Re: JCON's physics degree says qE(1 + v^2/c^2) is made up equation!

On Oct 5, 10:46 am, jcon <cirej...@xxxxxxxxx> wrote:
On Oct 5, 9:09 am, "gu...@xxxxxxxxxxx" <gu...@xxxxxxxxxxx> wrote:

at:http://groups.google.com/group/sci.physics.relativity/msg/b3adcd1867e...

I write:
"> we are solely using the equation F = q(E + v X B) = qE(1 + v^2/
c^2) to determine the distance in BOTH FRAMES."

JCON replies:
" It looks like you got the dimensions right, which is pretty good for
around here, but except for that, it's just a made up equation."

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I hope you don't cause a radiation explosion over there.

Well everyone qE(1 + v^2/c^2) is a made up equation. And please do
not worry they always reply they ment something else afterwards.

No, I meant what I said. It's not impossible that that
equation might be applicable to *some* situation, but the
expression you wrote F = q(E + v X B) = qE(1 + v^2/c^2)
is trivially incorrect as a general statement (consider the
case of zero E field). It is also trivially incorrect
when applied to two electrons.

Of course, it would be very easy to prove me wrong.
You just need to find this in a reference somewhere.
You know, in one of those books you don't own :)

I guess I should be honored that I now have thread with
my name on them, so everyone can see me make fun of you.

-jc

Yes ha ha, let's see who's making fun OF WHO by using not my words but
another person's.
And he will still be denying the words by using a twisted
interpretation of the words no matter how direct they are and they are
very direct.

And if it IS SO FUNNY then why "will" JC need to use all his brain
juice to "defend" himself.

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1st let us quote what JC just wrote above here:

but the
expression you wrote F = q(E + v X B) = qE(1 + v^2/c^2)
is trivially incorrect as a general statement (consider the
case of zero E field).

http://groups.google.com/group/sci.physics.relativity/msg/d9a53c99ed22186b

Not my words but Tom Roberts' words, quote:

"In general, it is not possible to find a frame in which E vanishes,
if E
is nonzero in some inertial frame. In general the same is true about
B
[#]. The difference is: for a single point charge moving inertially,
one
can find an inertial frame in which B=0 everywhere; there is no
corresponding frame for E.

[#] just consider two charges moving differently; one cannot
make both of them be at rest.


Tom Roberts "

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