Table of Contents

CM.0 Newtonian Mechanics

CM.0.1 Newton's Laws for a Single Particle

The goal of physics is to predict the future from present conditions. In classical mechanics, we usually want to predict the future position of particles given their current positions and velocities. Newton gave three laws of motion for doing so, which I am here quoting from [1]: (1) An object at rest remains at rest, and an object in motion remains in motion at constant speed and in a straight line unless acted on by an unbalanced force. (2) The acceleration of an object depends on the mass of the object and the amount of force applied. (3) Whenever one object exerts a force on another object, the second object exerts an equal and opposite force on the first, acting in the direction pointing from one object to the other.

We now want to state the mathematical implications of these laws and introduce the notion of conserved quantities, which are fundamental to physics. This Chapter will largely follow the first few pages of Goldstein, [2], since it is the simplest way to present this material.

CM.0.1.1 Conservation of Momentum

Newton's second law gives us our equation of motion for a particle:

$$𝐅=\frac{d𝐩}{dt},\text{(CM.0.1)}$$

where $𝐅$ is the total force on the particle, which can be a function of time or many other variables, $𝐩=m𝐯$ is the particle's momentum, $m$ is the particle's mass, $𝐯=d𝐫/dt$ is the particle's velocity, and $𝐫$ is the particle's position (bold quantities are vectors). We will assume that the mass of our particle is constant, but the velocity can be a function of time, $𝐯(t)$, whose derivative is the acceleration $𝐚=d𝐯/dt$. Moving the constant mass outside of the time derivative in Equation CM.0.1 gives the familiar form of Newton's Second Law,

$$𝐅=m\frac{d𝐯}{dt}=m𝐚.\text{(CM.0.2)}$$

A conservation law is an equation that states that a certain quantity, call it $Q$ is not a function of time, i.e. $dQ/dt=0$. From Equation CM.0.1, we can see that if the force $𝐅=0$, then the momentum of our particle is conserved, that is it is constant in time, $d𝐩/dt=0$.

CM.0.1.2 Conservation of Angular Momentum

The angular momentum $𝐋$ of a particle, with respect to the origin of coordinates being used to define the position $𝐫$, is defined as

$$𝐋=𝐫\times 𝐩,\text{(CM.0.3)}$$

where $\times$ is the vector cross product. We define the torque $\mathbf{\tau }$ as

$$\mathbf{\tau }=𝐫\times 𝐅.\text{(CM.0.4)}$$

How does the angular momentum of a particle with constant mass change with time? We need to use the product rule to calculate it:

$$\frac{d𝐋}{dt}=\frac{d𝐫}{dt}\times 𝐩+𝐫\times \frac{d𝐩}{dt}=m𝐯\times 𝐯+𝐫\times 𝐅.$$

The cross product of any vector with itself is zero, so we conclude that

$$\mathbf{\tau }=\frac{d𝐋}{dt},\text{(CM.0.5)}$$

where we have used the definition of the torque. Therefore, we can conclude that the angular momentum of our particle is conserved when there is no applied torque, $\mathbf{\tau }=0$.

CM.0.1.3 Conservation of Energy

The next familiar quantity we want to define is the work $W$. It is a measure of how much the force is pushing in the direction of the particle's movement. The work is the sum of the dot product of the force with all the small displacements of the particle along its path, $𝐅·d𝐫$. In the limit that the step size goes to zero, the sum turns into a line integral between two points 1 and 2 along the path of the particle:

$$W_{12}=\underset{1}{\overset{2}{\int }}𝐅·d𝐫.\text{(CM.0.6)}$$

In general, the work will depend on the path taken between the two points, not just on the endpoints themselves. When the mass is constant, we can use Newton's Law, Equation CM.0.2, to evaluate the integral defining the work:

$$\int 𝐅·d𝐫=\int m\frac{d𝐯}{dt}·\frac{d𝐫}{dt}dt=\frac{m}{2}\int \frac{d(𝐯·𝐯)}{dt}dt,$$

where we have also multiplied and divided by $dt$ inside the integral sign and identified the derivative $d(𝐯·𝐯)/dt=(1/2)(d𝐯/dt)·(d𝐫/dt)$. The quantity $𝐯·𝐯$ is just the usual magnitude squared of the velocity vector, $v^2$. The integral of the derivative of a function is just the function itself, so we have now shown that the work done on a particle by the force $𝐅$ moving it between points 1 and 2 is just

$$W_{12}=\frac{1}{2}m{v}_2^2-\frac{1}{2}m{v}_1^2=T_2-T_1,\text{(CM.0.7)}$$

where $T_i=m{v}_i^2/2$ is the particle's kinetic energy evaluated at point $i$.

