Author Archives: Sung Lee

About Sung Lee

I am a mathematician and a physics buff. I am interested in studying mathematics inspired by theoretical physics as well as theoretical physics itself. Besides, I am also interested in studying theoretical computer science.

Matter and Waves

Since Huygens and Newton, physicists have known that light is described by (electromagnetic) waves and particles (photons). Such peculiar nature of light is called wave-particle duality. What about material particles such as electrons? de Broglie proposed a bold hypothesis that what is true for light is also true for material particles i.e. they will also exhibit wave nature. (This was indeed confirmed by experiments.) So how do we mathematically model such a wave? What physicists thought of using to study material particles was a complex plane wave called de Broglie wave. It looks like $$\psi(x,t)=Ae^{i(kx-\omega t)}$$ for 1-dimensional case. For 3-dimensional case, it would be \begin{equation}\label{eq:planewave}\psi({\bf r},t)=Ae^{i({\bf k}\cdot{\bf r}-\omega t)}\end{equation} Before we continue, one may wonder how physicists came up with this kind of wave. I can only speculate but such a complex plane wave was already well-known to physicists as it is a solution of Maxwell’s equation in electromagnetism. Hence, complex plane wave may describe electromagnetic wave, and naturally it became the first candidate for modeling material particles. In fact, it worked out well as we shall see and consequently complex numbers played a crucial role in building quantum mechanics.

Let us first study some properties of plane waves. The plane wave \eqref{eq:planewave} describes a free particle, more accurately a free particle in a state. In order for a plane wave to behave like a particle, we want it to be localized i.e. the wave is defined in a tiny region. (There is a more mathematically subtle reason why we require this.) We can achieve this by redefining $\psi({\bf r},t)$ as $$\psi({\bf r},t)=\left\{\begin{array}{ccc}Ae^{i({\bf k}\cdot{\bf r}-\omega t)} & \mbox{for} & {\bf r}\ \mbox{within a volume}\ V=L^3\\0 & \mbox{for} & {\bf r}\ \mbox{outside a volume}\ V=L^3\end{array}\right.$$ Physicists call this box renormalization. Physically the state of a particle must not depend on a particular location of the tiny box, so we require the periodicity condition $$\psi(x,y,z,t)=\psi(x+L,y,z,t)=\psi(x,y+L,z,t)=\psi(x,y,z+L,t)$$ Here, $L$ is called wave length. The periodicity condition implies that ${\bf k}$ is quantized as $${\bf k}=\frac{2\pi}{L}{\bf n}$$ where ${\bf k}=(k_x,k_y,k_z)$, ${\bf n}=(n_x,n_y,n_z)$, and $n_i=0,1,2,\cdots$, $i=x,y,z$. The vector ${\bf k}$ is called wave vector and for 1-dimensional case, $k$ is called wave number. If the wave is periodic in time, say $\psi(x,t)=\psi(x,t+T)$, then we obtain $e^{-i\omega T}=1$. The nonzero minimum value of $T$ is $T=\frac{2\pi}{\omega}$. $\omega=\frac{2\pi}{T}$ is called angular frequency. $kx-\omega t$ is called phase and if $kx-\omega t$ is constant, the wave moves at the speed $v_p=\frac{dx}{dt}=\frac{\omega}{k}$. This $v_p$ is called phase velocity. For 3-dimensional case, \begin{align*}\psi({\bf r},t)&=Ae^{i({\bf k}\cdot{\bf r}-\omega t)}\\&=Ae^{i{\bf k}\cdot\left({\bf r}-\frac{\omega t}{|{\bf k}|^2}{\bf k}\right)}\\&=Ae^{i{\bf k}\cdot\left({\bf r}-\frac{\omega t}{|{\bf k}|}\hat{\bf k}\right)}\end{align*}So, the phase velocity would be $${\bf v}_p=\frac{d{\bf r}}{dt}=\frac{\omega}{|{\bf k}|}\hat{\bf k}$$

The image of wave function $\psi(x,t)=Ae^{i(kx-\omega t)}$ is a circle. We are in fact quite familiar with this kind of waves. On a beautiful day, you go to a lake. You would be then tempted to throw a rock into the cam water. When you do, you would see circular water waves spreading out from the point of impact.


