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

Sometimes it is said that a photon has zero invariant mass m yet it has a relativistic mass M=E/c^2. Are both correct? Is the photon's inertial mass also equal to M=E/c^2? What is the latest thinking about a photon's inertial mass?

When we say E=mc^2, here m is actually a relativistic mass, usually for any particles, m = m_0 * \gamma, where \gamma = 1/sqrt(1-v^2/c^2), and m_0 is rest mass or invariant mass. For photon, because v=c, so if m_0 is not zero, then m would be infinite. That's why we say a photon has zero invariant mass. yet since a phone has an energy of E=h*\mu and that's not zero, so its relativistic mass m is not zero. So basically you can treat it as m_0 = m*sqrt(1-c^2/c^2) = m* 0 = 0, where m is not zero.

Nice thread.

Hello Bengeng,

Yes, you've seen the problem. The invariant mass of a photon is zero but its relativistic mass is non-zero. (??) Somehow physicists live with both results.

Richard

In my sense from understanding the classical mechanic, mass of particle is an expression of interior motion of its constituents.

Dear everyone. BINGO.

Finally, I found a topic where more than one person thinks like me (2>1 or 3>1).

There are a few similar topics:

i. https://www.researchgate.net/post/Does_photon_has_mass (see my syllabus of photon masses and term apparent weight)

ii. https://www.researchgate.net/post/Do_photons_have_mass (see syllabus of Sergey Shevchenko and his problem with factor 2)

iii. https://www.researchgate.net/post/Is_the_concept_of_the_relativistic_mass_properly_founded_and_used#view=5f42aa571c1f56522a148ca1 (overall a lot to see)

At the beginning I wanted to show how mass approach analysis is hated even on Wikipedia:

iv. https://en.wikipedia.org/wiki/Talk%3APhoton%2FPhotons_and_Mass_Debate3

It is sad that physical knowledge can be a barrier to understanding the essence of physics. I have intentionally left out which side of the barricade I mean, so that everyone can choose theirs.

Part 2 of the debate was less interesting and part 1 a bit boring:

v. https://en.wikipedia.org/wiki/Talk%3APhoton%2FPhotons_and_Mass_Debate2

vi. https://en.wikipedia.org/wiki/Talk%3APhoton%2FPhotons_and_Mass_Debate1

Best regards,

Grzegorz M. Koczan.

Hello Grzegorz,

It's good to keep the discussion going until a clear and satisfying answer emerges.

If it deflected by gravity into a circular arc it has both gravitational charge and inertia. If it imparts an axial motion on impact to particles it has momentum hence inertia.

Any idea that says no is false.

If we chase photons, they turn red ("slow down"), if we run against them, they turn bluer ("accelerate"). Likewise, axial gravity can act on photons. Moreover, gravity can act perpendicularly on the photons and bend their path. So photons are subject to gravity just like other particles.

Dear Daniel Baldomir,

I usually agree with your comments on relativity (SR), but in this case your answer ("popular", 4 recommendations) is not correct.

Your claim "m=m0 gamma, and if m0=0 you always have zero" is wrong.

If m0=0 we obtain an indetermination because in gamma v=c, so we have 0/0. Besides, a photon fulfills p=E/c=mc, which means that (relativistic) m is not null.

I do not also agree with "The mass always is an inertial concept and this tell us that never is possible to produce an acceleration of a photon".

Since a photon has no null m and it is a particle, then we can obtain the acceleration with the general expression:

m.a=F - v.(F.v)/c2 (boldfaced means vectors)

Simple analysis of the last expression gives:

1 – The photon cannot have tangential acceleration (c=constant)

2 – If force is not collinear with velocity the photon suffers normal acceleration.

Regards

Dear Hugo Alberto Fernández ,

I generally support Your approach. However, three considerations:

A) When thinking about relativistic mass, let's use the unambiguous symbol M or mr. Then even if someone does not use m0 for the rest mass but m, there will be no collisions. The designation M was used by Lorentz when he was already an aged scientist.

B) According to Your proposed definition of the mass of the photon:

Ma=F-(F.v)v/c2

paradoxically, I expect M = E/c2 in every direction of motion - (except the tangent/parallel direction). For tangent direction, both sides of the equation are zero M*0=F-F. However, the limit of the indefinite symbol M = [0/0] is possible to calculate for a good mathematician - for the tangential direction. I can bet You a beer M=E/c2 would come out too.

