The Stefan-Boltzmann constant The Planck theory of thermal radiation tells us that in the (angular) frequency interval \(\omega\) to \(\omega+d \omega\), a black body of unit area radiates electromagnetically an amount of thermal energy per second equal to \(I(\omega)\) d \(\omega\), where $$ I(\omega)=\frac{\hbar}{4 \pi^{2} c^{2}} \frac{\omega^{3}}{\left(\mathrm{e}^{\hbar \omega / k_{a} T}-1\right)} $$ Here \(h\) is Planck's constant over \(2 \pi, c\) is the speed of \(h\) ght, and \(k_{B}\) is Boltzmann's constant. a) Show that the total rate at which energy is radiated by a black body per unit area, over all frequencies, is $$ W=\frac{k_{B}^{4} T^{4}}{4 \pi^{2} c^{2} h^{3}} \int_{0}^{\infty} \frac{x^{3}}{\mathrm{e}^{x}-1} \mathrm{~d} x . $$ b) Write a program to evaluate the integral in this expression. Explain what method you used, and how accurate you think your answer is. c) Even before Planck gave his theory of thermal radiation around the turn of the 20 th century, it was known that the total energy \(W\) given off by a black body per unit area per second followed Stefan's law: \(W=\sigma T^{4}\), where \(\sigma\) is the StefanBoltzmann constant. Use your value for the integral above to compute a value for the Stefan-Boltzmann constant (in SI units) to three significant figures. Check your result against the known value, which you can find in books or on-line. You should get good agreement.

Short Answer

Expert verified
Evaluate the integral using numerical methods and use it to calculate Stefan-Boltzmann constant. The computed value should match the known value \(5.67 \times 10^{-8}\).

Step by step solution

01

- Define the given expression

The power radiated per unit area by a black body across all frequencies is given by the integral: \[ W = \frac{k_{B}^{4} T^{4}}{4 \pi^{2} c^{2} h^{3}} \int_{0}^{\infty} \frac{x^{3}}{\mathrm{e}^{x}-1} \mathrm{~d} x \]
02

- Substitution

To simplify the integral, we use the substitution \( x = \frac{\hbar \omega}{k_{B} T} \). This leads to \[ x = \frac{\hbar \omega}{k_{B} T} \] which further implies that \( \omega = \frac{k_{B} T x}{\hbar} \) and \( d\omega = \frac{k_{B} T}{\hbar} dx \).
03

- Transform the integral

Substituting these values into the integral, we get: \[ W = \frac{1}{4 \pi^{2} c^{2} h^{3}} \int_{0}^{\infty} \frac{(\frac{k_{B} T x}{\hbar})^{3} \cdot \frac{k_{B} T}{\hbar} \mathrm{d} x}{\mathrm{e}^{x}-1} \]
04

- Simplification

Simplify the integral: \[ W = \frac{k_{B}^{4} T^{4}}{4 \pi^{2} c^{2} h^{3}} \int_{0}^{\infty} \frac{x^{3}}{\mathrm{e}^{x}-1} \mathrm{~d} x \]
05

- Write a Program to Evaluate the Integral

Write a simple Python program using numerical integration methods like Simpson's rule or the trapezoidal rule to evaluate the integral \( \int_{0}^{\infty} \frac{x^{3}}{\mathrm{e}^{x}-1} \mathrm{~d} x \). Ensure the chosen method provides high accuracy. Here's an example of using the scipy library:```pythonimport scipy.integrate as integrateintegral_value, error = integrate.quad(lambda x: (x**3)/(np.exp(x)-1), 0, np.inf)print('Integral Value:', integral_value)```
06

- Compute the Stefan-Boltzmann Constant

Using the calculated integral value, we can compute the Stefan-Boltzmann constant \( \sigma \) as follows: \[ \sigma = \frac{k_{B}^{4}}{4 \pi^{2} c^{2} h^{3}} \int_{0}^{\infty} \frac{x^{3}}{\mathrm{e}^{x}-1} \mathrm{~d} x \] Substitute the known values and the computed integral value to obtain: \[ \sigma \approx 5.67 \times 10^{-8} \text{ W/m}^2 \text{K}^4 \]
07

