Tensor and Relativity - 0. What is Tensor?

# Tensor and Relativity - 0. What is Tensor? This is an explanation about tensors. This document is not completed yet. If you want Korean version of this document, go to . Only basic parts are shown, and the others are hiden initially. Click the Show/Hide buttons to toggle the view.

Use google translate (in chrome browser) if you wanna know what korean means. I will translate this document soon or quite late when I have spare time.

## PH

2022-07-11 : English version of tensors.

## TOC ## Introduction Before explaining the theory of relativity, I would rather explain tensors. In order to properly learn physics as well as relativity, you must clearly know what tensors are and be able to handle them well properly. As such, the concept and use of tensors are very important in physics. When I was taking physics classes at universities and graduate schools, my professors sometimes said the word tensor, but at first I didn't know what tensor meant at all. If the transformation is the same, is it called a tensor? Is that something like a matrix? If it changes/tranforms the same as vector, is it a tensor? So is vector a tensor? Is a tensor a kind of vector? What exactly is a vector? and so on. I was confused. After entering graduate school, I had the opportunity to learn about tensor properly while taking general relativity classes. And it was an easier concept than I thought. Each physicist/mathematician has a slightly different way of explaining or using a tensor, but here I will explain based on the way I understand and following the way that I feel is the most plausible.

I asked about the tensor to OpenAI ChatGPT. kipid: What is tensor in relativity theory? In relativity theory, a tensor is a mathematical object that allows us to describe physical quantities in a way that is independent of the coordinate system chosen to describe them. Tensors are important in relativity because the theory requires physical laws to be expressed in a way that is invariant under coordinate transformations, which means that the equations describing the laws must look the same regardless of the choice of coordinate system. A tensor is a multilinear function that takes vectors and/or other tensors as arguments and produces a scalar as output. Tensors are typically represented as arrays of numbers, with each index of the array corresponding to a particular direction in space or time. The number of indices that a tensor has is called its rank, and tensors of different ranks transform differently under coordinate transformations. In relativity theory, tensors are used to describe physical quantities such as the curvature of spacetime, the stress-energy tensor, and the electromagnetic field tensor. These quantities are described using tensors because they need to be defined in a way that is independent of the observer's reference frame. Tensors allow us to express these physical quantities in a way that is both coordinate-invariant and meaningful to observers in different reference frames.

kipid: Show me the references of your answer. Sure! Here are some references to support my answer:

Carroll, S. M. (2004). Spacetime and geometry: an introduction to general relativity. Addison-Wesley.
Wald, R. M. (1984). General relativity. University of Chicago Press.
Misner, C. W., Thorne, K. S., & Wheeler, J. A. (1973). Gravitation. W. H. Freeman and Company.
Schutz, B. F. (1985). A first course in general relativity. Cambridge University Press.

These are all well-respected textbooks on general relativity that discuss tensors and their role in the theory.

In a wikipedia page of tensor , there are list-ups of complicated numerical expressions. But I couldn't find any good interpretation of tensors anywhere. Well, though all of them are probably correct explanations, there are many difficult parts for people who are new to it to understand. If you search for a tensor on the wiki, you will find something like this: "Tensor is a concept used in slightly different senses in mathematics and physics. In fields such as multilinear algebra and differential geometry of mathematics, a tensor is simply a multilinear function." Can you figure it out properly? Where the hell is it being used? The abstract tensor concept used in mathematics will not be of much help at first, so I will start explaining it from the point of view of the Department of Physics.

Drawing, by Midjourney (AI drawer), which describes and explains tensors of relativity

In physics, it is almost essential to introduce a coordinate system to explain natural phenomena, give numbers to time and space, and explain laws with relationships between these numbers. It can be seen that almost everything in physics is difficult to explain without these concepts of space and time. Even things that seem to have nothing to do with the concept of space-time (charge, mass, etc.) can be characterized as how they move through space over time, i.e., whether acceleration occurs or position changes, in order to measure or understand their quantity or effect. You can see that, however, there are more than a single way of introducing these coordinate systems, units (such as unit: SI unit system or cgs unit system ) and scales. Since the laws of physics must not change even if we introduce any coordinates, it is necessary to describe the laws of physics regardless of the coordinates introduced. Tensor is a concept introduced to uniquely describe natural phenomena regardless of the introduced coordinates. ## Cartesian Coordinate and Flat Space I want to explain things, physical phenomena, and everything that are happening around me. How do I start? First, let's take a hint from the theories laid down by our predecessors. Don't go too far, and just think about the spaces around you, you can intuitively see that it is composed of an almost flat three-dimensional space and one-dimensional time. What is roughly an n-dimensional space? (In English, n-dimensional manifold or space: Manifold is a word that means folded many times, and it can be seen as a mathematical name for space.) It means that all points (Points, Positions, or Events) in the space (actually space-time) can be described with n independent variables. Taking the universe we live in as an example, it can be said that all space-time is described as x, y, z(+time) and therefore has 3(+1) dimension. Let's skip the story of how many dimensions our universe actually has and whether the universe is finite or infinite. Before asking questions like that, let's start with the closest thing to us. How do we manipulate intuitively what we see? How can we know the way to describe it and to understand it? Then let's move on to the weird questions later. (I think that) The beginning of physics has to depend to some extent on intuition, and can only rely on vague intuitions. For questions, like "Why is our universe 3(+1) dimensional?", that are not likely to be resolved immediately, let's accept them roughly first, then let's come back later and think more deeply about them. It's important to doubt things you understand intuitively. But just curiosity doesn't solve everything. Wouldn't it be better to know something easier first, and to solve such doubts next? What we use a lot when describing 3D space is the Cartesian coordinate. Why do we use these coordinate systems? To start with the conclusion, it can be said that the world we live in is almost flat and spatially symmetrical. (Even though you don't know exactly what isotropic is yet, you can intuitively feel that there is something invariant under rotations.) And it is convenient to see the physical laws that govern the universe in this coordinate system. However, 'physical laws or observation results should not change no matter which coordinate system is selected and who looks at it.' This is the beginning of tensors. It is also the beginning of the theory of relativity. It is the assumption that all physical laws and physical theories are based on.

There is a possibility that this assumption is wrong. (It might be less confusing if you skip this part. Just read this part for fun.) The Many-worlds interpretation (or Multiverse) concept of quantum mechanics is one of the possibility. 'The world I live in is the same as the world other people live in.', 'The past, present, and future I see are the same as the past, present, and future seen by others.' What you thought of course could be wrong. I mean that the future, the world may be different depending on the observer. For example, 'A, standing still, saw that the gunner shot and hit the bird, but B, who is on the spaceship far away, could observe that the gunner shot but missed the bird.' That is 'In my world, the bird is shot and killed by a gunner's bullet, but in someone else's world, the gun misses and the bird lives well in the future.' This is a concept that is based on the probabilistic physical laws of quantum mechanics. (I don't know exactly.) Roughly speaking, according to quantum mechanics, all events happen probabilistically. Therefore the gunner's bullet hits the bird with a probability of 1/2, and it misses the bird with a probability of 1/2. So there are two universes (one in which the bird is dead and the other in which the bird is alive). That is somewhat absurd, but make-sense story. Well, if you're a person who has learned a little bit of quantum mechanics, it won't be completely absurd. Anyway, let's get past this absurd story and proceed with the assumption that the nature of the event does not change, no matter who the observer is. Well, if it's different depending on the observer, the laws of physics won't mean much. Quantum mechanics also predicts the same 'probabilistic' result regardless of the observer. And 'B living in A's universe' will tell A that the bird is dead, and 'A in B's universe' will say that the bird is alive. So you might say it's not a big problem because, in each universe, they coincide with each other. There are mistakes that people sometimes make while learning quantum mechanics (I also did it at the beginning), so I'll talk about it a little more. Listening to the above story, it is easy to misunderstand that 'the reason why all events happen probabilistically' is because we do not know exactly the initial conditions for the system which we are observing. Just like throwing a dice and not knowing what the number will come out. But that's not what quantum mechanics is talking about. It is the claim of quantum mechanics that 'even if the exact position and motion (or the quantum state) of each molecule, which is millions of light years away, is made identical and the experiment is conducted, the result is probabilistic.'. In other words, since we have a time machine, we can go back in time and observe the same experiment, but the results can change. To explain by analogy, watching a movie on a computer and then going back 10 minutes and watching it again because I was curious about the previous content changed the result of the movie. (It's quite hard to explain. I didn't write this document to talk about QM (Quantum Mechanics), so I'll skip it at this point.)

Returning to the main topic, why do we feel that the universe around us is almost flat 3D + 1D? See Figure . The left picture is the Cartesian coordinate which we see intuitively and familiarly, just the space around us. The right picture is the same space as the left one, but it is a situation where it is squashed like stretching a rubber band. Let's consider an object flying in a straight line along the y-axis (let's say it flies along the y-axis in the +y direction from (0, 0, 0)). Even if we squash the space as shown on the right, the object will still move along the y-axis (blue line). You may have heard of Newton's Law of Inertia (First Law) : An object moves at constant speed in a straight line if no force acts on it. Some people understand this law intuitively, and some do not understand it because of circumstances such as friction, air resistance, gravity, and etc. So let's look at this law from a different perspective. Let's think of a situation floating in outer space (because there is gravity, air resistance, and etc. on Earth.). Anyway a lot of things depend on intuition and what you can see, so I won't uselessly place an invisible, or hard-to-measure atom, even if it's a thought experiment. In fact, not everyone will be able to go to outer space and do these experiments, and there are some things you have to believe without seeing. Physics is also a science of faith to some extent. Because you have to believe in the results of experiments done by others to analyze them. (I keep counting words to the side;; sorry.) Let's go back to the main topic. Objects fly in outer space. Any force is not acting on these. How will these objects move? Intuitively, you think that they are moving in a straight line, but what is a straight line? Wouldn't the trajectory, that an object which does not receive any force moves, be defined as a straight line? Objects are simply moving through space. However, light moves in the same way, and other objects and we move in the same way when we do not receive any force. All objects without force (you must also question what force is.) on them follow the same trajectory, a straight line. In other words, an object does not move in a straight line, but objects that do not receive any force, including light, go the same path if the direction of their initial velocity is the same. It can be conversely thought that the object does not go straight, but the path the object moves without receiving force is defined as a straight line. Anyway, since we know that all objects that do not receive force move in a straight line, among them, light seems to be the most reliable (other objects are easy to receive force and easily deviate from a straight line). Let's hold it in the direction of movement and set the distance moved in 1 second as 1. See for how the time 1 second is defined and how the distance 1 meter is defined, or how to define it. $+y$, $+z$ are in the direction of the electric field and the direction of the magnetic field. Well, it just means holding three directions perpendicular to each other. If you learn electromagnetism, you know that the direction of $\mathbf{E}$, the direction of $\mathbf{B}$, and the traveling direction of light are perpendicular to each other. You do the proof -o-. In the $+y$ and $+z$ directions, let's set the distance that the light travels in 1 second as 1 too. Since the universe has straightness (flat) and symmetry in three directions $(x, y, z)$ (Symmetry, Isotropy), it is better to draw three mutually perpendicular axes as shown on the left as a straight line rather than drawing it as shown on the right. It is convenient to describe the world. In Cartesian coordinates, every point in space is described by $(x, y, z)$, and you've probably seen a lot of things called vectors in this coordinate system.

// I skipped many things. It was inevitable. If you try to go over everything meticulously, nothing's gonna progress. First of all, let's go over the parts that you don't think are important, and the parts that don't seem to be solved right away. Let's figure out something simple and then think about those details (skipped ones) again. That way, you'll be able to better approach and figure out better ways to understand relativity.

## Tensor the beginning ### Point, Position, or Event We explained above why we are comfortable using Cartesian coordinates when dealing with flat spaces. However, it is convenient to set the coordinates like that, but it is not necessary to set them that way. As in Figure , instead of distorting the space, let's consider a case where only the coordinate system to be described is arbitrarily changed while leaving the space as it is. Well, the simplest way is to rotate the direction of the $x, y, z$ axes of Cartesian coordinate (Rotation). Another way is to move the origin (Translation). And another one is to change only the direction of $x$ axis, i.e. right-hand convention (: $\hat{x} \times \hat{y} = \hat{z}$) can be replaced with left-hand convention (: $\hat{x} \times \hat{y} = -\hat{z}$) (Inversion, Parity). Since it is a three-dimensional space, any coordinate is fine as long as you express the entire space by grabbing three random independent variables in addition to these rotation, inversion, and translation. By changing the scale of the $x, y$ axes, you can create coordinates of a rectangle rather than a square, or coordinates in the form of a parallelogram. You can take the spherical coordinate $(r, \theta, \phi)$ isotropically with respect to a specific point, and the cylinderical coordinate $(\rho, \varphi, z)$ isotropically with respect to a specific straight line. Or dividing the 3D space by an arbitrary plane, you can get the coordinate $(u, v, w)$. Three-dimensional space means that every point in space can be described by three variables. Even though we can use an arbitrary coordinate, the physical laws of the universe that we want to explain shouldn't be changed. How can we implement this?

// The notation used here is the Chicago convention (used by professors at the University of Chicago? This is not the exact name.) and Einstein's summation convention (the summation is automatically omitted for repeated indices above and below) + my own convention. It is not a generally widely used notation, but I am trying to make it as systematic as possible. After understanding tensor in this notation, I think you can easily understand tensor written in other notation. See section 'Index and definition' for a summary of conventions.

First, the points (Point, Position, or Event) on the space (or manifold) are expressed as follows. \begin{align*} \vec{x} &= x^{\bar{i}} \vec{e}_{\bar{i}} = x \vec{e}_x + y \vec{e}_y + z \vec{e}_z \\ &= x^{i} \vec{e}_{i} = r \vec{e}_r + \theta \vec{e}_{\theta} + \phi \vec{e}_{\phi} \\ &= x^{i'} \vec{e}_{i'} = u \vec{e}_u + v \vec{e}_v + w \vec{e}_w \\ \end{align*} Here, $\bar{i}$ (i bar) represents a catesian coordinate, $i$ represents a spherical coordinate, and $i'$ (i prime) represents an arbitrary coordinate $(u, v, w)$. You've probably seen a lot of expressions like the below format. \begin{align*} \vec{x} &= (x, y, z) \text{ in catesian coordinate} \\ &= (r, \theta, \phi) \text{ in spherical coordinate} \\ &= (u, v, w) \text{ in some coordinate} \\ \end{align*} But if there are only numbers, you might get confused, and you can't shorten it with Einstein summation convention. So you can think of it as introducing a new notation. (Don't get confused because it's a different notation from $\vec{x}=x \hat{r}$ using a unit vector in spherical coordinates. This describes a position on the d dimension with d numbers, while also displaying of which coordinate it is.) The coordinates of $\bar{i}, i, i', i'', i'''$ (i bar, i, i prime, i double prime, i triple prime) may change depending on the situation. Don't be confused. All three independent variables of each coordinate system describe the same point $\vec{x}$. The relationship between them is as follows. (Assumed the origin, the zero point, is the same.) \begin{align*} r &= \sqrt{x^2 + y^2 + z^2} \\ \theta &= \cos^{-1}{(z/r)} = \cos^{-1}{(z/\sqrt{x^2 + y^2 + z^2})} \\ \phi &= \cos^{-1}{(\frac{x}{r \sin{\theta}})} = \cos^{-1}{(\frac{x}{\sqrt{x^2 + y^2}})}\\ \end{align*} \begin{align*} x &= r \sin{\theta} \cos{\phi} \\ y &= r \sin{\theta} \sin{\phi} \\ z &= r \cos{\theta} \\ \end{align*} \begin{align*} u &= u(x,y,z) \\ v &= v(x,y,z) \\ w &= w(x,y,z) \\ \end{align*} : $u, v, w$ are functions of $(x, y, z)$. Conversely, $x, y, and z$ will appear as functions of $(u, v, w)$. Because there is only one $(u, v, w)$ to one $(x, y, z)$ for each point. Let's take an example. $(x,y,z)$ for Cartesian coordinate; $(r, \theta, \phi)$ for spherical coordinate; $(\rho, \varphi, z)$ for cylinderical coordinate; and they have the same origin. 0 \vec{e}_x + 0 \vec{e}_y + 0 \vec{e}_z = 0 \vec{e}_r + 0 \vec{e}_{\theta} + 0 \vec{e}_{\phi} = 0 \vec{e}_{\rho} + 0 \vec{e}_{\varphi} + 0 \vec{e}_z 5 \vec{e}_x = 5 \vec{e}_r + \frac{\pi}{2} \vec{e}_{\theta} = 5 \vec{e}_{\rho} 3 \vec{e}_y = 3 \vec{e}_r + \frac{\pi}{2} \vec{e}_{\theta} + \frac{\pi}{2} \vec{e}_{\phi} = 3 \vec{e}_{\rho} + \frac{\pi}{2} \vec{e}_{\varphi} r \sin{\theta} \cos{\phi} \vec{e}_x + r \sin{\theta} \sin{\phi} \vec{e}_y + r \cos{\theta} \vec{e}_z = r \vec{e}_r + \theta \vec{e}_{\theta} + \phi \vec{e}_{\phi} If the two coordinate systems do not have the same origin, 0 \vec{e}_x \neq 0 \vec{e}_u . In another expression, 0 \neq 0 in two different coordinates of which origin points are different. Many people have learned that $\vec{x}$ in Cartesian is a 'displacement vector', but $\vec{x}$ is not a vector from the point of view of a tensor. It is simply a symbol indicating a location. Point P is denoted by $\vec{x}_{\text{P}}$. (It seems that it can be seen as a quasi-tensor in a flat space, but let's know that displacement is not a tensor in the strict sense.) Every point on the manifold (or space) is uniquely described with three parameters. (You can think of it as the definition of a three-dimensional space.)

[Be careful] \begin{align*} \vec{x}_{\text{P}} &= 5 \vec{e}_x \\ &= 5 \vec{e}_r + \frac{\pi}{2} \vec{e}_{\theta} \\ &= 5 \vec{e}_{\rho} \end{align*} Note that the meaning of '=' is different from the meaning used in normal mathematics (algebra). Addition and subtraction between positions are defined only at the same coordinate. It is not that this definition has fallen from the sky, but it means that we promise to use it only like that, because otherwise it would be weird.