C) This is just math, this can be calculated from the force equation:

F=dp/dt=d(E/c)/dt

The mentioned Daniel Baldomir had fallen (like many others) on the first unsigned symbol M = gamma * m=infinity*0. He probably understood his mistake later, but did not explicitly change his mind.

Dear Grzegorz:

You are wright, to avoid misunderstandings it is convenient to express the relativistic mass as mr

I just partially read this very interesting thread, where I found some very well-argued discussions. What really surprised me is that most researchers do not considered the possibility that mass, like energy, is a relativistic magnitude. Also, it is not logical for me to argue about mass using properties of SR's mathematical formulation. Besides, the tensor approach is not the only formulation and it has limitations, particularly with photons.

I really wish I had read this thread from the beginning.

"I really wish I had read this thread from the beginning." H. A. F.

There is nothing to regret, because it is not lost and You can make up for it :-). And then You can write a beautiful ending to the topic.

As a tease, I would like to add that Hugo made a similar mistake to Daniel - only much more subtle. Daniel thought M=0 because m=0. Hugo thought that M||=0 because a||=0. The second point is too complex for a forum - it's a good topic for an advanced research paper.

Dear Grzegorz,

The use of relativistic mass has been rejected by most researchers, without logical reasons. It is my opinion that this kind of discussion must continue.

I do not know what mistake you mean. Since the relativistic mass is a state variable, and any physical system has energy, whatever it is (point particle, real particle, body, photon, radiation), then mr>0.

I mean:

"1 – The photon cannot have tangential acceleration (c=constant)"

which I understood as (maybe You edited ?):

"1 – The photon cannot have tangential mass (c=constant)"

Otherwise, it is not logical to use the dynamics law to calculate the acceleration for this case. The usual kinematics is sufficient for this (a||=0 for speed c). In other words, flies are not hunted with a cannon.

Dear Grzegorz, I did not modify any post.

Okay, nevertheless Your post suggested that M|| = 0 - don't You think?

What do You think about it anyway?

PS. Besides, equation m*a||=0 has two solutions m=0(M||=0) or a||=0. Daniel would point out these first solutions.

Dear Grzegorz,

Considering that mr tangent and normal exist is a historical error that comes from accepting Newton's law F=ma. If the force is defined as F=dp/dt, the relativistic mass is isotropic, that is, the same in all directions. Therefore, arguing about tangent or normal mass only shows poor knowledge of the theory.

Dear Hugo,

let me refer to the sentences in a different order:

"Therefore, arguing about tangent or normal mass only shows poor knowledge of the theory."

This is not entirely true. On the contrary, M|| and M_|_ must be defined. Only then can it be stated that M|| = M_|_ or M||> M_|_. It may depend on the previous way of defining. In any case, if we do not define these two masses, we will never solve the problem.

" If the force is defined as F=dp/dt, the relativistic mass is isotropic, that is, the same in all directions."

You cannot see the masses (mass) in this equation, so believing that they are equal (isotropic) is just faith. I have not seen or heard of anyone else introducing an isotropic relativistic mass. Without bragging about this art, I managed to do it :-) :-) :-).

"Considering that mr tangent and normal exist is a historical error that comes from accepting Newton's law F=ma."

Not really, in the sense that Fi=Miai is simply a definition of masse in a given direction. I remind You of your teaching that the definition cannot be wrong. So to say that studying longitudinal and transverse mass is wrong is just as wrong as saying that relativistic mass is wrong.

PS. I was able to give such definitions and write down the laws that M|| =M_|_=M. Thank You, Hugo, for Your questions giving me the opportunity to speak about it.

Dear Grzegorz,

The historical procedure:

The Principle of Relativity provides a very important tool for the formulation and/or verification of relativistic laws. The procedure is as follows: once the quantities involved in a classical law have been defined, the Lorentz transformations are applied to determine how these quantities should be modified so that the law maintains its mathematical form. Then, it is verified that the relativistic law becomes the classical one for c tending to infinity. Finally, the convenience of such formulation is analyzed.

That was the case when trying to establish the fundamental law of relativistic mechanics. Einstein initially used Newton's Law expressed by F=ma. The way in which Force and acceleration are transformed when passing from one frame to another is different. Consequently, if Newton's law expressed in this way (F=ma) is intended to be relativistic, the mass must take different values ​​depending on whether a direction is parallel to or transversal to its velocity.

Einstein wrote (On The Electrodynamics of Moving Bodies.):

“With a different definition of force and acceleration we should naturally obtain other values for the masses.”

