relates to a change in momentum.

Now imagine Bert pushes a stone cube sitting on an ice rink. When he applies a force, it starts to move from rest and when he stops it moves with constant speed, say v. We might loosely call the inertia of the stone its resistance to change of motion. Now imagine Alice is walking towards the stone at constant speed v but some way behind Bert. From her point of view, the stone was initially moving towards her at speed v but before she reaches it, Bert has changed its velocity so after he stops pushing, it remains at a constant distance ahead of Alice. From her point of view, the action of Bert was to bring the moving stone to a halt, he removed its momentum.

I always think that "inertia" is nothing more than "momentum" described in a different inertial frame. "Inertial mass" in this sense is then simply the coefficient that relates change of velocity to change of momentum.

RG: The origin of inertial mass of objects, except as a constant in F=Ma for a particular material object such an electron or a photon, remains a mystery.

As I see it, "inertial mass" is exactly what you say, a constant in the equation, though F=ma is only valid at low speeds, the equation itself must be more complex to apply at all speeds.

RG: I think the fundamental question is "Why does energy have inertia in the first place?"

We are used to using Minkowski Diagrams with axes of space and time but they also apply if you plot energy versus momentum, the same hyperbolic geometry applies. that explains why, at the speed of light, the energy must be equal to the momentum, you are projecting from a line that lies at the same angle to both axes.

However, while that works nicely when considering energy and momentum, note that you cannot use F=ma because light never changes its speed.

Hi, I recommend you to read the article below written by the Nobel winner, the great Steven Weinberg.

Photons and Gravitons in s Matrix Theory: Derivation of Charge Conservation and Equality of Gravitational and Inertial Mass

Steven Weinberg (UC, Berkeley

1964

Phys.Rev. 135 (1964) B1049-B1056

Reprinted in *Noz, M.E. (ed.), Kim, Y.S. (ed.): Special relativity and quantum theory* 338-345DOI: 10.1103/PhysRev.135.B1049

Abstract:

We give a purely S-matrix-theoretic proof of the conservation of charge (defined by the strength of soft photon interactions) and the equality of gravitational and inertial mass. Our only assumptions are the Lorentz invariance and pole structure of the S matrix, and the zero mass and spins 1 and 2 of the photon and graviton. We also prove that Lorentz invariance alone requires the S matrix for emission of a massless particle of arbitrary integer spin to satisfy a "mass-shell gauge invariance" condition, and we explain why there are no macroscopic fields corresponding to particles of spin 3 or higher.

Hi George

Lorentz invariance of the laws of physics is satisfied as you certainly know if the laws of physics are cast in terms of four-vectors dot products which define the so called Lorentz scalars. The formula E=mc^2 is a Lorentz realization of this statement. So, you are absolutely right when you say that mass is a Lorentz scalar. A tensor in relativity is not an invariant quantity. The transformation law of a Tensor, of a Vector, are well know in SR. Only Lorentz scalars are invariants.

@Gauthier: "The fact of inertia remains. I think the fundamental question is "Why does energy have inertia in the first place?"

This question is roughly at the same level as the question "Why does mass have inertia in the first place?" In Newton's theory, it is mass that has inertia, in general relativity it is energy. Since general relativity is the more fundamental theory, it may be used to explain why mass is the quantity in Newton's theory that has inertia.

Of course, neither theory explains why the property of inertia as such exists. Although the property of having mass can be explained by coupling to (the) Higgs field(s). But that does not help a lot, because mass accounts only for part of the energy (in the case of the photon for none) and it is energy that has inertia, not mass.

@Vasconzellos: "The formula E=mc^2 is a Lorentz realization of this statement. So, you are absolutely right when you say that mass is a Lorentz scalar."

But in this formula, m is not a scalar, because the energy is not!

The formula, in which m is a scalar, is E2=m2c4+p2c2. Herein, E and p c are components of a four vector with squared norm E2-p2c2 =m2c4 so m is a Lorentz scalar, if c is a constant (and that is the case).

Dear Kassner,

It is not necessary to do the square of the momentum for obtaining the mass as a Lorentz scalar. You can do it directly on the moment

Pα=mvα

So far the mass was only took as a physical magnitude associated to the 4-vector momentum and that is true, but in field theory it is directly associated to the 2-tensor energy-momentum which is much more general because it includes de the effects of the Maxwell stress tensor.

