Drorbn - User contributions [en]

Notes for AKT-140303/0:21:43

2018-06-25T23:34:30Z

Donghao.ouyang: Created page with "Here we check for the cyclic symmetry of the structure coefficient <math>f</math>. First, we want to show that <math>f</math> is total antisymmetric. For the first antisymme..."

Here we check for the cyclic symmetry of the structure coefficient <math>f</math>. First, we want to show that <math>f</math> is total antisymmetric. For the first antisymmetry relation (i.e., the one with first two indices exchanged), we use antisymmetry relation of Lie algebra.

<math>f_{abc}=\langle\left[X_a,X_b\right],X_c\rangle=\langle -\left[X_b,X_a\right],X_c\rangle=-\langle\left[X_b,X_a\right],X_c\rangle=-f_{bac}.</math>

For the second antisymmetry relation (i.e., the one with last two indices exchanged), we use the fact the metric is invariant. A metric <math>\langle\cdot,\cdot\rangle:\mathfrak{g}\times\mathfrak{g}\rightarrow \mathbb{R}</math> for any finite-dimensional Lie algebra <math>\mathfrak{g}</math> is invariant

<math>\langle\left[z,x\right],y\rangle=-\langle x,\left[z,y\right]\rangle, \forall x,y,z\in\mathfrak{g}.</math>

Then, with the fact that the metric is symmetric, we write this relation in terms of structural coeffiecnt <math>f</math>.

<math>f_{abc}=\langle\left[X_a,X_b\right],X_c\rangle=-\langle X_b,\left[X_a,X_c\right]\rangle=-\langle \left[X_a,X_c\right],X_b\rangle=-f_{acb}.</math>

Thus, this shows the antisymmetry relations of <math>f</math>. Then, <math>f_{abc}=-f_{bac}=-\left(-f_{bca}\right)=f_{bca}</math>. Similarly, it shows <math>f_{bca}=f_{cab}</math>. Thus, the cyclic relation follows.

Notes for AKT-140108/0:24:50

2018-05-31T03:08:22Z

Donghao.ouyang:

I am still not clear on why the image of the unlink is the equator of the sphere. I am wandering why not any other circle above or below the equator. Do we homotop the image to the equator?

Since we have assumed that the embedding is invariant, we can keep the two circles in the same plane. Then, the directional vector would always be inside of the plane. Since we are free to choose which plane we want the unlink circles to be, without loss of generality, the two circles are chosen to be in the xy-plane. Then the directional vector will not have the z-component, hence the image of the map can only be on the equator (if we choose some other planes, the image would be a great circle on <math>S^2</math> by rotating the equator).

Notes for AKT-140117/0:28:51

2018-05-23T17:17:20Z

Donghao.ouyang: Created page with "To demonstrate how to use the Euler-Lagrange equation in classical mechanics, we solve brachistochrone problem as an example. The problem is described in the blackboard shot,..."

To demonstrate how to use the Euler-Lagrange equation in classical mechanics, we solve brachistochrone problem as an example. The problem is described in the blackboard shot, which is to find the path of a particle that minimizes the time <math>T</math> traveled from point <math>P_1</math> to point <math>P_2</math> in a uniform gravitational field. In this situation, we assume there is no friction along the path; thus the energy is conserved. Let <math>y</math> be the vertical coordinate. Then, by the conservation of energy, we have

<math>\frac{1}{2}mv^2-\frac{1}{2}mv_i^2=\frac{1}{2}mv^2=mgy.</math>

Thus, we have

<math>v=\sqrt{2gy}.</math>

Then, the time <math>T</math> may be described as

<math>T=\int_{P_1}^{P_2}\frac{ds}{v},</math>

where <math>ds</math> is the infinitesimal arclength of the path. Then, let <math>x</math> be the horizontal coordinate, we have <math>ds=\sqrt{dx^2+dy^2}=\sqrt{1+y'^2}dx.</math> Thus, the above equation would be

