This is just for clarication. In 4:50 you talked about rewriting the quantity x xdoubledot or xa and it became the parenthetical term in the next equation namely, d/dt(xxdot)-xdot^2. May I ask how? Because I am treating it as though the d/dt term is equal to va-v^2 = v(a-v) which is not the same as the original xa term?
Dear HQ, you evaluate d/dt(x dot(x)), using the product rule. That yields x(x dotdot) + (x dot)(x dot). You then move the last term (x dot)^2 to the other side and there you are. I hope this is clear.
Dear Sir, Thank you. However, there is a tiny mistake at 8.40, I think that there shouldn't be the friction constant(drag) \gamma - it is just cancelled. It makes also sense (in my logic) that there shouldn't be \gamma because at the previous video (ruclips.net/video/H9I0PmXwhdo/видео.html) we got that q=2 \gamma k_B T , where q is the "strength" of the random force and \gamma is the friction. Therefore, if t-->0 there is no random force so q=0 and therefore no drag force, \gamma=0. For me, it means that if there is no random force the big particle has no reason to feel any friction because q and \gamma are related in the Langevin equations - random force can be as strong as the friction enables it and friction is as big as the random force push it to be.
This is just for clarication. In 4:50 you talked about rewriting the quantity x xdoubledot or xa and it became the parenthetical term in the next equation namely, d/dt(xxdot)-xdot^2. May I ask how?
Because I am treating it as though the d/dt term is equal to va-v^2 = v(a-v) which is not the same as the original xa term?
Dear HQ, you evaluate d/dt(x dot(x)), using the product rule. That yields x(x dotdot) + (x dot)(x dot). You then move the last term (x dot)^2 to the other side and there you are.
I hope this is clear.
Oh yes.. I forgot the product rule. Thank you!
Dear Sir, Thank you.
However, there is a tiny mistake at 8.40, I think that there shouldn't be the friction constant(drag) \gamma - it is just cancelled.
It makes also sense (in my logic) that there shouldn't be \gamma because at the previous video (ruclips.net/video/H9I0PmXwhdo/видео.html) we got that
q=2 \gamma k_B T , where q is the "strength" of the random force and \gamma is the friction.
Therefore, if t-->0 there is no random force so q=0 and therefore no drag force, \gamma=0.
For me, it means that if there is no random force the big particle has no reason to feel any friction because q and \gamma are related in the Langevin equations - random force can be as strong as the friction enables it and friction is as big as the random force push it to be.
Dear Maria, Thanks for your remark. You're right, the gamma should go out, i.e. = kB T/m t^2.