3
$\begingroup$

I am extremely confused with the metric tensors. In my previous question, I figured out that tensors are "objective" entities that are independent of coordinate systems. However, metric tensors keep making troubles. In my book Ta-Pei Cheng's Relativity, Gravitation and Cosmology p. 198, metric tensors are defined as such :

In an $n$-dimensional space, for a set of coordinate basis vectors $\{\textbf{e}_i\}$, there are inverse basis vectors $\{\textbf{e}^j\}$ which are $\textbf{e}_i \cdot \textbf{e}_j=\delta^j_i$. In this coordinate system, metric functions are defined to be $\textbf{e}_i \cdot \textbf{e}_j= g_{ij}$ which are called metric and $\textbf{e}^i \cdot \textbf{e}^j= g^{ij}$ which are called inverse metric.

However, in my understanding, a metric is also a tensor which has its own entity regardless of coordinate systems. However, a metric tensor seems to be too much closely related to a given coordinate system. So this means that metric tensors exist not independently of coordinate systems? Also, the words "metric" and "inverse metric" also annoys me. For me $g_{ij}$ and $g^{ij}$ are just different components expression in one coordinate of the same metric tensor. Then, why do the words "metric" and "inverse metric" exist?

In addition, the transformation rules stated for metric tensors cause confusion. Given a coordinate transformation $L$ from an unprimed coordinate system to a primed coordinate system, $g'\;^{ij}= L^i\;_kL^j \;_l\;g^{kl}$ is said to hold. What is $g'\;^{ij}$? Is it the metric tensor for the unprimed coordiante system expressed in a different, primed coordinate system? If the primed coordinate system has basis $\{\textbf{e}'_i\}$ and inverse basis $\{\textbf{e}'^j\}$, what is the relation of $g'\;^{ij}$ and $\textbf{e}'^i \cdot \textbf{e}'^j$? I think $\textbf{e}'^i \cdot \textbf{e}'^j$ is the expression for "the metric tensor for the primed coordinate system", therefore must be different from $g'\;^{ij}$...

Could anyone please help me?

$\endgroup$
  • $\begingroup$ You're intertwining about three different formalisms. The metric tensor will look very different written using the "standard" formalism versus using the "tetrad" or "vierbein" formalism. Most of the other stuff you're discussing is just matrix multiplication and statemeents about the chain rule, ultimately. $\endgroup$ – Jerry Schirmer Mar 20 '18 at 16:23
  • $\begingroup$ What are the formalisms? How can I understand the metric tensor in each formalism? What is annoying me most is the relation between $g'^{ij}$ and $\textbf{e}'^i \cdot \textbf{e}'^j$. Could you explain the relation between them? $\endgroup$ – Keith Mar 20 '18 at 16:28
  • $\begingroup$ In concrete index notation, $g'^{ij}$ is the ij'th component of the metric, expressed in the primed coordinate system. In abstract index notation, we don't need the prime, we can't express the notion of a component, and $g^{ij}$ is just a symbol for the whole object. (The $ij$ are dummy symbols, sort of like when people talk about "the function f(x).") $\endgroup$ – user4552 Mar 20 '18 at 17:27
3
$\begingroup$

The confusion comes from the fact that apparently 2 concepts are involved in your question which makes the topic rather complicated. The 2 concepts are :

description of the metric tensor in coordinate dependent form (the classical way)

description of the metric tensor in coordinate free form (the modern way), which actually requires the use of differential forms.

It is so easy to write down an equation like $e_i \cdot e_j = \delta_{ij}$, and it looks so intuitive that one does not hesitate a moment of believing it, but behind there is the rather abstract theory of tangent vectors and differential forms on manifolds. In this theory the symbols $e_i$ are tangent vectors of a freely chosen point of a manifold, and these are actually written as expressions of partial derivatives and in your question they are chosen to be orthonormal. In a 2-dim. flat manifold there are for instance -- it's our choice -- 2 orthonormal tangent vectors $\frac{\partial}{\partial x}$ and $\frac{\partial}{\partial y}$ which have the (nice) property to orthonormal to their corresponding covectors ${ dx , dy}$. (Covectors are the basis of the dual space, here called cotangent space, which is defined pointwise, i.e. at each point of the manifold there is another tangent space and cotangent space etc. ... ) Explicitly:

$dx(\frac{\partial}{\partial x})=1$ and $dy(\frac{\partial}{\partial y})=1$ whereas $ dy(\frac{\partial}{\partial x})=0$ and $dx(\frac{\partial}{\partial y})=0$.

