Cold vs hot on the atomic scale

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SUMMARY

The discussion centers on the concepts of temperature at the atomic scale, specifically addressing the implications of 0 Kelvin and the behavior of photons in a vacuum. It is established that temperature is a collective phenomenon and cannot be attributed to a single atom or photon. Laser cooling techniques are employed to slow down atoms, which leads to a reduction in temperature, but this process does not involve the removal of warmer particles. Additionally, achieving temperatures lower than 1-10 μK requires evaporative cooling following laser cooling to reach states like Bose-Einstein condensation.

PREREQUISITES
  • Understanding of thermodynamics and temperature concepts
  • Familiarity with laser cooling techniques
  • Knowledge of Bose-Einstein condensation principles
  • Basic grasp of quantum mechanics and atomic behavior
NEXT STEPS
  • Research "Laser Cooling Techniques" for detailed methodologies
  • Explore "Bose-Einstein Condensation" and its experimental requirements
  • Study "Maxwell-Boltzmann Distribution" in the context of open quantum systems
  • Investigate the relationship between "Photon Energy Distribution" and temperature
USEFUL FOR

Physicists, researchers in quantum mechanics, and students studying thermodynamics and atomic physics will benefit from this discussion.

Fishpig
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TL;DR
slowest speed vs the fastest speed and temperature
So I've been hunting google for an answer but i cannot find a definitive one.

If 0 Kelvin is the coldest temperature and it is where atoms cease to move does this mean that a photon in a vacuum is the hottest particle because it is in turn moving at the fastest speed possible?

Second question is with cooling down the atoms they use a laser to push away the warmer particles until it's just the cold ones, so is it really "cooling" or are they just slowing them down? yes i do realize that it's both but it should be one or the other.

Thanks smart people.
 
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Fishpig said:
If 0 Kelvin is the coldest temperature and it is where atoms cease to move does this mean that a photon in a vacuum is the hottest particle because it is in turn moving at the fastest speed possible?
Temperature is a collective phenomenon. We cannot talk the temperature of one atom or one photon.

For a gas of photons, it is not their speed that is related to temperature, since they always move at ##c##. What we have is a distribution of the photons' energy (or frequency, or wavelength) that depends on temperature.

Fishpig said:
Second question is with cooling down the atoms they use a laser to push away the warmer particles until it's just the cold ones, so is it really "cooling" or are they just slowing them down? yes i do realize that it's both but it should be one or the other.
In laser cooling the atoms are slowed down. No experiment is perfect, but the cooling results from the slowing down, not from the loss of hot atoms.

However, there is a limit to the temperature that can be achieved by laser cooling (of the order of 1-10 μK), such that in experiments where lower temperatures are needed, for instance to achieve Bose-Einstein condensation, after the laser cooling phase there is an evaporative cooling phase, where cooling is obtained by letting the hottest atoms escape.
 
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That makes sense thank you, the laser is really just 1 part which i did know i just didn't know the order.
This site is brilliant because i have so many questions that i never knew who to ask.
 
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DrClaude said:
Temperature is a collective phenomenon. We cannot talk the temperature of one atom or one photon.
In principle that's right, but it can make well sense to consider the temperature of a single atom when it is coupled to some "heat bath", i.e., as an open quantum system. Then you expect the atomic states being Maxwell-Boltzmann distributed with the temperature given by the temperature of the heat bath.
 
vanhees71 said:
In principle that's right, but it can make well sense to consider the temperature of a single atom when it is coupled to some "heat bath", i.e., as an open quantum system. Then you expect the atomic states being Maxwell-Boltzmann distributed with the temperature given by the temperature of the heat bath.
I agree. I should have said that "We cannot talk the temperature of one atom or one photon in isolation."

I think however that it is important in the context of the OP to say that when an atom ceases to move, it does not mean that it is a zero temperature.
 
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