Problem 8 - A Modern Course in Statistical Physics

Question

Blackbody radiation in a box of volume \(V\) and at temperature \(T\) has internal energy \(U=a V T^{4}\) and pressure \(P=1 / 3 a T^{4}\), where \(a\) is the Stefan- Boltzmann constant. (a) What is the fundamental equation for blackbody radiation (the entropy)? (b) Compute the chemical potential.

Step-by-Step Solution

Answer

The fundamental equation for the entropy is \( S = \frac{4}{3} a V T^3 \). The chemical potential \( \text{ \( \text{ ( μ ) } } \) \) is zero.

Step 1 - Understand the Internal Energy and Pressure

Given the internal energy equation \[ U = a V T^4 \] and the pressure equation \[ P = \frac{1}{3} a T^4, \] where \( a \) is the Stefan-Boltzmann constant. Identify that these variables will be used to find entropy and chemical potential.

Step 2 - Use Thermodynamic Relations

Recall that the first law of thermodynamics is \[ dU = TdS - PdV. \]Given this relationship, express the entropy \( S \).

Step 3 - Specify the Entropy Equation

Recognize the total differential and solve for the entropy: \[ dU = T \frac{abla U}{abla T} dT + \frac{abla U}{abla V} dV = T dS - P dV. \] Using the provided relations, we can substitute to find: \[ dU = 4a V T^3 dT + a T^4 dV = T dS - \frac{1}{3} a T^4 dV. \]

Step 4 - Rearrange for the Entropy

Rearranging the equation to solve for the entropy yields: \[ dS = \frac{4a V T^3}{T} dT + a T^3 \frac{2}{3} dV. \]After integrating, the result is: \[ S = \frac{4}{3} a V T^3. \]

Step 5 - Calculate the Chemical Potential

The chemical potential \( \text{ ( μ ) } \) for photons is known to be zero: \[ \text{Since blackbody radiation consists of photons, which have zero rest mass and can't be created nor destroyed, their chemical potential is zero.} \]

Key Concepts

Internal Energy

In the context of blackbody radiation, internal energy is an important concept. For a given volume, the internal energy is represented by the equation: \( U = a V T^{4} \), where

• \( U \) is the internal energy
• \( V \) is the volume
• \( T \) is the temperature in Kelvin
• \( a \) is the Stefan-Boltzmann constant

This equation shows that internal energy depends on the volume and the fourth power of the temperature. This exponential increase signifies how quickly the internal energy of blackbody radiation rises with temperature.

Entropy

Entropy is a measure of disorder or randomness in a system. For blackbody radiation, finding the entropy involves understanding the relationship between internal energy, temperature, and volume. From the first law of thermodynamics, we know: \( dU = TdS - PdV \). Given the pressure \( P \) and internal energy \( U \), you can find entropy by solving: \( dU = 4a V T^3 dT + a T^4 dV = TdS - \frac{1}{3} a T^4 dV \). After simplifying and integrating, we get the entropy equation for blackbody radiation: \( S = \frac{4}{3} a V T^3 \).

• \( S \) represents the entropy
• \( a \) is the Stefan-Boltzmann constant
• \( V \) is the volume
• \( T \) is the temperature

This shows that entropy for blackbody radiation is directly proportional to volume and the cube of the temperature.

Chemical Potential

The chemical potential (\( \mu \)) describes the change in a system's energy when the number of particles changes. For blackbody radiation, which consists of photons, the chemical potential is zero. This is because:

• Photons are not conserved quantities—they are created or destroyed in processes without fixed numbers like in thermal radiation.
• Photons have zero rest mass.

Therefore, the chemical potential \( \mu \) for blackbody radiation can be directly stated as zero.

Stefan-Boltzmann Constant

The Stefan-Boltzmann constant (\( a \)) is a physical constant that plays a critical role in blackbody radiation. It is denoted by the symbol \( \sigma \) and has a value: \ \( 5.670374419 \times 10^{-8} W m^{-2} K^{-4} \). This constant is key in the Stefan-Boltzmann law which states that: \[ P = \sigma T^4\] Where \( P \) represents the power radiated per unit area of a blackbody and \( T \) is the absolute temperature. This demonstrates that the total energy radiated per unit surface area of a blackbody is proportional to the fourth power of its temperature. The same constant \( a \) reappears in the equations for internal energy and entropy, underlying its fundamental importance in thermodynamics and blackbody radiation.