Now we define what is meant by $\cdot$ the product between tangent vectors. For this we need the metric tensor $g$ which is a symmetrical tensor $e_i \cdot e_j := g(e_i, e_j)$. So if our basis is chosen to be orthonormal, then actually we get : $e_i \cdot e_j = g(e_i, e_j)=\delta_{ij}$. We will work out that a bit more. Our tensor $g$ actually is in the formalism of differential forms:

$g = dx \otimes dx + dy \otimes dy$

If we want to know its components we have evaluate it on the basis vectors (remember $dx(\frac{\partial}{\partial x})=1$ and $dy(\frac{\partial}{\partial y})=1$ whereas $ dy(\frac{\partial}{\partial x})=0$ and $dx(\frac{\partial}{\partial y})=0$. ):

$ e_x \cdot e_x = g(\frac{\partial}{\partial x}, \frac{\partial}{\partial x})= dx \otimes dx ( \frac{\partial}{\partial x}, \frac{ \partial}{\partial x} ) + dy \otimes dy( \frac{\partial}{\partial x}, \frac{ \partial}{\partial x} ) = g_{xx} = 1+0= 1.$

$ e_x \cdot e_y = g(\frac{\partial}{\partial x}, \frac{\partial}{\partial y})= dx \otimes dx ( \frac{\partial}{\partial x}, \frac{ \partial}{\partial y} ) + dy \otimes dy( \frac{\partial}{\partial x}, \frac{ \partial}{\partial y} ) = g_{xy} = 0 + 0 =0.$

$ e_y \cdot e_x = g(\frac{\partial}{\partial y}, \frac{\partial}{\partial x})= dx \otimes dx ( \frac{\partial}{\partial y}, \frac{ \partial}{\partial x} ) + dy \otimes dy( \frac{\partial}{\partial y}, \frac{ \partial}{\partial x} ) = g_{yx} =0 + 0 =0.$

$ e_y \cdot e_x = g(\frac{\partial}{\partial y}, \frac{ \partial}{\partial y})= dx \otimes dx ( \frac{\partial}{\partial y}, \frac{ \partial}{\partial y} ) + dy \otimes dy( \frac{\partial}{\partial y}, \frac{ \partial}{\partial y} ) = g_{yy}= 0 + 1 =1.$

We got the desired result, the basis vector are orthonormal as required. What happens if we change the metric ? Let's go to polar coordinates $(r,\phi)$. (Remember $(x,y) =(r cos\phi, r sin\phi)$, the derivatives below have to be carried out using this definition) With these coordinates we can construct the following tangent vectors $\left(\frac{\partial}{\partial r}, \frac{\partial}{\partial \phi}\right)$. The corresponding covectors are $(dr, d\phi)$:

The metric in polar coordinates looks like this: $g = dr \otimes dr + r^2 d\phi \otimes d\phi$

$g_{rr}= g(\frac{\partial}{\partial r}, \frac{ \partial}{\partial r})= dr \otimes dr ( \frac{\partial}{\partial r}, \frac{ \partial}{\partial r} ) + r^2 d\phi \otimes d\phi( \frac{\partial}{\partial r}, \frac{ \partial}{\partial r} ) = 1 + 0 =1. $

$g_{r\phi}= g(\frac{\partial}{\partial r}, \frac{ \partial}{\partial \phi})= dr \otimes dr ( \frac{\partial}{\partial r}, \frac{ \partial}{\partial \phi} ) + r^2 d\phi \otimes d\phi( \frac{\partial}{\partial r}, \frac{ \partial}{\partial \phi} ) = 0 + 0 =0.$

If $r$ and $\phi$ in the tangent vectors are swapped, the result is also zero: $g_{\phi r}=0$.

$g_{rr}= g(\frac{\partial}{\partial \phi}, \frac{ \partial}{\partial \phi})= dr \otimes dr ( \frac{\partial}{\partial \phi}, \frac{ \partial}{\partial \phi} ) + r^2 d\phi \otimes d\phi( \frac{\partial}{\partial \phi}, \frac{\partial}{\partial \phi} ) = 0 + r^2 =r^2$.

We actually find that our chosen tangent vectors are normal to each other, but not orthonormal. That's okay. That is our choice. The base system does not need to be orthonormal. We can actually easily fix the problem by choosing $e_\phi = \frac{1}{r}\frac{\partial}{\partial \phi}$. But there is a little caveat. Up to now our covectors (the dual vectors of the our tangent vectors) were total differentials. That is no longer possible for the new choice of coordinates. The covector of $e_\phi = \frac{1}{r}\frac{\partial}{\partial \phi}$ is $r d\phi$ which cannot longer be represented by a total differential. Such bases are called anholonom. They are extremely practical for computations, but kind of unnatural. Nevertheless in the modern formalism of differential forms you find them everywhere.

