Life with Physics

In this post, we will show how the wave optics becomes the geometrical optics in the limit of zero wavelength. When in the university, I only took one optics course at the level of college physics, which only mentioned the concept of Fermat's principle without too much discussions. Fermat's principle In geometrical optics, the Fermat's principle states that the path $\vec{r}$ of a light ray from the point A to B always minimizes the quantity: \begin{equation}\mathbb{S}[\vec{r}] = \int_A^B n(\vec{r})\,ds\,, \tag{1}\end{equation} where $n(\vec{r})$ is the distribution of reflective index of the medium in space and $ds$ is the differential arc length along the path \begin{equation}ds =\sqrt{d\vec{r}\cdot d\vec{r}}\,.\end{equation} The Fermat's principle provides an explicit connection between the geometrical optics and the classical mechanics in the form of principle of least action. Warm up: Snell's law Before deriving the general equation of motion from the Fermat...

There are three equivalent formalisms used in classical mechanics: Lagrangian, Hamiltonian and Hamilton-Jacobi equation.  I listed their main results as below: Lagrangian $L(q, \dot{q})$: \begin{equation}\frac{d}{dt}\frac{\partial L}{\partial \dot{q}}-\frac{\partial L}{\partial q}=0\,.\tag{1}\end{equation} Hamiltonian  $H(q, p)$: \begin{eqnarray} \dot{q}=\frac{\partial H}{\partial p}\,,\tag{2.1}\\ \dot{p}=-\frac{\partial H}{\partial q}\,.\tag{2.2}\end{eqnarray} Hamilton-Jacobi equation is about the "action" $S(t, q)$: \begin{equation}\frac{\partial S}{\partial t}+H\left(q, \frac{\partial S}{\partial q}\right)=0\,.\tag{3}\end{equation} This post is mainly about the derivation of Hamiltonian-Jacobi equation since it is less well known compared to the Lagrangian and Hamiltonian mechanics. A few words on the Lagrangian The Lagrangian $L(x, v, t)$ is a function of a coordinate $x$ and a velocity $v$ at a single time $t$. For example, in Newtonian mechanics, we have $L(x, v, t)...

新年元旦看了一个谈牛顿的报告： 视频链接 。这是中科院物理所 《学不分科》系列 中的第七期。 听了这个报告才知道，牛顿一直有自我暗示，觉得自己是天选之人，是有使命感的。毕竟他出生在圣诞节，是个遗腹子（玛利亚自己生了耶稣），读的学院叫三位一体，有和苹果相关的传说。牛顿甚至欣喜的发现自己姓名的拉丁文拼写Isaacus Newtonus刚好可以置换为Jeova Sanctus Unus（需要借助I和J，w和v的通假），意为"One True God"。 所以，牛顿并不是晚年才开始信神，人家一直都信，而且笃信自己生来就注定要荣耀上帝。这种天选之人的使命感或许也能解释牛顿糟糕的人际关系：牛顿真心觉得自己和周围凡夫俗子不一样吧。同样，牛顿极少主动发表自己的研究工作展示给同行。他的巨作原理还是被哈雷催着才出版的。 现在愈发觉得做大科学家多少要有些自负，因为这种自信，自负甚至使命感可以帮助他们在长期思考科学难题的痛苦中坚持下来而不放弃。记得在经济学家 凯恩斯纪念牛顿的演讲 里，是这样评价牛顿天赋的：His peculiar gift was the power of continuously holding in his mind a purely mental problem until he had seen straight through it. I fancy his preeminence is due to his muscles of intuition being the strongest and most enduring with which a man has ever been gifted. Anyone who has ever attempted pure scientific or philosophical thought knows how one can hold a problem momentarily in one’s mind and apply all one’s powers of concentrating to piercing through it, and how it will dissolve and escape and you find that what you are surveying ...

