Ulysses’ trip | Physics

From Picard iteration to Feynman path integral

2025-11-13T17:31:24-08:00

Discrete path integral

As we all know, the Schrödinger equation is $\fr\d{\d t}\ket{\fc\psi t}=-\i\fc Ht\ket{\fc\psi t},$ where $\ket\psi$ is the state vector and $H$ is the Hamiltonian operator (may be time-dependent). This is an ordinary differential equation (ODE), so we can express its solution as a sum $\ket{\fc\psi t}=\sum_{n=0}^\infty\ket{\fc{\psi^{\p n}}t},$ where each term in the sum is defined iteratively by (see also my past article) $\ket{\fc{\psi^{\p0}}t}\ceq\ket{\fc\psi0},\quad \ket{\fc{\psi^{\p{n+1}}}t}\ceq-\i\int_0^t\d t'\,\fc H{t'}\ket{\fc{\psi^{\p n}}{t'}}.$ This iteration is called the Picard iteration, which is most known as a method to prove the Picard–Lindelöf theorem.

Let us actually write out the general term in this sum as $\ket{\fc{\psi^{\p n}}t}=\fc{K^{\p n}}t\ket{\fc\psi0},$ where $\fc{K^{\p n}}t=\p{-\i}^n\int_0^t\d t_n\,\fc H{t_n}\int_0^{t_n}\d t_{n-1}\,\fc H{t_{n-1}}\cdots\int_0^{t_2}\d t_1\,\fc H{t_1}.$ Now with the trick of time-ordering, we can rewrite this as $\begin{align*} \fc{K^{\p n}}t&=\fr{\p{-\i}^n}{n!}\int_0^t\d t_n\int_0^t\d t_{n-1}\cdots\int_0^t\d t_1\,\bfc{\mcal T}{\fc H{t_n}\fc H{t_{n-1}}\cdots\fc H{t_1}}\\ &=\fr{\p{-\i}^n}{n!}\mcal T\p{\int_0^t\d t'\,\fc H{t'}}^n, \end{align*}$ where $\bfc{\mcal T}\cdots$ means to order the operators inside according to their time arguments. The factor of $1/n!$ appears because there are $n!$ ways to order $n$ time variables, but another way to see this is to note that the domain of integration is an $n$ -simplex, whose volume is $1/n!$ of the corresponding $n$ -parallelotope. When we then sum over all $n$ , we get the time-ordered exponential $\ket{\fc\psi t}=\fc Kt\ket{\fc\psi0},\quad \fc Kt=\sum_n\fc{K^{\p n}}t=\mcal T\fc\exp{-\i\int_0^t\d t'\,\fc H{t'}}.$ The operator $\fc Kt$ has a bunch of equivalent names, such as the time evolution operator, the propagator, the Green’s function, the Dyson operator, and the S-matrix (well, they are not entirely equivalent because they are used under different contexts).

The interesting part comes when we consider the matrix elements of $\fc Kt$ and how they relate to the matrix elements of $\fc Ht$ . We choose an orthonormal basis $\B{\ket x}$ , and then insert a $\sum_x\ket x\bra x$ between each pair of Hamiltonian operators in the expression of $\fc{K^{\p n}}t$ : $\bra{x_n}\fc{K^{\p n}}t\ket{x_0} =\fr{\p{-\i}^n}{n!}\int_0^t\d^nt\sum_{x_1,\ldots,x_{n-1}} \fc{h_{x_nx_{n-1}}}{t_n}\fc{h_{x_{n-1}x_{n-2}}}{t_{n-1}}\cdots\fc{h_{x_1x_0}}{t_1},$ where we have abbreviated $\fc{h_{xy}}t\ceq\bra x\fc Ht\ket y$ , and $t_1\le\cdots\le t_n$ are a specific ordering of the integrated time variables. We can pull out the sum over intermediate basis states and call the summand a contribution from a walk ¹ from $x_0$ to $x_n$ . In other words, $\bra y\fc{K^{\p n}}t\ket x =\sum_{\B{x_i}}^{\substack{x_n=y\\x_0=x}}\fc K{\B{x_i},t},$ where the sum is over all walks of length $n$ from $x$ to $y$ , and $\fc K{\B{x_i},t}\ceq\fr{\p{-\i}^n}{n!}\int_0^t\d^nt\, \prod_{i=1}^n\fc{h_{x_ix_{i-1}}}{t_i}.$ Now, the matrix elements of the full propagator is then the sum over contributions from all walks: $\bra y\fc Kt\ket x=\sum_{\text{walks }x\to y}\fc K{\text{walk},t}.$ We can imagine a “Hamiltonian graph” formed by taking the basis states as vertices and the Hamiltonian matrix elements as edge weights (the weight of the directed edge from $x$ to $y$ is $-\i\bra y\fc Ht\ket x$ ). Note that an edge can be a self-loop. Then, the propagator contribution from a walk is given by integrating the product of all the edge weights along the walk (for a trivial walk, which has zero length, the contribution is simply $1$ ). This formulation may be called the discrete path integral. There is a paper on arXiv that delivers the idea of the Hamiltonian graph. Its difference from the current article is that it only focuses on time-independent Hamiltonians and that it treats self-loops separately instead of just like normal edges. The following two sections (excluding self-loops and the Feynman path integral) largely follow from the ideas from this paper.

Excluding self-loops

In some cases, it may be hard to consider self-loops on the Hamiltonian graph. We may then benefit from counting only walks without self-loops. However, the contributions from each walk will now be different because we have to account for the same walk with self-loops inserted at various positions. In other words, for a walk $\B{x_i}$ without self-loops, instead of contributing $\fc K{\B{x_i},t}$ , we now want to find the contribution $\fc L{\B{x_i},t}$ that sums over all ways to insert self-loops into $\B{x_i}$ : $\fc L{\B{x_i},t}=\sum_{m_0,\ldots,m_n=0}^\infty \fc K{\B{x_i^{m_i}},t},$ where $m_i$ is the number of self-loops inserted at the vertex $x_i$ , and $\B{x_i^{m_i}}$ is an abbreviation of this walk: $\underbrace{x_0,\ldots,x_0}_{m_0}, \underbrace{x_1,\ldots,x_1}_{m_1},\ldots, \underbrace{x_n,\ldots,x_n}_{m_n}.$

I will show that we can find an expression for $\fc L{\B{x_i},t}$ for the case when the Hamiltonians at different times commute with each other. In this case, we have $\fc K{\B{x_i},t}=\fr{1}{n!}\prod_{i=1}^n \p{-\i\int_0^t\d t\,\fc{h_{x_ix_{i-1}}}t}.$ Therefore, by the definition of $\fc L{\B{x_i},t}$ , we have $\fc L{\B{x_i},t}=\fc K{\B{x_i},t}\sum_{m_0,\ldots,m_n} \fr{n!}{\p{n+\sum_i m_i}!} \prod_i\p{-\i\int_0^t\d t\,\fc{h_{x_ix_i}}t}^{m_i}.$ For abbreviation, for the rest of this section, we denote $S_i\ceq-\i\int_0^t\d t\,\fc{h_{x_ix_i}}t$ .

Now, we use a trick to replace the factorial in the denominator with an contour integral: $\fr1{N!}=\fr1{2\pi\i}\oint\d z\fr{\e^z}{z^{N+1}},$ where the contour is a counterclockwise simple closed curve around the origin in the complex plane. We then have $\fr{\fc L{\B{x_i},t}}{n!\,\fc K{\B{x_i},t}} =\fr1{2\pi\i}\sum_{m_0,\ldots,m_n} \oint\d z\fr{\e^z}{z^{n+\sum_im_i+1}} \prod_i S_i^{m_i} =\fr1{2\pi\i}\oint\d z\fr{\e^z}{z^{n+1}} \prod_i\sum_{m_i}\fr{S_i^{m_i}}{z^{m_i}}.$ We have thus separated the sums over different $m_i$ . Each sum is a geometric series that converges when we choose the contour large enough, so $\fr{\fc L{\B{x_i},t}}{n!\,\fc K{\B{x_i},t}} =\fr1{2\pi\i}\oint\d z\fr{\e^z}{z^{n+1}} \prod_i\fr1{1-S_i/z} =\sum_i\fr{\e^{S_i}}{\prod_{j\ne i}\p{S_i-S_j}},$ where the last step used the residue theorem for each pole at $z=S_i$ . This expression is exactly the expanded form of the divided difference of $\B{\e^{S_i}}$ , often denoted $\e^{\b{S_0,\ldots,S_n}}$ . Therefore, $\fc L{\B{x_i},t}=n!\,\fc K{\B{x_i},t}\e^{\b{S_0,\ldots,S_n}}.$

The discrete path integral can then be rewritten in a form that only involves collecting contributions from walks without self-loops: $\bra y\fc Kt\ket x=\sum_{\text{walks }x\to y}^{\text{no self-loops}}\fc L{\text{walk},t}.$

Feynman path integral

This discrete path integral formulation of the propagator already looks similar to the Feynman path integral, but we have to go a step further to take the continuum limit to actually get there. For simplicity, I will only consider a particle with unvarying mass $m$ moving in a time-independent potential in one dimension. Its Hamiltonian is $H=p^2/2m+\fc Vx$ , and the orthonormal basis is chosen to be the position basis, also denoted as $\B{\ket x}$ .

The more standard way to derive the Feynman path integral is to slice the time integral in Equation 1, to express the total exponentiation as a product as many small exponentiations, and then to insert $\int\d x\ket x\bra x$ between each pair of exponentiations (see, e.g., chapter 6 of Quantum Field Theory by Mark Srednicki). However, this approach does not make its connection to the discrete path integral clear. Instead, we will discretize the position space into a lattice with spacing $a$ , use the discrete path integral formulation on this lattice, and then take the continuum limit $a\to0$ at the end. Now, instead of $x\in\bR$ , we have $x\in a\bZ$ . Each basis vector $\ket x$ now has two nearest neighbors $\ket{x-a}$ and $\ket{x+a}$ .

In the position basis, the kinetic part of the Hamiltonian $p^2/2m$ is a second derivative operator. From numerical differentiation, we can approximate it on the lattice as $\fr{p^2}{2m}=\fr1{2ma^2}\p{2\ket x-\ket{x+a}-\ket{x-a}}\bra x.$ Therefore, the discretized Hamiltonian is $H=\sum_x\p{\fr1{2ma^2}\p{2\ket x-\ket{x+a}-\ket{x-a}}+\fc Vx\ket x}\bra x.$ Its matrix elements are then $h_{yx}=\fr1{2ma^2}\p{2\dlt_{y,x}-\dlt_{y,x+a}-\dlt_{y,x-a}}+\fc Vx\dlt_{y,x}.$ Conceptually, it consists of on-site energy $1/ma^2+\fc Vx$ and nearest-neighbor hops. The on-site energy looks bothersome, but we can remove it if we only consider walks without self-loops. Equation 2 becomes $\bra y\fc Kt\ket x =\sum_{n=0}^\infty\p{\fr\beta{ma^2}}^n \sum_{\B{x_i}}^{\substack{x_n=y\\x_0=x}}\e^{\b{S_0,\ldots,S_n}} \prod_{i=1}^n\p{\fr12\dlt_{x_i,x_{i-1}+a}+\fr12\dlt_{x_i,x_{i-1}-a}},$ where $\beta=\i t$ is the imaginary time, and $S_i\ceq-\beta\p{1/ma^2+\fc V{x_i}}$ is defined for the same abbreviation reason as the previous section. The terms proportional to $\dlt_{x_i,x_{i-1}}$ in the multiplicant are omitted because we only consider walks without self-loops. The rest of this section is done under a Wick rotation so that $\beta$ is assumed to be a positive real parameter.

First, let us tackle the divided difference $\e^{\b{S_0,\ldots,S_n}}$ . Define $\Dlt S_i\ceq S_i-\bar S$ , where $\bar S\ceq\sum_i S/\p{n+1}$ is the mean of $\B{S_i}$ . Then, $\Dlt S_i$ is of order unity (while $S_i$ is of order $\beta/ma^2$ , which is much larger than unity for small $a$ ). Then, from the expanded form of the divided difference, we can easily get $\e^{\b{S_0,\ldots,S_n}}=\e^{\bar S}\e^{\b{\Dlt S_0,\ldots,\Dlt S_n}}.$ Recalling how we initially derived the divided difference, we have $\e^{\b{\Dlt S_0,\ldots,\Dlt S_n}}=\sum_{m_0,\ldots,m_n}\fr1{\p{n+\sum_im_i}!}\prod_i\Dlt S_i^{m_i}.$ When $n$ is large (the reason of which will be explained in a minute), we have $n!\ll\p{n+1}!\ll\p{n+2}!$ etc., while $Q_i$ is of the order of unity, so we only need to consider the terms with the lowest $\sum_im_i$ . The leading term is the term with $\sum_im_i=0$ , which is trivially $1/n!$ , so we have $\e^{\b{S_0,\ldots,S_n}}=\fr1{n!}\e^{\i\bar S} =\fr1{n!}\fc\exp{-\fr{\beta}{ma^2}-\fr{\beta}{n+1}\sum_i\fc V{x_i}}.$ This contributes to the potential part of the action.

Substitute Equation 4 into Equation 3, and we have $\bra y\fc Kt\ket x =\sum_{n=0}^\infty\fr{\e^{-\lmd}\lmd^n}{n!} \sum_{\B{x_i}}^{\substack{x_n=y\\x_0=x}} \fc\exp{-\fr{\beta}{n+1}\sum_i\fc V{x_i}}\prod_{i=1}^n\p{\fr12\dlt_\cdots+\fr12\dlt_\cdots},$ where $\lmd\ceq\beta/ma^2$ is a large positive number when $a$ is small. Observe that the factor $\e^{-\lmd}\lmd^n/n!$ is the probability mass function of the Poisson distribution with mean $\lmd$ evaluated at $n$ . When $\lmd$ is very large, the Poisson distribution can be approximated by a delta distribution because the standard deviation $\sqrt\lmd$ is much smaller than $\lmd$ . In other words, $\e^{-\lmd}\lmd^n/n!\approx\dlt_{n,\lmd}$ . Therefore, $\bra y\fc Kt\ket x =\sum_{\B{x_i}}^{\substack{x_\lmd=y\\x_0=x}} \fc\exp{-\fr{\beta}{n+1}\sum_i\fc V{x_i}} \prod_{i=1}^\lmd\p{\fr12\dlt_{x_i,x_{i-1}+a}+\fr12\dlt_{x_i,x_{i-1}-a}}.$

Imaginary parameter Poisson distribution

The role of particle indistinguishability in statistical mechanics

2025-03-03T22:53:47-08:00

Classical vs. quantum statistical mechanics

Previously, I have written two blog articles (part 1 about thermal ensembles and part 2 about non-thermal ensembles) about a formalism of statistical ensembles. I will be using it as the formalism of classical statistical mechanics in this article.

In that formalism, the space of microstates of a system is a measure space $\mcal M$ , and the physical meaning of the measure is the number of microstates. A macrostate is described by the extensive quantities, which are a function of the microstate, so designating a macrostate restricts the microstates that can realize it to a subset of $\mcal M$ .

A state of the system is a probability density function $p$ on $\mcal M$ , whose physical meaning is an ensemble of microstates. The macroscopically measured extensive quantities of the system are defined to be the ensemble average, i.e., the measured value of $A:\mcal M\to\bR$ is $\int_{\mcal M}pA$ . Generally, any probability density function is a perfectly valid state, but the most important ones are those that are thermal equilibrium states, including the microcanonical ensembles, the thermal ensembles, and the non-thermal ensembles. The term about an ensemble being thermal or non-thermal is made up by me, but for most practical reasons, we only need to focus on thermal ensembles (because both canonical ensembles and grand canonical ensembles are thermal ensembles).

To avoid subtleties about measure theory and topology, in this article, we will only use counting measure and discrete spaces for the space of microstates and the space of extensive quantities.

Possible confusion of macrostate vs. state and an example

The notational convenience of imaginary time in the derivation of the metric in Poincaré coordinates

2024-09-10T14:19:33-07:00

Introduction

There are two major conventions for the metric signature: $\p{+,-,-,-}$ (west coast) and $\p{-,+,+,+}$ (east coast). However, the first convention that I have met in my journey of learning physics is neither of them: the imaginary time. Shortly after, I started using the west coast convention, so I never really used the imaginary time convention seriously. I personally dislike the imaginary time convention, and so do most people in the physics community and history, which is why most modern textbooks use either the west coast or the east coast convention. One of my past physics teachers deemed the imaginary time convention to be a heresy (异端邪说).