[1] Walter Greiner, Quantum Mechanics, An Introduction, 4th Edition, Springer, 2001

[2] Quantum Mechanics, H.-S. Song (in Korean)

Related Rates

Related rates problems often involve (context-wise) real-life applications of the chain rule/implicit differentiation. Here are some of the examples that are commonly seen in calculus textbooks.

Example. Car A is traveling west at 50mi/h and car B is traveling north at 60mi/h. Both are headed for the intersection of the two roads. At what rate are the cars approaching each other when car A is 0.3 mi and car B is 0.4 mi from the intersection?


Denote by $x$ and $y$ the distances from the intersection to car A and to car B, respectively. Then we have $\frac{dx}{dt}=-50$mi/h and $\frac{dy}{dt}=-60$mi/h. Let us denote $z$ the distance between $A$ and $B$. Then by Pythagorean law we have $$z^2=x^2+y^2$$ Differentiating this with respect to $t$, we obtain $$z\frac{dz}{dt}=x\frac{dx}{dt}+y\frac{dy}{dt}$$ and thus \begin{align*}\frac{dz}{dt}&=\frac{1}{z}\left[x\frac{dx}{dt}+y\frac{dy}{dt}\right]\\&=\frac{1}{0.5}[0.3(-50)+0.4(-60)]=-78\mathrm{mi/h}\end{align*}

Example. Air is being pumped into a spherical balloon so that its volume increases at a rate of $100\mathrm{cm}^3/\mathrm{s}$. How fast is the radius of the balloon increasing when the diameter is 50 cm?

Solution. Let $V$ and $r$ denote the volume and the radius of the spherical balloon. Then $V=\frac{4}{3}\pi r^3$. Differentiating this with respect to $t$, we obtain $$\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}$$ So, \begin{align*}\frac{dr}{dt}&=\frac{1}{4\pi r^2}\frac{dV}{dt}\\&=\frac{1}{4\pi(25)^2}100\\&=\frac{1}{25\pi}\mathrm{cm/s}\end{align*}

Example. Gravel is being dumped from a conveyor belt at a rate of $30 \mathrm{ft}^3/\mathrm{min}$ and its coarseness is such that it forms a pile in the shape of a cone whose base diameter and height are the same. How fast is the height of the pile increasing when the pile is 10 ft high?

Solution. The cross section of the gravel pile is shown in the figure below.

The amount of gravel dumped is the same as the volume of the cone. Let us denote the volume by $V$, its base radius by $r$, and its height by $h$. Then $V=\frac{1}{3}\pi r^2h$. Since $h=2r$, $V$ can be written as $$V=\frac{1}{12}\pi h^3$$ Differentiating this with respect to $t$, we obtain $$\frac{dV}{dt}=\frac{1}{4}\pi h^2\frac{dh}{dt}$$ So, we have \begin{align*}\frac{dh}{dt}&=\frac{4}{\pi h^2}\frac{dV}{dt}\\&=\frac{4}{\pi(10)^2}(30)=\frac{1.2}{\pi}\mathrm{ft/min}\approx 0.38\mathrm{ft/min}\end{align*}

Example. A ladder 10 ft long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 1 ft/s, how fast is the top of the ladder sliding down the wall when the bottom of the ladder is 6 ft from the wall?


Let us denote by $x$ and $y$ the distance from the wall to the bottom of the ladder and the distance from the top of the ladder to the floor, respectively. By Pythagorean law, we have $x^2+y^2=100$. Differentiating this with respect to $t$, we obtain $$x\frac{dx}{dt}+y\frac{dy}{dt}=0$$ Hence, we have \begin{align*}\frac{dy}{dt}&=-\frac{x}{y}\frac{dx}{dt}\\&=-\frac{6}{8}(1)=-\frac{3}{4}\mathrm{ft/s}\end{align*}

Example. A water tank has the shape of an inverted circular cone with base radius 2m and heigh 4 m. If water is being pumped into the tank at a rate of $2 \mathrm{m}^3/\mathrm{min}$, find the rate at which the water level is rising when the water is 3 m deep.

Solution. The cross section of the water tank is shown in the figure below.

