Does energy have mass?

23 May 2018 71 7K Report

If so, what means massless photon?

If no, why a moving photon has the mass equal to m=E/c^2?

'The theory of relativity deduces, from its fundamental assumption, a clear and convincing answer to this question, an answer again of a quantitative character: all energy resists change of motion; all energy behaves like matter; a piece of iron weighs more when red-hot than when cool; radiation traveling through space and emitted from the sun contains energy and therefore has mass; the sun and all radiating stars lose mass by emitting radiation. This conclusion, quite general in character, is an important achievement of the theory of relativity and fits all facts upon which it has been tested".

A. EINSTEIN AND L. INFELD, THE EVOLUTION OF PHYSICS, Page 208,

https://archive.org/stream/evolutionofphysi033254mbp/evolutionofphysi033254mbp_djvu.txt

Manuele Santoprete

This is to clarify something: according to special relativity the photon has rest-mass zero.

In special relativity, for a particle of rest-mass m0 and momentum p moving at velocity v we have the following two relationships:

1) (E/c)2=p2+m02 c2

2) p=E v/c2

(I use p, and v to denote the norm of the momentum and the velocity, respectively)

For a photon v=c, and so p=E/c. Substituting this in 1) we find that m02c2=0. Hence, in special relativity the photon has rest-mass 0.

For balancing the equation of conservation of energy one can introduce an equivalent moving mass m=E/c^2. I believe that text you quote refers to this latter type of I mass. The book Special Relativity by Albert Shadowitz has a short discussion on this. Maybe someone else can provide other references.

Stam Nicolis

No-the statement is meaningless. There isn't any such notion as a moving photon, since the photon can't stand still. The photon always moves at the speed of light.

It's wrong that a photon has mass. Mass is invariant under Lorentz transformations, energy isn't.

Energy and momentum are conserved in any given inertial frame. So to check whether an object, when emitting a photon, must change its mass, or not, it suffices to impose energy and momentum conservation in the frame in which the object is, initially, at rest.

Eric Lord

The famous equation E = mc2 applies only to matter at rest. Matter in motion has kinetic energy in addition to that and the formula for total energy is given by (E/c)2 = m2c2 + p2, where p is momentum. A photon is never “at rest” (its so-called “restmass” is m= 0) but it has momentum and energy: E = pc.

Forrest Noble

For calculation purposes a photon has a rest mass of zero. Of course in theory photons can never be at rest and must travel at the speed of light, c. A photon's energy at the speed of light in a vacuum is related to the frequency of the observed wave that contains it.

As to the energy of relative motion, force time distance equals the increased energy of a mass over its rest mass.

Energy itself is a meaningless term unless the type of energy is specified.

Stam Nicolis

Energy, of course is meaningful: it's the conserved quantity due to time translation invariance-which refers to a given, inertial, frame. The ``type of energy'' doesn't make sense, on the other hand, because it's not unambiguously defined.

Manuele Santoprete

Eric Lord gave an excellent answer. I was trying to say more or less the same thing, but he explained it much better. I just want to reiterate that there are two concept of mass in special relativity:

1) the rest mass (or intrinsic mass)

2) the relativistic mass (or equivalent moving mass).

The rest mass of the photon is zero, the relativistic mass is m=E/c2

The following Wikipedia article gives a decent explanation of the notion of mass in special relativity:

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

I believe the sentence " radiation traveling through space and emitted from the sun contains energy and therefore has mass" from The evolution of physics by EINSTEIN AND L. INFELD, refers to the relativistic mass.

Hossein Javadi

Dear Sydney Ernest Grimm

Thank you for attention and mentions.

I agree that there is no consensus on the mass of energy. Even the standard model needs to be completed.

But the issue the energy has mass is an important discussion.

Einstein in his article: "DOES THE INERTIA OF A BODY DEPEND UPON ITS ENERGY‐CONTENT?" in 1905, wrote: "The mass of a body is a measure of its energy-content; if the energy changes by L, the mass changes in the same sense by L/9 × 1020, the energy being measured in ergs, and the mass in grams".

https://www.fourmilab.ch/etexts/einstein/E_mc2/e_mc2.pdf

In general, there is a complete consensus among the physicists, and that is, the moving photon has the mass that is given by m=E/c^2.

The disagreement only concerns the rest mass of photon and the massless photon is just an assumption.

In addition to many papers on the upper limit rest mass of the photon is published,

Lawrence Berkeley National Lab keeps a list of the various methods to measure mass and charge of subatomic particles, including photons at:

http://pdg.lbl.gov/2017/listings/rpp2017-list-photon.pdf

And as far as I know, the concept of the mass of energy is the foundation of many modern physics problems that should be solved.

Hossein Javadi

I doubt that the rest mass of the photon is infinitesimal.

The new experiment shows photon has shape and is 4 meters long.

https://arxiv.org/pdf/1604.08020v1.pdf

Therefore, the photon will gradually lose its energy (and also its mass) when it absorbing by matter.

Which experiment can put the photon in a state of rest and determine whether it is massive or not?

Now consider in pair production an energetic photon is converted to two massive electron-positron.

