Life with Physics

I reviewed some basics of thermodynamics as well as the Boltzmann entropy in this post . Here, I continue to review the Gibbs entropy, the information entropy as well as the cross-entropy that is widely used in machine learning. In the modern view, entropy is a measure of the ignorance of an observer on the system. Boltzmann Entropy Physics students start to learn the concept of entropy from the Carnot engine and Boltzmann entropy. The Boltzmann entropy applies to a single system with the fixed energy: \begin{equation}S=k_B\log \Omega\,,\tag{1}\end{equation}where $k_B$ is the  Boltzmann constant  and $\Omega$ is the number of microstates corresponding to a system's macrostate. Boltzmann also derived another form of entropy that is very close to the modern form in mathematics. Consider a single system consisting of $N$ particles, we partition $N$ particles into $n$ groups. Let $N_i$ be the number of particles in the group $i$, then the number of microstates is \begin{equa...

In my junior year, I was taught thermodynamics and statistical physics in a one-semester course. In thermodynamics, there is a well-known equation that summarizes all the key concepts in thermodynamics: \begin{equation}dE=T\,dS-P\,dV+\mu\,dN\,\tag{1},\end{equation} where $E,T,S,P,V,\mu, N$ are a system's energy, temperature, entropy, pressure, volume, chemical potential and particle number, respectively. In statistical physics, there is also another famous equation on the microscopic interpretation of entropy: \begin{equation}S = k_B\,\log \Omega\,,\tag{2}\end{equation} where $k_B$ is the Boltzmann constant and $\Omega$ is the number of microstates corresponding to a system's macrostate.  The question is how to prove the entropy in thermodynamics (1) is the same as the Boltzmann entropy (2)? My course teacher might present the proof in class, but I really don't have any impression. In recent days after work, I browsed some textbooks on thermodynamics and statistical physic...

In mathematics, the first homotopy group of a pathwise-connected topological space $\cal{M}$, denoted by $\pi_1(\cal{M})$, is about the classification of all the loops in $\cal{M}$: The relations between loops that can be continuously deformed into one another is an equivalence relation. Then all loops in $\cal{M}$ can be partitioned into disjoint equivalent classes. Finally, all equivalent classes of loops form a group.  Note: Let $\alpha$ and $\beta$ be two loops in $\cal{M}$, the group multiplication between the equivalent classes $[\alpha]$ and $[\beta]$ is defined as $[\alpha]\cdot[\beta]=[\alpha * \beta]$, where $\alpha * \beta$ is a concatenated loop that we first traverse through the loop $\alpha$ and then through the loop $\beta$. In general, the n-th homotopy group, $\pi_n(\cal{M})$ is about the classification of the n-dimensional sphere $S^n$ in $\cal{M}$. Examples: $\pi_1\left(S^1\right)=\mathbb{Z}$ suggests that we can label each equivalent class of loops in $S^1...

What is a particle? S. Weinberg described a particle simply as "a physical system that has no continuous degrees of freedom except for its total momentum".  Recall that the spacetime admits the so-called Poincare symmetry \begin{equation} {x'}^{\mu} = \Lambda^{\mu}_{\,\,\nu}\,x^{\nu} + a^{\mu}\,.\end{equation} By Noether's theorem , each continuous symmetry indicates a conservation. Thus, the total momentum $p^{\mu}$ is a good quantum because of the spacetime translation invariance. As we shall see below, there are other discrete degrees of freedom, denoted by $\sigma$, that are associated with the Lorentz invariance. So the quantum state of a particle is described as $|p, \sigma\rangle$.  Under the Lorentz transformation $\Lambda$, the one-particle state $|p, \sigma\rangle$ changes to a new state $U(\Lambda)|p,\sigma\rangle$ by a unitary operator $U(\Lambda)$. Note that the Lorentz transformation only changes the reference frames, elementary particles should remain...

Problem:  Show that the Lorentz transformation $W$ that keeps $k^{\mu}\equiv[1, 0,  0,  1]^T$ unchanged forms a two-dimensional Euclidean group SE(2). Solution: As shown in this post , the general solution of the Lorentz transformation $W$ that satisfies $W^{\mu}_{\,\,\,\,\nu}\,k^{\nu}=k^{\mu}$ can be parametrized in three parameters $\alpha,\beta,\theta$ as \begin{equation}W(\alpha, \beta, \theta)=\left[\begin{matrix} 1+(\alpha^2+\beta^2)/2  & -\alpha & -\beta & -(\alpha^2+\beta^2)/2  \\  -\alpha & 1 & 0 & \alpha \\ -\beta & 0 &  1 & \beta \\ (\alpha^2+\beta^2)/2 & -\alpha & -\beta & 1-(\alpha^2+\beta^2)/2 \end{matrix}\right]\left[\begin{matrix} 1 &  &  & \\  & \cos\theta & -\sin\theta & \\ & \sin\theta & \cos\theta  &   \\ & &  & 1\end{matrix}\right]\,.\tag{1}\end{equation} To work out the Lie algebra, we expand $W(\...

