What do physicists mean by *coordinate transformation* exactly?

Question

Physics book always use the following sentence: Let $x\rightarrow x'$, then the vector/field/tensor... transform as ...

In the book Introducing Einstein’s Relativity by Ray d’Inverno and James Vickers chapter 6 p87, when discussing the Lie derivative, it offers me some insight.

This is a long paragraph, I quote it for completeness

My questions are

1. First line of equation 6.5. we have $$ \tilde{T}^{ab}(\tilde{x}) = \frac{\partial \tilde{x}^a}{\partial x^c} \frac{\partial \tilde{x}^b}{\partial x^d} T^{cd}(x).\tag{6.5} $$

What is the meaning of this equation? The LHS is a function (the component of the tensor, the two authors don't use abstract index notation) in chart $\tilde{x}$ and RHS is a function in chart $x$, so it seems to say two functions $f(x)=g(y)$, which is weird for me.

2. Equation 6.7 express the Lie derivative by the substraction between an actively transformed tensor $\tilde{T}$ and a tensor T which is not transformed. But in usual case, we define the Lie derivative by flow of a vector field. So whether this means coordiante transformation in physics is indeed some map related to flow of some vector field? (I guess maybe the vector field along coordinate curves)

Consider the transformation $$ \tilde{x}^a = x^a + \delta u \, X^a(x), \tag{6.3} $$ where $\delta u$ is small. This is called a point transformation and is to be regarded actively as sending the point $P$, with coordinates $x^a$, to the point $Q$, with coordinates $x^a + \delta u X^a(x)$, where the coordinates of each point are given in the same $x^a$-coordinate system, i.e. $$ P \to Q, \quad x^a \to x^a + \delta u X^a(x). $$ The point $Q$ clearly lies on the curve of the congruence through $P$ which $X^a$ generates (Fig. 6.4). Differentiating (6.3), we get $$ \frac{\partial \tilde{x}^a}{\partial x^b} = \delta^a_b + \delta u \, \partial_b X^a. \tag{6.4} $$ Next, consider the tensor field $T^{ab}$ at the point $P$. Then its components at $P$ are $T^{ab}(x)$ and, under the point transformation (6.3), we have the mapping $$ T^{ab}(x) \to \tilde{T}^{ab}(\tilde{x}). $$ i.e. the transformation `drags' the tensor $T^{ab}$ along from $P$ to $Q$. The components of the dragged-along tensor are given by the usual transformation law for tensors (see (5.30)), and so, using (6.4) $$ \tilde{T}^{ab}(\tilde{x}) = \frac{\partial \tilde{x}^a}{\partial x^c} \frac{\partial \tilde{x}^b}{\partial x^d} T^{cd}(x) $$ $$ = (\delta^a_c + \delta u \, \partial_c X^a)(\delta^b_d + \delta u \, \partial_d X^b) T^{cd}(x) $$ $$ = T^{ab}(x) + \left[ \partial_c X^a \, T^{cb}(x) + \partial_d X^b \, T^{ad}(x) \right] \delta u + O(\delta u^2). \tag{6.5} $$ Applying Taylor's theorem to first order, we get $$ T^{ab}(\tilde{x}) = T^{ab}(x^c + \delta u \, X^c(x)) = T^{ab}(x) + \delta u \, X^c \partial_c T^{ab}(x). \tag{6.6} $$ We are now in a position to define the Lie derivative of $T^{ab}$ with respect to $X^a$, which is denoted by $\mathcal{L}_X T^{ab}$, as $$ \mathcal{L}_X T^{ab} = \lim_{\delta u \to 0} \frac{T^{ab}(x) - \tilde{T}^{ab}(\tilde{x})}{\delta u}. \tag{6.7} $$

EDIT, lets consider another example, the metric tensor case, which is more familair and also usually used in physics book (like Conformal Field Theory for Particle Physicists, by Marc Gillioz, eq. 2.4), for a metric tensor $g$, we have $$ g'_{\mu \nu}(x') dx'^\mu dx'^\nu = g_{\mu \nu}(x) dx^\mu dx^\nu,\tag{2.4} $$

For me, everything here should be (let $\phi$ be the first chart with coordinate $x$, and $\psi$ another,with $y$) First we have $$g(p)=g(p)$$ then the aurthor seems to mean $$(g\circ \phi^{-1})(x)=(g\circ \psi^{-1})(y)$$ and then expand a tensor into its components and basis $$(g\circ \phi^{-1})_{\mu \nu}(x)dx^\mu \otimes dx^\nu=(g\circ \psi^{-1})_{\mu \nu}(y)dy^\mu \otimes dy^\nu$$ where $y$ behave like $x'$ and $(g\circ \psi^{-1})_{\mu \nu}$ behave like $g'_{\mu \nu}$

But I doubt do we really can say $(g\circ \phi^{-1})(x)=(g\circ \psi^{-1})(y)$? Because I think they're in different charts, and I only understand what do we mean by $f(x)=g(x)$, if someone tells me $f(x)=g(y)$, I will assume it means $f(x)=g(x)$, which is not the case above, I think.