- Compare with Known Value

Finally, compare the computed value of \( \sigma \) with the known value \( 5.67 \times 10^{-8} \text{ W/m}^2 \text{K}^4 \) to validate our result.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Stefan-Boltzmann Constant
The Stefan-Boltzmann constant is denoted by \( \sigma \). It describes the power radiated per unit area of a black body as a function of its temperature. This concept is crucial in understanding thermal radiation.
According to Stefan's law, the total energy \( W \) emitted per unit area of a black body is proportional to the fourth power of its absolute temperature. The law is expressed as: \[ W = \sigma T^4 \] Here, we calculated the value of \( \sigma \) using Planck's law by numerically evaluating an integral. Given known constants and integration, we confirmed that \( \sigma \) is approximately \( 5.67 \times 10^{-8} \text{ W/m}^2 \text{K}^4 \). This constant helps in predicting thermal radiation accurately.
Numerical Integration
Numerical integration is a method used to approximate the value of an integral when an analytical solution is difficult or impossible to find. In this exercise, we evaluated the integral \( \int_{0}^{\infty} \frac{x^{3}}{\mathrm{e}^{x}-1} \mathrm{~d} x \) using numerical integration techniques.
Some common methods of numerical integration include:
  • Simpson's Rule
  • Trapezoidal Rule
  • Midpoint Rule
We used Python's SciPy library for our calculations, a widely used tool for scientific computing. The function `quad` from SciPy allows us to perform numerical integration efficiently and accurately. The result of our integral provided a value that helped us compute the Stefan-Boltzmann constant precisely. Here's a brief snippet of the code used:
```python
import scipy.integrate as integrate
import numpy as np
integral_value, error = integrate.quad(lambda x: (x**3)/(np.exp(x)-1), 0, np.inf)
print('Integral Value:', integral_value)
```
This integral plays a vital role in deriving physical constants and understanding energy radiation.
Planck's Law
Planck's Law explains the electromagnetic radiation emitted by a black body in thermal equilibrium at a given temperature. It was a groundbreaking discovery by Max Planck that resolved issues with previous models of radiation.
The law states that the spectral radiance of a black body is a function of frequency \( \omega \) and temperature \( T \). The formula is given by:
\[ I(\omega) = \frac{\hbar}{4 \pi^{2} c^{2}} \frac{\omega^{3}}{\left(\mathrm{e}^{\hbar \omega / k_{B} T} - 1\right)} \]
Here,
  • \( \hbar \) is Planck's constant divided by \( 2 \pi \)
  • \( c \) is the speed of light
  • \( k_{B} \) is Boltzmann's constant
Planck's law is fundamental in quantum mechanics and provides the foundation for understanding black body radiation, including the derivation of Wien's Law and the Stefan-Boltzmann Law. It highlights the quantized nature of energy emission and absorption, marking a departure from classical physics predictions.
Thermal Radiation
Thermal radiation refers to the emission of electromagnetic waves from all matter that has a temperature greater than absolute zero. This phenomenon occurs due to the thermal motion of particles within matter.
Some key points about thermal radiation:
  • It is one of the modes of heat transfer
  • Emission depends on the temperature and nature of the material
  • A black body is an ideal emitter that completely absorbs and emits radiation uniformly
Key principles in thermal radiation include:
  • Stefan-Boltzmann Law: Total energy emitted is proportional to the fourth power of temperature
  • Planck's Law: Describes spectral radiance of radiation as dependent on temperature and frequency
  • Wien's Displacement Law: Peak wavelength of emission shifts inversely with temperature
Understanding thermal radiation is essential in fields like astronomy, climate science, and engineering, where heat and energy transfer play significant roles. It helps us predict and manipulate thermal properties in various technological and natural processes.

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Most popular questions from this chapter

Consider the integral $$ E(x)=\int_{0}^{x} \mathrm{e}^{-t^{2}} \mathrm{~d} t . $$ a) Write a program to calculate \(E(x)\) for values of \(x\) from 0 to 3 in steps of \(0.1 .\) Choose for yourself what method you will use for performing the integral and a suitable number of slices. b) When you are convinced your program is working, extend it further to make a graph of \(E(x)\) as a function of \(x\). If you want to remind yourself of how to make a graph, you should consult Section 3.1, starting on page 88 . Note that there is no known way to perform this particular integral analytically, so numerical approaches are the only way forward.

Rearranging Eq. (5.19) into a slightly more conventional form, we have: $$ \int_{a}^{b} f(x) \mathrm{d} x=h\left[\frac{1}{2} f(a)+\frac{1}{2} f(b)+\sum_{k=1}^{N-1} f(a+k h)\right]+\frac{1}{12} h^{2}\left[f^{\prime}(a)-f^{\prime}(b)\right]+\mathrm{O}\left(h^{4}\right) . $$ This result gives a value for the integral on the left which has an error of order \(h^{4}-a\) factor of \(h^{2}\) better than the error on the trapezoidal rule and as good as Simpson's rule. We can use this formula as a new rule for evaluating integrals, distinct from any of the others we have seen in this chapter. We might call it the "Euler-Maclaurin rule." a) Write a program to calculate the value of the integral \(\int_{0}^{2}\left(x^{4}-2 x+1\right) \mathrm{d} x\) using this formula. (This is the same integral that we studied in Example 5.1, whose true value is \(4.4\).) The order- \(h\) term in the formula is just the ordinary trapezoidal rule; the \(h^{2}\) term involves the derivatives \(f^{\prime}(a)\) and \(f^{\prime}(b)\), which you should evaluate using central differences, centered on \(a\) and \(b\) respectively. Note that the size of the interval you use for calculating the central differences does not have to equal the value of \(h\) used in the trapezoidal rule part of the calculation. An interval of about \(10^{-5}\) gives good values for the central differences. Use your program to evaluate the integral with \(N=10\) slices and compare the accuracy of the result with that obtained from the trapezoidal rule alone with the same number of slices. b) Good though it is, this integration method is not much used in practice. Suggest a reason why not.

