Defend the concept of mass

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pmb

Originally posted by quartodeciman
So where do we stand WRT your challenge:

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.
I said that F/q is not an electric field. In the primed frame there is no magnetic field by definition of the problem at hand and therefore in the primed frame there can be no magnetic force. The force is due soley to the electric field. Recall that S' is the rest frame of the particle and as such there can be no magnetic force anyway.

You made an error above when you wrote
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'
This relation is not valid. Why would you think you can do this?

Pete
 

jeff

Science Advisor
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Hi pete,

In your "A simple derivation of E = mc2" paper, the matter momentum terms in equation (2)

- m′ivex = - m′fvex + P′a + P′b.

show that you assumed the mass-energy equivalence that einstein proved. In particular, recall the final equation in einsteins argument,

K0 - K1 = (1/2)(L/c2)v2.

Note that he didn't write the kinetic energies explicitly in terms of mass and speed. This allows him to conclude simply that the rhs is a "kinetic energy" from which follows the identification of material mass with energy L/c2 of radiation.
 
Why do I think I can do it?
It's relativistic reciprocity:

interchange coordinate pairs x|ξ y|η z|ζ t|τ;
interchange component pairs X|X' Y|Y' Z|Z' L|L' M|M' N|N';
interchange velocity components +v|-v

to get the inverse transformations.

But Einstein does not exploit this so directly in §6. Instead, he asserts relativistic reciprocity of the empty-space Maxwell-Hertz equations. But first he derives likely candidates for the transformed components, then asserts they are the same as the components of the reciprocated MH equations (give or take a φ(v) factor). This requires inverse Lorentz transformations, which do not appear anywhere in OTEOMB, except once, and that is for the purpose of forcing φ(-v) = φ(v) = 1 and is actually for a third system K' with coordinates x',y'z',t', traveling at -v WRT system k along the common x|ξ line.

So maybe Einstein did NOT directly express the Lorentz force law early in §6, but he still seems to be aware of it at the end of §6, when he talks about electric force plus electromotive force (and describes the vector product that defines this last) as "the Old manner of expression" for his first three transformation equations.
 

pmb

Originally posted by quartodeciman
Why do I think I can do it?
It's relativistic reciprocity:

interchange coordinate pairs x|ξ y|η z|ζ t|τ;
interchange component pairs X|X' Y|Y' Z|Z' L|L' M|M' N|N';
interchange velocity components +v|-v


There's no such thing. You can't simply interchange components like that. It only works for the velocity of the coordinate system. And that's due to the Principle of relativity. Otherwise you'd always have the same field in each frame. For example: Suppose we start by chosing an inertial frame of referance S' in which there is only an electric field and no magnetic field. S is moving uniformly relative to S' as Einstein describes in that section. Then

X' = X
Y' = beta[Y - (v/c)N]
Z' = beta[Z + (v/c)M]

According to you we must also have

X = X'
Y = Y' + (v/c)N'
Z = Z' - (v/c)M'

But since N' = M' = 0 we must have

X = X'
Y = Y'
Z = Z'

Which is clearly a contradiction.

to get the inverse transformations.

Pete
 

X' = X
Y' = beta[Y - (v/c)N]
Z' = beta[Z + (v/c)M]

According to you we must also have

X = X'
Y = Y' + (v/c)N'
Z = Z' - (v/c)M'
"beta" is just shorthand for "1/(1 -(v/c)2)½". This is a factor of the transformations and must join in the reciprocation too. That would become "1/(1 -(-v/c)2)½". Call it "beta' ". Because (-v/c)2 = (v/c)2, then beta' = beta.

The reciprocated transformations (proposed inverse transformations) are:

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

Assume everywhere below that M' = N' = 0.

Solve for X, Y, Z:

X = X'
Y = betaY'
Z = betaZ'
.

OK. Now solve for M and N:

M = (v/c)Z
N = (v/c)Y
.

There might be trouble if M = N = 0, as well as M' = N' = 0. But that would mean that either v/c = 0 or Y = Z = 0.

if v/c = 0, that means v = 0, and these would be true:

X' = X
Y' = beta[Y]
Z' = beta[Z ]

X = X'
Y = beta[Y']
Z = beta[Z']

In this case, beta2 must be equal to 1. But we already know that from beta's definition. v/c = 0 yields beta = 1.
So, yes indeed,:

X = X'
Y = Y'
Z = Z'

in the case v/c = 0. That should be no surprise.

Next, suppose the case that M = N = 0 because Y = Z = 0.

Then the transformations for this case yield:

X' = X
Y' = beta[0 -(v/c)0] = 0
Z' = beta[0 - (v/c)0] = 0
.

So, Y = Y' = Z = Z' = M = M' = N = N' = 0. Every component is vanished, except maybe X, X', L, L'. This must be the case of electric and magnetic fields lined up along ±v, and I accept for this case:

X' = X
L' = L
.

