User talk:Anarchic Fox

Let's see...

1 $q_{f}(a,t)$ 2 $q(a,t)$ 3 $q_{d}(a,t)$ 3 $q_{d}(a',t)$ 3 $q_{f}(a',t)$ 3 $q(a',t)$

Moment of inertia

You wrote:

The expression for the inertia tensor for a continuous object should have used the 3x3 identity matrix, not the unit matrix.

Since Cayley introduced the unit matrix in 1858, this matrix is generally seen as identical to the identity matrix. See any book on linear algebra.

The tensor is diagonalizable because it is positive definite, not because it is real and symmetric.

You as a physicist should know that many a Hermitian operator has positive and negative eigenvalues. All Hermitian operators have an orthogonal (complete) set of eigenvectors. Translated to real matrices, this results reads: an n x n symmetric matrix has n orthogonal eigenvectors. Eigenvalues are real and may be all positive (positive definite matrix), of mixed sign (indefinite matrix) or all negative (negative definite matrix).

--P.wormer 07:58, 4 July 2007 (UTC)[reply]

Thank you.

Thank you for not being hardheaded, not all editors are! Please look here if you are interested in a discussion on the nomenclature of unit matrix.--P.wormer 12:03, 4 July 2007 (UTC)[reply]

Moment of inertia

Hi there AnarchicFox. You are eligible to nominate this article, since it is up to the second uninvolved editor (in this case me) to assess it. Oh and hopefully I too have not forgotten what I learnt in undergrad. Regards, Blnguyen (bananabucket) 07:52, 23 July 2007 (UTC)[reply]

Grammar

Thank you for correcting my grammar. As you may have guessed, English is a foreign language for me. Especially word order is difficult.--P.wormer 16:01, 30 July 2007 (UTC)[reply]

It's no problem. You're good enough at the language to get your points across effectively, which is what matters most. :) Anarchic Fox 07:41, 3 August 2007 (UTC)[reply]

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