Thanks for your question. The formula as it is used here is correct. If you would instead of an integral use a sum over time intervals Δt, you would need to divide by Δt. Note that a delta-function of a time argument has a dimension of 1/time.
Hi Jos, amazing video and thanks very much for the series you have up here on youtube! At ~17:00 you have that the correlation function for R(t) is q*delta(t1-t2). Should it not be (q / (Delta t)) *delta(t1-t2)?
Amazing lectures! Is the general rule for the delta-function presented in 24:10 used correctly in the transformation of the given delta-function? Why aren´t the remaining arguments divided by 1-gamma*delta t in the result for P(v,t+delta t) (in 24:14)? Can you please explain me how to transform the delta-function to obain your result?
@@subinsahu3350 yes it is obvious, but there are more terms of second order of (delta t)*2 which one of them belongs to the first term and he didnt use it! I dont know whats going on!
@@baharjafarizadeh9464 If you use the delta formula you get (v-R*delta t)/(1-gamma* delta t) using the taylor series for 1/1-x=1+x you can write it down as (v-R*delta t)*(1+gamma* delta t) by ignoring the delta t*2 you get v+(gamma*v-R)*delta t which is the result in that step
Dear Jos , Could you clear the step at 35:39 . When you took ensemble average , you let snuck inside d/dt . normally is not equal to d/dt . I am not able to follow the argument that why is one able to do it here. Thanks
Good point. The average of a quantity for a large number N of evolutions a can be written as = \sum_i a_i/N, where a_i is the realisation for a particular evolution (determined by what the random generator has spit out). A time derivative in front of that sum leads to d/dt = \sum_i \dot{a}_i/N. I hope this clarifies the point.
Hello sir, thank you for this video. At 39:00 there is a mistake, in the step where you take the taylor series of the exponential you forget to multiply -1 by m/gamma (but then proceed as if you did, so the result holds)
Thanks for your remark. The exponential is not expanded as t is large; it is neglected and only the term -1 survives in the parenthesis. Am I right or did I overlook something?
Thanks for your question. The formula as it is used here is correct. If you would instead of an integral use a sum over time intervals Δt, you would need to divide by Δt. Note that a delta-function of a time argument has a dimension of 1/time.
Hi Jos, amazing video and thanks very much for the series you have up here on youtube! At ~17:00 you have that the correlation function for R(t) is q*delta(t1-t2). Should it not be (q / (Delta t)) *delta(t1-t2)?
Thank you for the lecture! A question: why does pi appear when you take the continuum limit in 11:51?
The factor 2pi is wrong. Thanks for pointing this out. The factor which you keep is q, so it doesn't matter for the sequel
Thanks! Very useful lectures
Amazing lectures! Is the general rule for the delta-function presented in 24:10 used correctly in the transformation of the given delta-function? Why aren´t the remaining arguments divided by 1-gamma*delta t in the result for P(v,t+delta t) (in 24:14)? Can you please explain me how to transform the delta-function to obain your result?
I'm having trouble understanding the same step... Did you manage to figure it out, so you can help me?
He used the first-order expansion: 1/(1-gamma*delta t)= (1+gamma*delta t), which is valid for small delta t.
@@subinsahu3350 yes it is obvious, but there are more terms of second order of (delta t)*2 which one of them belongs to the first term and he didnt use it! I dont know whats going on!
@@baharjafarizadeh9464 If you use the delta formula you get (v-R*delta t)/(1-gamma* delta t) using the taylor series for 1/1-x=1+x you can write it down as (v-R*delta t)*(1+gamma* delta t) by ignoring the delta t*2 you get v+(gamma*v-R)*delta t which is the result in that step
Dear Jos ,
Could you clear the step at 35:39 . When you took ensemble average , you let snuck inside d/dt . normally is not equal to d/dt . I am not able to follow the argument that why is one able to do it here.
Thanks
Good point. The average of a quantity for a large number N of evolutions a can be written as = \sum_i a_i/N, where a_i is the realisation for a particular evolution (determined by what the random generator has spit out). A time derivative in front of that sum leads to d/dt = \sum_i \dot{a}_i/N. I hope this clarifies the point.
Hello sir, thank you for this video. At 39:00 there is a mistake, in the step where you take the taylor series of the exponential you forget to multiply -1 by m/gamma (but then proceed as if you did, so the result holds)
Thanks for your remark. The exponential is not expanded as t is large; it is neglected and only the term -1 survives in the parenthesis. Am I right or did I overlook something?
@@josthijssen6782 I mean as t goes to 0. 38:30 aprox.
I see, you're right. Thanks.
Hey i want to talk with you directly could I sir?