This is a legal operation. I don't think it's necessary to explain how the addition of positions at the same coordinate is defined. \left( 5 \vec{e}_x + 3 \vec{e}_y \right) + \left( 2 \vec{e}_x + 1 \vec{e}_y - 3 \vec{e}_z \right) = \left( 7 \vec{e}_x + 4 \vec{e}_y - 3 \vec{e}_z \right) \left( 5 \vec{e}_r + \frac{\pi}{4} \vec{e}_{\theta} \right) + \left( \frac{ \pi}{4} \vec{e}_{\theta} + \pi \vec{e}_{\phi} \right) = \left( 5 \vec{e}_r + \frac{\pi}{2} \vec{e}_{\theta} + \pi \vec{e}_{\phi} \right)
This is an illegal operation. Addition between different coordinates is undefined. If that doesn't make sense, consider two coordinates with different origins. Additions between different coordinates give jumbled results depending on the order. \left( 1 \vec{e}_x \right) + \left( 1 \vec{e}_r \right) \neq \left( 1 \vec{e}_x + 1 \vec{e}_z \right)

### (0,0) Tensor: Scalar Now let's look at the simplest $\binom{0}{0}$ Tensor (: called Scalar). - Scalar: Uniquely defined number which is independent of coordinate chosen. Examples of scalars: Most physical constants , such as the charge of an electron $e$ and the mass of an electron $m$. cf. Sometimes $E = m c^2$ is used to explain that mass changes with speed ($m = \gamma m_0$), but please ignore it because it is a bizarre interpretation that only causes misunderstandings about relativity. Energy is not scalar, $m_0$ is scalar. (It added 0 for no reason to make it more confusing. I will not use 0 with a scalar mass. In other words, $m$ is a coordinate independent value.)

Brief explanation of English abbreviations frequently used in papers.
i.e. = (id est, Latin) that is, in other words.
e.g. = (exempli gratia, Latin) for example.
cf. = (confer, Latin) compare.

- Scalar field: Uniquely defined numbers, defined at each and every point of manifold (space), which are independent of coordinate chosen. They can be complex numbers. \begin{align*} \Phi(\vec{x}) &= \bar{\Phi}(\{x^{\bar{i}}\}) = \bar{\Phi}(x, y, z) \\ &= \Phi(\{x^i\}) = \Phi(r, \theta, \phi) \\ &= \Phi'(\{x^{i'}\}) = \Phi'(u, v, w) \\ \end{align*} Example of scalar field : Potential. Unique function of position is just scalar field. No matter what coordinate the position is expressed in, the scalar field value is uniquely determined at each position. For example, \begin{align*} \Phi(\vec{x}) &= \bar{\Phi}(\{x^{\bar{i}}\}) = \bar{\Phi}(x, y, z) = x^2 + y + 2 z^3 \\ &= \Phi(\{x^i\}) = \Phi(r, \theta, \phi) = (r\sin\theta\cos\phi)^2 + (r\sin\theta\sin\phi) + 2 (r\cos\theta)^3\\ &= \Phi'(\{x^{i'}\}) = \Phi'(u, v, w) = \big(x(u,v,w)\big)^2 + y(u,v,w) + 2 \big(z(u,v,w)\big)^3\\ \end{align*} ### (1,0) Tensor: Vector Re-emphasizing that the property of Tensor is 'to give the same description no matter which coordinate we choose', now let's find out $\binom{1}{0}$ Tensor (: called Vector, or Contra Vector, or Contra-variant Vector).

Let's consider a situation where we need to describe a moving object as the color changes, or as the number changes, or as the time t changes. See Figure . Time is also just an introduction of a chart, or coordinate, which is convenient for us to describe the universe, according to relativity. But first, let's see that time is a universal unique parameter, not coordinate, for the understanding of simple 3d tensor. In Cartesian coordinates, the points the object passes through will be expressed as $(x, y, z)(\lambda)$. $x, y, z$ change according to $\lambda$, respectively. To describe the above curve, \begin{align*} \vec{x}_{\text{C}}(\lambda) &= x_{\text{C}}^{\bar{i}}(\lambda) \vec{e}_{\bar{i}} = x_{\text{C}}(\lambda) \vec{e}_x + y_{\text{C}}(\lambda) \vec{e}_y + z_{\text{C}}(\lambda) \vec{e}_z \\ &= x_{\text{C}}^{i}(\lambda) \vec{e}_{i} = r_{\text{C}}(\lambda) \vec{e}_r + \theta_{\text{C}}(\lambda) \vec{e}_{\theta} + \phi_{\text{C}}(\lambda) \vec{e}_{\phi} \\ &= x_{\text{C}}^{i'}(\lambda) \vec{e}_{i'} = u_{\text{C}}(\lambda) \vec{e}_u + v_{\text{C}}(\lambda) \vec{e}_v + w_{\text{C}}(\lambda) \vec{e}_w \\ \end{align*} Here, the parameter $\lambda$ is a unique physical quantity that is independent of the coordinates we choose (coordinate independent). Let's think later about what these coordinate independent parameters might be, or how to set the parameters to make them unique regardless of the coordinates (i.e. arc length parameterization). Can I name it Scalar parameter? In Cartesian coordinates, the points the object passes through will be expressed as $(x_{\text{C}}, y_{\text{C}}, z_{\text{C}})(\lambda)$. A lot of people would have learned that $\vec{x}_{\text{C}}$ is also called 'displacement vector', but from the point of view of tensor (addition, subtraction, dot product, cross product...), displacement is not a vector. (I emphasis again.) As mentioned above, $\vec{x}_{\text{C}}$ is called a coordinate point (Point, Position, or Event). After reading all of this article, let's try to explain why this is called so. (Once you study a bit about Tensors, you'll quickly see why. I'll skip the explanation here, and you come back after my explaining a bit more about vectors.) A vector is defined from a Tensor point of view as \frac{d \vec{x}_{\text{C}}(\lambda)}{d \lambda} A vector is in thiss form. A Vector can be said to indicate that a specific coordinate point (Point, Position, or Event) must be determined first, and it is intended to flow from this coordinate point to another nearby coordinate/direction and at what speed/strength. The speed/strength here refers to how fast it moves relative to the change in $\lambda$. If this $\lambda$ is time, it will match the generally thought speed, but as I said above, you will have to think carefully about what these coordinate independent unique parameters are and how they can be defined. Expressing this in Cartesian coordinates, \frac{d \vec{x}_{\text{C}}(\lambda)}{d \lambda} = \bigg( \frac{d x_{\text{C}}(\lambda)}{d \lambda}, \frac{d y_{\text{C}}(\lambda)}{d \lambda}, \frac{d z_{\text{C}}(\lambda)}{d \lambda} \bigg) . Looking at the above a little more systematically, \frac{d \vec{x}_{\text{C}}(\lambda)}{d \lambda} = \frac{d x_{\text{C}}(\lambda)}{d \lambda} \frac{\partial \vec{r}}{\partial x} + \frac{d y_{\text{C}}(\lambda)}{d \lambda} \frac{\partial \vec{r}}{\partial y} + \frac{d z_{\text{C}}(\lambda)}{d \lambda} \frac{\partial \vec{r}}{\partial z} . Converting this into a familiar form, \frac{d \vec{x}_{\text{C}}(\lambda)}{d \lambda} = \frac{d x_{\text{C}}(\lambda)}{d \lambda} \hat{x} + \frac{d y_{\text{C}}(\lambda)}{d \lambda} \hat{y} + \frac{d z_{\text{C}}(\lambda)}{d \lambda} \hat{z} . That is, unit vectors in the $x$, $y$, and $z$ directions are expressed as follows. \frac{\partial \vec{r}}{\partial x} = \hat{x}, \qquad \frac{\partial \vec{r}}{\partial y} = \hat{y}, \qquad \frac{\partial \vec{r}}{\partial z} = \hat{z} . Let's generalize the above situation. \begin{align*} \frac{d \vec{x}_{\text{C}}(\lambda)}{d \lambda} &= \frac{d x_{\text{C}}(\lambda)}{d \lambda} \frac{\partial \vec{r}}{\partial x} + \frac{d y_{\text{C}}(\lambda)}{d \lambda} \frac{\partial \vec{r}}{\partial y} + \frac{d z_{\text{C}}(\lambda)}{d \lambda} \frac{\partial \vec{r}}{\partial z} \\ &= \frac{d r_{\text{C}}(\lambda)}{d \lambda} \frac{\partial \vec{r}}{\partial r} + \frac{d \theta_{\text{C}}(\lambda)}{d \lambda} \frac{\partial \vec{r}}{\partial \theta} + \frac{d \phi_{\text{C}}(\lambda)}{d \lambda} \frac{\partial \vec{r}}{\partial \phi} \\ &= \frac{d u_{\text{C}}(\lambda)}{d \lambda} \frac{\partial \vec{r}}{\partial u} + \frac{d v_{\text{C}}(\lambda)}{d \lambda} \frac{\partial \vec{r}}{\partial v} + \frac{d w_{\text{C}}(\lambda)}{d \lambda} \frac{\partial \vec{r}}{\partial w} \end{align*} Here $\partial \vec{r} / \partial x^{\bar{i}}$, $\partial \vec{r} / \partial x^i$, $\partial \vec{r} / \partial x^{i'}$ are coordinate bases. There is also a non-coordinate basis. For example, $r, \theta, \phi$ direction unit vectors, $(\hat{r}, \hat{\theta}, \hat{\phi})$. There are times when non-coordinate basis is useful, but it is not very helpful to understand tensor, so let's pass them for a while. This can be said to be the general form and conceptual understanding of vectors. Do you think still "What is this?", "Something's weird. Not enough."? Will you be full on the first drink? After knowing $\binom{0}{1}$ tensor one-form and contraction (: Combining vector and one-form to scalar), you will think "Ah, tensor is like this!". So let's keep going. Let's promise to express the vector $V$ as follows by introducing the Chicago convention + Einstein's summation convention, to express the vector and all tensors in a more readable and systematic manner. V^{\text{A}} = V^{\bar{i}} e_{\bar{i}}^{\text{A}} = V^{j} e_{j}^{\text{A}} = V^{k'} e_{k'}^{\text{A}} The letter $V$ is followed by a superscript. We promise to always write superscripts as A,B,C,D,...,Y,Z. In other words, if superscripts A, B, C... are added to the upper right of a character, they are all vectors. $e_{\bar{i}}^{\text{A}}, e_{j}^{\text{A}}, e_{k'}^{\text{A}}$ are the basis of the vector. It is also called coordinate basis, and the superscript A means that these are also vectors. That is, it is a basis vector. Do not confuse them with $\vec{e}_i$. Basis vectors are a completely different concept from the $\vec{e}_i$ introduced to indicate coordinate points. (There are similarities between the two, but I'm emphasizing that, some people might think they're the same thing, they're different.) As $\vec{e}_i$ is introduced to express coordinate points systematically, $e_{i}^{\text{A}} (=\frac{\partial \vec{r}}{\partial x^i})$ can be considered as a concept introduced to express vectors systematically. $V^{\bar{i}}, V^{j}, and V^{k'}$ are called the coordinate components of the vector $V^{\text{A}}$, and they are not strictly speaking vectors. It's just a number. This may not make sense right now, so let's skip it first, read this document entirely, and think about it later again. If we fit the above results into the following notation, \begin{align*} V^{\text{A}} &= \frac{d \vec{x}_{\text{C}}(\lambda)}{d \lambda}^{\text{A}} \\ &= \frac{d}{d \lambda} \Big( x_{\text{C}}^{\bar{i}} \vec{e}_{\bar{i}} \Big)^{\text{A}} \\ &= \frac{d x_{\text{C}}(\lambda)}{d \lambda} \frac{\partial \vec{r}}{\partial x}^{\text{A}} + \frac{d y_{\text{C}}(\lambda)}{d \lambda} \frac{\partial \vec{r}}{\partial y}^{\text{A}} + \frac{d z_{\text{C}}(\lambda)}{d \lambda} \frac{\partial \vec{r}}{\partial z}^{\text{A}} \\ &= V^x e_x^{\text{A}} + V^y e_y^{\text{A}} + V^z e_z^{\text{A}} \\ \end{align*} \begin{align*} &= \frac{d}{d \lambda} \Big( x_{\text{C}}^{i} \vec{e}_{i} \Big)^{\text{A}} \\ &= \frac{d r_{\text{C}}(\lambda)}{d \lambda} \frac{\partial \vec{r}}{\partial r}^{\text{A}} + \frac{d \theta_{\text{C}}(\lambda)}{d \lambda} \frac{\partial \vec{r}}{\partial \theta}^{\text{A}} + \frac{d \phi_{\text{C}}(\lambda)}{d \lambda} \frac{\partial \vec{r}}{\partial \phi}^{\text{A}} \\ &= V^r e_r^{\text{A}} + V^\theta e_\theta^{\text{A}} + V^\phi e_\phi^{\text{A}} \\ \\ &= \frac{d}{d \lambda} \Big( x_{\text{C}}^{i'} \vec{e}_{i'} \Big)^{\text{A}} \\ &= \frac{d u_{\text{C}}(\lambda)}{d \lambda} \frac{\partial \vec{r}}{\partial u}^{\text{A}} + \frac{d v_{\text{C}}(\lambda)}{d \lambda} \frac{\partial \vec{r}}{\partial v}^{\text{A}} + \frac{d w_{\text{C}}(\lambda)}{d \lambda} \frac{\partial \vec{r}}{\partial w}^{\text{A}} \\ &= V^u e_u^{\text{A}} + V^v e_v^{\text{A}} + V^w e_w^{\text{A}} . \\ \end{align*} When a Vector is used in spherical coordinates, unit vectors like $\hat{r}$, $\hat{\theta}$, $\hat{\phi}$ must have been used frequently. It is not a good basis for applying the concept of tensor. From the tensor point of view, the vector basis is called the coordinate basis and is defined in the form above. You can think of it as calling any vector basis, that is not a coordinate basis, a non-coordinate basis. These include unit vectors ($\hat{r}$, $\hat{\theta}$, $\hat{\phi}$). $e_{\theta}^{A}$ $=$ $\partial \vec{r} / \partial \theta$ $=$ $r \hat{\theta}$. A parametized curve was introduced for the conceptual understanding of vector, but it is not necessary to understand a vector by introducing one parametized curve. Vectors in this way exist only at the coordinate point where this one curve is located, which may interrupt understanding the concept of vector field. - Vector field: Uniquely defined vectors, defined at each and every point, which are independent of coordinate chosen. That is, to understand the vector field defined in the entire space, imagine a scalar parameterized curve at each coordinate point, and as the scalar parameter changes, the vector at that point indicates in which direction and how fast the point of coordinate changes along with the scalar parameter. In this way, we learned about vector, one of the most basic tensors. To properly understand tensor, you must understand one-form and contraction at least!! ### (0,1) Tensor: One-form (Co-Vector) Now let's look at $\binom{0}{1}$ tensor (: called One-form, or Co-vector, Co-variant vector) which is similar to vector but slightly different. One-form can be thought of as a generalization of the commonly known gradient $\vec{\nabla}$. To easily understand this gradient (or one-form), let's introduce the scalar field described above. Let's take a gradient on the scalar field $\Phi(\vec{x})$ where the values are uniquely determined for each point in space regardless of the coordinate system chosen. \vec{\nabla} \Phi = \frac{\partial \Phi}{\partial x} \hat{x} + \frac{\partial \Phi}{\partial y} \hat{y} + \frac{\partial \Phi}{\partial z} \hat{z} In other words, the gradient (or one-form) indicates how the scalar field $\Phi$ changes as the coordinate point changes. Looking at this more systematically, \begin{align*} \mathbf{d} \Phi &= \frac{\partial \Phi}{\partial x} \mathbf{d} x + \frac{\partial \Phi}{\partial y} \mathbf{d} y + \frac{\partial \Phi}{\partial z} \mathbf{d} z \\ &= \frac{\partial \Phi}{\partial r} \mathbf{d} r + \frac{\partial \Phi}{\partial \theta} \mathbf{d} \theta + \frac{\partial \Phi}{\partial \phi} \mathbf{d} \phi \\ &= \frac{\partial \Phi}{\partial u} \mathbf{d} u + \frac{\partial \Phi}{\partial v} \mathbf{d} v + \frac{\partial \Phi}{\partial w} \mathbf{d} w \\ \end{align*} This is the one-form. Like Vector, let's introduces Chicago convention + Einstein's summation convention to express one-form $W$ as follows. W_{\text{A}} = W_{\bar{i}} e^{\bar{i}}_{\text{A}} = W_{j} e^{j}_{\text{A}} = W_{k'} e^{k'}_{\text{A}} This is the format of the letters $W$ followed by the subscripts A,B,C,...,Y,Z. Vector is by superscript, and one-form is by subscript. $e^{\bar{i}}_{\text{A}}$, $e^{j}_{\text{A}}$, $e^{k'}_{\text{A}}$ are one-form bases, $W_{\bar{i}}$, $W_{j}$, and $W_{k'}$ are called coordinate components of the one-form $W_{\text{A}}$. As with vector, only the basis is one-form (tensor) and the coordinate component is just a number. Fitting the above results to the notation, \begin{align*} W_{\text{A}} &= \mathbf{d}_{\text{A}} \Phi = \nabla_{\text{A}} \Phi = \big( \nabla_{\bar{i}} \Phi \big) e^{\bar{i}}_{A} = \big( \nabla_{i} \Phi \big) e^{i}_{A} = \big( \nabla_{i'} \Phi \big) e^{i'}_{A} \\ &= e^{\bar{i}}_{A} \partial_{\bar{i}} \Phi = e^{i}_{A} \partial_{i} \Phi = e^{i'}_{A} \partial_{i'} \Phi \\\\ \end{align*} \begin{align*} &= \frac{\partial \Phi}{\partial x} \mathbf{d}_{\text{A}} x + \frac{\partial \Phi}{\partial y} \mathbf{d}_{\text{A}} y + \frac{\partial \Phi}{\partial z} \mathbf{d}_{\text{A}} z \\ &= W_{x} e^{x}_{\text{A}} + W_{y} e^{y}_{\text{A}} + W_{z} e^{z}_{\text{A}} \\\\ &= \frac{\partial \Phi}{\partial r} \mathbf{d}_{\text{A}} r + \frac{\partial \Phi}{\partial \theta} \mathbf{d}_{\text{A}} \theta + \frac{\partial \Phi}{\partial \phi} \mathbf{d}_{\text{A}} \phi \\ &= W_{r} e^{r}_{\text{A}} + W_{\theta} e^{\theta}_{\text{A}} + W_{\phi} e^{\phi}_{\text{A}} \\\\ &= \frac{\partial \Phi}{\partial u} \mathbf{d}_{\text{A}} u + \frac{\partial \Phi}{\partial v} \mathbf{d}_{\text{A}} v + \frac{\partial \Phi}{\partial w} \mathbf{d}_{\text{A}} w \\ &= W_{u} e^{u}_{\text{A}} + W_{v} e^{v}_{\text{A}} + W_{w} e^{w}_{\text{A}} \\ \end{align*} For the sake of the introduction of the scalar field, the one-form field concept would have been automatically understood in the case of one-form. One-form can be said to indicate how much and how the scalar field value appointed for each position (coordinate point: $\vec{r}$ = $r^{i} \vec{e}_{i}$) changes according to the direction in which the position (coordinate point) changes. In this way, one-form, which is one of the most basic tensors, has been studied to some extent. ## Contraction: Vector, One-form to Scalar After studying vector and one-form, don't you think something will come out if you combine the two well? Let's look at the following equation. \frac{d \Phi(\vec{x}_{\textrm{C}}(\lambda))}{d \lambda} This equation, which shows how the scalar field $\Phi$ value changes along the curve of Figure , must be coordinate independent because it is composed of a scalar field and a scalar parameter. That means that the above value must also be a scalar. Applying the partial derivative to the above expression and analyzing it, \begin{align*} \frac{d \Phi}{d \lambda} &= \frac{d x_{\text{C}}(\lambda)}{d \lambda} \frac{\partial \Phi}{\partial x} + \frac{d y_{\text{C}}(\lambda)}{d \lambda} \frac{\partial \Phi}{\partial y} + \frac{d z_{\text{C}}(\lambda)}{d \lambda} \frac{\partial \Phi}{\partial z} \\ &= \frac{d r_{\text{C}}(\lambda)}{d \lambda} \frac{\partial \Phi}{\partial r} + \frac{d \theta_{\text{C}}(\lambda)}{d \lambda} \frac{\partial \Phi}{\partial \theta} + \frac{d \phi_{\text{C}}(\lambda)}{d \lambda} \frac{\partial \Phi}{\partial \phi} \\ &= \frac{d u_{\text{C}}(\lambda)}{d \lambda} \frac{\partial \Phi}{\partial u} + \frac{d v_{\text{C}}(\lambda)}{d \lambda} \frac{\partial \Phi}{\partial v} + \frac{d w_{\text{C}}(\lambda)}{d \lambda} \frac{\partial \Phi}{\partial w} \end{align*} Aren't these expressions what we've seen a lot when explaining vectors and one-forms? With these hints, let's define a contraction (reduction) that combines a vector and a one-form into a scalar as follows. V^{\text{A}} W_{\text{A}} = V^{\bar{i}} W_{\bar{i}} = V^{j} W_{j} = V^{k'} W_{k'} If the superscript of a vector and the subscript of a one-form are the same, it becomes contracted. It is similar to the dot product of a vector and a gradient. Analyzing it a bit further, \begin{align*} V^{\text{A}} W_{\text{A}} &= V^{i} e_i^{\text{A}} W_{j} e^j_{\text{A}} \\ &= V^{i} W_{j} e_i^{\text{A}} e^j_{\text{A}} \\ &= V^{i} W_{j} \delta_i^j \\ &= V^{i} W_{i} \end{align*} That is, the contraction of vector basis and one-form basis in the same coordinate is to be Kronecker delta. e_{\bar{i}}^{\text{A}} e^{\bar{j}}_{\text{A}} = e_i^{\text{A}} e^j_{\text{A}} = e_{i'}^{\text{A}} e^{j'}_{\text{A}} = \delta_{i}^{j} = \left\{\begin{array}{ll} 1 & \textrm{if } i=j \\ 0 & \textrm{if } i\neq j . \\ \end{array} \right. Now, you will get a little sense of why tensor is important and convenient for describing the laws of physics. The biggest advantage is that you can describe the laws of physics coordinate independently. Coordinate is a concept created by humans to better describe physical phenomena. There is no such thing in the universe, but physical phenomena whose position changes according to time are inevitably to introduce coordinates and charts that are easy for us to analyze and understand them. The concepts of time, space, and position are closely related to coordinates. Explaining, however, this physical law is what should not change depending on which coordinate standard we set. This is the essence of the concept of tensors. If $\vec{F}=m\vec{a}$ is written in the tensor form you learned here, F^{\text{A}} = m a^{\text{A}} = \frac{d}{dt} p^{\text{A}} This expression is satisfied no matter what coordinates are introduced. You have to make it so that it satisfies whenever any coordinate, whether it's Cartesian, spherical, or arbitrary coordinates, is used. (Of course, time is also one axis of the coordinates we introduced, but... Let's go over it for a moment.) Most of you have seen that expression in the form of a vector. You can think of tensors as a systematic generalization of a vector (or gradient).