Albert Manfredi

Hossein, we've debated this same question many times before. In "traditional" quantum mechanics, energy does not have mass, but there is an equivalency. Equivalency does not have to be sameness. Everything can be explained that way, and a particle that can only exist at v = c does not have to have any rest mass. It can still have momentum at v = c.

I'm not sure whether the supposed shape of a photon, slim and they now say about 4 meters long, has much to do with any discussion about its rest mass. To make any kind of convincing arguments on this, I would think the first step is to prove, through experimentation, that the speed of light, let's say in a vacuum, is variable. You need to show that photons with different energy content travel at different speeds.

Until you can show that, we are only conjecturing. Of course, there is no reason to religiously insist that photons have no mass, but the math and all experiments so far work out remarkably well that way. So to claim otherwise requires a certain burden of proof. And I mean proof, not "logical" arguments. Logical arguments can turn out to be entirely wrong, devoid of any link to reality.

Vikram Zaveri

Energy does not have a mass. But mass is a form of energy. Rest mass of a particle is one form of energy. The relation between two is like between water and ice. If energy is water then rest mass is ice. This condensation of energy happens according to Bose Einstein Condensate theory after big bang in following quantum gravity theory. Relativistic mass of a particle is another form of energy which happens to be the kinetic energy of the particle. Photon only has kinetic energy equivalent of relativistic mass. Therefore it is wrong to say that energy has a mass. But mass is a form of energy and charge is another form of energy. Entire universe is made up of only and only one thing which is energy.

https://www.researchgate.net/publication/262492889_Periodic_quantum_gravity_and_cosmology

Hans van Leunen

Nothing is so foolish as to apply a formula beyond its context. Typical examples are E = m c2 and E = h v. A am astonished why F = m a does not pose such problems. I never see it applied to massless particles. The most foolish formula is m c2 = h v. Still it makes sense by interpreting the Compton wavelength of an electron. However, if you think twice, then you know that an electron has no wavelength. The notion of Compton wavelength is just a play with formulas.

Manuele Santoprete

It is well known that the concept of relativistic mass is controversial. And many people agree with Mohamed Hassani rejecting this concept. In fact, the link I provided in my previous answer does include a nice discussion of the issue:

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

Moreover the link include links and quotes by several physicists that reject the concept of relativistic mass (including Okun). On the other hand, the question mentions a quote by Einstein and Infeld where they refer to the relativistic mass, so it makes sense to explain what it is.

Manuele Santoprete

Mohamed Hassani criticizes the derivation of the relation

E = c|p|, (1)

from the well-known relativistic energy-momentum relation

E2= (cp)2 + (mc2)2 (2)

I think he makes a valid point. However, at least from a mathematical point of view, relation (2) can be taken as fundamental. Indeed it is the square of the Hamiltonian of a relativistic particle. If one starts from the Hamiltonian formalism then one does not have to assume

p = γmv

as done by Mohamed Hassani. Hence, there is no mathematical problem at all. Moreover, if one thinks of the momentum as p=γmv, then Mohamed Hassani's derivation is also invalid. Of course one can always criticise the derivation using the energy-momentum relation from a physical or philosophical point of view.

Hossein Javadi

Dear Hans

Exactly, why do you want to put all the new questions in the old contexts?

I doubt on F=ma or F=dp/dt (relativistic Newtonian formula). In these formulas, the definition of acceleration is related to Newton's time and is even reviewed in relativity regardless of the structure of the particles.

If there is a limit speed in the universe, precisely because of the structure of particles.

Hossein Javadi, Graviton and Newton's second law, 2015

https://www.researchgate.net/publication/279185909_Graviton_and_Newton's_second_law

Hossein Javadi, et, at., Adaptive Review of Three Fundamental Questions in Physics, 2016

https://www.researchgate.net/publication/302313653_Adaptive_Review_of_Three_Fundamental_Questions_in_Physics

Hossein Javadi

Dear Albert

Yes, we have discussed a lot before, but you always repeat the same old opinions.

You have written:"You need to show that photons with different energy content travel at different speeds." Incidentally contrary to your claim, this new approach shows why photons with different energy content travel at same speed in the vacuum.

In addition, the photon speed in a medium is the function of the interaction of elements forming the photon in the medium.

You wrote:"I'm not sure whether the supposed shape of a photon, slim and they now say about 4 meters long,..."

You are not familiar with new experiments is not my fault.

I gave the link, I'll give you the new address again. Please see:

Victor Leong, et, at., Time-resolved scattering of a single photon by a single atom, Nature Communications volume 7, Article number: 13716 (2016) doi:10.1038/ncomms13716

https://www.nature.com/articles/ncomms13716

Herb Spencer

This thread illustrates the confusion confounding modern physics.

When two basic concepts, like mass and energy are confused because of a mathematical "proof" it shows how badly we have gone 'off course'.

Without clear concepts there can be no conceptual progress.

Hossein Javadi

Dear Jacinto Jesús

You will never move at a speed of light.

You need to use infinite energy to quickly get light speed.

See:

Hossein Javadi, Graviton and Newton's second law, 2015

https://www.researchgate.net/publication/279185909_Graviton_and_Newton's_second_law

Hossein Javadi

ِDear Herb

I agree with you, the reason is the massless photon is an assumption only,

that is based on a mathematical interpretation of Newton's second law.

