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What different between covariant metric tensor and contravariant metric tensor

 
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Aug3-06, 12:23 AM   #1
 

What different between covariant metric tensor and contravariant metric tensor


I read some books and see that the definition of covariant tensor and contravariant tensor.
Covariant tensor(rank 2)
A'_ab=(&x_u/&x'_a)(&x_v/&x'_b)A_uv
Where A_uv=(&x_u/&x_p)(&x_u/&x_p)
Where p is a scalar
Contravariant tensor(rank 2)
A'^uv=(&x'^u/&x^a)(&x'^v/&x^b)A^ab
Where A^ab=dx_a dx_b

Metric tensor sometimes g^uv,sometimes g_uv
ds^2=g_uv dx^u dx^v=g^uv dx_u dx_v
What different between them?
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Aug21-06, 02:36 PM   #2
 
Essentially, there is no difference between the covariant and contravariant forms of the metric in the sense that they both "measure" things. However, consider the following. If you have a metric [tex]g[/tex] on a manifold then it is usually regarded as being a map which takes two vectors into a real number. For example, you can calculate the squared length of a vector [tex]X[/tex] as being

[tex]g(X,X) = g_{ab}X^aX^b[/tex].

However, vectors aren't the only things which you may want to measure the length of. Another interesting class of objects are "one-forms." If you want to measure the "squared length" of a one-form [tex]\alpha[/tex] then you can do it thusly:

[tex]g(\alpha,\alpha) = g^{ab}\alpha_a\alpha_b[/tex]

where [tex]g^{ab}[/tex] is the covariant form of the metric. The covariant form of a metric can always be obtained from the contravariant form by virtue of something called a "musical isomorphism" (it's a technical point that you really don't need to worry about). The only restriction on the relationship between the covariant and contravariant forms of the metric are that they should satisfy the following:

[tex]g_{ab}g^{bc} = \delta^{a}_{c}.[/tex]
Aug23-11, 01:21 AM   #3
 
Quote by HeilPhysicsPhysics View Post
I read some books and see that the definition of covariant tensor and contravariant tensor.
Covariant tensor(rank 2)
A'_ab=(&x_u/&x'_a)(&x_v/&x'_b)A_uv
Where A_uv=(&x_u/&x_p)(&x_u/&x_p)
Where p is a scalar
Contravariant tensor(rank 2)
A'^uv=(&x'^u/&x^a)(&x'^v/&x^b)A^ab
Where A^ab=dx_a dx_b

Metric tensor sometimes g^uv,sometimes g_uv
ds^2=g_uv dx^u dx^v=g^uv dx_u dx_v
What different between them?
Friend,
I offer a simple explanation that clarifies in definitive the essential difference enters the meanings of covariante and contravariant. It considers a vector V in the space. It represents this vector in a system of oblique ortogonais axles.
1) It now calculates the components of vector V according to oblique axles (that is, it decomposes vector V throughout the oblique axles);
2) In the origin of the system of oblique axles it considers another system of ortogonais axles. Now, it decomposes vector V throughout the axles of this ortogonal system the same.
Then you have the same vector V represented in oblique components (system S) and in ortogonais components. (The vector the same continues being. But the ways to represent it are different).

Now:

1) The components of vector V in the ortogonal system are called covariantes components; 2) The components of the vector in the oblique system are called contravariant components!

An so on.

VictorNeto
Aug17-12, 02:20 PM   #4
 

What different between covariant metric tensor and contravariant metric tensor


Huh?
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