Mass-Energy-Equivalence: E = mc2 versus E2 = m2c4 + p2c2

Wolfgang G. Gasser

"uncertainty principle" – 1999-10-22

By Nathan Urban:

The uncertainty relation is between position and momentum, and though photons have a single speed, they can have any momentum.

At least the original photons (postulated by Einstein) have both position and momentum. This is easy to understand. Assume a small source of photons in a dark room and a small detector. If a photon is detected, we know not only its position but also its momentum.

The magnitude of the momentum is

p = v m = c (h f /c2) = h f / c

and its direction is exactly given by the vector from the source to the detector. To deny such simple reasonings is pure sophistry.

In some respect, QM was an attempt not to admit that Einstein was right after experiments (Compton 1923, Bothe and Geiger 1925) had shown that Bohr was wrong and Einstein right (e.g. Bohr, Kramers and Slater). The fathers of QM tried to save as much as possible from their previously advocated but now experimentally refuted positions, taking refuge with obscure mathematics.

By Matthew Konicki:

If E = m c2, then why does light have energy, despite being massless?

By Nathan Urban:

Because E = m c2 is a special case of a more general formula, and that special case only applies in the rest frame of a massive particle. The general formula valid in all frames for all particles is E2 = (m c2)2 + (p c)2.  For the case of a photon, it reduces to E = p c. Photons have energy and momentum, but no mass.

Modern physics has given up or never taken seriously the mass energy equivalence. The distinction between particles with mass and particles without mass leads to a lot of contradictions.

The formula E2 = (m c2)2 + (p c)2 does not make a lot of sense because the momentum p depends on the mass-energy equivalence itself:

p[v] = m[v]  v = m[0] / sqrt[1 - v2/c2]  v

So if rest mass m[0] = m we get [for normal matter]

E2 = (m c2)2 + (p c)2 = (m c2)2 + (m v c)2 / (1 - v2/c2) = ... = (m c2)2 / (1 - v2/c2)

The version E = m c2 / sqrt[1 - v2/c2] is simpler and does not implicitly assume something which is not explicitly stated.

In the case of photons, the momentum relation p = mv is fundamental because it is a necessary condition for consistent momentum conservation in all cases.

"Mass-energy-equivalence" – 1999-10-31

By  Francis Rey:

Very curious abundancy of mass! Three mass in the problem? m, m[v], and  m[0]

In fact it never existed any m[v]. In SR one writes merely: m = m[0] / sqrt[1-v2/c2]

By using m[v] I want to stress that the mass is a function of velocity. m[0] = m[velocity=0]

Also the detailed version

E[v]2 = (m[0] c2)2 + (c  p[v]  )2

= (m[0] c2)2 + (c v m[v])2

= (m[0] c2)2 + (c v m[0] gamma[v])2

shows that this formula is not more fundamental than the original

E[v] = m[0] c2 gamma[v]

By the way, I think that lots of errors are caused in modern theoretical physics because not enough attention is given to the distinction between variables expressing constants and variables expressing functions depending on other variables.

Isn't it obvious that the consistent application of the mass-energy-equivalence can only lead to the conclusion that massless particles are impossible?

"Comments on FAQ - Does light have mass" – 1999-11-04

Comments on FAQ - Does light have mass?

Extracts from corepower.com/~relfaq/light_mass.html:

The photon is a massless particle. According to theory it has energy and momentum but no mass and this is confirmed by experiment to within strict limits.

Experimentally confirmed massless-ness? What is the concrete meaning of 'massless' in this context?

Even before it was known that light is composed of photons it was known that light carries momentum and will exert a pressure on a surface. This is not evidence that it has mass since momentum can exist without mass.

What is the concrete evidence that momentum can exist without mass?

In modern terminology the mass of an object is its invariant mass which is zero for a photon.

What is exactly the invariant mass of 1024 molecules of hydrogen? Does invariant mass vary with temperature and chemical state?