This loss of isotropy of the mass was not "attractive" conceptually, and it was solved by proposing F=dp/dt as the law of mechanics, since this way of expressing Newton's Law retains its form under Lorentz transformations, without the mass lose isotropy.

Anyone can use Newton's law as F=ma in SR theory, since definitions are harmless, but all further development will have to be rewritten.

Dear Hugo,

Einstein (AdP 1905, pp. 891-921):

“With a different definition of force and acceleration we should naturally obtain other values for the masses.”

Thank You for this great quote. It shows that Einstein saw a need to redefine mass. He also talked about the redefinition of acceleration. However, the rest acceleration and the four-acceleration are not a real redefinition. There is another relativistic acceleration A.

H.A.F. wrote:

"This loss of isotropy of the mass was not "attractive" conceptually, and it was solved by proposing F=dp/dt as the law of mechanics, since this way of expressing Newton's Law retains its form under Lorentz transformations, without the mass lose isotropy."

The apparent loss of mass isotropy was not resolved by F = dp/dt, but only was swept under the carpet. Due to the fact that the matter was not named, the problem could not be solved for almost 115 years.

Dear Grzegorz,

I can understand that many young physicists and researchers without a deep knowledge of Modern Physics and the Theory of Relativity, repeat that relativistic mass is an "obsolete concept", although this has no logical reason.

What is inexplicable to me is that renowned researchers who are experts in relativity do not accept that mass can be a relative magnitude.

Best regards

Dear Hugo,

we agree on the subject - possibly we could elaborate on something deeper in the calculations.

On the other hand, the question of where are the opponents of the relativistic mass and the photon mass - were they afraid of the substantive arguments? I don't ask where Daniel Baldomir is, because he was probably a bit of a lonely "stoic" on this subject - so he may not feel confident as a sheep among "wild wolves." As a rule, there are only sheep everywhere - no wolves. And in this topic there are wolves, and the sheep ran away :-).

I call priests of the resting mass as "Stoics" (from the Stoicism philosophy). In Polish "stój" means "stop - rest", so stoicism semantically connects with the resting mass. Unfortunately, I do not know if this link is valid in English and other languages?

1. Some simplification of the notion of inertial mass of a point particle within SRT might be obtained if one assumes that the Newton-form (or: dynamical) equation of motion expressed in terms of a given arbitrary fixed IRF with coordinates [t,x] reads as follows:

$$ A(t) = m^{-1} F[t, x(t); v(t)], . . . . t \in R, v(t) := x'(t) := dx(t)/dt , $$

where A(t) is the proper 4-acceleration of the world-line {[t, x(t)]; t\in R }, F[t,x;v] is a 4-vector forces-field causing the motion, m > 0 is a constant. This makes the equation covariant wrt the Lorentz transformations.

2. Physical interpretation of the form 1. is that the forces depend on the points of the tangent space of the Minkowski time-space, at most. Moreover different types particles moving in this forces-field influence the equation by own factor m, which models the rest mass of a single particle.

3. The proper acceleration A used above is by assumption the second derivative of the world-line with respect to the proper time, i.e.

$$ A(t) = \gamma(t) d[ \gamma(t); \gamma(t) v(t)]/dt , . . . \gamma(t) := 1 / \sqrt{ 1 - v^2(t) }.$$

4. In case of electromagnetic fields, the proposed equation at 1. is consistent with the Planck-Einsteins equations (1905-1906) under the assumption that F is linear wrt the normed 4-velocity

$$ V(t) := [ \gamma(t); \gamma(t) v(t)] $$

via an anti-symmetric 1-1 tensor field as follows (abbreviated notion applied)

$$ Ft,x;v] = e V(t) \circ J \circ [ 0, -E[t,x] ; E[t,x] , M[t,x] ] $$

with the diagonal J metric tensor equal to diag{1,-1,-1,-1}.

Daer Joachim Domsta,

You read my mind telapatically - because I wanted to ask You to explain why You recommend me. I have the opportunity to thank You for Your recommendations.

I compiled Your formulas with Codecogs and added the notation collision explanation.

https://www.codecogs.com/latex/eqneditor.php

Dear Grzegorz, thanks for the decoding and showing the differences in notation. By the way, I have used personalized notation to make the content readable for some who prefer matrix symbols over the Einstein convention

Dear Joachim,

You gave the definition of the rest mass m := m0. Does that mean You are a Stoic? Or maybe You just decided to just start with a covariant approach?

