## Wednesday, October 31, 2012

### gravitationally bound systems

imagine you've assembled a self-gravitating bound equilibrated system from $n$ particles of mass $m$, such that the total mass is $M=nm$ and such that the system has a gravitational binding energy $E$. now you reassemble the system from $p$ times more particles of masses $m/p$ so that the total mass would be identical, $M = np \times m/p=nm$. would the gravitational binding still be $E$?

bonus question: why are the traces $\mathrm{tr}(A^n)$ of a square matrix $A$ taken to the power $n$ invariants? how are they related to the eigenvalues of $A$? would that be a viable way for computing the eigenvalues?

1. bonus question: the trace $\mathrm{tr}(A)$ is invariant under transformations with a matrix $R$ because of the cyclic property of the trace, $\mathrm{tr}(AB)=\mathrm{tr}(BA)$: $\mathrm{tr}(R^{-1}AR)=\mathrm{tr}((R^{-1}A)R)=\mathrm{tr}(R(R^{-1}A))=\mathrm{tr}(A)$ with $R^{-1}R=\mathrm{id}$. likewise, you can write for $B=R^{-1}AR$: $\mathrm{tr}(B^n) = \mathrm{tr}(R^{-1}A R \cdots R^{-1} A R) = \mathrm{tr}(R^{-1} A\cdots A R) = \mathrm{tr}(R^{-1} A^n R) = \mathrm{tr}(A^n)$, so the trace has the same value in all systems.
2. bonus question 2: transforming into the eigensystem of $A$ the trace $\mathrm{tr}(A)$ is just the sum of all eigenvalues. $\mathrm{tr}(A^n)$ is the sum of the $n$th-powers of the eigenvalues, giving a nonlinear system of equations for computing the eigenvalues, which of course does not make much sense in comparison to the much easier linear eigenvalue solution.