As far as I understand the relevant FAQ texts, invariant mass is a relative concept (not in the kinematic sense). Depending on the physical situation studied, the total mass is considered as composed of invariant mass and forms of energy. But it should be remembered that inertia and gravitational effects do not depend only on the invariant mass but on the sum of invariant mass and energy (i.e. total mass).

If we now return to the question "Does light have mass?" this can be taken to mean different things if the light is moving freely or trapped in a container.

This statement only confirms that invariant mass is a relative concept. If we consider the container and the radiation as a unity, we get only invariant mass and no energy. But we also can distinguish between invariant mass of the box and the energy corresponding to its contained radiation (and temperature).

The definition of the invariant mass of an object is m = sqrt(E2/c4 - p2/c2).

How is this formula applied in the case of a gas or in the case of a box containing very fast rotating objects?

Relativistic mass is equivalent to energy so it is a redundant concept.

It has only been renamed to total mass, which admittedly is a better name.

In the modern view mass is not equivalent to energy. It is just that part of the energy of a body which is not kinetic energy. Mass is independent of velocity whereas energy is not.

In the modern view invariant rest-mass is not equivalent to rest-energy. It is just that part of the rest-energy of a body which is not thermal energy. Invariant rest-mass is independent of temperature whereas rest-energy is not.

The mass is then independent of velocity and is closer to the old Newtonian concept.

Really? Most equations of classical physics (e.g. weight) depend rather on total mass than on invariant mass. E.g. inertia depends on total mass. Not even linear momentum of a rotating object depends on the object's invariant (non-rotational) mass.

The interpretation most widely used is a compromise in which mass is invariant and always has energy so that total energy is conserved but kinetic energy and radiation does not have mass. The distinction is purely a matter of semantic convention.

It's a compromise making sense in several situations, but not more.

A massless particle can have energy E and momentum p because mass is related to these by the equation m2 = E2/c4 - p2/c2 which is zero for a photon because E = pc for massless radiation.

What about photons in media with refraction coefficients n > 1. I suppose that the momentum of a photon is then p = f h / (c n) and that a photon transfers a part of its momentum to a lens when it gets into it. If this is true, the general validity of m2 = E2/c4 - p2/c2 is refuted, isn't it?

The energy and momentum of light also generates curvature of space-time, so according to theory it can attract objects gravitationally.

But why then is the question whether neutrinos have mass or not so important?

The above example of the box containing photons has shown that photons contribute in the same way to gravitational effects as other energy forms such as nuclear or thermal energy. If 'massless' particles generate curvature of space-time anyway, the question whether neutrinos are declared massless or not, becomes totally irrelevant to the dark matter problem.

"Comments on FAQ - Does light have mass" – 1999-11-08

By FAQ:

The photon is a massless particle. According to theory it has energy and momentum but no mass and this is confirmed by experiment to within strict limits.

By Wolfgang (post):

Experimentally confirmed massless-ness? What is the concrete meaning of 'massless' in this context?

By  John Anderson:

In SR:  E2 = (p c) 2 + (m c2) 2

You measure the energy and momentum and if E = p c, then the particle is massless.

No other concrete meaning than the one resulting from postulating the validity of an equation?

Neither energy nor momentum are frame independent. Could we be certain that the equation E = p c holds in all frames if its validity were experimentally confirmed in one frame?

By Wolfgang:

What is the concrete evidence that momentum can exist without mass?

By John Anderson:

Photons, if you use the SR relation above.

That's not concrete evidence but an argument from definition.

By FAQ:

If we now return to the question "Does light have mass?" this can be taken to mean different things if the light is moving freely or trapped in a container.

By Wolfgang:

This statement only confirms that invariant mass is a relative concept. If we consider the container and the radiation as a unity, we get only invariant mass and no energy. But we also can distinguish between invariant mass of the box and the energy corresponding to its contained radiation (and temperature).

By John Anderson:

No, it's invariant in the sense defined above.

Assume the emergence of a pair of photons propagating in opposite directions. Invariant mass is a relative (philosophical sense) concept insofar as it depends on us whether we consider the two photons as a unit having invariant mass, or whether we consider them as two separate photons without invariant mass.

