Consider a system of \(n\) different masses \(m_i\), interacting with each other and being acted on by external forces. We can write Newton's second law for the positions \(\vec{r}_i\) of each of these masses with respect to a fixed origin \(\cal{O}\), thereby obtaining a system of equations governing the motion of the masses. \begin{align} m_1 \frac{d^2\, \vec{r}_1}{dt^2} &=\vec{F}_1+\;\; 0\;\, +\vec{f}_{12}+\vec{f}_{13}+\;\,\dots\;\, +\vec{f}_{1n}\nonumber\\ m_2 \frac{d^2\, \vec{r}_2}{dt^2} &=\vec{F}_2+\vec{f}_{21}+\;\; 0\;\, +\vec{f}_{23}+\;\,\dots\;\, +\vec{f}_{2n} \label{NewtonSystem}\\ \vdots\nonumber\\ m_n \frac{d^2\, \vec{r}_n}{dt^2} &=\vec{F}_n+\vec{f}_{n1}+\vec{f}_{n2}+\dots+\vec{f}_{n(n-1)}+0\quad\nonumber \end{align} Here, we have chosen the notation \(\vec{F}_i\) for the net external forces acting on mass \(m_i\) and \(\vec{f}_{ij}\) for the internal force of mass \(m_j\) acting on \(m_i\).
In general, each internal force \(\vec{f}_{ij}\) will depend on the positions of the particles \(\vec{r}_i\) and \(\vec{r}_j\) in some complicated way, making \((\ref{NewtonSystem})\), a set of coupled differential equations. To solve \((\ref{NewtonSystem})\), we first need to decouple the differential equations, i.e. find an equivalent set of differential equations in which each equation contains only one variable.
The weak form of Newton's third law states that the force \(\vec{f}_{12}\) of \(m_2\) on \(m_1\) is equal and opposite to the force \(\vec{f}_{21}\) of \(m_1\) on \(m_2\). We see that each internal force appears twice in the system of equations \((\ref{NewtonSystem})\), once with a positive sign and once with a negative sign. Therefore, if we add all of the equations together, the internal forces will all cancel, leaving: \begin{equation} \sum_{i=1}^n m_i \frac{d^2 \vec{r}_i}{dt^2} =\sum_{i=1}^n\vec{F}_i\label{NewtonCOM} \end{equation}
Notice what a surprising equation \((\ref{NewtonCOM})\) is. The right-hand side directs us to add up all of the external forces, each of which acts on a different mass; something you were taught never to do in introductory physics.
The left-hand side of \((\ref{NewtonCOM})\) directs us to add up (the second derivatives of) \(n\) “weighted" position vectors pointing from the origin to different masses. We can simplify the left-hand side of \((\ref{NewtonCOM})\) if we multiply and divide by the total mass \(M=m_1+m_2+\dots+m_n\) and use the linearity of differentiation to “factor out” the derivative operator: \begin{align} \sum_{i=1}^n m_i \frac{d^2 \vec{r}_i}{dt^2} &=M\frac{d^2}{dt^2} \left(\sum_{i=1}^n \frac{m_i}{M}\, \vec{r}_i\right)\label{CenterOfMass1}\\ &=M\frac{d^2 \vec{R}_{cm}}{dt^2}\label{CenterOfMass2} \end{align} We recognize (or define) the quantity in the parentheses on the right-hand side of \((\ref{CenterOfMass1})\) as the position vector \(\vec{R}_{cm}\) from the origin to the “center of mass” of the system of particles, i.e. \begin{equation} \vec{R}_{cm}=\sum_{i=1}^n\frac{m_i}{M}\, \vec{r}_i\label{CenterOfMass3} \end{equation} With these simplifications, equation (\ref{NewtonCOM}) becomes: \begin{equation} M \frac{d^2 \vec{R}_{cm}}{dt^2} =\sum_{i=1}^n\vec{F}_i\label{NewtonCOM2} \end{equation} which has the form of Newton's 2nd Law for a fictitious particle with mass \(M\) sitting at the center of mass of the system of particles and acted on by all of the external forces from the original system.
We can define the momentum of the center of mass as the total mass times the time derivative of the position of the center of mass: \begin{equation} \vec{P}_{cm}=M\frac{d\vec{R}_{cm}}{dt} \end{equation} If there are no external forces acting, then the acceleration of the center of mass is zero and the momentum of the center of mass is constant in time (conserved). \begin{equation} M\frac{d^2 \vec{R}_{cm}}{dt^2}=\frac{d\vec{P}_{cm}}{dt}=0 \label{MomentumConservation} \end{equation}
Notice that the entire discussion above applies even if all of the internal forces are zero \(\vec{f}_{ij}=0\), i.e. none of the particles have any way of knowing that the others are even present. Such particles are called non-interacting. The position of the center of mass of the system will still move according to equation \((\ref{NewtonCOM2})\).
assignment Homework
(Straightforward algebra) Purpose: Discover the change of variables that allows you to go from the solution to the reduced mass system back to the original system. Practice solving systems of two linear equations.
