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Re: A question (fwd)



I suspect that the inquiry was probing the radiation reaction force.

If you include the "self force" of radiation reaction:

F_s = (q^2/6*PI*epsilon*c^3)*(da/dt), where the derivative is the "jerk",

then a lone electron follows the DE:

mx'' =(q^2/6*PI*epsilon*c^3) *x''' , where ' denotes a time derivative.

This has solutions in which the electron flies off to infinity almost
instantaneously.

This is an unsolved problem and points to the limitations and difficulties
inherent in the model of the electron and the consequent calculation of
Fs.

(Conceptually, one might ask if there could even be a radiation reaction
force in a single particle universe.)


Bob Sciamanda
Physics, Edinboro Univ of PA (em)
trebor@velocity.net
http://www.velocity.net/~trebor
----- Original Message -----
From: "brian whatcott" <inet@INTELLISYS.NET>
To: <PHYS-L@lists.nau.edu>
Sent: Wednesday, April 11, 2001 06:24 PM
Subject: Re: A question (fwd)


It is in the nature of tilting at windmills that at times
the windmills do not roll over and die. All the same,
remembering a long Indian tradition of advanced maths
I find it better to provide a supportable answer to a question
as asked, rather than using some other assumptions more plausible
to my modest insights. I thereby hope that some deeper meaning
may become clear over time. What second particular charged geometry
would you assume in answering this question Bob, as a matter
of interest?

Brian

At 10:14 4/11/01 -0400, you wrote:
Brian,
If (as I think you are doing) we read the question as asking about a
universe consisting ONLY of a single charged particle, then I would
answer
that this is a nonsense question. /snip/
Bob Sciamanda (W3NLV)


From: "Prof. S.D. Agashe" <eesdaia@ee.iitb.ac.in>

What is(are) the differential equation(s) governing
the "classical" motion of a single charged body?

S D Agashe


The quixotic answer is that the three orthogonal components
of velocity would be constant, and acceleration would be zero
/snip/

brian whatcott <inet@intellisys.net> Altus OK
Eureka!