Different observers will measure the same object to have the same mass.

This statement shows that measurements themselves are often rather meaningless whereas their interpretation is crucial. Modern theoretical physics like medieval theology (which like physics today was the 'hardest' science) is rather an interpretational than an experimental science.

By FAQ:

The energy and momentum of light also generates curvature of space-time, so according to theory it can attract objects gravitationally.

By Wolfgang:

But why then is the question whether neutrinos have mass or not so important?

By John Anderson:

Because it would tell you something about neutrinos, not massless particles in general.  There are experiments that measure neutrinos from the sun that would be simpler to explain if neutrinos were massive.  I can't think of any experiment that would be easier to explain if photons were massive.

According to classical physics, gravity and weight are attributes of mass and not of energy. Therefore photons have mass (corresponding to the total energy with respect to e.g. the CMBR frame) at least insofar as they generate curvature of space-time.

In the case of neutrinos however, it is assumed that their total mass (corresponding to total energy) does not generate curvature of space-time. Therefore the question whether they have also invariant mass is so important for current theories.

Concrete statements concerning total and invariant mass are:

·         In the case of PHOTONS, TOTAL MASS is responsible for gravitational effects.

·         In the case of NEUTRINOS, INVARIANT MASS is responsible for gravitational effects.

·         In the case of ordinary matter, the situation is very complicated and nobody really knows ...

"Comments on FAQ - Does light have mass" – 1999-11-16

By Wolfgang (post):

Experimentally confirmed massless-ness? What is the concrete meaning of 'massless' in this context?

By  Jim Carr:

That the measurement is consistent with zero to within experimental uncertainties.

If 'massless' has no concrete and transparent meaning, then it is irrelevant whether we say a neutrino or a photon has mass or not. I'm interested in the concrete physical properties which are associated with 'mass', and not whether it does or does not make sense to call 'mass' a physical quantity arising in a complicated and rather obscure theoretical framework.

At least in pre-Maxwellian physics, mass was considered proportional to inertia and to gravitational effects, and it was subject to a conservation law.

By FAQ:

If we now return to the question "Does light have mass?" this can be taken to mean different things if the light is moving freely or trapped in a container.

By Wolfgang:

This statement only confirms that 'invariant mass' is a relative concept.

By  Jim Carr:

No, it confirms that 'invariant mass' is an invariant, if you pay attention to the details or work out an example.

'Invariant mass' is misleading. 'Rest-mass' would be a better and more honest name because 'invariant' only refers to the motion of the mass center. 'Invariant mass' varies with temperature, rotation, chemical state, electric polarization and so on.

By FAQ:

Relativistic mass is equivalent to energy so it is a redundant concept.

By Wolfgang:

It has only been renamed to 'total mass', which admittedly is a better name.

By  Jim Carr:

Do you see Newton use mass in a way that makes it depend on v? Do chemists?  Engineers?  No.  Hence bad idea.

Do you see Newton use mass in a way that makes it depend on temperature? Do chemists? Engineers?

A big problem of physics since the creation of the concept energy in the nineteenth century is the split between absolute and relative energy forms. The mass-energy-equivalence has further aggravated this problem.

The currently 'official' solution states that only absolute energy forms are equivalent to mass.

I think that the best solution consists in assuming that also 'relative' energy forms such as kinetic energy are absolute insofar as their absolute values depend on the motion relative to the surrounding matter systems (particles) according to a simple quantified version of Mach's principle.

I'm quite sure that such a view will replace the currently prevailing one.

"Was QED intended as a joke" – 1999-11-11

By Wolfgang (post):

What about photons in media with refraction coefficients n > 1. I suppose that the momentum of a photon is then p = f h / (c n) and that a photon transfers a part of its momentum to a lens when it gets into it. If this is true, the general validity of m2 = E2/c4 - p2/c2 is refuted, isn't it?

The explanation of light refraction by Huygens' principle is a first-rate example of an effective (i.e. simple and elegant) physical explanation.