Hello George and all,

I think that you're right that inertial mass and momentum are closely related (but different). If we use F= ma = dp/dt (with vector quantities) we get inertial mass m = (dp/dt)/(dv/dt) = dp/dv. Now if p is defined as mv for a slowly moving object of mass m, then dp/dv = m , which is not very enlightening. But if p is expressed in terms of energy then it becomes more interesting. I started out some time ago trying to model photons and electrons by internally-superluminal energy flows. My resting electron model (and as it turned out some other persons' as well, as I later found out) was modeled as a circling photon-like object with rest energy Eo and therefore a rotating internal momentum vector p=Eo/c . When the dp/dt = omega p of this rotating momentum vector is divided by dv/dt which is the centripetal acceleration Acentrifugal of the circling velocity vector, the result is m = (dp/dt)/(dv/dt) = Eo/c^2 (with a few simple steps in between) , i.e. the inertial mass m of a resting electron came out (not surprisingly) given by Eo=mc^2 , where m is also the electron's invariant mass 0.511 MeV/c^2. But the inertial mass was derived by way of the changing momentum internal to the electron model. The inertial mass work is described in my article Working Paper Origin of the Electron's Inertia and Relativistic Energy‐Mom...

When I considered a relativistic electron model composed of a helically circulating photon-like object, described at Research Electrons are spin 1/2 charged photons generating the de Bro...

The above momentum equation got me thinking about Minkowski diagrams and led to a modification of traditional Minkowski diagrams for both space-time and for momentum-energy (momenergy), which simplified the diagrams as well as having the 4-vector momentum relations correspond to the momentum structure of my relativistic electron model. This is approach is described in my article at Article Relativity Simplified by Modified Minkowski Metric Spacetime...

Finally, when I learned that de Broglie had proposed around 1934 that a photon is a composite particle composed of two spin-1/2 half-photons, I realized that I should have been talking about half-photons composing my electron models all along. A photon can be described as a superluminal rotating oppositely-charged-dipole double-helix model. The photon model's inertial mass M=hf/c^2 = E/c^2 is calculated in the same way as I did with my electron model and its internal circulating momentum, even though a photon's invariant mass is m=0. Then I discovered that an Italian engineer Oreste Caroppo had discovered the same superluminal double-helix photon model in 2005. And it then turned out that I had developed the same photon model in 2002 but (after publishing it on a Dutch free-energy website) almost forgot about it when I turned my attention to non-composite photon and electron modeling. My latest double-helix photon model article is at Conference Paper Entangled Double-Helix Superluminal Photon Model Defined by ...

Richard

"A photon can be described as a superluminal rotating oppositely-charged-dipole double-helix model. The photon model's inertial mass M=hf/c^2 = E/c^2 is calculated in the same way as I did with my electron model and its internal circulating momentum, even though a photon's invariant mass is m=0. Then I discovered that an Italian engineer Oreste Caroppo had discovered the same superluminal double-helix photon model in 2005." -- What would be the physical meaning of this "model inertial mass M"? What would be the meaning of "inertial" in this context, has this been discussed by you or by Oreste Caroppo? Very generally speaking, the word "inertial" is only used in situations where an object reacts to an external influence (with the default behavior of no reaction for vanishing external influence), i.e. in a situation where an interaction of this object and its surrounding occurs. What would that interaction be for this "model inertial mass M"?

Very interesting Jan.

Dear Richard,

It is a pity, because you have made quite a lot work and basically is wrong. For instance, when you write:

My resting electron model (and as it turned out some other persons' as well, as I later found out) was modeled as a circling photon-like object with rest energy Eo and therefore a rotating internal momentum vector p=Eo/c

This cannot be a relationship between an electron energy Eo and linear momentum p. This disperssion law is only true for particles having rest mass zero as the photon.

On the other hand, when you follow writting:

the result is m = (dp/dt)/(dv/dt) = Eo/c^2 (with a few simple steps in between)

This formula is wrong for being applied to a photon because its rest mass is zero and remember that the timelife of a photon is infinite without having changes of time. My advice is that you clean your contributions to this question and try to the physics behind photons and electrons before applying them the basic classical mechanical concepts.

