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It all depends on the type of gas and on its temperature.

If the gas is made of single atoms like $^{87}Rb$ or $^{39}K$ (typical atoms that are Bose-Condensed, since you are asking), then it as $3$ degrees of freedom - motion along $x, y,$ and $z$.

If the constituents are diatomic molecules, like $H_2$, then the two atoms could also rotate about the centre of mass, which introduces $2$ new degrees of freedom - clockwise, counterclockwise.

They can also vibrate, as if they were connected by a spring - this introduces $2$ new modes, for in-phase and out-of-phase vibration.

Evidence for the $3$ translational, $2$ rotational and $2$ vibrational degrees of freedom is in the plot of heat capacities (below), where it can also be seen that, when present, the latter two modes only get unlocked (unfrozen) at higher temperatures. This is because the energy associated with that motion is on the order of several $\gg k_B T$:   

enter image description here

So even it you made a BEC of diatomic molecules, you will not see the rotational and vibrational modes, as you are too cold.
This is why a very active field is the one of cold molecules, where you bind two (cold) atoms together with, e.g., a Feshbach resonance, so as for the energy of the mode to be much lower and hence accessible.

It all depends on the type of gas and on its temperature.

If the gas is made of single atoms like $^{87}Rb$ or $^{39}K$ (typical atoms that are Bose-Condensed, since you are asking), then it as $3$ degrees of freedom - motion along $x, y,$ and $z$.

If the constituents are diatomic molecules, like $H_2$, then the two atoms could also rotate about the centre of mass, which introduces $2$ new degrees of freedom - clockwise, counterclockwise.

They can also vibrate, as if they were connected by a spring - this introduces $2$ new modes, for in-phase and out-of-phase vibration.

Evidence for the $3$ translational, $2$ rotational and $2$ vibrational degrees of freedom is in the plot of heat capacities (below), where it can also be seen that, when present, the latter two modes only get unlocked (unfrozen) at higher temperatures. This is because the energy associated with that motion is on the order of several $\gg k_B T$:  enter image description here

So even it you made a BEC of diatomic molecules, you will not see the rotational and vibrational modes, as you are too cold.
This is why a very active field is the one of cold molecules, where you bind two (cold) atoms together with, e.g., a Feshbach resonance, so as for the energy of the mode to be much lower and hence accessible.

It all depends on the type of gas and on its temperature.

If the gas is made of single atoms like $^{87}Rb$ or $^{39}K$ (typical atoms that are Bose-Condensed, since you are asking), then it as $3$ degrees of freedom - motion along $x, y,$ and $z$.

If the constituents are diatomic molecules, like $H_2$, then the two atoms could also rotate about the centre of mass, which introduces $2$ new degrees of freedom - clockwise, counterclockwise.

They can also vibrate, as if they were connected by a spring - this introduces $2$ new modes, for in-phase and out-of-phase vibration.

Evidence for the $3$ translational, $2$ rotational and $2$ vibrational degrees of freedom is in the plot of heat capacities (below), where it can also be seen that, when present, the latter two modes only get unlocked (unfrozen) at higher temperatures. This is because the energy associated with that motion is on the order of several $\gg k_B T$: 

enter image description here

So even it you made a BEC of diatomic molecules, you will not see the rotational and vibrational modes, as you are too cold.
This is why a very active field is the one of cold molecules, where you bind two (cold) atoms together with, e.g., a Feshbach resonance, so as for the energy of the mode to be much lower and hence accessible.

Source Link
SuperCiocia
SuperCiocia
  • 25.3k
  • 20
  • 90
  • 178

It all depends on the type of gas and on its temperature.

If the gas is made of single atoms like $^{87}Rb$ or $^{39}K$ (typical atoms that are Bose-Condensed, since you are asking), then it as $3$ degrees of freedom - motion along $x, y,$ and $z$.

If the constituents are diatomic molecules, like $H_2$, then the two atoms could also rotate about the centre of mass, which introduces $2$ new degrees of freedom - clockwise, counterclockwise.

They can also vibrate, as if they were connected by a spring - this introduces $2$ new modes, for in-phase and out-of-phase vibration.

Evidence for the $3$ translational, $2$ rotational and $2$ vibrational degrees of freedom is in the plot of heat capacities (below), where it can also be seen that, when present, the latter two modes only get unlocked (unfrozen) at higher temperatures. This is because the energy associated with that motion is on the order of several $\gg k_B T$: enter image description here

So even it you made a BEC of diatomic molecules, you will not see the rotational and vibrational modes, as you are too cold.
This is why a very active field is the one of cold molecules, where you bind two (cold) atoms together with, e.g., a Feshbach resonance, so as for the energy of the mode to be much lower and hence accessible.