How Does High Temperature Affect Boson Number Density in Quantum Statistics?

In summary: So the square root becomes just E, and the integral becomes the integral for the zeta function, giving us the desired result. In summary, the number density for bosons at high temperatures and low chemical energies can be simplified to n= ζ(3)/(π^2)gT^3, using the assumption that T>>m and dropping the m^2 term from the square root. This simplification can be justified by the fact that the distribution has most of its weight on values E>>m, making E^2-m^2 approximately equal to E^2.
  • #1
ChrisVer
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Homework Statement


Show that the number density for bosons if T (temperature) >>μ (chemical energy) is:
[itex] n= \frac{ζ(3)}{\pi^{2}} gT^{3}[/itex]

(T>>m too)

Homework Equations



[itex] n= \frac{g}{2 \pi^{2}} \int_{m}^{∞} f(E) (E^{2}-m^{2})^{1/2} EdE[/itex]
[itex] f(E)= \frac{1}{e^{\frac{E-μ}{T}} -1}[/itex]
[itex]ζ(s)= \frac{1}{Γ(s)} \int_{0}^{∞} dx \frac{x^{s-1}}{e^{x}-1}[/itex]

The Attempt at a Solution



I am taking:
[itex] n= \frac{g}{2 \pi^{2}} \int_{m}^{∞} \frac{1}{e^{\frac{E-μ}{T}} -1} (E^{2}-m^{2})^{1/2} EdE[/itex]

Then use μ<<T to drop the part of it in the exponential

[itex] n= \frac{g}{2 \pi^{2}} \int_{m}^{∞} \frac{1}{e^{\frac{E}{T}} -1} (E^{2}-m^{2})^{1/2} EdE[/itex]

If now I make a change of variables, [itex] x= \frac{E}{T}[/itex] I think I'll get something similar to the definition of the zeta function... By this I have:
[itex]x= \frac{E}{T}[/itex],
[itex]dE= T dx[/itex],
[itex] E^{2}= x^{2}T^{2}[/itex]
and [itex]E= x T[/itex]
The integration limits will be: [itex]x_{1}= \frac{m}{T}=0[/itex] and the upper one will be ∞...

[itex] n= \frac{g}{2 \pi^{2}} \int_{0}^{∞} \frac{1}{e^{x} -1} (x^{2}T^{2}-m^{2})^{1/2} xT^{2}dx[/itex]

Now I can see that dropping [itex]m^{2}[/itex] from the square root would give me the desired answer... However I don't know why can I?
alright m<<T, but T also gets multiplied by the variable...
If I'd drop it:

[itex] n= \frac{g}{2 \pi^{2}} \int_{0}^{∞} \frac{1}{e^{x} -1} xT(1-\frac{m^{2}}{x^{2}T^{2}})^{1/2} xT^{2}dx[/itex]

[itex] n= \frac{g}{2 \pi^{2}} \int_{0}^{∞} \frac{1}{e^{x} -1} x^{2} T^{3}dx[/itex]

[itex] n= \frac{g}{2 \pi^{2}} ζ(3) Γ(3) T^{3}dx[/itex]

[itex] Γ(3)= 2[/itex]
and I get the right result...By the assumption I could drop m^2 from the square root...
 
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  • #2
Do we have a question here?
 
  • #3
-haha that was a good one Dauto, by the fact that on paper I had done it wrong, while here I did it right-
But yes, there's still a question on...
"Now I can see that dropping m2 from the square root would give me the desired answer... However I don't know why can I?alright m<<T, but T also gets multiplied by the variable..."

Could I instead do:
[itex] (x^{2}T^{2}-m^{2})^{1/2}= T (x^{2}- m^{2}/T^{2})^{1/2}= T x[/itex]
?
 
  • #4
Since T>>m, the distribution has most of its weight on values E>>m. Therefore E^2 - m^2 ~ E^2 as far as that integral is concerned.
 
  • #5
But I don't know why...
I would say that the assumption of T>>m allows us to neglect the term m^2 in the square root because it is much smaller compared to the term x^2T^2. This is because m<<T, so m^2 is even smaller. Therefore, dropping m^2 from the square root will not significantly affect the overall result and we can make this simplification. This simplification also makes the integration easier and leads to the desired result.
 

FAQ: How Does High Temperature Affect Boson Number Density in Quantum Statistics?

What is the definition of Boson Number Density?

Boson Number Density is a measure of the number of bosons (particles with integer spin) present in a given volume of space. It is typically denoted by the symbol n and has units of particles per unit volume.

What does T>>m mean in the context of Boson Number Density?

T>>m refers to a scenario where the temperature (T) is much greater than the mass (m) of the boson particles. This condition is important because it affects the behavior and properties of the boson gas.

How is Boson Number Density related to temperature and mass?

In general, as the temperature increases, the Boson Number Density also increases. This is because at higher temperatures, more boson particles are excited and present in the given volume. On the other hand, as the mass of the boson particles increases, the Boson Number Density decreases due to the decrease in the available energy levels for the particles to occupy.

What are the applications of studying Boson Number Density at high temperatures?

Studying Boson Number Density at high temperatures has many applications in fields such as cosmology, particle physics, and condensed matter physics. It can help us understand the behavior of the early universe, the properties of particles at extreme temperatures, and the behavior of superfluids and superconductors.

What factors can affect the Boson Number Density in a system?

The Boson Number Density in a system can be affected by several factors, including temperature, mass of the boson particles, and the volume of the system. Additionally, external factors such as pressure, magnetic fields, and interactions between particles can also play a role in determining the Boson Number Density.

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