# Definition of metric tensor and line elements

I know that the metric tensor $$g_{ij}$$ is defined as:

$$g_{ij} = X_i \cdot X_j$$

where $$x_i$$ and $$x_j$$ are the covariant basis vectors ($$X_i = \frac{\partial X}{\partial X^i}$$) (the definition is from my book "Introduction to Tensor Analysis and the Calculus of Moving" by Pavel Grinfeld). Surfaces. Now, I know that the metric encodes all the information about the geometry of the space it specifies. My confusion is to relate the definition above to the line elements. Suppose for example that the line element of a 2d space is given by: $$dl^2 = 2(dx^1)^2 + 5(x^1) (dx^2)^2$$. Now from this definition I can see that $$g_{11} = 2$$ and $$g_{22} = 5(x^1)$$ (the other coefficients are zero). However, using the definition why is $$X_1 \cdot X_1 = 2$$?

• Where does Grinfeld write $X_i = \frac{\partial X}{\partial X^i}$? – Qmechanic Dec 17 '19 at 9:45
• The definition is taken from page 55 of the book, but perhaps I should've clarified that by $X$ is meant the position vector $\vec{R}(X)$ – daljit97 Dec 17 '19 at 14:51

The text book is a bit misleading as it uses the natural scalar product on $$\mathbb R^n$$ to define the metric but if we ignore this fact everything is good. Suppose we are talking about polar coordinates that is $$x= r \cos \theta$$ and $$y = r \sin \theta$$. Then the "covariant basis vectors" would be (choosing $$X_1 \leftrightarrow \hat r$$ and $$X_2 \leftrightarrow \hat \theta$$)

$$\hat r = \frac{\partial x}{ \partial r} \hat x + \frac{\partial y}{ \partial r} \hat y = \cos \theta \,\hat x + \sin \theta \,\hat y$$

and similarly

$$\hat \theta = \frac{\partial x}{ \partial \theta} \hat x+ \frac{\partial y}{ \partial \theta} \hat y = -r\sin \theta \,\hat x + r \cos \theta \,\hat y$$

The components of the metric tensor is then

$$g_{rr} = \hat r \cdot \hat r =1 \qquad g_{\theta\theta} = \hat \theta\cdot\hat \theta = r^2$$

and the other components $$g_{r\theta} = \hat r \cdot \hat \theta$$ are zero. So the total line element is then

$$\mathrm{d}l^2= \mathrm{d}r^2+ r^2 \mathrm{d}\theta^2$$

To get back to your example, first note that the metric you gave is not a good one because it has zero component for $$x_1=0$$ and even negative one for $$x_1<0$$. Suppose we chose the line element

$$\mathrm{d}l^2=2 \mathrm{d}X_1^2+ X_1^2 \mathrm{d}X_2^2$$

instead. You can clearly see the how the polar coordinate line element is related to this one. The book assumes that you know first an equation like the polar coordinate one $$x= r \cos \theta$$ and $$y = r \sin \theta$$ and then tells you how to compute the metric from this information, which is what I did above.

Now of course, line elements are much more general than that and a priori you can write down a line element without referring to any "covariant basis vectors" or stuff like that (hence my complaint in the beginning). I think this was your main confusion. You wrote down a metric and asked where are the "covariant basis vectors"? The upshot is you don't need them!

• What would be a more "fundamental" definition of the metric? – daljit97 Dec 11 '19 at 12:15
• I'm assuming you don't know much differential geometry. So you can think about a metric $g_{ij}$ as being a matrix with positive eigenvalues (i.e. positive definite). In practice the metric is usually diagonal so you can just say that the diagonal entries are positive. Remember the "purpose" of a metric is to define distances, you can do this just with a matrix ($g_{ij}$) without referring to any "fundamental vectors". – Gonenc Dec 11 '19 at 15:23

The metric tensor's components $$g_{\mu\nu}$$ is defined as the inner product $$(e_\mu,e_\nu)$$ where $$e_\mu$$ and $$e_\nu$$ are basis of some vector space $$V$$. The inner product defined for the vector space is linear. One can also show that $$\bf{g}=\it{g_{\mu\nu}e^\mu\otimes e^\nu}$$ linearly maps from $$V\times V$$ to $$\mathbb{R}$$. In differential geometry these basis vectors are defined as $$e_\mu=\partial_\mu\vec{R}\tag{1}$$ Where $$\vec{R}$$ is an arbitrary vector expressed in the coordinate system. Since a line element is defined as $$ds^2=(dx^\mu e_\mu, dx^\nu e_\nu)=\bf{g}\langle dx,dx \rangle$$ Using linearity $$ds^2=dx^\mu dx^\nu g_{\mu\nu}$$ Using $$(1)$$ $$ds^2=dx^\mu dx^\nu (\partial_\mu\vec{R},\partial_\nu\vec{R})$$ This is the relation between the line element and definition of metric in terms of the basis.

Why is $$(X_1,X_1)=2$$?

That is what comes when you take inner product of the basis. Let me take a well known example to demonstrate. Consider radial coordinate system $$(r,\theta)$$. We know that In Cartesian system $$x=r\cos(\theta)$$ and $$y=r\sin(\theta)$$. Let the vector $$\vec{R}=xe_x + ye_y$$. Solving for $$e_r$$ and $$e_\theta$$: $$e_r=\cos(\theta)e_x+\sin(\theta)e_y$$ $$e_\theta=-r\sin(\theta)e_x+r\cos(\theta)e_y$$ The metric components are $$g_{rr}=(e_r,e_r)=1$$ $$g_{r\theta}=g_{\theta r}=(e_r,e_\theta)=0$$ $$g_{\theta\theta}=r^2$$ Just like in polar coordinates $$(e_r,e_r)=1$$ a coordinate system can be defined such that it is $$2$$.

Hope this helps.

Note that in the notation I have used $$(,)$$ is the inner product and $$\langle,\rangle$$ is an ordered pair of vectors belonging to $$V\times V$$.

• This makes sense! As a curiosity, why is the line element defined as it is? – daljit97 Dec 11 '19 at 19:05
• @daljit97 It is a generalisation from the Euclidean line element. The line element must be invariant under any coordinate transformations. Applying an arbitrary coordinate transformation $x\rightarrow x'$ one can arrive at that definition. – Manvendra Somvanshi Dec 11 '19 at 22:20