Dear Jan

I have to agree with Daniel. You use the formula vector p=Eo/c. Daniel is rigth when he says that:

"This cannot be a relationship between an electron energy Eo and linear momentum p. This disperssion law is only true for particles having rest mass zero as the photon."

@Baldomir:

"It is not necessary to do the square of the momentum for obtaining the mass as a Lorentz scalar. You can do it directly on the moment Pα=mvα"

First, it is much more elegant to use a scalar product to show something is a scalar, which is what I did.

Second, your formula is wrong. At least, if vα is the velocity, i.e., the usual meaning of the symbol. If it is the velocity, then m is the inertial mass, which is not a scalar as it is not invariant. The correct formula using the symbol m for mass would then read

Pα=mγvα, where γ=1/(1-vα2/c2)1/2.

Hello Jan-Martin,

People often say that a photon can't show inertial mass because its speed is always c. But force and inertial mass are related to change of vector velocity and not just change of speed. When a photon is reflected in a mirror its velocity vector changes while its speed is c before and after. Similarly, when a photon scatters off of an electron in Compton scattering, the photon's velocity vector changes direction in the reaction's center of momentum frame (as well as in the lab frame), as does the electron's vector velocity. In both cases the formula F=Ma = M dv/dt = dp/dt can be applied in vector form and gives the result (assuming that the reflection doesn't occur instantaneously but requires even a minute finite time, so that dv/dt and dp/dt can be calculated for the reflection) that the photon's inertial mass M is M=E/c^2 . Please see my article on this at Article A Photon Has Inertial Mass in Mirror Reflection and Compton Scattering

The double-helix photon model yield a photon's inertia mass M=E/c^2 in direction calculation from the model's two helically moving components.

Dear Kassner,

I thought that I didn't understood you properly, because that you have used the square of the 4-momentum to say that the mass in such a case was one Lorentz scalar( Herein, E and p c are components of a four vector with squared norm E2-p2c2 =m2c4 so m is a Lorentz scalar ). From pure algebra product of p

pα pα= Lorentz scalar=m2vα vα

But obviously it is possible to write

pα pα= Lorentz scalar=m02γ2vα vα with m=m0γ. The gamma γ dosn't change the tensorial character of the mass! The linear momentum is a vector independently that the mass transform with a gamma or not! What changes the transformations are the 6 generators of the Lorentz group SO(3,1) and nothing more.

Kassner. I have forgotten that you are the wise person who can find Newton's equations in quantum mechanics thanks to Ehrenfest theorem and also to calculate the acceleration operator among states. Sorry, my memory is quite bad and I think that we don't waste more time in discussion.

Hello Daniel,

If it were easy to show that an resting electron with energy Eo has a circling inner linear momentum p= Eo/c that gives rise to the electron's spin hbar/2 and to the relativistic energy-momentum equation through its internal-external momentum relationships, it would have been done a long time ago. You are giving the standard objections to this idea, but have you considered that the idea MAY have some physical validity in it? If a photon has inertial mass M=E/c^2 (and I think that many physicists agree with this, equating inertial mass with energy in the context of general relativity), then a circling photon-like object composing an electron would also have this inertial mass M=Eo/c^2 as well as circling linear momentum p=Eo/c. The inertial mass of the electron (which has invariant mass m = Eo/c^2 = 0.511MeV) would also have inertial mass m=Eo/c^2 because it would have the inertial mass Eo/c^2 of the circling photon-like object composing it, which I now call a half-photon rather than a photon, because an e-p pair is created in e-p pair production by "splitting" a photon (which could actually be composite in the first place) with sufficient energy. If a photon could be reasonably considered to be possibly a composite particle (a number of physicists have researched this possiblility in peer-reviewed articles after de Broglie's initial suggestion to this effect) then the electron could be composed of a circling spin-1/2 charged half-photon produced in e-p pair production. Obviously this is all theoretical and speculative and requires some direct experimental proof to gain general acceptance. Anyway, this is not my full time job (I am physics teacher) so I am free to speculate about these matters without worrying too much about what my physics colleagues at school will say (though of course I am open to constructive criticism, and thank you for yours.)

