Tensor Indices Switch with Infinitesimals and Space-Time Derivatives

  • Context: Graduate 
  • Thread starter Thread starter waht
  • Start date Start date
  • Tags Tags
    Switch Tensor
Click For Summary
SUMMARY

The discussion centers on the validity of manipulating expressions involving infinitesimals and space-time derivatives in tensor calculus. Specifically, the transformation from the expression \delta\omega^{u}_{v} x^v \partial_u to \delta\omega_{u v} x^v \partial^u without using the metric tensor is confirmed as valid. The participants agree that this operation can be viewed as a double dot-product, affirming the mathematical correctness of the manipulation. Tim's clarification emphasizes the relationship between the components and the underlying geometry.

PREREQUISITES
  • Tensor calculus fundamentals
  • Understanding of infinitesimals in mathematical physics
  • Knowledge of space-time derivatives
  • Familiarity with metric tensors and their role in raising and lowering indices
NEXT STEPS
  • Study the properties of infinitesimals in differential geometry
  • Learn about the implications of tensor index manipulation in general relativity
  • Explore the use of metric tensors in raising and lowering indices
  • Investigate the concept of dot-products in tensor analysis
USEFUL FOR

This discussion is beneficial for physicists, mathematicians, and students engaged in advanced studies of tensor calculus, particularly those focusing on general relativity and differential geometry.

waht
Messages
1,502
Reaction score
4
Wondering if this is valid to do, if I start with the expression

\delta\omega^{u}_{ \singlespacing v} x^v \partial_u

where \delta\omega is an infinitesimal, and \partial a space-time derivative,

is it still valid to drop and raise the u to obtain

\delta\omega_{u v} x^v \partial^u

without involving the metric tensor?
 
Physics news on Phys.org
what said:
Wondering if this is valid to do, if I start with the expression

\delta\omega^{u}_{ \singlespacing v} x^v \partial_u

where \delta\omega is an infinitesimal, and \partial a space-time derivative,

is it still valid to drop and raise the u to obtain

\delta\omega_{u v} x^v \partial^u

without involving the metric tensor?

Hi what! :smile:

Yes, it's just a (double) dot-product:

\delta\omega^{u}_{ \singlespacing v} x^v \partial_u

=\ \delta\omega_{w\singlespacing v}g^u_w x^v \partial_u

=\ \delta\omega_{w\singlespacing v} x^v \partial_w :smile:
 
Thanks Tim, that cleared it up.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
5K
  • · Replies 12 ·
Replies
12
Views
5K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
7K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 22 ·
Replies
22
Views
4K
  • · Replies 13 ·
Replies
13
Views
2K