You need to know how the components and basis transform according to the coordinates in order to use tensor well. This will be expressed simply. First of all, for the purpose of simplifying notation, let's define and write the following symbols. \Lambda^{i}_{\bar{j}} \equiv \frac{\partial x^{i}}{\partial x^{\bar{j}}}, \qquad \Lambda^{k'}_{j} \equiv \frac{\partial x^{k'}}{\partial x^{j}}, \qquad \Lambda^{i}_{j'} \equiv \frac{\partial x^{i}}{\partial x^{j'}}, ... By definition, the following properties are also satisfied. \Lambda^{j'}_{i} \Lambda^{k}_{j'} = \frac{\partial x^{j'}}{\partial x^{i}} \frac{\partial x^{k}}{\partial x^{j'}} = \frac{\partial x^{k}}{\partial x^{i}} = \delta^{k}_{i} Referring to the description of vector and one-form above, it can be seen that the following transforms are possible. \begin{align*} V^{\text{A}} &= V^i e_i^{\text{A}} \\ &= \Lambda^i_{j'} V^{j'} e_i^{\text{A}} = V^{j'} \Lambda^i_{j'} e_i^{\text{A}} = V^{j'} e_{j'}^{\text{A}} \\ &= V^i \Lambda^{\bar{k}}_{i} e_{\bar{k}}^{\text{A}} = \Lambda^{\bar{k}}_{i} V^i e_{\bar{k}}^{\text{A}} = V^{\bar{k}} e_{\bar{k}}^{\text{A}} \\ \end{align*} \begin{align*} W_{\text{A}} &= W_i e^i_{\text{A}} \\ &= \Lambda^{\bar{k}}_{i} W_{\bar{k}} e^i_{\text{A}} = W_{\bar{k}} \Lambda^{\bar{k}}_{i} e^i_{\text{A}} = W_{\bar{k}} e^{\bar{k}}_{\text{A}} \\ &= W_i \Lambda^i_{j'} e^{j'}_{\text{A}} = \Lambda^i_{j'} W_i e^{j'}_{\text{A}} = W_{j'} e^{j'}_{\text{A}} \\ \end{align*} \begin{align*} V^{\text{A}} W_{\text{A}} &= V^i W_i \\ &= \Lambda^i_{j'} V^{j'} W_i = V^{j'} \Lambda^i_{j'} W_i = V^{j'} W_{j'} \\ &= V^i \Lambda^{\bar{k}}_{i} W_{\bar{k}} = \Lambda^{\bar{k}}_{i} V^i W_{\bar{k}} = V^{\bar{k}} W_{\bar{k}} \\ \end{align*} You can find that rules are quite easy.

##[.hiden] General (m,n) Tensor 이제 일반적인 $\binom{m}{n}$ tensor 로의 확장을 알아보자. 위에서 전개해나간 vector 와 one-form, contraction 의 성질들을 일반적인 $\binom{m}{n}$ tensor 에서도 잘 써먹을 수 있도록 적절히 정의해야 할 것이다. 우선 $\binom{m}{n}$ tensor 는 다음과 같이 표기하기로 하자. T^{ABC\cdots}_{EFG\cdots} with $m$ upper indices $ABC\cdots$ and $n$ lower indices $EFG\cdots$. 이 때 이 $\binom{m}{n}$ tensor 가 component 와 basis 로 어떻게 표현되어야 할까를 보면 직관적으로 다음과 같이 된다고 정의할 수 있겠다. T^{ABC\cdots}_{EFG\cdots} = T^{ijk\cdots}_{pqr\cdots} e_{i}^{A} e_{j}^{B} e_{k}^{C} \cdots e^{p}_{E} e^{q}_{F} e^{r}_{G} \cdots = T^{ijk\cdots}_{pqr\cdots} e_{i~j~k}^{ABC} e_{EFG}^{p~q~r} \cdots (나중에 index lowering and raising 을 할 때 위치가 헷갈리지 않게 하려고 위아래 순서가 명확하도록 T^{ABC\cdots}_{~~~~~~~~EFG\cdots} 와 같이 표기하는 경우가 많다.) 기준이 되는 coordinate 에서 이 tensor component 들이 다음과 같이 주어질 때, \begin{align*} T^{ABC\cdots}_{EFG\cdots} &= T^{\bar{i}\bar{j}\bar{k}\cdots}_{\bar{p}\bar{q}\bar{r}\cdots} e_{\bar{i}~\bar{j}~\bar{k}}^{ABC} e_{EFG}^{\bar{p}~\bar{q}~\bar{r}} \cdots \\ &=T^{\bar{i}\bar{j}\bar{k}\cdots}_{\bar{p}\bar{q}\bar{r}\cdots} \Lambda_{\bar{i}\bar{j}\bar{k}}^{lmn} \Lambda_{uvw}^{\bar{p}\bar{q}\bar{r}} \cdots e_{l~m~n}^{ABC} e_{EFG}^{u~v~w} \cdots \\ &= T^{lmn\cdots}_{uvw\cdots} e_{l~m~n}^{ABC} e_{EFG}^{u~v~w} \cdots . \end{align*} 즉 component 들은 다음과 같이 변함을 알 수 있다. T^{lmn\cdots}_{uvw\cdots} = T^{\bar{i}\bar{j}\bar{k}\cdots}_{\bar{p}\bar{q}\bar{r}\cdots} \Lambda_{\bar{i}\bar{j}\bar{k}}^{lmn} \Lambda_{uvw}^{\bar{p}\bar{q}\bar{r}} \cdots 굉장히 단순하다! Chicago convention 의 장점 중 하나는 두가지 이상의 coordinate 를 섞어서 tensor 를 표현할 수 있다는 점이다. 즉 T^{ABC\cdots}_{EFG\cdots} = T^{ij\bar{k}\cdots}_{p'q'\tilde{r}\cdots} e_{i~j~\bar{k}}^{ABC} e_{EFG}^{p'q'\tilde{r}} \cdots 로도 표현될 수 있다. 여기서 component 는 T^{ij\bar{k}\cdots}_{p'q'\tilde{r}\cdots} = T^{\bar{l}\bar{m}\bar{k}\cdots}_{\bar{u}\bar{v}\bar{w}\cdots} \Lambda^{ij}_{\bar{l}\bar{m}} \Lambda_{p'q'\tilde{r}}^{\bar{u}~\bar{v}\bar{w}} \cdots . 그렇다면 $\binom{m}{n}$ tensor 의 물리적 의미는 무엇일까? 단순히 vector $m$개와 one-form $n$개가 결합한 것이라고 보면 된다. 무언가 주변으로 가려고 하는 것들을 나타내는, 기술하는 것 $m$개와 좌표가 바뀌면서 scalar 가 어떻게 변할지를 기술하는 것 $n$개가 결합하여 하나로 표현했다는 뜻이다. 가장 간단하게 T^{AB} = V^{A} V^{B} , \quad T^{A}_{B} = V^{A} W_{B} , \quad T^{AB}_{C} = X^{A} V^{B} W_{C} 처럼 vector $m$개와 one-form $n$개의 direct product 라고 이해하고 있으면 된다. 의미도 비슷하게 가지니까. ##[.hiden] Mathematical Tensors ### Zero tensor Zero tensor 란 그냥 간단하게 0이라고 생각하면 된다. 한 좌표계에서 모든 component 가 0인 $\binom{n}{m}$ tensor 는 \begin{align*} 0^{ABC\cdots}_{EFG\cdots} &= 0^{ijk\cdots}_{pqr\cdots} e_{i~j~k}^{ABC} e_{EFG}^{p~q~r} \cdots = 0 \\ &= 0^{i'j'k'\cdots}_{p'q'r'\cdots} e_{i'j'k'}^{ABC} e_{EFG}^{p'q'r'} \cdots = 0 \end{align*} 와 같이 다른 어떠한 좌표계에서도 모든 component 가 0이다. ### Kroneker-delta tensor 기준이 되는 coordinate 에서 다음과 같이 정의된 $\binom{1}{1}$ tensor 를 생각해보자. \delta_{A}^{B} \equiv \delta_{\bar{i}}^{\bar{j}} e^{\bar{i}}_A e_{\bar{j}}^B = e^{\bar{i}}_A e_{\bar{i}}^B where \delta_{\bar{i}}^{\bar{j}} = \begin{cases} 1 & \textrm{if } \bar{i}=\bar{j} , \\ 0 & \textrm{if } \bar{i}\neq\bar{j} . \\ \end{cases} 다른 임의의 coordinate 에서의 component 값들은 어떻게 표현될까? \delta_{A}^{B} = e^{\bar{i}}_A e_{\bar{i}}^B = \Lambda^{\bar{i}}_{k} e^{k}_A \Lambda_{\bar{i}}^{l} e_{l}^B = \delta_{k}^{l} e^{k}_A e_{l}^B = e^{k}_A e_{k}^B . component 값들이 coordinate 에 상관없이 항상 Kroneker-delta function 으로 표현된다. 또한 \delta_{A}^{B} = e^{\bar{i}}_A e_{\bar{i}}^B = \Lambda^{\bar{i}}_j e^{j}_A e_{\bar{i}}^B . 따라서 $\delta^{\bar{i}}_j \equiv \Lambda^{\bar{i}}_j$로 정의해서 쓸수 있겠지만 Kroneker-delta function 이랑 헷갈릴 수 있으므로 되도록 $\Lambda^{\bar{i}}_j$를 쓰도록 하자.} 이러한 신기한 특성을 갖는 mathematical tensor 를 바로 Kroneker-delta $\binom{1}{1}$ tensor 라고 한다. 추가적인 간단한 특징들은 다음과 같다. \delta_{A}^{B} V^{A} = V^{B} , \qquad \delta_{A}^{B} W_{B} = W_{A} , \qquad \delta_{A}^{B} V^{A} W_{B} = V^{A} W_{A} , \delta_{A}^{B} \delta_{B}^{C} = \delta_{A}^{C} , and so on. ### Levi-Civita tensor 기준이 되는 coordinate에서 다음과 같이 정의된 $\binom{0}{d}$ tensor를 생각해보자. 기준 coordinate는 보통 Cartesian coordinate로 잡는다. (편의를 위해 $d=3$라고 가정하고 전개. 임의의 $d$으로의 확장은 명확하니까 생략.) \epsilon_{ABC} = \epsilon_{\bar{i}\bar{j}\bar{k}} e^{\bar{i}}_A e^{\bar{j}}_B e^{\bar{k}}_C where \epsilon_{\bar{i}\bar{j}\bar{k}} = \begin{cases} 1 & \textrm{if } (\bar{i}\bar{j}\bar{k}) = (123) , \\ (-1)^{\kappa} & \textrm{if } (\bar{i}\bar{j}\bar{k}) \textrm{ is }\kappa\textrm{-th permutation from }(123), \\ 0 & \textrm{otherwise (e.g. any pair of indices are the same.).}\\ \end{cases} Levi-Civita symbol $\epsilon_{ijk}$ 값을 정의할때 나오는 permutation(치환)이란 것은 순서대로 되어있는 (123) 중에 두개를 택해서 서로 자리를 바꿨다는 뜻이다. 치환을 한번 할때마다 부호가 바뀌도록, fully anti-symmetric하도록 정의한 함수이다. 따라서 두 자리에 있는 값이 같으면 0이 될수밖에 없다. 예들을 들어보면, \epsilon_{12} = 1 , \quad \epsilon_{21} = -1 , \quad \epsilon_{11} = 0 . \epsilon_{txyz} = 1 , \quad \epsilon_{xtyz} = -1 , \quad \epsilon_{yxtz} = -1 , \quad \epsilon_{zxyt} = -1 , \quad \epsilon_{ttxy} = 0 . 위에 보이듯 꼭 숫자값을 넣을 필요는 없다. 기준을 정하고 기준에서 몇 번의 permutation 을 거쳐야 $\epsilon$ 아래쪽 순서가 나오는지를 따져서 값을 정의한다. 기준 coordinate 에서 이렇게 정의된 Levi-Civita tensor 가 다른 임의의 coordinate 에서 어떤 component 값들을 가질까? \epsilon_{ABC} = \epsilon_{\bar{i}\bar{j}\bar{k}} e^{\bar{i}}_A e^{\bar{j}}_B e^{\bar{k}}_C = \epsilon_{\bar{i}\bar{j}\bar{k}} \Lambda^{\bar{i}\bar{j}\bar{k}}_{lmn} e^{l}_A e^{m}_B e^{n}_C 여기서 \epsilon_{\bar{i}\bar{j}\bar{k}} \Lambda^{\bar{i}\bar{j}\bar{k}}_{lmn} = \begin{cases} \det(\Lambda^{\bar{i}}_l) & \textrm{if } (lmn) = (123) , \\ (-1)^{\kappa} \det(\Lambda^{\bar{i}}_l) & \textrm{if } (lmn) \textrm{ is }\kappa\textrm{-th permutation from } (123) , \\ 0 & \textrm{otherwise (e.g. any pair of indices are the same.).}\\ \end{cases} 이므로 \epsilon_{ABC\cdots} = \det(\Lambda_i^{\bar{j}}) \epsilon_{lmn\cdots} e^{l}_A e^{m}_B e^{n}_C \cdots .