Problem:  Let $p^{\mu}\equiv[|\mathbf{p}|, p^1, p^2, p^3]^T$ be the  four-momentum  of a massless particle, assuming $c=1$. In some inertial frame, such four-momentum becomes to $k^{\mu}\equiv[\kappa, 0,  0,  \kappa]^T$.  The problem is to find a Lorentz transformation (matrix) $L^{\mu}_{\,\,\nu}(p)$ such that $p^{\mu} =L^{\mu}_{\,\,\nu}(p)\,k^{\nu}$. (This problem is about the proof of Eq. (2.5.44) in Weinberg's QFT book, Volume I) Solution: To transform $k^{\mu}$ to $p^{\mu}$, we can first boost $k^{\mu}$ to $[|\mathbf{p}|, 0, 0, |\mathbf{p}|]^T$ followed by a spatial rotation to $p^{\mu}$.  Recall the Lorentz boost on the energy and momentum, \begin{equation}\begin{split}E'&=&\gamma\,(E-v\,p^3)\,,\\{p'}^3&=&\gamma\,(p^3-v\,E)\,,\end{split} \end{equation} with $\gamma \equiv 1/\sqrt{1-v^2}$. In the matrix form, \begin{equation}\left[\begin{matrix}E' \\ 0 \\ 0 \\ {p'}^3\end{matrix}\right]\equiv B(v)\, \left[\begin{matrix}E \\ 0 \\...

  Problem:  Let $k^{\mu}\equiv[1, 1,  0,  0]^T$, find a Lorentz transformation (matrix) $W^{\mu}_{\,\,\,\,\nu}$  such that $ W^{\mu}_{\,\,\,\,\nu}\,k^{\nu}=k^{\mu}$. (This problem is about the proof of Eq. (2.5.28) in Weinberg's QFT book, Volume I) Solution: Let $t^{\mu}\equiv[1, 0,  0,  0]^T$, the relations \begin{equation}\begin{split}(Wt)^{\mu}(Wt)_{\mu}&=&t^{\mu}t_{\mu}=-1\,,\\ (Wt)^{\mu}k_{\mu}&=&t^{\mu}k_{\mu}=-1\,, \end{split}\end{equation} suggest that $(Wt)^{\mu}$ is of the form \begin{equation} (Wt)^{\mu}= [1+\zeta\,,\zeta\,, \alpha\,, \beta]^T\,,\end{equation} with \begin{equation}\zeta = (\alpha^2+\beta^2)/2\,.\tag{1}\end{equation} As a result, the Lorentz matrix must take the form of \begin{equation}W = \begin{pmatrix} \begin{matrix} 1+\zeta & -\zeta \\ \zeta & 1-\zeta \end{matrix}  & \mathbf{X} \\  \begin{matrix} \alpha & \quad -\alpha \\ \beta & \quad -\beta \end{matrix}  & \m...

Problem: Let $p^{\mu}\equiv[E, p^1, p^2, p^3]^T$ be the  four-momentum of a particle with the rest mass $M$, assuming $c=1$. In the rest frame , such four-momentum becomes to $k^{\mu}\equiv[M, 0,  0,  0]^T$.  The problem is to find a Lorentz transformation (matrix) $L^{\mu}_{\,\,\nu}(p)$ such that $p^{\mu} =L^{\mu}_{\,\,\nu}(p)\,k^{\nu}$. (This problem is about the proof of Eq. (2.5.24) in Weinberg's QFT book, Volume I) Solution: Recall that the vector form of Lorentz boost of four-momentum in two inertial frames with relative velocity $\mathbf{v}$: \begin{equation}\begin{split} E'&=&\gamma\left(E - v\,\mathbf{n}\cdot\mathbf{p}\right)\,,\\ \mathbf{p}'&=&\mathbf{p} + (\gamma-1)\left(\mathbf{n}\cdot\mathbf{p}\right)\mathbf{n}-\gamma\,E\, v\,\mathbf{n}\,,\end{split}\tag{1}\end{equation} where $v$ is the magnitude of $\mathbf{v}$ and $\mathbf{n}\equiv \mathbf{v}/v$ is the unit vector of  $\mathbf{v}$. $\gamma\equiv \frac{1}{\sqrt{1-v^2}}$ is the L...

To illustrate the concept of relativity of simultaneity in special relativity, Einstein proposed a thought experiment where lightning strikes at the head and tail of a running train (points H and T, in short). There are two observers, Alice on the ground and Bob on the train exactly at the middle between H and T. The two events that lightning strikes H and T occur simultaneously to Alice, but not to Bob. In the analysis of this thought experiment, it is important to distinguish the time that the observers receive the lights ( received time ) and the time that the light sent from both ends of the train ( sent time ). Both Newton and Einstein agree that Bob should receive the light from H earlier than the light from T since Bob is moving toward H and faraway from T. The disagreement is the sent time. Suppose in Alice's frame (ground),  the train's length is $2L$; the train's speed is $v$ and the speed of light is $c$; Bob receives the light sent from H (T) at the time $t_H$ (...

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...