Albert Manfredi

Hossein, I take you are saying that the speed of light need not be the speed of these massive photons? Because I don't see any way for massive photons, with different energy content, to travel at precisely the same speed. Unless we change the equation

E2 = (pc)2 + (moc2)2

I think most of the time, the assumption is that the speed of light is the speed of photons. Going by that assumption, if photon rest mass is not 0, then won't photons with different energy content have to have different velocities in that p term?

p = mov / sqr(1 - v2/c2)

The term moc2 won't change, if photon rest mass is some constant value, so what has to change is the pc term, incorporating the velocity of the massive photon. Now you can have massive photons of different energy content, but their speed will be different.

Not just me, of course. I've attached a couple of links that also make this point.

Are you maybe saying that the speed of light could be the interaction of these massive photons, as the speed of electricity is related to the interaction between the electrons being dislodged, and not the same as the speed of the electrons?

Issam Mohanna

Manuele Santoprete

Please , if you wouldn't mind , I would like to know the difference between the inertial mass and the rest mass.

Hossein Javadi

ِDear Mohamed

Thank you for interesting mentions.

It is true that my claim is affected od SRT.

But I got it through other reasons and analyzing.

When Einstein was presented SRT, the theory of quantum mechanics was in its early stages.

In the last century, with the advent of quantum mechanics, the mind of man has changed a lot.

One of these developments is the quantum mechanics approach to the concept of force and the theory of exchange particles.

My definition of acceleration is not the Newtonian definition.

Using the particle structure with a new approach, I reviewed relativistic Newton's second law, and concluded that only particles (including photons) can move at the speed of light in which there is a color-charges symmetry in their structure. Any particle that does not have such a symmetry (including charged particles) can never reach the speed of light.

Hossein Javadi, et, at., Generalization of the Dirac’s Equation and Sea, 2016

https://www.researchgate.net/publication/304082658_Generalization_of_the_Dirac's_Equation_and_Sea

Hossein Javadi, et, at., Adaptive Review of Three Fundamental Questions in Physics, 2016

https://www.researchgate.net/publication/302313653_Adaptive_Review_of_Three_Fundamental_Questions_in_Physics

Hossein Javadi

Albert Manfredi

I do not claim that E2 = (pc)2 + (moc2)2 needs to change.

In this relation, you can choice m0=0 fo photon and use E=mc2. Because you cannot find a photon at rest condition.

But note that when a particle takes energy is absorbed (absorbs a photon), the mass of the photon is added to the particle mass.

The reason for the speed of light is also explained in the previous comment.

Andrew Worsley

Energy and relativistic mass are clearly interchangeable.

Article The formulation of harmonic quintessence and a fundamental e...

Hossein Javadi

Dear Jacinto Jesús

Thank you for sharing your idea.

I saw your page on Twitter.

Hope more success for you.

Hossein Javadi

Dear Mohamed

Thank you for sharing this important work.

Albert Manfredi

Hossein and Mohamed, I will look more closely at Mohamed's equations soon. For now, my response to Hossein is that when you say we do not need to change the total energy equation, and at the same time you say that one can use mo = 0 for the photon's rest mass in that equation, then you are ignoring my point entirely. If people want to posit that the photon does have rest mass, then that total energy equation will show differing speeds for photons, depending on their energy content. Using mo = 0 would be inaccurate, in that total energy equation, according to this new theory.

If we can accept that the speed of light is independent of the speed of photons, then that would be a clear statement to make, at this point.

And Mohamed, you seem to be saying that the speed of light is a constant by convention, and that the speed of those photons is in fact not constant. Your last equation says this, does it not? But again, I have to go through your numbers in more detail.

In short, claims that photons have rest mass must be followed by clear explanations of the consequences. Such as, do not equate the speed of light with the speed of photons, do not assume that the speed of photons is constant, and/or, do not assume that the "true" speed of light is a constant. This is the kind of claim that will cause people to wake up and take notice.

Saying tha the energy of the photon depends on nu, or even that the mass depends on nu (which one can derive using algebra that I may not find terribly convincing), is not enough of a striking comment, to wake people up.

Hans van Leunen

Physics has not yet solved the problem that due to expansion of space at the instant of emission of a photon the geometrical conditions differ from those that reign at the instant of absorption of the photon. After a long range trip of the photon the same kind of atom can no longer synchronize with the exchanged photon. We call this phenomenon red shift, but we do not comprehend what really happens.

In the event of the collision of two neutron stars that was observed by LIGO, Virgo and many other observatoria the blue light, the red light, the gamma rays and the microwaves all appeared to arrive at different instants. The reason was not that this radiation was emitted at different instants! The gravitational wave affected the travelling time of the radiation.

In my personal view photons are carriers of pure energy. They don't possess mass. However, they must follow the deformations of their carrier field.

Hossein Javadi

Dear Albert

Light is a collection of photons. Photons are the particles of light.

You wrote:"If people want to posit that the photon does have rest mass, then that total energy equation will show differing speeds for photons."

Your approach to this equation is a classic approach.

If we want to increase photon speed in a vacuum, how is this done?

The answer is simple and clear.

We have to accelerate to the photon.

How can we do that?