The amount of water $V$ when the water level is $h$ and the surface radius is $r$ is $V=\frac{1}{3}\pi r^2h$. From the above figure we have the following ratio holds $$\frac{2}{4}=\frac{r}{h}$$ i.e. $r=\frac{h}{2}$. SO $V$ can be written as $$V=\frac{1}{12}\pi h^3$$ Differentiating this with respect to $t$, we obtain $$\frac{dV}{dt}=\frac{1}{4}\pi h^2\frac{dh}{dt}$$ Hence, \begin{align*}\frac{dh}{dt}&=\frac{4}{\pi h^2}\frac{dV}{dt}\\&=\frac{4}{\pi(3)^2}(2)\\&=\frac{8}{9\pi}\mathrm{m/min}\approx 0.28\mathrm{m/min}\end{align*}

The Klein-Gordon Equation

For the Schrödinger equation $$i\hbar\frac{\partial\psi}{\partial t}=\hat H\psi({\bf x},t),$$ the Hamiltonian $$\hat H=-\frac{\hbar^2}{2m_0}\nabla^2+V({\bf x})$$ corresponds to the nonrelativistic energy-momentum relation $$\hat E=\frac{\hat p^2}{2m_0}+V({\bf x})$$ where $$\hat E=i\hbar\frac{\partial}{\partial t},\ \hat p=-i\hbar\nabla$$ So, naturally considering the relativistic energy-momentum relation $$\frac{E^2}{c^2}-{\bf p}\cdot{\bf p}=m_0^2c^2$$ would be the starting point to obtain a relativistic generalization of the Schrödinger equation. Replacing $E$ and ${\bf p}\cdot{\bf p}$ by operators $$\hat E=i\hbar\frac{\partial}{\partial t}\ \mbox{and}\ \hat p\cdot\hat p=-\hbar^2\nabla^2$$ acting on a wave function $\psi$, we obtain the Klein-Gordon equation for a free particle \begin{equation}\label{eq:k-g}\left(\Box-\frac{m_0^2c^2}{\hbar^2}\right)\psi=0\end{equation} where $$\Box=-\frac{1}{c^2}\frac{\partial^2}{\partial t^2}+\nabla^2$$

Free solutions of the Schrödinger equation with $V({\bf x})=0$ are of the form $$\psi=\exp\left[\frac{i}{\hbar}(-Et+{\bf p}\cdot{\bf x})\right]$$ They are also free solutions of the Klein-Gordon equation \eqref{eq:k-g} with the energy condition $$E=\pm c\sqrt{m_0^2c^2+p^2}$$ The solutions yielding negative energies appear to be unphysical and initially considered so by physicists, but later they were interpreted as antiparticles. Antiparticles are indeed seen in nature. In reality, antiparticles also have positive energies. Antiparticles as wave functions with negative energies is merely an interpretation of the mathematical representation of the energy condition. If antiparticles weren’t discovered, the negative energy condition would have been still thought to be unphysical.

Other than allowing solutions with negative energies, there was another issue with the Klein-Gordon equation noted by physicists. The conservation of four-current density $$j_\mu=\frac{i\hbar}{2m_0}(\psi^\ast\nabla_\mu\psi-\psi\nabla_\mu\psi^\ast),$$ where $\psi^\ast$ denotes the complex conjugate of $\psi$ and $\nabla_\mu=\left(-\frac{1}{c^2}\frac{\partial}{\partial t},\nabla\right)$, implies that the quantity $$\rho=\frac{i\hbar}{2m_0c^2}\left(\psi^\ast\frac{\partial\psi}{\partial t}-\psi\frac{\partial\psi^\ast}{\partial t}\right)$$ can be considered as a probability density. However, the problem is that $\rho$ can be negative. This is due to the appearance of first-order partial derivative $\frac{\partial\psi}{\partial t}$, which is the consequence of the Klein-Gordon equation being of second-order in time. Because of this, the Klein-Gordon equation was not regarded as a physically viable relativistic generalization of the Schrödinger equation and physicists were instead looking for a relativistic generalization of first-order in time like the Schrödinger equation. Such an equation was finally discovered by P. A. M. Dirac and is called the Dirac equation. On the other hand, the Klein-Gordon equation drew attention of physicists again after they realized that $\rho$ can be interpreted as charge density, and indeed charged pions $\pi^+$ and $\pi^-$ were discovered. Today, the Klein-Gordon equation is an important relativistic equation that describe charged spin-0 particles.


