Rank 2 covariant tensors and dimesionality

In summary, the conversation is about finding the value of alpha in a rank-2 covariant tensor in d-dimensions. The tensor can be written in the form of t sub i,j + alpha*metric tensor*T super k, sub k and it is also known that t super i, sub i = 0. There is a suggestion to multiply the whole thing by the inverse metric tensor and take the trace to find the value of alpha. The discussion also mentions a relation to the dimensional coefficients of the Ricci tensors and curvature scalars in the Weyl tensor.
  • #1
SIlasX
2
0
I've already handed in my (I can only assume) incorrect solution, but I just felt like posting, though I'm not sure if anyone will be able to help.

I have a rank-2 covariant tensor, T sub i,j. This can be written in the form of t sub i,j + alpha*metric tensor*T super k, sub k (I hope my notation makes sense). I am also told that t super i, sub i = 0. I need to find alpha in d-dimensions.

I was thinking that I multiply the whole thing by the inverse metric tensor. That would give me a scalar on the right hand side of the equation and, I believe, would raise one index on t sub i,j so that I could use the given. I'm lost after that.

If it helps, it seems that this is somehow related to the dimensional coefficients of the Ricci tensors and curvature scalars in the Weyl tensor.
 
Physics news on Phys.org
  • #2
Is that T arbitrary?
[tex] T_{ij}=t_{ij}+\alpha g_{ij}T^{k} \ _{k} [/tex] (1)

That means u decomposed the initial tensor into a sum of a tracless tensor (t) and the the trace of the initial tensor (surely,metaphorically speaking,the assertion needs to be understood in terms of space dimentionality).
Take trace of the (1)...You'll find alpha immediately.

Daniel.
 
  • #3
Thanks! :shy:
 

1. What is a rank 2 covariant tensor?

A rank 2 covariant tensor is a mathematical object that describes the transformation of a vector from one coordinate system to another. It is represented by a 2-dimensional matrix and is used in fields such as physics and engineering to describe the relationship between physical quantities.

2. How is the dimensionality of a covariant tensor determined?

The dimensionality of a covariant tensor is determined by the number of indices it has. A rank 2 covariant tensor has two indices, and the dimensionality is determined by the number of possible combinations of these indices. For example, a rank 2 covariant tensor in 3-dimensional space would have a dimensionality of 3x3=9.

3. What is the significance of rank 2 covariant tensors in physics?

Rank 2 covariant tensors are used in physics to describe the relationship between physical quantities in different coordinate systems. They are important in fields such as general relativity and electromagnetism, where they are used to describe the curvature of spacetime and the electromagnetic field, respectively.

4. How do you manipulate rank 2 covariant tensors?

Rank 2 covariant tensors can be manipulated using matrix multiplication rules. This involves multiplying the elements of the tensor by the corresponding elements of another tensor and then summing the results. It is important to keep track of the indices and make sure they are in the correct order.

5. Can rank 2 covariant tensors be used to describe rotations?

Yes, rank 2 covariant tensors can be used to describe rotations in 3-dimensional space. However, they are not the most efficient way to do so, as they involve a lot of calculation. Other mathematical objects such as matrices and quaternions are better suited for describing rotations.

Similar threads

A Problems with the interpretation of the Torsion tensor and the Lie Bracket
    • Differential Geometry
Replies
16
Views
3K
I About Covariant Derivative as a tensor
    • Differential Geometry
Replies
6
Views
2K
B Torsion tensor definition doubt
    • Differential Geometry
Replies
3
Views
1K
Showing that the gradient of a scalar field is a covariant vector
    • Advanced Physics Homework Help
Replies
5
Views
2K
A Question about covariant derivatives
    • Differential Geometry
Replies
7
Views
3K
A On the dependence of the curvature tensor on the metric
    • Differential Geometry
Replies
6
Views
2K
I Purpose of Tensors, Indices in Tensor Calculus Explained
    • Special and General Relativity
Replies
10
Views
2K
B Attempted proof of the Contracted Bianchi Identity
    • Differential Geometry
Replies
1
Views
1K
I A couple questions about the Riemann Tensor, definition and convention
    • Differential Geometry
Replies
2
Views
2K
I Reconciling 2 expressions for Riemann curvature tensor
    • Differential Geometry
Replies
34
Views
2K

Hot Threads

Recent Insights

Back
Top