We now want to identify the special condition under which we have a conserved quantity, which we will call the energy $E$, involving $T$. That special condition is when the work is path-independent. That is, the line integral in Equation CM.0.6 is only a function of the endpoints 1 and 2. We now want a mathematical statement that is equivalent to the work only being a function of the endpoints. Take any two paths $A$ and $B$ between the points 1 and 2. If the particle takes path $A$, then the work is $W_{12}$. However, if it takes path $B$, then the work must be the same, $W_{12}$, because the work only depends on the endpoints. However, consider now the reversed path $\overline{B}$ taken by going from point 2 to point 1. The work along $\overline{B}$ must be $W_{21}=-W_{12}$, because integrals are odd functions under exchanging their bounds. The combined path $A\overline{B}$ is a closed path from point 1 to point 2 and back to point 1; the work done on it must be $W_{11}=W_{12}+W_{21}=W_{12}-W_{12}=0$. However, the work only depends on the endpoints, so we could have taken any closed path that went from point 1 to point 1 and gotten the same answer, $W_{11}=0$ (If we wanted to be slick, we could have said that the path where we start at point 1 and do not move immediately gives us $W_{11}=0$).

We have just shown that if the work only depends on the endpoints and not on the path taken, then the work on a closed path must always be zero:

$$\oint 𝐅·d𝐫=0.\text{(CM.0.8)}$$

The converse is also true. If Equation CM.0.8 holds, then the work must be independent of path. The proof is the reverse of the argument we just gave. If $W_{11}=0$ for any path, then we can choose any other point 2, and any paths $A$ from 1 to 2 and $\overline{B}$ from 2 to 1, and find $W_{11}=W_{12,A}+W_{21,\overline{B}}=0$. Therefore, $W_{12,A}=-W_{21,\overline{B}}=W_{12,B}$ by reversing the path $\overline{B}$ to $B$ using the minus sign to flip the integration bounds. This argument holds for any points 1 and 2 and any paths $A$ and $B$ between these points, so the work only depends on the endpoints 1 and 2, not the paths.

When Equation CM.0.8 holds, the force $𝐅$ is called a conservative force. If we pick a convenient point $𝐎$, we can define the potential energy function $V(𝐩)$ as the negative of the work from $𝐎$ to $𝐩$:

$$V(𝐩)=-\underset{𝐎}{\overset{𝐩}{\int }}𝐅·d𝐫.\text{(CM.0.9)}$$

Finally, we can write the work $W_{12}$ in terms of the potential energy:

$$W_{12}=\underset{1}{\overset{2}{\int }}𝐅·d𝐫=\underset{1}{\overset{𝐎}{\int }}𝐅·d𝐫+\underset{𝐎}{\overset{2}{\int }}𝐅·d𝐫=V_1-V_2.\text{(CM.0.10)}$$

Combining with Equation CM.0.7, we find

$$V_1-V_2=W_{12}=T_2-T_1\to V_1+T_1=V_2+T_2.\text{(CM.0.11)}$$

We can define the energy $E=T+V$, which is conserved along the particle's path when the force is conservative.

We could have immediately found the above results using multivariable calculus, as you probably know. Stokes' Theorem says that conservative forces satisfy

$$\mathbf{\nabla }\times 𝐅=0,\text{(CM.0.12)}$$

their curl is zero, so that they can be written as the gradient of a scalar function $V$, the exact potential energy that we defined above:

$$𝐅=-\mathbf{\nabla }V.\text{(CM.0.13)}$$

CM.0.2 Many Particles

The real world has more than one particle. Newton's second law holds for each particle individually. It is convenient to separate the total force ${𝐅}_i$ on each particle into those that result from external sources, leading to a total force ${𝐅}_i^{(e)}$ on particle $i$, and those that result from interactions between pairs of particles in our system, leading to total forces ${𝐅}_{ji}$ on particle $i$ from particle $j$. Then, each particle in our system, with position ${𝐫}_i$ and corresponding velocity, mass, and momentum, satisfies