In classical mechanics, we need to use Newton's second law F=ma=dp/dt. Suppose right side of F=dp/dt, we have a photon. You should define left side a force (F) that be able to change the momentum of the photon.

According to quantum mechanics, you cannot do that. Because force is applied by the discrete amount of energy that moves with speed of light. It means we need to review relativistic Newton's second law.

The speed increase in a black hole is not specific to quantum mechanics. Even in general relativity, the speed of light in the gravitational field is not identical from the viewpoint of different observers.

See picture

In special relativity, the speed of light is the upper limit for the speeds. But in the presence of gravity, the speed of light is not same for all observers. "In the presence of gravity the speed of light becomes relative. To see the steps how Einstein theorized that the measured speed of light in a gravitational field is actually not a constant but rather a variable depending upon the reference frame of the observer." [1] Also, Einstein wrote the paper in 1911 in German” 'On the Influence of Gravitation on the Propagation of Light', Annalen der Physik, 35, 1911. [1] To see the Einstein paper in German Language see:

http://myweb.rz.uni-augsburg.de/~eckern/adp/history/einstein-papers/1911_35_898-908.pdf

1 - Speed of Light in Gravity, Effect of gravity on measured Speed of Light

http://www.speed-light.info/speed_of_light_gravity.htm

Herb Spencer

All this discussion about waves or particles misses the point.

Newton showed (and everyone knows) that gravity acts at a distance ('Far Action') but his commonsense observation was disputed by small minds who emphasize their personal experience with human touch ('Local Action') as the only source of action. Hence these silly Field Theories that hide their mysticism (faith in the Ancient Aether, like Maxwell) with confusing, difficult complex continuum mathematics; revived in Einstein's failed Gravity Theory.

The Optical War (of the 18th century) is another example of this type of confused thinking. We do NOT need a moving ENTITY (independent existent) to cross the Gap betwen remote interacting electrons (UET6).

We are not in a Scientific Debate here but a hidden, medieval Theological Dispute that has always tried to defeat skeptical thinking (Philosophy).

Albert Manfredi

Okay, from my understanding, Mohamed and Hossein, you are both going beyond the narrower point, of the constant nature of the speed of light in a vacuum (or within other homogeneous media), for different photon energy content, or the associated question of whether the speed of photons themselves is c, or whether photon speed is different, and variable, compared with c. I ALSO have no problem accepting that gravity affects light, nor do I have any issue with c being a different value in different media. These are well known phenomena, none of which rely on photons having any rest mass.

Mohamed, I get that the energy of a photon is E = hf. That's not a problem for me. And it's not too hard to see what the consequences of that are.

For (let's say "possibly") massless photons, E = hf.

For massive particles, we have E2 = (pc)2 + (moc2)2.

You can see that if the rest mass of a photon is claimed to be zero, that second equation cannot apply to photons, in their photon state of existence.

We can simplistically enough say that when a photon is absorbed by an atom, that the mass of that atom can increase by some amount, while the photon vanishes entirely. Since we are only talking about increase in mass,

If E = hf for the photon and E = pc for the massive particle.

hf = mvc, and

m = hf/vc

We also know that v = f * lambda, so we can rewrite that equation

m = h/(lambda * c)

where lambda is the wavelength of that photon.

The only thing this equation is telling us, from my perspective, is that when a photon of energy content E is absorbed by an atom, the mass of that atom is increased by h/(lambda * c). This happens because the energy of the (very possibly) zero-rest-mass photon is entirely converted into mass, as that photon, its speed, and its wavelength, vanish. I do not think this simple derivation insists that the photon itself has any rest mass of m.

On the other hand, as my two references, many other references, and I think I also indicated, if the photons themselves have any amount of rest mass, go ahead and use the total energy equation for massive particles, and you will have to conclude that the speed of photons is much lower than c, and even more amazingly, the speed of photons must be variable, depending on their energy content. It's an inevitable outcome. I think experiments to determine the veracity of these consequences might tell us more convincingly, if photons do indeed have rest mass.

This is my only issue. I have no big problems with gravity affecting photons, because at their speed, if massless, there is that Lorentz transformation singularity.

Herb Spencer

Electron scattering or X-rays scattering off electron containing matter can be explained by a particle ontology combined with a fluctuating interaction model; as I will show soon. This obsoletes all 'field' fluctuation 'explanations' (existential waves).

Please be patient.

Albert Manfredi

Mohamed, thank you for your careful derivations. I probably used some confusing words, but we are not really saying much that is different. And yet, nothing suggests, in Compton scattering, that the photon must have rest mass. That's where we differ.

I agree that we have the expression E = hf for the photon.

In the attached, recent paper on Compton scattering, everything is derived assuming that photons have no rest mass. It does does point out, equation (3), that we have the equivalency E = cp for the massless photon. They conclude this, in equation (2), because in the equation

E2 = (pc)2 + (moc2)2

that last term becomes 0 for the massless particle, including a photon, in their view. The rest mass of the photon is assumed zero, so the rest energy is also 0. We are left with E = cp, for photons. No rest energy. So, they disagree with you on this point.

Now, this is along the lines of what I showed you previously, but it's not my own derivation, it is the article's.