This classical explanation of refraction requires that waves propagate at different velocities depending on the refraction coefficients of the media and that the waves can freely move over distances larger than their wavelength.

In QED, the velocity of photons does not depend on constitutive constants of the medium as in Maxwell's theory but is always c. The propagation delays corresponding to refraction coefficients are explained by regular absorption and reemission of photons.

The refraction coefficient of water at 20 degree Celsius is n = 1.333 for visible light of a wave length of 590 nanometer. So according to common sense physics, the wave length of photons shrinks from around 590 nm to around 590 nm / 1.333 = 440 nm when the photons penetrate water. But the diameter of water molecules is only around 0.3 nm. So how can Huygens' principle work, if photons are continuously absorbed and reemitted by molecules?

The situation is especially strange as the path-integral method of QED is essentially the same as Huygens principle. So in this respect, QED is not even self-consistent!

A further problem is that water molecules are assumed to absorb (and reemit at exactly the same frequency) a continuous spectrum and not only discrete lines corresponding to discrete energy levels.

Yes, I know, the work of Bohr, Heisenberg, Dirac, Feynman, and others has demonstrated that only a fool can be naive enough to believe that ordinary logic, conservation laws and similar common-sense principles are relevant to the strange world of quanta.

Sometimes I wonder whether Feynman, a gifted entertainer with common sense, really believed in what he taught mankind. Maybe virtual particles were at least partially intended as a good joke. In claiming that particles can propagate backwards in time (from the future to the past!), Feynman either ignores our philosophical heritage or he makes fun of it (and of us).

But in proposing his path-integral method as an alternative to Heisenberg's and Schroedinger's formulations of QM, Feynman has shown that QM is (apart from the Planck-Einstein-Bohr quantum concept and from integrated experimental facts) essentially not much more than an unwarranted generalization of the old principle of Huygens.

"Momentum of waves in general" – 1999-11-22

By Paul B. Andersen:

Given two masses joined by a (massless) rigid rod:

____         ____

|    |   L   |    |   M = M1 + M2

| M1 |-------| M2 |

|____|       |____|

The assembly is stationary in an inertial frame.

A "wave complex" is emitted from M1 and is absorbed by M2. If the energy of the complex is E, then the momentum is according to Maxwell p = E/c.

When the complex is emitted, the speed of the M1/M2 assembly will be v = p/M = E / (M * c)

The wave complex will be absorbed after a time t = L/c, at which time the assembly will be stopped.

The assembly has thus moved a distance d = t * v = L * E / (M * c2)

But the center of mass must remain at the same position, and thus a mass m = M * d /L must have moved from M1 to M2.

Thus m = E/c2

Note that this quite literally means that the mass M1 must have decreased by E/c2 and the mass M2 must have increased by the same. We know that the energy E is "taken from" the body with mass M1, (for example by lowering its temperature), and "given to" the body with mass M2 thus increasing its temperature.

So we have shown that a change of mass must indeed accompany the change of energy content of the bodies, and that the relationship between change of energy content and the change of mass is E = mc2.

I have used first order approximations all the way, m << M, d << L, v << c, etc.

And there is of course no such thing as a rigid rod in the real world. (But this is Newton's and Maxwell's world.)

As far as I can understand, this shows that "constancy of mass" is in fact incompatible with Newton's laws of motion and Maxwell's theory.

The above considerations could well have been made before 1905. I don't know if they were, though.

Do you know a similar simple and elegant way to show why p = E/c in Maxwell's theory. What kind of reasoning has led Maxwell to this conclusion?

What are the analogues of electromagnetic momentum in the case of other wave forms? I suppose that in the case of sound an energy transfer of E from an emitter to a receiver is accompanied by a momentum transfer of p = E/c2 v, where v is the speed of sound. If I'm right, then Maxwell's derivation of p = E/c must either be based on something outside general wave theory or p = E/c2 v must be a consequence of general wave theory.

"Momentum of waves in general" – 1999-11-23

By Wolfgang (post):

Do you know a similar simple and elegant way to show why p = E/c in Maxwell's theory.

