Tensor Product, Basis Vectors and Tensor Components

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  • Thread starter nigelscott
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  • #1
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I am trying to figure how to get 1. from 2. and vice versa where the e's are bases for the vector space and θ's are bases for the dual vector space.

1. T = Tμνσρ(eμ ⊗ eν ⊗ θσ ⊗ θρ)

2. Tμνσρ = T(θμν,eσ,eρ)

My attempt is as follows:

2. into 1. gives T = T(θμν,eσ,eρ)(eμ ⊗ eν ⊗ θσ ⊗ θρ)

Now if I assume that (θμν,eσ,eρ) Ξ (θμ ⊗ θν ⊗ eσ ⊗ eρ) this becomes:

T = T(θμ ⊗ θν ⊗ eσ ⊗ eρ)(eμ ⊗ eν ⊗ eσ ⊗ θρ)

= θμeμ ⊗ θνeν ⊗ eσθσ ⊗ eρθρ

Now using θνeμ = δνμ this becomes:

T = T(I ⊗ I ⊗ I ⊗ I)

So T = T

This seems to work but I'm not sure if this is the correct way to do it. I'm shaky on the tensor product stuff and my interpretation of T(_,_,_,_). Does this look right?
 

Answers and Replies

  • #2
28
1
I am trying to figure how to get 1. from 2. and vice versa where the e's are bases for the vector space and θ's are bases for the dual vector space.

1. T = Tμνσρ(eμ ⊗ eν ⊗ θσ ⊗ θρ)

2. Tμνσρ = T(θμν,eσ,eρ)

My attempt is as follows:

2. into 1. gives T = T(θμν,eσ,eρ)(eμ ⊗ eν ⊗ θσ ⊗ θρ)

Now if I assume that (θμν,eσ,eρ) Ξ (θμ ⊗ θν ⊗ eσ ⊗ eρ) this becomes:

T = T(θμ ⊗ θν ⊗ eσ ⊗ eρ)(eμ ⊗ eν ⊗ eσ ⊗ θρ)

= θμeμ ⊗ θνeν ⊗ eσθσ ⊗ eρθρ

Now using θνeμ = δνμ this becomes:

T = T(I ⊗ I ⊗ I ⊗ I)

So T = T

This seems to work but I'm not sure if this is the correct way to do it. I'm shaky on the tensor product stuff and my interpretation of T(_,_,_,_). Does this look right?

It looks all right to me!

Let's try to intuitively understand what you did.

##T## is the generalization of a vector, in the sense that, ##T## is simply the sum of a bunch of components ##{T^{\mu\nu}}_{\sigma\rho}## multiplied by basis vectors ##e_{\mu} \otimes e_{\nu} \otimes \theta^{\sigma}\otimes \theta^{\rho}##. This is the interpretation of equation ##1## in your post.

Therefore, in order to get the component ##{T^{\mu\nu}}_{\sigma\rho}##, you would naively want to multiply the basis vector ##e_{\mu} \otimes e_{\nu} \otimes \theta^{\sigma}\otimes \theta^{\rho}## with itself. But then, you realize that ##T## exists not in flat space, but on a curved manifold. Therefore, you multiply the basis vector ##e_{\mu} \otimes e_{\nu} \otimes \theta^{\sigma}\otimes \theta^{\rho}## not with itself, but by its dual basis vector ##\theta^{\mu} \otimes \theta^{\nu} \otimes e_{\sigma} \otimes e_{\rho}##. That's exactly the interpretation of equation ##2## in your post.

Your check of the consistency of equations ##1## and ##2## is simply a mathematical way of rewriting my above two paragraphs.
 
  • #3
28
1
Let me know if my answer is clear, or if there's anything that you would need clarification with.
 

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