Simple metric tensor question

  • #1
JTC
100
6
Good Day,

Another fundamentally simple question...

if I go here;

http://www-hep.physics.uiowa.edu/~vincent/courses/29273/metric.pdf

I see how to calculate the metric tensor. The process is totally clear to me.

My question involves LANGUAGE and the ORIGIN

LANGUAGE: Does one say "one calculates the metric tensor that relates the Cartesian coordinate system to the spherical coordinate system?" How should I say that?

ORIGIN: So I can calculate the metric tensor (and, again, please correct my language), that relates the Cartesian and Cylindrical, the Cylindrical and Spherical, etc...

But how does one get the original metric tensor for the Cartesian (the identity matrix)? What doe that relate?
It seems to me I need two coordinate systems to do this process.

As I begin to calculate derivatives (see the fourth equation on that PDF above), what am I taking the derivative of and with respect to what? What gets me the metric tensor for the Cartesian that "starts the ball rolling?" (so to speak)

And while you are at it: Go to Frankel "GEOMETRY OF PHYSICS" and go to the bottom of page xxxiv.
What is he doing? I see x1, u1, x and theta. I see no organization to this. Could someone elaborate what derivatives he is taking?
 

Answers and Replies

  • #2
14,170
8,149
Doesn't the cartesian metric tensor come from simply expressing the pythagorean theorem in tensor form?
 
  • #3
WWGD
Science Advisor
Gold Member
6,291
8,135
I think you can say that you calculate the metric tensor _in_ so-and-so coordinates. Basically , non-identity coefficients in the metric tensor show the existence of curvature in the space. Maybe if you give us a link from, e.g., Google books, we can read Frankel's book, otherwise I, at least, don't have access to the book. In the treatment I am familiar with , one uses a parametrization ##u=f(x_1,x_2,..,x_n) ## and then just applies the exterior differential to each x_i and subs in into ## dx^2 =dx_1^2+...+dx_n^2 ##.
 
  • #4
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
20,004
10,647
I think you can say that you calculate the metric tensor _in_ so-and-so coordinates.
What you compute are usually the components of the metric tensor in some given coordinates.

Basically , non-identity coefficients in the metric tensor show the existence of curvature in the space.
Do not confuse the metric tensor with the curvature tensor. I can introduce a curvilinear and non-orthogonal coordinate system in a flat space and obtain a metric tensor that looks rather strange. What tells you if a space is curved or not is the curvature tensor. If you adopt the Levi-Civita connection, the curvature tensor will be directly given by how the metric tensor depends on the coordinates (the Christoffel symbols, and hence the components of the curvature tensor, contain derivatives of the metric components).

In the treatment I am familiar with , one uses a parametrization u=f(x1,x2,..,xn)u=f(x1,x2,..,xn)u=f(x_1,x_2,..,x_n) and then just applies the exterior differential to each x_i and subs in into dx2=dx21+...+dx2ndx2=dx12+...+dxn2 dx^2 =dx_1^2+...+dx_n^2 .
Yes, the easiest way of finding the components of the metric tensor is to look at how the line element depends on the coordinates. One of the easier examples being the line element in polar coordinates in the two-dimensional Euclidean plane such that ##x = r \cos(\theta)## and ##y = r \sin(\theta)##, which leads to
$$
dx = \cos(\theta) dr - r \sin(\theta) d\theta, \quad dy = \sin(\theta) dr + r \cos(\theta) d\theta \quad \Longrightarrow \quad
ds^2 = dx^2 + dy^2 = dr^2 + r^2 d\theta^2 = g_{rr} dr^2 + 2 g_{r\theta} dr\, d\theta + g_{\theta\theta} d\theta^2
$$
from which you can immediately identify ##g_{rr} = 1##, ##g_{r\theta} = g_{\theta r}= 0##, ##g_{\theta\theta} = r^2##. Note that this metric tensor is not proportional to identity, but that it still describes the Euclidean plane, which is flat.
 
  • #5
WWGD
Science Advisor
Gold Member
6,291
8,135
What you compute are usually the components of the metric tensor in some given coordinates.


Do not confuse the metric tensor with the curvature tensor. I can introduce a curvilinear and non-orthogonal coordinate system in a flat space and obtain a metric tensor that looks rather strange. What tells you if a space is curved or not is the curvature tensor. If you adopt the Levi-Civita connection, the curvature tensor will be directly given by how the metric tensor depends on the coordinates (the Christoffel symbols, and hence the components of the curvature tensor, contain derivatives of the metric components).


