Evaluating T(A,A) and Components of T(A,__) in Minkowski Spacetime

Click For Summary
SUMMARY

The discussion focuses on evaluating the tensor operation T(A, A) and the components of T(A,__) in Minkowski spacetime, as presented in Exercise 1.6 of Roger Blandford and Kip Thorne's textbook. The correct evaluations are T(A, A) = -9 and T(A,__) = (1, -4, 0, 0). A common mistake identified was failing to lower the indices of vector A before performing the multiplication, which led to incorrect intermediate results of 7 and (7, 0, 0).

PREREQUISITES
  • Understanding of Minkowski spacetime and its properties
  • Familiarity with tensor operations and index notation
  • Knowledge of vector and tensor components in physics
  • Ability to perform matrix multiplication in a relativistic context
NEXT STEPS
  • Study the properties of Minkowski spacetime in detail
  • Learn about tensor index lowering and raising techniques
  • Explore the implications of tensor operations in general relativity
  • Review matrix multiplication methods specific to physics applications
USEFUL FOR

Students of physics, particularly those studying general relativity, mathematicians working with tensors, and educators teaching advanced classical physics concepts.

Rasalhague
Messages
1,383
Reaction score
2
exercise Exercise 1.6 of Roger Blandford and Kip Thorne's online textbook Applications of Classical Physics:

"In Minkowski spacetime, in some inertial reference frame, the vector A and second rank
tensor T have as their only nonzero components A0 = 1, A1 = 2, A2 = A3 = 0. T00 = 3, T01 = T10 = 2, T11 = −1. Evaluate T(A, A) and the components of T(A,__) and A\otimesT."

http://www.pma.caltech.edu/Courses/ph136/yr2008/

The given answers to the first two questions are T(A, A) = -9, and T(A,__) = (1, -4, 0, 0). But I get 7, and (7, 0, 0, 0).

\left( \begin{matrix} 1 & 2 & 0 & 0 \end{matrix} \right) \left(\begin{matrix} 3 & 2 & 0 & 0 \\ 2 & -1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{matrix} \right) = \left( \begin{matrix} 7 & 0 & 0 & 0 \end{matrix} \right)

\left( \begin{matrix} 7 & 0 & 0 & 0 \end{matrix} \right) \left( \begin{matrix} 1 \\ 2 \\ 0 \\ 0 \end{matrix} \right) = 7
 
Physics news on Phys.org
You forgot to lower the indices of A before doing the multiplication.
 
Thanks! Yes, that's better:

\left( \begin{matrix} -1 & 2 & 0 & 0 \end{matrix} \right) \left(\begin{matrix} 3 & 2 & 0 & 0 \\ 2 & -1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{matrix} \right) = \left( \begin{matrix} 1 & -4 & 0 & 0 \end{matrix} \right)

\left( \begin{matrix} 1 & -4 & 0 & 0 \end{matrix} \right) \left( \begin{matrix} -1 \\ 2 \\ 0 \\ 0 \end{matrix} \right) = -9
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
4K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 21 ·
Replies
21
Views
4K
  • · Replies 4 ·
Replies
4
Views
5K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
Replies
2
Views
3K
  • · Replies 17 ·
Replies
17
Views
5K
  • · Replies 5 ·
Replies
5
Views
2K