Your Turn 10C = L, Verify that the function P (x) = A (e−(x-L)ƒ /kBT - 1) vanishes at x = solves the Smoluchowski equation with the potential Vtot (x) = fx, and resembles the curve sketched in Figure 10.12 between 0 and L. (Here A is some positive constant.) Substitute into Equation 10.3 to find that j(1d) (x) is everywhere constant and positive. dP |j (1d) = net number crossing per time == -MD + 1 dx KBT P dUtot dx). (10.3)
Your Turn 10C = L, Verify that the function P (x) = A (e−(x-L)ƒ /kBT - 1) vanishes at x = solves the Smoluchowski equation with the potential Vtot (x) = fx, and resembles the curve sketched in Figure 10.12 between 0 and L. (Here A is some positive constant.) Substitute into Equation 10.3 to find that j(1d) (x) is everywhere constant and positive. dP |j (1d) = net number crossing per time == -MD + 1 dx KBT P dUtot dx). (10.3)
Classical Dynamics of Particles and Systems
5th Edition
ISBN:9780534408961
Author:Stephen T. Thornton, Jerry B. Marion
Publisher:Stephen T. Thornton, Jerry B. Marion
Chapter4: Nonlinear Oscillations And Chaos
Section: Chapter Questions
Problem 4.15P
Related questions
Question
Don't provide hand writing solution
![Your Turn 10C
= L,
Verify that the function P (x) = A (e−(x-L)ƒ /kBT - 1) vanishes at x =
solves the Smoluchowski equation with the potential Vtot (x) = fx, and
resembles the curve sketched in Figure 10.12 between 0 and L. (Here A is some
positive constant.) Substitute into Equation 10.3 to find that j(1d) (x) is
everywhere constant and positive.
dP
|j (1d) = net number crossing per time
==
-MD
+
1
dx KBT
P
dUtot
dx).
(10.3)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4b7eba7e-d29f-42c7-9b0b-214fcff5d41f%2F31a29193-1ae5-4091-b97b-9467c6d52e88%2Ffk2xapj_processed.png&w=3840&q=75)
Transcribed Image Text:Your Turn 10C
= L,
Verify that the function P (x) = A (e−(x-L)ƒ /kBT - 1) vanishes at x =
solves the Smoluchowski equation with the potential Vtot (x) = fx, and
resembles the curve sketched in Figure 10.12 between 0 and L. (Here A is some
positive constant.) Substitute into Equation 10.3 to find that j(1d) (x) is
everywhere constant and positive.
dP
|j (1d) = net number crossing per time
==
-MD
+
1
dx KBT
P
dUtot
dx).
(10.3)
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