Planck Spectrum and wavelength subsitution

In summary, Schroeder asks you to substitute \lambda = \frac{hc}{\epsilon} into the equation for the Planck spectrum, but when you do the substitution the function doesn't peak at 2.82 like it should. He asks you to solve for the energy distribution function in terms of lambda to find the maximum.
  • #1
FortranMan
30
0
All the below is presented in latex format. In Schroeder's intro to thermal, problem 7.38, he asks you to substitute

[EQ] \lambda = \frac{hc}{\epsilon} [/EQ]

into the following equation

[EQ] \frac{U}{V} = \int^{\infty}_{0} \frac{8 \pi \epsilon^{3}/(hc)^{3}}{e^{\epsilon / kT}-1} d \epsilon [/EQ]

from which you should get the Planck spectrum as a function of wavelength. So far I've been careful in replacing the differential of epsilon with

[EQ] d \epsilon = - \frac{hc}{\lambda^2} d \lambda [/EQ]

In preparation to plot and solve the integrand, he asks you to express the above equation in terms of hc/kT, which I'm guessing can be done by substituting the value x below.

[EQ] x = \frac{hc}{kT \lambda} [/EQ]

Again being careful to replace the differential of lambda,

[EQ] d \lambda = - \frac{hc}{kT x^2} dx [/EQ]

However as soon as arrange everything in terms of x, it looks exactly like this equation

[EQ] \frac{U}{V} = \frac{8 \pi (kT)^4}{(hc)^3} \int^{\infty}_{0} \frac{x^3}{e^{x}-1} dx [/EQ]

which, according to Schroeder, it shouldn't because when you plot the spectrum in terms of the wavelength the function should NOT peak at 2.82 and look exactly like the spectrum in terms of epsilon. So, what stupid differential mistake am I making?
 
Physics news on Phys.org
  • #2
I'm going to take your first expression, [EQ] \lambda = \frac{hc}{\epsilon} [/EQ] and substitute "[ tex ]" for [EQ] and "[ /tex ]" for [/EQ], and here's what I get [tex] \lambda = \frac{hc}{\epsilon} [/tex]. Eliminate the spaces in the tex /tex pair to get it to actually work.
 
Last edited:
  • #3
FortranMan said:
All the below is presented in latex format. In Schroeder's intro to thermal, problem 7.38, he asks you to substitute

[tex] \lambda = \frac{hc}{\epsilon} [/tex]

into the following equation

[tex] \frac{U}{V} = \int^{\infty}_{0} \frac{8 \pi \epsilon^{3}/(hc)^{3}}{e^{\epsilon / kT}-1} d \epsilon [/tex]

from which you should get the Planck spectrum as a function of wavelength. So far I've been careful in replacing the differential of epsilon with

[tex] d \epsilon = - \frac{hc}{\lambda^2} d \lambda [/tex]

In preparation to plot and solve the integrand, he asks you to express the above equation in terms of hc/kT, which I'm guessing can be done by substituting the value x below.

[tex] x = \frac{hc}{kT \lambda} [/tex]

Again being careful to replace the differential of lambda,

[tex] d \lambda = - \frac{hc}{kT x^2} dx [/tex]

However as soon as arrange everything in terms of x, it looks exactly like this equation

[tex] \frac{U}{V} = \frac{8 \pi (kT)^4}{(hc)^3} \int^{\infty}_{0} \frac{x^3}{e^{x}-1} dx [/tex]

which, according to Schroeder, it shouldn't because when you plot the spectrum in terms of the wavelength the function should NOT peak at 2.82 and look exactly like the spectrum in terms of epsilon. So, what stupid differential mistake am I making?

rephrased. Now can anyone answer my question?
 
  • #4
Subtle difference

Your differentials are perfect. You just took them too far. Dont substitute with x. You want to solve for the energy distribution function in terms of lambda. That way you can take the derivative of the energy distribution function and set to zero to find maximum. Check out the wiki on Wien's Displacement Law and the derivation at the bottom.

en.wikipedia.org/wiki/Wien's_displacement_law
 

FAQ: Planck Spectrum and wavelength subsitution

What is the Planck Spectrum and how is it related to wavelength substitution?

The Planck Spectrum, also known as the blackbody spectrum, is a theoretical distribution of electromagnetic radiation emitted by a perfect blackbody at a given temperature. Wavelength substitution refers to the process of replacing a given wavelength with a different one in the spectrum. This can be done to study the effects of varying wavelengths on the emitted radiation.

Why is the Planck Spectrum important in scientific research?

The Planck Spectrum is important in scientific research because it provides a theoretical baseline for the distribution of radiation emitted by objects at different temperatures. It also allows scientists to study the effects of varying wavelengths on the emitted radiation, which has many applications in fields such as astrophysics, materials science, and climate research.

How is the Planck Spectrum derived?

The Planck Spectrum is derived from the Planck's law, which describes the relationship between the energy of radiation and its wavelength at a given temperature. This law was developed by German physicist Max Planck in 1900 and is considered one of the cornerstones of quantum mechanics.

Can the Planck Spectrum be observed in real life?

While the Planck Spectrum is a theoretical concept, it can be approximated and observed in real life. Perfect blackbodies do not exist, but many objects in nature, such as stars, come close to behaving like one. By studying the emitted radiation from these objects, scientists can observe and analyze the Planck Spectrum.

How does the Planck Spectrum relate to the study of the cosmic microwave background?

The cosmic microwave background (CMB) is a remnant of the early universe and is often referred to as the "afterglow" of the Big Bang. The CMB is a perfect blackbody radiation with a temperature of about 2.7 Kelvin, and its spectrum closely follows the Planck Spectrum. By studying the CMB, scientists can gain insights into the early universe and its evolution, as well as the validity of the Planck Spectrum.

Back
Top