Remember that physical reality is not as simple and straight as the highway. If someone wants to describe a winding forest road with roots and ravines in the language of the highway, he will fail. You can straighten the road with a roller, but it won't be this road anymore. In other words, uniforms can be sewn for citizens - and those who do not fit into these uniforms lose their citizenship.

Dear Grzegorz, You asked me:

" maybe You just decided to just start with a covariant approach? "

YES, I did. I am not opposing any reason to consider curly world, but like anyone tending to understand the whole road, I prefere to start with a small piece where everything is more simple. Afterwards I hope to get ability to study the next parts:-)

[Reedited after an hour:]

By the way, the presented standard construction with the embedded rest mass seems to be not opposing your very interesting way of distiguishing new notions (e.g. the relativistic acceleration and/or the nonisotropic mass). My entrance' aim was just to remind the basis, which I think might be expanded or even changed but with a great carefulness that would help in preserving the already accepted and confirmed features of the resulting model of the reality.

Dear Joachim,

four-acceleration and four-force are not elementary concepts even if we write them down more simply than You did.

Can You define the rest mass somehow more general and more directly - so that it is known that mass is mass and not to put Your hands v = 0, and not as a coefficient?

Dear Grzegorz,

Your question enters the theory of Theories. The notion "degree of directness of a definition" (of rest mass in this case) is off my abilities and I am giving up. I am interested in correctness and consistency with observations. The reminded by me Minkowski's definition is consistent with the SRT and simple according to my private criteria. As it follows from Your kind comments, our evaluations are different. I see your definitions presented in the paper of 2019 as very interesting and nice, but any rigorous statement on their simplicity is impossible since there are no really useful criteria. The more that I am not going to do research on the Theory of Simple, Direct, Elementary and General Theories :-)

Best regards, JoaD

Dear Joachim,

I think You are aware of Your cloudy speech. You gave useless definitions from the point of view of the topic - see, the topic is about photons for which v = c :-).

If You cannot correct it - then please - do not think (do not write) that the given definitions are complete and correct.

PS. Sorry, but I am a very demanding debater - even for allies.

The well known equations given in sections 7-8. are quantitatively expressing the following opinions mentioned by Grzegorz M. Koczan within this thread on August 26 and 27,

"axial gravity can act on photons. Moreover, gravity can act perpendicularly on the photons and bend their path."

" M=E/c2 would come out too."

and the following one by Hugo Alberto Fernández of August 27 (related to the perpendicular acceleration)

"Since a photon has no null m and it is a particle, then we can obtain the acceleration with the general expression:

m.a=F- v. (F.v)/c^2$ (boldfaced means vectors)}"

In the enclosed 2-page essay the equations for the luminal motions are derived just for showing their consistency with the Einstein-Planck-Minkowski formulas presented within 1--4. for the relativistic subluminal counterpart of the Newton second law, via suitable limit procedure.

After reedition::

The current version of the file is attached to the next answer. JoaD

Dear Followers,

due to substantial misprint in my last pdf file, I am enclosing the corrected version b. The first one will be deleted soon. Joachim Domsta

Dear Joachim Domsta,

bravo, could say You've derived the transverse mass M of the photon.

At the same time, You did not prove that m = 0, but assumed that m -> 0 by definition.

Your methodology does not calculate the longitudinal mass of the photon. I showed how to do it based on the idea of Hugo Alberto Fernández.

Dear Grzegorz M. Koczan

And what is the definition of the longitudinal mass of the photon, please?

to GMK continued:

. . . . . . . . . . . . . . .

I didn't derive any formula for the mass. But I understand, that Your definition of the transverse mass is the coefficient M in the equality \dot{v}..|..= M-1 E..|.. which can be inferred from the presented essay.

. . . . . . . . . . . . . . .

What I have done is a presentation of an equation of motion of luminal particles as a consequence of some particular limit procedure applied to Einstein-Planck-Minkowski, assumed as the relativistic counterpart of the Newton equation of motion. There is no mass in the final relation, since, as Hugo stated already some days before, the result is a differential equations for two parameters of the state: the energy {\cal E}(t) and the 3D unit vector of the velocity w(t) := \dot{x}(t)$, provided the 3D field E[t,x] of forces is given.

. . . . . . . . . . . . . . .

The derivation is not my, but a result of reconstruction of some notes seen/heard long time ago in many different places; so I cannot say which sources were decisive.

. . . . . . . . . . . . . . .

