# Defend the concept of mass

pmb

## Main Question or Discussion Point

Since I defend the concept of mass when it comes up I thought it might be of interest to others to note something very interesting. In Einstein's paper "On the Electrodynamics of Moving Bodies," of 1905 (http://www.fourmilab.ch/etexts/einstein/specrel/www/) he made an error on the value of transverse mass (in 1906 Planck published a paper showing that one can write Force as F = dp/dt where p = mv and m = relativistic mass = m_o/sqrt[1-(v/c)^2]). This has already been noted in the physics literature. Most notably in two papers -

In his book "Albert Einstein's Special Theory of Relativity," Arthur Miller, pp. 328-331, the author explains the transverse mass error. This error was also noted in the now well known paper "Does mass really depend on velocity, dad?" Carl G. Adler, Am. J. Phys., 55(8), Aug 1987 page 742
--------------------------------------------
It should be noted that Einstein's original formula for transverse
mass was incorrect. It was corrected by Planck in 1906. Planck was the
first to introduce the formula m_o*v/sqrt[1-(v/c)^2].
--------------------------------------------

But it seems there is some misinformation being passed around the web on this subject (notably by someone who was banned from here but is still posting misinformation elsewhere). Clarity is in order.

In the 1905 paper Einstein examines the motion of a charged particle which is instantaneously at rest in the primed system in which there is a pure electric field. At t' = 0 the charge is at the origin of the primed system S'. In the unprimed frame S the particle is moving in the +x direction and at t = 0 the charge is at the origin of S. Einstein writes for S'

Einstein uses (X,Y,Z) for (E_x, E_y, E_z) and (L,M,N) for (B_x, B_y, B_z). I use the regular E_x, etc. m = rest mass

First Einstein considers the particle to be at rest in S

m*d^2x/dt^2 = qE_x
m*d^2y/dt^2 = qE_y
m*d^2z/dt^2 = qE_z

That's the first equation. It holds since m*d^x/dt^2 is the force since the particle is instantaneously at rest and therefore gamma = 1. Then he addresss the moving partricle which is at rest in S' and therefore (I use x', y', z' while Einstien uses Greek letters). Remember in S' there is only an electric field

m*d^2x'/dt^2 = qE'_x
m*d^2y'/dt^2 = qE'_y
m*d^2z'/dt^2 = qE'_z

Now Einstein transforms to S

d^2x/dt^2 = (q/m*gamma^3)E_x
d^2y/dt^2 = (q/m*gamma)(E_y - vB_z)
d^2z/dt^2 = (q/m*gamma)(E_z + vB_y)

rewritten

gamma*md^2x/dt^2 = qE_x = x-component of Lorentz Force
gamma*m*d^2y/dt^2 =q(E_y - vB_z) = y- component of Lorentz Force
gamma*m*d^2z/dt^2 = q(E_z + vB_y) = z- component of Lorentz Force

These are the correct equations of motion.

Einstein makes the error of equationg m*gamma^2 d^2/dt^2 = q*gamma(E_y - vB_z) with qE'_y and thus he gets the wrong value for the transverse mass (same with the other components). In effect what he does is fail to transform the force. He writes an expression like

MA = F' and then calls M transverse mass. This of course is incorrect since to call the M 'mass' one has to have all components in the same system. So if F = MA then one can call M a 'mass.'

For details see -
http://www.geocities.com/physics_world/sr/ae_1905_error.htm

Pete

Related Other Physics Topics News on Phys.org
Aside from the interest in the history of special relativity and physics in general (which can be quite interesting and with which I have no qualms), it should be noted that when we are *not* talking history, we should use modern notions.
The notion of mass, e.g., has advanced since the time of Einstein and Feynman. The modern use of the word "mass" with no adjectives included is taken to mean "rest mass" (or "invariant mass" if you want). Two particular arguments for using this notion of mass are:
(1) In gravitation, one might conclude that if the mass of an object increases as you speed up toward c, you could possibly reach a mass resulting in a black hole (an observer traveling with the object might want to disagree with this). This is incorrect.
(2) In quantum field theory, one might conclude that if you could speed a stable elementary particle up fast enough, its mass would be large enough to be unstable to decay into less massive elementary particles (an observer traveling with the particle would disagree with this conclusion). This is incorrect.

Much confusion and some apparent paradoxes can be avoided by using the notion of mass to be rest mass, and nothing else.