However, in some cases, the imaginary time convention can be convenient due to the use of multi-index notation (which is more concise and feature-rich than the Einstein notation). Here is one of such cases: the derivation of the metric in Poincaré coordinates for the anti-de Sitter space.

The $d$ -dimensional anti-de Sitter space $\mrm{AdS}_d$ of scale $l$ is defined as the hyperboloid $-l^2=-T_1^2-T_2^2+\sum_{i=1}^{d-1}\p{X^i}^2$ in $M^{d-1,2}$ (the analogue of the Minkowski space, but with signature $d-1,2$ ). The Poincaré coordinates are defined as $\begin{align*} z&\ceq\fr{l^2}{T_1+X^{d-1}},\\ t&\ceq\fr{lT_2}{T_1+X^{d-1}},\\ x^i&\ceq\fr{lX^i}{T_1+X^{d-1}},&i=1,\ldots,d-2. \end{align*}$

The derivation

Define $T\ceq T_1$ and $X\ceq X^{d-1}$ just for fun. Then, define two $\p{d-1}$ -dimensional multi-indices $Y\ceq\p{\i T_2,X^1,\ldots,X^{d-2}},\quad y\ceq\p{\i t,x^1,\ldots,x^{d-2}}.$

The hyperboloid constraint and the metric (east coast convention) are then $X^2-T^2+Y^2=-l^2,\quad \d s^2=\d X^2-\d T^2+\d Y^2,$ which are equivalently $\p{X+T}\p{X-T}=-l^2-Y^2,\quad\d s^2=\p{\d X+\d T}\p{\d X-\d T}+\d Y^2.$ The definition of the Poincaré coordinates can be written as $z=\fr{l^2}{X+T},\quad y=\fr zlY,$ or equivalently $X+T=\fr{l^2}z,\quad Y=\fr{ly}z.$

Substitute Equation 2 into the first equation in Equation 1. Then, we have $X-T=-z-\fr{y^2}z.$ Differentiate Equation 2 and 3, and we have $\d X+\d T=-\fr{l^2}{z^2}\,\d z,\quad \d X-\d T=-\d z+\fr{y^2}{z^2}\,\d z-\fr{2y}z\,\d y,\quad \d Y=l\p{\fr{\d y}z-\fr y{z^2}\,\d z}.$ Substitute this into the second equation in Equation 1, and we have $\d s^2=-\fr{l^2}{z^2}\d z\p{-\d z+\fr{y^2}{z^2}\,\d z-\fr{2y}z\,\d y}+l^2\p{\fr{\d y}z-\fr y{z^2}\,\d z}^2 =\fr{l^2}{z^2}\p{\d y^2+\d z^2}.$

Finally, substitute back the definition of $y$ , and we have the result $\d s^2=\fr{l^2}{z^2}\p{-\d t^2+\sum_{i=1}^{d-2}\p{\d x^i}^2+\d z^2}.$

The smallest wave packet in the lowest Landau level

2024-07-01T01:01:46-07:00

Introduction

Exercise 12.5 from Modern Condensed Matter Physics (Girvin and Yang, 2019) asks to construct a Gaussian wave packet in the lowest Landau level in the Landau gauge, such that it is localized as closely as possible around some point $\mbf R\ceq\p{R_x,R_y}$ .

Actually, we can prove that the smallest wave packet is a Gaussian wave packet. Here is the derivation.

The problem

First, for readers who are not familiar with the Landau levels, here is a brief introduction. For an electron confined in the $xy$ plane under a magnetic field $\mbf B=B\bhat z$ , its Hamiltonian is $H=\fr1{2m_e}\p{p_x^2+\p{p_y-\fr{eB}cx}^2}$ under the Landau gauge $\mbf A=Bx\bhat y$ . Its eigenstates in the position representation are $\fc{\psi_{nk}}{x,y}=\e^{\i ky}\fc{H_n}{\fr xl-kl} \e^{-\p{x-kl^2}^2/2l^2}$ labeled by $n\in\bN$ and $k\in\bR$ , where $H_n$ is the Hermite polynomial of degree $n$ and $l\ceq\sqrt{\hbar c/eB}$ . States with the same $n$ are degenerate in energy ( $E_n=\p{n+1/2}\hbar eB/m_ec$ ) and make up the $n$ th Landau level. The Landau level with $n=0$ is called the lowest Landau level.

The problem, now, is this optimization problem: $\begin{align*} \min_{a_k}\quad&\mel{\Psi}{x^2+y^2}{\Psi}\\ \st\quad&\braket{\Psi}{\Psi}=1,\\ &\mel{\Psi}{x}{\Psi}=R_x,\\ &\mel{\Psi}{y}{\Psi}=R_y \end{align*}$ (optimizing $\a{x^2+y^2}$ is equivalent to optimizing $\sgm_x^2+\sgm_y^2$ because $\a x$ and $\a y$ are both fixed), where $\ket\Psi$ is defined as the state whose position representation is $\fc\Psi{x,y}=\int\d k\,a_k\e^{\i ky}\e^{-\p{x-kl^2}^2/2l^2}.$

The solution

Consider the moment-generating function $\begin{align*} \fc M{u,v}&\ceq\mel{\Psi}{\e^{ux+vy}}{\Psi}\\ &=\iint\d x\d y\,\e^{ux+vy} \int\d k\,a_k^*\e^{-\i ky}\e^{-\fr1{2l^2}\p{x-kl^2}^2} \int\d k'\,a_{k'}\e^{\i k'y}\e^{-\fr1{2l^2}\p{x-k'l^2}^2}\\ &=\iint\d k\d k'\,a_k^*a_{k'}\int\d x\,\e^{ ux-\fr1{2l^2}\p{x-kl^2}^2-\fr1{2l^2}\p{x-k'l^2}^2 }\underbrace{\int\d y\,\fc\exp{vy+\i\p{k'-k}y}}_{2\pi\fc\dlt{k'-k-\i v}}\\ &=2\pi\int\d k\,a_k^*a_{k+\i v}\underbrace{\int\d x\,\fc\exp{ ux-\fr1{2l^2}\p{x-kl^2}^2-\fr1{2l^2}\p{x-\p{k+\i v}l^2}^2 }}_{l\sqrt\pi\fc\exp{\fr14l^2\p{4ku+u^2+2\i uv+v^2}}}\\ &=2\pi^{3/2}l\fc\exp{\fr14l^2\p{u^2+2\i uv+v^2}} \int\d k\,a_k^*a_{k+\i v}\e^{kl^2u}\\ &=2\pi^{3/2}l\int\d k\,a_k^*\left( a_k+kl^2a_ku+\i a_k'v+\fr14l^2\p{1+2k^2l^2}a_ku^2 \right.\\&\qquad\qquad\qquad\qquad\left. {}+\fr14\p{l^2a_k-2a_k''}v^2 +\fr\i2l^2\p{a_k+2ka_k'}uv+\cdots \right), \end{align*}$ where $a_k'\ceq\d a_k/\d k$ and $a_k''\ceq\d^2a_k/\d k^2$ . On the other hand, we have $\fc M{u,v}=\mel{\Psi}{1+ux+uy+\fr12u^2x^2+\fr12v^2y^2+uvxy+\cdots}{\Psi}.$ Compare the expansion coefficients, and we have $\begin{align*} \braket{\Psi}{\Psi}&=2\pi^{3/2}l\int\d k\,a_k^*a_k,\\ \mel{\Psi}{x}{\Psi}&=2\pi^{3/2}l^3\int\d k\,a_k^*ka_k,\\ \mel{\Psi}{y}{\Psi}&=2\i\pi^{3/2}l\int\d k\,a_k^*a_k',\\ \mel{\Psi}{x^2}{\Psi}&=\fr12\pi^{3/2}l^3\int\d k\,a_k^*\p{1+2k^2l^2}a_k,\\ \mel{\Psi}{y^2}{\Psi}&=\fr12\pi^{3/2}l\int\d k\,a_k^*\p{l^2-2a_k''}a_k. \end{align*}$

Define $\fc\vphi k\ceq a_k\sqrt{2\pi^{3/2}l}$ . Define fictitious position and momentum operators acting on $\vphi$ as $\Xi\vphi:k\mapsto k\fc\vphi k,\quad \Pi\vphi:k\mapsto-\i\fc{\vphi'}k.$ Using the constraints of the original optimization problem and abusing the bra–ket notation on $\vphi$ , we have $\braket{\vphi}{\vphi}=1,\quad\mel\vphi\Xi\vphi=\fr{R_x}{l^2},\quad \mel\vphi\Pi\vphi=-R_y.$ The objective function then becomes $\mel{\Psi}{x^2+y^2}{\Psi}=\fr12l^2+\mel{\vphi}{\mcal H}{\vphi},$ where $\mcal H\ceq \Pi^2/2+l^4\Xi^2/2$ is a fictitious Hamiltonian, which is the Hamiltonian of a harmonic oscillator with mass $1$ and angular frequency $\omg\ceq l^2$ .

The optimization problem can now be re-stated in terms of $\ket\vphi$ as $\begin{align*} \min_{\ket\vphi}\quad&\mel\vphi{\mcal H}{\vphi}\\ \st\quad&\braket\vphi\vphi=1,\quad\mel\vphi\Xi\vphi=R_x/\omg,\quad\mel\vphi\Pi\vphi=-R_y. \end{align*}$ Physically, this means that we want to find the state of a harmonic oscillator with the given expectation values of position and momentum and the lowest energy. To find it, we can use Hisenberg’s uncertainty principle: $\begin{align*} \a{\mcal H}&=\fr12\a{\Pi^2}+\fr12\omg^2\a{\Xi^2}\\ &=\fr12\p{\a\Pi^2+\sgm_\Pi^2}+\fr12\omg^2\p{\a{\Xi^2}+\sgm_\Xi^2}\\ &=\fr12\sgm_\Pi^2+\fr12\omg^2\sgm_\Xi^2+\fr12R_y^2+\fr12 R_x^2\\ &\ge\omg\sgm_\Pi\sgm_\Xi+\fr12R^2 \ge\fr12\omg+\fr12R^2. \end{align*}$ The equality in the first “ $\ge$ ” is achieved when $\sgm_\Pi=\omg\sgm_\Xi$ , and that in the second “ $\ge$ ” is achieved when the uncertainty principle is saturated. As we know from quantum mechanics, the coherent state of a harmonic oscillator satisfies both conditions. The wavefunction of this state is $\fc\vphi k=\p{\fr\omg\pi}^{1/4} \fc\exp{-\fr12\omg\p{k-\fr{R_x}{\omg}}^2-\i R_yk}.$ Express the final result in terms of $a_k$ : $a_k=\fr1{\sqrt2\pi}\e^{-\i kR_y}\e^{-\fr1{2l^2}\p{R_x-kl^2}^2}.$ We may work out the integral to get the wave function of the wave packet: $\fc{\Psi}{x,y}=\fr1{\sqrt{2\pi}l}\fc\exp{-\fr1{4l^2}\p{ \p{x-R_x}^2+\p{y-R_y}^2-2\i\p{x+R_x}\p{y-R_y} }}.$ This is a Gaussian wave packet centered at $\mbf R$ with covariance matrix $\opc{Diag}{l^2,l^2}$ .

Further problems

The optimal wave packet is indeed Gaussian. This makes me curious about whether this is a coincidence or not.

Another thing worth noting is that this result is actually the Dirac delta wave function peaking at $\mbf R$ projected into the lowest Landau level. This was actually my first idea to solve the problem. I was like: well, isn’t the Dirac delta the smallest possible wave packet by all means? If the basis is complete, I can surely combine them into a Dirac delta, and it would be very easy to work out $a_k$ in this case. Then, I was like: nah, merely a single Landau level is not complete, so I cannot do that anyway. I then did not even bother to proceed with this approach and went on to trying other methods. It turns out that this approach is actually correct—at least it gives the same result as the correct approach.

Regularizing the partition function of a hydrogen atom

2024-06-30T21:18:12-07:00

The partition function of a hydrogen atom diverges (only considering bound states). However, we can regularize it to get finite answers. Different regularizations give the same result. They largely agree with the physical arguments for the case of the hydrogen atom at room or cold temperature, but this should be considered a mere coincidence. The results from regularized partition functions cannot generally be trusted.

The duality between two plane trajectories related by a conformal map

2023-12-22T11:19:04-08:00

I always feel amazed about how 2D physics can often be fascinating due to theorems in complex analysis. This article is about one among such cases.

Theorem. The conformal map $\fc wz$ transforms the trajectory with energy $-B$ in potential $\fc Uz\ceq A\v{\d w/\d z}^2$ into the trajectory with energy $-A$ in potential $\fc Vw\ceq B\v{\d z/\d w}^2$ .

This result is pretty amazing in that it reveals a quite implicit duality between the two potentials, and it looks very symmetric as written.

This theorem, as I know of, was first introduced in the appendix of V. I. Arnold’s book Huygens and Barrow, Newton and Hooke. Part of this article is already covered in the relevant part of the book.

Power-law central-force potentials

Before I show the proof of it, let me first introduce it by a much more well-known example.

As we all know, Bertrand’s theorem states that the only two types of central-force potentials where all bound orbits are closed are $U\propto r^{-1}$ (the Kepler problem) and $U\propto r^2$ (the harmonic oscillator). How the two potentials are special among all sorts of different central-force potentials makes people wonder if there is any connection between them. Fortunately, there is one, and it is obvious once we notice that the complex squaring transforms any center-at-origin ellipses into focus-at-origin ellipses. Inspired by this, it is easy to see that trajectories in the Kepler problem can be transformed into trajectories of harmonic oscillators under complex squaring.

You may ask, how can we notice complex squaring does the said transformation on ellipses? The observation is noticing the simple algebra $\p{z+\fr1z}^2=z^2+\fr1{z^2}+2,$ which means that the Joukowski transform $z\mapsto z+1/z$ of a unit circle simply translates under complex squaring. We can then try to generalize this to circles of other radii, whose Joukowski transformations are just ellipses! (If you remember, this is the second time Joukowski transformation appears in my blog. The first time was here.)

Then, are the Kepler problem and the harmonic oscillator the only two central-force potentials whose trajectories can be transformed into each other by a complex function? The answer is no. In fact, for any trajectory in almost any power-law central-force potential, we can take some power of it to get a trajectory in another power-law central-force potential.

This result can be summarized as follows. Taking the $\p{\alp/2+1}$ th power of a trajectory with energy $E$ in the potential $U=ar^\alp$ ( $\alp\ne-2$ ) gives a trajectory with energy $F$ in the potential $V=br^\beta$ , where $\p{\alp+2}\p{\beta+2}=4,\quad b=-\fr14\p{\alp+2}^2E,\quad F=-\fr14\p{\alp+2}^2a.$ To prove this, we just need to reparameterize the transformed trajectory in a new time coordinate $\tau$ defined as $\d\tau=\v z^\alp\,\d t$ , where $z$ is the complex position of the original trajectory. Then, by some calculation and utilizing the energy conservation, we can show that the parameter equation in terms of the new time coordinate satisfy the equation of motion we expect. I will not show the details here because they would be redundant once I prove the more general case using the same methods.

Corollaries and applications

There is an interesting special case, which is $\alp=-2$ . There is no potential that is dual to $U\propto r^{-2}$ . Another interesting case is $\alp=-4$ , which is dual to itself ( $\beta=-4$ ). It kind of means that the coefficient in the potential is “interchangeable” with the energy, and the trajectories can be derived from each other by taking the complex reciprocal.

We can get some interesting results with $a=0$ , which is just the case of a free particle, whose trajectories are all straight lines. Since in this case we necessary have $F=0$ , we can say that the zero-energy trajectory in any power-law potential is related to a straight line by a power. From this result, we can derive some interesting corollaries. For example, the zero-energy trajectory in the Kepler problem is a parabola (square of a straight line), which is well-known. The zero-energy trajectory in $U\propto-r^{-4}$ is a circle passing through the origin (reciprocal of a straight line), which is a pretty interesting not-so-well-known result.

Another interesting result is that, the deflection angle of an incident zero-energy particle scattered by the potential $U\propto-r^\alp$ is $\tht$ under paraxial limit, if $\alp=\fr{2\vphi}{\pi-\vphi},\quad\vphi=\pm\tht-2k\pi,\quad k\in\bN.$ This result can be easily derived by using the conformal transform of the real line (actually, a straight line that approaches the real line). The crucial part here is that $k$ cannot take negative integers because we need $\alp>-2$ . The reason is that, when $\alp\le-2$ , paraxial zero-energy particles are bound to sink into the origin, and thus no scattering actually happens. This small pitfall indicates that the trajectory in the dual potential is not a two-side infinite straight line, either, in that limit, in contrast to being seemingly a free particle.

