Derive Eqs. 15.36 and 15.37 .

Name: Problem 16, Chapter 15: Modern Physics
Uploaded: 2021-10-22T13:41:05-08:00
Duration: 8 min 37 s
Channel: Eduard Sanchez
Description: Derive Eqs. 15.36 and 15.37 .

Key Concepts

Underlying Physical Principles

Even when the derivation is mainly mathematical, the process is generally guided by underlying physical laws or assumptions (such as conservation laws, symmetry properties, or approximations) that dictate the form of the equations being derived.

Application of Boundary and Initial Conditions

In many derivations, particularly in physics, the application of appropriate boundary or initial conditions is essential. These conditions help determine constants that appear during integration and ensure that the solution is specific to the physical situation being modeled.

Mathematical Derivation

This concept involves the step?by?step manipulation of equations using algebraic and calculus techniques. Derivations require applying operations such as differentiation, integration, and algebraic rearrangement systematically to obtain a result from underlying principles.

Use of Differential Equations

Often, the derivation of key equations in many physical contexts relies on setting up and solving differential equations. Understanding how to differentiate functions, integrate terms, and apply methods for solving differential equations is fundamental to deriving such results.

Transcript

00:01 Hi there.

00:02 So for this problem, we need to doer divide equation 15 .36 and 15 .37.

00:10 So we're going to start with the first one.

00:14 So we know that the number with energies that are superior to the ground energy divided by the volume is equal to the volume to the minus 1, the integral from the energy 0 to the infinite of the number in function of the energy and with the differential of energy.

00:48 So in here we just simply substitute the expression for the function n.

00:53 So we will obtain.

00:54 We just take out everything that is constant.

00:57 So we have 8 times pi times the time plums constant times the speed of light to the 3.

01:05 We have the integral from e .0 to infinity of the energy to the square times the exponential of the energy divided by minus m bolzman constant times the temperator.

01:22 And in here we will have then, if we make a change in variables, we will have the following.

01:36 We will have that this, where we have done the change in here.

01:42 We have done the change that adds is equal to the energy divided by boltzman constant times its imperator.

01:54 So we are left with the following.

01:56 So now these are the limits.

02:03 And the integral is simply x to the squared times the exponential of minus x, the differential in x.

02:16 So after we have done that integral, we will left with the following.

02:21 8 times pi times the product between plus constant speed of line to the 3 times, equals my constant times the imperator to the 3.

02:31 And from the integral, we obtain that that is equal to minus x to the square times the exponential of minus x, minus 2 times x times the exponential of minus x, minus 2 times x, minus 2 times the add 2 times the exponential of minus 2 times the exponential of minus x.

02:57 And this evaluated from the energy sub -zero divided by bosom constant times the temperature to the infinity.

03:09 So after evaluating this, we will obtain the following, and that is the expression that we wanted to obtain for the first part.

03:17 That is a times pi times the product between plants constant and the speed of light to the three, a times bolzman constant, times the temperateur and that to the three, the exponential of minus the energy sub 0 divided by boltzman constant times the imperator and all of this times the energy sub zero divided by boltzman constant times the temperator and all of that elevated to the two.

03:52 This plus two times the energy this plus 2.

04:02 So that's the solution for the first part of this problem.

04:08 Now for the second part, we need to find the expression for the frequency...

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