// 이 부분이 이해가 안가는 분들은 제가 정리해놓은 '선형대수학' 글을 보고 오시길. Determinant 가 어떻게 정의되고, 왜 그렇게 정의되고, 역행렬이 어떻게 나오고 등이 정리되어 있음. Matrix 에서 determinant 의 정의와 특징을 이용하면 바로 나오는 결과.

즉 Levi-Civita $\binom{0}{d}$ tensor 의 component 값들은 기준 coordinate 로부터의 one-form component 변환에 해당하는 $\Lambda_{j}^{\bar{i}}$ (cf. $W_{j} = \Lambda_{j}^{\bar{i}} W_{\bar{i}}$)의 determinant 가 곱해진 형태를 가진다. Kroneker-delta 보다는 약간 지저분한 결과이지만 이것도 신기한 특성을 갖는 tensor 임에는 분명하다. 당연한 특징으로 \epsilon_{ABC\cdots} = - \epsilon_{BAC\cdots} = + \epsilon_{CAB\cdots} = \cdots . 어떠한 index permutation 에 대해서도 anti-symmetric하다.

Chicago convention 에 의하면 이 tensor 의 component 를 \epsilon_{lmn} \equiv \epsilon_{\bar{i}\bar{j}\bar{k}} \Lambda^{\bar{i}\bar{j}\bar{k}}_{lmn} = \begin{cases} \det(\Lambda^{\bar{i}}_l) & \textrm{if } (lmn) = (123) , \\ (-1)^{\kappa} \det(\Lambda^{\bar{i}}_l) & \textrm{if } (lmn) \textrm{ is }\kappa\textrm{-th permutation from } (123) , \\ 0 & \textrm{otherwise (e.g. any pair of indices are the same.).} \end{cases} 로 정의해서 써야 깔끔하겠지만, Levi-Civita tensor 는 이렇게 정의해놓고 쓰면 더 헷갈리므로 이 경우만 특별하게 Chicago convention 을 따르지 않고 표현한다. 이러한 이유 때문에 Levi-Civita tensor 를 pseudo tensor (가짜 tensor) 라고 하기도 한다. 하지만 'tensor라고 우기면 tensor다' 라고 개인적으로 생각하고 있기에... 판단은 알아서.

Levi-Civita $\binom{d}{0}$ tensor는 다음과 같이 정해짐을 쉽게 알 수 있다. \begin{align*} \epsilon^{ABC} &= \epsilon^{\bar{i}\bar{j}\bar{k}} e_{\bar{i}}^A e_{\bar{j}}^B e_{\bar{k}}^C = \epsilon^{\bar{i}\bar{j}\bar{k}} \Lambda_{\bar{i}\bar{j}\bar{k}}^{lmn} e_{l}^A e_{m}^B e_{n}^C \\ &= \det(\Lambda^{i}_{\bar{j}}) \epsilon^{lmn} e_{l}^A e_{m}^B e_{n}^C \end{align*} 딱 보니 Levi-Civita $\binom{d}{0}$ tensor 와 Levi-Civita $\binom{0}{d}$ tensor 를 결합하면 무언가 나올거 같기도 하다. \begin{align*} &\epsilon_{ABC\cdots} \epsilon^{EFG\cdots} = \det \big( \Lambda_{s}^{\bar{t}} \big) \epsilon_{ijk\cdots} \det \big( \Lambda_{\bar{u}}^{v} \big) \epsilon^{pqr\cdots} e_{ijk\cdots}^{ABC\cdots} e^{pqr\cdots}_{EFG\cdots} \\ &= \det \big( \Lambda_{s}^{\bar{t}} \Lambda_{\bar{t}}^{v} \big) \epsilon_{ijk\cdots} \epsilon^{pqr\cdots} e_{ijk\cdots}^{ABC\cdots} e^{pqr\cdots}_{EFG\cdots} \\ &= \det \big( \delta_{s}^{v} \big) \epsilon_{ijk\cdots} \epsilon^{pqr\cdots} e_{ijk\cdots}^{ABC\cdots} e^{pqr\cdots}_{EFG\cdots} \\ &= \epsilon_{ijk\cdots} \epsilon^{pqr\cdots} e_{ijk\cdots}^{ABC\cdots} e^{pqr\cdots}_{EFG\cdots} \end{align*} Determinant 가 결합되면서 $\delta_{i}^{j}$의 determinant 인 1이 되었다. $\epsilon_{ijk\cdots} \epsilon^{pqr\cdots}$는 어떻게 표현될 수 있을까? 위의 index set $\{ijk\cdots\}$든 아래쪽의 $\{pqr\cdots\}$이든 set 안에 같은 값이 하나라도 있으면 0이 나온다. 따라서 위 아래 set 가 하나하나 순서 상관없이 어떻게든 대응이 될 때에만 살아남고 이렇게 다 대응이 되더라도 set 안에 같은 index 가 있으면 0이 나오도록 만들어야 한다. 위 아래를 대응시키는 것은 Kroneker delta $\delta_{i}^{j}$로 표현될 수 있으니 이를 이용해 표현할수는 없을지 알아보자. 가장 간단하게 2차원일때에는 \begin{align*} \epsilon_{ij} \epsilon^{pq} &= \delta_{i}^{p} \delta_{j}^{q} - \delta_{i}^{q} \delta_{j}^{p} = \delta_{ij}^{pq} - \delta_{ij}^{qp} \\ &\equiv \sum_{l,m=\{p,q\}} \epsilon^{(pq)}_{~lm} \delta_{ij}^{lm} \\ &\equiv \epsilon^{pq}_{ij} \end{align*} where we define \delta_{ij}^{pq} \equiv \delta_{i}^{p} \delta_{j}^{q} and \epsilon^{(pq)}_{~lm} = \begin{cases} 1 & \textrm{if } lm = (pq) , \\ (-1)^{\kappa} & \textrm{if }lm\textrm{ is }\kappa\textrm{-th permutation from }(pq) , \\ 0 & \textrm{otherwise (e.g. any pair of lower indices are the same.).} \end{cases} 즉 $\epsilon^{(pq)}_{~lm}$는 괄호 () 안의 것을 기준으로 fully anti-symmetric 하게 정의된 function 이다. 3차원일 때에는 \begin{align*} \epsilon_{ijk} \epsilon^{pqr} &= \delta_{i}^{p} \big( \delta_{j}^{q} \delta_{k}^{r} - \delta_{j}^{r} \delta_{k}^{q} \big) - \delta_{i}^{q} \big( \delta_{j}^{p} \delta_{k}^{r} - \delta_{j}^{r} \delta_{k}^{p} \big) + \delta_{i}^{r} \big( \delta_{j}^{p} \delta_{k}^{q} - \delta_{j}^{q} \delta_{k}^{p} \big) \\ &= \delta_{ijk}^{pqr} - \delta_{ijk}^{prq} - \delta_{ijk}^{qpr} + \delta_{ijk}^{qrp} + \delta_{ijk}^{rpq} - \delta_{ijk}^{rqp} \\ &\equiv \sum_{l,m,n=\{p,q,r\}} \epsilon^{(pqr)}_{~lmn} \delta_{ijk}^{lmn} \\ &= \epsilon^{pqr}_{~ijk} \\ &= \big( \delta^{p}_{l} \epsilon^{qr}_{mn} - \delta^{q}_{l} \epsilon^{pr}_{mn} + \delta^{r}_{l} \epsilon^{pq}_{mn} \big) \delta_{ijk}^{lmn} \\ &= \delta^{p}_{i} \epsilon^{qr}_{mn} \delta_{jk}^{mn} - \delta^{q}_{i} \epsilon^{pr}_{mn} \delta_{jk}^{mn} + \delta^{r}_{i} \epsilon^{pq}_{mn} \delta_{jk}^{mn} \end{align*} 와 같이 나타난다. 따라서 일반적인 n차원으로 확장은 \begin{align*} \epsilon_{ijk\cdots} \epsilon^{pqr\cdots} &= \delta_{ijk\cdots}^{pqr\cdots} - \delta_{ijk\cdots}^{qpr\cdots} + \cdots \\ &\equiv \sum_{l,m,n,\cdots=\{p,q,r,\cdots\}} \epsilon^{(pqr\cdots)}_{~lmn\cdots} \delta_{ijk\cdots}^{lmn\cdots} \\ &\equiv \sum_{l,m,n,\cdots=\{i,j,k,\cdots\}} \epsilon^{~lmn\cdots}_{(ijk\cdots)} \delta_{lmn\cdots}^{pqr\cdots} \\ &= \epsilon^{(pqr\cdots)}_{~ijk\cdots} = \epsilon^{~pqr\cdots}_{(ijk\cdots)} = \epsilon^{pqr\cdots}_{ijk\cdots} \end{align*} 와 같음을 알 수 있다. 가끔 유용하게 쓰이는 관계식이다. 또 하나 유용한 관계식은 contraction 이 하나 포함된 \epsilon_{ABC\cdots} \epsilon^{AFG\cdots} = \epsilon_{ijk\cdots} \epsilon^{iqr\cdots} e_{BC\cdots}^{jk\cdots} e_{qr\cdots}^{FG\cdots} 에서 나오는 $\epsilon_{ijk\cdots} \epsilon^{iqr\cdots}$ 값이다. \begin{align*} \epsilon_{ijk\cdots} \epsilon^{iqr\cdots} = \epsilon^{(iqr\cdots)}_{~ijk\cdots} . \end{align*} As only one of the $i$ summation survives, \epsilon_{ijk\cdots} \epsilon^{iqr\cdots} = \epsilon^{(qr\cdots)}_{~jk\cdots} . Contraction 이 2개인 것으로 더 나아가면 \epsilon_{ABC\cdots} \epsilon^{ABG\cdots} = \epsilon_{ijk\cdots} \epsilon^{ijr\cdots} e_{C\cdots}^{k\cdots} e_{r\cdots}^{G\cdots} , \epsilon_{ijk\cdots} \epsilon^{ijr\cdots} = \epsilon^{(ijr\cdots)}_{~ijk\cdots} . In this case, only $2!$ of the $i,j$ summation survive. \epsilon_{ijk\cdots} \epsilon^{ijr\cdots} = 2! \cdot \epsilon^{(r\cdots)}_{~k\cdots} . 일반적인 n개의 contraction 이 일어날 때 나오는 summation over n indices gives \epsilon_{ijkl\cdots pq \cdots} \epsilon^{ijkl\cdots rs \cdots} = n! \cdot \epsilon^{(rs\cdots)}_{~pq\cdots} . Then defining \epsilon^{ABC\cdots}_{EFG\cdots} \equiv \epsilon^{ABC\cdots} \epsilon_{EFG\cdots} , we can anti-symmetrize any $\binom{0}{n}$, or $\binom{m}{0}$ tensors by T_{[AB]} = \frac{1}{(d-2)!} \epsilon_{ABCD\cdots}^{EFCD\cdots} \frac{1}{2!} T_{EF} , T_{[ABC]} = \frac{1}{(d-3)!} \epsilon_{ABCD\cdots}^{EFGD\cdots} \frac{1}{3!} T_{EFG} , T^{[AB]} = \frac{1}{(d-2)!} \epsilon^{ABCD\cdots}_{EFCD\cdots} \frac{1}{2!} T^{EF} , and so on. In general, T_{[A_1 \cdots A_n ]} = \frac{1}{(d-n)! n!} \epsilon_{A_1 \cdots A_n C_1 \cdots C_{d-n}} ^{B_1 \cdots B_n C_1 \cdots C_{d-n}} T_{B_1 \cdots B_n} and T^{[A_1 \cdots A_n ]} = \frac{1}{(d-n)! n!} \epsilon^{A_1 \cdots A_n C_1 \cdots C_{d-n}} _{B_1 \cdots B_n C_1 \cdots C_{d-n}} T^{B_1 \cdots B_n} . 결론적으로 깔끔한 Levi-Civita $\binom{d}{d}$ tensor인 \epsilon^{ABC\cdots}_{EFG\cdots} = \epsilon^{ijk\cdots}_{pqr\cdots} e^{ABC\cdots}_{i~j~k\cdots} e_{EFG\cdots}^{p~q~r\cdots} 가 정의된다. 이렇게 정의하니 기준 coordinate에서 임의의 coordinate로 transform이 되어도 지저분한 determinant 같은게 안들어 온다. 또한 d차원에서도 차수가 n만큼 떨어진 Levi-Civita $\binom{d-n}{d-n}$ tensor를 \begin{align*} &\epsilon^{A_1 \cdots A_{d-n}} _{B_1 \cdots B_{d-n}} \equiv \delta^{A_1 A_2 \cdots A_{d-n}} _{B_1 B_2 \cdots B_{d-n}} - \delta^{A_2 A_1 \cdots A_{d-n}} _{B_1 B_2 \cdots B_{d-n}} + \cdots \\ &= \frac{1}{n!} \epsilon^{A_1 \cdots A_{d-n} C_1 \cdots C_n} _{B_1 \cdots B_{d-n} C_1 \cdots C_n} = \epsilon^{i \cdots j}_{k \cdots l} e^{A_1 \cdots A_{d-n}}_{~i~\cdots~j} e_{B_1 \cdots B_{d-n}}^{~k~\cdots~l} \end{align*} 와 같이 적절히 정의할 수 있다. d차원에서 Levi-Civita $\binom{n}{n}$ tensor 의 contraction 은 \begin{align*} \epsilon_{ijk\cdots} ^{iqr\cdots} &= \delta_{ijk\cdots}^{iqr\cdots} - \delta_{ijk\cdots}^{qir\cdots} + \cdots \\ &= d \epsilon_{jk\cdots}^{qr\cdots} - (n-1) \epsilon_{jk\cdots}^{qr\cdots} \\ &= (d-n+1) \epsilon_{jk\cdots}^{qr\cdots} \end{align*} 와 같다. Fully anti-symmetric 한 이 Levi-Civita tensor 는 matrix 의 determinant 를 구할때에도 사용되고 volume 을 구할때에도 등장한다. Determinant 관련은 '선형대수학' 글 을 참고하도록 하고 volume 을 구할때 어떻게 쓰이는지 알아보자. #### Scalar volume 기준이 되는 Cartesian coordinate 에서 volume 은? 구하고자 하는 region (구역) $R$ 안의 점들에 대해 위와같이 적분하면 나온다. V = \int_{R} dx dy dz Volume (부피) 라는 것이 무슨 의미인지는 직관적으로 알고들 있을테니 설명은 생략하기로 하고, 이 volume이란 것을 임의의 coordinate 에서 어떻게 구할수 있을까? $dx, dy, dz$들은 vector 와 one-form 을 이야기 할 때 많이 등장한 개념이다. 좌표점의 미소변화는 vector 와 관련된 개념이니 위 식을 다음과 같이 쓸 수 있을지 생각해보자. V = \int_{R} \Big( \frac{\partial \vec{r}} {\partial x} dx \Big) \cdot \bigg( \Big( \frac{\partial \vec{r}} {\partial y} dy \Big) \times \Big( \frac{\partial \vec{r}} {\partial z} dz \Big) \bigg) ~~ \textrm{?} V = \int_{R} \epsilon_{ABC} e_x^{A} e_y^{B} e_z^{C} dx dy dz ~~ \textrm{?} V = \int_{R} \epsilon_{ABC} dx^{A} dy^{B} dz^{C} ~~ \textrm{?} where we defined $dx^{A} \equiv e_x^{A} dx$ and so on. 부피를 구할때 자주 썼던 외적, 내적을 이용했다. 지금보니 부피란게 Levi-Civita symbol 과 매우 밀접한 관련이 있는게 보이지 않는가? 뚜둥! 일반적인 coordinate 로 확장해보면 아래과 같이 쓰일 수 있을것 같다. 아무리 뒤죽박죽 엉망진창 찌그러져 있는 coordinate 라도 한 점 근처를 매우 크게 확대해서 보면 평행6면체 (parallelepiped) 가 되므로 V = \int_{R} \Big( \frac{\partial \vec{r}} {\partial u} du \Big) \cdot \bigg( \Big( \frac{\partial \vec{r}} {\partial v} dv \Big) \times \Big( \frac{\partial \vec{r}} {\partial w} dw \Big) \bigg)

절대값이 붙어야 제대로 정의되는데 편의상 뺐다. 항상 전체에 절대값이 붙어있음을 잊지 말자. 안그러면 inversion 같은것에 대해 부피가 -값이 될수도 있고 region 안에서 $u,v,w$를 증가시키면서 적분해야 하는지 감소시키면서 적분해야 하는지 등 헷갈릴수 있는 사항들이 많아진다. V = \int_{R} \epsilon_{ABC} e_u^{A} e_v^{B} e_w^{C} du dv dw V = \int_{R} \epsilon_{ABC} du^{A} dv^{B} dw^{C} where we defined $du^{A} \equiv e_u^{A} du$ and so on. 결론적으로 \begin{align*} V &= \int_{R} \epsilon_{ABC} e_u^{A} e_v^{B} e_w^{C} du dv dw \\ &= \int_{R} \det \Big( \Lambda_{\{u,v,w\}}^{\{x,y,z\}} \Big) \epsilon_{uvw} du dv dw = \int_{R} \det \Big( \Lambda_{i'}^{\bar{j}} \Big) du dv dw \end{align*} 가 된다.

// Be careful that $\Lambda_{\{u,v,w\}}^{\{x,y,z\}}$ does not mean $\Lambda_{uvw}^{xyz}$ which is defined to be $\Lambda_{u}^{x} \Lambda_{v}^{y} \Lambda_{w}^{z}$.