E = hf, and E = cp. Therefore, for these massless photons,

p = E / c and it follows p = hf / c for photons. No mass terms in the equation, at all.

Since we know generically that v = f * lambda, and the assumption here is that v = c for these massless photons, then it must follow that

f = c / lambda for massless photons, and therefore

p = h / lambda.

Note: still no mass terms in that photon momentum equation. Then, when this photon's momentum strikes an electron that is (comparatively) at rest, with the electron's momentum pe = 0, the momentum of the photon becomes imparted on the electron, accelerating it in one direction, and a gamma ray is scattered in a different direction.

The electron that we assumed more or less at rest now has added kinetic energy, but the scattered gamma ray still does not have mass, or kinetic energy, just as the original moving photon did not. That gamma ray is just as you would expect of a photon that has lost some of its energy when bouncing off the electron. Like billiard balls, except that one of the balls is massless.

So, any hf = moc2 is an equivalency expression. I'm not making this up. Or at least, I'm not alone in stating this. It need not be interpreted as photons having that rest mass mo.

My only, and continuing objection, remains. These effects, which you and Hossein mention, including Compton scattering, including how gravity affects photons, including how photons absorbed by atoms increase the mass of the atom, do not need to depend on photons having rest mass, which would in turn create other problems. Such as, photon speed depending on their energy content, not a constant value, and not equal to c.

Ultimately, photons may indeed be proven to have rest mass. Many articles have proposed the maximum possible rest mass of photons, but the minimum possible continues to be zero. Do read the attached link. It is dated 2015. (Sorry that I seem unable to insert links in these posts.)

Albert Manfredi

Hidouri, in short, when you say "correlated to the mass," it does not imply that the photon has that mass. It is only an equivalency. "IF the energy of a photon were to become the rest energy of a massive particle, THEN the rest mass of that massive particle would be m."

There is NO implication, in that equation alone, that the photon has any rest mass. None.

Hossein Javadi

Dear Hidouri

Relation to E=mc2, at least shows a moving photon has a mass equal to m=E/c^2.

But there is no general consensus on the mass of photon at rest.

I have not seen any empirical evidence that a photon is massless. Even Einstein assumed that the photon is massless.

But empirical evidence shows that photons have a mass limit, which is not zero.

Lawrence Berkeley National Lab keeps a list of the various methods to measure mass and charge of subatomic particles, including photons at:

http://pdg.lbl.gov/2017/listings/rpp2017-list-photon.pdf

Read more; Are photons really massless?

https://www.researchgate.net/publication/321589753_Are_photons_really_massless

Beemnet Mengesha Kassahun

In physics, mass–energy equivalence states that anything having mass has an equivalent amount of energy and vice versa, with these fundamental quantities directly relating to one another by Albert Einstein's famous formula:E = mc2. Despite Dr. Albert Einstein derived one of the world's most famous equations, E = mc2,, didn't receive the Nobel prize in physics for this discovery. This equation helped develop the modern nuclear age.

https://www.reddit.com/r/Physics/comments/4pb8qg/how_does_energy_have_mass_andor_viceversa/

http://www.desy.de/user/projects/Physics/Relativity/SR/light_mass.html

Herb Spencer

There are NO photons: there is simply momentum/energy being exchanged in an interaction between TWO remote electrons. [Observable FACTS.]

Please stop inventing hypothetical entities (independent existents), like EM 'waves' or 'photons'; there really is a difference between Science and Magic.

Nikolay Krobka

Why not?

Herb Spencer

@Nikolay Krobka

Dear Nikolay:

Assuming your vague (undirected) question was addressed to me; I will help you with your metaphysical education.

The Original Thinker, William of Ockham gave good advice 700 years ago: - "Keep it Simple". Inventing hypotheticals is an endless task providing no new insights; this is why Newton wrote about gravity: "Hypothesi non Fingo". Newton still advanced our views on gravity WITHOUT the need for gravitons (or any of the explosion of modern hypotheticals).

Hanna Edwards

I think the key to your question lays in the understanding of what energy and mass stands for .

a) I understand energy as a measure of change over space and time . This applies to mechanics or thermodynamics but also to a "wave" which describes a certain amount of change over space and time . The term "energy" describes "how much" .

b) I understand mass as another word for inertia , which measures the resistance to a force or say the resistance against change happening to it .

A wave gains its nature as a freely propagating disturbance - so it has no inertia based on what it is - but a) holds . A photon freely propagates at speed c - so I think this applies here .

A restmatter particle (not "free" (to propagate at c)) however i personally picture as a standing disturbance - and such has inertia - so a) plus b) holds .

I think that whilst energy content gets maintained during physical processes - inertia can be lost or gained during the changes that happen to matter which are linked to the absorption or emission of a photon .

Albert Manfredi

But, Mohamed, I think maybe the whole crux of the problem is that you insist that a massless photon cannot have momentum, and that therefore

E = cp

cannot apply to massless particles. But the literature is full of references to the momentum of massless photons, so your diagreement is what is unusual. Not my skepticism. My skepticism only comes because the effects we have been talking about don't mandate that photons have mass. (But I'm open to verified experiments that mandate massive photons.) And furthermore, your massive photon would cause many wide-open eyes, if you would mention the other consequences of photons having mass. I think too often, these other consequences are not made abundantly clear. People would sit up and take notice, if they were made clear.

