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.

  3. gravity reaches to infinity and therefore, the gravitational binding energy per particle does not necessarily converge if more and more particles are added to the system. this property of gravity is referred to as the non-extensivity of the gravitational binding energy, it is *not* proportional to the number of particles of a system.