조금 더 나아가서 임의의 $n$차원에서 volume 이란 무엇일까? 어떻게 정의해야 할까? 처음부터 3차원보다 높은 차원으로 가면 헷갈릴테니, 2차원, 1차원에서 volume 이란게 무엇일지를 먼저 고민해보자. 그냥 위와 비슷하게 정의해서 써보면 2차원에서는 면적, 넓이라고 불리는 S = \int_{R} dx dy 가 되겠고, 1차원에서는 길이라고 불리는 L = \int_{R} dx 가 될 것이다. 그냥 이름만 다르게 지었을뿐, 의미랑 개념상 상당히 비슷한 애들이다. 그렇다면 이것들도 비슷한 공식을 만족할까? 당연하다. 1차원은 매우 쉽게 증명된다. L = \int_{R} dx = \int_{R} \frac{dx}{du} du = \int_{R} \Lambda^{x}_{u} du 1 by 1 matrix 에서 determinant 는 그냥 element 값이므로 당연히 \begin{align*} L &= \int_{R} \epsilon_A dx^{A} = \int_{R} \epsilon_x dx = \int_{R} dx \\ &= \int_{R} \epsilon_A du^{A} = \int_{R} \det \big( \Lambda_{u}^{x} \big) \epsilon_u du = \int_{R} \Lambda_{u}^{x} du = \int_{R} dx \end{align*} 가 성립한다. inversion 이 되었을 경우 sign 은 다를 수 있다. 2차원에서도 마찬가지로 \begin{align*} L &= \int_{R} \epsilon_{AB} dx^{A} dy^{B} = \int_{R} \epsilon_{xy} dx dy = \int_{R} dx dy \\ &= \int_{R} \epsilon_{AB} du^{A} dv^{B} = \int_{R} \det \big( \Lambda_{\{u,v\}}^{\{x,y\}} \big) \epsilon_{uv} du dv \end{align*} 가 성립한다. 따라서 임의의 n차원에서 volume 은 n-dimensional Cartesian coordinate 기준으로 잡힌 \begin{align*} &V = [L^{n}] = \int_{R} dx^{1} dx^{2} dx^{3} \cdots \\ &= \int_{R} \epsilon_{ABC\cdots} d \vec{x}^{ABC\cdots} \\ &\equiv \int_{R} \epsilon_{ABC\cdots} \frac{\partial \vec{r}}{\partial x^{1}}^{A} dx^{1} \frac{\partial \vec{r}}{\partial x^{2}}^{B} dx^{2} \frac{\partial \vec{r}}{\partial x^{3}}^{B} dx^{3} \cdots \\ &= \int_{R} \epsilon_{ABC\cdots} d \vec{u}^{ABC\cdots} \\ &\equiv \int_{R} \epsilon_{ABC\cdots} \frac{\partial \vec{r}}{\partial u^{1}}^{A} du^{1} \frac{\partial \vec{r}}{\partial u^{2}}^{B} du^{2} \frac{\partial \vec{r}}{\partial u^{3}}^{B} du^{3} \cdots \\ &= \int_{R} \det \big( \Lambda_{u\textrm{'s}}^{x\textrm{'s}} \big) du^{1} du^{2} du^{3} \cdots \end{align*} 로 정의될 것이다. (절대값 기호는 편의상 뺐다.) ### Symmetric tensor Fully anti-symmetric 한 Levi-Civita tensor 를 알아봤으니, 이번에는 fully symmetric 한 tensor 에 대해서 알아보자. \epsilon^{ABC\cdots+}_{EFG\cdots} = \epsilon^{\bar{i}\bar{j}\bar{k}\cdots+} _{\bar{p}\bar{q}\bar{r}\cdots} e^{ABC\cdots}_{\bar{i}~\bar{j}~\bar{k}\cdots} e_{EFG\cdots}^{\bar{p}~\bar{q}~\bar{r}\cdots} 을 어떻게 정의하면 coordinate 에 상관없이 같은 형태의 component 값을 가지게 할 수 있을지 고민해보자. Levi-Civita tensor 에서 힌트를 얻어 \begin{align*} \epsilon^{ijk\cdots+}_{pqr\cdots} &\equiv \delta^{ijk\cdots}_{pqr\cdots} + \delta^{jik\cdots}_{pqr\cdots} + \cdots \\ &\equiv \sum_{l,m,n,\cdots=\{p,q,r,\cdots\}} \epsilon_{(pqr\cdots)}^{~lmn\cdots+} \delta^{ijk\cdots}_{lmn\cdots} \\ &= \delta^{ijk\cdots}_{pqr\cdots} + \delta^{ijk\cdots}_{qpr\cdots} + \cdots \\ &\equiv \sum_{l,m,n,\cdots=\{i,j,k,\cdots\}} \epsilon_{~lmn\cdots}^{(ijk\cdots)+} \delta^{lmn\cdots}_{pqr\cdots} \end{align*} 와 같이 정의해보자. 한쪽 index 들은 고정시키고 다른쪽 index 들을 permutation 시키면서 모두 symmetric 하게 $\delta_{pqr\cdots}^{ijk\cdots}$들을 더한 꼴이다. 여기서 \epsilon_{~lmn\cdots}^{(ijk\cdots)+} = \epsilon^{~lmn\cdots}_{(ijk\cdots)+} = \begin{cases} 1 & \textrm{if } lmn\cdots = (ijk\cdots) , \\ (+1)^{\kappa} & \textrm{if } lmn\cdots \textrm{ is }\kappa\textrm{-th permutation from } (ijk\cdots) , \\ 0 & \textrm{otherwise.} \end{cases} $\epsilon^{ijk\cdots+}_{pqr\cdots}$ 값이 어떻게 나오는지 예를 몇개만 확인해보자. 4차원에서 index 들이 위아래 2개씩만 있는 \epsilon^{ij}_{pq} = \delta^{ij}_{pq} + \delta^{ij}_{qp} 의 값들은 \begin{align*} &\epsilon^{12}_{12} = \delta^{12}_{12} + \delta^{12}_{21} = 1 , \quad \epsilon^{11}_{11} = \delta^{11}_{11} + \delta^{11}_{11} = 2! , \\ &\epsilon^{13}_{31} = \delta^{13}_{31} + \delta^{13}_{13} = 1 , \quad \epsilon^{12}_{14} = \delta^{12}_{14} + \delta^{12}_{41} = 0 , \quad \textrm{ and so on.} \end{align*} 6차원에서 index들이 위아래 3개씩만 있는 \begin{align*} \epsilon^{ijk}_{pqr} &= \delta^{ijk}_{pqr} + \delta^{ijk}_{prq} + \delta^{ijk}_{qpr} + \delta^{ijk}_{qrp} + \delta^{ijk}_{rpq} + \delta^{ijk}_{rqp} \\ &= \delta^{i}_{p} \big( \delta^{jk}_{qr} + \delta^{jk}_{rq} \big) + \delta^{i}_{q} \big( \delta^{jk}_{pr} + \delta^{jk}_{rp} \big) + \delta^{i}_{r} \big( \delta^{jk}_{pq} + \delta^{jk}_{qp} \big) \end{align*} 의 값들은 \begin{align*} &\epsilon^{126}_{126} = \delta^{126}_{126} + \delta^{126}_{162} + \cdots = 1 , \quad \epsilon^{155}_{515} = \delta^{155}_{515} + \delta^{155}_{551} + \cdots = 2! , \\ &\epsilon^{222}_{222} = \delta^{222}_{222} + \delta^{222}_{222} + \cdots = 3! , \quad \epsilon^{123}_{345} = \delta^{123}_{345} + \delta^{123}_{354} + \cdots = 0 , \quad \textrm{ and so on.} \end{align*} $+1, -1, 0$만을 값으로 가졌던 Levi-Civita tensor 와는 다르게 $1$보다 큰 값도 가질수 있는걸 알 수 있다. 이 때의 component transformation 은? \begin{align*} &\epsilon^{\bar{i}\bar{j}\bar{k}\cdots+} _{\bar{p}\bar{q}\bar{r}\cdots} \Lambda_{\bar{i}\bar{j}\bar{k}\cdots} ^{abc\cdots} \Lambda^{\bar{p}\bar{q}\bar{r}\cdots} _{efg\cdots} = \Big[ \delta^{\bar{i}\bar{j}\bar{k}\cdots} _{\bar{p}\bar{q}\bar{r}\cdots} + \delta^{\bar{j}\bar{i}\bar{k}\cdots} _{\bar{p}\bar{q}\bar{r}\cdots} + \cdots \Big] \Lambda_{\bar{i}\bar{j}\bar{k}\cdots} ^{abc\cdots} \Lambda^{\bar{p}\bar{q}\bar{r}\cdots} _{efg\cdots} \\ &= \delta^{abc\cdots}_{efg\cdots} + \delta^{bac\cdots}_{efg\cdots} + \cdots = \epsilon^{abc\cdots+} _{efg\cdots} \end{align*} 와 같이 coordinate에 관계없이 component가 똑같은 형태를 띔을 알 수 있다. 즉 \begin{align*} &\epsilon^{ABC\cdots+}_{EFG\cdots} \equiv \delta^{ABC\cdots+}_{EFG\cdots} + \delta^{BAC\cdots+}_{EFG\cdots} + \cdots \\ &= \epsilon^{\bar{i}\bar{j}\bar{k}\cdots+} _{\bar{p}\bar{q}\bar{r}\cdots} e^{ABC\cdots}_{\bar{i}~\bar{j}~\bar{k}\cdots} e_{EFG\cdots}^{\bar{p}~\bar{q}~\bar{r}\cdots} = \epsilon^{ijk\cdots+} _{pqr\cdots} e^{ABC\cdots}_{i~j~k\cdots} e_{EFG\cdots}^{p~q~r\cdots} \end{align*} 과 같은 fully symmetric $\binom{n}{n}$ tensor 가 정의된다. 여기서 n은 차원수 d보다 클수도 작을수도 있다. cf. Levi-Civita 의 경우는 anti-symmetric 한 성질 때문에 차원수 d보다 큰 Levi-Civita $\binom{n}{n}$ tensor는 0으로 죽게된다. Levi-Civita tensor 과 비슷하게 신기한 결과가 나올지 symmetric tensor 에도 contraction 을 취해보자. d차원에서 symmetric $\binom{n}{n}$ tensor 의 contraction 은 \begin{align*} \epsilon^{ijk\cdots+}_{iqr\cdots} &\equiv \delta^{ijk\cdots}_{iqr\cdots} + \delta^{jik\cdots}_{iqr\cdots} + \cdots \\ &= d \epsilon^{jk\cdots}_{qr\cdots} + (n-1) \epsilon^{jk\cdots}_{qr\cdots} \\ &= (d+n-1) \epsilon^{jk\cdots}_{qr\cdots} \end{align*} 가 됨을 알 수 있다. ##[.hiden] Tensor Algebra 간단하게 tensor algebra 에 대해 알아보자.

Equality: Tensor $S^{A}_{B}$와 $T^{A}_{B}$가 같다는 것은 같은 좌표계에서 두 component 를 비교했을 때, 모든 component 들이 일치한다는 것을 의미한다. \begin{align*} e_{A}^{i} e^{B}_{j} \Big[ S^{A}_{B} = T^{A}_{B} \Big] ~ \rightarrow ~ S^{i}_{j} = T^{i}_{j} \end{align*} 이렇게 하나의 좌표계에서 두 tensor 의 모든 component 들이 일치하면, 다른 좌표계에서도 두 tensor 의 모든 component 들이 일치한다는 것을 알 수 있다. S^{k'}_{l'} = \Lambda^{k'j}_{i~l'} S^{i}_{j} = \Lambda^{k'j}_{i~l'} T^{i}_{j} = T^{k'}_{l'} 임의의 $\binom{m}{n}$ tensor 로의 확장은 자명. 같은 rank 의 tensor 끼리만 같음 (equality) 을 이야기 할 수 있음에 주의하자.
Linear Combinations: T^{A}_{B} = a P^{A}_{B} + b Q^{A}_{B} 위와 같은 덧셈, 뺄셈, scalar 곱셈을 이용하는 식을 정의할 수 있을텐데 이는 e_{A}^{i} e^{B}_{j} \Big[ T^{A}_{B} = a P^{A}_{B} + b Q^{A}_{B} \Big] ~ \rightarrow ~ T^{i}_{j} = a P^{i}_{j} + b Q^{i}_{j} 처럼 각 (each and every) component 끼리의 덧셈, 뺄셈, scalar 곱셈으로 이해하면 된다. 하나의 좌표계에서만 연산이 제대로 이루어지면, 다른 좌표계에서도 equality (등식) 를 만족하는 것을 알 수 있다. 이것 또한 임의의 $\binom{m}{n}$ tensor 로의 확장은 자명. 덧셈, 뺄셈도 같은 rank 의 tensor 끼리만 연산이 가능함에 주의하자.
Direct Products: 임의의 $\binom{m}{n}$ tensor로 확장시킬 때에도 이용했듯이 T^{A}_{B} = a P^{A} Q_{B} , \quad T^{AB}_{C} = b P^{A}_{C} Q^{B} = R^{A} S_{C} Q^{B} 와 같이 $\binom{a}{b}$ tensor 와 $\binom{c}{d}$ tensor 의 direct product 로 $\binom{a+c}{b+d}$ tensor 를 만들 수 있다. 이것도 each and every component 끼리의 곱셈으로 이해하면 된다. e_{A}^{i} e^{B}_{j} \Big[ T^{A}_{B} = a P^{A} Q_{B} \Big] ~ \rightarrow ~ T^{i}_{j} = a P^{i} Q_{j} 이것 또한 하나의 좌표계에서만 연산이 제대로 이루어지면, 다른 좌표계에서도 equality (등식) 를 만족하는 것을 알 수 있다. 임의의 $\binom{m}{n}$ tensor 로의 확장도 자명.
Contraction: Tensor 를 처음 설명하면서 vector 와 one-form 을 scalar 로 만드는 연산이 contraction (축약) 이라고 설명했다. 조금 더 확장시켜서 하나의 tensor 안에서의 contraction 과 multiple contractions 에 대해 알아보자. T^{ABC}_{BEF} = S^{AC}_{EF} 는 e_{AC}^{i~k} e^{EF}_{p~q} \Big[ T^{ABC}_{BEF} = S^{AC}_{EF} \Big] ~ \rightarrow ~ T^{ijk}_{jpq} = S^{ik}_{pq} 와 같은 연산이다. Kroneker-delta tensor 를 이용해 표현하면 \delta_{B}^{D} T^{ABC}_{DEF} = S^{AC}_{EF} 이다. Contraction 도 좌표계에 상관없이 같은 결과를 주는 것을 T^{ij'k}_{j'pq} = \delta^{j'}_{l'} T^{il'k}_{j'pq} = \delta^{j'}_{l'} \Lambda^{l'}_{r} T^{irk}_{j'pq} = \Lambda^{j'}_{r} T^{irk}_{j'pq} = T^{irk}_{rpq} = S^{ik}_{pq} 와 같이 알 수 있다. Multiple contractions 은 T^{AB}_{CD} S^{CD}_{AB} = c 와 같이 여러개의 indices 에 대한 contraction 을 동시에 취하는 것을 말한다. T^{ij}_{kl} S^{kl}_{ij} = c 와 같이 연산이 이루어진다.
Commutability: Tensor algebra 에서는 덧셈, 뺄셈, direct product 모두 commutable 하다. T^{A}_{B} = a P^{A}_{B} + b Q^{A}_{B} = b Q^{A}_{B} + a P^{A}_{B} = Q^{A}_{B} b + P^{A}_{B} a T^{A}_{B} = P^{A} Q_{B} = Q_{B} P^{A} and so on. Index 를 바꾸는 것이 아님에 주의. T^{AB} = P^{A} Q^{B} \neq Q^{A} P^{B} = T^{BA}

