Life with Physics

I occasionally browsed a youtube video on "2023 MIT Integration Bee". I thought it is a competition between students taking the course of calculus. I took a try and it turns out that I was not able to even solve the first problem in the allowed 4 mins.  The first problem is to solve \begin{equation} I \equiv \int_0^{\infty} \frac{\sqrt[3]{\tan x}}{(\sin x + \cos x)^2}dx\,.\tag{1}\end{equation} It is obvious that \begin{eqnarray}I &=&  \int_0^{\infty} \frac{\sqrt[3]{\tan x}}{(\tan x + 1)^2} \frac{dx}{\cos^2 x} =  \int_0^{\infty} \frac{\sqrt[3]{x}}{(x+1)^2}dx\\&=&-\left. \frac{\sqrt[3]{x}}{x+1}\right|_0^{\infty} +  \int_0^{\infty} \frac{d\,\sqrt[3]{x}}{x+1}\\&=&  \int_0^{\infty}\frac{dx}{x^3+1}\,.\tag{2}\end{eqnarray} I know how to solve the indefinite integral $\int\frac{dx}{x^3+1}$, but I was not able to accomplish the entire calculation within 4 mins. Here is another youtube video  that provides step-by-step calculat...

I recently listened to a public lecture on the electrodynamics  at the middle night on New Year's Eve. This is the first time I heard that the Maxwell equation is initially described by James Clerk Maxwell  himself with the help of quaternion , a mathematics created by  Sir William Rowan Hamilton  in 1843. But in today's textbooks, Maxwell equations are taught in the language of vector analysis and the concept of quaternion has been completely eliminated. A brief review of this history, for example, can be found in the paper "Development of Vector Analysis from Quaternions" .  Quaternion definition As the generalization of the complex number $z=a+i\,b$ with $i^2=-1$, quaternion is defined in the form of \begin{equation} \mathcal{Q} = q_0+q_1\,\mathbf{i}+q_2\,\mathbf{j}+q_3\,\mathbf{k}\,\tag{1}\end{equation} where $q_0, q_1, q_2,q_3\in \mathbb{R}$ and $\mathbf{i}, \mathbf{j},\mathbf{k}$ satisfy the multiplication rule \begin{equation}\mathbf{i}^2=\mathbf{j}^...

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 \\...