By Paul B. Andersen:

The momentum in a wave is D x B integrated over the volume of the wave. Thus p = E n /c

Yes, but what I'm interested in, is why "D x B integrated over the volume of the wave" results in mechanical momentum.

By Wolfgang:

What kind of reasoning has led Maxwell to this conclusion?

By Paul B. Andersen:

I am not sure which reasonings Maxwell did himself. But to state it very simplified:

You can find the energy stored in an electric field by considering work done on electric charges, and you can find energy stored in a magnetic field in a similar manner (using electric currents instead of charges.)

In a propagating wave, you can find the energy flow by considering the change in stored energy.

The momentum can be found through the force exerted on a conducting surface when a wave is reflected off it.

The problem is that electric and magnetic fields are perpendicular to the momentum transfer direction. I don't see how these fields could be responsible for the needed forces.

Maybe it was only a lucky guess of Maxwell (or of someone else) based on simple dimensional considerations. I'm sure that Maxwell's theory is completely inconsistent. Because of his totally unjustified claim that all electro­magnetic effects propagate at the velocity of electro­magnetic radiation, it was a necessity for him to postulate that electromagnetic fields carry momentum. But the so defined momentum can never be enough to explain magnetic and electrostatic attraction and repulsion. This problem has never been resolved (in a transparent and consistent way).

By Wolfgang:

What are the analogues of electromagnetic momentum in the case of other wave forms? I suppose that in the case of sound an energy transfer of E from an emitter to a receiver is accompanied by a momentum transfer of p = E/c2 v, where v is the speed of sound. If I'm right, then Maxwell's derivation of p = E/c must either be based on something outside general wave theory or p = E/c2 v must be a consequence of general wave theory.

By Paul B. Andersen:

p = E/c is a consequence of Maxwell's equations, you do not need anything else.

But acoustic waves will indeed have both energy and momentum.

However, I do not think your equation p = E/c2 v is right, my guess is rather p = E/v. But I am only guessing, and I don't feel like looking into acoustic waves right now.

As far as your guess entails that p = E/v is valid in general, where v is the propagation speed of the wave, it cannot be a lucky one. Think about a sea wave power station!

I suppose your reasoning is that the momentum of the wave is as if a mass E/c2 were moving with the speed v.

Yes, it is the same reasoning you have used to show that photons transfer mass. Instead of deriving mass/energy transfer from momentum transfer, we can also do the inverse.

I can however not see why that should be true for acoustic waves, and it isn't even true for EM-waves in a medium different from vacuum.

Why? The by far simplest and most elegant assumption is that the momentum of a photon is p = E/c2 v = E/c2 (c/n) = f h /c /n, where n is the refractive index.

"Momentum of waves in general" – 1999-11-25

By Paul B. Andersen:

I suppose your reasoning is that the momentum of the wave is as if a mass E/c2 were moving with the speed v.

By Wolfgang (post):

Yes, it is the same reasoning you have used to show that photons transfer mass. Instead of deriving mass/energy transfer from momentum transfer, we can also do the inverse.

By Paul B. Andersen:

You can? How?

The reasoning you have used to show that photons transfer inertia corresponds to the reasoning of Einstein's The principle of Conservation of Motion of the Center of Gravity and the Inertia of Energy, published in June 1906. The principle of such reasoning does neither depend on the transfer velocity nor on whether inertia is transferred in the form of electromagnetic radiation or in the form of e.g. acoustic waves.

The more I think about it, the more I become convinced that the formula p = E/c2 v is generally valid for all wave forms, provided that no matter (protons, neutrons, electrons) is (durably) transferred.

Your guess "p = E/v" cannot be true because it leads to the impossible conclusion that momentum is inversely proportional to velocity. It is true that the momentum of an object is inversely proportional to the object's kinetic energy [velocity at constant kinetic energy]. But in this (classical physics) case, momentum results from the transfer of rest-mass, and not from transfer of inertia corresponding to kinetic energy.