$$\frac{d{𝐩}_i}{dt}={𝐅}_i={𝐅}_i^{(e)}+\sum _j{𝐅}_{ji}.\text{(CM.0.14)}$$

For simplicity, we will assume that all the masses remain constant. We will also assume Newton's third law holds for all forces, so that ${𝐅}_{ji}=-{𝐅}_{ij}$, and that the forces act along the direction pointing from one object to the other, even though the third law is not true in general (e.g. for magnetic forces). We can also note that $F_{ii}=0$, since particles do not exert forces on themselves.

CM.0.2.1 Conservation of Momentum

We now define the center of mass $𝐑$ of our collection of particles:

$$𝐑=\frac{\sum _i{m}_i{𝐫}_i}{M},\text{(CM.0.15)}$$

where $M=\sum m_i$ is the total mass of our system of particles. The center of mass is the (mass-weighted) average position of particles in our collection. Let's try to treat our center of mass as the position of a particle with mass $M$. What would its equation of motion be? Since all of the masses of our particles are constant, taking two time derivatives of $𝐑$ only hits the individual particles' position vectors in the sum. Therefore, we find that Newton's law for the center of mass is

$$M\frac{d^2𝐑}{d{t}^2}=\sum _i{m}_i\frac{d^2{𝐫}_i}{d{t}^2}=\sum _i{𝐅}_i^{(e)}+\sum _{i,j}{𝐅}_{ji}.$$

There are two force sums. The first is the sum over all of the external forces to give the total external force, ${𝐅}^{(e)}=\sum {𝐅}_i^{(e)}$. The second sum is the sum over all of the pairwise internal forces. By Newton's third law, this sum should be zero, since each force has an equal and opposite force canceling it. Mathematically, ${𝐅}_{ji}+{𝐅}_{ij}=\mathbf{0}$. So, the right-hand side just becomes the total external force.

We consider now the total momentum $𝐏$ of our system, defined as the sum of the individual momenta of each particle:

$$𝐏=\sum _i{m}_i\frac{d{𝐫}_i}{dt}=M\frac{d}{dt}\left(\frac{{\sum }_i{m}_i{𝐫}_i}{M}\right)=M\frac{d𝐑}{dt},\text{(CM.0.16)}$$

where we multiplied and divided by the total mass $M$ and moved masses in and out of derivatives as necessary. Combining the previous two results, we have shown that

$$\frac{d𝐏}{dt}={𝐅}^{(e)},\text{(CM.0.17)}$$

the total momentum of the system and the total external force on all the particles give the usual form of Newton's second law. Therefore, the total momentum of our collection of particles is conserved if the total external force is zero.

CM.0.2.2 Conservation of Angular Momentum

We have defined the total momentum and found a conservation law for it. Now we will define the total angular momentum $𝐋$ and find its conservation law. The total angular momentum is

$$𝐋=\sum _i{𝐋}_i=\sum _i{𝐫}_i\times {𝐩}_i.\text{(CM.0.18)}$$

We should now take the time derivative of the total angular momentum. Using the product rule gives

$$\frac{d𝐋}{dt}=\sum _i\frac{d{𝐫}_i}{dt}\times {𝐩}_i+\sum _i{𝐫}_i\times \frac{d{𝐩}_i}{dt}=\sum _i{𝐫}_i\times \frac{d{𝐩}_i}{dt},$$

since the individual momenta and velocities point in the same direction, meaning their cross-product is zero. We can expand the derivatives of the momenta using Newton's law, Equation CM.0.14, so that

$$\sum _i{𝐫}_i\times \frac{d{𝐩}_i}{dt}=\sum _i{𝐫}_i\times {𝐅}_i^{(e)}+\sum _{i,j}{𝐫}_i\times {𝐅}_{ji}.$$

