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Notes on AKT-140117: [edit, refresh]
Euler-Lagrange problems, Gaussian integration, volumes of spheres.
# | Week of... | Notes and Links |
---|---|---|
1 | Jan 6 | About This Class (PDF). ![]() Tricolourability without Diagrams ![]() ![]() Friday Introduction (the quantum pendulum) |
2 | Jan 13 | Homework Assignment 1. ![]() ![]() ![]() |
3 | Jan 20 | Homework Assignment 2. ![]() ![]() ![]() Class Photo. |
4 | Jan 27 | Homework Assignment 3. ![]() ![]() ![]() ![]() |
5 | Feb 3 | Homework Assignment 4. ![]() The Fulton-MacPherson Compactification (PDF). ![]() ![]() ![]() ![]() |
6 | Feb 10 | Homework Assignment 5. ![]() ![]() ![]() |
R | Feb 17 | Reading Week. |
7 | Feb 24 | ![]() ![]() ![]() ![]() From Gaussian Integration to Feynman Diagrams (PDF). |
8 | Mar 3 | Homework Assignment 6 (PDF) ![]() ![]() ![]() Graph Cohomology and Configuration Space Integrals (PDF) ![]() Mar 9 is the last day to drop this class. |
9 | Mar 10 | Homework Assignment 7 (PDF) ![]() ![]() ![]() ![]() |
10 | Mar 17 | Homework Assignment 8 (PDF) ![]() ![]() ![]() ![]() Gaussian Integration, Determinants, Feynman Diagrams (PDF). |
11 | Mar 24 | Homework Assignment 9 (PDF) ![]() ![]() ![]() Friday: class cancelled. |
12 | Mar 31 | Monday, Wednesday: class cancelled. ![]() |
E | Apr 7 | ![]() |
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Dror's Notebook | ||
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$\frac{d}{d\epsilon}f(\epsilon)\mid_{\epsilon = 0} = \frac{mg}{s}\int_{-L}^{L}y_q\sqrt{1+\dot{y}_q^2}+y_c(1+\dot{y}_c^2)^{\frac{-1}{2}(\dot{y}_c\dot{y}_q)}dx = 0$
Thus, $0 = \frac{mg}{s}(I_1 + I_2)$, where $I_1 = \int_{-L}^L y_q\sqrt{1+\dot{y}_c^2}dx$ and $I_2 = \int_{-L}^L y_c\dot{y}_c\dot{y}_q(1+\dot{y}_c^2)^{\frac{-1}{2}}dx$.
Integrating $I_2$ by parts with $u = y_c\dot{y}_c(1+\dot{y}_c^2)^{\frac{-1}{2}}$ and $dv = \dot{y}_q dx$, and applying boundary conditions $y_q(-L) = y_q(L) = 0$, we obtain
$I_2 = \int_{-L}^L y_q\big(y_c\dot{y}_c^2\ddot{y}_c(1+\dot{y}_c^2)^{\frac{-3}{2}} - (\dot{y}_c^2 + y_c\ddot{y}_c)(1+\dot{y}_c^2)^{\frac{-1}{2}})\big)dx$
From $I_1 + I_2 = 0$, factoring out the $y_q$ and the fundamental lemma of the calculus of variations, we obtain an ODE (replacing $y_c$ with $y$):
$0 = (1+\dot{y}^2)^\frac{-3}{2}((1+\dot{y}^2)^2 - (\dot{y}^2 + y\ddot{y})(1+\dot{y}^2) + y\dot{y}^2\ddot{y})$
Dividing through by $(1+\dot{y}^2)^\frac{-3}{2}$, expanding, and simplifying, we obtain our final ODE:
$1 + \dot{y}^2 - y\ddot{y} = 0$
A solution to this is $y(x) = \frac1{\lambda}cosh(\lambda x + c)$, where $\lambda$ and c are determined by physical constants and the boundary values of $y$ (at $x = -L$ and $x = L$). It turns out that this ODE is the same one you get when solving the soap bubble problem (HW 2, problem 3), since the Lagrangians of the two systems are the same up to constants.
Let . This now becomes a single variable minimum/maximum problem. We set
, and solve for
. First, simplifying
, we compute
+ higher order terms).
Thus,
Integrating by parts with , this is equal to
The first term is equal to 0 by boundary conditions of , so we obtain the equality
, exactly as stated in the conclusion of Lemma 3.4. Solving this ODE with initial conditions gives the desired result. Explicitly, the solution of this ODE (with
) is
Plugging in
and
, we have
and
, implying that
, as claimed.
where are the initial time and final time, respectively. The integrand
is known as the Lagrangian and is assumed to be time-independent for convenience. The idea here is to find the path that minimize the action
. Now, we introduction the idea of variation, which can be viewed as an infinitesimal shift from the original path; however, it does not change the terminal points. Since the path we are interested is the path that minimizes the action, then the variation of the action should be 0 and that is
Then, by the integration by parts, we have that
since the boundary term does not vary so that . Thus, we arrive at the point where the classical particle must obey the path where the equation
This equation is known as the Euler-Lagrange equation.
Thus, we have
Then, the time may be described as
where is the infinitesimal arclength of the path. Then, let
be the horizontal coordinate, we have
Thus, the above equation would be
Now, let , we apply the Euler-Lagrange equation and obtain
If we rearrange the equation and integrate, we obtain the equation
where is some constant. Then, we rearrange the equation and obatin
Then, we can solve this equation with parameterization and obtain the final result