For systems of particles, we used the formulas \begin{align} \vec{R}_{cm}&=\frac{1}{M}\left(m_1\vec{r}_1+m_2\vec{r}_2\right) \nonumber\\ \vec{r}&=\vec{r}_2-\vec{r}_1 \label{cm} \end{align} to switch from a rectangular coordinate system that is unrelated to the system to coordinates adapted to the center-of-mass. After you have solved the equations of motion in the center-of-mass coordinates, you may want to transform back to the original coordinate system. Find the inverse transformation, i.e. solve for: \begin{align} \vec{r}_1&=\\ \vec{r}_2&= \end{align} Hint: The system of equations (\ref{cm}) is linear, i.e. each variable is to the first power, even though the variables are vectors. In this case, you can use all of the methods you learned for solving systems of equations while keeping the variables vector valued, i.e. you can safely ignore the fact that the \(\vec{r}\)s are vectors while you are doing the algebra as long as you don't divide by a vector.
assignment Homework
At low temperatures, a diatomic molecule can be well described as a rigid rotor. The Hamiltonian of such a system is simply proportional to the square of the angular momentum \begin{align} H &= \frac{1}{2I}L^2 \end{align} and the energy eigenvalues are \begin{align} E_{\ell m} &= \hbar^2 \frac{\ell(\ell+1)}{2I} \end{align}
What is the energy of the ground state and the first and second excited states of the \(H_2\) molecule? i.e. the lowest three distinct energy eigenvalues.
At room temperature, what is the relative probability of
finding a hydrogen molecule in the \(\ell=0\) state versus finding it
in any one of the \(\ell=1\) states?
i.e. what is
\(P_{\ell=0,m=0}/\left(P_{\ell=1,m=-1} + P_{\ell=1,m=0} + P_{\ell=1,m=1}\right)\)
At what temperature is the value of this ratio 1?
assignment Homework
Give the general solution of the differential equation: \[\frac{d^2 y}{dx^2}+Ay=0\] Make sure that you can give the solution of this equation regardless of the geometric names of the dependent and independent variables and for either sign for the constant \(A\).
It is NOT necessary to show any work. You may NOT, however, give a solution that has a negative number inside a square root. I am testing whether you can recognize this equation and remember its solution. This equation comes up over and over again in physics, but disguised by different symbols. I am also testing whether you recognize that the geometric character of the equation changes depending on the sign of \(A\).
assignment Homework
Find \(N\).
assignment Homework
In class, you measured the isolength stretchability and the isoforce stretchability of your systems in the PDM. We found that for some systems these were very different, while for others they were identical.
Show with algebra (NOT experiment) that the ratio of isolength stretchability to isoforce stretchability is the same for both the left-hand side of the system and the right-hand side of the system. i.e.: \begin{align} \frac{\left(\frac{\partial {x_L}}{\partial {F_L}}\right)_{x_R}}{\left(\frac{\partial {x_L}}{\partial {F_L}}\right)_{F_R}} &= \frac{\left(\frac{\partial {x_R}}{\partial {F_R}}\right)_{x_L}}{\left(\frac{\partial {x_R}}{\partial {F_R}}\right)_{F_L}} \label{eq:ratios} \end{align}
group Small Group Activity
30 min.
keyboard Computational Activity
120 min.
finite difference hamiltonian quantum mechanics particle in a box eigenfunctions
Students implement a finite-difference approximation for the kinetic energy operator as a matrix, and then usenumpy to solve for eigenvalues and eigenstates, which they visualize.
assignment Homework
assignment Homework
The goal of this problem is to show that once we have maximized the entropy and found the microstate probabilities in terms of a Lagrange multiplier \(\beta\), we can prove that \(\beta=\frac1{kT}\) based on the statistical definitions of energy and entropy and the thermodynamic definition of temperature embodied in the thermodynamic identity.
The internal energy and entropy are each defined as a weighted average over microstates: \begin{align} U &= \sum_i E_i P_i & S &= -k_B\sum_i P_i \ln P_i \end{align}: We saw in clase that the probability of each microstate can be given in terms of a Lagrange multiplier \(\beta\) as \begin{align} P_i &= \frac{e^{-\beta E_i}}{Z} & Z &= \sum_i e^{-\beta E_i} \end{align} Put these probabilities into the above weighted averages in order to relate \(U\) and \(S\) to \(\beta\). Then make use of the thermodynamic identity \begin{align} dU = TdS - pdV \end{align} to show that \(\beta = \frac1{kT}\).