best wishes,

Richard

Dear Richard,

First of all spin is related with rotations and no with translations which could have p as a generator. On the other hand, you don't understand and it seems that other physicists do the same that for a photon you cannot have the equivalence of mass with the rest energy E=m c2, that is wrong the same in special or in general relativity! In general relativity if the photons move curving close to a gravitational potential is not due to the gravitational force or attraction on they. That is a pure Newtonian image, what happens is that the energy-momentum just creates geodesics in spacetime which curves the the trajectories of the photons. That is all! No inertial mass, gravitational mass or so on for a photon. The photon is always with zero mass!!!!

CAVZ: "The formula E=mc^2 is a Lorentz realization of this statement. So, you are absolutely right when you say that mass is a Lorentz scalar."

KK: But in this formula, m is not a scalar, because the energy is not!

Energy is a scalar, as you say it is just one of the components of the 4-vector obtained by projecting the vector onto the time axis, which gives a scalar result.

KK: The formula, in which m is a scalar, is E2=m2c4+p2c2. Herein, E and p c are components of a four vector with squared norm E2-p2c2 =m2c4 so m is a Lorentz scalar, if c is a constant (and that is the case).

So is E and each of the components of p.

The attached diagram might help with this. It is like a Minkowski diagram but with energy on vertical axis and the momentum in the direction of motion on the horizontal axis. Hopefully it will make the relationship between the various parameters clearer as there seems to be a lot of common understanding between the participants but still some slight confusion somewhere.

Hi George,

The attached file shows a comparison between the normal Minkowski diagram showing the relation of energy, momentum and mass compared with my modified Minkowski diagram. This is from my article at Article Relativity Simplified by Modified Minkowski Metric Spacetime...

Sorry Richard, I don't follow your top diagram. You have p=4 and E=5 hence m=3 and that would usually be drawn as the vector (px, E) from (0,0) to (4,5) but you seem to have it drawn along the line from (4,0) to (0,5). As drawn from the origin in my diagram, a Lorentz Boost then becomes a hyperbolic rotation of the mass line, the norm of the vector which is labelled "Mass" remains constant but a change of frame rotates the line about the origin by the rapidity and you then read off energy and momentum directly.

Also you have only shown the red section as part of the length. Remember this is a Euclidean projection (due to the nature of monitor screens) of a non-Euclidean geometry so the mass should be drawn as the full length of your line. Your red line is only shown as m=1.8, the hypotenuse is 3.0 long.

In the top caption, you describe that as the "invariant momentum" but momentum is never invariant. The length of the entire line shows the invariant mass but momentum is only p=mv in the Newtonian version, not SR, and that obviously doesn't work for light where m=0.

Hi George,

In the top diagram I am following Taylor and Wheeler's standard book Spacetime Physics, 2nd edition for momenergy diagrams. In his diagrams, the invariant mass (or the invariant momentum quantity mc) is only a fractional part of the Pythagorean hypotenuse produced by the E and the p sides of the triangle (Taylor and Wheeler indicate this in their book by a shorter rectangle draw on top of the longer hypotenuse. The relativistic energy-momentum formula (with c's removed) is m^2 = E^2 - p^2. But my modified Minkowski diagram plots the invariant quantity mc (which is a constant momentum quantity corresponding to the particle's mass m) on the vertical axis and the momentum p on the horizontal axis, giving E/c as the Pythagorean hypotenuse rather than E/c on the vertical axis as in the usual Minkowski diagram. So no hyperbolic rotation is involved in interpreting the second (modified Minkowski) diagram, and a single particle's mass m (or mc) remains constant on the vertical axis as the particle's energy E and momentum p are changed. That's why I call it a modified Minkowski diagram. In a two-particle collision problem, the invariant masses are plotted on the vertical axis, the particle momenta on the horizontal axis and the energies as diagonals. This is all explained in the Modified Minkowski diagram article I inserted in my previous comment.

