Verifying Tensor Math: Is Step 0 True for Any Tensor A?

  • Thread starter Thread starter redstone
  • Start date Start date
  • Tags Tags
    Tensor
Click For Summary
SUMMARY

The discussion centers on the verification of tensor math, specifically the validity of Step 0 in the equation {{A}^{ab}}=\frac{1}{D}{{g}^{ab}}{{g}_{cd}}{{A}^{cd}}. It is established that this equation is not universally true for any tensor A, as it implies that any rank two tensor is a multiple of the metric tensor, which is incorrect. The participants clarify that while the contractions of two tensors may be equal, this does not imply the original tensors are equal, thus invalidating the assumption made in Step 0.

PREREQUISITES
  • Understanding of tensor notation and operations
  • Familiarity with metric tensors and their properties
  • Knowledge of tensor contraction and its implications
  • Basic principles of linear algebra as applied to tensors
NEXT STEPS
  • Study the properties of metric tensors in differential geometry
  • Learn about tensor contraction and its significance in tensor calculus
  • Explore the implications of tensor equality and uniqueness
  • Investigate advanced topics in tensor analysis, such as rank and type of tensors
USEFUL FOR

Mathematicians, physicists, and students in fields involving tensor calculus, particularly those working with general relativity or differential geometry.

redstone
Messages
26
Reaction score
0
Looking for a check on my tensor math to make sure I've done this correctly...
Where D equals the dimension of the metric -
Step 0: [tex]{{A}^{ab}}=\frac{1}{D}{{g}^{ab}}{{g}_{cd}}{{A}^{cd}}[/tex]
Step 1: [tex]{{g}_{ab}}{{A}^{ab}}={{g}_{ab}}\frac{1}{D}{{g}^{ab}}{{g}_{cd}}{{A}^{cd}}[/tex]
Step 2: [tex]{{g}_{ab}}{{A}^{ab}}=\frac{1}{D}{{g}_{ab}}{{g}^{ab}}{{g}_{cd}}{{A}^{cd}}[/tex]
Step 3: [tex]{{g}_{ab}}{{A}^{ab}}=\frac{1}{D}g_{a}^{a}{{g}_{cd}}{{A}^{cd}}[/tex]
Step 4: [tex]{{g}_{ab}}{{A}^{ab}}=\frac{1}{D}D{{g}_{cd}}{{A}^{cd}}[/tex]
Step 5: [tex]{{g}_{ab}}{{A}^{ab}}={{g}_{cd}}{{A}^{cd}}[/tex]
Step 6: [tex]{{g}_{ab}}{{A}^{ab}}={{g}_{ab}}{{A}^{ab}}[/tex]

So I know that the equation in step 0 is true for any tensor A, is that correct?
 
Physics news on Phys.org
No. Step 0 says A is equal to g*trace(A)/D. If that were true, then it would say any rank two tensor is a multiple of the metric tensor. That doesn't sound right, does it? It isn't true for any tensor. You can't go backward from Step 1 to Step 0. Just because the contractions of two tensors are equal, you can't say the original two tensors are equal.
 
Last edited:

Similar threads

Replies
1
Views
3K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
2
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
12
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K