<math>T=\int_{P_1}^{P_2}\sqrt{\frac{1+y'^2}{2gy}}dx.</math>

Now, let <math>L=\sqrt{\frac{1+y'^2}{2gy}}</math>, we apply the Euler-Lagrange equation and obtain

<math>\frac{\partial L}{\partial y}=-\frac{1}{2y}\sqrt{\frac{1+y'^2}{2g}}=\frac{d}{dx}\left(\frac{\partial L}{\partial y'}\right)=\frac{1}{\sqrt{2g}}\frac{2yy''-y'^4-y'^2}{2(y(1+y'^2))^{3/2}}.</math>

If we rearrange the equation and integrate, we obtain the equation

<math>C=\frac{1}{\sqrt{2gy\left(1+y'^2\right)}}</math>

where <math>C</math> is some constant. Then, we rearrange the equation and obatin

<math>\left(1+y'^2\right)y=\frac{1}{2gC^2}=k^2.</math>

Then, we can solve this equation with parameterization and obtain the final result

<math>x\left(t\right)=\frac{k^2}{2}\left(t-\sin t\right),y\left(t\right)=\frac{k^2}{2}\left(1-\cos t\right)</math>

Notes for AKT-140117/0:26:27

2018-05-23T16:35:49Z

Donghao.ouyang: Created page with "Just some generalization for the least action principle and Euler-Lagrange equation for the classical cases. In the calculus of variation, we have developed a tool for descri..."

Just some generalization for the least action principle and Euler-Lagrange equation for the classical cases. In the calculus of variation, we have developed a tool for describing various physics situation. In general, let <math>q</math> be the coordinate of a particular configuration space and <math>\dot{q}</math> be its time derivative. Then, the action is described as

<math>\mathcal{L}=\int_{t_i}^{t_f}dt L\left(q,\dot{q}\right),</math>

where <math>t_i,t_f</math> are the initial time and final time, respectively. The integrand <math>L</math> 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 <math>\mathcal{L}</math>. 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

<math>0=\delta \mathcal{L}=\delta \int_{t_i}^{t_f}dt L\left(q,\dot{q}\right)=\int_{t_i}^{t_f}dt \delta L\left(q,\dot{q}\right)</math>
<math>=\int_{t_i}^{t_f}dt \left(\frac{\partial L}{\partial q}\delta q+\frac{\partial L}{\partial\dot{q}}\delta \dot{q}\right)=\int_{t_i}^{t_f}dt \left(\frac{\partial L}{\partial q}\delta q+\frac{\partial L}{\partial\dot{q}}\frac{d}{dt}\left(\delta q\right)\right).</math>

Then, by the integration by parts, we have that

<math>0=\int_{t_i}^{t_f}dt \left(\frac{\partial L}{\partial q}-\frac{d}{dt}\left(\frac{\partial L}{\partial\dot{q}}\right)\right)\delta q+\int_{t_i}^{t_f}\frac{d}{dt}\left(\frac{\partial L}{\partial\dot{q}}\delta q\right)=\int_{t_i}^{t_f}dt \left(\frac{\partial L}{\partial q}-\frac{d}{dt}\left(\frac{\partial L}{\partial\dot{q}}\right)\right)\delta q+\frac{\partial L}{\partial\dot{q}}\delta q\big|_{t_i}^{t_f}=\int_{t_i}^{t_f}dt \left(\frac{\partial L}{\partial q}-\frac{d}{dt}\left(\frac{\partial L}{\partial\dot{q}}\right)\right)\delta q,</math>

since the boundary term does not vary so that <math>\delta q\left(t_i\right)=\delta q\left(t_f\right)=0</math>. Thus, we arrive at the point where the classical particle must obey the path where the equation

<math>\frac{d}{dt}\left(\frac{\partial L}{\partial\dot{q}}\right)-\frac{\partial L}{\partial q}=0.</math>

This equation is known as the Euler-Lagrange equation.