If we insist, I think the right expression is $$(g\circ \phi^{-1})(x)=(g\circ \psi^{-1})(y(x))=(g\circ \psi^{-1} \circ \psi \circ \phi^{-1})(y(x))=(g\circ \phi^{-1})(x),$$ which is trival

Answer 1

That equation describes the transformation of a tensor under coordinate transformations. The way it says it is somewhat confusing, but the whole $f(x)=g(y)$ is just saying “this tensor in coordinates $\tilde x$ is equal to the expression for the tensor in coordinates $x$ times the derivative expressions between $\tilde x$ and $x$”.

There is a more artful explanation (or, depending on your opinions on math, less artful) that I learned for this while I was studying the basics. Let $\phi$ be the transformation mapping $x_1\to y_1$. Then, arbitrary tensors are changed by this transformation as well; a tensor $X$ with $p$ contravariant and $q$ covariant indices then transforms under $\phi$ into another tensor $Y$, and the two are related by

$$Y^{y_1y_2\dots y_{p-1}y_p}_{y_{p+1}y_{p+2}\dots y_{q-1}y_q}=X^{x_1x_2\dots x_{p-1}x_p}_{x_{p+1}x_{p+2}\dots x_{q-1}x_q}\left(\prod_{n=1}^p\frac{\partial y_n}{\partial x_n}\right)\left(\prod_{m=1}^{q}\frac{\partial x_m}{\partial y_m}\right);$$

in simpler terms, contravariant indices transform like

$$Y^y=X^x\frac{\partial y}{\partial x}$$

and covariant indices transform like

$$Y_y=X_x\frac{\partial x}{\partial y}.$$

These relations are immediately useful when you want to change coordinate systems, e.g. when you have something in spherical coordinates that you want to express in rectangular coordinates.

Answer 2

Here I am only answering the part "...it seems to say two functions f(x)=g(y), which is weird for me."

It is more than one function because they are tensors, so there is one equation for each different combination of indexes. But in essence, you are right. It is simpler to see this with a scalar function. Let us say you have the temperature along a rod, which is a scalar field, and let us assume a specific function: $T(x)=x^2$. Now, if you change your system of coordinates, let us say $y=x/2$. In your new coordinates you will have a different function on different coordinates $T'(y)$, with $T'(y)=T(x)$. this is not weird, it means that y and x are related. You have $T(x)=T(2y)=(2y)^2=4y^2=T'(y)$.

Answer 3

Well, it seems that you need a mathematical translation of physical manipulation of symbols.

Regarding your first issue, we have two possible interpretations of (6.5).

(A) [active interpretation] Mathematically speaking, the used notion is here the pushforward of vector fields, associated to the (local) diffeomorphism implicit in the active coordinate transformations $x\to x’$. (6.5) is the coordinate expression of the action of the pushforward. In this interpretation the coordinate system remains fixed but the point $p$ of coordinates is transformed to the point $p’$ of coordinates $x’$ in the same local chart. (6.5) describes the associated natural action on tensor fields in the considered local chart.

Alternatively,

(B) [passive interpretation] $p$ remains fixed but we change the local coordinate system around it. And we pass from a local chart where $p$ has coordinates $x$ to a local chart where $p$ has coordinates $x’$. From this perspective (6.5) just accounts for the relation between the components of the same tensor $T$ applied at $p$ when changing local coordinates.

Concerning your second issue, yes the book is using the active interpretation and focuses on the action of the flow of a vector field. So, a unique coordinate system is given and the family of local diffeomorphisms parametrised by the parameter $u$ are defined by the integral curves of the vector field $X$ in the said local chart. In other words the book is employing the flow of $X$ written in coordinates. (6.7) is the, in my view misleading, expression of the Lie derivative of $T$ at $x$. Misleading because it is not emphasised in the formula that one is working at $x$ fixed.

Regarding your last issue, that identity makes sense if used with some precautions. Here $g$ has to be viewed as a smooth map $$g: M\ni p \mapsto T_p^*M\otimes T_p^*M\subset \Gamma(T^*M\otimes T^*M).$$ Hence, if $(U,\phi)$ and $(V,\psi)$ are two local charts on $M$, an identity like $$(g\circ \phi^{-1})(x)=(g\circ \psi^{-1})(y)$$ is meaningful in $U\cap V$ provided $x:= \phi(p)$ and $y:= \psi(p)$ for any $p\in U\cap V$.