Proving tensor symmetry under transformation

Click For Summary
SUMMARY

The discussion focuses on proving that a 2nd order symmetric tensor D remains symmetric under transformation into any coordinate system using indicial notation. The transformation law for tensors is defined as D'_{pq} = a_{pr}a_{qs}D_{rs}. The proof demonstrates that D'_{pq} equals D'_{qp} by applying the properties of transposition and the symmetry of the tensor D, confirming that the transformed tensor retains its symmetry. The conclusion is that the transformation preserves the symmetric nature of the tensor.

PREREQUISITES
  • Understanding of tensor notation and operations
  • Familiarity with the properties of symmetric tensors
  • Knowledge of linear algebra, specifically matrix transposition
  • Grasp of the tensor transformation law
NEXT STEPS
  • Study the properties of higher-order tensors and their transformations
  • Explore applications of symmetric tensors in physics and engineering
  • Learn about the implications of tensor symmetry in continuum mechanics
  • Investigate the role of coordinate transformations in tensor calculus
USEFUL FOR

Students and professionals in mathematics, physics, and engineering who are working with tensor analysis and need to understand the implications of tensor symmetry under transformations.

Hendrick
Messages
41
Reaction score
0

Homework Statement


Using indical notation, prove that a 2nd order symmetric tensor D remains symmetric when transformed into any other coordinate system.


Homework Equations


Tensor law of transformation (2nd order):
D'_{pq} = a_{pr}a_{qs}D_{rs}


The Attempt at a Solution


I think I'm required to prove that D'_{pq} = D'_{qp} (where D is a symmetric 2nd order tensor)


D'_{pq} = a_{pr}a_{qs}D_{rs}
D'^{T}_{pq} = (a_{pr}a_{qs}D_{rs})^T
D'_{qp} = a_{qs}a_{pr}D_{sr} (can someone please explain why when you transpose this, the a's swaps position but the D swaps indices?)

D_{rs} = D_{sr} (as D is symmetric)

=>D'_{pq} = a_{pr}a_{qs}D_{rs} = a_{qs}a_{pr}D_{rs} = D'_{qp}

Thus D'_{pq} = D'_{qp}


Did I prove it correctly?

Thank you
 
Last edited:
Physics news on Phys.org
Did I not write my question properly or does no one know?
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 22 ·
Replies
22
Views
4K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 6 ·
Replies
6
Views
1K
Replies
6
Views
1K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 7 ·
Replies
7
Views
936
  • · Replies 1 ·
Replies
1
Views
3K