# Is there any difference between Metric, Metric Tensor, Distance Func?

1. Apr 26, 2014

### devd

From what I've understood,
1) the metric is a bilinear form on a space
2) the metric tensor is basically the same thing

Is this correct?

If so, how is the metric related to/different from the distance function in that space?
Some other sources state that the metric is defined as the distance function. This is where i'm getting confused.

With reference to SR, should i consider the Space-Time interval as the metric, or the metric tensor denoted as g?

I don't know if i'm being able to communicate efficiently, though. :uhh:

2. Apr 26, 2014

### WannabeNewton

A metric is a distance function defined on a metric space. A metric tensor is a specific bilinear form defined on smooth manifolds. They are entirely different from one another although one can use a metric tensor to define a metric on Riemannian manifolds if the manifold is nice enough (e.g. connected); this Riemannian distance function so derived from the metric tensor is rarely if ever used in GR. Rather, in GR, the term "metric" is used interchangeably with the term "metric tensor" which is just a very unfortunate abuse of terminology.

To summarize, in GR the term "metric" means metric tensor whereas in mathematics the term "metric" refers to a distance function on a metric space.

3. Apr 26, 2014

### devd

Ok, that seems to have cleared up my confusion. Thanks! In SR, then, what one means by 'metric' is basically a metric tensor as opposed to the space-time interval, right?

The space-time interval would be some sort of a metric in the mathematical sense?

4. Apr 26, 2014

### pervect

Staff Emeritus
In SR, the metric tensor gives the space-time interval. So this would be backwards. However, points that are light-like separated (a Lorentz interval) are not necessarily "neighbors". So one might define a distance function on $\mathbb{R}^4$ as $x_0^2 + x_1^2 + x_2^2 + x_3^2$ (note: no minus sign, so it's not the Lorentz interval) to define open balls as per http://mathworld.wolfram.com/OpenBall.html or "neighborhoods"http://mathworld.wolfram.com/Neighborhood.html.

5. Apr 27, 2014

### DrGreg

The metric tensor is a type of tensor, in this case a bilinear form. As such, it can be represented, relative to a given coordinate system, as a 4×4 matrix, e.g.$$g=\begin{bmatrix} c^2 && 0 && 0 && 0 \\ 0 && -1 && 0 && 0 \\ 0 && 0 && -1 && 0 \\ 0 && 0 && 0 && -1 \end{bmatrix}$$Within the context of relativity, the name "metric tensor" is often shortened to just "metric".

The "spacetime interval" is the name given to the quantity usually denoted as "ds", e.g.$$ds^2 = c^2dt^2 - dx^2 - dy^2 - dz^2$$However, the two equations above are related to each other; if you know one, then you can write down the other. So sometimes the second equation is referred to as just "the metric", even though, technically, the two equations aren't quite the same thing.

As others above have pointed out, in mathematics there is also something called a "metric space" which has a distance function called "the metric", but that's not relevant in relativity (except in some technicalities of setting up a rigorous definition of a "smooth manifold").

6. Apr 27, 2014

### HallsofIvy

To convert from the metric tensor,
$$\begin{bmatrix}c^2 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1\end{bmatrix}$$
to the space-time interval, $ds^2= c^2dt^2- dx^2- dy^2- dz^2$, you do a "double multiplication":
$$\begin{bmatrix} dt & dx & dy & dz\end{bmatrix}\begin{bmatrix}c^2 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1\end{bmatrix}\begin{bmatrix}dt \\ dx \\ dy \\ dz\end{bmatrix}$$

7. Aug 31, 2014

### victorneto

Difference between the metric and the metric tensor.

1) Metric: is a quadratic relationship involving the squares of the differences between the coordinates of two nearby points in space-time, which allows calculating the geometrical properties of spacetime, as distances (or intervals between events) and angles between neighboring points.

2) metric tensor: the set of coefficients of the squares of the differences of the same type coordinates.

To put things in a simple and intuitive perspective, let's look at some examples:

Exemple 1

1) Measure of a two-dimensional flat space in Cartesian coordinates:
ds^2 = dx^2 + dy^2 (Pythagorean theorem).