In this total energy equation:

E2 = (pc)2 + ((moc2)2,

the term to the right is dedicated to the rest energy of a massive particle. The pc term can be expanded, for massive particles, using the Lorentz transformation, just as you describe, if the particle has rest mass and relativistic speeds are going to exist. (And I think it is important to note that no massive particle can reach v = c, while it still has mass.)

Or, that p term can consist of classical physics momentum, for slower moving particles. In either case above, yes, the kinetic energy adds to the rest energy.

Or that p term can consist of the momentum of a massless particle. For this last case, the term to the right, that moc2 term, becomes zero, and the momentum p is instead related to h and lambda, as derived before.

p = h / lambda, for massless photons. It is not zero. There are plenty of references to this, in the literature.

What you describe is only the first example of those three I listed: the particle is a massive particle, and it travels at relativistic speeds. So, that pc term can indeed blow up:

pc = (m / sqr(1 - v2/c2) * c

which is what you wrote, for the momentum of a massive particle moving at relativistic speeds. But this is not the term to the right, which is dedicated only to the rest energy.

So there is no 0 = 0 problem, unless you believe that the p term must absolutely incorporate m.

And then parenthetically, I derived this short, one-page curiosity that I showed here before, and attach again, for your amusement. This shows that for a massive particle to be accelerated to v = c, the amount of energy needed is equal to moc2. Which says to me, that massive particle has to become a massless photon, because its entire mass must be converted to energy.

Hossein Javadi

Dear Ali

Thank you for comment. How can momentum be acceptable without mass?

Even Einstein has not claimed photon is massless, he assumed that the massless photon.

I suggest see:

Hossein Javadi, Are photons really massless?

https://www.researchgate.net/publication/321589753_Are_photons_really_massless

Hossein Javadi

Dear Hanna

Thank you for comment and some mentions about mass and energy.

In field equation of Einstein's General Relativity, the photon has inertial properties. This is the reason why light obeys of space-time curvature.

This question is beyond of it, as quantum mechanics shows, a moving photon has the mass equal to m=E/c^2,

The question is, when an electron absorbs a photon, what happens to the mass of the photon?

Paul Gradenwitz

Dear All,

has a photon mass? Does a photon exist? When I talk about a photon then I talk about a well defined amount of energy. A photon starts its existence through the emission out of matter. The well defined properties of that matter allow only the emission in these units of energy. A photon is detected through absorption by matter. The well defined properties of matter allow only the absorption in well defined units or energy. It seems possible to make emissions of single photons. That means that we can control the number of photons we emit in a field. If we emit N photons into a field and absorb 1 photon out of that field, then for sure we have added N defined amounts of energy to the field and also taken one defined amount of energy from the field. However we have no way to tag each photon that was added to the field so to prove that the energy that was taken from the field in the amount of 1 photon came exclusively from the addition of that same photon in an earlier emission event. Paraphrased we can add buckets of water to a pond and take buckets of water from a pond but the water in one bucket can be a mix of water from many buckets that were added to the pond. As long as this question is not solved it makes little sense to insist that photons in their travel from emitter to absorber remain well defined units that can be attributed with a property of mass.

Regards,

Paul Gradenwitz

Manuele Santoprete

This is to answer Issam Mohanna.

There are several notion of mass, and several way of thinking about them.

Rest or invariant mass can be defined

from the Minkowski norm of the energy-momentum four vector:

p=(p0,p1,p2,p3)=(E/c,px,py,pz),

that is,

||p||^2=-E2/c2+|p|2=-m0^2c^2

where |p| denotes the Euclidean norm and ||p|| the Minkowski norm. Inertial mass is invariant and thus does not depend on the reference frame.

The relativistic mass is instead defined from the time component of the energy-momentum four vector:

mrel=p0/c=E/c2

Inertial mass in special relativity is the relativistic mass:

https://cds.cern.ch/record/489609/files/0103008.pdf

Hossein Javadi

Dear Mohamed

I agree with you that we need a new definition of rest condition of the photon.

In addition, we need to explain the mechanism of absorption and radiation photon by the electron.

Albert Manfredi

It seems to me, an argument that goes something like: "It is not logical for a massless particle to have momentum, and therefore photons must have mass," is not terribly convincing. If certain particles can only exist at v = c, then the notion that they can't have momentum, and therefore they do not, is just weak. "Unphysical idea" is emotion, or logic from what we think we know, but not more than that.

Of course, before QM existed, people might not have been familar with the possible existence of massless particles, and must have found them hard to conceptualize. But since then, we've seen them postulated many times. It's hardly a new idea anymore.

There are plenty of mentions of the total energy equation applying to massless particles too, in the literature. I'm not a physicist, so I can't easily debate all of the finer nuances, but I can read. Many physicists have stated time and again, that formula can work when particles have no mass.

In response to Mohamed's three points:

1. Yes, initially I use indefinite integrals, then at the end I use a definite integral. The assumption going in is that the massive particle is initially at rest. Hence, k = 0.