I have chosen the example of the sea wave power station exactly for the reason that the velocity of surface waves on the sea is very low (in the order of 1 m/s). If you were right, a huge force parallel to the sea wave propagation would act on the power station during the transformation of wave energy into electric energy.

By Wolfgang:

The by far simplest and most elegant assumption is that the momentum of a photon is p = E/c2 v = E/c2 (c/n) = f h /c /n, where n is the refractive index.

By Paul B. Andersen:

But that assumption is wrong.

p = h / λ,  E = h f, thus  p = E / (λ f) = E/v  = f h n/c

[Comment 2014-11-18: Now I assume that Paul confused two different changes of wavelength λ = c/f. As photons of higher frequency have more energy, p = h/λ is valid in vacuum. In our case here, the wavelength of photons of a given frequency is smaller because the photons have lower speed in a medium with n > 1.]

My assumption would be wrong only if Maxwell were right. But Maxwell's theory cannot be right because it is inconsistent.

Whereas my (relativistic) common sense physics leads to the conclusion that the momentum of photons is proportional to velocity

p = (f h) v/c2       v = c/n

your assumption entails that the lower the velocity of a photon, the higher its momentum

p = (f h) /v           v = c/n

Let us examine the case where photons cross a plane of glass at rest, perpendicular to the photon propagation. Let us assume that the kinetic energy of the glass does not change as a result of the interaction with the photons (i.e. the mass of the glass must be substantially bigger than the mass (corresponding to the energy) of the photons).

Either the energy content of the photons change when getting into the glass, or their energy content does not. Because the photons cannot exchange kinetic energy with the glass and they leave the plane with the same energy they had before getting in, the by far most obvious assumption is that  E = f h  is valid generally and not only in empty space.

If equation p = f h /v were indeed correct and momentum of photons higher within the glass than outside of it, then momentum conservation would entail:

The glass suffers an impulse from incoming photons which is opposite to the photon propagation, and an impulse from outgoing photons which is in the propagation direction.

I hope that nobody will object that Heisenberg's authority invalidates such straightforward reasonings.

Why do physicists react to simple and consistent reasonings showing that Maxwell's theory cannot be correct like medieval theologians to a logical refutation of the Immaculate Conception? Are possibly the theologians of the past the physicists of today?

"Momentum of waves in general" – 1999-11-26

By Wolfgang (post):

Your guess "p = E/v" cannot be true because it leads to the impossible conclusion that momentum is inversely proportional to velocity.

By Tom Roberts:

Only if E is independent of velocity.

We are dealing here with waves. Does the velocity of waves (substantially) depend on energy (per some unit)? Do you think that p = E/v is correct, where E is wave energy transferred from an emitter to a receiver and p the corresponding momentum transfer?

By Wolfgang:

It is true that the momentum of an object is inversely proportional to the object's kinetic energy.

By Tom Roberts:

Where do you get that? In Newtonian mechanics, E = p2 /2m, which implies momentum is proportional to the square-root of energy.

Sorry for the confusion, but the context could have made it clear:

Your guess "p = E/v" cannot be true because it leads to the impossible conclusion that momentum is inversely proportional to velocity. It is true that the momentum of an object [of variable mass] is inversely proportional to the object's velocity at constant kinetic energy. But in this (classical physics) case, momentum results from the transfer of rest-mass, and not from transfer of inertia corresponding to kinetic energy.

Ekin = ˝ m v2  = ˝ p v       p = 2 Ekin /v

By Wolfgang:

Let us assume that the kinetic energy of the glass does not change as a result of the interaction with the photons

By Tom Roberts:

You just assumed away the effect you are looking for. A much better assumption is that the increase in kinetic energy is small, and to keep only the lowest-order terms in its recoil velocity. To keep the problem tractable, you then need to consider a plane wave with a given energy per area, and consider the glass's mass per area, etc.

The typical method of modern physics: trying to obfuscate instead of aiming at a maximum of simplicity. I have stated that the plane of glass is at rest. My reasoning is correct because kinetic energy (relative to the rest frame) of the glass is proportional to the square of its momentum, and the momentum change due to interaction with photons can be made as small as one likes by increasing the thickness of the glass and reducing the number of photons.