The sum of all natural numbers One may see the following mysterious identity that the sum of all natural numbers equals to $-\frac{1}{12}$:   \begin{equation} 1 + 2 + 3 + \cdots = -\frac{1}{12}\,.\tag{1}\end{equation} The rigorous mathematical interpretation relies on the so-called Riemann zeta function $\zeta(s)$ and its analytic continuation. This is not the focus of this post, so we just highlight the main conclusions: \begin{equation}\zeta(s) = \sum_{n=1}^{\infty}\frac{1}{n^s}\tag{2}\end{equation} is only valid when $\mathfrak{Re} (s) > 1$. After analytic continuation, we can compute $\zeta(-1)=-\frac{1}{12}$. Therefore, mathematically, Eq.(1) should be interpreted in the language of analytic continuation. In this post, we explain Eq. (1) by applying a scheme that is commonly used in physics when treating a divergent series $\sum_{n=1}^{\infty}a_n$:  Multiply each term $a_n$ by some regulator such as $e^{-n\epsilon}$ to make the series converge.  Compute the r...

We proved the hyperbolic motion viewed in the lab frame of a relativistic particle with constant proper acceleration as in this post : \begin{equation}t(\tau) = \frac{c}{a}\sinh\left(\frac{a\tau}{c}\right)\,,\quad x(\tau) = \frac{c^2}{a}\cosh\left(\frac{a\tau}{c}\right)\,,\tag{1} \end{equation} where $\tau$ is the proper time of the particle. The next question is: What is the time and space coordinates transformation between the lab frame $(t, x)$ and the particle frame? The time in the particle frame is just the proper time $\tau$ and we denote the space coordinate in the particle frame by $\xi$. The transformation between $(t, x)$ and $(\tau, \xi)$ is obviously not Lorentz transformation since the particle frame is not an inertial frame because of the acceleration. Instead, Eq. (1) is a special case of the transformation at $\xi=0$. The key insight is that although the particle frame is not an inertial frame, there exists a series of instantaneous co-moving inertial frames at each ti...

Considering special relativity, what is the world line of a moving particle with constant proper acceleration $a$ observed in the still world frame? The proper acceleration is defined as the acceleration measured by the inertial observer to which the particle looks rest during the measurement. Lorentz transformation of acceleration The problem is about the Lorentz transformation of acceleration. For this purpose, we start by the Lorentz transformations in special relativity between a still frame $(t, x)$ and a moving frame $(t', x')$ with constant speed $v$ relative to the still frame: \begin{eqnarray} x'&=&\frac{x-vt}{\sqrt{1-{v^2}/{c^2}}}\,,\tag{1}\\ t'&=& \frac{t-vx/c^2}{\sqrt{1-{v^2}/{c^2}}}\,.\tag{2}\end{eqnarray} A motion with speeds $u_x\equiv \frac{dx}{dt}$ in the still frame and $u'_x\equiv \frac{dx'}{dt'}$ in the moving frame has the transformation: \begin{eqnarray} u'_x = \frac{u_x - v}{1-{vu_x}/{c^2}}\,,\tag{3} \end{eqnarray} ...

 The formula of the monthly mortgage payments for a fixed-rate loan can be found online as \begin{equation} M = P\frac{r(1+r)^n}{(1+r)^n-1}\,,\tag{1}\end{equation} where  $M$ is the mortgage payment.   $P$ is the principal, i.e., the initial amount borrowed.  $r$ is the monthly interest rate. For annual interest rate $2.5\%$, $r=2.5\% / 12$. $n$ is the number of payments. For 30-years fixed rate mortgage, $n=30 * 12 = 360$. Here we provide a derivation of the above formula. Let $b_i$ be the owned balance in the $i$-th month. Initially, $b_0=P$ and we choose the constant monthly payment $M$ such that $b_n=0$. We then have the recursion relation between two consecutive month as \begin{equation}b_{i+1}=b_i(1+r) - M\,.\tag{2}\end{equation} A trick to solve (2) is to write Eq. (2) into the following form \begin{equation}b_{i+1}+C = \left(b_i+C\right)(1+r)\,,\tag{3}\end{equation} so that $b_i+C$ is a geometric sequence: \begin{equation}b_i + C = (b_0 + C) (1+r)^i\,.\end{eq...