Mathematical Physics Methods on Haifei's Home

Are Christoffel Symbols Tensors After All?

hfwang132@gmail.com (hfwang132) — Sun, 27 Apr 2025 23:12:31 +0800

My favorite way to understand this is to regard $\Gamma$ as a connection on the frame bundle. Then, the transformation law of the Christoffel symbols

$\boxed{ \begin{aligned} \bar{\Gamma}^i_{\mu j} = \frac{\partial \bar{x}^i}{\partial x^k}\frac{\partial x^l}{\partial \bar{x}^j} \frac{\partial x^\nu}{\partial \bar{x}^\mu} \Gamma^k_{\nu l} \color{red}{ + \frac{\partial \bar{x}^i}{\partial x^k} \frac{\partial^2 x^k}{\partial x^\mu \partial \bar{x}^j}} \end{aligned} }$

is nothing more than a gauge transformation:

$\boxed{ \bar{A} = g^{-1}A g \color{red}{+ g^{-1} \mathrm{d}g} }$

The extra inhomogeneous term $ g^{-1} \mathrm{d}g$ is inherently part of a gauge transformation.

From Quantum Field Theory to Cavity Quantum Electrodynamics [Higher and More Subtle Electrodynamics · 6]

hfwang132@gmail.com (hfwang132) — Sun, 13 Apr 2025 01:52:37 +0800

In this article, we start from the QED Lagrangian:

\[\begin{aligned} \mathcal{L}= \bar{\psi}(\mathrm{i}\gamma^\mu \partial_\mu -m)\psi - eA_\mu \bar{\psi} \gamma^\mu \psi -\frac{1}{4} F_{\mu\nu} F^{\mu\nu} \end{aligned} \]

This Lagrangian is highly complex: it not only takes into account the electron’s antiparticle—the positron—but the coupling term $- eA_\mu \bar{\psi} \gamma^\mu \psi$ is also a cubic term, capable of describing various processes such as electron-positron pair creation/annihilation. Due to the presence of the cubic term, this Lagrangian has no analytic solution and can only be solved using perturbation theory in quantum field theory.

Dirac Magnetic Monopoles—Chern Classes and Chern Numbers [Higher and More Wonderful Electrodynamics · 4]

hfwang132@gmail.com (hfwang132) — Sun, 30 Mar 2025 01:13:18 +0800

Previous article in this series:

Godfly: The electromagnetic field is more than just electric and magnetic fields—The AB effect and Berry connection [Higher and More Wonderful Electrodynamics · 3]

In the previous article, we mentioned that if a manifold is topologically trivial, then integrating the Berry curvature over a closed surface $\Sigma$ should yield zero. This is because, according to Stokes’ theorem:

\[\int_\Sigma F = \int_{\partial \Sigma } A = 0\]

However, on a topologically nontrivial manifold, the integral $\int_\Sigma F$ need not be zero. This is because a globally single-valued connection $A$ cannot be defined in this case, so Stokes’ theorem no longer applies. What does this mean?

Electrodynamics from the Perspective of Gauge Field Theory [Higher and More Elegant Electrodynamics · 2]

hfwang132@gmail.com (hfwang132) — Mon, 17 Feb 2025 17:15:55 +0800

If you have not yet read the previous article, please see:

Electrodynamics from the Perspective of Differential Geometry [Higher and More Elegant Electrodynamics · 1]

This article uses many concepts from the previous article, so please make sure you have read it.

This article continues to use the metric convention of $(-,+,+,+)$.

Why is electrodynamics a gauge field theory?

To understand what this statement means, we must first understand what electrodynamics is and what gauge field theory is.

Reviewing Groups, Rings, Fields, Modules, and Vector Spaces Through an Elementary School Math Problem

hfwang132@gmail.com (hfwang132) — Thu, 16 Mar 2023 21:54:23 +0800

0. Introduction to the Problem

With this single college entrance examination mock question, we can review concepts from abstract algebra such as groups, rings, fields, modules, and vector spaces. The problem is shown below:

A college entrance examination mock question from somewhere; in fact, you can figure it out with a little trial and error—it is not difficult.

Simply put, it is a matter of “pulling one hair and moving the surrounding area.” Changing one cell also rotates the surrounding cells. I believe many people have played this puzzle game. Even elementary school students can understand this game.

Highly Useful Formulas

hfwang132@gmail.com (hfwang132) — Sat, 01 Jan 2022 19:21:09 +0800

The following formulas can make problem solving much more convenient.

I. Gaussian Integrals

\[I_n = \int_{0}^{\infty}x^n e^{-b x^2}\mathrm{d}x %= %\left\{\begin{aligned} %\end{aligned}\right. \]

Here is a convenient way to remember it:

First, remember

\[I_0 = \frac{1}{2}\sqrt{\frac{\pi}{b}},\quad I_1 = \frac{1}{2}\frac{1}{b}\]

Next,

\[I_{n+2}=\frac{n+1}{2b}I_n\]

This formula applies to both odd and even cases of $n$.

If you are dealing with Gaussian functions (for example, in probability theory), then this integral will appear very frequently!

II. Powers of Trigonometric Functions

\[I_n=\int_{0}^{\frac{\pi}{2}}\sin^{n}x\mathrm{d}x = \int_{0}^{\frac{\pi}{2}}\cos^{n}x\mathrm{d}x\]

The following expression gives $I_n$

Foundations of Axiomatic Set Theory (Part I): The Empty Set and Russell's Paradox

hfwang132@gmail.com (hfwang132) — Sat, 23 Jan 2021 17:41:35 +0800

Preview

After reading this article, you will understand the first three axioms of ZF axiomatic set theory—the Axiom of Existence, the Axiom of Extensionality, and the Axiom Schema of Separation—and use them to resolve Russell’s paradox and define the empty set.

Introduction

We often hear statements such as “set theory is the foundation of modern mathematics.” In the eyes of many people, the set theory learned in secondary school consists merely of sets and simple operations—how could that be foundational? To understand this, we must understand how set theory was axiomatized.