Vector fields, flows and tensor fields

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Vector fields generate flows that are crucial in various physical contexts, such as fluid dynamics and electromagnetism, while the generalization to tensor fields, particularly rank-2 tensors, raises questions about their ability to produce similar flows. The discussion highlights that metric tensors do not generate isometries directly, as isometries are defined as diffeomorphisms that preserve the metric. The stress-energy tensor can be interpreted as influencing the curvature of spacetime, but its relationship to flow is complex and context-dependent, particularly in general relativity. The Frobenius theorem is referenced to explore the potential for defining flows in terms of fields of planes rather than vectors, suggesting a more nuanced approach to understanding tensor fields. Ultimately, the notion of flow remains primarily one-dimensional, complicating the direct application to higher-rank tensors.
  • #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?
 
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  • #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...
 

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