One might ask, then, why in special relativity inertia seems to change with speed. The interpretation is that our notions of space and time are modified by SR (not the intrinsic properties of a particle e.g.), and that spacetime geometry in turn tells an object how to move. As a result, we observe an object to have "strange" inertial properties relative to classical considerations. The intrinsic nature of the particle is a separate thing (ie, it has a unique mass value that all reference frames should agree on).

This viewpoint is clearly preferred when you want to extend to general relativity which also holds that there is an intrinsic mass of objects and spacetime geometry tells those objects how to move (in this case, since space*time* can be curved, a particle initially at rest will inevitably accelerate according to an observer who is in a different reference frame than the object).

Really!!!??? Ever since I started reading popular physics book, the expression for relativistic mass is m_*/[squ][1-(v/c)^2].

I found the same expression on

http://
and
http://.

Surely it cannot possibly be wrong??? Or does transverse mass differ from relativistc mass?

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pmb
Originally posted by Javier
Aside from the interest in the history of special relativity and physics in general (which can be quite interesting and with which I have no qualms), it should be noted that when we are *not* talking history, we should use modern notions.
The notion of mass, e.g., has advanced since the time of Einstein and Feynman. The modern use of the word "mass" with no adjectives included is taken to mean "rest mass" (or "invariant mass" if you want).
Your assumption on what mass means when unqualified is just plain wrong. Counter examples abound! There is no "modern" notation since different physicists use different conventions. For example: A cosmologist would mean something different than a particle physicists would if they used the word "mass."

For example: The textbook used in Edmund Bertschinger's class in cosmology (at MIT) is "Cosmological Physics," by John A. Peacock. Part of this text is online
http://assets.cambridge.org/0521422701/sample/0521422701WS.pdf

On page 18 when Peacock uses the term "mass density" he does not mean "rest mass density" but simply "energy per unit volume"/c^2 which is what some people call relativistic mass density. Many new SR/GR texts use mass to mean rel-mass too.

We've already discussed this several times in this forum. Please take this to a thread in which this is discussed. This is not the topic here. For example: If you wish to discuss what mass means then post under the "mass of light" thread.

Two particular arguments for using this notion of mass are:
(1) In gravitation, one might conclude that if the mass of an object increases as you speed up toward c, you could possibly reach a mass resulting in a black hole (an observer traveling with the object might want to disagree with this). This is incorrect.
If someone thought that the gravitational field of an object increases with speed then they'd be thinking about "active gravitational mass" and they'd be right. If someone thinks that the object would then become a black hole then they'd be wrong. That's not what a black hole is. A black hole is an object NOT with a certain mass - but when an objects mass ***is confined to a certain region of space in the objects rest frame." Simply put - if a object is not a black hole it its rest frame then its not a black hole period. Teaching mass as rest mass will incorrectly lead someone to believe that the intensity of a gravitational field is independant of the objservers frame of referance and that is certainly ***not*** true.

The physics of this (moving object) has all been down out in the following paper

"Measuring the active gravitational mass of a moving object," by D. W. Olson and R. C. Guarino, Am. J. Phys. 53, 661 (1985)

The authors only use the term "active gravitational mass" once or twice - after that they simply call it "mass."

(2) In quantum field theory, one might conclude that if you could speed a stable elementary particle up fast enough, its mass would be large enough to be unstable to decay into less massive elementary particles (an observer traveling with the particle would disagree with this conclusion). This is incorrect.
A particle is not unstable due to its mass. A black hole is the most stable object in the universe and some are smaller than an atom. Why would anyone come to such a conclusion???? I know I wouldn't.

One might ask, then, why in special relativity inertia seems to change with speed. The interpretation is that our notions of space and time are modified by SR (not the intrinsic properties of a particle e.g.), and that spacetime geometry in turn tells an object how to move.
That's incorrect. The increase in inertia/mass is a direct result of time dilation and length contraction.

Pmb

pmb
Originally posted by Hyperreality
Really!!!??? Ever since I started reading popular physics book, the expression for relativistic mass is m_*/[squ][1-(v/c)^2].

I found the same expression on

http://
and
http://.

Surely it cannot possibly be wrong??? Or does transverse mass differ from relativistc mass?
No. This is a popular misconception that is being passed around. While it's true that more people tend not to use the concept of relativistic mass its certainly not true that a large majority doesn't use it. There's a reason that transverse mass equals relativistic mass too. It's because if it didn't then the rest mass of a body would not increase with energy as E = mc^2. This follows from the fact that if you were accelerating an object and the particles were moving to the side the the sum of the forces on your hand (or whatever) would equal the rest mass of the object and times the rate of acceleration. Since this is a sum then the particles moving to the side must exert a force in keeping with the relation for relativistic mass .