[1] Walter Greiner, Relativistic Quantum Mechanics, 3rd Edition, Springer-Verlag, 2000

The Curvature, the Einstein Equations, and the Black Hole II: The Curvature

In this lecture, we study different notions of curvatures of a Riemannian or a pseudo-Riemannian $n$-manifold $M$ with metric tensor $g_{ij}$. We will discuss only local expressions of curvatures as those are the ones we actually use for doing physics in general relativity.

First we need to introduce the Christoffel symbols $\Gamma_{ij}^k$. The Christoffel symbols are associated with the differentiation of vector fields in a Riemannian or a pseudo Riemannian manifold $M$, called the Levi-Civita connection. The Levi-Civita connection $\nabla$ is a generalization of the covariant derivative of vector fields in the Euclidean space. Locally the Levi-civita connection is defined by $$\nabla_{\frac{\partial}{\partial x^i}}\frac{\partial}{\partial x^j}=\sum_{k}\Gamma_{ij}^k\frac{\partial}{\partial x^k}$$ and the Christoffel symbol is given by $$\Gamma_{ij}^k=\frac{1}{2}\sum_\ell g^{k\ell}\left\{\frac{\partial g_{j\ell}}{\partial x^i}+\frac{\partial g_{\ell i}}{\partial x^j}-\frac{\partial g_{ij}}{\partial x^\ell}\right\}$$ where $g^{k\ell}$ is the inverse of the metric tensor.

Locally the Riemann curvature tensor $R_{ijk}^\ell$ is given by $$R_{ijk}^\ell=\frac{\partial}{\partial x^j}\Gamma_{ik}^\ell-\frac{\partial}{\partial x^k}\Gamma_{ij}^\ell+\sum_p\left\{\Gamma_{jp}^\ell\Gamma_{ik}^p-\Gamma_{kp}^\ell\Gamma_{ij}^p\right\}$$

Locally the sectional curvature $K(X,Y)$ of $M$ with respect to the plane spanned by tangent vectors $X,Y\in T_pM$ is given by \begin{equation}\label{eq:sectcurv}K_p(X,Y)=g^{ii}R_{iji}^j\end{equation} assuming that $X,Y\in\mathrm{span}\left\{\frac{\partial}{\partial x^i},\frac{\partial}{\partial x^j}\right\}$. The sectional curvature is a generalization of the Gaußian curvature of a surface in 3-space. To see this, let $\varphi: M^2\longrightarrow M^3$ be a conformal parametric surface $M^2$ immersed in 3-space $M^3$ with metric $e^{u(x,y)}(dx^2+dy^2)$. The Gaußian curvature $K$ of $\varphi$ can be calculated using the formula (due to Karl Friedrich Gauß) $$K=\frac{\ell n-m^2}{EG-F^2}$$ where \begin{align*}E&=\langle\varphi_x,\varphi_x\rangle,\ F=\langle\varphi_x,\varphi_y\rangle,\ G=\langle\varphi_y,\varphi_y\rangle,\\\ell&=\langle\varphi_{xx},N\rangle,\ m=\langle\varphi_{xy},N\rangle,\ n=\langle\varphi_{yy},N\rangle\end{align*} Here, $\langle\ ,\ \rangle$ stands for the inner product induced by the conformal metric $e^{u(x,y)}(dx^2+dy^2)$ and $N$ is the unit normal vector field on $\varphi$. The Gaußian curvature is then obtained as the Liouville’s partial differential equation \begin{equation}\label{eq:liouville}\nabla^2 u=-2Ke^u\end{equation} On the other hand, using \eqref{eq:sectcurv} we find the sectional curvature of $\varphi$ to be $$g^{11}R_{121}^2=-\frac{e^{u(x,y)}}{2}\nabla^2u$$ which coincides with the Gaußian curvature $K$ from \eqref{eq:liouville}

Example. Let us compute the sectional curvature of the hyperbolic plane $$\mathbb{H}^2=\{(x,y)\in\mathbb{R}^2: y>0\}$$ with metric $$ds^2=\frac{dx^2+dy^2}{y^2}$$