Coefficients of the squares above terms: 1.1.

Including the cross terms in the metric above, 0.dx.dy dy.dx and 0, we have:
ds^2 = 1.dx^2 + 0.dxdy + 0.dydx +1. dy^2

The set of coefficients is now: 1,0,0, 1

Arranging these numbers in a matrix, we have:

[ (1 & 0 @ 0 & 1)]

This matrix, called the coefficients of the metric matrix, which sets up the call:

Metric Tensor.

Where the non-zero coefficients (dx^2 = dx.dx terms), are in the main diagonal and zero coefficients (cross terms: dxdy, dydx) turn off the main diagonal.

Example 2.

2) Robertson-Walker metric:

ds^2 = c^2 dt^2 - a (t)^2 [dr^2 / (1-kτ^2) + r^2 (+ sin^2 dΘ^2 dφ^2 ), or

ds^2 = c^2 .dt^2 - [a (t)^2 / (1-kτ^2)] dr^2 + [a (t) ^2 r^2] dΘ^2 + [a (t)^2sin^2 Θ] dφ^2

metric coefficients are:

c 2 ,
[a (t)^2 / (1-kτ^2)],
[a (t)^2 r^2] and
[a (t)^2 sin^2 Θ

Following the procedures of Example 1, the metric tensor associated with the metric above is (after completing the cross-terms), we have the matrix that sets the metric tensor of the metric in question, whose elements are:

a)in the main diagonal: c2, [a (t) 2 / (1-kτ2)], [a (t) 2r2] and [a (t) 2sin2 Θ]
b)off-diagonal: all elements are null:

And so on.

Victor

8. Aug 31, 2014

### victorneto

Dears,

Excuse me, how to use Latex in reply?

Sds,
VictorNeto

9. Sep 1, 2014

### DrGreg

10. Sep 1, 2014

### victorneto

Thanks, friend.

11. Sep 2, 2014

### victorneto

Thanks, DrGreg.

Now, the correct text.

Difference between the metric and the metric tensor.

1) Metric: is a quadratic relationship involving the squares of the differences between the coordinates of two nearby points in space-time, which allows calculating the geometrical properties of spacetime, as distances (or intervals between events) and angles between neighboring points.

2) metric tensor: the set of coefficients of the squares of the differences of the same type coordinates.

To put things in a simple and intuitive perspective, let's look at some examples:

Exemple 1

1) Measure of a two-dimensional flat space in Cartesian coordinates:
ds^2 = dx2 + dy2 (Pythagorean theorem).

Coefficients of the squares above terms: 1.1.

Including the cross terms in the metric above, 0.dx.dy and 0.dy.dx, we have:
ds2 = 1.dx2 + 0.dxdy + 0.dydx +1. dy2

The set of coefficients is now: 1,0,0, 1

Arranging these numbers in a matrix, we have:

$$\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$$

This matrix, called the coefficients of the metric matrix, which sets up the call:

Metric Tensor.

Where the non-zero coefficients (dx^2 = dx.dx terms), are in the main diagonal and zero coefficients (cross terms: dxdy, dydx) turn off the main diagonal.

Example 2.

2) Robertson-Walker metric:

ds2 = c2 dt2 - a (t)2 [dr2 / (1-kτ2) + r2 (+ sin222 ), or

ds2 = c2 .dt2 - [a (t)2 / (1-kτ2)] dr2 + [a (t)2 r2] dΘ2 + [a (t)2sin2 Θ] dφ2

metric coefficients are:

c2 ,
[a (t)2 / (1-kτ2)],
[a (t)2 r2] and
[a (t)2 sin2 Θ

Following the procedures of Example 1, the metric tensor associated with the metric above is (after completing the cross-terms), we have the matrix that sets the metric tensor of the metric in question, whose elements are:

a)in the main diagonal: c2, [a (t)2 / (1-kτ2)], [a (t) 2r2] and [a (t)2sin2 Θ]
b)off-diagonal: all elements are null:

And so on.

12. Sep 2, 2014

### ChrisVer

So they call $ds^2$ a metric? In my Dif.Geom. class we called it 1st fundamental form.