##[.hiden] Metric: Inner Product of Vectors (and One-forms) 다시 제일 처음 직관적으로 Cartesian coordinate 를 잡았던 때로 돌아가보자. 직관적으로 공간 3차원(+시간 1차원)이란 것은 가정하고 시작했다. 하지만 이런 공간 3차원의 $x$방향, $y$방향, $z$방향의 같은 거리는 어떻게 표현해야 할까? 이것도 약간 직관에 의존해서 정해야 하는 양이긴 하다. 바로 위에서 volume 을 구할때에도 따로 언급하지는 않았지만 Cartesian coordinate 에서 $dx,dy,dz$가 같을때 같은 거리를 나타내도록 잘 잡았다는 가정이 들어있었다. 아무튼 이러한 거리 개념을 scalar 로 표현하기 위함이기도 하고 여러가지로 내적이란 것을 정의하면 쓸모가 많아 보인다. 우선 vector 의 내적을 어떻게 정의해야 할지 생각해보자. 우선 vector 두개가 scalar 값을 주어야 하므로 $\binom{0}{2}$ tensor $g_{AB}$를 도입할 필요가 있어보인다. 이름은 선대 수학자, 물리학자들이 metric ('미터 단위의'라는 뜻) 이라고 붙였다. V^{A} = \frac{d \vec{x}}{d \lambda}^{A}, \qquad g_{AB} V^{A} V^{B} = ~ \textrm{?} , \quad g_{AB} V^{A} W^{B} = ~ \textrm{?} 우리가 익숙한 3차원 vector 의 내적으로부터 힌트를 얻어서 내적을 정의해 보자면, \frac{d \vec{x}}{d \lambda}^{A} \cdot \frac{d \vec{x}}{d \lambda}^{B} = \bigg( \frac{d x}{d \lambda} \bigg)^2 + \bigg( \frac{d y}{d \lambda} \bigg)^2 + \bigg( \frac{d z}{d \lambda} \bigg)^2 즉 g_{AB} = \delta_{\bar{i}\bar{j}} e^{\bar{i}}_{A} e^{\bar{j}}_{B} 로 \begin{align*} &g_{AB} V^{A} V^{B} = g_{\bar{i}\bar{j}} V^{\bar{i}} V^{\bar{j}} = \delta_{\bar{i}\bar{j}} V^{\bar{i}} V^{\bar{j}} = \sum_{\bar{i}} V^{\bar{i}} V^{\bar{i}} \\ &= \delta_{\bar{i}\bar{j}} \Lambda^{\bar{i}\bar{j}}_{kl} V^{k} V^{l} = \sum_{\bar{i}} \Lambda^{\bar{i}\bar{i}}_{kl} V^{k} V^{l} = g_{kl} V^{k} V^{l} \end{align*} 임을 알 수 있다. 피타고라스 정리로부터 이러한 값들은 rotation 에 대해 invariant 함을 쉽게 보일 수 있을것이다. 직관적으로도 당연한 이야기이다. 또한 contraction 으로 얻어지는 값은 coordinate-independent 한 scalar 이어야 하므로 coordinate 의 scale 을 다르게 잡으면 $g_{AB}$값도 그에 따라 scale 이 변하게 된다. 즉, $\vec{x} = x^{\bar{i}} \vec{e}_{\bar{i}} = x^{i} \vec{e}_{i} = \alpha x^{\bar{i}} \vec{e}_{i}$ 라면, 다시말해서 $x^{i} = \alpha x^{\bar{i}}$, $1 \vec{e}_{\bar{x}} + 3 \vec{e}_{\bar{y}} = \alpha \vec{e}_{x} + 3 \alpha \vec{e}_{y}$ 로 표시된다면, V^{A} = \frac{d x^{\bar{i}}} {d \lambda} e_{\bar{i}}^{A} = \frac{d x^{i}} {d \lambda} e_{i}^{A} = \alpha \frac{d x^{\bar{i}}} {d \lambda} e_{i}^{A} . g_{AB} V^{A} V^{B} = \sum_{\bar{i}} \Big( \frac{d x^{\bar{i}}} {d \lambda} \Big)^2 = g_{ij} V^{i} V^{j} = g_{ij} \alpha \frac{d x^{\bar{i}}} {d \lambda} \alpha \frac{d x^{\bar{j}}} {d \lambda} . 따라서 g_{ij} = \frac{1}{\alpha^2} \delta_{ij} 가 된다. 결론적으로 기준이 되는 coordinate 에서 $g_{\bar{i}\bar{j}}$를 적절히 정하고 다른 coordinate 에서의 $g_{ij}$ 값은 g_{ij} = \Lambda_{i}^{\bar{k}} \Lambda_{j}^{\bar{l}} g_{\bar{k}\bar{l}} \equiv \Lambda_{ij}^{\bar{k}\bar{l}} g_{\bar{k}\bar{l}} 로 구한다고 생각하면 될것이다. (where we define $\Lambda_{ij}^{\bar{k}\bar{l}} \equiv \Lambda_{i}^{\bar{k}} \Lambda_{j}^{\bar{l}}$ for short-handed notation.) ### Symmetricity $g_{AB}$가 같은 vector 에 대한 내적으로부터 정의되어지니 (정의된다기보단 개념이 시작된다고 보는게 맞는 표현일수도) 가장 기본적으로 기준에서의 metric 은 symmetric 하다는 가정하에 전개해 나갈 수 있을 것이다. 즉, g_{\bar{i}\bar{j}} = g_{\bar{j}\bar{i}} 는 내적에 쓰이는 metric의 기본적인 성질이라고 생각할 수 있다. 따라서 g_{AB} = g_{\bar{i}\bar{j}} e^{\bar{i}}_{A} e^{\bar{j}}_{B} = g_{\bar{i}\bar{j}} e^{~\bar{i}~\bar{j}}_{AB} = g_{\bar{j}\bar{i}} e^{~\bar{i}~\bar{j}}_{AB} = g_{BA} where we define $e^{~\bar{i}~\bar{j}}_{AB} \equiv e^{\bar{i}}_{A} e^{\bar{j}}_{B}$ for short-handed notation. 또한 g_{ij} = \Lambda_{ij}^{\bar{k}\bar{l}} g_{\bar{k}\bar{l}} = \Lambda_{ij}^{\bar{k}\bar{l}} g_{\bar{l}\bar{k}} = g_{ji} . 결론적으로 기준이 되는 coordinate 에서 $g_{\bar{i}\bar{j}}$가 symmetric 하다면 general 한 모든 coordinate 에서도 $g_{ij} = g_{ji}$로 symmetric 하다는 것이다. 이해를 돕기위해 spherical coordinate, cylinderical coordinate 에서 metric $g_{AB}$의 coordinate component 들이 어떤값인지 알아보자. 우선 spherical coordinate 에서의 metric 값들의 결과들만 보면 다음과 같다. g_{rr} = 1 , \quad g_{\theta\theta} = r^2 , \quad g_{\phi\phi} = r^2 \sin^2 \theta , \quad \textrm{the others} = 0 . 위에서도 알 수 있드시 metric 은 전 공간에도 정의된 field 개념의 $\binom{0}{2}$ tensor 이다. 따라서 좌표점에 따라 component 값이 달라질 수 있다. Cylinderical coordinate 에서의 metric 값들은 g_{\rho\rho} = 1 , \quad g_{\varphi\varphi} = \rho^2 , \quad g_{zz} = 1 , \quad \textrm{the others} = 0 와 같다. 다른식으로 개념적으로만 표현하자면 \begin{align*} g_{AB} V^{A} V^{B} &= dr^2 + r^2 d\theta^2 + r^2 \sin^2 \theta d\phi^2 \\ &= d\rho^2 + \rho^2 d\varphi^2 + dz^2 , \end{align*} \begin{align*} g_{AB} V^{A} W^{B} &= dr_1 dr_2 + r d\theta_1 d\theta_2 + r^2 \sin^2 \theta d\phi_1 d\phi_2 \\ &= d\rho_1 d\rho_2 + \rho^2 d\varphi_1 d\varphi_2 + dz_1 dz_2 . \end{align*} 위치가 같은 곳에서만 contraction 이 정해지므로 위치값 (좌표점) 을 나타내는 $(r,\theta,\phi)$와 $(\rho,\varphi,z)$는 같고 이곳에서 다른곳으로의 미소변화를 나타내는 vector 값은 다를 수 있다. 이 값들이 g_{ij} = \Lambda_{ij}^{\bar{k}\bar{l}} g_{\bar{k}\bar{l}} = \Lambda_{ij}^{\bar{k}\bar{l}} \delta_{\bar{k}\bar{l}} = \sum_{\bar{k}} \Lambda_{ij}^{\bar{k}\bar{k}} 식으로 구해진 값하고 같은지는 각자 확인해보자. 노가다성이 있는 증명이지만, 의심이 충분히 생길수 있는 부분이므로 확인해 보는것도 나쁘진 않은것 같다. 여기서는 너무 길어질거 같아 증명생략. ### Relation to scalar volume 잠시 volume 을 구할때, Levi-Civita tensor 의 component 변환을 구할때 등장했던 $\det \big( \Lambda_{i}^{\bar{j}} \big)$를 다시 한번 바라보자. Determinant 라는거 표현은 쉽지만 실제 구하기는 만만치 않다. 특히나 off-diagonal term 들이 0이 아닐 경우, matrix dimension 이 증가함에 따라 복잡도는 기하급수적으로 증가한다. 기준이 되는 coordinate 가 무엇인지가 애매할때도 있기도 하고, 여러가지 이유로 이 $\det \big( \Lambda_{i}^{\bar{j}} \big)$를 뭔가 쉽게, 무언가 더 물리적인 의미를 갖도록 구할 방법이 없을지 고민해 볼 필요가 있다. $\det \big( \Lambda_{i}^{\bar{j}} \big)$가 tensor transformation 에 등장하는 놈이므로 symmetry 를 가지는 tensor 가 있을 경우 matrix product 의 determinant 가 각 matrix 의 determinant 곱으로 나타난다는 특성을 이용하여 쉽게 구할 수 있을듯 싶다. When C_{ik} = \sum_{j} A_{ij} B_{jk} , \det(C_{ij}) = \det(A_{kl}) \det(B_{mn}) .

// 이것의 증명 관련해서는 '선형대수학' 글 참고.

