Question

Consider the Langevin equation $$ \frac{d v}{d t}=-\gamma v+\frac{1}{m} F^{\prime}(t) $$ where the first term on the right is a phenomenological expression for the slowly varying part of the interaction force, whose rapidly fluctuating part is denoted by $F^y(t)$ If $F^{\prime}$ is neglected, the solution of the resulting equation is $v=u \exp (-\gamma t)$ where $u$ is a constant. In the general case where $F^{\prime} \neq 0$, assume a solution of the same form with $u=u(t)$ and ehow that the solution of the Langevin equation gives for the velocity at time $t$ the result $$ v=v_0 e^{-\gamma^t}+\frac{1}{m} e^{-\gamma^1} \int_0^t e^{\gamma^{\prime \prime}} F^{\prime}\left(t^{\prime}\right) d t^{\prime} $$ where $v_0=v(\mathbf{0})$.

   Consider the Langevin equation

$$
\frac{d v}{d t}=-\gamma v+\frac{1}{m} F^{\prime}(t)
$$

where the first term on the right is a phenomenological expression for the slowly varying part of the interaction force, whose rapidly fluctuating part is denoted by $F^y(t)$ If $F^{\prime}$ is neglected, the solution of the resulting equation is $v=u \exp (-\gamma t)$ where $u$ is a constant. In the general case where $F^{\prime} \neq 0$, assume a solution of the same form with $u=u(t)$ and ehow that the solution of the Langevin equation gives for the velocity at time $t$ the result

$$
v=v_0 e^{-\gamma^t}+\frac{1}{m} e^{-\gamma^1} \int_0^t e^{\gamma^{\prime \prime}} F^{\prime}\left(t^{\prime}\right) d t^{\prime}
$$

where $v_0=v(\mathbf{0})$.
Show more…
Fundamentals of Statistical and Thermal Physics
Fundamentals of Statistical and Thermal Physics
Rief F. 1st Edition
Chapter 15, Problem 4 ↓

Instant Answer

verified

Step 1

We're told that when $F'(t) = 0$, the solution is $v = u e^{-\gamma t}$ where $u$ is a constant.  Show more…

Show all steps

lock
AceChat toggle button
Close icon
Ace pointing down

Please give Ace some feedback

Your feedback will help us improve your experience

Thumb up icon Thumb down icon
Thanks for your feedback!
Profile picture
Consider the Langevin equation $$ \frac{d v}{d t}=-\gamma v+\frac{1}{m} F^{\prime}(t) $$ where the first term on the right is a phenomenological expression for the slowly varying part of the interaction force, whose rapidly fluctuating part is denoted by $F^y(t)$ If $F^{\prime}$ is neglected, the solution of the resulting equation is $v=u \exp (-\gamma t)$ where $u$ is a constant. In the general case where $F^{\prime} \neq 0$, assume a solution of the same form with $u=u(t)$ and ehow that the solution of the Langevin equation gives for the velocity at time $t$ the result $$ v=v_0 e^{-\gamma^t}+\frac{1}{m} e^{-\gamma^1} \int_0^t e^{\gamma^{\prime \prime}} F^{\prime}\left(t^{\prime}\right) d t^{\prime} $$ where $v_0=v(\mathbf{0})$.
Close icon
Play audio
Feedback
Powered by NumerAI
Need help? Use Ace
Ace is your personal tutor. It breaks down any question with clear steps so you can learn.
Start Using Ace
Ace is your personal tutor for learning
Step-by-step explanations
Instant summaries
Summarize YouTube videos
Understand textbook images or PDFs
Study tools like quizzes and flashcards
Listen to your notes as a podcast
Continue solving this problem
Create a free account to:
  • View full step-by-step solution
  • Ask follow-up questions with Ace AI
  • Save progress and study later
Continue Free
Numerade

Get step-by-step video solution
from top educators

Continue with Clever
or



By creating an account, you agree to the Terms of Service and Privacy Policy
Already have an account? Log In

A free answer
just for you

Watch the video solution with this free unlock.

Numerade

Log in to watch this video
...and 100,000,000 more!


EMAIL

PASSWORD

OR
Continue with Clever