"John Schoenfeld" wrote in message
om...
"Bill Hobba" wrote in message
...
"N:dlzc D:aol T:com (dlzc)" N: dlzc1 D:cox wrote in
message news:Vbqoc.118739$Jy3.11513@fed1read03...
Dear Bill Hobba:

"Bill Hobba" wrote in message
...

"John Schoenfeld" wrote in message
om...
SCHOENFELD-EINSTEIN RELATIVISTIC FORCE:
...
QED

I thought you said you were leaving this group. Why the about face?

Asperger's syndrome. He can't help it.

You know I used to believe it was a bit moor complex than that. I now
suspect it is not.

It would appear from your babble that there is some sort of error in
the mathematics which I generously posted, but in usual Bill Bubba
style, Bubby fails to come through with the math.

One guy says I am a mere mathematician, another says I can't even do that.

F = dp/dt
= d(m(v) v)/dt
= m(v) (dv/dt) + v (dm/dv)(dv/dt)
= (m(v) + v dm/dv)a

Learn the product rule
http://archives.math.utk.edu/visual....roduct_rule.5/

The answer is F = m(v) a + v dm(v)/dt.

Thanks
Bill

"N:dlzc D:aol T:com \(dlzc\)" N: dlzc1 D:cox wrote in message news:Vbqoc.118739$Jy3.11513@fed1read03...
Dear Bill Hobba:

"Bill Hobba" wrote in message
...

"John Schoenfeld" wrote in message
om...
SCHOENFELD-EINSTEIN RELATIVISTIC FORCE:
...
QED

I thought you said you were leaving this group. Why the about face?

Asperger's syndrome. He can't help it.

You don't actually believe in all that hocus pocus, do you? Half those
disorders are mere inventions to keep the unemployabe (like you)
employed. Go get a job!

David A. Smith

Sam Wormley wrote in message ...
John Schoenfeld wrote:

Sam Wormley wrote in message ...
John Schoenfeld wrote:

SCHOENFELD-EINSTEIN RELATIVISTIC FORCE:

F = (m + y^3 B/c m0 v)a


Oh man, you broke my bull**** meter!

It simplifies to

F = (paY)/(bc)

which can be rewritten as

F = y^2m0a which is similar to the well known result of F = y^3m0a,
but that would requre you to know some physics which you clearly
don't.


Relativistic Force is indeed F_x = m_o gamma^3 a_x
F_y = m_o gamma a_y
F_z = m_o gamma a_z

So what the hell is your F = y^2m0a ?

The "SCHOENFELD-EINSTEIN FORCE".

Crank Information
http://www.google.com/search?q=Schoe...ers.pandora.be

You are giving the impression to unwary posters that my derivation is
wrong, care to point out the error? Didn't think so.

I see you didn't answer my question. If it is "bull****" as you
claimed it is, then at least do your credibility a favour and show
why.

"Bill Hobba" wrote in message
...

"John Schoenfeld" wrote in message
om...
"Bill Hobba" wrote in message
...
"N:dlzc D:aol T:com (dlzc)" N: dlzc1 D:cox wrote in
message news:Vbqoc.118739$Jy3.11513@fed1read03...
Dear Bill Hobba:

"Bill Hobba" wrote in message
...

"John Schoenfeld" wrote in message
om...
SCHOENFELD-EINSTEIN RELATIVISTIC FORCE:
...
QED

I thought you said you were leaving this group. Why the about
face?

Asperger's syndrome. He can't help it.

You know I used to believe it was a bit moor complex than that. I now
suspect it is not.

It would appear from your babble that there is some sort of error in
the mathematics which I generously posted, but in usual Bill Bubba
style, Bubby fails to come through with the math.

One guy says I am a mere mathematician, another says I can't even do that.

F = dp/dt
= d(m(v) v)/dt
= m(v) (dv/dt) + v (dm/dv)(dv/dt)
= (m(v) + v dm/dv)a

Learn the product rule
http://archives.math.utk.edu/visual....roduct_rule.5/

The answer is F = m(v) a + v dm(v)/dt.

Whoops I forgot to add that since v is a vector you can not apply the chain
rule to m(v) because then its derivative would be a vector - you must apply
it to m/sqrt(1-(v/c)2) and we take v2 as v.v.

Thanks
Bill

John Schoenfeld wrote:
"N:dlzc D:aol T:com \(dlzc\)" N: dlzc1 D:cox wrote in message news:Vbqoc.118739$Jy3.11513@fed1read03...

Dear Bill Hobba:

"Bill Hobba" wrote in message
...

"John Schoenfeld" wrote in message
e.com...

SCHOENFELD-EINSTEIN RELATIVISTIC FORCE:

...

QED

I thought you said you were leaving this group. Why the about face?

Asperger's syndrome. He can't help it.


You don't actually believe in all that hocus pocus, do you? Half those
disorders are mere inventions to keep the unemployabe (like you)
employed. Go get a job!

So, you not only think that you know physics better than the physicists,
but also psychology better than the psychologists?

You are really full of yourself.


Bye,
Bjoern

John Schoenfeld wrote:
Sam Wormley wrote in message ...

John Schoenfeld wrote:

Sam Wormley wrote in message ...

John Schoenfeld wrote:

SCHOENFELD-EINSTEIN RELATIVISTIC FORCE:

F = (m + y^3 B/c m0 v)a


Oh man, you broke my bull**** meter!