Some straightforward proofs

Let’s go back to the theorem I stated at the beginning of this article.

Proof. Consider a new time coordinate $\tau$ defined as $\d\tau=\v{\d w/\d z}^2\,\d t$ . Then, the motion of $w$ satisfies $\begin{align*} m\fr{\d^2w}{\d\tau^2} &=m\fr{\d t}{\d\tau}\fr{\d}{\d t}\p{\fr{\d t}{\d\tau}\fr{\d w}{\d t}}\\ &=m\v{\fr{\d z}{\d w}}^2\fr{\d}{\d t}\p{\v{\fr{\d z}{\d w}}^2\fr{\d w}{\d z}\fr{\d z}{\d t}}\\ &=m\fr{\d z}{\d w}\p{\fr{\d z}{\d w}}^*\p{\p{\fr{\d^2z}{\d w^2}\fr{\d w}{\d z}\fr{\d z}{\d t}}^*\fr{\d z}{\d t} +\p{\fr{\d z}{\d w}}^*\fr{\d^2 z}{\d t^2}}. \end{align*}$ Here we need to substitute $\d^2 z/\d t^2$ by the equation of motion for $z$ . By computing the real and imaginary parts separately, we can derive that for any holomorphic function $f$ , the gradient of $\v f^2$ expressed as a complex number is $\nabla\v f^2=2\p{\d f/\d z}^*f$ . Therefore, the equation of motion for $z$ is $m\fr{\d^2z}{\d t^2}=-2A\fr{\d w}{\d z}\p{\fr{\d^2w}{\d z^2}}^*.$ According to series reversion, we have $\d^2 w/\d z^2=-\p{\d w/\d z}^3\d^2 z/\d w^2$ . Therefore, the equation of motion for $z$ can also be written as $m\fr{\d^2z}{\d t^2}=2A\v{\fr{\d w}{\d z}}^2\p{\fr{\d w}{\d z}}^{*2}\p{\fr{\d^2 z}{\d w^2}}^*.$ Substitute this, and we have $m\fr{\d^2w}{\d\tau^2}=\fr{\d z}{\d w}\p{\fr{\d^2z}{\d w^2}}^* \p{m\v{\fr{\d z}{\d t}}^2+2A\v{\fr{\d w}{\d z}}^2}.$ Substitute the energy conservation of the motion of $z$ : $\fr12m\v{\fr{\d z}{\d t}}^2+A\v{\fr{\d w}{\d z}}^2=-B,$ and we have $m\fr{\d^2w}{\d\tau^2}=-2B\fr{\d z}{\d w}\p{\fr{\d^2z}{\d w^2}}^*,$ which is the equation of motion for $w$ that we expect.

To get the energy of the motion of $w$ , we calculate $\begin{align*} \fr12m\v{\fr{\d w}{\d\tau}}^2+B\v{\fr{\d z}{\d w}}^2 &=\fr12m\v{\fr{\d w}{\d z}\fr{\d z}{\d t}\fr{\d t}{\d\tau}}^2+B\v{\fr{\d z}{\d w}}^2\\ &=\v{\fr{\d w}{\d z}}^2\p{-B-A\v{\fr{\d w}{\d z}}^2}\v{\fr{\d z}{\d w}}^4+B\v{\fr{\d z}{\d w}}^2\\ &=-A, \end{align*}$ which is the energy conservation of the motion of $w$ in the potential $V$ that we expect. $\square$

Noticing that we are only interested in the trajectory, we can just use Maupertuis’ principle to get a simpler proof.

Proof. $\mcal S_0=\int\v{\d z}\sqrt{2m\p{-B-A\v{\fr{\d w}{\d z}}^2}}=\int\v{\d w}\sqrt{2m\p{-A-B\v{\fr{\d z}{\d w}}^2}}.$ The abbreviated action is then exactly the same for the motion of $z$ and the motion of $w$ . Therefore, by Maupertuis’ principle, for any physical trajectory of $z$ , the trajectory of $w$ is also physical. $\square$

Details worth noting

Invertibility of the conformal map

There are two different definitions of a conformal transformation in two dimensions. One is that a function defined on an open subset of $\bC$ is conformal iff it is holomorphic and its derivative is nowhere zero. The other is that a function is conformal iff it is biholomorphic (is bijective and has a holomorphic inverse).

You may think here I have adopted the second definition because when I say $\fc Vw\ceq B\v{\d z/\d w}^2$ , I am implicitly assuming that I can take the inverse of $\fc wz$ to get the function $\fc zw$ and then take the derivative of it. However, if that is the case, an immediate problem is that then the duality between the Kepler problem and the harmonic oscillator, from which I introduced the more general result in the first place, would not be actually covered by the “more general” result. This is because $z\mapsto z^2$ is not biholomorphic (because it is not injective).

Then, why did this never become a problem when we were studying the duality between the Kepler problem and the harmonic oscillator? All we have talked about is how we can derive a trajectory in the Kepler problem by squaring the trajectory of a harmonic oscillator, but we have not discussed about how we can reverse this process, as an essential part of the duality. You may think the reverse of the process would be totally natural given how symmetric our theorem is regarding the two potentials. However, the reverse is not actually well-defined since the inverse of squaring, i.e., taking the square root, is not a single-valued function. Nevertheless, it is still well-defined in some sense: starting with whichever branch we like, tracing one point on the trajectory of the Kepler problem, and moving it along this trajectory for two cycles, we will end up with a trajectory of the harmonic oscillator if we take the square root of the position and ensure we always choose the branch so that the mapping is continuously done.

What about other power-law central potentials? In those cases, we have non-closed trajectories, so we cannot just move along the trajectory for two cycles. For example, if we take $w=z^3$ , then the potential would be $U=9A\v z^4$ . For any non-closed trajectory, we can uniquely map it to a trajectory of the potential $V=B\v w^{-4/3}/9$ . However, we cannot uniquely do the reverse mapping. There would be three different trajectories in the potential $U$ that can be mapped to the same trajectory in $V$ , and we can in turn map the trajectory in $V$ to any of the three trajectories in $U$ depending on which branch we choose.

Therefore, to generalize this for more general potentials, we can use similar arguments. Because $z\mapsto w$ has non-zero derivative everywhere in our considered region, it is everywhere locally invertible by the Lagrange inversion theorem. We can then bijectively map the trajectories in the two dual potentials locally for every small (and finite) segment and then patch them together to get the global correspondence between the two trajectories. This mapping may not be well-defined globally, but the trajectories can still be considered dual to each other. If the potential also becomes multi-valued due to the mapping $w\mapsto z$ being multi-valued, then we should imagine this situation like this: at some point, the potential may be different when the particle visit here for the second time. This case does not happen if we only look at power-law potentials, but it does happen for more general cases.

What makes this sense of duality weaker is that one trajectory can be dual to multiple different trajectories. A case worth noting is that sometimes one trajectory can be mapped to infinitely many different trajectories. This happens when the trajectory runs around a logarithmic branch point. However, we can gain the sense of duality back if we can also consider the case where $z\mapsto w$ is multi-valued. The notion of conformal transformation is now too limited to cover this case, a better notion is a global analytic function, which generalizes the notion of analytic function to allow for multiple branches.

Requirements for the potential

Not any potential can be expressed as $A\v{\d w/\d z}^2$ . How can we determine whether a potential can be expressed in this form?

Theorem. A continuous potential $U$ can be expressed in the form of $A\v{\d w/\d z}^2$ (where $\fc wz$ is a conformal transformation) iff one of the following conditions is met:

$U$ is zero everywhere, or
$\ln\v U$ is a harmonic function on the domain of $U$ .

Proof. First, prove the necessity.

An obvious requirement is that the potential must be positive everywhere or negative everywhere (or zero everywhere, but that is trivial). The sign is determined by the sign of $A$ . Therefore, without loss of generality, we can assume $A=1$ because we can always absorb a factor of $\sqrt{\v A}$ into $w$ and adjust the overall sign of $U$ accordingly.

We can decompose $\p{\d w/\d z}^2$ in the polar form $\p{\d w/\d z}^2=\v{\d w/\d z}^2\e^{\i\vphi}=U\e^{\i\vphi},$ where $\vphi$ is a real function of $z$ . Applying the Cauchy–Riemann equations to $\p{\d w/\d z}^2$ gives $\i\partial_x\p{\fr{\d w}{\d z}}^2=\partial_y\p{\fr{\d w}{\d z}}^2 \implies\i\p{\e^{\i\vphi}\partial_xU+\i U\e^{\i\vphi}\partial_x\vphi} =\e^{\i\vphi}\partial_yU+\i U\e^{\i\vphi}\partial_y\vphi.$ Equate the real and imaginary parts, and we have $\begin{cases}U\partial_x\vphi=-\partial_yU,\\U\partial_y\vphi=\partial_xU.\end{cases}$ Use the symmetry of second derivatives on $\vphi$ , and we have $\partial_x\partial_y\vphi-\partial_y\partial_x\vphi=0 \implies\partial_x\fr{\partial_xU}U+\partial_y\fr{\partial_yU}U=0.$ In the language of vector analysis, this is just $\nabla^2\ln U=0$ .

Considering the case where $U$ is negative everywhere, we have that $\ln\v U$ is a harmonic function.

Then, prove the sufficiency.

The case where $U$ is zero everywhere is trivial. Otherwise, because $\ln\v U$ is defined everywhere on the domain of $U$ , we must have $U$ is non-zero everywhere. Because $U$ is continuous, we have $U$ is either positive everywhere or negative everywhere.

Without loss of generality, assume $U$ is positive everywhere. Let $\vphi$ be the harmonic conjugate of $\ln U$ . Then, $\ln U+\i\vphi$ is a holomorphic function. We can then define $\fr{\d w}{\d z}=\sqrt U\e^{\i\vphi/2},$ which is also a holomorphic function. $\square$

From now on, we will call this requirement on $U$ as being log-harmonic for obvious resons.

We should notice that whether $U$ is log-harmonic does not respect that any potential can have an additive constant and still be essentially the same potential. An immediate example is that a function that is positive everywhere may be negative somewhere if we add a constant to it. We may then want to ask whether $U$ can be log-harmonic if we allow it to be added an additive constant. It is easy to do this: we can just apply the same test to $U+C$ , and see if there is some $C$ that makes it work. To illustrate, solve the equation $\nabla^2\ln\v{U+C}=0$ for $C$ , and then see whether it is a constant over the whole complex plane.

A property of log-harmonic functions is that the product of two log-harmonic functions is also log-harmonic.

Trajectories that run out of the domain

Trajectories often run out of the domain of the potential. For example, in the discussions about power-law potentials before, though not emphasized, the origin is outside the domain of the potential because it is either a pole or a zero of $\d w/\d z$ (except the trivial case where $w$ is simply proportional to $z$ ). Another example that is rather overlooked is that unbound trajectories go to infinity while infinity is often not in the domain of the potential, either.

What need to take care of is that, when the trajectories run out of the domain, the trajectory is cut off there, and the rest of the trajectory is never considered (even if it may come back to the domain again later). Take the Kepler problem ane the harmonic oscillator as an example. If a trajectory of the harmonic oscillator passes through the origin, which is outside the domain, the trajectory degrades from a closed ellipse to a segment. If you take the square of a segment passing through the origin, you will get a broken line folded into itself, which looks like a particle in the Coulomb field may sink into the origin and then goes back along the exact path it came along. This would confusing if it were physical.

Arbitrariness in the construction of the conformal map

The construction of $z\mapsto w$ is not unique for a given $U$ .

Rotation and translation

First, we can observe that the substitution $w\to w'\ceq w\e^{\i\tht}+w_0$ does not change $\v{\d w/\d z}$ (nor thus $U$ ). The real number $\tht$ is a function of $z$ in principle, but if we want $w$ to be holomorphic on a connected region, then $\tht$ must be a constant (except the trivial case where $w=0$ ).

The dual trajectory does change, though, but the dual potential $V$ is also changed, too. Because $\v{\d z/\d w'}=\v{\d z/\d w}$ , we have $\fc{V'}{w'}=\fc Vw=\fc V{\p{w'-w_0}\e^{-\i\tht}}.$ Therefore, the dual trajectory and the dual potential are also rotated and translated by the same amount.

Scaling

Before introducing scaling, I need to add some words about the unit systems. In the above discussions, I have never mentioned what units or dimensions do $z,w,A,B$ have. The natural way of thinking is to let $z,w$ have the dimension of length and let $A,B$ have the dimension of energy. However, this is not the only way of thinking. We will later see that the $z$ -space and the $w$ -space can have totally different dimensions.

The dimensions or units of variables in a physical formula can be totally different from what they were originally intended to be. For example, when a particle is rotating, its motion needs to satisfy $\dot{\mbf r}=\bs\omg\times\mbf r$ , where $\bs\omg$ is the angular velocity. However, although $\mbf r$ has the dimension of length when it is first introduced, this formula is satisfied by any rotating vectors. A typical example is that the angular momentum changes according to this formula when a rigid body is doing precession. For another example, in classical mechanics and general relativity, the coordinates used to describe the motion of a particle are often not in the dimension of length, but have all sorts of dimensions. For another example that is less well-known, just because the Berry connection has the same gauge transformation as the electromagnetic potential, a bunch of formulas that are useful in electromagnetic theory can be applied to the Berry connection to define all sorts of interesting quantities with rich physical implications. The units of Berry connection are, however, very unimportant because they are literally arbitrary.

Therefore, what does a unit system actually bring us in a physical theory? The only thing it brings us is the ability to conveniently see in what aspects our theories are invariant under the scaling of some quantities. For example, in classical mechanics, we can scale the mass and the potential of any system with the same factor, and then the system will still behave the same in terms of the time-dependent length-based motion. This is because the part of the dimension of energy that is independent of length and time is to the first power of the dimension of mass. For similar reasons, we can derive another two scaling invariances, one about length-scaling and the other about time-scaling. In quantum mechanics, we suffer one less such scaling invariances because of the existence of $\hbar$ ; in special relativity, we suffer one less such scaling invariances because of the existence of $c$ ; and in general relativity, we suffer two less such scaling invariances because of the existence of $G$ and $c$ . This is the incentive of introducing natural units in physics: they give us a more clear image of how our theory can be scaled leaving the physics invariant.

As for dimensional analysis, the essence of it is to find the required form of theory so that it satisfies some sort of scaling invariance. For example, we can use dimensional analysis to derive that the frequency of a harmonic oscillator is proportional to the square root of the ratio of the stiffness to the mass. We know this must be correct because this is the only theory that is consistent with the three scaling invariances that must be satisfied by any theories under the framework of classical mechanics.

Now, consider the scaling in $w$ , i.e., $w\to w'\ceq w/C$ for some non-zero real number $C$ . The potential $U$ can be kept invariant by scaling $A\to A'\ceq C^2A$ . However, we cannot change $B$ if we want to leave the trajectory of $z$ unchanged because it is determined by the energy of the trajectory of $z$ . Therefore, the dual potential $V$ would be scaled to $\fc{V'}{w'}=C^2\fc Vw=C^2\fc V{Cw'}.$ This means that physics is unchanged if length is scaled by $C$ and energy and potential are both scaled by $C^2$ . This corresponds to one of the three scaling invariances in classical mechanics that we talked about before.

What is interesting here is that the length-scaling in the $w$ -space is done independently of that in the $z$ -space. This means that the length dimension in the two systems are independent of each other, so the two systems can have totally different unit systems.

Canonical transformation of time

The transformation from $z$ to $w$ seems like a coordinate transformation, which is covered by canonical transformations. However, here we have an additional requirement about the form of the Hamiltonian: $H=\fr{p_z^2}{2m}+\fc Uz,\quad K=\fr{p_w^2}{2m}+\fc Vw,$ where $K$ is the transformed Hamiltonian (or called the Kamiltonian in the jargon of canonical transformations). This is not generally true because the transformation in the generalized momentum is restrictively determined when the transformation in the generalized coordinate is already given. From the proof of the original theorem, we can see that a transformation in time is a must, which is given by $\d\tau=\v{\d w/\d z}^2\,\d t$ .

The problem is that the canonical transformations covered in most textbooks usually do not allow for a transformation in time, but only for a transformation in the canonical variables. Therefore, I need to first address the problem of integrating the transformation of time into the theory of canonical transformations. I will not do this for the most general case, but only for the case general enough for the purpose of explaining the case interesting this article.