The metric tensor is $(g_{ij})=\begin{pmatrix}\frac{1}{y^2} & 0\\0 & \frac{1}{y^2}\end{pmatrix}$. The Riemann curvature tensor $R_{121}^2$ is \begin{align*}R_{121}^2&=\frac{\partial}{\partial y}\Gamma_{11}^2-\frac{\partial}{\partial x}\Gamma_{12}^2+\sum_p\{\Gamma_{2p}^p\Gamma_{11}^p-\Gamma_{1p}^2\Gamma_{12}^p\}\\&=\frac{\partial}{\partial y}\Gamma_{11}^2-\frac{\partial}{\partial x}\Gamma_{12}^2+\Gamma_{21}^2\Gamma_{11}^1-\Gamma_{11}^2\Gamma_{12}^1+\Gamma_{22}^2\Gamma_{11}^2-\Gamma_{12}^2\Gamma_{12}^2\end{align*} We find the Christoffel symbols $$\Gamma_{11}^2=\frac{1}{y},\ \Gamma_{12}^1=-\frac{1}{y},\ \Gamma_{12}^2=0,\ \Gamma_{21}^2=0,\ \Gamma_{22}^2=-\frac{1}{y}$$ Thus we obtain $R_{121}^2=-\frac{1}{y^2}$ and hence $\mathbb{H}^2$ has the constant negative sectional curvature $$K=g^{11}R_{121}^2=y^2\left(-\frac{1}{y^2}\right)=-1$$ What is the shortest path connecting two points $(x_1,y_1)$ and $(x_2,y_2)$ in $\mathbb{H}^2$? Such shortest paths are called geodesics in differential geometry. To find out what a geodesic in $\mathbb{H}^2$ looks like, let $$J=\int_{(x_1,y_1)}^{(x_2,y_2)}ds=\int_{(x_1,y_1)}^{(x_2,y_2)}\frac{\sqrt{1+y_x^2}}{y}dx$$ where $y_x=\frac{dy}{dx}$. The shortest path would satisfy the Euler-Lagrange equation \begin{equation}\label{eq:E-L}\frac{\partial f}{\partial x}-\frac{d}{dx}\left(f-y_x\frac{\partial f}{\partial y_x}\right)=0\end{equation}with $f(y,y_x,x)=\frac{\sqrt{1+y_x^2}}{y}$. Since $f$ does not depend on $x$, $\frac{\partial f}{\partial x}=0$ and the Euler-Lagrange equation \eqref{eq:E-L} becomes $$\frac{d}{dx}\left[\frac{1}{y\sqrt{1+y_x^2}}\right]=0$$ i.e. \begin{equation}\label{eq:E-L2}\frac{1}{y\sqrt{1+y_x^2}}=C\end{equation} where $C$ is a constant. The equation \eqref{eq:E-L2} results in a separable differential equation $$\frac{dy}{dx}=\frac{\sqrt{r^2-y^2}}{y}$$ where $r^2=\frac{1}{C}$. The solution of this equation is $$(x-a)^2+y^2=r^2$$ where $a$ is a constant. Since $y>0$, the solution represents an equation of upper semi circle centered at $(a,0)$ with radius $r$, that is the shortest path (geodesic) between two points $(x_1,y_1)$ and $(x_2,y_2)$ in $\mathbb{H}^2$ is a part of an upper semicircle joining them. In particular, if $x_1=x_2$, the geodesic between $(x_1,y_1)$ and $(x_2,y_2)$ is the vertical line passing through the two points. Such a vertical line can still be considered as an upper semicircle with radius $\infty$.

Geodesics in Hyperbolic Plane

Two other notions of curvatures are Ricci and scalar curvatures. The Ricci curvature tensor is given by $$\mathrm{Ric}_p\left(\frac{\partial}{\partial x^i},\frac{\partial}{\partial x^j}\right)=\sum_kR_{ikj}^k$$ We usually denote $\mathrm{Ric}_p\left(\frac{\partial}{\partial x^i},\frac{\partial}{\partial x^j}\right)$ simply by $R_{ij}$. The scalar curvature $\mathrm{Scal}(p)$ is given by $$\mathrm{Scal}(p)=\sum_{i}g^{ii}R_{ii}$$ The scalar curvature can be given, in terms of the sectional curvature, by $$\mathrm{Scal}(p)=\sum_{i\ne j}K_p\left(\frac{\partial}{\partial x^i},\frac{\partial}{\partial x^j}\right)$$ The scalar curvature is usually denoted by $R$ in general relativity.

Definition. A Riemannian or a pseudo-Riemannian manifold $(M,g)$ is said to be maximally symmetric if $(M,g)$ has constant sectional curvature $\kappa$.

