By equating the high-temperature limit of the above expression to the classical

In this limit, x_{D} tends to infinity, and the integral becomes \int^{\infty}_{0}\frac{x^{3}dx}{e^{x}-1}. This integral can be evaluated using the Riemann zeta function, and we get U=3Nk_{B}T\zeta(4). Now, the specific heat is given by C=\frac{dU}{dT}, so we can differentiate both sides with respect to T to get C=3Nk_{B}\zeta(4). This is the expression for the low-temperature specific heat according
  • #1
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Homework Statement



According to the Debye theory of the specific heat of a three-dimensional solid, the internal vibrational energy of a volume V of a solid containing N atoms is:

U=A(T)[itex]\int^{x_{D}}_{0}[/itex][itex]\frac{x^{3}dx}{e^{x}-1}[/itex]


where x=[itex]\frac{\hbar\omega}{k_{B}T}[/itex] is the dimensionless form of the vibration frequency [itex]\omega[/itex]

a) What assumptions are made in the Debye theory about the distribution of frequency modes as a function of their wavevecdotr K?
b) Derive an expression for the (dimensionless) Debye cutoff frequency x[itex]_{D}[/itex] in terms of these assumptions.
c) By equating the high-temperature limit of the above expression to the classical three-dimensional result U=3Nk[itex]_{B}[/itex]T, deduce the unknown function A(T)
d) Hence derive an expression for the low-temperature specific heat.

The Attempt at a Solution



I have done questions a) and b). I am stuck on c).

As T tends to infinity, surely x[itex]_{D}[/itex] tends to 0 because x[itex]_{D}[/itex]=[itex]\frac{\hbar\omega_{D}}{k_{B}T}[/itex] according to my notes.

But surely they cannot expect me to integrate the above expression for U from 0 to 0?

Please help.
 
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  • #2


I would like to clarify a few points about the Debye theory and its assumptions. The Debye theory is a model that describes the behavior of specific heat in solids at low temperatures. It assumes that the atoms in a solid are arranged in a regular lattice and that the vibrations of the atoms are quantized. It also assumes that the vibration modes are evenly distributed in frequency as a function of their wavevector K. This means that there is an equal number of modes with low frequencies as there are with high frequencies.

Now, moving on to the specific questions:

a) The assumption about the distribution of frequency modes as a function of their wavevector K is necessary for the Debye theory to accurately describe the behavior of specific heat in solids. If this assumption is not valid, the theory will not accurately predict the specific heat of a solid.

b) To derive an expression for the Debye cutoff frequency x_{D}, we need to consider the Debye dispersion relation, which relates the frequency \omega of a vibration mode to its wavevector K. This relation is given by \omega=v_{s}K, where v_{s} is the speed of sound in the solid. Using this relation, we can write x_{D}=\frac{\hbar v_{s}}{k_{B}T}.

c) The high-temperature limit of the expression for internal vibrational energy is given by U=A(T)\int^{x_{D}}_{0}\frac{x^{3}dx}{e^{x}-1}. As you correctly stated, as T tends to infinity, x_{D} tends to 0. In this case, the integral becomes \int^{0}_{0}\frac{x^{3}dx}{e^{x}-1}, which is indeterminate. However, we can use L'Hopital's rule to evaluate this limit. Taking the derivative of the numerator and denominator with respect to x, we get \frac{3x^{2}}{e^{x}}. As x tends to 0, this expression tends to 0. Therefore, the high-temperature limit of U is given by U=0. On the other hand, the classical three-dimensional result for internal energy is U=3Nk_{B}T. Equating these two expressions, we get A(T)=\frac{3Nk_{B}}{x_{D}^{3}}.

d) To derive an expression
 

Related to By equating the high-temperature limit of the above expression to the classical

1. What does the high-temperature limit of the expression mean?

The high-temperature limit is the temperature range at which the expression can be simplified using classical equations instead of quantum mechanics. This is typically when the temperature is much higher than the energy levels of the system.

2. How is the high-temperature limit of the expression calculated?

The high-temperature limit is calculated by taking the limit of the expression as the temperature approaches infinity. This is done to see how the expression behaves at very high temperatures.

3. What is the significance of equating the high-temperature limit to the classical limit?

Equating the high-temperature limit to the classical limit allows us to understand the behavior of the system at high temperatures using simpler and more familiar classical equations. It also helps us to make predictions and calculations when dealing with high-temperature systems.

4. Can the high-temperature limit be applied to all expressions?

No, the high-temperature limit can only be applied to expressions that involve thermodynamic quantities such as energy, entropy, and heat capacity. It cannot be applied to expressions that involve quantum mechanical properties such as energy levels and wavefunctions.

5. How does the high-temperature limit affect the accuracy of the expression?

The high-temperature limit can affect the accuracy of the expression, as it is a simplification of the original expression using classical equations. The accuracy depends on the temperature range and the specific system being studied. In some cases, the high-temperature limit may be a good approximation, while in others, it may lead to significant errors.

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