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Apply the result of Problem 6.18 to obtain a formula for the standard deviation of the energy of a system of N identical harmonic oscillators (such as in an Einstein solid), in the high-temperature limit. Divide by the average energy to obtain a measure of the fractional fluctuation in energy. Evaluate this fraction numerically for N = 1, 104, and 1020. Discuss the results briefly.

Short Answer

Expert verified

Therefore,

σEE¯=1NσEE¯N=1=1σEE¯N=104=0.01σEE¯N=104=1.0×10-10

Step by step solution

01

Given information

A formula for the standard deviation of the energy of a system of N identical harmonic oscillators (such as in an Einstein solid), in the high-temperature limit. Divide by the average energy to obtain a measure of the fractional fluctuation in energy.

02

Explanation

In terms of the heat capacity CV, the standard deviation σEis given by:

σE=TCvk

Consider a system of N harmonic oscillators operating at high temperatures, where CV=Nkis the heat capacity at constant volume.

σE=TNk2σE=kTN

The standard deviation divided by the average energy E¯=NkTequals the fractional fluctuation in energy, so:

σEE¯=kTNNkTσEE¯=1N

The energy fluctuation for N=1

σEE¯N=1=11=1σEE¯N=1=1

The energy fluctuation for N=104 is:

σEE¯N=104=1104=0.01σEE¯N=104=0.01

The energy fluctuation for N=1020 is:

σEE¯N=1020=11020=1.0×10-10σEE¯N=1020=1.0×10-10

Because the energy fluctuation diminishes as the number of subsystems increases, we can utilise the energy E instead of the average energy E¯for large N.

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