Theorem. If a Riemannian or a pseudo-Riemannian manifold $(M,g)$ is maximally symmetric, then $$R_{ii}=\kappa(n-1)g_{ii}$$ where $\kappa$ is the constant sectional curvature of $(M,g)$ and $n=\dim M$.

Corollary. If $(M,g)$ has the constant sectional curvature $\kappa$, then $$\mathrm{Scal}(p)=n(n-1)\kappa$$ where $n=\dim M$.

The Curvature, the Einstein Equations, and the Black Hole I: Riemannian and Pseudo-Riemannian Manifolds

This is the first part of the lecture note that is an extended version of the series of lectures I have given in the physics seminar at the University of Southern Mississippi. The majority of the audience were graduate students who have never had any prior encounter with differential geometry. Therefore, I tried to maintain mathematical rigor and technicalities at a minimum when discussed differential geometric concepts, instead mostly used hand-waving and rudimentary arguments with emphases on physical ideas and intuition.

In order to study general relativity, we need to get familiar with (pseudo-)Riemannian manifolds. But first, what is a manifold? A manifold is, very roughly speaking, a space which is locally looks like our space (Euclidean space). In other words, for any point $p$ in a manifold $M$ there exists a neighborhood (called a coordinate neighborhood) $U$ of $p$ such that $U\cong\mathbb{R}^n$. Here $\cong$ means they are homeomorphic i.e. topologically indistinguishable. Such a property is said to be locally Euclidean and a space which is locally $\mathbb{R}^n$ is called an $n$-dimensional manifold. Actually being locally Euclidean is not the only condition for a space to be a manifold but that is the most important property of a manifold for physicists.

Figure 1, A manifold

Figure 1 shows a manifold $M$, two coordinate neighborhood $U$ and $V$ with homeomorphisms $\phi$ and $\psi$, respectively. Why do we need manifolds by the way? In order to do physics, we need coordinates. Without coordinates we can’t write equations of motion. Unfortunately, even for a simple familiar space there is no guarantee that there will be a global coordinate system. Here is an example.

Example. The points $(x,y,z)$ on the 2-sphere $S^2$ are represented in terms of the spherical coordinates $(\theta,\phi)$ as $$x=\sin\phi\cos\theta,\ y=\sin\phi\sin\theta,\ z=\cos\phi,\ 0\leq\phi\leq\pi,\ 0\leq\theta\leq 2\pi$$ Using the chain rule, we can write the standard basis $\frac{\partial}{\partial\theta}$, $\frac{\partial}{\partial\phi}$ for the tangent space $T_\ast S^2$ in spherical coordinates in terms of the standard basis $\frac{\partial}{\partial x}$, $\frac{\partial}{\partial y}$, $\frac{\partial}{\partial y}$ in rectangular coordinates as \begin{align*}\frac{\partial}{\partial\theta}&=-\sin\phi\sin\theta\frac{\partial}{\partial x}+\sin\phi\cos\theta\frac{\partial}{\partial y}\\\frac{\partial}{\partial\phi}&=\cos\phi\cos\theta\frac{\partial}{\partial x}+\cos\phi\sin\theta\frac{\partial}{\partial y}-\sin\phi\frac{\partial}{\partial z}\end{align*} This frame field is not globally defined on $S^2$ because $\frac{\partial}{\partial\theta}=0$ at $\phi=0,\pi$ i.e. at the north pole $N=(0,0,1)$ and at the south pole $S=(0,0,-1)$ as also seen in Figure 2.

Figure 2. The 2-sphere with frame field

The 2-sphere $S^2$ is covered by two coordinates neighborhoods $U=S^2\setminus{N}$ and $V=S^2\setminus{S}$, each of which is identified with $\mathbb{R}^2$, the Euclidean plane via the stereographic projection. Figure 3 shows the stereographic projection from the north pole $N$, which is a one-to-one correspondence from $U$ to $\mathbb{R}^2$.

Figure 3. The Stereographic Projection

A global coordinate system exists in the flat Euclidean space (or a flat pseudo-Euclidean space including Minkowski spacetime), however general relativity has taught us that a physical space is not necessarily a flat space (vaccum spacetime). This is where a manifold comes in. A manifold guarantees the existence coordinate system at least locally and for most cases that is good enough to do physics in particular we write physical equations in a coordinate independent way, so that if a physical equation holds in on coordinate neighborhood, it should also hold in another coordinate neighborhood in the same way.