Change in the time variable in the stationary-action principle

Before diving into the general canonical transformation, let’s first consider the case where the transformation is only in the time variable.

Consider a system with the Lagrangian $\fc L{q,\dot q}$ (not explicitly dependent on time). Then, the action can be expressed as $S=\int_{t_1}^{t_2}\fc L{q,\dot q}\d t.$ The same integral can be expressed in terms of a new time variable $\tau$ as $S=\int_{\tau_1}^{\tau_2}\fc L{q,\mathring q\dot\tau}\fr{\d\tau}{\dot\tau},$ where $\mathring q\ceq\d q/\d\tau$ is the generalized velocity in the new time variable. The transformed Lagrangian, or what I want to call the Magrangian ¹, is then $\fc M{q,\mathring q}\ceq\fc L{q,\mathring q\dot\tau}\fr1{\dot\tau}.$ For the case that we are concerning, $\dot\tau$ is a positive real function of $q$ but does not (explicitly) depend on $t$ . The limits $\tau_1,\tau_2$ satisfy the condition $\tau_2-\tau_1=\int_{t_1}^{t_2}\fc{\dot\tau}q\,\d t.$ This relation is crucial. When finding the variation $\dlt S$ , we are fixing $t_1,t_2$ . However, we cannot fix both $\tau_1,\tau_2$ because their difference is dependent on the path $\fc qt$ . What we can do is to fix $\tau_1$ and to let $\tau_2$ have a variation given by $\dlt\tau_2=\int_{t_1}^{t_2}\fc{\dot\tau'}q\dlt q\,\d t =\int_{\tau_1}^{\tau_2}\fr{\fc{\dot\tau'}q}{\fc{\dot\tau}q}\dlt q\,\d\tau,$ where $\dot\tau'$ is the derivative (or gradient, in higher dimensions) of $\dot\tau$ as a function of $q$ . As can be seen, only if $\dot\tau$ is a constant (i.e., $\tau$ is simply an affine transform of $t$ ) does $\dlt\tau_2$ vanish for any $\dlt q$ .

Using the well-known variation of the action when there is variation in the time coordinate, we have $\dlt S=\int_{\tau_1}^{\tau_2} \p{\fr{\partial M}{\partial q}-\fr{\d}{\d\tau}\fr{\partial M}{\partial\mathring q}}\dlt q\,\d t -\fc{K}{\fc q{\tau_2},\fc{\mathring q}{\tau_2}}\dlt\tau_2,$ where $\fc K{q,\mathring q}\ceq\mathring q\fr{\partial M}{\partial\mathring q}-M$ is the energy (or the Kamiltonian, but as a function of generalized coordinates and velocities) of the system.

A quick check of this variation

You can replace $\mathrm i$ with $-\mathrm i$ in the Schrödinger equation?

2023-10-18T09:57:54-07:00

This article is adapted from the letter that I wrote to my professor of quantum mechanics. Background: the professor asked the class why the infinitesimal translation $I-\mathrm i P\varepsilon$ instead of $I+\mathrm i P\varepsilon$ (here $P$ is the momentum operator). I pointed out immediately that this is not a legitimate question to ask because we can freely replace $\mathrm i$ with $-\mathrm i$ in the Schrödinger equation. The original letter was sent at 2023-10-10 16:42 -0700.

Hi! I said in today’s class that it is just a random choice whether we use $\mathrm i$ or $-\mathrm i$ . Here is the justification:

First, mathematically, conjugation is an automorphism of $\mathbb C$ (in the sense of being a field). This fact can be easily verified. It can be easily understood by considering $\mathbb C$ as the extension field $\mathbb R[X]/(X^2+1)$ . Furthermore, due to this fact, all theorems in complex analysis are still valid if we replace every number by its conjugate.

Then, consider replacing $-\mathrm i$ with $\mathrm i$ in the SE, namely changing $\psi' = -\mathrm iH\psi$ into $\psi' = \mathrm iH\psi$ . Due to the mathematical fact above, the new SE should lead to exactly the same theory as our familiar QM because all physically meaningful quantities are real (so that their conjugate are still themselves). The solution to the SE will be $\psi = \psi_0\exp(\mathrm iHt)$ instead of $\psi = \psi_0\exp(-\mathrm iHt)$ , and they are exactly the same except an opposite phase (which does not matter) (given that $\psi_0$ also becomes its original counterpart’s conjugate in the new theory, where by saying “conjugate” here I mean taking the conjugate of all of its coordinates under the basis of eigenvectors of $H$ ).

What about time reversal? The time reversal is $t\to-t$ in the SE, which is actually slightly different from $\mathrm i\to-\mathrm i$ because when doing the latter I also assume that we make $\psi_0$ its conjugate, while $t\to-t$ leaves $\psi_0$ unchanged. However, the close connection between conjugate and time reversal does give us a hint about what the T-symmetry looks like in QM: QM does have T-symmetry, but $T$ cannot be a linear operator because it unavoidably involves conjugation. Actually, conjugation often does look like time reversal. For example, $[X,P]=\mathrm i$ becoming $[X,P]=-\mathrm i$ can be either due to conjugation (the $\mathrm i\to-\mathrm i$ here) or due to time reversal ( $P\to-P$ while $X$ unchanged).

Other than saving some minus signs here or there, there is actually a benefit (though minor) about replacing our familiar QM with its conjugate: this makes equations in QM have the same convention as in electrical engineering. Specifically, QM uses $\exp(-\mathrm i E t)$ while EE uses $\exp(\mathrm i\omega t)$ . I don’t know why, but conventions in EM seem to be the same as in QM because they also use $\exp(-\mathrm i\omega t)$ . It seems strange that EE does not use the same conventions in EM.

Back to where this topic was brought up: why is infinitesimal translation identity minus $\mathrm i P \varepsilon$ instead of plus? The answer to this question is the choice we made when we wrote the SE, which is just a matter of convention. The question that can be genuinely asked is this: why is infinitesimal translation identity minus $\varepsilon \mathrm d/\mathrm dx$ instead of plus? The arguments made in class are then valid to answer this question.

A measure-theoretic formulation of statistical ensembles (part 2)

2023-05-01T16:26:42-07:00

This article follows part 1.

Introduction

In part 2, I will focus on non-thermal ensembles.

Before I proceed, I need to clarify that almost all ensembles that we actually use in physics are thermal ensembles, including the microcanonical ensemble, the canonical ensemble, and the grand canonical ensemble (the microcanonical ensemble can be considered as a special case of thermal ensemble where $\vec W^\parallel$ is the trivial).

The theory of thermal ensembles is built by letting the system in question be in thermal contact with a bath. Similarly, if we let the system in question be in non-thermal contact with a bath, we can get the theory of non-thermal ensembles. An example of non-thermal ensembles that is actually used in physics is the isoenthalpic–isobaric ensemble, where we let the system in question be in non-thermal contact with a pressure bath.

However, we will see that it is harder to measure-theoretically develop the theory of non-thermal ensembles if we continue to use the same method as in the theory of thermal ensembles.

Introducing non-thermal contact with an example

A thermal contact is a contact between thermal system that conducts heat (while exchanging some extensive quantities). A non-thermal contact is a contact between thermal system that does not conduct heat (while exchanging some extensive quantities). For reversible processes, thermodynamically and mathematically, heat is equivalent to a form of work, where the entropy is the displacement and where the temperature is the force. However, this is not true for non-reversible processes because of the Clausius theorem. This should have something to do with the fact that entropy is different from other extensive quantities (as is illustracted in part 1).

First, I may introduce how we may cope with the reversible processes of two subsystems in non-thermal contact in thermodynamics. As an example, consider a tank of monatomic ideal gas separated into two parts by a thermally non-conductive, massless, incompressible plate in the middle that can move. The two parts can then adiabatically exchange energy ( $U$ ) and volume ( $V$ ) but not number of particles ( $N$ ). For one of the parts, we have $0=\delta Q=\mathrm dU+p\,\mathrm dV=\mathrm dU+\frac{2U}{3V}\,\mathrm dV,$ which is good and easy to deal with because it is simply a differential 1-form.

However, this convenience is not possible for non-reversible processes because then we do not have the simple relation $p=2U/3V$ . Actually, the pressure is only well-defined for equilibrium states, and it is impossible to define a pressure that makes sense during the whole non-reversible process, which involves non-equilibrium states. Therefore, although it seems that the “thermally non-conductive” condition imposes a stronger restriction on what states can the composite system reach without external sources, it actually does not because the energy exchanged by the subsystems when they exchange volume is actually arbitrary (as long as it does not violate the second law of thermodynamics) if the process is not reversible.

The possible states of the non-thermally composite system then cannot be simply described by a vector subspace of $W^{(1)}\times W^{(2)}$ . If we try to use the same approach as constructing the thermally composite system to construct the non-thermally composite system, the attempt will fail.

Continuing with our example of a tank of gas. Although the pressure is not determined in the non-reversible process, there is one thing that is certain: the pressure on the plate by the gas on one side is equal to the pressure on the plate by the gas on the other side. This is because the plate must be massless (otherwise its kinetic energy would be an external source of energy; also, remember that it is incompressible: this means that it cannot be an external source of volume). Therefore, the relation between the volume exchanged and the energy exchanged is determined as long as at least one side of the plate is undergoing a reversible process because then the reversible side has determined pressure, which determines the pressure of the other side.

This is the key idea of formulating the non-thermal ensembles without formulating the non-thermally composite system. In a thermal or non-thermal ensemble, the composite system consists of two subsystems, one of which is the system in question, and the other is the bath which we are in control of. We can let the bath have zero relaxation time (the time for it to reach thermal equilibrium) so that any process of it is reversible. Then, the pressure (or generally, any other intensive quantities that we are in control of times the temperature) is determined (and actually constant), and we can express the non-conductivity restriction as $\mathrm dU+p\,\mathrm dV=0,$ where $p$ is the pressure, which is a constant. This is a homogeneous linear equation on $\vec W^{\parallel(1)}$ (whose vectors are denoted as $(\mathrm dU,\mathrm dV)$ in our case) which defines a vector subspace of $\vec W^{\parallel(1)}$ , which we call $\vec W^{\parallel\parallel(1)}$ . The dimension of $\vec W^{\parallel\parallel(1)}$ is that of $\vec W^{\parallel(1)}$ minus one. The physical meaning of $\vec W^{\parallel\parallel(1)}$ in this example is the hyperplane of fixed enthalpy.

Note that our bath actually has the fixed intensive quantities $i=\left(1/T,p/T\right)\in\vec W^{\parallel(1)\prime}$ , we can rewrite the above equation as $\vec W^{\parallel\parallel(1)} =\left\{s_1\in\vec W^{\parallel(1)}\,\middle|\,i\!\left(s_1\right)=0\right\}.$ Wait! What does $T$ do here? It is supposed to mean the temperature of the bath, but the temperature of the bath is irrelevant since the contact is non-thermal. Actually, it is. The temperature of the bath serves as an overall constant factor of $i$ , which does not affect $\vec W^{\parallel\parallel(1)}$ as long as it is not zero or infinite. So far, this means that the temperature of the bath is not necessarily fixed, so the actual number of fixed intensive quantities is the dimension of $\vec W^{\parallel(1)\prime}$ minus one, which is the same as the dimension of $\vec W^{\parallel\parallel(1)}$ . Later we will see that anything that is relevant to the temperature of the bath will finally be irrelevant to our problem. This seems magical, but you will see the sense in that after we introduce another way of developing the non-thermal ensembles (that do not involve baths and non-thermal contact) later.

We can define a complement of $\vec W^{\parallel\parallel(1)}$ in $\vec W^{\parallel(1)}$ as $\vec W^{\parallel\perp(1)}$ . Then, we have $\vec W^{\parallel(1)}=\vec W^{\parallel\parallel(1)}+\vec W^{\parallel\perp(1)}$ . The space $\vec W^{\parallel\perp(1)}$ is a one-dimensional vector space.

For convenience, define $W^{\star\perp(1)}\coloneqq W^{\perp(1)}+\vec W^{\parallel\perp(1)}$ . The vector space $\vec W^{\star\perp(1)}$ associated with it is a complement of $\vec W^{\parallel\parallel(1)}$ in $\vec W^{(1)}$ . To make the notation look more consistent, we can use $\vec W^{\star\parallel(1)}$ as an alias of $\vec W^{\parallel\parallel(1)}$ . They are the same vector space, but $\vec W^{\star\parallel(1)}$ emphasizes that it is a subspace of $\vec W^{(1)}$ , and $\vec W^{\parallel\parallel(1)}$ emphasizes that it is a subspace of $\vec W^{\parallel(1)}$ . Then, we have $W^{(1)}=W^{\star\perp(1)}+\vec W^{\star\parallel(1)}$ . Every point in $W^{(1)}$ can be uniquely written as a sum of a point in $W^{\star\perp(1)}$ and a vector in $\vec W^{\star\parallel(1)}$ . We can describe the decomposition by a projection $\pi^{\star(1)}:W^{(1)}\to W^{\star\perp(1)}$ .

We will heavily use the “ $\star$ ” on the superscripts of symbols. Any symbol labeled with “ $\star$ ” is dependent on $i$ (but independent on an overall constant factor on $i$ ). You can regard those symbols to have an invisible “ $i$ ” in the subscript so that you can keep in mind that they are dependent on $i$ .

Example. Suppose we have a tank of gas with three extensive quantities $U,V,N$ . It is in non-thermal contact with a pressure bath with pressure $p$ so that it can exchange $U$ and $V$ with the bath. Then, the projection $\pi^{\star(1)}$ projects macrostates with the same enthalpy and number of particles into the same point. Because a complement of a vector subspace is not determined, there are multiple possible ways of constructing the projection. One possible way is $\pi^{\star(1)}\!\left(U,V,N\right)\coloneqq\left(U+pV,0,N\right).$ Here the fixed intensive quantity $p$ is involved. Note that this projection is still valid for different temperatures of the bath, so an overall constant factor of $i$ does not affect the projection.

Non-thermal contact with a bath

Now, after introducing non-thermal contact with an example, we can now formulate the non-thermal contact with a bath.

Suppose we have a system $\left(\mathcal E^{(1)},\mathcal M^{(1)}\right)$ . The main approach is constructing a composite system out of the composite system for the $\vec W^{\parallel(1)}$ -ensemble.

The composite system for the $\vec W^{\parallel(1)}$ -ensemble was introduced in part 1. We denote the bath that is in contact with our system as $\left(\mathcal E^{(2)},\mathcal M^{(2)}\right)$ .

Consider this projection $\pi^\star:W\to W^{\star\perp}$ (where $W^{\star\perp}$ is an affine subspace of $W$ and the range of $\pi^\star$ ): $\pi^\star\!\left(e_1,e_2\right) \coloneqq\left(\pi^{\star(1)}\!\left(e_1\right), \rho_{\pi(e_1,e_2)}\!\left(\pi^{\star(1)}\!\left(e_1\right)\right)\right).$ To ensure that it is well-defined, we need to guarantee that $\pi^{\star(1)}\!\left(e_1\right)\in W^{\parallel(1)}_{\pi(e_1,e_2)}$ for any $e_1,e_2$ , and this is true.

The two spaces $W^{\star\perp}$ and $W^{\perp}$ do not have any direct relation. The only relation between them is that the dimension of $W^{\star\perp}$ is one plus the dimension of $W^{\perp}$ (if they are finite-dimensional).

What is good about the projection $\pi^\star$ is that it satisfies $\vec W^{\star\parallel(1)}=\vec c^{(1)}\!\left(\vec\pi^\star(0)\right)$ . This makes our notation consistent if we construct another composite system out of $\pi^\star$ . Now, consider the composite system of $\left(\mathcal E^{(1)},\mathcal M^{(1)}\right)$ and $\left(\mathcal E^{(2)},\mathcal M^{(2)}\right)$ under the projection $\pi^\star$ . In the notation of the spaces and mappings that are involved in the newly constructed composite system, we write “ $\star$ ” in the superscript.

Just like how $\vec W^{\star\parallel(1)}$ is a subspace of $\vec W^{(1)}$ , $\vec W^{\star\parallel(2)}$ is also a subspace of $\vec W^{(2)}$ . This means that both $\vec\rho^{-1}\circ\vec\rho^\star$ and $\vec\rho\circ\vec\rho^{\star-1}$ are well-defined. The former maps $\vec W^{\star\parallel(1)}$ to another subspace of $\vec W^{(1)}$ , and the latter maps $\vec W^{\star\parallel(2)}$ to another subspace of $\vec W^{(2)}$ .