I got the idea for the modified Minkowski diagram from the arrangement of the internal and external momenta in my relativistic internally-helically-moving electron model (Figure 20 in my article) where the model's internal transverse linear momentum component mc together with the electron's external longitudinal linear momentum component p = gamma mv give the total linear momentum E/c = gamma mc of the circulating charged spin-1/2 half-photon (as I call it now) composing the relativistic electron model.

Richard

RG: In the top diagram I am following Taylor and Wheeler's standard book Spacetime Physics, 2nd edition for momenergy diagrams

Cool, I have the paperback version.

RG: In his diagrams, the invariant mass (or the invariant momentum quantity mc) is only a fractional part of the Pythagorean hypotenuse produced by the E and the p sides of the triangle (Taylor and Wheeler indicate this in their book by a shorter rectangle draw on top of the longer hypotenuse.

First look at Figure 7-1, it's on page 193 in the paperback, the hard cover pages differ. Check the caption and you will see that they say the mass is the constant magnitude of the vector, it is the whole of its length. If you draw the base of the arrow at the origin of an energy/momentum plot, you get my figure. They have a 3D picture showing x and y momentum components along with energy in Figure 7-2.

The diagram you are referring to is Figure 7-3 where for some reason they show a fixed length "handle" overlaid on the vector which is very odd because the value of the mass is still the length of the entire vector, that looks longer because of the hyperbolic geometry of reality while the pages in the paperback obey Euclidean geometry. There's an acknowledgement to William Shurcliff for this idea at the end of the section under Table 7-1 at the bottom of page 213 but I think it is perhaps more confusing than helpful.

Anyway, the relevance to your original question lies in Figure 7-6. That shows the collision of two particles but it is equally appropriate to considering two photons. Note that the centre diagram shows the total system parameters and the effective mass is sqrt(302-102)=28.28. For consistency, their arrow in the centre diagram should have a "handle" with that value written in it.

If you draw a diagram like either of those on either side but with energy=momentum for each of the two arrows, the vector sum will have a non-zero mass, and if the component arrows are equal, the mass is 2E. If you then vary the length of each photon momenergy vector as they would be altered by Doppler shift for a moving observer, that resultant arrow from the vector addition also describes a hyperbola identical to that for a massive particle with m=2E. That is then the answer to your question I think, that value is the inertial mass of the pair of photons.

My own diagram is similar to their Figure 7-6 but also shows how other common terms (relativistic mass and kinetic energy) relate to the vector components, I've attached it again for convenience.

The important aspect is that in the Taylor and Wheeler diagrams, you can use normal vector addition to find the combination of different particles but note that in their example, the 2 invariant masses of 8 and 12 combine to give an effective system mass of 28.28, you can't add invariant masses directly, so I'm not sure why you think you drawing method works better than theirs because you have invariant mass on your vertical axis. Multiplying by c doesn't give you any real-world momentum by the way.

P.S. I've added two more sketches I drew last year showing first the vectors for two photons, one red and one blue and then their combination showing the invariant mass as black and the hyperbolic locus obtained by applying a variable Lorentz boost in green.

Hello George,

It is well-known that the invariant mass of two passing particles is not the same as the sum of the individual invariant masses of the particles (whether photons or particles with mass), and in fact the total invariant mass of the two particles can be larger than the sum of the invariant masses of the two individual particles. But the calculated total invariant mass of these two passing particles is independent of the inertial reference frame from which the two passing particles is observed. That's what it means to be an invariant mass (as you know). The total invariant mass of two (or more) passing particles is however determined by the total momentum of the system of particles and the total energy of the system of particles in any particular inertial frame, by the formula Mtotal^2= Etotal^2 - Ptotal^2 . In different inertial frames, Etotal and Ptotal for the pair of passing particles will be different, but Mtotal will be constant (as indicated by a single value of Mtotal on the vertical axis of the Modified Minkowski diagram, with an array of values of Etotal and Ptotal (as measured from different inertial frames) drawn to form different right triangles on the modified Minkowski diagram.