We would be needing more than topological manifolds to do physics. For an obvious reason we need differentiable manifolds. I am not going to delve into this except for just saying that a differentiable manifold is a manifold on which the differentiability of functions and vector fields can be defined and also to which tangent space at each point can be considered. (If we can’t differentiate fields, we cannot do physics.) In addition, we need Riemannian manifolds. A Riemannian manifold is a differentiable manifold with a Riemannian metric. So what is a Riemannian metric? A Riemannian metric $g$ is a positive definite bilinear symmetric form $g_p: T_pM\times T_pM\longrightarrow\mathbb{R}$, which induces a positive definite inner product on each tangent space $T_pM$. In a  coordinate neighborhood, the metric $g$ can be locally given by \begin{equation}\label{eq:metric}g=g_{ij}dx^i\otimes dx^j\end{equation}Here we are using the Einstein’s summation convention. The  $n\times n$ matrix $(g_{ij})$ is called a metric tensor and physicists often simply write $g_{ij}$ for the metric tensor, not for the component. Since $g_{ij}$ is a symmetric tensor, it can be diagonalized. Since the metric is preserved under diagonalization (which amounts to a change of coordinates), without loss of generality we may assume that $g_{ij}=0$ if $i\ne j$ so that the metric tensor \eqref{eq:metric} is written as\begin{equation}\label{eq:metric2}g=g_{ii}dx^i\otimes dx^i\end{equation}Let the dimension of $M$ be $n$. Then each tangent space $T_pM$ is an $n$-dimensional vector space with the canonical orthonormal basis $\left(\frac{\partial}{\partial x^1}\right)_p,\cdots,\left(\frac{\partial}{\partial x^n}\right)_p$. Thus any tangent vector $v\in T_pM$ can be written as $$v=v^j\left(\frac{\partial}{\partial x^j}\right)_p$$ The differential 1-forms $d^i$ are the duals of $\frac{\partial}{\partial x^i}$, respectively. $$dx^i\left(\frac{\partial}{\partial x^j}\right)=\delta_{ij}$$ and hence $$dx^i(v)=v^i$$ For any two tangent vectors $v,w\in T_pM$ using \eqref{eq:metric2} we obtain \begin{equation}\label{eq:metric3}g(v,w)=g_{ii}dx^i\otimes dx^i(v,w)=g_{ii}dx^i(v)dx^i(w)=g_{ii}v^iw^i\end{equation}\eqref{eq:metric3} shows how the metric $g$ induces an inner product on each tangent space $T_pM$. In doing physics, in particular general relativity, the physical space is often a pseudo-Riemannian manifold rather than a Riemannian manifold. A pseudo-Riemannian manifold is equipped with a pseudo-Riemannian metric which is an indefinite symmetric bilinear form. So the induced inner product is indefinite. A good example is the Minkowski spacetime $\mathbb{R}^{3+1}$ which is $\mathbb{R}^4$ with the Minkowski metric or the Lorentz-Minkowski metric \begin{equation}\label{eq:minkowski}g=-dt^2+dx^2+dy^2+dz^2\end{equation} The Minkowski metric \eqref{eq:minkowski} induces the inner product on $\mathbb{R}^{3+1}$ (The Minkowski spacetime has a single coordinate neighborhood $\mathbb{R}^{3+1}$ itself and every tangent space $T_p\mathbb{R}^{3+1}$ is isomorphic to $\mathbb{R}^{3+1}$, hence $\mathbb{R}^{3+1}$ is a manifold and at the same time it is also a vector space.) $$\langle v,w\rangle=-v^0w^0+v^1w^1+v^2w^2+v^3w^3$$ where $v=(v^0,v^1,v^2,v^3)$ and $w=(w^0,w^1,w^2,w^3)$ are four-vectors in $\mathbb{R}^{3+1}$.

I am ending this lecture with saying that the metric tensor $g_{ij}$ is the most important ingredient of a Riemannian or a pseudo-Riemannian manifold. You can literally find out everything about the geometry of a Riemannian or a pseudo-Riemannian manifold with the metric tensor. With the metric tensor, you can also find out about what gravity does when there is matter (the source of gravity) present in the manifold.