Life with Physics

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 $t_H$ (...

Like most people, my first time to know the word of topology is from the context like a coffee mug has the same topology as a donut because they can be deformed “continuously” to each other if made in plasticine. Such example is very intuitive and satisfying. But when I was a senior in college, I was taught the rigorous definition of topology in a course on differential geometry for physicists. Since then I haven't really understood such rigorous definition and how it relates the picture of mug and donut. The course started with the definition of topology space via open set , followed by some examples like trivial, discrete and usual topology. Although I learned a lot from the course, I was not able to see the mug and donut in the rigorous definition. Later during the graduate study, I tried multiple times to understand some topology, for example, by reading the top rated book “ geometry, topology and physics ”. My Ph.D. advisor really loves reading this book. Although the book is ...

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)...