Saturday, December 20, 2014

A Look At General Relativity and Tensors (Part 2)

Continued from earlier instalment:

3.Diagonalizing tensors.

This is analogous to obtaining the eigenvalues for a matrix in linear algebra. Consider the object below for which we want the principal axis.
No automatic alt text available.

Fig. 3. We wish to find the principal axis, a i j.

And we have a i j

 With T’ =  A × I × At

Where t denotes the transpose.  Then we obtain, T’ =

(15 …..0……..0)
(0…….11 ....-3Ö2)
(0……..-3Ö2…..8 )

Which is to be diagonalized.  Writing this out:

(15 - l…...0…….....0)
(0……..11- l ....-..3Ö2)
(0……..--3Ö2…....8l)


This leads to a cubic - i.e. with triple roots -  which are:

l1 = 15,  l2 =  5, and l3 =  14

Substituting l1 in the matrix we get:

(0  ….....0……......0)   ( c x  )
(0……....-4 ....-..(-3Ö2))  ( c y  )
(0……..-(-3Ö2)…..(-7) )   (  c z )

For which the separate equations, e.g. in  c x, c y and c z can be solved. For example,

-4 y   -   3Ö z  =  0 

 -   3Öy   -7   =  0


After working through all the solutions, we obtain:

C  =  - Ö (2/3) e2^   +  1/ Ö3  (e3^)


Example  Problem:

In a certain rectangular coordinate system, the directions of whose axes are given by the unit vectors i, j and k, the inertia tensor of an object is given by:

I = K x

(1….0…..0)
(0….1…..1)
(0….1… ..1)

a) What are the principal moments of inertia of the object (the moments of inertia along the principal axis) relative to the origin of the above coordinate system?

b) What is the direction of the principal axis corresponding to the principal moment of inertia and equal to K?

 c)  If the origin of the above rectangular coordinate system is at the center of mass of the object and the total mass of the object is M, what is the change in the inertial tensor of the object if the rectangular coordinate system is displaced parallel to itself a distance ro in the direction:

 (1/ Ö2)j +  (1/Ö2)k?



Solutions:

a) We have:  I = K x

(1….0…..0)
(0….1…..1)
(0….1… ..1)

We write out the determinant with eigenvalue l:

(1 - l….0…..0)
(0….1 - l   ..1)
(0….1… ..1 - l)

Leading to the characteristic equation:

(1 - l)3 – (1 - l) = 0

Factoring:

(1 - l) [ ((1 - l)2 – 1] = 0

Or:

(1 - l) (l2 –  2l) = 0

Yielding eigenvalues: l= 0, l = 2

Then:

T = Kl,  so:

 T1 = 0, T2 = K, and T3 =2K

Or: (0, K, 2K)


b) We have to take:  (I T1)C

So that:

K [(1….0…..0)
  [(0….1…..1)
  [(0….1… ..1)


K (1….0…..0)](x)
  (0….1….. 0)] (y)
  (0….0… ..1)] (z)

=
 (0….0…..0) (x)
  (0….0…..1) (y)
  (0….1… ..0) (z)   =   0

So that:  0  =

(0)
(z)
(y)   

Which 0 for the x component implies the answer is i.



c) By the analog of the parallel axis theorem:

I o = IG  M(R2 I – RR)

D I =  I o -  IG   =    M(R2 I – RR)


RR =   r o 2     x

(0….0………0)
(0….1/2…..1/2)
(0….1/2… ..1/2)

D I =   M  r o 2     x

 [(1….0…..0)
  [(0….1…..0)
  [(0….0… ..1)

 (0….0……..0)]
  (0….1/2.. 1/2)]
  (0….1/2… ..1/2)]

= M  r o 2     x

(1….0………..0)
(0….1/2…..-1/2)
(0….-1/2… ..1/2)


Problems:

1)  For the example problem given, use the remaining two eigenvalues, l2 and l3, to diagonaliize the matrix and obtain solutions in c x, c y and c z.

2) Thereby obtain the principal axes in terms of: e^’x,        e ^’y and e^’ z .

3) Show in particular that:

  e^’ z    = - 1/ Ö3   e1^ -  1/ Ö3   e2^ + 1/ Ö3   e3^

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