@Sean Havlin The original comment by Jernej is correct. The lecturer incorrectly wrote down p dot at 18:39. He did not acknowledge this mistake, but he wrote down the correct expression at 20:05. The canonical momentum p is differentiating w.r.t q dot, not q. You are incorrect in saying that p dot refers to the derivative w.r.t position. By definition, a dot above a function denotes a TIME derivative. Another mistake that he made at 18:39 was that he forgot to write down a dot on top of the first r vector. Once again, he wrote down the correct form of this expression at 20:05.
@Sean Havlin I agree that the time derivative of the canonical momentum [d/dt(dL/dqdot)] is usually equal to the derivative w.r.t q [dL/dq], but we are not working with the usually form of Lagrange's equation here. We are working with the equation that he wrote down at 14:30 which takes into account the generalized force. In that equation, [dT/dq]=0. Therefore, the time derivative of [dT/dqdot] is equal to Q, not to [dT/dq]. One way to check that he meant to write down p instead of p dot at 18:39 is by looking at the expression on the right hand side sum(m*rdot*[drdot/dqot]). He wrote down r instead of r dot, but this was a mistake on his part because T has no dependence on r(T=sum(m*rdot^2). You could ONLY derive this expression by differentiating T w.r.t to q dot, not q, because T has no dependence on q. It is also clear that he did not take the time derivative of this expression. Another way to see that he meant to write down p is the fact that he rewrote the expression at 20:05, but this time he correctly used p instead of p dot. If you still don't believe that writing p dot is an error, then I recommend taking the momentum expression at 20:05 and differentiating it w.r.t time. You will clearly not get the same expression as p dot at 18:39; this is because the expression at 18:39 is NOT p dot.
@Sean Havlin I have found a pdf of the textbook for this course: online.kottakkalfarookcollege.edu.in:8001/jspui/bitstream/123456789/357/1/introduction-to-lagrangian-hamiltonian-mechanics%20%281%29.pdf. Page 32 and equation 2.37 gives the correct expression
Really bad cameraman here... The important part is what written in chalkboard, not the prof face/body -.- too much moving and zoom in zoom out... I'm feeling motion sick just from watching
I'm sorry... but the time derivative of the momentum you wrote at 18:39 is probably just the momentum p, non the p dot, right?
@Sean Havlin The original comment by Jernej is correct. The lecturer incorrectly wrote down p dot at 18:39. He did not acknowledge this mistake, but he wrote down the correct expression at 20:05. The canonical momentum p is differentiating w.r.t q dot, not q. You are incorrect in saying that p dot refers to the derivative w.r.t position. By definition, a dot above a function denotes a TIME derivative.
Another mistake that he made at 18:39 was that he forgot to write down a dot on top of the first r vector. Once again, he wrote down the correct form of this expression at 20:05.
@Sean Havlin I agree that the time derivative of the canonical momentum [d/dt(dL/dqdot)] is usually equal to the derivative w.r.t q [dL/dq], but we are not working with the usually form of Lagrange's equation here. We are working with the equation that he wrote down at 14:30 which takes into account the generalized force. In that equation, [dT/dq]=0. Therefore, the time derivative of [dT/dqdot] is equal to Q, not to [dT/dq].
One way to check that he meant to write down p instead of p dot at 18:39 is by looking at the expression on the right hand side sum(m*rdot*[drdot/dqot]). He wrote down r instead of r dot, but this was a mistake on his part because T has no dependence on r(T=sum(m*rdot^2). You could ONLY derive this expression by differentiating T w.r.t to q dot, not q, because T has no dependence on q. It is also clear that he did not take the time derivative of this expression. Another way to see that he meant to write down p is the fact that he rewrote the expression at 20:05, but this time he correctly used p instead of p dot. If you still don't believe that writing p dot is an error, then I recommend taking the momentum expression at 20:05 and differentiating it w.r.t time. You will clearly not get the same expression as p dot at 18:39; this is because the expression at 18:39 is NOT p dot.
@Sean Havlin I have found a pdf of the textbook for this course: online.kottakkalfarookcollege.edu.in:8001/jspui/bitstream/123456789/357/1/introduction-to-lagrangian-hamiltonian-mechanics%20%281%29.pdf. Page 32 and equation 2.37 gives the correct expression
@@tomasvirgen8275 Nice, got a link to the second half of the course?
The cameraman was dancing, I think, while recording. 😵😵
Really bad cameraman here... The important part is what written in chalkboard, not the prof face/body -.-
too much moving and zoom in zoom out... I'm feeling motion sick just from watching
And, actually, very bad audio quality.
You have a low attention span
U R LECTURES R EXCELENT BUT TOO LONG - AMARJIT ADVOCATE DELHI HIGH COURT INDIA
ADHD