Einstein

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

quartodeciman

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 β. 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 ψ(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

quartodeciman

Incorrect!

Gee! That's too bad! It leads so nicely to Einstein's transformations.

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 βs {β² = 1/(1 - (v/c)²)}:
Gussy up the equations:

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

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

L' = L
(v/c)M' = Z' - Z/β = βZ +β(v/c)M - Z/β = β(1 - 1/β²)Z + β(v/c)M
(v/c)N' = Y/β - Y' = Y/β - βY + β(v/c)N = β(1/β² - 1)Y + β(v/c)N
.

Now:

1 - 1/β² = 1 - (1 - (v/c)²) = +(v/c)²
;

so:

L' = L
(v/c)M' = β(v/c)²Z + β(v/c)M
(v/c)N' = β(-(v/c)²)Y + β(v/c)N

Finally,

L' = L
M' = β(M + (v/c)Z)
N' = β(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, β! :)

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

E' = β(E + v/c x B)
E = β(E' - v/c x B')
B' = β(B - v/c x E)
B = β(B' + v/c x E')
.

:(
No! Belay that! It yields X' = βX, X = βX', L' = βL and L = βL'. 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

quartodeciman

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

quartodeciman

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.

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

Hi pete,

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

- m′_ive_x = - m′_fve_x + 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.

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

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
[FONT=times new roman][SIZE=3]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

quartodeciman

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.

---

pmb

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

Pete

quartodeciman

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

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.

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

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