The first term on the right-hand side looks like the total external torque. We need to do some work on the second term. As before, the trick is to pair term $ij$ with term $ji$, since ${𝐅}_{ji}=-{𝐅}_{ij}$. Then, term $ij$ is ${𝐫}_i\times {𝐅}_{ji}$, while term $ji$ can be rewritten as $-{𝐫}_j\times {𝐅}_{ji}$. Adding these two terms and factoring out the common force gives the cross product $({𝐫}_i-{𝐫}_j)\times {𝐅}_{ji}$. While this form may not look any better, it is actually the key. Newton's third law not only guaranteed that the forces were equal and opposite, but it also stipulated that they acted on the line connecting the two particles. That is, ${𝐅}_{ji}$ is parallel to the vector $({𝐫}_i-{𝐫}_j)$, since the difference in position vectors is the vector pointing between the two particles. Therefore, by pairing up terms and noticing that they are zero, we have found

$$\sum _{i,j}{𝐫}_i\times {𝐅}_{ji}=\mathbf{0}.$$

Putting everything together and defining the total external torque ${\mathbf{\tau }}^{(e)}=\sum {\mathbf{\tau }}_i^{(e)}=\sum {𝐫}_i\times {𝐅}_i^{(e)}$, we find

$$\frac{d𝐋}{dt}={\mathbf{\tau }}^{(e)}.\text{(CM.0.19)}$$

Therefore, if the total external torque is zero, then the total angular momentum of our collection of particles is conserved.

Optional: Total angular momentum in terms of the center of mass. We have found the conservation law for angular momentum, but it would be nice to also know an expression for the total angular momentum in terms of the center of mass $𝐑$. For the total linear momentum, the expression was as simple as Equation CM.0.16. To find the corresponding equation for the total angular momentum, we will have to directly calculate from Equation CM.0.18. First, we rewrite the position vector of each particle with respect to the center of mass:

$${𝐫}_i={\mathbf{\rho }}_i+𝐑,\text{(CM.0.20)}$$

where we have defined the vectors ${\mathbf{\rho }}_i={𝐫}_i-𝐑$, the position of particle $i$ measured in a coordinate system where the center of mass is at the origin. By taking the time derivative, we get a similar equation for the velocities, defining the new quantity ${𝐰}_i=d{\mathbf{\rho }}_i/dt$ and the center of mass velocity $𝐕=d𝐑/dt$:

$${𝐯}_i={𝐰}_i+𝐕.\text{(CM.0.21)}$$

Using Equations CM.0.20 and CM.0.21 in Equation CM.0.18 and distributing the cross product gives

$$𝐋=\sum _i{m}_i{𝐫}_i\times {𝐯}_i=\sum _i(m_i{\mathbf{\rho }}_i\times {𝐰}_i+m_i{\mathbf{\rho }}_i\times 𝐕+m_i𝐑\times {𝐰}_i+m_i𝐑\times 𝐕);.$$

The summations can now commute with some of the cross products. In the final term, the sum just collects all of the masses into the total mass $M$. In the second term, we have $(\sum m_i{\mathbf{\rho }}_i)\times 𝐕$. In the third term, we have $𝐑\times d(\sum m_i{\mathbf{\rho }}_i)/dt$. Both of these latter terms rely on the quantity $\sum m_i{\mathbf{\rho }}_i$, which we can calculate directly from Equation CM.0.20:

$$\sum _i{m}_i{\mathbf{\rho }}_i=\sum _i{m}_i{𝐫}_i-\sum _i{m}_i𝐑=M𝐑-M𝐑=\mathbf{0},$$

where we have used the definition of the center of mass position. We could have anticipated this result, since $\sum m_i{\mathbf{\rho }}_i$ is (proportional to) the center of mass's position, but calculated in the coordinate system where the center of mass is at the origin.

Finally, we can state the result:

$$𝐋=M𝐑\times 𝐕+\sum _i{m}_i{\mathbf{\rho }}_i\times {𝐰}_i.\text{(CM.0.22)}$$

In words, the total angular momentum is the angular momentum of the center of mass around the origin plus the total angular momentum of the particles moving around the center of mass. The value of the first term, the angular momentum of the center of mass, depends on the origin of our coordinate system. The second term, the total angular momentum of the particles about the center of mass, does not depend on the choice of coordinate system, because it is always calculated in the coordinate frame where the center of mass is at the origin. Therefore, in all coordinate systems where the first term is zero---for example, when the origin of coordinates is moving with the center of mass's velocity $𝐕$---the total angular momentum $𝐋$ has the same value, equal to the total angular momentum measured in the frame where the center of mass is at the origin.