In my Modified Minkowski diagram, Ptotal would be plotted on the horizontal axis and Etotal would be the (usually) diagonal line segment with one end of the Etotal segment on the end of the Ptotal line segment on the horizontal axis and the other end of Etotal line segment leaning against the Mtotal axis, giving the value of Mtotal, the invariant mass for that particular combination of Ptotal and Etotal for the two passing particles. This is just the diagram of the energy-momentum formula above, when Ptotal is on the horizontal axis, Mtotal is on the vertical axis, and Etotal is the Pythagorean hypotenuse given by Etotal^2 = Ptotal^2 + Mtotal^2. No hyperbolic transformations are needed here.

The point of the modified Minkowski approach for momenergy diagrams is that the the invariant quantity m (or mtotal) does not have to be diagramed as a hyperbolic length relating E and p as in the standard Minkowski diagram. The relativistic energy momentum formula E^2 = p^2 +m^2 (with the c's left out that would make all the terms carry momentum: mc, p and E/c as in 4-momentum vectors) is more simply represented as a Pythagorean right triangle with perpendicular sides p and m and hypotenuse E. The invariant mass m is the vertical side, p is the horizontal side and E is the hypotenuse. (And the value mc IS physically meaningful in my relativistic electron model, as is E/c.)

As an analogy, consider a picture of the sun shining from the upper left of vertical rod stuck in the ground. The rod has an invariant height H. The sun casts a shadow of length S which is horizontal and to the right of the base of the stick. The light beam segment from the top of the stick to the end of the stick's shadow has length L. As the sun moves across the sky, both the shadow length S and the light beam segment's length L change while the height H of the stick remains invariant and is given by H^2 = L^2 -S^2, similar to the relativistic energy-momentum equation above.

The modified Minkowski diagram also applies to relativistic spacetime calculations with the invariant spacetime interval sigma = c delta tau, where c delta tau is plotted on the vertical axis, the distance v delta t is plotted on the x axis, and the coordinate time-difference measure c delta t is plotted as the hypotenuse, rather than being plotted on the y-axis as in the normal Minkowski diagram. This is explained in my modified Minkowski diagrams article mentioned previously.

I hope this is helpful. The above approach is just a simpler way of diagraming the relativistic energy-momentum equation and also the invariant spacetime interval equation of special relativity. (This modified diagram method was discovered independently by several others as well as myself, the first being Lewis Carroll Epstein, as mentioned in my article.) It is only tangentially related to the inertial mass issue, but the idea of this approach arose during the course of my inertial mass studies. By the way, 45-degree light cones are not needed in this modified Minkowski approach since c delta tau is plotted on the vertical axis rather than c delta t for light beams (where delta x is plotted on the horizontal axis), since c delta tau is zero for a single light beam traveling a distance delta x in time c delta t.

Richard

RG: In my Modified Minkowski diagram, Ptotal would be plotted on the horizontal axis and Etotal would be the (usually) diagonal line segment

Ah, thanks, I see what you are doing now. Yes, that works fine, you just have to do the vector addition using those non-perpendicular axes.

In that case, you might be interested in this "Insight" article if you haven't come across it before.

https://www.physicsforums.com/insights/relativity-rotated-graph-paper/

Hi George,

Thanks for this. I've just joined Physics Forums to hopefully get some feedback on the Modified Minkowski approach (which was independently discovered by several others, as mentioned in my article Article Relativity Simplified by Modified Minkowski Metric Spacetime...

best wishes,

Richard

Richard Gauthier

Dear Richard,

A photon has a transverse inertness (transverse mass). All answers to your question suggest that the photon is a point object. But there are no points in nature. The photon has dimensions. Therefore, we must expand the equation

E^2-p^2*c^2=m^2*c^4 in three axes. All speak about motion along the axis OZ only.

Applying equation to the translational motion of the photon along the axis OZ, we find that the photon's longitudinal momentum is pz = Ez/c, and the photon's longitudinal inertness mz is zero. Apply the formula to annular rotation of the photon. When pφ = 0, the photon will have angular inertness mφ. Photon energy of the toroidal rotation determine the angular inertness. We can say the same about transverse inertia along the R-axis.

You can look at the pictures in the book "Electromagnetic Gravity. Part 1. p.3.5" at my profile. Please see the animation of photon on my site http://gravity.spb.ru

Yours

Valeriy Pakulin

Hello Valery,

Thank you very much. I could not find the animation of the photon at your site. Could you please give specific directions to find it. Thanks.

Richard