I'm not explaining this every well right now but I'm in a bit of a hurry. See
http://www.geocities.com/physics_world/gr/weight_move.htm

for more on the weight of a moving body.

See
http://www.geocities.com/physics_world/gr/uniform_move_trans.htm

for more on the gravitational field in a moving frame

Pete

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It seems to me that if mv = &gamma;m0 is given up, then we loose the lovely simplicity of E = mc2, except in the case &gamma; = 1 (v = 0). The whole world has to learn "E = &gamma;mc2".

pmb
Originally posted by quartodeciman
It seems to me that if mv = &gamma;m0 is given up, then we loose the lovely simplicity of E = mc2, except in the case &gamma; = 1 (v = 0). The whole world has to learn "E = &gamma;mc2".
Yes! My sentiments exactly. This recent trend of doing away with the concept of inertial mass is a very bad idea in my opinion. And I truly do not think it's logical.

Pmb

marcus
Gold Member
Dearly Missed
Originally posted by quartodeciman
It seems to me that if mv = &gamma;m0 is given up, then we lose the lovely simplicity of E = mc2, except in the case &gamma; = 1 (v = 0). The whole world has to learn "E = &gamma;mc2".
why be hasty about cultural decisions
take time to get a deep feeling about what is involved
before making decisions about language (and also
realize that our decisions are in some sense meaningless
since language is its own organism evolving somewhat outside
of conscious control)

i am not arguing with you quartodeciman but thinking aloud
about what you just said while listening to the Bach bminor Kyrie

A day or two ago I read an english translation of the
1905 paper ("Ist die Traegheit eines Koerpers.....")
that gave the first version of that equation (so far as I know)
and it was for a body at rest

I think that E=mc2 was never meant to be for anything but a body at rest. (tho pete will disagree of course!)

Anyway the picture I get from that 1905 paper is there is a warm object with a certain Traegheit (inertia, sluggishness) and Einstein watches it sitting their radiating off some of its heat
and after a while he observes that it has less Traegheit than it did earlier

and the inertia it has lost is proportional to the infrared heat energy which it has radiated off into space, and he writes

m = L/c2

the equation that got later rearranged by changing the symbol for energy from L to E and writing it

E = mc2

(a number of authoritative sources make the point that this is only true in the rest frame, e.g. the online textbook that Tom suggested for Special Relativity by a guy at Princeton IAS, where there is a funny story about it)

Now they are singing the Gratias Agimus, do you know the bminor?

marcus
Gold Member
Dearly Missed
Quart,
I want to make sure you are aware of something: inertia as a force/acceleration quantity and as an IDEA is directionless and
therefore not defined for moving objects.

a moving object's resistance to acceleration depends on the direction in which force is applied and so sporadically over the past century one hears people talking about
"longitudinal inertia" and "transverse inertia"

but basically it is a can of worms to apply the concept inertia to moving objects

Pete uses words in strange ways sometimes and he just said there was a "trend to discard inertial mass"

I am in accord with contemporary usage which is very much in
favor of PRESERVING the idea of mass as inertia.
A body's inertia is only well-defined when it is at rest
and therefore the inertial mass is the rest mass.
This (rest) mass is identical with the inertia and for majority speakers is the default
One need not say "rest" mass because that is redundant---mass means rest mass.
One need not say "inertial" mass because that is redundant---mass is the body's inertia at rest.

There is a long tradition of thinking of mass this way----it is not (as pete would suggest) a recent "trend" and it does not constitute (as pete suggests) the abandonment of the "inertial" idea of mass but, on the contrary, a confirmation of that simple and time tested idea.
The word that comes to mind is "Restoration" as in the Meiji Restoration or English R.-----in the culture of modern physics, in this little edy of language, there is an unexpected "restoration" element, perhaps an isolated case. I favor it, dont know about you.

pmb
I think that E=mc2 was never meant to be for anything but a body at rest. (tho pete will disagree of course!)
marcus - I wish you wouldn't try to put words in my mouth. When I say something I say it with what I think (and hope) to be the most precise statement that there is and as such to be completely accurate. And I do not think Einstein was always consistent on this point. From everything I know this is what I believe to be true

For a particle with non-zero rest mass - "mass" meant "rest mass"
For a photon with energy E - "mass" meant E/c^2
For a continuous system - mass meant "energy momentum tensor"