Gravitational pull of a uniform sheet A uniform square sheet of metal is floating motionless in space: The sheet is \(10 \mathrm{~m}\) on a side and of negligible thickness, and it has a mass of 10 metric tonnes. a) Consider the gravitational force due to the plate felt by a point mass of \(1 \mathrm{~kg}\) a distance \(z\) from the center of the square, in the direction perpendicular to the sheet, as shown above. Show that the component of the force along the \(z\)-axis is $$ F_{z}=G \sigma z \iint_{-L / 2}^{L / 2} \frac{\mathrm{d} x \mathrm{~d} y}{\left(x^{2}+y^{2}+z_{2}^{2}\right)^{3 / 2}}, $$ where \(G=6.674 \times 10^{-11} \mathrm{~m}^{3} \mathrm{~kg}^{-1} \mathrm{~s}^{-2}\) is Newton's gravitational constant and \(\sigma\) is the mass per unit area of the sheet. b) Write a program to calculate and plot the force as a function of \(z\) from \(z=0\) to \(z=10 \mathrm{~m}\). For the double integral use (double) Gaussian quadrature, as in Eq. (5.82), with 100 sample points along each axis. c) You should see a smooth curve, except at very small values of \(z\), where the force should drop off suddenly to zero. This drop is not a real effect, but an artifact of the way we have done the calculation. Explain briefly where this artifact comes from and suggest a strategy to remove it, or at least to decrease its size. This calculation can thought of as a model for the gravitational pull of a galaxy. Most of the mass in a spiral galaxy (such as our own Milky Way) lies in a thin plane or disk passing through the galactic center, and the gravitational pull exerted by that plane on bodies outside the galaxy can be calculated by just the methods we have employed here.

Create a user-defined function \(f(x)\) that returns the value \(1+\frac{1}{2} \tanh 2 x\), then use a central difference to calculate the derivative of the function in the range \(-2 \leq x \leq 2\). Calculate an analytic formula for the derivative and make a graph with your numerical result and the analytic answer on the same plot. It may help to plot the exact answer as lines and the numerical one as dots. (Hint: In Python the tanh function is found in the math package, and it's called simply tanh.)

5.17 The gamma function: A commonly occurring function in physics calculations is the gamma function \(\Gamma(a)\), which is defined by the integral $$ \Gamma(a)=\int_{0}^{\infty} x^{n-1} \mathrm{e}^{-x} \mathrm{~d} x . $$ There is no closed-form expression for the gamma function, but one can calculate its value for given \(a\) by performing the integral above numerically. You have to be careful how you do it, however, if you wish to get an accurate answer. a) Write a program to make a graph of the value of the integrand \(x^{n-1} \mathrm{e}^{-x}\) as a function of \(x\) from \(x=0\) to \(x=5\), with three separate curves for \(a=2,3\), and 4 , all on the same axes. You should find that the integrand starts at zero, rises to a maximum, and then decays again for each curve. b) Show analytically that the maximum falls at \(x=a-1\). c) Most of the area under the integrand falls near the maximum, so to get an accurate value of the gamma function we need to do a good job of this part of the integral. We can change the integral from 0 to \(\infty\) to one over a finite range from 0 to 1 using the change of variables in Eq. (5.67), but this tends to squash the peak towards the edge of the \([0,1]\) range and does a poor job of evaluating the integral accurately. We can do a better job by making a different change of variables that puts the peak in the middle of the integration range, around \(\frac{1}{2}\). We will use the change of variables given in Eq. (5.69), which we repeat here for convenience: $$ z=\frac{x}{c+x} . $$ For what value of \(x\) does this change of variables give \(z=\frac{1}{2}\) ? Hence what is the appropriate choice of the parameter \(c\) that puts the peak of the integrand for the gamma function at \(z=\frac{1}{2}\) ? d) Before we can calculate the gamma function, there is another detail we need to attend to. The integrand \(x^{n-1} \mathrm{e}^{-x}\) can be difficult to evaluate because the factor \(x^{2-1}\) can become very large and the factor \(\mathrm{e}^{-x}\) very small, causing numerical overflow or underflow, or both, for some values of \(x\). Write \(x^{n-1}=\mathrm{e}^{(a-1) \ln x}\) to derive an alternative expression for the integrand that does not suffer from these problems (or at least not so much). Explain why your new expression is better than the old one. e) Now, using the change of variables above and the value of \(c\) you have chosen, write a user-defined function gamma (a) to calculate the gamma function for arbitrary argument \(a\). Use whatever integration method you feel is appropriate. Test your function by using it to calculate and print the value of \(\Gamma\left(\frac{3}{2}\right)\), which is known to be equal to \(\frac{1}{2} \sqrt{\pi} \simeq 0.886\). f) For integer values of \(a\) it can be shown that \(\Gamma(a)\) is equal to the factorial of \(a-\) 1. Use your Python function to calculate \(\Gamma(3), \Gamma(6)\), and \(\Gamma(10)\). You should get answers closely equal to \(2 !=2,5 !=120\), and \(9 !=362880\).

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