We can regard the construction of the new composite system as replacing the “plate” between the subsystems in the original composite system from a “thermally conductive plate” to a “thermally non-conductive plate”. Suppose that in the new situation, the intensive quantities “felt” by subsystem 1 is $i^\star\in\vec W^{\star\parallel(1)\prime}$ . Then, because the bath is still the same bath in the two situations, we have $-i^\star\circ\vec\rho^{\star-1}=-i\circ\vec\rho^{-1}.$ Therefore, $i^\star\coloneqq i\circ\vec\rho^{-1}\circ\vec\rho^\star$ would be a good definition of $i^\star$ . However, actually $i^\star$ is trivial: $i^\star=0.$ This is because 2 shows that $\rho\!\left(W^{\star\parallel(1)}_e\right)=W^{\star\parallel(2)}_e$ , and thus $\vec\rho^{-1}\!\left(\vec\rho^\star\!\left(\vec W^{\star\parallel(1)}\right)\right) =\vec W^{\star\parallel(1)},$ which is the kernel of $i$ by definition.

Because $i^\star$ is trivial, it is irrelevant to the temperature of the bath because it is zero no matter what temperature the bath is at.

Example. Suppose a system described by $U_1,V_1,N_1$ is in non-thermal contact with a pressure bath, and they can exchange energy and volume. The projection $\pi$ is $\pi\!\left(U_1,V_1,N_1,U_2,V_2,N_2\right) =\left(\frac{U_1+U_2}2,\frac{V_1+V_2}2,N_1,\frac{U_1+U_2}2,\frac{V_1+V_2}2,N_2\right).$ Then, the projection $\pi^\star$ can be $\pi^\star\!\left(U_1,V_1,N_1,U_2,V_2,N_2\right) =\left(U_1+pV_1,0,N_1,U_2-pV_1,V_1+V_2,N_2\right).$ By choosing a different $\pi^{\star(1)}$ or a different $\pi$ , we can get a different $\pi^\star$ . They physically mean the same composite system.

The space $W^\perp$ is four-dimensional, and the space $W^{\star\perp}$ is five-dimensional. We can denote the five degrees of freedom as $U,V,H_1,N_1,N_2$ , where $U\coloneqq U_1+U_2$ is the total energy, $V\coloneqq V_1+V_2$ is the total volume, and $H_1\coloneqq U_1+pV_1$ is the enthalpy of subsystem 1. Then, the projection $\pi^\star$ can be written as $\pi^\star\!\left(U_1,V_1,N_1,U_2,V_2,N_2\right) =\left(H_1,0,N_1,U-H_1,V,N_2\right).$ We can get $W^{\star\parallel}_e$ by finding the inverse of the projection, where $e\coloneqq\left(H_1,0,N_1,U-H_1,V,N_2\right)$ : $W^{\star\parallel}_e\coloneqq\pi^{\star-1}\!\left(e\right) =\left\{\left(H_1-pV_1,V_1,N_1,U-H_1+pV_1,V-V_1,N_2\right)\middle|\,V_1\in\mathbb R\right\}.$ Because it is parameterized by one real parameter $V_1$ , it is a one-dimensional affine subspace of $W$ . Projecting it under $c^{(1)}$ and $c^{(2)}$ will respectively give us $W^{\star\parallel(1)}_e$ and $W^{\star\parallel(2)}_e$ : $W^{\star\parallel(1)}_e \coloneqq\left\{\left(H_1-pV_1,V_1,N_1\right)\middle|\,V_1\in\mathbb R\right\},$ $W^{\star\parallel(2)}_e \coloneqq\left\{\left(U-H_1+pV_1,V-V_1,N_2\right)\middle|\,V_1\in\mathbb R\right\}.$

The affine isomorphism $\rho^\star_e$ is then naturally $\rho^\star_e\!\left(H_1-pV_1,V_1,N_1\right)=\left(U-H_1+pV_1,V-V_1,N_2\right).$ Its vectoric form is then $\vec\rho^\star\!\left(-p\,\mathrm dV_1,\mathrm dV_1,0\right) =\left(p\,\mathrm dV_1,-\mathrm dV_1,0\right).$

Our fixed intensive quantities are $i$ , which is defined as $i\!\left(\mathrm dU_1,\mathrm dV_1,0\right)=\frac1T\,\mathrm dU_1+\frac pT\,\mathrm dV_1$ . We can then get $i^\star$ by $i^\star\coloneqq i\circ\vec\rho^{-1}\circ\vec\rho^\star =\left(-p\,\mathrm dV_1,\mathrm dV_1,0\right)\mapsto0.$ This is consistent with Equation 4.

Non-thermal ensembles (bath version)

Now, we can define the non-thermal contact with a bath to be the same as the thermal contact with a bath under $\pi^\star$ . Utilizing this definition, we can define the composite system for non-thermal ensembles.

Definition. A composite system for the non-thermal $\vec W^{\parallel(1)}$ -ensemble of the system $\left(\mathcal E^{(1)},\mathcal M^{(1)}\right)$ with fixed intensive quantities $i$ is the same as the composite system for the thermal $\vec W^{\star\parallel(1)}$ -ensemble with fixed intensive quantities $i^\star=0$ (given by Equation 4), where $\vec W^{\star\parallel(1)}$ is defined by Equation 1.

This definition looks very neat. Also, just like how we define the domain of fixed intensive quantities of a thermal ensemble, we can define the domain of fixed intensive quantities of a non-thermal ensemble to consist of those values that make the integral in the definition of the partition function converge.

Because we already derived the formula of the partition function in part 1 that does not involve information about the bath anymore, we can drop the “ $(1)$ ” in the superscripts. The partition function of the non-thermal ensemble is then $Z^\star\!\left(e,i^\star\right)=\int_{s\in\vec E^{\star\parallel}_e} \Omega\!\left(e+s\right) \mathrm e^{-i^\star\left(s\right)}\,\mathrm d\lambda^{\parallel}\!\left(s\right),\quad e\in E^{\star\perp},\quad i^\star\in I^\star_e\subseteq\vec W^{\star\parallel\prime}.$ Here, the $i^\star$ is not fixed at the trivial value $0$ (I abused the notation here) but actually is an independent variable serving as one of the arguments of the partition function that takes values in $I^\star_e$ (which is not the domain of fixed intensive quantities of the non-thermal ensemble that was mentioned above).

However, the only meaningful information about this non-thermal ensemble is in the behavior of $Z^\star$ at $i^\star=0$ instead of any arbitrary $i^\star\in I^\star_e$ , but we do not know whether $0\in I^\star_e$ or not. This is then a criterion of judge whether $i$ is in the domain of fixed intensive quantities of the non-thermal ensemble or not. To be clear, we define $J\coloneqq\left\{i\in\vec W^{\parallel\prime}\,\middle|\, \exists e\in E^{\star\perp}:0\in I^\star_{e}\right\}.$ A problem about this formulation is that it is possible to have two $i$ ’s that share the same thermal equilibrium state. In that case, the non-thermal ensemble is not defined.

Because $i^\star=0$ , the observed extensive quantities in thermal equilibrium are just $\varepsilon^\circ =e+\left.\frac{\partial\ln Z^\star\!\left(e,i^\star\right)}{\partial i^\star}\right|_{i^\star=0} =e+\frac{\int_{s\in\left(E-e\right)\cap\vec W^{\star\parallel}} s\Omega\!\left(e+s\right)\mathrm d\lambda^{\parallel}\!\left(s\right)} {\int_{s\in\left(E-e\right)\cap\vec W^{\star\parallel}} \Omega\!\left(e+s\right)\mathrm d\lambda^{\parallel}\!\left(s\right)},$ and the entropy in thermal equilibrium is just $S^\circ=\ln Z^\star\!\left(e,0\right) =\ln\int_{s\in\left(E-e\right)\cap\vec W^{\star\parallel}} \Omega\!\left(e+s\right)\mathrm d\lambda^{\parallel}\!\left(s\right).$ We can cancel the parameter $e$ by Equation 5 and 6 to get $S^\circ=\ln Z^\star\!\left(\pi^\star\!\left(\varepsilon^\circ\right),0\right) =\ln\int_{s\in\left(E-\varepsilon^\circ\right)\cap\vec W^{\star\parallel}} \Omega\!\left(\varepsilon^\circ+s\right)\mathrm d\lambda^{\parallel}\!\left(s\right).$

What is interesting about Equation 7 is that it actually does not guarantee the intensive variables to be defined in $\vec W^\parallel$ . Physically this means that the temperature is not necessarily defined, unlike the case of thermal ensembles (this is because the thermal contact makes the temperature the same as the bath and thus defined). The thing that is guaranteed is that the intensive variables are defined in $\vec W^{\star\parallel}$ and they must be zero. Therefore, whenever the intensive variables are defined in $\vec W^\parallel$ , it must be parallel to $i$ (and remains the same if we scale $i$ by an arbitrary non-zero factor). Physically, this means that the system must have the same intensive variables as the bath up to different temperatures.

Non-thermal ensembles (non-bath version)

It may seem surprising that we can define non-thermal ensembles without a bath. How is it possible to fix some features about the intensive variables without a bath? The inspiration is looking at Equation 1. We can make a guess here: if we contract the system along $\vec W^{\star\parallel}$ , the contraction satisfy the equal a priori probability principle. We make this guess because of the following arguments:

Mathematically, contraction is a legal new system, so it should also satisfy the axioms that we proposed before.
Physically, because the temperature of the bath is arbitrary, the different accessible macrostates should not be too different because otherwise the temperature would matter (as appears in the expression of the partition function).

After finding the equilibrium state of the contraction, we can use the contractional pullback to find the equilibrium state of the original system.

If you do it right, you should get the same answer as Equation 7.

Summary

The only axiom that we used is the equal a priori probability principle. Then, we formulated three types of ensembles: microcanonical, thermal, and non-thermal.

A measure-theoretic formulation of statistical ensembles (part 1)

2023-03-30T21:49:51-07:00

I feel that the process of using statistical ensembles to find properties of thermal system is not rigorous enough. There are some operations that need to be defined precisely. Also, it is not generalized enough. Currently, the only generally used statistical ensembles are the microcanonical ensemble, the canonical ensemble, and the grand canonical ensemble, but there are actually other possible ensembles that are potentially useful. Therefore, I feel it necessary to try to have a mathematical formulation.

Mathematical tools and notations

Suppose $(\Omega,\sigma(\Omega),P)$ is a probability space. Suppose $W$ is an affine space. For some map $f:\Omega\to W$ , we define the $P$ -expectation of $f$ as $\mathrm E_P\!\left[f\right]\coloneqq\int_{x\in\Omega}\left(f(x)-e_0\right)\mathrm dP(x)+e_0,$ where $e_0\in W$ is arbitrary. Here the integral is Pettis integral. The expectation is defined if the Pettis integral is defined, and it is then well-defined in that it is independent of the $e_0$ we choose.

Suppose $X,Y$ are Polish spaces. Suppose $(Y,\sigma(Y),\mu),(X,\sigma(X),\nu)$ are measure spaces, where $\mu$ and $\nu$ are σ-finite Borel measures. Suppose $\pi:Y\to X$ is a measurable map so that $\forall A\in\sigma(X):\nu(A)=0\Rightarrow\mu\!\left(\pi^{-1}\!\left(A\right)\right)=0.$ Then, for each $x\in X$ , there exists a Borel measure $\mu_x$ on the measurable subspace $\left(\pi^{-1}(x),\sigma\!\left(\pi^{-1}(x)\right)\right)$ , such that for any integrable function $f$ on $Y$ , $\int_{y\in Y}f\!\left(y\right)\mathrm d\mu(y) =\int_{x\in X}\mathrm d\nu(x)\int_{y\in\pi^{-1}(x)}f\!\left(y\right)\mathrm d\mu_x(y).$

Proof

Even solutions to bound states in an odd number of $\delta$ potential wells

2023-02-02T12:43:33-08:00

This article is translated from a Chinese article on my Zhihu account. The original article was posted at 2021-01-13 15:33 +0800.

We try solving the even function solutions to the time-independent Schrödinger equation for the potential $V=-\alpha\sum_{j=-n}^n\delta\!\left(x-ja\right)$ such that $E<0$ (bound states).

Obviously, the solutions have the form $\psi= \begin{cases} A_{\left\lfloor\frac{\left|x\right|}{a}\right\rfloor}\mathrm e^{\kappa\left|x\right|} +B_{\left\lfloor\frac{\left|x\right|}{a}\right\rfloor}\mathrm e^{-\kappa\left|x\right|}, &\left|x\right|na, \end{cases}$ where $A_j,B_j$ ( $j=0,1,\ldots,n$ ) are constants of integration, and $\kappa\coloneqq\frac{\sqrt{-2mE}}{\hbar}$ .

Noting that we are finding bound states, we should have $\lim_{x\to\infty}\psi=0$ . Therefore, $A_n=0.$ Function $\psi$ is naturally continuous at $x=0$ . Considering the continuity of $\psi$ at $\left|x\right|=ja$ ( $j=1,2,\ldots,n$ ), we have $A_{j-1}\mathrm e^{\kappa ja}+B_{j-1}\mathrm e^{-\kappa ja}=A_j\mathrm e^{\kappa ja}+B_j\mathrm e^{-\kappa ja}.$

For $j=-n,\ldots,n$ , integrate both sides of the time-independent Schrödinger equation over interval $\left[ja-\varepsilon,ja+\varepsilon\right]$ and let $\varepsilon\to0$ , and we have $\left.\frac{\mathrm d\psi}{\mathrm dx}\right|_{ja^-}^{ja^+}=-\beta\left.\psi\right|_{ja},$ where $\beta\coloneqq\frac{2m\alpha}{\hbar^2}$ .

For $j=0$ , the formula above gives $\left(A_0\kappa-B_0\kappa\right)-\left(-A_0\kappa+B_0\kappa\right)=-\beta\left(A_0+B_0\right).$

For $j=1,2,\ldots,n$ , on the other hand, $\left(A_j\kappa\mathrm e^{\kappa ja}-B_j\kappa\mathrm e^{-\kappa ja}\right) -\left(A_{j-1}\kappa\mathrm e^{\kappa ja}-B_{j-1}\kappa\mathrm e^{-\kappa ja}\right) =-\beta\left(A_j\mathrm e^{\kappa ja}+B_j\mathrm e^{-\kappa ja}\right).$

Equations 1, 2, 3, and 4 together form a homogeneous linear equation w.r.t. $A_j,B_j$ ( $j=0,1,\ldots,n$ ). To require that the equation has non-zero solutions, the determinant of the coefficient matrix should be zero, and we can find $\kappa$ by this property. However, the equation is transcendental for $n>0$ .

If we found the value of $\kappa$ , the solution space for the homogeneous linear equation should be one-dimensional, and then we can determine all the constants by normalizing $\psi$ .

Defend our earth against aliens’ bullets!

2023-02-01T11:11:11-08:00

This article solves the first part of the problem proposed in a Chinese article on my Zhihu account. The original article was posted at 2020-08-30 18:27 +0800.

The aliens intiated their attack to the earth! They shoot bullets with mass $m$ and speed $v$ from a far-awar planet. To defend, humans built a field $U=\alpha/r$ that can repel the bullets. What regions are safe?

Every possible trajectory of the bullet is parameterized by $b$ , the impact parameter. The bullet has energy $E=\frac12mv^2$ and angular momentum $M=mvb$ , which are conserved. According to the well-known results of Kepler problem, the trajectory is a hyperbola $-\frac pr=1+e\cos\varphi,$ where $p\coloneqq\frac{M^2}{m\alpha},\quad e\coloneqq\sqrt{1+\frac{2EM^2}{m\alpha^2}}.$ For convenience, denote the radius of the hyperbola as $a\coloneqq\frac\alpha{2E},$ then we can write the equation of the trajectory as $-\frac{b^2}{ar}=1+\sqrt{1+\frac{b^2}{a^2}}\cos\varphi.$ Rotate the trajectory so that the incident direction is always towards the positive $x$ direction: $0=F\!\left(r,\varphi,b\right)\coloneqq\frac{b^2}{ar}+1+\cos\varphi+\frac ba\sin\varphi.$ To find the envelope of the family of trajectories, solve $0=\frac{\partial F}{\partial b}=\frac{2b}{ar}+a\sin\varphi,$ and we have $b=-\frac12r\sin\varphi.$ Substitute back into Equation 1, and we have finally the equation of the envelope: $\frac{4a}r=1-\cos\varphi,$ which is a parabola with the the semi-latus rectum being $4a$ . Therefore, the safe regions are the interior of a circular paraboloid.