Note that energy has it's complete description as a tensor - thus mass does as well. This is what Einstein seemed to be saying when he said in his 1916 paper on GR
The special theory of relativity has led to the conclusion that inert mass is nothing more or less than energy, which finds its complete mathematical expression in a symmetrical tensor of second rank, the energy-tensor.
Note that the (first) 1905 paper was his first paper which said anything about mass. The next 1905 paper was based on the first paper on mass and refer to the coefficient of kinetic energy for a slowly moving body. That of course is the rest mass of the body. His next work was in 1906 in a paper called "The Principle of Conservation of Energy of Motion of the Center of Gravity and the Inertia of Energy," Annalen der Physik 20 (1906). The first part is the "photon in a box" gerdankin experiment you hear about all the time. The second part one never hears of. In that part Einstein clearly states that if the energy density of a and EM field is u then the mass density is u/c^2. This notion is what many modern relativity/EM texts mean when they speak of the center of mass - E.g. Jackson's EM text, etc.

The next paper seems to read that Einstein is heading toward a deeper understanding of energy and mass and towards the fact the what is really needed is a tensor. This paper was called "On the Inertia of Energy Required by the Relativity Principle," Einstein, Annalen Der Physik (1907). Even then he knew that this was just in the early part of understanding mass for he wrote
The 'general' answer to the question posed is not yet possible becuase we do not yet have a complete world view that would correspond to the principle of relativity.
(I've omitted the question he refered to and the associated details)

Anyway the picture I get from that 1905 paper is there is a warm object with a certain Traegheit (inertia, sluggishness) and Einstein watches it sitting their radiating off some of its heat
and after a while he observes that it has less Traegheit than it did earlier

and the inertia it has lost is proportional to the infrared heat energy which it has radiated off into space, and he writes

m = L/c2
He really didn't write the m. But he did mean that the rest mass decreased.

However that derivatition has been heavily critisized in the physics literature for being very obscure and even not logical. I disagree with the claim that it was illogical. I agree that it was very unclear. I wrote up one that Einstein should have used. It makes more sense to be at least. See
http://arxiv.org/abs/physics/0308039

E = mc2

(a number of authoritative sources make the point that this is only true in the rest frame, )
And a number of authoritative sources make the point that its true in all frames. You have to be careful on that point. MTW mean two different things in "Gravitation" depending on their derivation.

Pmb

pmb
Pete uses words in strange ways sometimes and he just said there was a "trend to discard inertial mass"
What do you mean "strange ways"?

This (rest) mass is identical with the inertia and for majority speakers is the default
You constantly say this but you've also constantly refused to prove it or state what you're basing this impression of yours on.

Many relativists, probably the majority (but I haven't asked them all) mean the ratio of momentum to velocity (magnitudes of course) when they refer to inertia mass. One of the most popular GR texts that is used is that of Schutz "A first course in general relativity" and in it Schutz clearly states that inertial mass is velocity dependant.

There is a long tradition of thinking of mass this way----it is not (as pete would suggest) a recent trend ...
Of course it is. People have always had a difference of opinion on this but they've remained in the minority. At the end of the 90s that trend ceased and took a different direction. The trend then shifted the other way with a paper by Carl Adler and one later by Lev Okun. However both these papers have serious errors in them.

The debate dates back as far as Einstein with a comment he made to Lincoln Barnett and then more recently in 1968 with the paper

"The Advantage of Teaching Relativity with Four-Vectors," Robert W. Brehme. Am. J. Phys. 36 (10), October 1968

However there were papers which were also published which were intended to head ofd this terrible idea of abandoning inertial mass and reverting to rest mass. It also has had the undesireable effect of leading recent students to believe that this is what everyone does - far from the truth. At best particle physicists do one thing and cosmologists/GRists do another (which varying opinions within both groups.

It also has led to a poor understanding of gravity and how mass acts are the source - for details see the section of Peacock I posted.

Pmb

marcus,

MASS

Still crestfallen about it. It is probably because I began learning my relativity some 50 years ago. If we have proper and improper time intervals, then why not proper and improper mass values. It seemed OK back then. And the good old energy-mass equation becomes a one-way affair only. You got mass, then you got a total energy; you got NO mass, then the equation is useless to you. The light quantum gets its energy from E = h&nu;. We also have to swallow momentum with NO mass. Whatever happened to p = mv? No good if v = c. Even p = &gamma;mv is NO use for light quanta. But I guess that comes along with intrinsic angular momentums (spins), with nothing really spinning and neither mass distribution nor action arm, the price of the quantum world.

general relativity: doesn't use mv. Does anyone calculate the gravitational situation of a tight cluster of rapidly-moving supergiant stars?

quantum field theory: that which I understand 0 about. No improper mass there, I'm told. But the equations are supposed to preserve Lorenz invariance. So doesn't that mean 'm' and '&gamma;' often appear together?

