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We know that at 0 K, particles of matter almost stop vibrating/moving completely. Also, as we increase temperature of the system, the particles gain energy (in the form of the heat supplied) which gets converted to kinetic energy, and the motion of the particles become faster.

Since it is practically impossible for an object to get to the speed of light, it is just possible for it to have a speed near the speed of light.

My question is,

What is the temperature, at which, particles of matter gain so much energy that they start vibrating/moving with a speed near the speed of light?

Also, what would our observation be, on looking at such a piece of matter? What would it look like?

Mrigank Pawagi
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    How would you define "near the speed of light"? – enumaris Jul 20 '18 at 16:34
  • @enumaris maybe, as something, which is almost the speed of light, but just slightly slower. – Mrigank Pawagi Jul 20 '18 at 16:36
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    Have you searched for "relativistic plasma"? – JEB Jul 20 '18 at 16:40
  • @JEB No! What is it about? Is it related to my question? – Mrigank Pawagi Jul 20 '18 at 16:43
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    That still doesn't answer the question. What do you mean by "almost" and "just slightly lower"? Is $v=0.9c$ "near" the speed of light? or does it need to be $v=0.999999c$ to count? What about $v=0.5c$ or $v=0.1c$? – Emilio Pisanty Jul 20 '18 at 17:17
  • @EmilioPisanty I see your point that we cannot very much define an almost. However, what I meant was as slightly lower than speed of light, as it can be. – Mrigank Pawagi Jul 20 '18 at 17:47
  • You may find the following useful/helpful: https://physics.stackexchange.com/q/216819/59023 or https://physics.stackexchange.com/q/12344/59023 or https://physics.stackexchange.com/q/262382/59023 or https://physics.stackexchange.com/q/148617/59023 . – honeste_vivere Jul 24 '18 at 13:58

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Since you don't provide an actual speed, I can only give an order of magnitude type estimate for you. A particle is said to become "relativistic" if it's kinetic energy becomes comparable (or greater than) its rest mass-energy $mc^2$. Particles in thermal equilibrium have kinetic energies of order $kT$ where $k$ is the Boltzmann constant and $T$ is the temperature. This gives us, for matter of mass $m$ the temperature at which the individual particles would become relativistic at a temperature of $T\approx mc^2/k$. If $m$ is the electron mass (the electrons in the matter will go relativistic first), this corresponds to temperatures of order $10^{10}K$. If $m$ is an atomic mass unit, this corresponds to temperatures of order $10^{13}K$.

For a particle which has kinetic energy equal to the rest mass, the gamma factor would be $\gamma=2$ which corresponds to a velocity of $\sqrt{3}c/2\approx .87c.$

EDIT (per user request): Only the most massive of stars will ever achieve temperatures close to these (and even then, only in their cores). And those stars will run into pair production instabilities due to the spontaneous production of electron-positron pairs in their cores from the high energy gamma radiation around. As an isolated object (i.e. not massive like a star), an object of this temperature wouldn't be bound in any way. One would have to design apparatus to contain such an object. It would look essentially like a very hot plasma. (Since that's exactly what this object would be - a very hot plasma)

enumaris
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  • Thanks for that Answer. But what would be our observation on looking at such an object? – Mrigank Pawagi Jul 20 '18 at 17:48
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    Only the most massive of stars will ever achieve temperatures close to these (and even then, only in their cores). And those stars will run into pair production instabilities due to the spontaneous production of electron-positron pairs in their cores from the high energy gamma radiation around. As an isolated object (i.e. not massive like a star), an object of this temperature wouldn't be bound in any way. One would have to design apparatus to contain such an object. It would look essentially like a very hot plasma. (Since that's exactly what this object would be - a very hot plasma) – enumaris Jul 20 '18 at 17:59
  • Thank You enumaris. You should Add that to your answer so that I can accept your answer! Thanks once again! – Mrigank Pawagi Jul 21 '18 at 05:42
  • simple and elegant reasoning, clear explanation, thanks! – niels nielsen Jul 23 '18 at 08:48
  • @MrigankPawagi done. – enumaris Jul 23 '18 at 20:37