Raising Indices with Minkowski Metric: Solving Weak Field Approx

Click For Summary
SUMMARY

The discussion focuses on the manipulation of indices in the context of the Minkowski metric, specifically addressing the expression ##g_{ab}=\eta_{ab}+\epsilon h_{ab}##. The participant seeks clarification on the origin of the minus sign in the raised indices expression ##g^{ab}=\eta^{ab}-\epsilon h^{ab}##. It is established that the Minkowski metric ##\eta_{ab}## has a signature of ##(-,+,+,+)##, and the relationship between covariant and contravariant delta functions is crucial, particularly when ##\epsilon## is treated as an infinitesimal.

PREREQUISITES
  • Understanding of Minkowski metric and its properties
  • Familiarity with tensor notation and index manipulation
  • Knowledge of covariant and contravariant indices
  • Basic grasp of infinitesimal quantities in mathematical physics
NEXT STEPS
  • Study the properties of the Minkowski metric in detail
  • Learn about tensor index raising and lowering techniques
  • Explore the implications of infinitesimals in tensor calculus
  • Investigate the relationship between covariant and contravariant tensors
USEFUL FOR

This discussion is beneficial for theoretical physicists, mathematicians specializing in differential geometry, and students studying general relativity or advanced tensor analysis.

binbagsss
Messages
1,291
Reaction score
12
I have the expression ##g_{ab}=\eta_{ab}+\epsilon h_{ab}##,

The indices on ##h^{ab}## are raised with ##\eta^{ab}## to give ##g^{ab}=\eta^{ab}-\epsilon h^{ab}##

I am not seeing where the minus sign comes from.

So I know ##\eta^{ab}\eta_{bc}=\delta^{a}_{c}## and ##h^{ab}=\eta^{ac}\eta^{bd}h_{cd}## ,
And that ##\eta_{ab}## has signature ##(-,+,+,+)##,

Any help greatly appreciated.
 
Physics news on Phys.org
If ##\epsilon## is an infinitessimal then the sign changes between the covariant and contravariant delta so that the contraction is unchanged to order ##\epsilon##
 
Mentz114 said:
If ##\epsilon## is an infinitessimal then the sign changes between the covariant and contravariant delta so that the contraction is unchanged to order ##\epsilon##

Thanks .
 

Similar threads

  • · Replies 37 ·
2
Replies
37
Views
3K
  • · Replies 17 ·
Replies
17
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 9 ·
Replies
9
Views
827
  • · Replies 15 ·
Replies
15
Views
2K
Replies
9
Views
2K
  • · Replies 8 ·
Replies
8
Views
5K
Replies
9
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K