OK. I guess that everywhere I see 'm' I can satisfy myself by reading 'm0', and everywhere I see an 'm' that is obviously dependent on 'v', then I can read '&gamma;m0'. yes, I guess I can do that. :(

B-MOLL MESSE

It has been a long time since I heard it. I tried to make a connection with our topic. Your mention was probably just incidental, but I finally derived an explanation. Pronounce it: "be minor mass!", in other words, only assume the smallest value, the rest mass. :)

"Gratias Agimus"- you got past the long Kyrie and into the Gloria. As I recall the counterpoint and rhythm of the Credo used to get my blood rushing, but its been a long time. Probably with my attitude about mass, I belong back with Guillaume de Machaut (Messe de Notre Dame) :)

BACK TO MASS

To prepare a reply to another topic thread, I searched for explanations of synchrotron bending and focusing systems. The only useful one was on the CERN site and they blamed the stiffening of charged particle trajectories on higher kinetic energy. GRRR!

TFYP!

quart

{EDIT}this vB uses the same glyph as 'v' for '& nu ;'. GRRR-squared!

{EDIT}Someone once told me that a most glorious rendition of the Bach B-minor mass could be heard by the Bethlehem Pennsylvania Chorus, because they had been performing it annually for many decades and knew it so well.

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pmb
Still crestfallen about it. It is probably because I began learning my relativity some 50 years ago. If we have proper and improper time intervals, then why not proper and improper mass values.
This is an excellent point. It was for this reason that Sir Arthur Eddington refered to "rest mass" as "proper mass" in "The Mathematical Theory of Relativity" Cambridge University Press, (1924) pg 30.

The light quantum gets its energy from E = h&nu;. We also have to swallow momentum with NO mass. Whatever happened to p = mv? No good if v = c. Even p = &gamma;mv is NO use for light quanta.
This is an inappropriate use of the formula m = m_o/sqrt[1-(v/c)^2] which is what I assume that you're refering to. This relation is derived only for tardyon's (particles with non-zero rest mass which travel at v < c). The definition of "inertial mass" is its the 'm' in p = mv. In this case light has mass m = p/v = p/c. This is how A.P. French defines inertial mass in his famous text "Special Relalativity." In fact all definitions of inertial mass that I'm aware of define mass in this way (it's a bit more complicated but that's what it boils down to).

The reason people say light has no mass is due to the relation

E^2 - (pc)^2 = m_o^2 c^4

For a photon or EM radiation E = pc which gives m_o = 0. However if one defines mass in this way then a circularity arises when the particle is a tardyon. E is defined as the sum of the particles rest energy and it's kinetic energy. If you try to calculate mass using the above then you'd have to first know both the kinetic energy and the rest mass energy. However if you know the rest mass energy then you need not do anything except divide by c. You can't really measure E either. Particle accelerator labs use calorimeters to measure kinetic energy - not E. So defining mass as the m in

E^2 - (pc)^2 = m^2 c^4

is circular.

general relativity: doesn't use mv.
Since when?

OK. I guess that everywhere I see 'm' I can satisfy myself by reading 'm0', and everywhere I see an 'm' that is obviously dependent on 'v', then I can read '&gamma;m0'. yes, I guess I can do that. :(
Use extreme caution. It's not always obvious as to what someone means. For example: Take a look at

http://assets.cambridge.org/0521422701/sample/0521422701WS.pdf

Go to page 18 and read it. When the author uses the term "mass density" Do you think the author means "rest mass density" or do you think he means "relativistic mass density"?

Go here -
http://www.geocities.com/physics_world/sr/mass_energy_equiv.htm

Read up to Eq. (12). What do you think I meant by delta m' in that equation from what I've written above it. This is what you'll see in many many many places. Quite often in the physics literature.

Here are some online examples
http://www.geocities.com/physics_world/relativistic_mass.htm

This page contains some important papers on this subject in the physics literature

http://www.geocities.com/physics_world/mass_articles.htm

Pmb

Pete,

My statements past "It seemed OK back then" are all in emulation of the "rest mass only" discipline and its apparent consequences, undesirable ones IMO. NO rest mass and NO rest for light quanta means NO mass for light quanta at all (in this discipline). But, YES, p = h/&lambda;. GRUMBLE! GRUMBLE!

"NO mv in GR" I found used somewhere in an argument I can't find right now. I suspect that is because general relativists tend to pick their gravitational sources carefully as rather stodgy mass sloths, thereby avoiding genuine horrific messes. E. T. Bell said that Tullio Levi-Civita once tried to do a true two-body problem in GR and gave it up.

