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Message from discussion Relativity and Aharonov-Bohm
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Pmb  
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 More options Apr 29 2003, 5:32 am
Newsgroups: sci.physics.relativity
From: "Pmb" <physics_wo...@yahoo.coom>
Date: Tue, 29 Apr 2003 09:32:42 GMT
Local: Tues, Apr 29 2003 5:32 am
Subject: Re: Relativity and Aharonov-Bohm
Since you're being an asshole and since you REFUSE TO RESPOND TO THE
CORRECTIONS THAT  POINTED OUT TO YOU REGARDING YOUR SO-CALLED "DERIVATION"
and since you can't even FATHOM the idea that it's impossible for me to post
using my newsreader and NOT that I don't know how (how dumb are you asshole?
I even explained it all to you) and since you're impossible to discuss
physics with given your scumbag personality then this conversation is
terminated

advice for bilgey - don't be such a fucking asshole all the time and try not
to be such an arrogant prick - admit your mistakes when you make them
asshole

Pmb

"Bilge" <dubi...@radioactivex.lebesque-al.net> wrote in message

news:slrnbash7k.51.dubious@radioactivex.lebesque-al.net...
> Pmb:

>  >"Bilge" <dubi...@radioactivex.lebesque-al.net> wrote

>  >>    When you learn to use your newsreader or get a newsreader that
doesn't
>  >> insert 3 or 4 blank lines between every paragraph and turn a 30 line
>  >> post into one which is 120 lines, which I have to reformat.

>  >Then don't respond!! Nobody is forcing you to.

>   And no one is forcing you to whine and ask me a question about
> why I think you should learn to use a newsreader. If you don't want
> an answer, don't ask a question.

>  >      In fact if you're going tto
>  >have this attitude because I'm having a software problem then it's
highly
>  >undesireable - So in the future please just don't respond until you
check
>  >the attitude at the door!

>   Tough luck. If you don't like it, learn to post.

> [...]

>  >Actually this point is moot.

>   Naturally.

>  >The vector potential is fully defined in frame O.

>   Sheeesh. No it isn't. Are you really on the same planet as the
> rest of us. I can add the four-gradient of any scalar function
> whatsoever to the potentials and get the same physical result,
> classically and I already proved that by showing F^{uv} didn't
> change when one does that.

>  >The fact that it satisfies the "grad Phi' =  - (1/c) &A'/&t'" is simply
a
>  >matter of fact. It has nothing to do with the physical meaning of Phi

>   You think not, eh?


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