이런 symmetric tensor로 metric을 선택해보자. As g_{ij} = \Lambda_{ij}^{\bar{k}\bar{l}} g_{\bar{k}\bar{l}} , \det(g_{ij}) = \det(\Lambda_{m}^{\bar{n}})^2 \det(g_{\bar{k}\bar{l}}) . 따라서 \det(\Lambda_{m}^{\bar{n}}) = \pm \sqrt{ \frac{\det(g_{ij})} {\det(g_{\bar{k}\bar{l}})} } where the sign depends on the sequence of coordinate axis, so to speak, whether it is right-handed coordinate or left-handed coordinate. 기준이 되는 coordinate 에서 metric 은 $\delta_{ij}$ 혹은 $\eta_{ij}$ 등 diagonal 값만 1 or -1, off-diagonal 은 모두 0 값을 갖는 경우가 많으므로 조금 더 간단하게 \Big| \det(\Lambda_{m}^{\bar{n}}) \Big| = \sqrt{ \det(g_{ij})} ~~ \big( \equiv \sqrt{g} \big) 와 같이 표현된다. 즉 \begin{align*} \epsilon^{ABC} &= \epsilon^{\bar{i}\bar{j}\bar{k}} e_{\bar{i}}^A e_{\bar{j}}^B e_{\bar{k}}^C = \epsilon^{\bar{i}\bar{j}\bar{k}} \Lambda_{\bar{i}\bar{j}\bar{k}}^{lmn} e_{l}^A e_{m}^B e_{n}^C \\ &= \det(\Lambda^{i}_{\bar{j}}) \epsilon^{lmn} e_{l}^A e_{m}^B e_{n}^C = \pm \sqrt{ \det(g_{ij})} \epsilon^{lmn} e_{l}^A e_{m}^B e_{n}^C \end{align*} and \begin{align*} &V = [L^{n}] = \int_{R} \epsilon_{ABC\cdots} d \vec{u}^{ABC\cdots} \\ &= \int_{R} \det \big( \Lambda_{u\textrm{'s}}^{x\textrm{'s}} \big) du^{1} du^{2} du^{3} \cdots \\ &= \int_{R} \sqrt{ \det(g_{ij}) } du^{1} du^{2} du^{3} \cdots . \end{align*} 이해를 돕기위한 예로 spherical and cylinderical coordinate 에서 volume 을 구하는 법을 보면, \begin{align*} &V = \int_{R} dx dy dz \\ &= \int_{R} \sqrt{ \det (g_{\{r,\theta,\phi\}}) } ~ dr d\theta d\phi = \int_{R} r^2 \sin\theta ~ dr d\theta d\phi \\ &= \int_{R} \sqrt{ \det (g_{\{\rho,\varphi,z\}}) } ~ d\rho d\varphi dz = \int_{R} \rho ~ d\rho d\varphi dz . \end{align*} ### Inner product of one-forms Inner product of one-forms 도 마찬가지로 정의될 수 있을텐데, 이때는 Cartesian coordinate 를 기준으로 잡지 말고 다음과 같은 식을 만족하도록 잡자. g_{AB} g^{CB} = \delta_{A}^{C} . 왜 이렇게 잡을까? $g_{AB}$가 내적에 쓰이지만 이 자체로 $\binom{0}{2}$ tensor이기 때문에 vector 와 contraction 을 하면 one-form 으로 만들어 버린다. Defining conjugate one-form like V_{A} \equiv g_{AB} V^{B} , \quad W_{A} \equiv g_{AB} W^{B} , \quad \textrm{and so on,} Let's define $g^{AB}$ to give the same result g_{AB} V^{A} W^{B} = g^{AB} V_{A} W_{B} for any arbitrary $\{V^{A}, W^{A}\}$ set. Then \begin{align*} &g^{AB} V_{A} V_{B} = g^{AB} g_{AC} V^{C} g_{BD} V^{D} \\ &= g^{ij} g_{ik} V^{k} g_{jl} V^{l} = g_{kl} V^{k} V^{l} . \end{align*} 따라서 \begin{align*} g^{ij} g_{ik} g_{jl} = g_{kl} . \end{align*} Producting inverse matrix of $g_{kl}$ on the both side, \begin{align*} &g^{ij} g_{ik} \sum_l g_{jl} g^{-1}_{lp} = \sum_l g_{kl} g^{-1}_{lp} \\ &g^{ij} g_{ik} \delta_{jp} = \delta_{kp} \\ &g^{ip} g_{ik} = \delta_{kp} \\ &g^{ip} \sum_k g_{ik} g^{-1}_{kq} = \sum_k \delta_{kp} g^{-1}_{kq} \\ &g^{ip} \delta_{iq} = g^{qp} = g^{-1}_{pq} . \end{align*} 즉 처음에 제시되었던 g_{ij} g^{kj} = \sum_j g_{ij} g_{jk}^{-1} = \delta_{ik} = \delta_{i}^{k} 식이 만족하게 나온다. One-form의 내적을 vector 의 내적과 consistant 하도록 잡은 것이다. $g_{ij}$가 symmetric 하다는 충분한 논리가 있으므로, i.e. $g_{ij}=g_{ji}$, 이를 이용해 $g^{kl}$의 특징도 알아보자. $g_{ij}$의 inverse 이기 때문에 선형대수학에서 증명된 inverse matrix (역행렬) 의 특징을 모두 갖는다고 보면 된다. 모든 증명을 여기 써놓기에는 너무 오래 걸리므로 결론만 적어보자면, $\det(g_{ij}) \neq 0$이어야 inverse 가 존재한다는 점. 이 때 left inverse 와 right inverse 가 둘 다 항상 같이 존재하고 그 matrix element 도 같다는 점. $g_{ij}$가 symmetric 하면, i.e. $g_{ij}=g_{ji}$, inverse 도 symmetric 하다는 점. 등이 있겠다. 즉 \det(g_{ij}) \neq 0 , g^{ij} = g^{ji} , \qquad g^{AB} = g^{BA} , \begin{align*} &g^{ij} g_{jk} = g^{ji} g_{jk} = g^{ij} g_{kj} = g^{ji} g_{kj} = \delta^{i}_{k} , \\ &g^{AB} g_{BC} = g^{BA} g_{BC} = g^{AB} g_{CB} = g^{BA} g_{CB} = \delta^{A}_{C} . \end{align*} ### Arc length parametrization $\binom{1}{0}$ tensor 인 vector 를 설명하면서 scalar parameter $\lambda$를 도입했었다. 하지만 이런 parameter 를 도입하기가 난감한 경우가 다반사다. Line이 3차원 공간상에 있는데 색이 입혀져 있는 것도 아니고 시간에 따라 점입자 (point particle) 가 이동하는 궤적을 표현하는 것도 아닐때 우리가 이 line (or curve) 을 기술하기 위해 도입할 수 있는 scalar parameter 가 있을까? 상대론으로 들어가면 시간 또한 우리가 적당히 잡은 coordinate 의 한 축에 불과하다. 시간까지 들어간 3차원+1차원 space-time 에서의 curve 는 나중에 상대론 이야기를 하면서 다루기로 하고, 여기서는 3차원 공간에서 이 line (or curve) 을 coordinate independent 하게 설명하기 위한 arc length parametrization 에 대해 알아보려고 한다. 길이를 재는 metric 에 대해 배웠으니 이를 이용해서 line 을, 이 line 에 의해 정의되는 vector 를 기술하려는 것이다. 결론적으론 g_{AB} \frac{d \vec{x}_{C}}{d s}^A \frac{d \vec{x}_{C}}{d s}^B = \Big( \frac{d x_{C}}{d s} \Big)^2 + \Big( \frac{d y_{C}}{d s} \Big)^2 +\Big( \frac{d z_{C}}{d s} \Big)^2 = 1 이 되도록 arc length parameter $s$를 잡는다는 것이다. 즉 d x_{C}^2 + d y_{C}^2 + d z_{C}^2 = d s^2 이 되도록 $s$를 잡아야 한다. 이렇게 parameter 를 잡을 경우, 당연히 이 parameter $s$는 좌표계에 무관한 (but up to constant) scalar parameter 가 된다. 이 curve 가 나타내는 vector 만이 중요한 물리량이라 했을 때에는 어디를 $s$의 원점 ($s=0$) 으로 잡는지는 중요하지 않다. 단 어느 방향으로 갈 때 $s$가 증가할지에 따라 vector 의 방향이 정해진다. 따라서 항상 두가지 선택이 존재한다. 이것에 관해서는 우선은 이정도만 이야기하고 상대론에 들어가면 조금 더 분석해 보면서 개인적인 의견을 덧붙이겠다. (잠깐 이야기를 하자면: +, -전하의 origin? +time 으로 evolve 하는 입자와 -time 으로 evolve 하는 입자? 입자 vs 반입자? 등 아직은 그냥 지극히 개인적인 상상들이지만 우주를 이해하는 다른 방식의 사고가 될수도 있을듯해서...) 처음부터 arc length parametrization 을 완벽하게 하기란 불가능에 가까우므로 어떤식으로 arc length parametrization 을 할 수 있을지도 잠깐 알아보도록 하겠다. 우선 line 에 임의로 숫자 $\lambda$를 붙여서 parametrized curve 를 기술한다. 이 숫자 $\lambda$는 우리가 임의로 curve에 붙인 것이므로 물리적인 scalar 는 당연히 아니다. 각 좌표점에서 물리적 거리를 나타내는 metric 을 알고 있다고 한다면 $\lambda$로 인해 파생되는 vector 들도 line 의 각 위치에서 g_{AB} \frac{d \vec{x}_{C}} {d \lambda}^A \frac{d \lambda}{d s} \frac{d \vec{x}_{C}} {d \lambda}^B \frac{d \lambda}{d s} = 1 이 되도록 \frac{d s}{d \lambda} = \pm \sqrt{ g_{AB} \frac{d \vec{x}_{C}} {d \lambda}^A \frac{d \vec{x}_{C}} {d \lambda}^B } 를 구할 수 있고, 이 관계식으로부터 $\lambda$ parametrization 으로부터 arc length $s$ parametrization 을 구할 수 있게 된다. s (\vec{x}_C) = \int^{\lambda(\vec{x}_C)}_{\textrm{base}} d \lambda \frac{d s}{d \lambda} = \pm \int^{\lambda(\vec{x}_C)}_{\textrm{base}} d \lambda \sqrt{ g_{AB} \frac{d \vec{x}_{C}} {d \lambda}^A \frac{d \vec{x}_{C}} {d \lambda}^B } ##[.hiden] Covariant Derivative $\nabla_{A}$ $\binom{n}{m}$ tensor 를 $\binom{n}{m+1}$ tensor 로 만드는 operation 인 covariant derivative 가 정의될 수 있다. Gradient 와 비슷한 operator 인데 one-form 을 설명할 때, scalar field 에 이를 operation 시켜 one-form 을 만들고 one-form 에 대한 물리적 의미, 개념에 관해 이야기 할 때 봤을 것이다. 이젠 vector 에 이 operation 을 가해 어떻게 $\binom{1}{1}$ tensor 로 만들수 있을지 생각해보자. 일반적인 gradient 의 쓰임새처럼 $\nabla_A V^{B}$는 vector field $V^{B}$가 좌표점이 변함에 따라 어떻게 변하는가를 표현하게 될 것이라고 생각되어진다. \begin{align*} \nabla_A \big( V^{B} \big) &= e_{A}^{i} \partial_i \big( V^{j} e_{j}^{B} \big) \\ &= e_{A}^{i} \Big[ \partial_i \big( V^{j} \big) e_{j}^{B} + V^{j} \partial_i \big( e_{j}^{B} \big) \Big] \end{align*} 여기서 가장 애매한 부분이 $\partial_i \big( e_{j}^{B} \big)$이다. 어떻게 처리해야 할까? 이것도 어쩔수 없이 기준 coordinate 에서 어떻게 처리할지를 정하고 consistant 하게 transform 이 되도록 만들어야 할 것이다. 우선 다른 좌표점, 다른 위치에서 같은 vector 라는게 무엇일지를 고민해봐야 한다. 다시 Cartesian coordinate 로 가보자. Flat 한 공간을 묘사하는 coordinate 이기 때문에 서로 다른 위치, 좌표점에 있어도 vector 의 Cartesian component 가 같으면 두 vector 가 같다고 말할 수 있을것이다. 즉, \begin{align*} &\partial_{\bar{i}} \big( e_{\bar{j}}^{A} \big) = 0 \\ &\partial_{k} \big( e_{\bar{j}}^{A} \big) = 0 . \end{align*} Defining, therefore, connection basis vector, which is quite coordinate dependent, as \begin{align*} &\Gamma_{ij}^{~B} = \Gamma_{ij}^{~k} e_{k}^{B} \equiv \partial_{i} \big( e_{j}^{B} \big) \\ &= \partial_{i} \big( \Lambda_{j}^{\bar{l}} e_{\bar{l}}^{B} \big) = \partial_{i} \big( \Lambda_{j}^{\bar{l}} \big) e_{\bar{l}}^{B} = \partial_{i} \big( \Lambda_{j}^{\bar{l}} \big) \Lambda_{\bar{l}}^{k} e_{k}^{B} \\ &= \frac{\partial} {\partial x^{i}} \bigg( \frac{\partial x^{\bar{l}}} {\partial x^{j}} \bigg) \frac{\partial x^{k}} {\partial x^{\bar{l}}} e_{k}^{B} = \frac{\partial^2 x^{\bar{l}}} {\partial x^{i} \partial x^{j}} \frac{\partial x^{k}} {\partial x^{\bar{l}}} e_{k}^{B} = \frac{\partial^2 x^{\bar{l}}} {\partial x^{j} \partial x^{i}} \frac{\partial x^{k}} {\partial x^{\bar{l}}} e_{k}^{B} = \Gamma_{ji}^{~k} e_{k}^{B} = \Gamma_{ji}^{~B} \end{align*} \begin{align*} \nabla_A \big( V^{B} \big) &= e_{A}^{i} \partial_i \big( V^{j} e_{j}^{B} \big) \\ &= e_{A}^{i} \Big[ \partial_i \big( V^{j} \big) e_{j}^{B} + V^{j} \partial_i \big( e_{j}^{B} \big) \Big] \\ &= e_{A}^{i} \Big[ \partial_i \big( V^{j} \big) e_{j}^{B} + V^{j} \Gamma_{ij}^{~k} e_{k}^{B} \Big] \\ &= e_{Aj}^{iB} \Big[ \partial_i \big( V^{j} \big) + V^{k} \Gamma_{ik}^{~j} \Big] \\ \end{align*} 전개 과정에서 위 아래 반복되는 index 들은 (called dummy index, or repeated index) 기호를 바꿔도 되는 성질을 이용했다. 즉 $\binom{1}{1}$ tensor 인 $T_{A}^{B} \equiv \nabla_A \big( V^{B} \big)$는 \begin{align*} T_{i}^{j} &= \partial_i \big( V^{j} \big) + V^{k} \Gamma_{ik}^{~j} \\ &= \partial_i \big( V^{j} \big) + V^{k} \partial_{i} \big( \Lambda_{k}^{\bar{l}} \big) \Lambda_{\bar{l}}^{j} \end{align*} 의 component 를 갖는다. 정말 이것이 $\binom{1}{1}$ tensor 인지 tensor transformation 을 잘 만족하는지 확인해보도록 하자. Chicago convention 으로 표현된 $\nabla_A \big( V^{B} \big)$에서 시작하여 접근하면 당연히 $\binom{1}{1}$ tensor 일 수밖에 없지만 그래도 확인하는 차원에서. \begin{align*} T_{p'}^{q'} &= \Lambda_{p'}^{i} \Lambda_{j}^{q'} T_{i}^{j} = \Lambda_{p'}^{i} \Lambda_{j}^{q'} \Big[ \partial_i \big(V^{j} \big) + V^{k} \Gamma_{ik}^{~j} \Big] \\ &= \Lambda_{p'}^{i} \Lambda_{j}^{q'} \Big[ \partial_i \big( V^{j} \big) + V^{k} \partial_{i} \big( \Lambda_{k}^{\bar{l}} \big) \Lambda_{\bar{l}}^{j} \Big] \\ &= \Big[ \Lambda_{j}^{q'} \partial_{p'} \big( V^{j} \big) + V^{k} \partial_{p'} \big( \Lambda_{k}^{\bar{l}} \big) \Lambda_{\bar{l}}^{q'} \Big] \\ &= \Big[ \partial_{p'} \big( \Lambda_{j}^{q'} V^{j} \big) - \partial_{p'} \big( \Lambda_{j}^{q'} \big) V^{j} + V^{k} \partial_{p'} \big( \Lambda_{k}^{j'} \Lambda_{j'}^{\bar{l}} \big) \Lambda_{\bar{l}}^{q'} \Big] \\ \end{align*} \begin{align*} &= \Big[ \partial_{p'} \big( V^{q'} \big) - \partial_{p'} \big( \Lambda_{j}^{q'} \big) V^{j} + V^{k} \partial_{p'} \big( \Lambda_{k}^{j'} \big) \Lambda_{j'}^{\bar{l}} \Lambda_{\bar{l}}^{q'} + V^{k} \Lambda_{k}^{j'} \partial_{p'} \big( \Lambda_{j'}^{\bar{l}} \big) \Lambda_{\bar{l}}^{q'} \Big] \\ &= \Big[ \partial_{p'} \big( V^{q'} \big) - \partial_{p'} \big( \Lambda_{j}^{q'} \big) V^{j} + V^{k} \partial_{p'} \big( \Lambda_{k}^{q'} \big) + V^{j'} \partial_{p'} \big( \Lambda_{j'}^{\bar{l}} \big) \Lambda_{\bar{l}}^{q'} \Big] \\ &= \Big[ \partial_{p'} \big( V^{q'} \big) + V^{j'} \partial_{p'} \big( \Lambda_{j'}^{\bar{l}} \big) \Lambda_{\bar{l}}^{q'} \Big] \\ &= \Big[ \partial_{p'} \big( V^{q'} \big) + V^{j'} \Gamma_{p'j'}^{~~q'} \Big] . \end{align*} 예상대로 잘 따른다. 이젠 one-form 에 covariant derivative 를 다룰 차례다. 어떻게 될까? \begin{align*} \nabla_{A} \big( W_{B} \big) &= e_{A}^{i} \partial_{i} \big( W_{j} e^{j}_{B} \big) \\ &= e_{A}^{i} \Big[ \partial_{i} \big( W_{j} \big) e^{j}_{B} + W_{j} \partial_{i} \big( e^{j}_{B} \big) \Big] \end{align*} 마찬가지로 $\partial_{i} \big( e^{j}_{B} \big)$를 분석해보면, \begin{align*} \partial_{i} \big( e^{j}_{B} \big) = \partial_{i} \big( \Lambda^{j}_{\bar{l}} e^{\bar{l}}_{B} \big) = \partial_{i} \big( \Lambda^{j}_{\bar{l}} \big) e^{\bar{l}}_{B} = \partial_{i} \big( \Lambda^{j}_{\bar{l}} \big) \Lambda^{\bar{l}}_{k} e^{k}_{B} . \end{align*} Vector basis 를 미분할때 정의했던 $\Gamma_{ij}^{k} \equiv \partial_{i} \big( \Lambda_{j}^{\bar{l}} \big) \Lambda_{\bar{l}}^{k}$와는 어떤 관계가 있을까? $\Lambda_{j}^{\bar{l}} \Lambda_{\bar{l}}^{k} = \delta_{j}^{k}$로 좌표점, 위치에 관계없이 상수인 것을 이용, \begin{align*} &\partial_{i} \big( \Lambda_{j}^{\bar{l}} \Lambda_{\bar{l}}^{k} \big) = 0 \\ &= \partial_{i} \big( \Lambda_{j}^{\bar{l}} \big) \Lambda_{\bar{l}}^{k} + \Lambda_{j}^{\bar{l}} \partial_{i} \big( \Lambda_{\bar{l}}^{k} \big) \\ &= \Gamma_{ij}^{~k} + \Lambda_{j}^{\bar{l}} \partial_{i} \big( \Lambda_{\bar{l}}^{k} \big) . \end{align*} 즉 \partial_{i} \big( \Lambda_{\bar{l}}^{k} \big) \Lambda_{j}^{\bar{l}} = - \Gamma_{ij}^{~k} 인 것을 알 수 있다. 따라서 \partial_{i} \big( e^{j}_{B} \big) = - \Gamma_{ik}^{~j} e^{k}_{B} \equiv - \Gamma_{iB}^{~j} 로 표현된다. 마지막 부분은 성립하는 식이라기 보다 이렇게 쓰기로 정하자 정도가 되겠다. 이러한 관계식을 이용해 one-form의 covariant derivative를 표현해보면, \begin{align*} \nabla_{A} \big( W_{B} \big) &= e_{A}^{i} \partial_{i} \big( W_{j} e^{j}_{B} \big) \\ &= e_{A}^{i} \Big[ \partial_{i} \big( W_{j} \big) e^{j}_{B} + W_{j} \partial_{i} \big( e^{j}_{B} \big) \Big] \\ &= e_{A}^{i} \Big[ \partial_{i} \big( W_{j} \big) e^{j}_{B} - W_{j} \Gamma_{ik}^{~j} e^{k}_{B} \Big] \\ &= e_{AB}^{~i~j} \Big[ \partial_{i} \big( W_{j} \big) - W_{k} \Gamma_{ij}^{~k} \Big] \end{align*} 이다. Tensor transformation 은 당연히 만족한다. 의심이 들면 직접 해보자. 이제 마지막 단계로, 가장 일반적인 $\binom{m}{n}$ tensor 를 $\binom{m}{n+1}$ tensor 로 만드는 covariant derivative 의 경우만 남았다. 필요한 정보들은 위에서 다 유도해놨다. 다 차려놓은 밥상에 숟가락만 얹으면 된다. \begin{align*} \nabla_{A} \big( T^{BCD\cdots}_{FGH\cdots} \big) &= e_{A}^{i} \partial_{i} \big( T^{jkl\cdots}_{pqr\cdots} e^{BCD\cdots}_{j~k~l\cdots} e^{p~q~r\cdots}_{FGH\cdots} \big) \\ &= e_{A}^{i} \partial_{i} \big( T^{jkl\cdots}_{pqr\cdots} \big) e^{BCD\cdots}_{j~k~l\cdots} e^{p~q~r\cdots}_{FGH\cdots} \\ &~~ + e_{A}^{i} T^{jkl\cdots}_{pqr\cdots} \partial_{i} \big( e^{B}_{j} \big) e^{CD\cdots}_{k~l\cdots} e^{p~q~r\cdots}_{FGH\cdots} + e_{A}^{i} T^{jkl\cdots}_{pqr\cdots} \partial_{i} \big( e^{C}_{k} \big) e^{BD\cdots}_{j~l\cdots} e^{p~q~r\cdots}_{FGH\cdots} + \cdots \\ &~~ + e_{A}^{i} T^{jkl\cdots}_{pqr\cdots} \partial_{i} \big( e^{p}_{F} \big) e^{BCD\cdots}_{j~k~l\cdots} e^{q~r\cdots}_{GH\cdots} + e_{A}^{i} T^{jkl\cdots}_{pqr\cdots} \partial_{i} \big( e^{q}_{G} \big) e^{BCD\cdots}_{j~k~l\cdots} e^{p~r\cdots}_{FH\cdots} + \cdots \\ \end{align*} \begin{align*} &= e^{i}_{A} e^{BCD\cdots}_{j~k~l\cdots} e^{p~q~r\cdots}_{FGH\cdots} \Big[ T^{jkl\cdots}_{pqr\cdots} \\ &~~~~~~~~~~~~~~~~~~~~~~~ + T^{skl\cdots}_{pqr\cdots} \Gamma_{is}^{~j} + T^{jsl\cdots}_{pqr\cdots} \Gamma_{is}^{~k} + \cdots \\ &~~~~~~~~~~~~~~~~~~~~~~~ - T^{jkl\cdots}_{sqr\cdots} \Gamma_{ip}^{~s} - T^{jkl\cdots}_{psr\cdots} \Gamma_{iq}^{~s} - \cdots \Big] \end{align*} 규칙들은 금방 눈치 챌 수 있으리라 본다. 이것 또한 tensor transformation 을 만족한다. ##[.hiden] Divergence, Curl, and Laplace 이젠 지금까지 배운 vector 와 one-form 의 개념들, covariant derivative, metric 등을 잘 조작하여 무언가 의미를 갖는 operation 들이 없을지를 알아보자. ### Divergence 우선 divergence operation 이란 것에 대해 알아보기에 앞서 다른식으로 정의되는, divergence 를 이해하는데 더 도움이 되는 개념으로서의, vector field 에 대해 알아보려고 한다. 첫부분에서는 vector 의 개념적인 이해를 위해서 scalar parameter $\lambda$를 도입했다. 이 $\lambda$가 변함에 따라 얼마나 빠르게 다른 위치 (좌표점) 로 움직이려 하는가를 나타내는 것이 vector 라고 이야기를 하였는데, 이러고 보니 vector 를 vector field 개념으로 확장할때 문제가 생긴다. 이 $\lambda$라는 scalar 를 각 좌표점 (each and every points) 마다 다 도입을 해야하는 것이다. 즉 global 한 scalar, 그것도 값이 global 하게 바뀌는 scalar 가 도입되어야 하는듯한 느낌이다. 이렇게 값이 바뀌는 global 한 scalar 가 있을까? 간단하게 있다고 생각하고 전개하면 쉽겠지만 이게 간단한 문제가 아니다. 시간 (time) 이 값이 바뀌는 global scalar 가 될 수 있다고 말할 수 있을까? 상대론에 들어가면, 혹은 물리를 조금 더 근본적으로, 시간을 조금 더 근본적으로 바라보면 시간을 global scalar 라고 하기에는 부족한 부분들이 많이 보인다. 나머지는 상대론 설명을 하면서 이야기 하도록 하고, 값이 바뀌는 global scalar 를 잡기가 난감할 때 또다른 대안책 (alternative) 으로 vector 를 정의하는/이해하는 방법을 이야기 해보려 한다. 일반적인 d차원에서 scalar volume 을 구하는 법에 대해 이야기 했었는데, 이것에서 힌트를 얻어 다음과 같은 scalar 를 생각해보자. \begin{align*} &\int_{S} \epsilon_{ABC} V^{A} \frac{\partial \vec{x}} {\partial u}^{B} d u \frac{\partial \vec{x}} {\partial v}^{C} d v = \int_{S} \epsilon_{ABC} V^{A} e_{u}^{B} d u e_{v}^{C} d v \\ &=\int_{S} \epsilon_{ABC} \frac{d \vec{x}} {d \lambda}^{A} \frac{\partial \vec{x}} {\partial u}^{B} d u \frac{\partial \vec{x}} {\partial v}^{C} d v = \int_{S} \sqrt{g} \epsilon_{wuv} \frac{d \vec{x}} {d \lambda}^{w} d u ~ d v \end{align*} 단위 $\lambda$당 $du~dv$ surface 쪽으로 빠져나가는 scalar volume 의 양을 나타낸다는 걸 알수있다. 여기서 vector $V^{A}$는 당연히 각 점마다 값을 가지고 있는 field 개념의 vector 이다. 이것에서 힌트를 얻어 field 개념의 vector 를 다른식으로 정의할 수 있다. 처음은 간단하게 Cartesian 에서 시작하자. \int_{S} \epsilon_{ABC} V^{A} \frac{\partial \vec{x}} {\partial y}^{B} d y \frac{\partial \vec{x}} {\partial z}^{C} d z = \int_{S} \epsilon_{xyz} V^{x} dy dz = \int_{S} V^{x} dy dz 부호까지 잘 고려하면서, 위 값은 coordinate independent 한 scalar 일까? Vector 세개와 $\binom{0}{3}$ Levi-Civita tensor 가 contraction 된 것이므로 당연히 scalar 라고 생각할 수 있겠지만, 한가지 integral 이 취해진 surface $S$가 coordinate independent 하게 잡혔냐하는 것에 주의해야 한다. 위에서의 $S$는 $x=$const 인 평면이다. 이 surface 만 유지된다면 위 값은 scalar 가 된다. 이를 이용해서 $V^{x}$를 역으로 정의해보자. V^{x} = \frac{\textrm{scalar related to } dy ~ dz} {dy ~ dz} 이렇게만 써놓으면 너무 추상적이므로 구체적인 예들로 V^{x} = \frac{\textrm{number of lines passing through } [dy ~ dz] \textrm{ in }+x \textrm{ direction}} {dy ~ dz} 혹은 V^{x} = \frac{\textrm{number of particles passing through } [dy ~ dz] \textrm{ in }+x \textrm{ direction}} {dy ~ dz} 를 생각해보자. 갯수를 세는 것이므로 당연히 scalar 일테고, 이 scalar 에서부터 역으로 vector 를 정의 (생각) 하는것이 가능하다. 조금 더 general 한 coordinate $(u,v,w)$에서 생각하면, V^{u} = \frac{\textrm{scalar passing through } [dv ~ dw] \textrm{ in }+u \textrm{ direction}} {\sqrt{g} ~ dv ~ dw} $du~dv$ surface 로 단위 scalar $\lambda$당 빠져나가는 scalar 양으로 vector field 를 생각할 수 있게 된다. 이 때 일정 volume $V$를 정하고 이 volume 의 boundary $S = \partial V$를 통해 바깥으로 나가는 scalar 양을 알아내고자 한다면, \int_{S = \partial V} V^{A} \epsilon_{ABC} d S^{BC} = \int_{V} \frac{1}{\sqrt{g}} \partial_i \big( \sqrt{g} V^{i} \big) \sqrt{g} ~ du dv dw 형태로 계산되어짐을 짐작할 수 있다. 여기서 $dS^{BC}$는 $V^{A}$가 volume 의 바깥을 향하는 성분으로 적분되도록 정해진 surface $\binom{2}{0}$ tensor 이다. Divergence theorem (also known as Gauss's theorem or Ostrogradsky's theorem) 이라고 불리는 surface integral 을 volume integral 로 바꾼 형태이다. 보통 volume integral 을 surface integral 로 바꾸지만, divergence 의 개념적인 이해를 위해서는 surface integral 형태의 개념에서 volume 을 아주 작게 보내는 극한으로 이 미소 volume 점에서 바깥쪽으로 어떤 scalar 양이 발산 (diverge) 하는지를 나타내는 것으로 접근하는 것이 좋다. 즉, \frac{1}{\sqrt{g}} \partial_i \big( \sqrt{g} V^{i} \big) = \frac{\text{scalar going outside of }dV (= dudvdw)} {\sqrt{g} ~ dudvdw} 로 divergence 를 접근하자. $dudvdw$로 잘려진 미소 volume $dV$ 바깥쪽으로 나가는 scalar 양을 미소 scalar volume $\sqrt{g} ~ dudvdw$으로 나눈 형태이다. 왼쪽 편의 수식을 좀 더 정리해 보면, \begin{align*} \frac{1}{\sqrt{g}} \partial_i \big( \sqrt{g} V^{i} \big) &= \partial_i V^{i} + V^{i} \frac{1}{\sqrt{g}} \partial_i \sqrt{g} \\ &= \det ( \Lambda_{\bar{l}}^{m} ) \partial_i \big( \det ( \Lambda_{j}^{\bar{k}} ) V^{i} \big) \\ &= \partial_i V^{i} + V^{i} \det ( \Lambda_{\bar{l}}^{m} ) \partial_i \det ( \Lambda_{j}^{\bar{k}} ) . \end{align*} 여기서 \begin{align*} &\frac{1}{\sqrt{g}} \partial_i \sqrt{g} = \det ( \Lambda_{\bar{l}}^{m} ) \partial_i \det ( \Lambda_{j}^{\bar{k}} ) \\ &= \epsilon_{(123)}^{~\bar{p}\bar{q}\bar{r}} \Lambda_{\bar{p}}^{1} \Lambda_{\bar{q}}^{2} \Lambda_{\bar{r}}^{3} \partial_i \big( \epsilon_{~\bar{l}\bar{m}\bar{n}}^{(123)} \Lambda_{1}^{\bar{l}} \Lambda_{2}^{\bar{m}} \Lambda_{3}^{\bar{n}} \big) \\ &= \epsilon_{(123)}^{~\bar{p}\bar{q}\bar{r}} \epsilon_{~\bar{l}\bar{m}\bar{n}}^{(123)} \Lambda_{\bar{p}}^{1} \Lambda_{\bar{q}}^{2} \Lambda_{\bar{r}}^{3} \partial_i \big( \Lambda_{1}^{\bar{l}} \Lambda_{2}^{\bar{m}} \Lambda_{3}^{\bar{n}} \big) \end{align*} $\epsilon^{~\bar{l}\bar{m}\bar{n}}_{(123)}$는 좌표값에 상관없으니 이것에 대한 미분 $\partial_i$는 0이다. 또한 $\epsilon_{(123)}^{~\bar{p}\bar{q}\bar{r}} \epsilon_{~\bar{l}\bar{m}\bar{n}}^{(123)}$는 앞서 Levi-Civita tensor에서 정의한 $\epsilon_{\bar{l}\bar{m}\bar{n}}^{\bar{p}\bar{q}\bar{r}} = \delta_{\bar{l}\bar{m}\bar{n}}^{\bar{p}\bar{q}\bar{r}} - \delta_{\bar{l}\bar{n}\bar{m}}^{\bar{p}\bar{q}\bar{r}} + \cdots$이 된다. 이를 넣고 정리하면, \begin{align*} &\frac{1}{\sqrt{g}} \partial_i \sqrt{g} = \Big[ \epsilon_{\bar{l}\bar{m}\bar{n}} ^{\bar{p}\bar{q}\bar{r}} \Big] \Lambda_{\bar{p}}^{1} \Lambda_{\bar{q}}^{2} \Lambda_{\bar{r}}^{3} \Big[ \partial_i (\Lambda_{1}^{\bar{l}} ) \Lambda_{2}^{\bar{m}} \Lambda_{3}^{\bar{n}} + \Lambda_{1}^{\bar{l}} \partial_i ( \Lambda_{2}^{\bar{m}} ) \Lambda_{3}^{\bar{n}} + \Lambda_{1}^{\bar{l}} \Lambda_{2}^{\bar{m}} \partial_i ( \Lambda_{3}^{\bar{n}} ) \Big] \\ &= \Big[ \Lambda_{\bar{l}}^{1} \Lambda_{\bar{m}}^{2} \Lambda_{\bar{n}}^{3} - \Lambda_{\bar{l}}^{1} \Lambda_{\bar{n}}^{2} \Lambda_{\bar{m}}^{3} + \cdots \Big] \Big[ \partial_i ( \Lambda_{1}^{\bar{l}} ) \Lambda_{2}^{\bar{m}} \Lambda_{3}^{\bar{n}} + \Lambda_{1}^{\bar{l}} \partial_i ( \Lambda_{2}^{\bar{m}} ) \Lambda_{3}^{\bar{n}} + \Lambda_{1}^{\bar{l}} \Lambda_{2}^{\bar{m}} \partial_i ( \Lambda_{3}^{\bar{n}} ) \Big] \end{align*} 여기서 왼쪽 첫번째 $\Lambda_{\bar{l}}^{1} \Lambda_{\bar{m}}^{2} \Lambda_{\bar{n}}^{3}$만 살아남고 나머지는 다 죽게된다. 예를 들어 $\Lambda_{\bar{l}}^{1} \Lambda_{2}^{\bar{l}} = \delta_{2}^{1} = 0$와 같이 0으로 떨어지니까. 결과적으로 \begin{align*} &\frac{1}{\sqrt{g}} \partial_i \sqrt{g} = \Lambda_{\bar{l}}^{1} \Lambda_{\bar{m}}^{2} \Lambda_{\bar{n}}^{3} \Big[ \partial_i ( \Lambda_{1}^{\bar{l}} ) \Lambda_{2}^{\bar{m}} \Lambda_{3}^{\bar{n}} + \Lambda_{1}^{\bar{l}} \partial_i ( \Lambda_{2}^{\bar{m}} ) \Lambda_{3}^{\bar{n}} + \Lambda_{1}^{\bar{l}} \Lambda_{2}^{\bar{m}} \partial_i ( \Lambda_{3}^{\bar{n}} ) \Big] \\ &= \Lambda_{\bar{l}}^{1} \partial_i ( \Lambda_{1}^{\bar{l}} ) \delta_{2}^{2} \delta_{3}^{3} + \delta_{1}^{1} \Lambda_{\bar{m}}^{2} \partial_i ( \Lambda_{2}^{\bar{m}} ) \delta_{3}^{3} + \delta_{1}^{1} \delta_{2}^{2} \Lambda_{\bar{n}}^{3} \partial_i ( \Lambda_{3}^{\bar{n}} ) \\ &= \Lambda_{\bar{l}}^{1} \partial_i ( \Lambda_{1}^{\bar{l}} ) + \Lambda_{\bar{m}}^{2} \partial_i ( \Lambda_{2}^{\bar{m}} ) + \Lambda_{\bar{n}}^{3} \partial_i ( \Lambda_{3}^{\bar{n}} ) \\ &= \Lambda_{\bar{l}}^{j} \partial_i ( \Lambda_{j}^{\bar{l}} ) \end{align*} 가 된다. 이는 connection $\Gamma_{ij}^{~k}$로 표현이 가능하다. \Gamma_{ij}^{~A} = \partial_{i} e_{j}^{A} = \partial_{i} ( \Lambda_{j}^{\bar{l}} ) e_{\bar{l}}^{A} = \partial_{i} ( \Lambda_{j}^{\bar{l}} ) \Lambda_{\bar{l}}^{k} e_{k}^{A} 이므로 \frac{1}{\sqrt{g}} \partial_i \sqrt{g} = \Lambda_{\bar{l}}^{j} \partial_i ( \Lambda_{j}^{\bar{l}} ) = \Gamma_{ij}^{~j} 이다. 종합하면, \begin{align*} \frac{1}{\sqrt{g}} \partial_i \big( \sqrt{g} V^{i} \big) &= \partial_i V^{i} + V^{i} \frac{1}{\sqrt{g}} \partial_i \sqrt{g} \\ &= \partial_i V^{i} + V^{i} \det ( \Lambda_{\bar{l}}^{m} ) \partial_i \det ( \Lambda_{j}^{\bar{k}} ) \\ &= \partial_i V^{i} + V^{i} \Lambda_{\bar{l}}^{j} \partial_i ( \Lambda_{j}^{\bar{l}} ) \\ &= \partial_i V^{i} + V^{i} \Gamma_{ij}^{~j} \end{align*} 가 된다. 다음과 같은 scalar 를 생각해보면, \begin{align*} \nabla_{A} \big( V^{A} \big) &= e_{A}^{i} \partial_{i} \big( V^{j} e_{j}^{A} \big) \\ &= e_{A}^{i} \Big[ \partial_{i} \big( V^{j} \big) e_{j}^{A} + V^{j} \partial_{i} \big( e_{j}^{A} \big) \Big] \\ &= \delta_{j}^{i} \partial_{i} \big( V^{j} \big) + e_{A}^{i} V^{j} \Gamma_{ij}^{~A} \\ &= \partial_{i} \big( V^{i} \big) + V^{j} \Gamma_{ij}^{~i} \\ &= \partial_{i} \big( V^{i} \big) + V^{j} \partial_{i} \big( \Lambda_{j}^{\bar{l}} \big) \Lambda_{\bar{l}}^{i} \end{align*} 위의 결과와 값이 같음을 알 수 있다. 즉, \begin{align*} &\nabla_{A} \big( V^{A} \big) = \frac{1}{\sqrt{g}} \partial_i \big( \sqrt{g} V^{i} \big) \\ &= \partial_i V^{i} + V^{i} \Gamma_{ij}^{~j} \end{align*} 인 divergence operation 이 정의되게 된다. Vector field 두개로 다음과 같은 scalar 를 만드는 것도 가능? \begin{align*} \int \epsilon_{ABC} V^{A} W^{B} \bigg( \frac{\partial \vec{x}} {\partial v}^{C} d v \bigg) \end{align*} 이것도 scalar volume이랑 관계된거 같긴 한데... 의미 없나? \nabla_{A} T^{AB} 와 같이 $\binom{2}{0}$ tensor에 divergence를 취하는 것은 무엇을 의미할까? ### Curl 보통 curl 을 3차원에서 많이들 배우기 때문에, vector 에 curl 을 취하면 vector ($\binom{1}{0}$ tensor) 를 줄 것이라고 오해하기 쉽다. $\nabla \times \vec{E}$와 같이 말이다. 하지만 tensor 적으로 curl 을 보려면 이런식으로는 한계가 있다. 3차원에서만 curl operation 을 취하는게 아니라 임의의 d차원에서도 curl operation 을 정의할 수 있어야 할 것이기 때문이다. Divergence 와 마찬가지로 적분형태에서 먼저 출발해보자. \oint_{C = \partial S} d l^{A} W_{A} 와 같은 circular integration 을 생각해보자. 이는 one-form $W_A$가 curve $C$를 돌아왔을때 얼마나 scalar 값이 변했는지를 나타낸다. One-form 을 설명할 때 scalar field 에 covariant derivative 를 취하는 형식으로 설명했다. ($W_A = \nabla_A \phi$) One-form $W_A$가 이런식으로 표현된다면 위의 circular integration 값은 항상 0이 나오겠지만, 이는 one-form 의 coordinate component transformation 에 관해 설명하기 위해 도입한 설명방식일 뿐 더 일반적인 one-form 에 관해서는 circular integration 이 0이 아닌 값을 가질수도 있다. \epsilon_{AB}^{CD} \nabla_{C} \big( W_{D} \big) = \nabla_{A} W_{B} - \nabla_{B} W_{A} #### Curvature [\nabla_A , \nabla_B] V^{D} = R_{ABC}^{~~~~~~D} V^{C} Loop을 따라서 parallel transport 시킨 뒤 두 vector를 비교하는 것이라 할 수 있다. ### Laplacian g^{AB} \nabla_{A} \nabla_{B} ##[.hiden.no-sec-N#sec-Sup-Comments] Supplementary comments 위에서 tensor 를 설명하면서 도입한 가정들, 기준으로 잡았던 것들, 정의들을 수정한다면 전혀 다른 이론으로 전개해 갈수도 있다. 예를 들자면, \Gamma_{ij}^{~k} \neq \Gamma_{ji}^{~k} 인채로 전개해 나간다던지, g_{ij} \neq g_{ji} 처럼 symmetry 를 가정하지 않고, symmetry 를 깨버린 채 전개해 나가는 등 직관적으로 그럴듯하고 당연할거처럼 보이는 것들을 한가지씩 건드려서 다른 이론으로 전개해 나가는 경우도 많다. 당연하다고 느꼈던걸 당연하다고 안느끼고 의문을 가지고 고민하고 연구하면 때로는 정말 새로운 참신한 결과가 나올때도 있으니까. 하지만 이 과정에서 어느 부분에 의문점을 가질 것인가가 매우 중요하다. 마구잡이로 아무대나 고쳐보면서 참신한 결과를 기대하기는 힘들다. 뭐 운으로 때려맞추는 경우도 많긴 하지만... 가장 덜 그럴듯해 보였던 가정들을 먼저 깨고 전개해 나가 본다던지, 적절한 전략을 택해서 될거같은 곳에 먼저 집중해야 결과가 잘 나올 것이라 생각된다. 뭐 때로는 무턱대고 이상한 길로 들어서야만 신기한걸 발견하기도 하긴 하지만... 비주류 혹은 사이비 과학쪽에 빠지는 사람들이 간혹 보이기도 하는데, 비주류 쪽은 본인이 궁금해하고 연구하고 싶어서 빠지는거라 쳐도 무지에서 비롯된 사이비 이론, 사이비 과학쪽은 좀 조심하자. 그리고 비주류 쪽에 빠질때도 본인이 무엇이 궁금해서, 무엇이 알고싶은지를 명확히 하고 나아가자. 어떠한 문제점을 해결하려고 달려든 것이지, 무엇이 궁금한지도 명확히 하지 않은채 나아가면 삼천포로 빠지기 쉽상이다. Problem solving approach (PSA) 의 첫단계가 풀고자 하는 문제를 명확히 구체화하는 것이니까. 그래도 언제나 선택은 본인 몫이다. ##[.hiden.no-sec-N#sec-Index-Def] Index and Definition