I always figure SR for tardyons because I can only derive the transformations for the tardyon case. The tachyon case is just a free extension with no other justification than "what if?". Likewise, it is often forgot: "E0 = m0c2" doesn't arrive from SR postulates; it is the kinetic energy equation "K = (&gamma; - 1)m0c2" that does so arrive. "E0 = m0c2" just "looks good".

quart

pmb
Hi quartodeciman
"NO mv in GR" I found used somewhere in an argument I can't find right now. I suspect that is because general relativists tend to pick their gravitational sources carefully as rather stodgy mass sloths, thereby avoiding genuine horrific messes.
Not all of them. In fact there is a popular general relativity text that does not adhere to this "mass = rest mass" concept. That text is "Basic Relativity," Richard A. Mould, Springer Verlag (1994). One of my favorite texts.

You might enjoy this derivation too. It shows that a moving body weighs more than the same body at rest. Therefore the passive gravitational mass is velocity dependant. See -
http://www.geocities.com/physics_world/gr/weight_move.htm

"E0 = m0c2" doesn't arrive from SR postulates ..
I disagree. I've derived this equation based on the postulates of special relativity. See -
http://www.geocities.com/physics_world/sr/mass_energy_equiv.htm

Pete

Pete,

Your mass_energy_equiv demonstration assumes more than classical SR and mechanical lore.

(1) E = h&nu; and p = h&nu;/c are borrowed from the quantum hypothesis. In fact, the p equation is usually derived from the E equation for photons using E = mc2 backwards, going from E to m. Since E = pc (which is what Einstein used in his demonstration, as I recall) is assured, and allowing for a mass conjoined with a velocity, c, in a momentum p, then E = mc2 is a sure thing for photons. The whole demonstration boils down to charging the energy of the photons to mass losses in the body, while kinetic energy and momentum of the body remain unchanged. In fact, that is what the photon masses mean- two masslets escaping from the body.

(2) There are no potential energy terms in the body. "There must have been an internal change that resulted in a physical state having lower value of energy". One might think that represents two drops in internal potential energy, with no change to the kinetic energy of the body. Indeed, the emission of photons should be charged to transitions that drop the potential energy state of some electrons. But no potential energy or change of potential energy is included in the demonstration, so it automatically charges the mass of the body instead.

Just a few quibbles with the demonstration itself. Since the velocity of the body v' = -v, then I would think the photon nearer to the Y' axis would have the high (blue-shifted) frequency and the photon farther from the Y' axis would have the lower (red-shifted) frequency. So maybe the &nu;- and &nu;+ labels ought to be switched. But it all gets straightened out in the frequency shift equations. The &Delta;m'v equations should have a minus sign between the two h&nu;s, and it does get straightened out in the next equate (that uses the frequency shift terms).

(3) The finale concludes that a small drop in body mass successfully provides the energy of two photons. But the grand generalization that arrives at E0 = m0c2 is tantamount to a blanket assumption that the whole body might evaporate into photons, which is not really true.

regards,
quart

pmb
Hi quart
Your mass_energy_equiv demonstration assumes more than classical SR and mechanical lore.
Sorry. I was thinking of my other derivation. See
http://arxiv.org/abs/physics/0308039

(2) There are no potential energy terms in the body. "There must have been an internal change that resulted in a physical state having lower value of energy".
There is a decrease in the energy of the body. Such a decrease can occur when there is a change in the internal configurartion of the body. It could be that two charged capacitor plates undergo a partial collapse and as a result come closer together. The volume of space inbetween the capacitor plates will decrease as a result. However the electric field will reamain about the same. Since the energy density associated with the field is proportioanl to the square of the magnitude of the field this means that the energy will decrease (A complete description of this from an arbitrary frame requires the unse of the stress-energy tensor). This energy is carried away by radiation which is generated by the collapsing capacitor. Radiation is generated becuase the plates are charged and when the capacitor partially collapses the plates will accelerate and thus the charges on the plate will accelerate.

Regardless of what happens inside the change in rest mass depends only on the amount of the change in energy and not the mechanism. So you can imagine anything you want. Perhaps something inside cooled off and that's where the radiation came from - i.e. thermal radiation etc. If dE is the amount of energy radiated then dE is the energy the body decreases and dm = dE/c^2 is the amount the rest mass of the body decreases.