Your derivation allegedly based on idea by Hugo Alberto Fernández does not possess sufficiently plain scheme, since in particular it requires an answer of the question:

What are the assumed equalities and which one is the result?

Dear Joachim,

1. The transverse mass was defined by Lorentz and Abraham. So did Einstein, but he got a little tangled up in her. So please don't call this definition mine.

2. If You want to know my photon mass derivation (especially the longitudinal mass - sic :-)) You have to be patient. This is material for publication, but my preprint has not been published yet:

https://arxiv.org/abs/1909.09084v3

3. I know the derivation is not Yours - you quoted it without specifying the source, so I called it that. The statement "well known equations" is not "diplomatic" - I guess it was a Landau and Lifszyc textbook or similar.

Continuing, J. D. wrote:

"I didn't derive any formula for the mass."

So maybe I congratulated You unnecessarily, since You think that You have not deduced that the transverse mass of the photon is M=E/c2=hf/c2=h/(Lc).

All in all, that's not clear to Your goal. It is not known whether You are a Stoic (supporter of a rest mass) or a walking relativist Socrates (an advocate of a relativistic mass)? You seem interested but undecided as the title of Your file is:

"A trial of defining the relativistic mass of luminal particles" J. D.

So Your declaration that You did not derive any formula for the mass are to be understood as a failure of the title goal?

Wow: if You are not a Stoic or a Socratic then You are a Cynic. :-)

PS. In addition to the walking Socrates, the view of the variable mass can be attributed to Heraclitus (panta rhei).

Dear Grzegorz,

What I can do for answering your personal questions is to repeat the paragraph of my latest answer, beginning with " What I have done is a presentation of an equation of motion of luminal particles as a consequence of some particular limit procedure applied to Einstein-Planck-Minkowski, assumed as the relativistic counterpart of the Newton equation of motion. . "

That's your way of interpreting the result as a formula for the transverse mass, since you accepted that it is the coefficient M in the equality M a..|.. = F..|.. . In that sense you are stating that my trial was succesful with respect to this question.

With respect to the longitudinal mass, if I were You, for the sake of consistency, I would accept that this is the coefficient L in the equality L a|| = F|| , which would imply NONEXISTENCE of this quantity (since a|| = 0 always). But I am not You, therefore let me repeat a kind request: tell Your definition if we should discuss how calculate this quantity from suitable equations of motion for luminal particles (not necessarily those from my sections 7--8).

Dear Grzegorz and Joachim , allow me to interrupt your interesting comments.

First Joachim (nice to see you here). I will refer about the 4-vector treatment of SR. As I told several times, such approach, which mathematicians like so much, has logical inconsistencies when working with radiation and massless particles. So, if we are dealing with photons, there are no correct way to apply the usual 4-vector representation, which requires the "proper time". The main reason (between others) is that 4-velocity, 4-acceleration, and 4-anything involving proper time, has not physical meaning with photons. A reference system with light velocity (c) is out of the SR theory. The (Minkowski) 4-vector approach is not general and, obviously, cannot be the best SR formulation.

Grzegorz, why complicate the simple, if that is correct?

Using F=dp/dt, the relativistic mass becomes isotropic. Forget about longitudinal and transversal mass.

Regards

Dear Hugo,

Thanks for your kind comments. I agree with all of them. Let me mention, however that the reason of starting with subluminal case was to show that dispite the singularity of the luminal case it is possible to get some features via a suitable limit as m->0 consistent with the equation F=dp/dt.

PS. Obviously, for luminal particles mass is simply absent unless we extend (by convention) the formula M=E/c2 to the case of the rest-mass=0.

PS2. Temporally I have lost possibility of using super and sub - indices etc. therefore the text is less precise. Sorry.

@ Joachim Domsta

You have written more clearly now. I understand Your point now. The equation L*0=F|| is indefinite for F||= 0, and for F|| >0 it means "physically infinite" L. This is the limit of m||=m*gamma^3=E/c^2*gamma^2 which is infinite for photons.

So I am actually talking about something more general than Lorentz defined. I am talking about the inertial mass M which I defined in 10+1 ways for v

A REMARK. For luminal motions the equations \dot{\cal E} {\bf v} + {\cal E } {\a}_{\perp } = {\bf F} and \dot[ {\cal E} {\bf v} ] = {\bf F } are OBVIOUSLY EQUIVALENT and imply isotropy of M in the definition M:={\cal E}/( c^2 ). I think a new definition of relativistic mass is not needed. Though any further analysis of many new VALID formulas might be interesting.