Finally, if you apply a coordinate transformation, the components of the metric tensor $g(e_i,e_j)$ transform according to the rule

$g(e'_i,e'_j) = \frac{\partial x^k}{\partial x'^i} \frac{\partial x^l}{\partial x'^j} g(e_k,e_l)$. The summation is carried out over double appearing indices.

Transformation from polar (unprimed) coordinates to cartesian (primed) coordinates: First we know from our computations above (we'll use the holonom coordinates $(r,\phi)$): $g_{rr}=1$, $g_{r\phi}=0$, and $g_{\phi\phi}=r^2$. With this in mind we set up the transformation equations:

$g_{xx} = \frac{\partial r}{\partial x} \frac{\partial r}{\partial x} g_{rr} + 2 \frac{\partial r}{\partial x} \frac{\partial \phi}{\partial x} g_{r\phi} + \frac{\partial \phi}{\partial x} \frac{\partial \phi}{\partial x} g_{\phi\phi} = cos^2\phi g_{rr} + 0+ \frac{(-sin\phi)^2}{r^2} g_{\phi\phi} = cos^2\phi + sin^2\phi =1. $

Remember, the metric tensor is symmetric, so we put together the 2 mixed terms into one and also realise that as $g_{r\phi}= g_{\phi r} =0$, we can forget the mixed terms altogether.

$g_{xy} = \frac{\partial r}{\partial x} \frac{\partial r}{\partial y} g_{rr} + 2 \frac{\partial r}{\partial x} \frac{\partial \phi}{\partial y} g_{r\phi} + \frac{\partial \phi}{\partial x} \frac{\partial \phi}{\partial y} g_{\phi\phi} = cos\phi sin\phi g_{rr} + 0 + \frac{cos\phi}{r} \frac{-sin\phi}{r} g_{\phi\phi} = cos \phi sin\phi - cos\phi sin\phi =0. $

$g_{xx} = \frac{\partial r}{\partial y} \frac{\partial r}{\partial y} g_{rr} + 2 \frac{\partial r}{\partial y} \frac{\partial \phi}{\partial y} g_{r\phi} + \frac{\partial \phi}{\partial y} \frac{\partial \phi}{\partial y} g_{\phi\phi} = sin^2\phi g_{rr} + \frac{cos\phi}{r} \frac{cos\phi}{r} g_{\phi\phi} = sin^2\phi + cos^2\phi =1. $

We can confirm that the formula for the transformation of the metric tensor in case of the coordinate transformation from polar to cartesian coordinates works correctly.
Actually, one can also do this with anholonom coordinates, may be there is a slight change in the transformation law, but a priori it should also work out. I hope this helps, but may be it would be necessary to learn something more about differential forms to make this answer even clearer.

$\endgroup$
2
$\begingroup$

OK, so the confusion I see here is basically terminology.

In most thoughts, we have the metric tensor $g_{ab}$, which is a generalization of the normal dot product to a vector space. In order to have a consistent set of rules for vector and one-form transformations and to have raising and lowering to be invertible operations, it's necessary that, in components, the inner product of two one-forms be given by the inverse matrix of $g_{ab}$, which we call $g^{ab}$ by convention. When people say "inverse metric" they literally mean that it is the inverse matrix, so that $g_{ab}g^{bc} = \delta_{a}{}^{c}$ by definition.

OK, so now what is the deal with these tetrad vectors ${\bf e}^{a}$?, well, it's kind of best to think of them as a set of vectors that form an orthonormal basis of the vector space spanned by $g_{ab}$ Therefore, there must be four of them, which we can label with a lower index (I will use capital Latin letters). By construction, we have:

$${\bf e}_{I}^{a}{\bf e}_{J}^{b}g_{ab} = \eta_{IJ}$$

where $eta$ is the Minkowski metric. Multiply on the left by $\eta^{JK}$, and we get:

$$\eta^{JK}{\bf e}_{I}^{a}{\bf e}_{J}^{b}g_{ab} = \delta_{I}{}^{J}$$

from which the easiest thing to conclude is that

$$\eta^{JK}{\bf e}_{I}^{a}{\bf e}_{J}^{b} = \frac{1}{4}g^{ab}\delta_{I}{}^{K}$$

or, more simply

$$\eta^{IJ}g_{I}^{a}g_{J}^{b} = g^{ab}$$

which is the relationship you cite in the question written more explicitly. One way of thinking about this is to imagine the tetrad as a "square root" of the metric tensor

It turns out that we can completely reformulate all of General Relativity in terms of the tetrad (sometimes this is called by it's german name, the 'vierbein') vectors $\bf e$ without ever directly referencing the metric tensor at all, and in fact, this is the ONLY way we can incorporate spinors into general relativity.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.