How to construct mechanics in higher dimensions?

2022-11-20T17:56:52-08:00

This article is translated from a Chinese article on my Zhihu account. The original article was posted at 2020-05-28 21:24 +0800.

The spacetime of higher dimensions that we mention here refers to a Galileo universe with $\iota$ time dimensions and $\chi$ space dimensions. It has affine structure, and we manually define a coordinate system on it. Galilean transformations include uniform-velocity motion (note that velocities are matrices (see below)), spacetime translation, and space rotation. The principle of Galilean relativity still holds.

The universe is $\left(\iota+\chi\right)$ -dimensional, manually separated into an $\iota$ -dimensional subspace and a $\chi$ -dimensional subspace, where the former is called the time space, and the latter is called the space space. World points (events) are described by the combination of time coordinates and space coordinates.

Note that, when we find the derivative of a function w.r.t. time, we need to distinguish the total partial derivative and the partial partial derivative. The former regards all variables as functions of time, while the latter one does not. We denote the total partial derivative of function $F$ w.r.t. the $j$ th time coordinate as $\partial_jF$ , and the partial partial derivative as $\frac{\partial F}{\partial t_j}$ .

To avoid confusion, there is an example. For example, the total partial derivative of $F\!\left(q,t\right)$ w.r.t. $t_j$ is $\partial_jF=\frac{\partial F}{\partial q}\partial_jq+\frac{\partial F}{\partial t_j}$ . We can see that the total partial derivative has an extra term in addition to the partial partial derivative, which originates from the change of other independent variables of the function due to the change of time.

A system with $s$ DOF needs $s$ multivariable functions $q_k\!\left(t\right)$ to describe, where $k$ is the subscript, and $t\in\mathbb R^\iota$ . The numbers $q_k$ are called generalized coordinates. Generalized coordinates are mappings from the time space to the space space.

Then, the generalized velocities become matrices, whose each component represents each generalized coordinate changes w.r.t. each component of time coordinates. Written explicitly, it is $\partial_jq_k$ . It can be regarded as the Jacobian matrix of generalized coordinates. We may think that the generalized velocities span a space called the velocity space.

Just like traditional universe with one time dimension, we define the action $\mathcal S$ as the volume integral of the function $\mathcal L$ in the time space, where the Lagrangian $\mathcal L$ is a scalar function defined on the space-velocity-time phase space. Written explicitly, it is $\mathcal S\coloneqq\int_M\mathcal L\\,\mathrm dV_t$ , where $M\subseteq\mathbb R^\iota$ is a region in the time space, and $\mathrm dV_t$ is the volume element in the time space.

Now, Hamilton’s principle still holds. What it says is that, if we regard $\mathcal S$ as a functional of the function $q$ , then the problem of finding the actual motion $q$ of the system is equivalent to solve the optimization problem: constraint the value of $q$ on $\partial M$ , and minimize $\mathcal S$ .

In this case, the Euler–Lagrange equation is (according to a previous post) $\sum_j\partial_j\frac{\partial\mathcal L}{\partial\!\left(\partial_jq_k\right)}=\frac{\partial\mathcal L}{\partial q_k}.$ Therefore, the momentum is defined to be the matrix $p_{j,k}\coloneqq\frac{\partial\mathcal L}{\partial\!\left(\partial_jq_k\right)}.$ Note that now the Euler–Lagrange equation is a set of second-order PDEs.

We try performing Legendre transformation on $\mathcal S$ and get Hamiltonian $\mathcal H$ .

As a function of space coordinates and velocities, the total derivative of $\mathcal L$ is $\mathrm d\mathcal L=\sum_{j,k}\frac{\partial\mathcal L}{\partial\!\left(\partial_jq_k\right)}\,\mathrm d\!\left(\partial_jq_k\right) +\sum_k\frac{\partial\mathcal L}{\partial q_k}\,\mathrm dq_k.$ Substitute the Euler–Lagrange equation and the definition of momenta, and we have $\mathrm d\mathcal L=\sum_{j,k}p_{j,k}\,\mathrm d\!\left(\partial_jq_k\right)+\sum_{j,k}\partial_jp_{j,k}\,\mathrm dq_k.$ By the product rule, the first term in the formula above can be written as $p_{j,k}\,\mathrm d\!\left(\partial_jq_k\right)=\mathrm d\!\left(p_{j,k}\partial_jq_k\right)-\partial_jq_k\,\mathrm dp_{j,k},$ and then we have $\mathrm d\!\left(\sum_{j,k}p_{j,k}\partial_jq_k-\mathcal L\right) =-\sum_{j,k}\partial_jp_{j,k}\,\mathrm dq_k+\sum_{j,k}\partial_jq_k\,\mathrm dp_{j,k}.$ If we let $\mathcal H\coloneqq\sum_{j,k}p_{j,k}\partial_jq_k-\mathcal L$ , then we have $\partial_jq_k=\frac{\partial\mathcal H}{\partial p_{j,k}},\quad\sum_j\partial_jp_{j,k}=-\frac{\partial\mathcal H}{\partial q_k}.$ This is the new Hamiltonian equations, or canonical equations. We may find that it lacks the beauty of the form in one-dimensional time.

Problem 1: Prove that if we add the Lagrangian by the “total divergence” w.r.t. time of some function defined on space and time, the new Lagrangian describes the same mechanical system as the original. In other words, $\mathcal L'\coloneqq\mathcal L+\sum_j\partial_jf$ has the same equation of motion as $\mathcal L$ , where $f$ is an arbitrary function defined on the spacetime.

Problem 2: Prove by principle of Galilean relativity that the Lagrangian of a single free particle system is $\mathcal L=\sum_{j,k}\frac{m_j}2\left(\partial_jq_k\right)^2$ , where $m_j$ are constants (their physical meaning is mass, which means that mass is not scalar in time of higher dimensions), and $q_k$ are Cartesian coordinates. Find and solve its equation of motion, and hence derive the law of inertia.

Problem 3: Does the conservation of energy still hold?

A problem that I am too lazy to consider: Consider the Minkowski universe. Poincaré transformations are defined as those affine transformations that preserve the spacetime distance between events, and the spacetime distance is defined as the difference of the square of Euclidean distance in time space and the square of Euclidean distance in space space. The principle of special relativity guarantees that the equation of motion of closed systems is invariant under Poincaré transformations. Find the Lagrangian of a single free particle system.

When a supersonic airplane flies over your head

2022-11-18T14:11:10-08:00

This article is translated from a Chinese article on my Zhihu account. The original article was posted at 2020-02-17 18:49 +0800.

(Continue to review middle school knowledge!)

Suppose a supersonic airplane has Mach number $M$ (Mach number is the speed divided by sound speed). It flies horizontally. At some time, it flies past over your head at height $h$ . What is the distance between you and it when you have just heard it?

Suppose the time when the airplane flies past over your head is $t=0$ ; the sound speed is $v$ . Then, the motion of the airplane is $x=Mvt.$ The distance between you and the airplane is $l=\sqrt{x^2+h^2}.$

Let $\tau$ be the time at which you hear the sound emitted by the airplane at $t$ , then $\tau=t+\frac lv=t+\sqrt{\tau_0^2+M^2t^2},$ where $\tau_0\coloneqq\frac hv$ .

Then, we can find the minimum value of $\tau$ by letting $\frac{\mathrm d\tau}{\mathrm dt}=0$ . However, here we use a middle school technique to reduce calculation.

Write the relationship between $\tau$ and $t$ as a quadratic equation w.r.t. $t$ $\left(M^2-1\right)t^2+2\tau t+\tau_0^2-\tau^2=0.$ It must have real solutions, so the determinant is non-negative, i.e. $\tau^2-\left(M^2-1\right)\left(\tau_0^2-\tau^2\right)\ge0.$

We can solve the inequality to get $\tau^2\ge\left(1-M^{-2}\right)\tau_0^2.$ Here we can see why do we require the airplane to be supersonic (otherwise the right-hand side of the inequality is negative, and $\tau$ does not have a minimum value).

By this, we have $\tau_{\mathrm{min}}=\tau_0\sqrt{1-M^{-2}}.$ Therefore, the answer to our question is $\left.l\right|_{t=\tau_{\mathrm{min}}}=Mh.$

Elegant!

Finding curvature radius physically

2022-11-16T17:54:36-08:00

This article is translated from a Chinese article on my Zhihu account. The original article was posted at 2020-03-15 11:50 +0800.

I occasionally saw a picture of the blackboard contents on a physics class from one of my schoolmates, and I found that they have learnt about decomposing the force into the tangential force and the normal force. I then thought that they should realize that they have gained a tool to find the curvature radius of a curve at some point.

In this article, from now on, I only use knowledge from high school.

First, consider an example question: a circle of radius $\rho$ can cover every point when it moves arbitrarily within the region interior to the boundary defined by $x^2=2py$ . Illustrate by calculation that $\rho\le p$ .

(The original problem comes from Tong Ji Advanced Mathematics (note of translation: the calculus textbook used by many Chinese universities for non-mathematics students), asking about the condition for a circular object to be used to be used to burnish a parabola shaped component.)

Obviously, the purpose of the problem is to ask you to find the curvature radius of the curve at the point where the curvature radius is smallest. By observing the graph, we can see that the point with smallest curvature radius is the vertex of the parabola (I just run around now, but I will prove this argument later).

Consider the motion $\vec r=ut\vec i+\frac{u^2t^2}{2p}\vec j$ ( $u$ here is introduced just for consistency in dimension, but actually we can just let $u\coloneqq1$ ).

Obviously the object is doing a uniform acceleration motion, and the acceleration is as large as $\frac{u^2}p$ , and its direction is $+y$ . The trajectory of the motion is the parabola $x^2=2py$ .

Consider the motion at $t=0$ . At this time, it is located at the vertex of the parabola. The velocity is $u$ in the direction of $+x$ . Therefore, obviously the normal acceleration is just its acceleration $\frac{u^2}p$ , so the curvature radius of the parabola at the vertex is $p$ .

Therefore, we can see that some problems that seem to require calculus can actually be solved using physics knowledge without using calculus.

Prove that the point on the parabola with smallest curvature radius is the vertex using physical method.

By the formula for uniform acceleration motion, the motion $\vec r=ut\vec i+\frac{u^2t^2}{2p}\vec j$ has velocity $\vec v=u\vec i+\frac{u^2t}{p}\vec j$ and acceleration $\vec a=\frac{u^2}{p}\vec j.$ By easy calculation, we can get the tangential acceleration $a_\tau=\frac{u^3t}{p\sqrt{p^2+u^2t^2}}$ and the normal acceleration $a_\nu=\frac{u^2}{\sqrt{p^2+u^2t^2}}.$ The curvature radius $\rho=\frac{v^2}{a_\nu}=\frac{\left(p^2+u^2t^2\right)^\frac32}{p^2}\ge p,$ where the equality holds iff $t=0$ .

A whirling point charge

2022-11-16T16:47:57-08:00

This article is translated from a Chinese article on my Zhihu account. The original article was posted at 2020-02-17 18:49 +0800.

In the vacuum, inside a fixed ring of radius $R$ with fixed charge $Q$ uniformly distributed, there is a point charge with charge $q$ and mass $m$ moving in the plane of the ring due tue the electrostatic force. It moves in the small region around the center of the ring, and the motion is periodic along a closed curve. The area of the region enclosed by the curve is $S$ . Denote the distance from the center to the point charge as $r$ , and $r\ll R$ . Find the magnetic induction $B$ at the center of the ring.

By using the cosine law, we can write the electrical potential in the plane of the ring inside the ring as the integral $U=\int_0^{2\pi}\frac{\frac{\mathrm d\theta}{2\pi}Q}{4\pi\varepsilon_0\sqrt{R^2+r^2-2Rr\cos\theta}}.$ Note that the elliptic integral of the first kind is $\begin{align*} K\!\left(\mu\right)&\coloneqq\int_0^{\frac\pi2}\frac{\mathrm d\varphi}{\sqrt{1-\mu\sin^2\varphi}}\\ &=\int_0^{\frac\pi2}\frac{\mathrm d\varphi}{\sqrt{1-\mu\frac{1-\cos2\varphi}{2}}}\\ &=\int_0^\pi\frac{\mathrm d\theta}{\sqrt{4-2\mu+2\mu\cos\theta}}&\theta\coloneqq2\varphi\\ &=\frac1{2\sqrt{4-2\mu}}\int_0^{2\pi}\frac{\mathrm d\theta}{1+\frac\mu{2-\mu}{\cos\theta}}.\\ \end{align*}$ On the other hand, the potential $U=\frac{Q}{8\pi^2\varepsilon_0\sqrt{R^2+r^2}}\int_0^{2\pi}\frac{\mathrm d\theta}{1-\frac{2Rr}{R^2+r^2}\cos\theta}.$ By comparing the two equations, we are motivated to find $\mu$ such that $\frac\mu{2-\mu}=-\frac{2Rr}{R^2+r^2},$ and we may solve to get $\mu=-\frac{4Rr}{\left(R-r\right)^2}.$ Therefore, $\begin{align*} U&=\frac{Q}{8\pi^2\varepsilon_0\sqrt{R^2+r^2}}\cdot\frac{4\sqrt{R^2+r^2}}{R-r}K\!\left(-\frac{4Rr}{\left(R-r\right)^2}\right)\\ &=\frac Q{2\pi^2\varepsilon_0\left(R-r\right)}K\!\left(-\frac{4Rr}{\left(R-r\right)^2}\right). \end{align*}$ We can expand $U$ in terms power series of $r$ (how?), and we get $U=\frac Q{4\pi\varepsilon_0R}+\frac{Q}{8\pi\varepsilon_0R^3}r^2+O\!\left(r^4\right).$ Then, the potential energy $E_\mathrm p=qU$ (and omit constant term and higher order terms) is $E_\mathrm p=\frac{qQ}{8\pi\varepsilon_0R^3}r^2.$ To make the trajectory a closed curve, the second derivative of the potential at the equilibrium should be positive, so $qQ>0$ , i.e. the ring and point charge have the same sign of charge.

As we all know, for the potential energy $E_\mathrm p=\frac12m\omega^2r^2$ , the motion is $\begin{cases}x=a\cos\omega t,\\y=a\sin\omega t,\end{cases}$ where $a$ and $b$ are determined by the initial conditions.

Then, we can solve the equation $\frac{qQ}{8\pi\varepsilon_0R^3}=\frac12m\omega^2$ to get $\omega=\frac1{2R}\sqrt{\frac{qQ}{\pi\varepsilon_0mR}}.$ Because the trajectory is an ellipse, the area is $S=\pi ab.$ We can take the derivate of the coordinates w.r.t. $t$ to get the velocity $\begin{cases}v_x=-a\omega\sin\omega t,\\v_y=a\omega\cos\omega t,\end{cases}$ By Biot–Savart law, the magnetic induction $B$ at the center of the ring is $\begin{align*} B&=\frac{\mu_0q}{4\pi r^3}\left|\begin{matrix}x&v_x\\y&v_y\end{matrix}\right|\\ &=\frac{\mu_0q}{4\pi r^3}\omega ab\left(\cos^2\omega t+\sin^2\omega t\right)\\ &=\frac{\mu_0qS}{4\pi^2r^3}\omega\\ &=\frac{\mu_0qS}{8\pi^2Rr^3}\sqrt{\frac{qQ}{\pi\varepsilon_0mR}}. \end{align*}$

An example of non-uniform elements: heavy elastic rope

2022-11-13T10:15:12-08:00

This article is translated from a Chinese article on my Zhihu account. The original article was posted at 2019-12-18 11:20 +0800.

Review middle school contents!

Suppose a uniform heavy elastic rope has mass $m$ , original length $L_0$ , and stiffness $k$ . Find the mass distribution and length of it when hung vertically.

When the elastic rope has its original length, divide it into $n$ equal segments, where $n$ is large so that when the elastic rope is stretched, the mass distribution within every segment is uniform.

