Decomposing Rank-2 Tensors in Dirac's "General Theory of Relativity

  • Context: Graduate 
  • Thread starter Thread starter Kostik
  • Start date Start date
  • Tags Tags
    Tensors
Click For Summary
SUMMARY

Dirac's "General Theory of Relativity" establishes that a general rank-2 tensor can be decomposed into a sum of outer products, specifically expressed as $$T^{\mu\nu} = A^\mu B^\nu + A'^\mu B'^\nu + A''^\mu B''^\nu + \cdots$$. This decomposition is reiterated in the context of the covariant derivative on page 18, where Dirac states that a tensor $$T_{\mu\nu}$$ can be represented as a sum of terms like $$A_\mu B_\nu$$. The discussion clarifies that this representation is straightforward when using the definitions of tensors and the properties of summation.

PREREQUISITES
  • Understanding of rank-2 tensors in differential geometry
  • Familiarity with the concepts of outer products and tensor decomposition
  • Knowledge of covariant derivatives in the context of general relativity
  • Proficiency in LaTeX for mathematical notation
NEXT STEPS
  • Study the properties of rank-2 tensors in differential geometry
  • Learn about the covariant derivative and its applications in general relativity
  • Explore the use of outer products in tensor algebra
  • Practice writing and interpreting LaTeX for mathematical expressions
USEFUL FOR

Mathematicians, physicists, and students of general relativity who are looking to deepen their understanding of tensor decomposition and its applications in theoretical physics.

Kostik
Messages
274
Reaction score
32
TL;DR
Dirac says that a general rank-2 tensor ## T^{\mu\nu} ## can be decomposed as ## A^\mu B^\nu + A'^\mu B'^\nu + A''^\mu B''^\nu + \cdots\, ##. Is this obvious?
Dirac's book "General Theory of Relativity" says on p. 2 that a general rank-2 tensor can be written as a sum of outer products: $$ T^{\mu\nu} = A^\mu B^\nu + A'^\mu B'^\nu + A''^\mu B''^\nu + \cdots $$ Importantly, he repeats this on p. 18, in developing the covariant derivative, where he mentions that a tensor ## T_{\mu\nu} ## is "expressible as a sum of terms like ## A_\mu B_\nu ##".

Is this obvious? Can someone show or explain this?
 
Last edited:
Physics news on Phys.org
Kostik said:
TL;DR Summary: Dirac says that a general rank-2 tensor can be decomposed: ##T^\mu\nu = A^\mu B^\nu + A'^\mu B'^\nu + A''^\mu B''^\nu + ...##. Is this obvious?

Dirac's book "General Theory of Relativity" says on p. 2 that a general rank-2 tensor can be written as a sum of outer products:

$$T^\mu\nu = A^\mu B^\nu + A'^\mu B'^\nu + A''^\mu B''^\nu + ...$$

Importantly, he repeats this on p. 18, in developing the covariant derivative, where he mentions that a tensor ##T_\mu\nu$ is "expressible as a sum of terms like $A_\muB_\nu##".

Is this obvious? Can someone show or explain this?
By definition, a tensor of rank two can be written as
$$
T = T^{\mu\nu} e_\mu \otimes e_\nu
$$
We can introduce the vectors ##A^\nu = T^{\mu\nu} e_\mu## and ##B_\nu = e_\nu## (note that here ##\nu## is being used as a counter rather than a component index) and therefore
$$
T = A^\nu \otimes B_\nu
$$
 
  • Like
Likes   Reactions: vanhees71
I'm not familiar with your notation, I wonder if Dirac's decomposition can be explained using only his definition of tensors.
 
Last edited:
Kostik said:
I added the missing braces, but the LaTex still doesn't seem to be working in the original post.
It's a known issue when you make the first post to use LaTeX (OP or reply) on a page. The parser doesn't get loaded until you refresh the page. Your LaTeX looks fine to me, and will look fine to you once you've hit refresh.
 
  • Like
Likes   Reactions: Kostik
Ibix said:
It's a known issue when you make the first post to use LaTeX (OP or reply) on a page. The parser doesn't get loaded until you refresh the page. Your LaTeX looks fine to me, and will look fine to you once you've hit refresh.
Aha, yes, I see it now.
 
Oh, I think it's actually fairly straightforward. Write (showing the summation explicitly): $$T^{\mu\nu}=\sum_{\lambda,\kappa}T^{\lambda\kappa}{\delta_\lambda}^\mu{\delta_\kappa}^\nu \,\,\,\,\text{(no Einstein summation)}$$ Then ##A^\mu = T^{\lambda\kappa}{\delta_\lambda}^\mu## (not summed over ##\lambda##) and ##B^\nu={\delta_\kappa}^\nu##. (Regard ##\lambda## and ##\kappa## as fixed.) Since everything in sight is a tensor, the ##A^\mu## and ##B^\nu## are obviously vectors (no need to worry about constructing a non-vector).
 

Similar threads

  • · Replies 38 ·
2
Replies
38
Views
1K
  • · Replies 16 ·
Replies
16
Views
2K
  • · Replies 1 ·
Replies
1
Views
685
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 9 ·
Replies
9
Views
1K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 50 ·
2
Replies
50
Views
3K
  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 13 ·
Replies
13
Views
955