2. I agree. I mixed classical with relativistic, but I used the Lorentz transformation in doing so. Same as you have done, in redefining momentum in your post, it looks like, in your number 4 response. So, this is what you wrote in your item 4, and I will follow your equation with some simple algebra, and show you that we are saying the very same thing:

You say, the indefinite integral of momentum at relativistic speeds computes to this equation:

Integral wrt v of momentum = -moc2(1 - v2/c2)1/2 + k.

Okay! Continuing,

= -moc2((c2 - v2) / c2)1/2 + k (simply make c2 the denominator in that radical term)

= -moc2(1 / c)(c2 - v2)1/2+ k (just take that c2 denominator out of the radical)

= -moc(c2 - v2)1/2 + k

So, this is the solution of the indefinite integral of momentum, of a mass traveling at relativistic speeds.

Now, look at my penultimate equation, in the paper. It says precisely the same thing! Having computed that indefinite integral of momentum, k = 0 as my initial condition, because the mass is at rest initially, we come to the same result. Then, I compute the definite integral, as the mass is accelerated from 0 to c. And came up with a very interesting result indeed.

3. It is clear that I am not familiar with SRT formalism. Yes, true, I'm not a physicist. But in this case, I don't think it makes a difference?

4. Yes, I simply added a little bit of algebra to your result, and computed the definite integral, with v starting at 0 and ending at c.

Issam Mohanna

Manuele Santoprete

Great thanks to your great and physically practical and applicable answer.

In fact,mass-energy equivalence is valid in both proper or rest frame and the inertial frame which with respect to the object moves , by means of Lorentz Transformation for energy applied between the proper frame where E0= m0c2 (m0 is the rest and non-conserved mass) ,and the (improper) inertial frame ,yielding the mass-energy equivalence E = mc2, where E is the total energy and m is the relativistic conserved mass since relativistic effects due to special relativity are considered unlike Newtonian mechanics where relativistic effects are ignored. As a consequence, the greater the total energy is or the greater the velocity is, the heavier the object is relativistically.

Actually , in particle accelerators, the kinetic energy of incident high-speed particles is transformed into mass by which new particles are created, which is, for example, the case of proton-proton collision resulting in two protons and an additional pion, where the total energy equivalent to relativistic mass is taken into account and not the proper mass or rest energy alone, else, in case relativistic effects are ignored, the calculation leads to wrong results when compared to the experimental results. Thus, when writing the conservation law of (total) energy as an equation and later shifting to the conservation law of mass, the relativistic mass should be considered which is equal to the proper mass in case the particle is at rest.

Please check the attached document from the department of physics at The University of Virginia by Professor Michael Fowler, describing the p-p scattering yielding a pion or an antiproton. http://galileo.phys.virginia.edu/classes/252/particle_creation.pdf

Albert Einstein in his 1916 paper titled "On the Special and General Theory of Relativity", Section "General Results of the Theory", mentioned that the most important result of special relativity is that it has unified both the conservation law of energy and the conservation law of mass into one law (which is actually the law of conservation of mass-energy by means of the relativistic equivalence-of-mass-and-energy relation), unlike classical mechanics where both laws are apparently independent. In fact, from that 1916 paper the relativistic mass can be derived.

Thus, in special relativity there is only one conservation law regarding (relativistic)mass or energy, which when combined to the conservation law of momentum, results in the conservation law of four-momentum.

One of the consequences of E = mc2 or of the relativistic mass is that the former or the latter can replace the second postulate of special relativity yielding a relativity theory that, in every case, must be special relativity, according to relativist Wolfgang Rindler.

Comparing to proper time (which is invariant thus a tensor of rank zero) and the coordinate time (which is not a tensor of rank zero) , the rest mass is invariant thus a tensor of rank zero but not conserved whilst the relativistic mass is conserved but because it is frame dependent it is not a tensor of rank zero. Since both masses are measures of inertia, both are inertial masses.

Albert Manfredi

Mohamed, while you were writing here, I was adding a response to your comments on my short paper. They are in my previous post, so you might have missed them.

Short answer: we agree on our math.

Albert Manfredi

Mohamed, there is something very basic that we are going past one another and not seeing, although I thank you for providing the proper terminology.

Let's start, in your post above, where you say "Conclusion." From there on, you are repeating most of what is in my paper, except for the very last result. (I do not think I made any double replacement, I think you simply misread my explanation).

Take a look at your own expression for the relativistic lagrangian:

L = -mc2(1 - v2/c2)1/2 + k

I copied what you wrote, above, verbatim. (Note, that m is rest mass.) This is the solution of the indefinite integral of momentum with respect to velocity, in your own post, agreed? Now look at my paper. Setting k = 0, because starting from rest, I get this same result. You can simplify that relativistic lagrangian, that you wrote, to this:

L = -moc(c2 - v2)1/2

Just do the algebra, as I showed you already in my previous post. Straightforward.

So, getting to the bottom of my short paper, I computed the definite integral of momentum with respect to velocity, of a "relativistic mass" (if you insist), from v = 0 to v = c. That's what my paper does, and your own post confirms almost everything I wrote. Without actually solving the definite integral. So you did most of it, but you didn't finish. You missed the punch line.

Either we are both wrong, or we are both right.

