What are relativistic particles? I got it in a question of mechanics.
So, what is it about a particle that makes it "relativistic"?
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1$\begingroup$ Does this answer your question? Relativistic effects $\endgroup$– hftCommented Apr 19 at 4:23
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3 Answers
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Relativistic particles are usually considered to be those whose kinetic energy is close to or greater than their rest mass energy.$^1$ That is, if we call the rest mass $m_0$ then a relativistic particle has kinetic energy $$K\ge m_0 c^2$$ where $c$ is the speed of light.
By no means does this suggest that massless particles are not relativistic. Particles with no mass are necessarily relativistic and they travel at the speed of light (it is a requirement of relativity that massless particles move with speed $c$). And when we look at the full relativistic dispersion relation, $$E^2=p^2c^2+m_0^2c^4\tag1$$ where as explained we require $m_0=0$, the relativistic dispersion relation reduces to $$E=pc$$ and energy is now a linear function of momentum$^2$.
$^1$ See here for more information:
In experiments, massive particles are relativistic when their kinetic energy is comparable to or greater than the energy $E=m_0c^2$ corresponding to their rest mass. In other words, a massive particle is relativistic when its total mass-energy is at least twice its rest mass. This condition implies that the speed of the particle is close to the speed of light. According to the Lorentz factor formula, this requires the particle to move at roughly 85% of the speed of light. Such relativistic particles are generated in particle accelerators, as well as naturally occurring in cosmic radiation. In astrophysics, jets of relativistic plasma are produced by the centers of active galaxies and quasars.
$^2$ This can also roughly apply to massive particles moving at speeds close to $c$ so that the momentum term on the right of equation (1) dominates. i.e., $$E\approx pc$$
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Relativistic particles are any particles that are travelling near (or at) the speed of light (c). Their velocity is so quick that they do not behave classically, so we must treat them relativistically.
Examples include photons, muons, neutrinos, etc.
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If to be very technical, then relativistic particles are those which have speeds , satisfying condition :
$$\tag 1 \Gamma\left( \frac {c^2}{(c−v)^2} \right) \gg 1,$$
where $\Gamma(x)$ is Gamma function, so called factorial in real and complex domain.
For example, when particle speed reaches about $6~\text{AU/h}$ (light speed is about $7.2~\text{AU/h}$), so particle is traveling at about $0.83~c$,- then (1) expression will go to $\approx 10^{38}$, which is a lot greater than $\mathbb 1$.
Declaring conditions in this way is preferred in physics , instead of mentioning exact percentages or ratio limits, because this makes formulas/theory approximation methods simpler.
answered Apr 19 at 7:03
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$\begingroup$ It's standard definition of integral from wiki : $$ \Gamma (z)=\int _{0}^{\infty }t^{z-1}e^{-t}{\text{ d}}t, $$ $\endgroup$ Commented Apr 19 at 7:23
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1$\begingroup$ That is a very weird way of expressing that $v$ becomes comparable to $c$. $\endgroup$ Commented Apr 19 at 7:25
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$\begingroup$ What is "weird" is subjective and depends on ones view. As I've said in Theoretical Physics conditions are usually expressed as limit boundaries, like $\text{when}~x \to \infty$ or $y \gg 1$, etc. Why it's bad to use same methodology for expressing $v$ approaching $c$ ? $\endgroup$ Commented Apr 19 at 7:31
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1$\begingroup$ It is weird in the sense that I have not seen it ever in the physics literature. You could express the same thing simpler by saying "$v/c$ is no longer \approx 0", or that $\gamma(v)>2$, or something like that. $\endgroup$ Commented Apr 19 at 8:34