It simplifies to

F = (paY)/(bc)

which can be rewritten as

F = y^2m0a which is similar to the well known result of F = y^3m0a,
but that would requre you to know some physics which you clearly
don't.


Relativistic Force is indeed F_x = m_o gamma^3 a_x
F_y = m_o gamma a_y
F_z = m_o gamma a_z

So what the hell is your F = y^2m0a ?


The "SCHOENFELD-EINSTEIN FORCE".

Nice. Now present evidence that the formulas of SR above are wrong and
your formula is right, please.

You could start by reading the introductory document on accelerators I
gave you a link to.


Bye,
Bjoern

John Schoenfeld wrote:
Sam Wormley wrote in message ...

John Schoenfeld wrote:

SCHOENFELD-EINSTEIN RELATIVISTIC FORCE:

F = (m + y^3 B/c m0 v)a


Oh man, you broke my bull**** meter!


It simplifies to

F = (paY)/(bc)

I supposed that p = y m0 v, but what are Y and b? Did you perhaps mean
y and B instead? Can't you keep your own notation consistent?


which can be rewritten as

F = y^2m0a which is similar to the well known result of F = y^3m0a,

Nice. Now please present evidence that that well-known result is wrong,
and your formula is right.

[snip]


What is the experimental basis of Special Relativity?
http://math.ucr.edu/home/baez/physic...periments.html


So you are claiming that my proposed formula is empircally falsified
by SR? It's a pitty you can't show why.

I already told you that linear accelerators are built using the
predictions of SR for the mass increase with velocity.



Crank Information
http://www.google.com/search?q=Schoe...ers.pandora.be


You are giving the impression to unwary posters that my derivation is
wrong, care to point out the error? Didn't think so.

Since your end result above is different from the well known result of
SR, there is *obviously* something wrong with your derivation.


Bye,
Bjoern

John Schoenfeld wrote:
"Franz Heymann" wrote in message ...

"John Schoenfeld" wrote in message
.com...

SCHOENFELD-EINSTEIN RELATIVISTIC FORCE:

F = (m + y^3 B/c m0 v)a

This is the relativistic mass-varying force.

PROOF:
let y = gamma = 1/sqrt(1-v^2/c^2)
let B = beta = v/c

Consider relativistic mass as a function of v.
m(v) = y(v) m0

Since
y(v) = 1 / sqrt(1 - v^2/c^2)
= 1 / sqrt(u) { where u(v) = 1 - v^2/c^2 }
= u^-1/2
Then,
dy/dv = -1/2 u^-3/2 (du/dv)
= -1/2 u^-3/2 2v/c^2
= v/[c^2 sqrt(u^3)]
= v/c^2 y^3 { since 1/sqrt(u^3) = y^3 }
= y^3 B/c

Consider the relativistic mass rate of change w.r.t velocity,
m(v) = y(v) m0
dm/dv = d(y(v) m0)/dv
= m0 (dy/dv)
= m0 y^3 B/c

Since relativistic mass is velocity-dependent, the mass-varying

force

law is derived as:
F = dp/dt
= d(m(v) v)/dt
= m(v) (dv/dt) + v (dm/dv)(dv/dt)
= (m(v) + v dm/dv)a

Since the relativistic mass rate of change w.r.t velocity is,
dm/dv = m0 y^3 B/c

Then the relativistic mass-varying force (SCHOENFELD-EINSTEIN FORCE)
is,
F = (m(v) + y^3 B/c m0 v)a

All of that is pure horse dung.


Had you competence you would realize that the SCHOENFELD-EINSTEIN
FORCE reduces to F = (ypa)/(Bc) or F = y^2m0a and is consist with SR
by design (dm=dv = (dym)/dv).

Err, SR gives another relation between force and acceleration than yours
here, if you didn't notice - so obviously your result is *not*
consistent with SR.



All particle accelerators were designed by utilising only one
equation, namely

dp/dt = e(E + v x B)


Is F = y^2m0a inconsistent with this?

F = y^2 m0 a is inconsistent with
F_x = m_o gamma3 a_x, F_y = m_o gamma a_y, F_z = m_o gamma a_z
(for a velocity vector in x-direction)


No particle accelerator has ever been built which has not worked
acording to specifications anywhere in the world.


Is this your proof that my equation is wrong?

Essentially, yes.



Accelerators have
been built in which the particles move a speeds of approximately
{1-10^-11) c.
The design of the ISR was based on this equation. It was possible to
keep two proton beams on track to within a couple of millimetres for a
distance of 6 light months in that machine. So, not to be too polite
about it, to hell with your nonsense.


Is this your proof then?

Essentially, yes.


Bye,
Bjoern

"Bjoern Feuerbacher" wrote in message
...
John Schoenfeld wrote:
"N:dlzc D:aol T:com \(dlzc\)" N: dlzc1 D:cox wrote in
message news:Vbqoc.118739$Jy3.11513@fed1read03...

Dear Bill Hobba:

"Bill Hobba" wrote in message
...

"John Schoenfeld" wrote in message
e.com...

SCHOENFELD-EINSTEIN RELATIVISTIC FORCE:

...

QED

I thought you said you were leaving this group. Why the about face?

Asperger's syndrome. He can't help it.


You don't actually believe in all that hocus pocus, do you? Half those
disorders are mere inventions to keep the unemployabe (like you)
employed. Go get a job!

So, you not only think that you know physics better than the physicists,
but also psychology better than the psychologists?

You are really full of yourself.

Almost by definition crackpots must be, after all how else can they maintain
the positions they do.

Thanks
Bill