As to TERMINOLOGY: Lorentz transformation(s). As I said, I'm not a physicist, but I've always seen the term below called Lorentz transformation. In the link attached to this post, they also call the expression

1 / sqr(1 - v2/c2)

if applied to length, mass, time, the "Lorentz" or "Lorentz-Fitzgerald" transformations. Take a look. It's right towards the top. But, whatever.

Herb Spencer

This is all Old Math; for some new ideas see below.

It also contains a historical review of how Einstein's SRT paper was accepted over the next 45 years.

Albert Manfredi

Whew, that's your own paper, Herb. Impressive. Before attempting to go through it, are there actual experimental results that "traditional" QM just ain't explaining right? Like, the answers are coming out all wrong?

Herb Spencer

Dear Albert:

No, there are no NEW extra experiments because my whole approach to physics is that there is far too much "Jumping to Orthodoxy": one single theory is picked as the SOLE explanation for a phenomena; it is then taught as the Truth in our educational institutions and rarely challenged thereafter as most people recognize challenging 'The Truth' is a Career-Limiting move; further promoting the Orthodoxy - just like monster religions.

Thus, I research the History of Science for 'Alternatives' that may have been dropped too quickly and then I develop them further.

Experiments are ambiguous ("thought-Experiments" are worse). It requires acute intuition to see that an alternative is viable.

This takes Courage and Imagination - both extremely rare in modern physics. There are NO easy ways like a New Experiment. Most "proofs" are bogus.

I stated my Research Plan in my first paper [10 years ago - see attached].

Sergey Shevchenko

“…The famous equation E = mc2 applies only to matter at rest….” It is well known that the concept of relativistic mass is controversial. … .”

That isn’t so, and such rather questionable interpretations of the notions “rest mass” and “relativistic mass” appear only in the SR, because of it postulates that its mathematical formalism is totally adequate to the objective reality.

Thus in the SR so much attention is taken to so called “relativistic invariants”, including to “invariant mass”, so that rather evidently important even in the SR notion/phenomenon “relativistic mass” was claimed as quite inessential physical parameter.

However that surely isn’t so.

This, i.e. Minkowski’s, formalism is adequate only provided that the basic SR postulates that Matter’s spacetime is absolute and all/every inertial reference frames are totally and completely equivalent and legitimate. Only at this condition rather strange premise as that real Matter’s spacetime is some [mathematically] imaginary “Minkowski space” becomes be “grounded”; and the premise that the SR mathematics absolutely correctly “determinates” physics becomes be “grounded” as well.

However from the postulates above any number of evidently absurd logical and physical consequences follow, from what follows also that practically totally mathematical approach that is the base of the SR’s application in the mainstream physics, has, in the reality, some limitations, including fundamental. First of all Minkowski space isn’t the real Matter’s spacetime. Matter’s spacetime is the absolute [5]4D Euclidian “empty container”/spacetime, where every material object [a body, particle, etc.] and Matter as a whole exist and constantly change.

Since all changes in Matter proceed with a stable frequency, at that every change is a step in the 4D sub-spacetime and, in parallel, in the 5-th, i.e. “true time” dimension, every object moves in the 4D sub-spacetime and in the 1D true time with absolute 4D speeds of light and 1D speed of light correspondingly;

having always at the 4D motion 4D momentums P=mc [bold connotes 4D Euclidian vector] and energies E=Pc=mc2;

where the variable “m” characterizes the resistance of material object to attempts to change its current state, which was, seems by Newton, called the “inertia” and “m” was called as the “measure of the inertia”.

Inertia is an absolutely fundamental property of everything [of the information, when all/everything what exist is/are some informational patterns/systems of the patterns], and it’s “ideological” realization principally is the same at any 4D speed and of any material object. Every object absolutely fundamentally has some inertia and so has some “inertial mass” [every material object has “the gravitational mass” also, but that is peculiar property of Matter’s objects only], including every material object,

either it is a “having rest mass” body, particle, etc., which always moves in the 4D sub-spacetime having non-zero value of “zero” [the “coordinate time”] component of its 4D speed and 4D momentum,

or “restmassless” photons that move only in 3D space of the spacetime with the speed of light and so have zero values of zero components of their so 3D speeds and momentums,

have some inertial masses “m”.

Having rest mass objects, when they are at absolute 3D rest, move in the coordinate time only, having minimal energy, E0, minimal 1D [only zero component] momentum,. P0, and so minimal inertia for which corresponds minimal mass, which is the “rest mass”, m0. If, after an impact when such object obtains some 3D momentum, Ps, [what is standard situation in mechanics, 4D momentums appear really only in high energy physics, when new particles are produced] this object moves in the 3D space also, and its momentum increases as P=P0+Ps . Since the coordinate time coordinate is orthogonal to any spatial line, P=Ps(1-V2/c2)-½ , i.e. the inertial mass increases in the Lorentz factor, be equal thus to the “transverse mass”.

A spatial impact on photons cannot change their speeds and so result in changing of photons’ energy E=Pc and momentums P, and, correspondingly, of their inertias and inertial masses m=P/c=E/c2.

That is another thing that the changes of inertias, and so of masses, of the objects at large energies differ from “Newton’s” ones, when, for example. inertia and the mass become be dependent on the angle between the object’s motion and acting force directions, however that isn’t principal in this thread’s question.