Conservation of Total Angular Momentum Proof

For this post, I want to prove that in the absence of external forces, the total angular momentum of an N-particle system is conserved.

I start with  \vec{P} = \displaystyle \sum_{\alpha} \vec{p}_{\alpha} which is the total momentum of an N-particle system. Now I can vectorially multiple the total momentum by \vec{r} which is the position vector measured from the same origin O for each particle. This will give me the total angular momentum of the system which can be written as \vec{L} = \displaystyle \sum_{\alpha = 1}^{N} \vec{\ell}_{\alpha} = \displaystyle\sum_{\alpha = 1}^{N} \vec{r}_{\alpha} \times \vec{p}_{\alpha} . After differentiating with respect to t , I obtain \dot{\vec{L}} = \displaystyle \sum_{\alpha} \dot{\vec{\ell}}_{\alpha} = \displaystyle \sum_{\alpha} \frac{d}{dt} (\vec{r} \times \vec{p}) = \displaystyle \sum_{\alpha} ( \dot{\vec{r}} \times \vec{p}) + (\vec{r} \times \dot{\vec{p}}) . In the first cross product, I can substitute \vec{p} with m\dot{\vec{r}} , and since the cross product of any two parallel vectors is zero, the first term becomes zero. This leaves implies that my sum becomes \displaystyle\sum_{\alpha}\vec{r}_{\alpha} \times \vec{F}_{\alpha} . Now, I can rewrite the net force on particle \alpha as \vec{F}_{\alpha} = \displaystyle \sum_{\beta \neq \alpha} \vec{F}_{\alpha \beta}, where \vec{F}_{\alpha \beta} represents the force exerted on particle \alpha by particle \beta . Now I can make a substitution for \vec{F}_{\alpha} to give me \dot{\vec{L}} = \displaystyle \sum_{\alpha} \displaystyle \sum_{\beta \neq \alpha} \vec{r}_{\alpha} \times \vec{F}_{\alpha \beta}



…I will finish the rest of this at some point. I seem to have misplaced the book.


Matrix Proof

I want show that e^{(At)} = Se^{(\Lambda t)} S^{-1} , where S = (v_{1} v_{2}) ,  A is a 2 \times 2 matrix, and \Lambda = \begin{pmatrix} \lambda_{1} & 0 \\0 & \lambda_{2} \end{pmatrix}

I start by writing the middle sum in summation notation which gives me Se^{(\Lambda t)}S^{-1} = S( \displaystyle\sum_{k = 0}^{\infty} \frac{1}{k!}(\Lambda t)^{k})S^{-1} . Now I can use the identity S^{-1}AS = \Lambda which will then give me S(\displaystyle\sum_{k = 0}^{\infty} \frac{1}{k!}(S^{-1}AS)^{k}t^{k})(S^{-1}) . After pulling terms out of the sum, I will get SS^{-1} (\displaystyle \sum_{k = 0}^{\infty} \frac{t^{k}A^{k}}{k!})SS^{-1} . The SS^{-1} terms create an identity matrix and the middle sum is equivalent to e^{At} as shown below.

Matrix Proof



Given the matrix equation \dot{u} = Au , where u = \begin{pmatrix} x \\ y \end{pmatrix} and A is a 2 \times 2 matrix, I want to show that \dot{u} = e^{At}u_{0} is the solution where e^{At} = I + At + \frac{1}{2!}A^{2}t^{2} + \frac{1}{3!}A^{3}t^{3} + ... and I is the identity matrix.

I start out by writing the above series in summation notation which gives me \displaystyle\sum_{k = 0}^{\infty} \frac{t^{k}A^{k}}{k!}. I can now take a time derivative of the sum to give me \frac{d}{dt}e^{At} = \displaystyle\sum_{k = 0}^{\infty}\frac{kt^{k - 1}A^{k}}{k!} . Since the first term of the series after differentiation is 0 , I can rewrite and reduce the sum to give me \displaystyle\sum_{k = 1}^{\infty}\frac{t^{k - 1}A^{k}}{(k - 1)!} . Now, I can pull out a single matrix term to give me A \displaystyle\sum_{k = 1}^{\infty}\frac{t^{k - 1}A^{k - 1}}{(k - 1)!} . I can now simplify the sum once again to give me A \displaystyle\sum_{k = 0}^{\infty} \frac{t^{k}A^{k}}{k!} . This is now equivalent to Ae^{At} . Assuming that u = e^{At}u_{0}  is the solution, I can differentiate it with respect to time to give me \dot{u} = \frac{d}{dt}[e^{At}]u_{0} . I just showed that \frac{d}{dt} e^{At} = Ae^{At} , which I can plug in to give me the equation \dot{u} = Ae^{At}u_{0} . Since e^{At}u_{0} = u , I am left with my original matrix equation \dot{u} = Au