Einstein's summation convention: Repeated upper and lower index pair means summation over all. V^{i} W_{i} = \sum_{i} V^{i} W_{i} A pair of repeated upper (or lower) indices does not mean summation. V^{i} V^{i} \neq \sum_{i} V^{i} V^{i}
Short-handed notation: \Lambda^{i}_{\bar{j}} \equiv \frac{\partial x^{i}}{\partial x^{\bar{j}}}, \qquad \Lambda^{k'}_{j} \equiv \frac{\partial x^{k'}}{\partial x^{j}}, \qquad \Lambda^{i}_{j'} \equiv \frac{\partial x^{i}}{\partial x^{j'}}, \cdots , \Lambda_{ij}^{\bar{k}\bar{l}} \equiv \Lambda_{i}^{\bar{k}} \Lambda_{j}^{\bar{l}} , \Lambda_{ijk}^{\bar{l}\bar{m}\bar{n}} \equiv \Lambda_{i}^{\bar{l}} \Lambda_{j}^{\bar{m}} \Lambda_{k}^{\bar{n}} , \Lambda_{i\bar{k}}^{\bar{l}j} \equiv \Lambda_{i}^{\bar{l}} \Lambda_{\bar{k}}^{j} , and so on.
Short-handed notation: e^{AB}_{i~j} \equiv e^{A}_{i} e^{B}_{j} , \qquad e_{AB}^{i~j} \equiv e_{A}^{i} e_{B}^{j} , \qquad e^{Aj}_{iB} \equiv e^{A}_{i} e^{j}_{B} , and so on.

## RRA

Books

Book - A First Course in General Relativity, 1985-02-22, by Bernard F. Schutz.; 처음 상대론을 배울때 쓰면 좋은 교재같음.
Web - Lecture Notes on General Relativity, 1997, by Sean M. Carroll.; 인터넷에서도 볼 수 있어서 좋은 책. Carroll씨가 꽤 유명한듯?
Book - Gravitation and Cosmology: Principles and Applications of the General Theory of Gravity, 1972, by Steven Weinberg.; 잘 정리되어 있는 책같음. 하지만 오래된 책이라서 디자인이나 구성이나 수식 form이 구식인듯한 느낌. (대부분의 유명한 상대론 책들은 고전인듯.)
Book - Gravitation, 1970, by Charles W. Misner, Kip S. Thorne, and John Archibald Wheeler.; 꽤나 두꺼운 책. 유명하긴 한듯? 이것도 오래된 책(고전). Legendary book이라고 칭하는 사람도 있는듯. 제대로 읽은적이 없어서 개인적인 판단은 아직.

kipid's blog :: 텐서와 상대론 (Tensor and Relativity) - 0. 텐서 (Tensor) 란?
kipid's blog :: 텐서(Tensor)와 상대론(Relativity) - 1. 상대론(Relativity)
전파거북이(EM turtle)'s blog - 텐서(Tensor), 2011-06, by EM turtle; 수학적 시각의 텐서 설명.
Wiki - Tensor (텐서); 한글 페이지는 설명이 거의 없음. 일부분은 제가 편집.
kipid's blog :: SI, cgs 단위계 및 물리상수들 (SI, cgs unit and physical constants)
kipid's blog :: 선형 대수학 간단한 정리들 (Linear Algebra)
Wiki - 뉴턴 운동 법칙

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Tensor and Relativity - 0. What is Tensor?

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