Thermal Probability and Trig integrals <3

In summary: Perhaps it would be helpful to list all the steps involved in a typical problem like this?Integrating from 0 to \pi overcounts the number of particles by a factor of 2. You only integrate from the normal to the plane, which is 0 to \pi/2. Your normalization integral actually vanishes as written.Thank you.
  • #1
sugar_scoot
7
0
Given the number of molecules hitting unit area of a surface per second with speeds between v and v +dv and angles between [tex]\theta[/tex] and d[tex]\theta[/tex] to the normal is

[tex]\frac{1}{2} v n f(v)dv sin \theta cos \theta d\theta[/tex]

show that the average value of cos [tex]\theta[/tex] for these molecules is [tex]\frac{2}{3}[/tex].

I have convinced myself the answer is 4/3 instead. Can anyone show me where I am wrong?
I used P(cos [tex]\theta[/tex]) = sin [tex]\theta[/tex] cos [tex]\theta[/tex]

Then I normalized:
1 = c [tex]\int^{\pi}_{0} sin \theta cos \theta d\theta[/tex]
so that:
c = 2

<cos [tex]\theta[/tex]> = 2 [tex]\int^{\pi}_{0} sin \theta cos^{2}\theta d \theta[/tex] = 2 (2/3) = 4/3
 
Physics news on Phys.org
  • #2
sugar_scoot said:
Given the number of molecules hitting unit area of a surface per second with speeds between v and v +dv and angles between [tex]\theta[/tex] and d[tex]\theta[/tex] to the normal is

[tex]\frac{1}{2} v n f(v)dv sin \theta cos \theta d\theta[/tex]

show that the average value of cos [tex]\theta[/tex] for these molecules is [tex]\frac{2}{3}[/tex].

I have convinced myself the answer is 4/3 instead. Can anyone show me where I am wrong?
I used P(cos [tex]\theta[/tex]) = sin [tex]\theta[/tex] cos [tex]\theta[/tex]

Then I normalized:
1 = c [tex]\int^{\pi}_{0} sin \theta cos \theta d\theta[/tex]
so that:
c = 2

<cos [tex]\theta[/tex]> = 2 [tex]\int^{\pi}_{0} sin \theta cos^{2}\theta d \theta[/tex] = 2 (2/3) = 4/3

Integrating from 0 to [tex]\pi[/tex] overcounts the number of particles by a factor of 2. You only integrate from the normal to the plane, which is 0 to [tex]\pi/2[/tex]. Your normalization integral actually vanishes as written.
 
  • #3
Thank you.

Is there an intuitive reason why normalization is unnecessary in this case? Should I continue to attempt normalization as a first step in problems like these?
 
  • #4
Actually I just did the problem over again with the new integration limits and although my <cos[tex]\theta[/tex]> is now correct, I still found a normalization constant of 2.
 

1. What is thermal probability?

Thermal probability is the likelihood of a specific thermal energy state occurring in a system. It is based on the Boltzmann distribution, which describes the distribution of thermal energy among particles in a system.

2. How is thermal probability calculated?

Thermal probability is calculated using the Boltzmann factor, which is equal to e^(-E/kT), where E is the energy of a particular state, k is the Boltzmann constant, and T is the temperature of the system. The higher the energy of a state, the lower its probability of occurring.

3. What is the relationship between thermal probability and temperature?

The relationship between thermal probability and temperature is inverse. As temperature increases, the thermal probability of higher energy states also increases. This means that at higher temperatures, there is a greater chance of particles having higher thermal energies.

4. What are trigonometric integrals?

Trigonometric integrals are integrals that involve trigonometric functions, such as sine, cosine, tangent, etc. These integrals often require the use of trigonometric identities and substitution techniques for evaluation.

5. How are trigonometric integrals used in thermal probability?

Trigonometric integrals are used in the calculation of partition functions, which are important in determining thermal probability. These integrals are also used in the derivation of thermodynamic properties, such as specific heat capacity and entropy, which are related to thermal probability.

Similar threads

  • Calculus and Beyond Homework Help
Replies
5
Views
619
  • Calculus and Beyond Homework Help
Replies
3
Views
488
Replies
14
Views
942
  • Calculus and Beyond Homework Help
Replies
3
Views
838
  • Calculus and Beyond Homework Help
Replies
7
Views
882
  • Calculus and Beyond Homework Help
Replies
2
Views
590
  • Calculus and Beyond Homework Help
Replies
8
Views
829
  • Calculus and Beyond Homework Help
Replies
6
Views
276
  • Calculus and Beyond Homework Help
Replies
28
Views
1K
  • Calculus and Beyond Homework Help
Replies
5
Views
997
Back
Top