Vector fields, flows and tensor fields

Click For Summary
SUMMARY

Vector fields generate flows, which are one-parameter groups of diffeomorphisms, extensively utilized in physics, such as in fluid dynamics and electromagnetism. The discussion explores the generalization of this concept to rank-2 tensor fields, particularly focusing on the metric tensor and the stress-energy tensor. It concludes that while the stress-energy tensor can suggest a flow-like behavior in terms of space-time curvature, it does not conform to the traditional definition of flow due to the absence of directional arrows in tensor fields. The Frobenius theorem is highlighted as a critical concept for understanding the relationship between vector fields and tensor fields.

PREREQUISITES
  • Understanding of vector fields and their role in generating flows.
  • Familiarity with rank-2 tensor fields, particularly metric and stress-energy tensors.
  • Knowledge of diffeomorphisms and isometries in differential geometry.
  • Basic comprehension of the Frobenius theorem and its implications in geometry.
NEXT STEPS
  • Study the implications of the Frobenius theorem in differential topology.
  • Explore the role of the stress-energy tensor in general relativity and its relationship to space-time curvature.
  • Investigate the concept of integral curves and surfaces in the context of tensor fields.
  • Learn about antisymmetric tensors and their applications in defining flows in physics.
USEFUL FOR

Mathematicians, physicists, and students of differential geometry interested in the interplay between vector fields and tensor fields, particularly in the context of general relativity and fluid dynamics.

  • #31
TrickyDicky said:
Right, I knew this proof, that is why I said that I agreed a general order two tensor is not necessarily the tensor product of two vectors, a general second order tensor can be constructed with the sum of ##n^2## tensor products.

But I think you might have misunderstood what this proof implies from what Schutz write in the Appendix. You seemed to imply that the tensor product of two vectors has at most 2n independent components, and therefore by counting the independent components of a tensor one might deduce if it is the result of a tensor product or not. This is not correct. Just do the calculation, multiply two vectors using the outer product, you get a second order tensor with ##n^2## independent components. Granted you have only used 2n components to build it and that is why a general tensor is not necessarily the outer product of two vectors, but still after the product the ##n^2## components formed from the 2n are considered to be independent.

This doesn't make sense to me...If I can use the counting to prove that a general rank 2 tensor cannot be expressed as a dyad, why can't I use this counting to prove that a specific rank 2 tensor cannot be expressed as a dyad?

Can you come up with an example where my reasoning fails? A rank two tensor with more than 8 independent components that can be expressed as the direct product of 2 vectors alone?
 
Physics news on Phys.org
  • #32
Matterwave said:
This doesn't make sense to me...If I can use the counting to prove that a general rank 2 tensor cannot be expressed as a dyad, why can't I use this counting to prove that a specific rank 2 tensor cannot be expressed as a dyad?
You can. What you can't is take an arbitrary rank 2 tensor and decide whether it is a dyad or a sum of dyads just based on the number of independent components of the tensor, this is using the proof backwards.
Can you come up with an example where my reasoning fails? A rank two tensor with more than 8 independent components that can be expressed as the direct product of 2 vectors alone?
The best way is to compute it as I said, and realize that the fact that you use only 8 initial components to construct 16 components doesn't mean that just 8 of the final 16 are independent, if this was so you should be able to say which 8 are the independent ones, but you can't.
 
  • #33
I guess it is true the whole thing doesn't make much sense with tensor fields, certainly not at the level of smooth manifolds where the usual flows for vector fields are defined. Perhaps after introducing a metric and curvature it could make more sense for certain geometries' isometry groups.
 
  • #34
I don't see any way to make it work...
 

Similar threads

  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
5K
Replies
3
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 7 ·
Replies
7
Views
869
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 8 ·
Replies
8
Views
3K