In any case, if you had a better understanding of the actually fundamental principles of physics, you would not have written what you have. But you are not the only one, most modern physicists lack such an understanding. And that's not surprising, because training in physics consists primarily in learning how to deal with formulae one cannot understand.

By Tom Roberts:

Your usage of "photons" shows that you are not arguing about Maxwell's theory.

Don't you see that the same reasoning is possible without using photons?  Why should I prefer a clumsy description using energy density and so on to a straightforward one using photons?

DOES electromagnetic radiation transfer more inertia in a medium with a refractive index n > 1 than in empty space, despite having a smaller velocity (c/n instead of c), or DOES it NOT? That's the question!

"Momentum of waves in general" – 1999-11-30

Einstein's E=mc2 derivation of 1905 simplified

Let us assume a perfect flat mirror in empty space at rest in an arbitrary frame of reference (REST FRAME) and a beam of light perpendicular to the mirror (consisting e.g. of a fixed number of photons of the same frequency).

||

||         <<<<          -->

||       light beam     vmirror

mirror

Even if we do not know whether the reflection of the beam has an effect on the momentum of the mirror, we can assume that the mirror is so massive and the beam so weak that a possible frequency shift of the radiation, resulting from a change in momentum of the mirror during reflection, is negligibly small.

Now let us consider the same situation in a frame of reference where the mirror, having mass M, moves at infinitesimally small velocity vmirror = v to the right.  In this MOVING FRAME the mirror has the kinetic energy Emirror = 0.5 M v2.

If the beam has the energy Ebeam (in both directions) in the REST FRAME, then its energy changes in the MOVING FRAME during reflection from Ebeam (c-v) /c to Ebeam (c+v) /c. This increase in energy by dE = Ebeam 2 v /c is not only a result from the relativistic Doppler shift but also a well-confirmed empirical fact.

Unlike in the REST FRAME, there is a non-negligible energy exchange between the mirror and the beam in the MOVING FRAME. The kinetic energy of the mirror must decrease by the beam's energy increase Eafter = Emirror - dE.

The momentum change dp of the mirror is given by

dpmirror = M dv = M ( vafter - v )

where vafter is the mirror's velocity corresponding to its kinetic energy after reflection of the beam. Because of E = M v2 /2 we get

vafter = sqrt [2            /M   (Emirror - dE)]

= sqrt [2 Emirror /M (1 - dE/Emirror)]

= sqrt [        v2         (1 - dE/Emirror)]

= v sqrt[1 - dE/Emirror]

= v - 0.5 dV dE/Emirror

= v - 0.5 dV (Ebeam 2 dV /c) / (0.5 M dV2)

= v - 2 Ebeam /c / M

and

dpmirror = -2 Ebeam / c

This momentum change of the mirror occurs in every frame (whose velocity to the mirror is small). The REST FRAME, however, is the only frame where the corresponding change in kinetic energy can be ignored, because only in this frame, dE is proportional to dp2.

If we further assume that the motion of the inertia center of the mirror-beam-system does not change during reflection, we conclude from the situation in the REST FRAME that the mass Ebeam /c2 must correspond to the beam.

Then the original velocity vcenter of the inertia center with respect to the REST FRAME

vcenter = ( 0 M  –  c Ebeam /c2) / (M + Ebeam /c2)

is the same as its velocity after beam reflection

vcenter = (dv M  +  c Ebeam /c2) / (M + Ebeam /c2)

This derivation of the mass-energy-equivalence has been inspired by Einstein's "Does the Inertia of a Body Depend on its Energy Content?" (published in November 1905) where he writes:

"The mass of a body is a measure of its energy content; if the energy changes by E, then its mass changes in the same sense by E/(9x1020 cm2/s2)".

By Wolfgang:

DOES electromagnetic radiation transfer more inertia in a medium with a refractive index n > 1 than in empty space, despite having a smaller velocity (c/n instead of c), or DOES it NOT?

