What is a classical relativistic particle? Is this a classical particle, one used in elementary physics classes and problems, traveling at relativistic speeds?

I think so. It might be a good idea to clarify if it is in the general or special relativistic sense though.

Typically if we're talking about particles, we're neglecting gravity. I am a bit confused, though. In what context was this term used, amadeus?

"Relativistic" might mean "travelling at a reasonable fraction of the speed of light", or it might just mean "we're treating this particular using relativistic mechanics, rather than non-relativistic mechanics". "Classical" just means "we're not using quantum mechanics in looking at the behaviour of this particle".

Sorry for the lateness of my reply: The expression comes from the title of a paper from Physics Review, "Presymmetry of Classical Relativistic Particles". I tried to wade through this paper while in college--I didn't make it very far--and recently, I came across it. I was going to spend a little (or possibly a lot) time trying to understand it. However, it has been a long time since I have studied any thing of this caliber and so this may be a lost cause. (I had to find the notes I had pertaining to this paper to refresh my memory of what "presymmetry" means. I believe I have a vague, if somewhat incomplete and possibly incorrect, notion of its definition.) Also, I believe quantum mechanics is not discussed in the paper and the particle is treated using special relativity. I am unclear as to why this is true. If I remember correctly, in elementary physics classes point particles were the first things studied. Thanks to all for the responses, Amadeusboy

The effects of gravity on a charged particle are about 10^{large number ~40} times greater than gravity. I'll check out your paper when I get to the office tomorrow.

\(\left| \frac{F_{\mathrm{Electromagnetism}}}{F_{\mathrm{Gravity}}} \right| = \frac{\frac{1}{4 \pi \epsilon_0}\frac{q_e^2}{r^2}}{G\frac{m^2}{r^2}} = \frac{q_e^2}{4 \pi \epsilon_0 G m^2} \approx \frac{3.457 \times 10^{-18} kg^2}{m^2}\) For two electrons, this ratio is about: \(4.166 \times 10^{42}\) For two protons, this ratio is about: \(1.236 \times 10^{36}\) For two singly charged U-238 ions, this ratio is about: \(2.213 \times 10^{31}\) http://www.google.com/search?q=what is (electron charge)^2/(4*pi*G*electric constant) ?