# Some help understanding the metric tensor

#### mysearch

Gold Member
Hi,
I am in the process of trying to teach myself GR Maths, at the A101 level, and have been working through the idea of tensors as scalars, vectors and matrices, i.e. rank-0, 1 and 2 tensors. Think I have also acquired some idea of the concept of contravariance and covariance, which then seems to define the superscripting and subscripting on the tensor. Have also worked through the basic coordinate transforms associated with these concepts and have now arrived at the idea of a metric tensor, which seems pretty important in terms of GR, so wanted to try to clarify some issues before plodding on.

[1] $$ds^2 = g_{\alpha \beta} da^\alpha db^\beta$$

I am using (da) and (db), not the usual (dx), to identify what I am assuming to be differential displacement vectors because I want to used (x), in the description below, to identify the unit vectors, which I believe in this context are called the basis vectors. As such, I am also assuming that (da) and (db) can be related to (a) and (b) in [2] below, which might be a mistake. Anyway, [1] above appears, to me, to be a generic description of any metric, e.g. 2D flat space through to 4D curved spacetime. For fairly obvious reasons I would like to start by trying to interpret this equation in terms of the simplest form, i.e. normal 2D flat space, as described in terms of Pythagoras’ theorem and Cartesian coordinates:

[2] $$s^2 = a^2 + b^2$$

As indicated, the superscripted terms in [1] have been assumed to be contravariant displacement vectors [da, db] or rank-1 tensors, while (g) is an n-dimensional array or rank-2 tensor. In the 2D example associated with [2], (g) is a 2D array:

[3] $$g= \left(\begin{array}{cc}1&0\\0&1\end{array}\right)$$

I will defer my questions about the nature of (g) to another time, as this post will probably end up being too detailed and too long to solicit any response. Anyway, if I expand [1] in terms of the multiplication of 2 contravariant 2D vectors, where the basis vectors are [x1, x2], I appear to end up with:

[4] $$a^\alpha b^\beta = a^1 b^1 x^1 x^1 + a^1 b^2 x^1 x^2 + a^2 b^1 x^2 x^1 + a^2 b^2 x^2 x^2$$

Now it would seem that [4] can be simplified by multiplying through by [3], which seems correct for the 2D flat space metric in [2]:

[5] $$s^2 = g_{\alpha \beta} a^\alpha b^\beta = a^1 b^1 + a^2 b^2$$

At this point, I wanted to try to anchor the notation in [5] to an actual geometric example, which would be a specific solution of [2], i.e. [a=3] and [b=4] giving [s=5]. My assumption was that the magnitude of (a1) is defined with respect to the basis axis (x1). Therefore, given that (a,b) in [2] actually align to the basis vector (x1, x2), I assumed the following values: (a1=3, a2=0, b1=0, b2=4). However, these values do not appear to prove any obvious equivalence between [2] and [5], therefore I must assume that I have misunderstood some fundamental concept and would appreciate any help on offer. Thanks

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#### Mentz114

Gold Member
$$ds^2 = g_{\alpha \beta} da^\alpha db^\beta$$
This does not make sense to me. If you are using a chart with coordinate names say (t,x,y,z) then conventionally dx0 is dt, dx1 is dx, dx2 is dy and so on.

The metric is used to raise and lower tensor indexes, so vector $t_\mu=g_{\rho\mu}t^\rho$, for instance.

The displacement equation for 2D Euclidean space is

$$ds^2 = g_{\alpha \beta} dx^\alpha dx^\beta$$

where $\alpha$ and $\beta$ range over 1,2 say ( or 0,1) and g is given by your equation (3).