---

I have already conceded that Einstein reciprocated the Maxwell-Hertz equations instead of the transformation equations, and set them equal to likely candidates for the transformed components, with a generality factor φ(v) that was later determined to be 1. But I would guess he first thought it through using the Lorentz force as if it were the electric field in the second system k. It's just a guess.

---
 
Last edited:

pmb

Never mind. I goofed on that transformation. I must be tired. :-)

Pete
 
I've pretty much talked this to death already. Just one more comment or two and I'll quit.

I originally read paragraph 2 of §6 (Transformations of the Maxwell-Hertz equations...) incorrectly.

"If we apply to these equations the transformation developed in §3, by referring the electromagnetic processes developed there introduced, moving with the velocity v, we obtain the equations"

[§6.2]-[§6.7] {I won't write them}

"where β="...... {the usual}
"."

I thought it meant Einstein claimed to get it all from the KINEMATICAL PART of his paper. But that couldn't be true for the electric/magnetic components, so he must have presaged what they would need to be from somewhere, hence my cookery with the Lorentz force equation.

I realized after one or two of our exchanges that I had missed the import of the word "obtain". He did what I call a physics-teacher move: knowing already what the component transformations need to be, he worked out [§6.2]-[§6.7] deliberately to make the terms stand out, all for the purpose of identifying them with components in the Maxwell-Hertz equations for the moving system.

A more interesting criticism of my thoughts would be to ask about a charge moving in the rest system K at a velocity u parallel to v, but not equal in magnitude or orientation. The velocity addition/subtraction theorems would be needed, I reckon.

Regards,
quart
 
An answer to the question whether Einstein knew the F = q[E + v X B] relation ->

Pais {SITL,OUP(1982)} states on p. 124 that Lorentz in his 1895 paper included a set of corresponding-state equations:

x' = x - vt
t' = t - vx/c2 --- {that's a scalar product, I guess}
E' = E + v X H/c --- {'X' is vector product operator}
H' = H - v X E/c
P' = P --- {'P' for electric polarization vector}
.

On p. 125 Pais states that Lorentz included K = e(E + v X H/c) for a moving ion and called it "electrische Kraft".

On p. 133 Pais states that Einstein knew the 1895 paper of Lorentz before 1905.
 

pmb

Originally posted by quartodeciman
An answer to the question whether Einstein knew the F = q[E + v X B] relation ->

Pais {SITL,OUP(1982)} states on p. 124 that Lorentz in his 1895 paper included a set of corresponding-state equations:

x' = x - vt
t' = t - vx/c2 --- {that's a scalar product, I guess}
E' = E + v X H/c --- {'X' is vector product operator}
H' = H - v X E/c
P' = P --- {'P' for electric polarization vector}
.

On p. 125 Pais states that Lorentz included K = e(E + v X H/c) for a moving ion and called it "electrische Kraft".

On p. 133 Pais states that Einstein knew the 1895 paper of Lorentz before 1905.
Thanks - Great info. I know an historian of relativity. I'll be meeting with him in the future. I'll get back with his comments.

Pete
 
In 1899 Lorentz wrote a paper including the true Lorentz transformations, a name that Poincaré invented.

x' = εγ(x - vt)
y' = εy, z' = εz
t' = εγ(t - vx/c2)

{"ε" is a scale factor}
.

{all this is included in ibid., ch. 6}

In a 1904 paper (included in The Principle Of Relativity anthology) Lorentz applied the Lorentz factor (β this time!) to the electric/magnetic component transformations, but showed a queer form of space-time transformations:

x' = βlx, y' = ly, z' = lz
t' = (l/β)t - βlv/c2

{"l" a scale factor}
.

Of course, these transformations were just a mathematical formalism to Lorentz, with no "real" space/time significance. Einstein evidently didn't know these papers when he wrote his first relativity paper. He must have divined the necessity of a multiplier by himself, or from what he knew of Poincaré's reviews of the subject.

It strikes me as odd that the co-producer of the Fitzgerald-Lorentz contraction idea didn't use the factor in his earlier (e.g.1895) papers.
 
Something suddenly sticks out. Einstein DID know the Lorentz 1904 paper after all. The reason is notation.

Up until at least 1900, Lorentz used the letter "γ" to represent the transformation factor, following Voldemar Voigt (1887), the first researcher to consider transformations involving functions of v/c. During the early 1900s, Lorentz was publishing work about electron theory and, specifically,calculating the electron mass. In this research, he was running against the theory of Max Abraham, and the letter "β" was used to represent v/c there. I doubt Lorentz would have reused "β" to also represent the transformation factor in his electrodynamics at that time. It is (apparently) not until the Lorentz 1904 paper that "β" suddenly started getting used (don't know why) for the transformation factor, rather than "γ". Einstein used "β" the same way in his first relativity paper (1905). That is a big coincidence, if Einstein didn't know Lorentz's 1904 paper.

Miller {AESTOR,Addison-Wesley(1981)} states (section 1.15.1) that Einstein "maybe" knew this paper of Lorentz.
 

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