By Tom Roberts:

Offhand I don't know. E&M in polarizable media is something I haven't used for decades and don't remember very well. And I am uninspired to bother to look it up for you.

You don't recognize the crucial importance of this question. If Paul B. Andersen is right and, according to Maxwell's theory, the momentum of a plane monochromatic wave in a medium is p = E/v and not p = E/c2 v, then Maxwell's theory is even inconsistent in the case of transversal electromagnetic radiation.

(The assumption that the momentum transfer corresponding to attraction and repulsion of particles and macroscopic objects is transmitted by electro­magnetic fields propagating at c has no relationship to reality and logical reasoning at all.)

"Momentum of waves in general" – 1999-12-06

I understand that you must declare correct reasonings and empirical facts wrong, if they contradict orthodoxy. But isn't it exaggerated to declare simple relativistic reasonings wrong, only because they are presented by someone not agreeing with essential points of current mainstream physics?

By Tom Roberts:

Note that the usual approach in physics is to make an exact computation and then investigate its behavior in various limiting cases. Your approach here is fraught with dangers, and in fact you make numerous errors below.

In particular, you use non-relativistic formulas in an attempt to analyze a situation involving light. This gives numerous inconsistencies, but you happened not to notice them.

My derivation is even EXACTLY valid within a relativistic framework, if the velocity vmirror and the beam/mirror inertia-ratio become infinitesimally small.

Based on the same principles, it is possible to derive momentum change from kinetic-energy considerations in classical mechanics, using elastic collision:

Let us assume a perfect mirror in empty space at rest in an arbitrary frame of reference (REST FRAME) and an object with mass m moving towards the mirror.

||                <--

||                 o                  -->

||      Object with velocity u            vmirror << u

mirror

Even if we do not know whether the reflection of the object has an effect on the momentum of the mirror, we can assume that the mirror is so massive and the object so small that a possible change in the object's kinetic energy, resulting from a change in momentum of the mirror during reflection, is negligibly small.

Now let us consider the same situation in a frame of reference where the mirror, having mass M, moves at infinitesimally small velocity vmirror = vm [to the right]. In this MOVING FRAME the mirror has the kinetic energy Em = 0.5 M v2.

If the object has the energy Eobject = 0.5 m u2 (in both directions) in the REST FRAME, then its energy changes in the MOVING FRAME during reflection from 0.5 m (u-vm)2 to 0.5 m (u+vm)2.

This results in an energy increase of the object of

dE = 0.5 m (u + vm) 2               - 0.5 m (u - vm) 2

= 0.5 m (u2 + 2uvm + vm2) - 0.5 m (u2 - 2uvm + vm2)

= 2 m u vm

Unlike in the REST FRAME, there is a non-negligible energy exchange between the mirror and the object in the MOVING FRAME. The kinetic energy of the mirror must decrease by the object's energy increase

Eafter = Em - dE

The momentum change dpm of the mirror is given by

dpm = M  (vafter - v)

where vafter is the mirror's velocity corresponding to its kinetic energy after reflection of the object

vafter = sqrt [2            /M    (Em - dE)  ]

= sqrt [2 Emirror /M  (1 - dE/Em)]

= sqrt [        vm2        (1 - dE/Em)]

= vm  sqrt[1 - dE/Em]

= vm  (1 - 0.5 dE / Em)

= vm - 0.5 V (2mu vm) / (0.5 M vm2)

= vm - 2u m/M

We get

dpm = - 2 u m

This momentum change of the mirror occurs in every frame. The REST FRAME, however, is the only frame where the corresponding change in kinetic energy can be ignored, because only in this frame, dE is proportional to dp2.

By Wolfgang:

dE = Ebeam 2 v /c is not only a result from the relativistic Doppler shift [...]

By Tom Roberts:

Not true. This is the NON-RELATIVISTIC Doppler shift.

There is more involved than a pure Doppler shift. In any case, for the situation we are dealing with, it is possible to derive my formula from Einstein's formula given in On the Electrodynamics of Moving Bodies, "8. Transformation of the Energy of Light Beams.

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