So if we have two points, (2,3) and (6,4), the distance between them is

$$ds^2= g_{11}dx^1dx^1 + g_{22}dx^2dx^2$$

which evaluates to
$$(2-6)^2+(3-4)^2$$

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#### mysearch

Gold Member
This does not make sense to me. If you are using a chart with coordinate names say (t,x,y,z) then conventionally dx0 is dt, dx1 is dx, dx2 is dy and so on. The metric is used to raise and lower tensor indexes, so vector $t_\mu=g_{\rho\mu}t^\rho$, for instance. The displacement equation for 2D Euclidean space is

(A) $$ds^2 = g_{\alpha \beta} dx^\alpha dx^\beta$$

where $\alpha$ and $\beta$ range over 1,2 say ( or 0,1) and g is given by your equation (3). So if we have two points, (2,3) and (6,4), the distance between them is

(B) $$ds^2= g_{11}dx^1dx^1 + g_{22}dx^2dx^2$$

which evaluates to

(C) $$(2-6)^2+(3-4)^2$$
Hi,
I have inserted labels (A), (B) and (C) into the quote above purely for reference and please bear in mind that much of this does make sense to me either, which is why I am asking questions to rectify the misunderstanding. So, clearly, there is an implication that my replacement of [dx] in [A] with [da] and [db] is nonsensical. This may well be the case, I had simply assumed that [dx] was representative of a differential displacement vector that could be renamed to better align its application to 2D Euclidean space:

[1] $$s^2 = a^2 + b^2$$

Therefore, could you or anybody elaborate a little further on the scope/interpretation of dx, which would then explain the difference in implication such that [2] below is inappropriate and not representative of (A):

[2] $$ds^2 = g_{\alpha \beta} da^\alpha db^\beta$$

While I understand the inference of (C) being a solution of the Pythagoras form in [1] above, it is unclear as to how you substituted your values in (C) into (B), i.e. is there a specific expansion of $$dx^\alpha dx^\beta$$ along the lines:

$$a^\alpha b^\beta = a^1 b^1 x^1 x^1 + a^1 b^2 x^1 x^2 + a^2 b^1 x^2 x^1 + a^2 b^2 x^2 x^2$$

Anyway, appreciated the helpful insights. Thanks

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#### Mentz114

Gold Member
...it is unclear as to how you substituted your values in (C) into (B),
dx1=2-6
dx2=3-4

g11=1
g22=1

$$ds^2= g_{11}dx^1dx^1 + g_{22}dx^2dx^2=1 (2-6)(2-6)+ 1 (3-4)(3-4)$$

This,

$$a^\alpha b^\beta = a^1 b^1 x^1 x^1 + a^1 b^2 x^1 x^2 + a^2 b^1 x^2 x^1 + a^2 b^2 x^2 x^2$$

The term on the left looks like the product of 2 contravariant vectors. On the right is something else which I can't identify.

#### mysearch

Gold Member
$$ds^2= g_{11}dx^1dx^1 + g_{22}dx^2dx^2=1 (2-6)(2-6)+ 1 (3-4)(3-4)$$

$$a^\alpha b^\beta = a^1 b^1 x^1 x^1 + a^1 b^2 x^1 x^2 + a^2 b^1 x^2 x^1 + a^2 b^2 x^2 x^2$$

The term on the left looks like the product of 2 contravariant vectors. On the right is something else which I can't identify.
Again, much appreciate the help. I think one of the things I lost sight of is that the following equation represents a summation (?), i.e.

$$ds^2 = g_{\alpha \beta} dx^\alpha dx^\beta$$

$$(ds)^2 = (dx)^2 + (dy)^2= \sum (dx^i)^2$$ where i=1..2

This then seem to explain your equation:

$$ds^2= g_{11}dx^1dx^1 + g_{22}dx^2dx^2=1 (2-6)(2-6)+ 1 (3-4)(3-4)$$

The second equation in the quote above, where you didn’t understand the term on the right was taken from an articlehttp://www.tommangan.us/Tensors.pdf" [Broken] on page 8, which seemed to imply it was the multiplication of 2 rank-1 tensors. Obviously, I will need to read this more carefully as it wasn't 'gentle' enough. Thanks again.

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#### Mentz114

Gold Member
Yes, there is implied summation.

The thing on the left is a product of 2 rank-1 tensors ( i.e. vectors), and the result will be a rank-2 tensor, whose components are just the products of the components of the two multiplicands.

Keep going, you are moving forward.

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