When the rope is hung vertically, the tension experienced by every segment is $F_j=\frac{mg}n\left(n-j\right).$ Therefore, the length of every segment is $\Delta l_j=\frac{F_j}{nk}+\frac{L_0}n=\frac{mg\left(n-j\right)}{n^2k}+\frac{L_0}n.$ Therefore, the position of the end of every segment is $x_j=\sum_{s=1}^j\Delta l_s=\frac{2nL_0kj+2nmgj-mgj-mgj^2}{2n^2k}.$ The equation can be solved to get $j=\frac{2nmg+2nL_0k-mg\pm\sqrt{\left(mg-2nL_0k-2nmg\right)^2-8n^2mgkx_j}}{2mg}.$ The (linear) mass density of every segment is $\rho_j=\frac{m}{n\Delta l_j}=\frac{m}{n\left(\frac{mg\left(n-j\right)}{n^2k}+\frac{L_0}n\right)}.$ Express $j$ here with $x_j$ , and we have $\rho_j=\frac{2nmk}{mg\mp\sqrt{\left(2nL_0k+mg\left(2n-1\right)\right)^2-8n^2mgkx_j}}.$ The length of the rope is $L=x_n=\frac{mg\left(n-1\right)}{2nk}+L_0.$ Take the limit $n\to\infty$ , and we have $L=\frac{mg}{2k}+L_0,$ $\rho\!\left(x\right)=\pm\frac{mk}{\sqrt{L_0^2k^2+m^2g^2+2mgk\left(L_0-x\right)}}.$ The plus-minus sign here should take plus sign.

To test, calculate $\int_0^L\rho\!\left(x\right)\mathrm dx=m$ .

The conclusion can be written in a more beautiful way: $\rho\!\left(x\right)=\frac{m}{\sqrt{\left(2L-L_0\right)^2-4\left(L-L_0\right)x}}.$

Kinetic energy, momentum, and angular momentum of rigid bodies

2022-11-12T19:29:40-08:00

This article is translated from a Chinese article on my Zhihu account. The original article was posted at 2019-11-03 20:02 +0800.

Some notations:

$a$ indexes particles. $j,k,l$ are indices of vector components. $\mathbf q,\mathbf s,\mathbf t$ are arbitrary vectors.
$\mathbf I$ is the unit matrix. Matrix $\mathbf q\otimes\mathbf s\coloneqq\mathbf q\mathbf s^\mathrm T$ is the outer product of $\mathbf q$ and $\mathbf s$ , and $\mathbf q^{\otimes 2}\coloneqq\mathbf q\otimes\mathbf q$ .
A rigid body consists of some particles with distances between each other fixed. $\mu_a$ is the mass of a particle; $\mathbf r_a$ is the position of a particle relative to the center of the rigid body ( $\sum_a\mu_a\mathbf r_a=\mathbf 0$ ); and $\mathbf v_a$ is the velocity of a particle.
The total mass of the rigid body is $m=\sum_a\mu_a$ ; the velocity of its mass center is $\mathbf V$ ; and its angular velocity is $\mathbf\Omega$ . $\theta_a$ is the angle between $\mathbf r_a$ and $\mathbf\Omega$ .
$\mathbf m\coloneqq m\mathbf I$ , $\mathbf J\coloneqq\sum_a\mu_a\left(\mathbf r_a^2\mathbf I+\mathbf r_a^{\otimes 2}\right)$ .

Talk about homogeneous scalar fields of degree 2. Homogeneous scalar fields of degree 2 are function $f$ that satisfies $f\!\left(\lambda\mathbf q\right)=\lambda^2f\!\left(\mathbf q\right).$ It has the form $f\!\left(\mathbf q\right)=\frac12\sum_{j,k}A_{j,k}q_jq_k,$ where the coefficient matrix $\mathbf A$ is a symmetric matrix (this is why there is a factor of $\frac12$ in front).

Actually, such a function can be written in matrix multiplication as $f\!\left(\mathbf q\right)=\frac{\mathbf A}2\mathbf q\cdot\mathbf q$ (note that it is not the same as $\frac{\mathbf A}2\mathbf q^2$ , but is $\left(\frac{\mathbf A}{2}\mathbf q\right)\cdot\mathbf q$ instead).

Similarly, homogeneous vector field with degree 1 has the property $\mathbf g\!\left(\lambda\mathbf q\right)=\lambda\mathbf g\!\left(\mathbf q\right),$ and they have the form $\mathbf g\!\left(\mathbf q\right)=\mathbf A\mathbf q$ (now the coefficient matrix $\mathbf A$ is not necessarily symmetric).

Introduce outer product. If the matrix $\mathbf A\coloneqq\mathbf q\otimes\mathbf s$ , then $A_{j,k}=q_js_k.$ Specially, if $\mathbf A\coloneqq\mathbf q^{\otimes 2}$ , then $A_{j,k}=q_jq_k.$ The perfect square expansion of outer product is $\left(\mathbf q+\mathbf s\right)^{\otimes 2}= \mathbf q^{\otimes 2}+\mathbf q\otimes\mathbf s+\mathbf s\otimes\mathbf q+\mathbf s^{\otimes 2}.$

First, according to geometric relations, we can prove a theorem $\mathbf v_a=\mathbf V+\mathbf\Omega\times\mathbf r_a.$ Then, we can discuss the kinetic energy $T\coloneqq\sum_a\frac{\mu_a}2\mathbf v_a^2$ of the rigid body. We will discover that the kinetic energy of the rigid body can be represented as the sum of a homogeneous function of degree 2 w.r.t. $\mathbf V$ and a homogeneous function of degree 2 w.r.t. $\mathbf\Omega$ , i.e. $T=T_{\mathrm{tra}}+T_{\mathrm{rot}},$ where $T_{\mathrm{tra}}\coloneqq\frac{\mathbf m}2\mathbf V\cdot\mathbf V,$ $T_{\mathrm{rot}}\coloneqq\frac{\mathbf J}2\mathbf\Omega\cdot\mathbf\Omega.$

The proof is not difficult: $\begin{align*} T&\coloneqq\sum_a\frac{\mu_a}2\mathbf v_a^2\\ &=\sum_a\frac{\mu_a}2\left(\mathbf V+\mathbf\Omega\times\mathbf r_a\right)^2\\ &=\sum_a\frac{\mu_a}2\left( \mathbf V^2+2\mathbf V\cdot\left(\mathbf\Omega\times\mathbf r_a\right) +\left(\mathbf\Omega\times\mathbf r_a\right)^2 \right)\\ &=\frac12\mathbf V^2\underbrace{\sum_a\mu_a}_m +\mathbf V\cdot\left(\mathbf\Omega\times\underbrace{\sum_a\mu_a\mathbf r_a}_{\mathbf 0}\right) +\frac12\sum_a\mu_a\left(\mathbf\Omega\times\mathbf r_a\right)^2\\ &=T_{\mathrm{tra}}+\frac12\sum_a\mu_a\left(\mathbf\Omega\times\mathbf r_a\right)^2, \end{align*}$ while $\begin{align*} &\phantom{=}~\,\left(\mathbf\Omega\times\mathbf r_a\right)^2\\ &=\mathbf\Omega^2\mathbf r_a^2\sin^2\theta_a\\ &=\mathbf\Omega^2\mathbf r_a^2-\mathbf\Omega^2\mathbf r_a^2\cos^2\theta_a\\ &=\mathbf\Omega^2\mathbf r_a^2-\left(\mathbf\Omega\cdot\mathbf r_a\right)^2\\ &=\left(\sum_jr_{a,j}^2\right)\left(\sum_j\Omega_j^2\right) -\left(\sum_jr_{a,j}\Omega_j\right)\left(\sum_jr_{a,j}\Omega_j\right)\\ &=\sum_{j,l}r_{a,l}^2\Omega_j^2-\sum_{j,k}r_{a,j}r_{a,k}\Omega_j\Omega_k\\ &=\sum_{j,l}r_{a,l}^2\Omega_j\sum_kI_{j,k}\Omega_k -\sum_{j,k}r_{a,j}r_{a,k}\Omega_j\Omega_k\\ &=\sum_{j,k}\underbrace{\left(\sum_{l}r_{a,l}^2\right)}_{\mathbf r_a^2} I_{j,k}\Omega_j\Omega_k -\sum_{j,k}r_{a,j}r_{a,k}\Omega_j\Omega_k\\ &=\sum_{j,k}\left(\mathbf r_a^2I_{j,k}-r_{a,j}r_{a,k}\right)\Omega_j\Omega_k\\ &=\left(\mathbf r_a^2\mathbf I-\mathbf r_a^{\otimes2}\right)\mathbf\Omega\cdot\mathbf\Omega. \end{align*}$ Therefore, $\begin{align*} T&=T_{\mathrm{tra}}+\frac12\sum_a\mu_a\left(\mathbf r_a^2\mathbf I-\mathbf r_a^{\otimes2}\right)\mathbf\Omega\cdot\mathbf\Omega\\ &=T_{\mathrm{tra}}+T_{\mathrm{rot}}. \end{align*}$

Then, we may discuss the momentum of the rigid body $\mathbf P\coloneqq\sum_a\mu_a\mathbf v_a.$ We will discover that it is a homogeneous function of degree 1 w.r.t. $\mathbf V$ , i.e. $\mathbf P=\mathbf m\mathbf V.$

The proof is easier: $\begin{align*} \mathbf P&\coloneqq\sum_a\mu_a\mathbf v_a\\ &=\sum_a\mu_a\left(\mathbf V+\mathbf\Omega\times\mathbf r_a\right)\\ &=\underbrace{\left(\sum_a\mu_a\right)}_m\mathbf V +\mathbf\Omega\times\underbrace{\sum_a\mu_a\mathbf r_a}_{\mathbf 0}\\ &=\mathbf m\mathbf V. \end{align*}$

Finally, we can discuss the angular momentum of the rigid body $\mathbf M\coloneqq\sum_a\mathbf r_a\times\left(\mu_a\mathbf v_a\right).$ We will discover that it is a homogeneous function of degree 1 w.r.t. $\mathbf\Omega$ , i.e. $\mathbf M=\mathbf J\mathbf\Omega.$

The proof is also easy: $\begin{align*} \mathbf M&\coloneqq\sum_a\mathbf r_a\times\left(\mu_a\mathbf v_a\right)\\ &=\sum_a\mu_a\mathbf r_a\times\left(\mathbf V+\mathbf\Omega\times\mathbf r_a\right)\\ &=\underbrace{\left(\sum_a\mu_a\mathbf r_a\right)}_{\mathbf 0}\times\mathbf V +\sum_a\mu_a\mathbf r_a\times\left(\mathbf\Omega\times\mathbf r_a\right)\\ &=\sum_a\mu_a\left(\mathbf r_a^2\mathbf\Omega-\mathbf r_a\cdot\mathbf\Omega\mathbf r_a\right) \end{align*}$ (note the order of computation in the notation $\mathbf q\cdot\mathbf s\mathbf t$ : it is $\left(\mathbf q\cdot\mathbf s\right)\mathbf t$ ). Therefore, $\begin{align*} M_j&=\sum_a\mu_a\mathbf r_a^2\underbrace{\Omega_j}_{\sum_kI_{j,k}\Omega_k} -\sum_a\mu_a\left(\sum_kr_{a,k}\Omega_k\right)r_{a,k}\\ &=\sum_a\mu_a\sum_k\left(\mathbf r_a^2I_{j,k}-r_{a,j}r_{a,k}\right)\Omega_k\\ &=\sum_k\sum_a\mu_a\left(\mathbf r_a^2I_{j,k}-r_{a,j}r_{a,k}\right)\Omega_k\\ &=\sum_kJ_{j,k}\Omega_k, \end{align*}$ i.e. $\mathbf M=\mathbf J\mathbf\Omega.$

We discover that, it seems that the kinetic energy is the sum of “translational kinetic energy” and “rotational kinetic energy”, while the momentum only contains “translational momentum”, and the angular momentum only contains “rotational angular momentum”.

We may discover that $\mathbf J$ seems to represent some inertia related to rotation of the rigid body, so we call it the inertia matrix (or, in tensor language, the inertia tensor). By this means we introduced an important physical quantity. Now we study some of its properties.

Note that $\mathbf J$ is a real symmetric matrix, so it has real eigenvaluese $\mathbf J_{\mathrm{pri}}$ (this is a vector containing all the eigenvalues as its components), and its (normalized) eigenvectors $\mathbf e_{\mathrm{pri}}$ (this is a matrix, with its columns being the eigenvectors) orthogonal to each other. These eigenvalues are called the principal moments of inertia, and the axes parallel to the eigenvectors and passing through the center of mass are called the principal axes of inertia.

$\mathbf J$ is diagonalizable, i.e. $\mathbf e_{\mathrm{pri}}^{-1}\mathbf J\mathbf e_{\mathrm{pri}} =\operatorname{\mathbf{diag}}\mathbf J_{\mathrm{pri}}.$ Because the eigenvectors of $\mathbf J$ are orthogonal to each other, the matrix $\mathbf e_{\mathrm{pri}}$ is an orthogonal matrix, i.e. $\mathbf e_{\mathrm{pri}}^{\mathrm T}=\mathbf e_{\mathrm{pri},j},$ so $\begin{align*} \mathbf q_{\mathrm{pri}}&=\left(\mathbf q^\mathrm T\mathbf e_{\mathrm{pri}}\right)^\mathrm T\\ &=\mathbf e_{\mathrm{pri}}^\mathrm T\mathbf q\\ &=\mathbf e_{\mathrm{pri}}^{-1}\mathbf q, \end{align*}$ i.e. $\mathbf e_{\mathrm{pri}}\mathbf q_{\mathrm{pri}}=\mathbf q.$ We may find that $\begin{align*} \mathbf M&=\mathbf J\mathbf\Omega\\ &=\left(\mathbf e_{\mathrm{pri}}\mathbf e_{\mathrm{pri}}^{-1}\right)\mathbf J \left(\mathbf e_{\mathrm{pri}}\mathbf e_{\mathrm{pri}}^{-1}\right)\mathbf\Omega\\ &=\mathbf e_{\mathrm{pri}}\left(\mathbf e_{\mathrm{pri}}^{-1}\mathbf J\mathbf e_{\mathrm{pri}}\right) \left(\mathbf e_{\mathrm{pri}}^{-1}\mathbf\Omega\right)\\ &=\mathbf e_{\mathrm{pri}}\left(\operatorname{\mathbf{diag}}\mathbf J_{\mathrm{pri}}\right)\mathbf\Omega_{\mathrm{pri}}. \end{align*}$ On the other hand, we have $\mathbf M=\mathbf e_{\mathrm{pri}}\mathbf M_{\mathrm{pri}},$ so $\mathbf M_{\mathrm{pri}}=\left(\operatorname{\mathbf{diag}}\mathbf J_{\mathrm{pri}}\right)\mathbf\Omega_{\mathrm{pri}},$ i.e. $M_{\mathrm{pri},j}=\mathbf J_{\mathrm{pri},j}\Omega_{\mathrm{pri},j}.$ This means that the component of the angular momentum along some certain principal axis of inertia is the product of the principal inertia along this axis and the component of the angular velocity along this axis. This can simplify the calculation of the angular momentum a lot when the principal inertia can be calculated.

Now, we may study the rotational kinetic energy $\begin{align*} T_{\mathrm{rot}}&\coloneqq\frac{\mathbf J}2\mathbf\Omega\cdot\mathbf\Omega\\ &=\frac12\mathbf M\cdot\mathbf\Omega\\ &=\frac12\mathbf M\cdot\mathbf\Omega\\ &=\frac12\mathbf e_{\mathrm{pri}}\left(\operatorname{\mathbf{diag}}\mathbf J_{\mathrm{pri}}\right) \mathbf\Omega_{\mathrm{pri}}\cdot\mathbf\Omega\\ &=\frac12\left(\operatorname{\mathbf{diag}}\mathbf J_{\mathrm{pri}}\right) \left(\mathbf e_{\mathrm{pri}}\mathbf\Omega\right)\cdot\mathbf\Omega_{\mathrm{pri}}\\ &=\frac12\left(\operatorname{\mathbf{diag}}\mathbf J_{\mathrm{pri}}\right) \mathbf\Omega_{\mathrm{pri}}\cdot\mathbf\Omega_{\mathrm{pri}}\\ &=\frac12\sum_jJ_{\mathrm{pri},j}\Omega_{\mathrm{pri},j}^2. \end{align*}$ This also makes the calculation simplified when the principal inertia can be calculated.

Study how $\mathbf J$ will change if $\mathbf r_a$ experience some transformations.

First, consider translation. Let $\mathbf r_a'\coloneqq\mathbf r_a+\mathbf d,$ $\mathbf J'\coloneqq\sum_a\mu_a\left(\mathbf r_a'^2\mathbf I-\mathbf r_a'^{\otimes2}\right)$ (we may regard $\mathbf J'$ as the inertia matrix of the rigid body w.r.t. the point with $\mathbf d$ distance away from the center of mass, but not (in usual cases) w.r.t. the center of mass), and then we will discover that $\mathbf J$ can be calculated from $\mathbf J'$ by $\mathbf J=\mathbf J'-m\left(\mathbf d^2\mathbf I-\mathbf d^{\otimes2}\right).$ This provides us a useful way to calculate $\mathbf J$ when $\mathbf J'$ is easy to calculate while $\mathbf J$ is not.

