Generating Function of a Canonical Transformation | Examples and the Big Picture | Lecture 7

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  • เผยแพร่เมื่อ 6 พ.ย. 2024

ความคิดเห็น • 12

  • @mayabeleznay9910
    @mayabeleznay9910 2 ปีที่แล้ว +4

    You are such an amazing teacher. Thank you so much!!

  • @BheraRam-ib8ec
    @BheraRam-ib8ec ปีที่แล้ว +1

    Great Lectures with simplicity.

  • @giovannidelbufalo3589
    @giovannidelbufalo3589 2 ปีที่แล้ว +4

    VERY CLEAR!!!

  • @폴바-x8q
    @폴바-x8q 2 ปีที่แล้ว

    PLA requires only q to be fixed at t1 and t2, but qdot is not fixed. How can we make sure that p,Q,P are fixed at t1and t2?

  • @danielvolinski8319
    @danielvolinski8319 3 ปีที่แล้ว +1

    Please give an example of the relationship between the Lagrangian and a generating function

  • @kingplunger1
    @kingplunger1 9 หลายเดือนก่อน

    What did you do around 11:45 ? How do you come to that result from that integral

    • @Eta_Carinae__
      @Eta_Carinae__ 2 หลายเดือนก่อน

      Distribute the variation $\delta$ operator into the integral, and then into the time derivative, where you end up with:
      $$\delta \int_{t_1} ^{t_2} \frac{d}{dt} F(q, Q, t) dt = \int_{t_1} ^{t_2} \delta \frac{d}{dt} F(q, Q, t) dt = \int_{t_1} ^{t_2} \frac{d}{dt} \delta F(q, Q, t) dt$$
      Expand using the chain rule, where:
      $$\delta F(q, Q, t) = \frac{\partial F}{\partial q} \delta q + \frac{\partial F}{\partial Q} \delta Q$$
      $$\implies \int_{t_1} ^{t_2} \frac{d}{dt} \delta F(q, Q, t) dt = \int_{t_1} ^{t_2} \frac{d}{dt} \frac{\partial F}{\partial q} \delta q + \frac{\partial F}{\partial Q} \delta Q dt$$
      $$\implies \frac{\partial F}{\partial q} (q(t_2), Q(t_2), t_2) \delta q(t_2) + \frac{\partial F}{\partial Q} (q(t_1), Q(t_1), t_1) \delta Q(t_1)$$

  • @leejim3609
    @leejim3609 3 ปีที่แล้ว

    Hello professor, at 22:30, the three terms in the brackets are all zeros. Does this mean that dq/dt and dQ/dt are independent? Since Q = Q(q, p, t), it seems to me that q and Q are not independent, so that dq/dt and dQ/dt may not be independent. If so, how can we claim that the terms in the three brackets are all zeros? Thank you.

    • @nathanborak2172
      @nathanborak2172 11 หลายเดือนก่อน

      I think the idea is that in the Hamiltonian formalism, you think of the phase space s containing 2n independent variables to start with (q1 ... qn ; p1 ... pn) = (q, p). When you make a canonical transformation, you are essentially using a different coordinate system for phase space, (Q, P) instead of (q, p). However, in some circumstances the set (q, Q) may be enough to be a complete coordinate system on phase space, and from that perspective they are independent.