Just a few quibbles with the demonstration itself. Since the velocity of the body v' = -v, then I would think the photon nearer to the Y' axis would have the high (blue-shifted) frequency and the photon farther from the Y' axis would have the lower (red-shifted) frequency.
They do. Look at Eq. (1)

(3) The finale concludes that a small drop in body mass successfully provides the energy of two photons. But the grand generalization that arrives at E0 = m0c2 is tantamount to a blanket assumption that the whole body might evaporate into photons, which is not really true.
That is incorrect. Just because a body emits radiation at one time by a given amount it doesn't mean that it can emit radiation at all times until the body is gone. E.g. consider a white hot body. It will emit tons of thermal radiation. The amount of radiation will decrease and thus the body will loose rest mass. However that doesn't mean that the body will disappear when the body cools to absolute zero. It just means that it will radiate until it can't radiate anymore.

Pete

pmb
Back to the ssubject. Einstein clearly intends the following expression to hold and be interperated as he does.

Regarding the y-component of the force (i.e. the transverse part of the force) Einstein has this equation

Force in S' = (something)x(Acceleration in S)

He calls "something" the transverse mass. This is why his result is different then it is today. To correctly identify the transverse mass one must write

Force in S = (something)x(Acceleration in S)

Then the "something" can be said to be the transverse mass.

For a moment let's back up - I'd still like to discuss whether Einstein knew what the force on a moving charge was at that stage. I.e. if he knew that F = q[E + vxB] for a movin charge. On page 54 he says

"If a unit of electric point charge is in motion in an electromagentic field, the force acting on it is equal to the electric force which is present at the locality of the charge, and which we ascertain by transformation of the field to the system of coordinates at rest relatively to the electrical charge (New manner of expression)."

To me this seems to indicate that he is not aware of the relation F = q[E + vxB]. Perhaps it wasn't discovered yet. From Einstein's comment on page 54 (above) it seems that he's not even addressing the magnetic force in that paragraph. Since the charge is moving there will be a magnetic force in general. Notice right below it he writes

"The analogy holds with 'magnetomotive forces. etc"

I'm too lazy to type it all in. For those reading on his 1905 relativity paper is online at
http://www.fourmilab.ch/etexts/einstein/specrel/www/

Pete

The Lorentz force is implicit in the first item listed:

""
1. If a unit electric point charge is in motion in an electromagnetic field, there acts upon it, in addition to the electric force, an "electromotive force" which, if we neglect the terms multiplied by the second and higher powers of v/c, is equal to the vector-product of the velocity of the charge and the magnetic force, divided by the velocity of light. (Old manner of expression.)
""

"electric force plus an electromotive force": qE' = q(E + v/c x B) .

In fact, A.E. doesn't really explain his E-B component transformations well. In paragraph 2 of article §6 (II. ELECTRODYNAMICAL PART), his citation of §3 to justify his transformations yielding E'-B' components doesn't really help tell us how electric/magnetic components should be transformed, except maybe for the novelty of occurrences of &beta;. He seems to get the rest implicitly from the Lorentz force equation, without saying so.

I don't know what "terms multiplied by the second and higher powers of v/c" is supposed to mean, unless it is from the mysterious &psi;(v).

Maybe the term "Lorenz force" was invented later by someone else; up to that point it is just "electric plus electromotive force". Maybe!

In summary: the old manner of expression is to call the electric field in a reference frame (with a moving charge) the electric field in the reference frame (with the charge stationary) PLUS an extra electromotive force; the new manner of expression is a transformation of electric/magnetic vectors that incorporates it in the transformation instead. That echoes nicely Einstein's introductory motivational commentary about magnet and conductor symmetry. That may have been what impressed Max Planck so much with A. E.'s paper.

pmb

qE' = q(E + v/c x B) .
That equation is incorrect. The force on a charge is

F = [E + v/c x B]

However it is incorrect to set F/q and call it an electric field. This holds true only when the force purely electric in nature. For example. :Le there be only a magnetic field in frame S. If a charged particle is moving in S then then the force per unit charge is not an electric field since there is no electric field in this frame. However if you're moving relative to this frame then there will be an electric field.

Pete

Incorrect!

Suppose we accept E' = E + v/c x B and its correspondent E = E' + v'/c x B' = E' - v/c x B'. (humor me!)

Then, using Einstein notation for electric and magnetic components:

v = <v, 0, 0>
E = <X, Y, Z>
B = <L, M, N>
v' = <-v, 0, 0>
E' = <X', Y', Z'>
B' = <L', M', N'>
.

Then:

X' = X + 0(N) - 0(M) = X

Y' = Y + 0(L) - (v/c)N = Y - (v/c)N

Z' = Z + (v/c)M - 0(L) = Z + (v/c)M
.

