Measuring Temperature of Particles in Motion

Edi
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Temperature is basically the movement/ vibrations of particles. They vibrate only because they are in interaction with other particles/ fields - otherwise they would just fly in a straight line.
As they have kinetic energy. Temperature = internal kinetic energy.

How can one measure particles temperature, if it is just lone flying in a straight line? It does not have internal kinetic energy. (unless it's non-fundamental)
It has RELATIVE velocity, by witch one could measure its temperature, but only relative.
Relative - there is a frame, where the particle doesn't move at all. Meaning its temperature is 0K ? (witch is thought not to be possible)
 
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Temperature is not the same as kinetic energy. Temperature is a concept which arises in a statistical averaged description of a system. In a system with lot's of particles, the particles interact and exchange energy and momentum. Any given particles keeps on moving from one (quantum) state to another. The system may have a fixed energy: the distribution of this energy among the particle is constantly changing. By statistical averaging over "state occupied by a particle" x "relative time it occupies this state" you can still say meaningful things about the properties of the system as a whole -- concepts like pressure, temperature, and so on arise from this statistical description.

Saying that "Temperature = internal kinetic energy" is therefore not true, I'm afraid. The reason where this intuitive, but false statement comes from, is that for a gas of particles at some temperate T, the average kinetic energy of a particle is propertional to the temperature T, E\propto T. So for a gas the statement "Temperature = internal kinetic energy" makes more sense, but this brakes down in a lot of other systems.

In particularly, the system you are describing is not a system in thermal equilibrium. It actually doesn't have a temperature. The reason is that you cannot assign a temperature to a system consisting out of one particle. The particle has a particular energy, and this energy "distribution" does not fluctuate, since it is all assigned to one particle.
 
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"They vibrate only because they are in interaction with other particles/ fields"

false,,in fact most of your post is factually incorrect...why not read Wikipedia or another source and then ask some questions with a foundation...even a particle at absolute zero in free space has zero point motion (vibrations that never disappear)...

A single isolated particle, if confined to an ever small volume exhibits increasing frequency(more oscillations) due to its confinement...according to quantum mechanics and Heisnenerg uncertainty...

However, also recognize that "temperature" is a complex issue in physics (like too many issues) and so has layer upon layer of complexity. If you are really interested in temperature, also look at information based temperature...the theory of information has many connections with thermodynamics...maybe even information IS thermodynamics...
 
Toponium is a hadron which is the bound state of a valance top quark and a valance antitop quark. Oversimplified presentations often state that top quarks don't form hadrons, because they decay to bottom quarks extremely rapidly after they are created, leaving no time to form a hadron. And, the vast majority of the time, this is true. But, the lifetime of a top quark is only an average lifetime. Sometimes it decays faster and sometimes it decays slower. In the highly improbable case that...
I'm following this paper by Kitaev on SL(2,R) representations and I'm having a problem in the normalization of the continuous eigenfunctions (eqs. (67)-(70)), which satisfy \langle f_s | f_{s'} \rangle = \int_{0}^{1} \frac{2}{(1-u)^2} f_s(u)^* f_{s'}(u) \, du. \tag{67} The singular contribution of the integral arises at the endpoint u=1 of the integral, and in the limit u \to 1, the function f_s(u) takes on the form f_s(u) \approx a_s (1-u)^{1/2 + i s} + a_s^* (1-u)^{1/2 - i s}. \tag{70}...

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