The proof is not difficult: $\begin{align*} \mathbf J'&\coloneqq\sum_a\mu_a\left(\mathbf r_a'^2\mathbf I-\mathbf r_a'^{\otimes2}\right)\\ &=\sum_a\mu_a\left(\left(\mathbf r_a+\mathbf d\right)^2\mathbf I-\left(\mathbf r_a+\mathbf d\right)^{\otimes2}\right)\\ &=\sum_a\mu_a\left(\left(\mathbf r_a^2+2\mathbf r_a\cdot\mathbf d+\mathbf d^2\right)\mathbf I -\left(\mathbf r_a^{\otimes2}+\mathbf d\otimes\mathbf r_a+\mathbf r_a\otimes\mathbf d+\mathbf d^{\otimes 2}\right)\right)\\ &=\underbrace{\sum_a\mu_a\left(\mathbf r_a^2\mathbf I-\mathbf r_a^{\otimes2}\right)}_\mathbf J +\underbrace{\left(\sum_a\mu_a\right)}_m\left(\mathbf d^2\mathbf I-\mathbf d^{\otimes2}\right) +\sum_a\mu_a\left(2\mathbf r_a\cdot\mathbf d\mathbf I-\mathbf d\otimes\mathbf r_a-\mathbf r_a\otimes\mathbf d\right)\\ &=\mathbf J+m\left(\mathbf d^2\mathbf I-\mathbf d^{\otimes2}\right) +2\underbrace{\left(\sum_a\mu_a\mathbf r_a\right)}_\mathbf 0\cdot\mathbf d\mathbf I -\mathbf d\otimes\underbrace{\left(\sum_a\mu_a\mathbf r_a\right)}_\mathbf 0 -\underbrace{\left(\sum_a\mu_a\mathbf r_a\right)}_\mathbf 0\otimes\mathbf d\\ &=\mathbf J+m\left(\mathbf d^2\mathbf I-\mathbf d^{\otimes2}\right), \end{align*}$ i.e. $\mathbf J=\mathbf J'-m\left(\mathbf d^2\mathbf I-\mathbf d^{\otimes2}\right).$

The special case of it is called the parallel axis theorem in general physics.

Now, consider rotation. Suppose the rotational matrix $\mathbf R$ represents the rotation around the unit vector $\mathbf u$ . Let $\mathbf r_a'\coloneqq\mathbf R\mathbf r_a,$ $\mathbf J'\coloneqq\sum_a\mu_a\left(\mathbf r_a'^2\mathbf I-\mathbf r_a'^{\otimes2}\right),$ and we will discover $\mathbf J'=\mathbf R\mathbf J\mathbf R^{-1}.$

Now we will prove the conclusion. First, we now that $\mathbf R$ is an orthogonal matrix, i.e. $\mathbf R^\mathrm T=\mathbf R^{-1}.$ By this we can get other properties like $\left(\mathbf R\mathbf q\right)^2=\mathbf q^2,$ $\left(\mathbf R\mathbf q\right)^{\otimes2}=\mathbf R\mathbf q^{\otimes2}\mathbf R^{-1}.$ Therefore, $\begin{align*} \mathbf J'&\coloneqq\sum_a\mu_a\left(\mathbf r_a'^2\mathbf I-\mathbf r_a'^{\otimes2}\right)\\ &=\sum_a\mu_a\left(\mathbf r_a^2\mathbf I-\mathbf R\mathbf r_a^{\otimes2}\mathbf R^{-1}\right)\\ &=\mathbf R\left(\sum_a\mu_a\left(\mathbf r_a^2\mathbf I-\mathbf r_a^{\otimes2}\right)\right)\mathbf R^{-1}\\ &=\mathbf R\mathbf J\mathbf R^{-1}. \end{align*}$

After getting this conclusion, we may be interested in non trivial (i.e. $\mathbf R\ne\mathbf I$ ) cases of $\mathbf J'=\mathbf J$ . We will prove that in this case $\mathbf u$ is a principal axis of the rigid body.

By $\mathbf J'=\mathbf R\mathbf J\mathbf R^{-1}$ , we can get $\mathbf J=\mathbf R\mathbf J\mathbf R^{-1},$ i.e. $\mathbf R\mathbf J=\mathbf J\mathbf R.$ Multiply $\mathbf u$ on both sides, and we have $\mathbf R\mathbf J\mathbf u=\mathbf J\mathbf R\mathbf u,$ i.e. $\mathbf R\left(\mathbf J\mathbf u\right)=\mathbf J\left(\mathbf R\mathbf u\right).$ By Euler’s rotation theorem, the only real eigenvalue of nontrivial $\mathbf R$ is $1$ , and its corresponding eigenvector is $\mathbf u$ , i.e. $\mathbf R\mathbf u=\mathbf u.$ Substitute, and we have $\mathbf R\left(\mathbf J\mathbf u\right)=\mathbf J\mathbf u,$ i.e. $\mathbf J\mathbf u$ is an eigenvector of $\mathbf R$ .

Because the only independent (real) eigenvector of notrial $\mathbf R$ is $\mathbf u$ , $\mathbf J\mathbf u$ is parallel to $\mathbf u$ , so $\mathbf u$ is an eigenvector of $\mathbf J$ , i.e. $\mathbf u$ is a principal axis of the rigid body.

This leads to an important conclusion: if the rigid body maintains its original inertia matrix after it rotates around an axis passing its center of mass by a nontrivial angle (an angle that is not a multiple of $2\pi$ ), then this axis is a principal axis of the rigid body.

By this we can also get another important conclusion: if a rigid body is rotational symmetric around some axis, then this axis is a principal axis of the rigid body. This conclusion simplifies the determination of principal axes of a rigid body with some symmetries.

The image of a circular object through a thin lens

2022-11-09T09:37:41-08:00

This article is translated from a Chinese article on my Zhihu account. The original article was posted at 2019-07-12 15:40 +0800.

The center of the circular luminous object is $C$ , and its radius is $r$ . The center of the thin lens is $O$ , and its focal length is $f$ . The object is in the same plane as $C$ , and $OC$ is perpendicular to the thin lens, and $\left\|OC\right\|=2f$ .

Set up Cartesian coordinate with $O$ being the origin and $CO$ being the $x$ -axis. Then, $C\left(-2f.0\right)$ . The luminous object is described by the parametric equations $\begin{cases} x=-2f+r\cos t,\\y=r\sin t. \end{cases}$ Pick point $P\left(-2f+r\cos t,r\sin t\right)$ on the object. According to the formula for imaging of thin lenses $\begin{cases} -\frac1x+\frac1{x'}=\frac1f,\\ \frac x{x'}=\frac y{y'}, \end{cases}$ the point $P$ is transformed to $P'\left(f\left(1+\frac f{f-r\cos t}\right),-f\frac{r\sin t}{f-r\cos t}\right)$ . Therefore, we can have the parametric equations of the image: $\begin{cases} x=f\left(1+\frac f{f-r\cos t}\right),\\y=-f\frac{r\sin t}{f-r\cos t}. \end{cases}$ Cancel $t$ , and we have $y^2=\left(\frac rf\right)^2\left(x-f\right)^2-\left(x-2f\right)^2,$ which means that the image is a conic section with the focus being $\left(2f,0\right)$ , the directrix being line $x=f$ , and the eccentricity being $\frac rf$ .

Alternatively, let $\rho\coloneqq\sqrt{\left(x-2f\right)^2+y^2}$ , and we have $\rho=\frac r{1-\frac rf\cos t},$ and we may have the same conclusion through this equation.

Mapping from Kepler problem to free particle on 3-sphere

2022-11-07T16:12:09-08:00

This article is adapted from the letter that I wrote to my professor of classical mechanics. Background: the professor said in class that he knew there is a way to transform the Kepler problem into the system of a free particle on a 3-sphere, but he did not know what is the explicit form of the mapping and asked us to try to research and find out the mapping. The original letter was sent at 2022-10-13 21:56 -0700.

After searching on the internet, I found something about what the mapping from the 3-dimensional Kepler problem to the 3-sphere is.

We choose the canonical coordinates as the 3 Cartesian coordinates and the 3 associated momenta. Then, use stereographic projection to map the 3-dimensional vector $\mathbf p/p_0$ (where $p_0\coloneqq\sqrt{-2mE}$ ) to a point on an 3-sphere (where we should already have built a coordinate system $\mathbf u$ for us to calculate with). The explicit expression for the projection is $\mathbf u\coloneqq\frac{p^2-p_0^2}{p^2+p_0^2}\hat{\mathbf n}+\frac{2p_0}{p^2+p_0^2}\mathbf p,$ where $\mathbf u$ is a unit 4-dimensional vector, and $\hat{\mathbf n}$ is a unit vector perpendicular to the hyperplane where $\mathbf p$ lies. I’m sorry for the abuse of notation, but the $\mathbf u$ in this email also stands for the (3-spherical) coordinates of it on the 3-sphere.

Now, we can use the coordinates on the 3-sphere to express the momenta $\mathbf p(\mathbf u)$ . Then, write down the one-form $\mathbf p\cdot\mathrm d\mathbf x$ in terms of the 3-dimensional Cartesian coordinates $\mathbf x$ and the coordinates on the 3-sphere, and convert it into the form $\mathrm d(\mathbf p\cdot\mathbf x)-\mathbf P(\mathbf x,\mathbf u)\cdot\mathrm d\mathbf u$ . Then, we get the canonical transform $(\mathbf x,\mathbf p)\mapsto(\mathbf u,\mathbf P)$ . The hamiltonian will be correspondingly transformed into exactly the same as the hamiltonian of a free particle on the 3-sphere.

The explicit calculation for the 2-dimensional case can be found in this paper. It also covered the case where $E=0$ and $E>0$ .

Figuring out this mapping is actually more straightforward if we consider it as a problem in quantum mechanics. Here is how Fock derived this historically. The Schrödinger equation of the Kepler problem, after a Fourier transformation, is $\left(\frac{p^2}{2m}-E\right)\Phi(p)=\frac{k}{2\pi^2\hbar}\int\mathrm d^3q\frac{\Phi(q)}{|\mathbf p-\mathbf q|^2}.$ Looking into it keenly enough, one may find that the factor $1/|\mathbf p-\mathbf q|^2$ is actually the Jacobian of the stereographic projection mentioned above. Some calculation should reduce this equation to the Schrödinger equation of a free particle on the 3-sphere.

Some online sources: several Wikipedia pages, Baez 2022, Egan 2013.

Thoughts on a middle school thermal physics problem

2022-11-06T09:21:56-08:00

This article is translated from a Chinese article on my Zhihu account. The original article was posted at 2019-03-23 12:49 +0800.

A field construction requires heating an aluminum alloy component of mass $4.2\,\mathrm{kg}$ from $10^\circ\mathrm{C}$ to $63^\circ\mathrm{C}$ or above. Now, we only have $1.2\,\mathrm{kg}$ of water of $90^\circ\mathrm{C}$ in the thermos. Ignoring heat dissipation, design a method to heat the component.

These texts are adapted from some competition problem from Zhejiang. Just throwing the component into the water is not the correct method. The correct answer is to divide the hot water into $3$ (or more) parts of equal mass, and heat the components for $3$ times.

The more parts we separate the hot water into, the higher the final temperature of the component is. This can trigger our thoughts: what is the final temperature if the number of parts tends to infinity?

Suppose the heat capacity of the component is $C_0$ , the initial temperature is $T_0$ ; the heat capacity of hot water is $C$ , the temperature is $T$ . Separate the hot water into $n$ parts of equal heat capacity (i.e. equal mass). Suppose the temperature of the component is $T_m$ after reaching thermal equilibrium with the $m$ th part of the water. Then, the final temperature is $T_n$ .

We now want to find the final temperature at $n\to\infty$ : $T'\coloneqq\lim_{n\to\infty}T_n.$ According to the formula of thermal equilibrium, $C_0\left(T_{m+1}-T_m\right)+\frac Cn\left(T_{m+1}-T\right)=0.$ Solve for $T_{m+1}$ : $T_{m+1}=\frac{C_0T_m+\frac CnT}{C_0+\frac Cn}.$ Let $k\coloneqq\frac{C_0}{C_0+\frac Cn}$ , and $b\coloneqq\frac{\frac CnT}{C_0+\frac Cn}$ , and we have $T_{m+1}=kT_m+b.$ Let $m\coloneqq0,1,\ldots,n-1$ respectively, and we have $\begin{align*} T_1&=kT_0+b,\\ T_2&=kT_1+b,\\ &\vdots\\ T_n&=kT_{n-1}+b. \end{align*}$ Transform the $n$ equations a little, and we have $\begin{align*} k^{n-1}T_1&=k^nT_0+k^{n-1}b, k^{n-2}T_2&=k^{n-1}T_1+k^{n-2}b, &\vdots\\ T_n=kT_{n-1}+b. \end{align*}$ Sum up all the $n$ equations, and we have $T_n=k^nT_0+\sum_{j=0}^{n-1}k^jb.$ According to the formula for summation of geometric progression, we have $T_n=k^nT_0+\frac{1-k^n}{1-k}b$ (this is actually the formula for the $n$ th recursion of linear function.)

Because $\begin{align*} b&\coloneqq\frac{\frac CnT}{C_0+\frac Cn}\\ &=\frac{C_0}{C_0+\frac Cn}\frac{\frac CnT}{C_0}\\ &=\frac{C_0}{C_0+\frac Cn}\left(\frac{C_0+\frac Cn}{C_0}-1\right)T\\ &=\left(1-\frac{C_0}{C_0+\frac Cn}\right)T, \end{align*}$ then $\begin{align*} T_n&=k^nT_0+\frac{1-k^n}{1-k}\left(1-k\right)T\\ &=T+k^n\left(T_0-T\right). \end{align*}$ Now calculate the limit of $k^n$ . We have $\begin{align*} \lim_{n\to\infty}k^n&=\lim_{n\to\infty}\left(\frac{C_0}{C_0+\frac Cn}\right)^n\\ &=\lim_{n\to\infty}\frac1{\left(\frac{C_0+\frac Cn}{C_0}\right)^n}\\ &=\lim_{n\to\infty}\frac{1}{\left(1+\frac{\frac C{C_0}}{n}\right)^n}\\ &=\mathrm e^{-\frac C{C_0}}. \end{align*}$ Therefore, $\begin{align*} T'&\coloneqq\lim_{n\to\infty}T_n\\ &=T+\mathrm e^{-\frac C{C_0}}\left(T_0-T\right). \end{align*}$ This is a intuitive result.

(Gonna take exams for high school entrance…)

Ulysses’ trip | Physics

From Picard iteration to Feynman path integral

Excluding self-loops

Feynman path integral

The role of particle indistinguishability in statistical mechanics

The notational convenience of imaginary time in the derivation of the metric in Poincaré coordinates

Introduction

The derivation

The smallest wave packet in the lowest Landau level

Introduction

The problem

The solution

Further problems

Regularizing the partition function of a hydrogen atom

The duality between two plane trajectories related by a conformal map

Power-law central-force potentials

Corollaries and applications

Some straightforward proofs

Details worth noting

Invertibility of the conformal map

Requirements for the potential

Trajectories that run out of the domain

Arbitrariness in the construction of the conformal map

Rotation and translation

Scaling

Canonical transformation of time

Change in the time variable in the stationary-action principle

You can replace i\mathrm i with −i-\mathrm i in the Schrödinger equation?

A measure-theoretic formulation of statistical ensembles (part 2)

Introduction

Introducing non-thermal contact with an example

Non-thermal contact with a bath

Non-thermal ensembles (bath version)

Non-thermal ensembles (non-bath version)

Summary

A measure-theoretic formulation of statistical ensembles (part 1)

Mathematical tools and notations

Even solutions to bound states in an odd number of δ\delta potential wells

Defend our earth against aliens’ bullets!

How to construct mechanics in higher dimensions?

When a supersonic airplane flies over your head

Finding curvature radius physically

A whirling point charge

An example of non-uniform elements: heavy elastic rope

Kinetic energy, momentum, and angular momentum of rigid bodies

The image of a circular object through a thin lens

Mapping from Kepler problem to free particle on 3-sphere

Thoughts on a middle school thermal physics problem

You can replace $\mathrm i$ with $-\mathrm i$ in the Schrödinger equation?

Even solutions to bound states in an odd number of $\delta$ potential wells