Now use some v' = -v symmetry:

X = X'

Y = Y' - (v'/c)N' = Y' + (v/c)N'

Z = Z' + (v'/c)M' = Z' - (v/c)M'
.

We have six equations; perhaps we can derive magnetic field component relations.

L,L' is hopeless; let's just say L' = L. (Just for the L of it!)

(v/c)M' = Z' - Z = Z + (v/c)M - Z = (v/c)M
M' = M
;

(v/c)N' = Y - Y' = Y - Y + (v/c)N
N' = N
.

That's just too symmetric. Maybe we need some of those &beta;s {&beta;2 = 1/(1 - (v/c)2)}:
Gussy up the equations:

X' = X
Y' = &beta;(Y - (v/c)N)
Z' = &beta;(Z + (v/c)M)
X = X'
Y = &beta;(Y' + (v/c)N')
Z = &beta;(Z' - (v/c)M')
.

OK. That's complicated enough. Now repeat the derivations:

L' = L
(v/c)M' = Z' - Z/&beta; = &beta;Z +&beta;(v/c)M - Z/&beta; = &beta;(1 - 1/&beta;2)Z + &beta;(v/c)M
(v/c)N' = Y/&beta; - Y' = Y/&beta; - &beta;Y + &beta;(v/c)N = &beta;(1/&beta;2 - 1)Y + &beta;(v/c)N
.

Now:

1 - 1/&beta;2 = 1 - (1 - (v/c)2) = +(v/c)2
;

so:

L' = L
(v/c)M' = &beta;(v/c)2Z + &beta;(v/c)M
(v/c)N' = &beta;(-(v/c)2)Y + &beta;(v/c)N

Finally,

L' = L
M' = &beta;(M + (v/c)Z)
N' = &beta;(N - (v/c)Y)
.

I'm not going to solve for M and N, but I suspect those will turn out nice too!

Thank you, &beta;! :)

It is looking like there is a set of four 3-vector equations that determine the electric/magnetic transformations:

E' = &beta;(E + v/c x B)
E = &beta;(E' - v/c x B')
B' = &beta;(B - v/c x E)
B = &beta;(B' + v/c x E')
.

:(
No! Belay that! It yields X' = &beta;X, X = &beta;X', L' = &beta;L and L = &beta;L'. Better wait for Minkowski's 4-by-4-tensor.

---

One more detail!

""
What led me more or less directly to the special theory of relativity was the conviction that the electromotive force [u/c xB] acting in a body in motion in a magnetic field was nothing else but an electric field."

Einstein quoted by Shankland, reproduced in Miller, A. I., AE'sSTOR,Addison-Wesley(1981), p. 163

---

Just a delusion?

pmb
Suppose we accept E' = E + v/c x B
Assume the region under investigation is charge free. Take the divergence. You'll notice that you don't get Div*E = 0 as you must in a charge free region. Also note that if you write E = F/q then you get a relation for a "field" which then depends on what the charge is doing and as such it doesn't tell you what's going on at the point of interest. I.e. I can't simply ask "What is the electric field at point X" since you'd have to ask me "What is the particle doing at X" before you can give a meaningful answer.

"
What led me more or less directly to the special theory of relativity was the conviction that the electromotive force [u/c xB] acting in a body in motion in a magnetic field was nothing else but an electric field."
That's absolutely true. Suppose you're moving with the charge. Then in your frame the charge is at rest and there is still a force. Therefore you can give a meaningful answer.

Haven't you ever wondered what the "v" is in the Lorentz force? Note that there is no absolute V and yet force either exists or in doesn't in inertial frames in relative motion.

Pete

Haven't you ever wondered what the "v" is in the Lorentz force? Note that there is no absolute V and yet force either exists or in doesn't in inertial frames in relative motion.
I would just say that electric (and magnetic) fields are relative, and that sometimes includes having a value of zero.

pmb
Originally posted by quartodeciman
I would just say that electric (and magnetic) fields are relative, and that sometimes includes having a value of zero.
That is 100% true. I hope I didn't post something which would indicate otherwise?

Pete

So where do we stand WRT your challenge:
...whether Einstein knew what the force on a moving charge was at that stage. I.e. if he knew that
F = q[E + vxB] for a movin[g] charge.
...this seems to indicate that he is not aware of the relation F = q[E + vxB].
I replied that Einstein DID know that law, and that he identified it as the electric force in the primed coordinate system. Then you said that F was NO electric force. I rejoined that Einstein thought so and must have divined his transformations from it (where else?)
---
Maybe someone else will climb in and I can go back to just reading this thread.