Yes, the easiest way of finding the components of the metric tensor is to look at how the line element depends on the coordinates. One of the easier examples being the line element in polar coordinates in the two-dimensional Euclidean plane such that ##x = r \cos(\theta)## and ##y = r \sin(\theta)##, which leads to
$$
dx = \cos(\theta) dr - r \sin(\theta) d\theta, \quad dy = \sin(\theta) dr + r \cos(\theta) d\theta \quad \Longrightarrow \quad
ds^2 = dx^2 + dy^2 = dr^2 + r^2 d\theta^2 = g_{rr} dr^2 + 2 g_{r\theta} dr\, d\theta + g_{\theta\theta} d\theta^2
$$
from which you can immediately identify ##g_{rr} = 1##, ##g_{r\theta} = g_{\theta r}= 0##, ##g_{\theta\theta} = r^2##. Note that this metric tensor is not proportional to identity, but that it still describes the Euclidean plane, which is flat.
Yes, my bad, I was kind of loose with terms, maybe just loose-enough to be wrong; it has been a while, thanks for the refresher.
 
  • #6
JTC
100
6
I think you can say that you calculate the metric tensor _in_ so-and-so coordinates. Basically , non-identity coefficients in the metric tensor show the existence of curvature in the space. Maybe if you give us a link from, e.g., Google books, we can read Frankel's book, otherwise I, at least, don't have access to the book. In the treatment I am familiar with , one uses a parametrization ##u=f(x_1,x_2,..,x_n) ## and then just applies the exterior differential to each x_i and subs in into ## dx^2 =dx_1^2+...+dx_n^2 ##.

Here is the excerpt from Frankel. I just do not understand what he is doing in this section I talk about. The designated symbols make no sense (bottom of the page and top of the next) HERE it is known as page 6 and 7

http://www.math.ucsd.edu/~tfrankel/the_geometry_of_physics.pdf

I understand how he gets equation (7) (for polar coordinates). But how does he get equation just above it? How does he get the identity matrix? I mean, I expect it to be that! But using his equation, how does he get it? What is he taking derivatives of? What are the subscripts?

I mean: to be honest, I do understand it all. I am just being really really picky. He gave an equation to calculate gij and I expect him to use it to get gij for Cartesian.
 
Last edited:
  • #7
WWGD
Science Advisor
Gold Member
6,291
8,135
I guess Bitlocker is not 3rd-party, right? Sorry, could not find a definitive answer.
 
  • #8
strangerep
Science Advisor
3,486
1,766
I understand how he gets equation (7) (for polar coordinates). But how does he get equation just above it?
? It's just 2D Cartesian coordinates. ##ds^2=dx^2+dy^2##.
 
  • #9
JTC
100
6
? It's just 2D Cartesian coordinates. ##ds^2=dx^2+dy^2##.

Yes, I see that.

But I also see a definition to get the metric tensor by taking derivatives.

But I cannot see what I take derivatives OF to get the metric tensor

I can see that if I want the polar, I take derivatives of the x with respect to theta or r. So then it seems to me that the gij for POLAR is DERIVED from the coordinates for Cartesian.

I think I am thinking myself into a box.
 
  • #10
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
20,004
10,647
I can see that if I want the polar, I take derivatives of the x with respect to theta or r. So then it seems to me that the gij for POLAR is DERIVED from the coordinates for Cartesian.
This presumes that you have an underlying Cartesian metric on a Euclidean space. If you do, you already know the inner product of the Cartesian basis vectors and you can define the vector basis ##\vec E_a = \partial \vec x/\partial x^a## and compute the components of the metric tensor according to
$$
g_{ab} = \vec E_a \cdot \vec E_b = \frac{\partial \vec x}{\partial x^a} \cdot \frac{\partial \vec x}{\partial x^b}.
$$
 
  • #11
JTC
100
6
AH HA!
BINGO.
That was what I was hoping to hear. Now I get it!

Never in all my learning has ONE SENTENCE explained it.

>>This presumes that you have an underlying Cartesian metric on a Euclidean space.


No one ever says that!

I am sorry, but waxing silly.. .but... O.M.G: Thank you.

That one sentence explained it all.
 

Suggested for: Simple metric tensor question

  • Last Post
Replies
11
Views
5K
  • Last Post
Replies
12
Views
4K
Replies
7
Views
59
Replies
3
Views
4K
Replies
2
Views
671
Replies
37
Views
6K
Replies
2
Views
364
  • Last Post
Replies
9
Views
1K
Replies
36
Views
2K
Top