CM.0.2.3 Conservation of Energy

Finally, we want to determine when energy is conserved in our many particle system. We start again with the work done on all of the particles, which is just the sum of the work done on the individual particles:

$$W_{12}=\sum _i\underset{1}{\overset{2}{\int }}{𝐅}_i·d{𝐫}_i=\sum _i\underset{1}{\overset{2}{\int }}{𝐅}_i^{(e)}·d{𝐫}_i+\sum _{i,j}\underset{1}{\overset{2}{\int }}{𝐅}_{ji}·d{𝐫}_i.\text{(CM.0.23)}$$

We can use the same trick of multiplying and dividing by $dt$, then substituting out the forces ${𝐅}_i$ using Newton's law, as we did in Equation CM.0.6, to find that the right-hand side of Equation CM.0.23 turns into the change in total kinetic energy, where the total kinetic energy is a sum over the individual kinetic energies of $m_i{v}_i^2/2$. So, we can write the work as

$$W_{12}=T_2-T_1,\text{(CM.0.24)}$$

where

$$T=\sum _i\frac{m_i{v}_i^2}{2}.\text{(CM.0.25)}$$

To have conservation of energy, we need conservative forces. First, we require that the external forces on each particle are conservative. That is, each $F_i^{(e)}$ is the negative gradient of a potential function $V_i$, so that

$$\underset{1}{\overset{2}{\int }}{𝐅}_i^{(e)}·d{𝐫}_i=-\underset{1}{\overset{2}{\int }}{\mathbf{\nabla }}_i{V}_i·d{𝐫}_i=V_i(1)-V_i(2),\text{(CM.0.26)}$$

where the index on the gradient means we are only taking derivatives with respect to the coordinates ${𝐫}_i$.

The pairwise forces are slightly more difficult. The external forces came from a potential that only depended on the position of particle $i$. We expect the pairwise forces to come from a potential that can depend on two particles $i$ and $j$. We must also require that the forces satisfy Newton's third law of being equal and opposite, lying on the line connecting the two particles:

$${𝐅}_{ji}=-{𝐅}_{ij}=({𝐫}_i-{𝐫}_j)f,\text{(CM.0.27)}$$

where $f$ is a function, so that the directon of the force vector is set by the difference in position vectors. We need a potential function whose negative gradient with respect to either set of coordinates $i$ or $j$ produces this type of force. If the potential is a function of only the magnitude of the separation vector ${𝐫}_{ij}={𝐫}_i-{𝐫}_j$ between the particles, then it gives the right form for the forces. Start by taking the gradient using the chain rule:

$$-{\mathbf{\nabla }}_i{V}_{ij}(|{𝐫}_i-{𝐫}_j|)=-V{\prime }_{ij}(|{𝐫}_i-{𝐫}_j|){\mathbf{\nabla }}_i|{𝐫}_i-{𝐫}_j|.$$

We can calculate the gradient of the length of the separation vector using its definition. In three dimensions, we know

$$|{𝐫}_i-{𝐫}_j|=\sqrt{(x_i-x_j{)}^2+(y_i-y_j{)}^2+(z_i-z_j{)}^2},\text{(CM.0.28)}$$

so that, for example, the derivative with respect to $x_i$ gives $(x_i-x_j)/(\sqrt{\dots })$. Therefore, we have shown that

$${\mathbf{\nabla }}_i|{𝐫}_i-{𝐫}_j|=\frac{{𝐫}_i-{𝐫}_j}{|{𝐫}_i-{𝐫}_j|},\text{(CM.0.29)}$$

so that we have proven what we wanted to show, namely that the force resulting from the pairwise potential has the form of Equation CM.0.27

$$-{\mathbf{\nabla }}_i{V}_{ij}(|{𝐫}_i-{𝐫}_j|)=-V{\prime }_{ij}(|{𝐫}_i-{𝐫}_j|)\frac{{𝐫}_i-{𝐫}_j}{|{𝐫}_i-{𝐫}_j|}=({𝐫}_i-{𝐫}_j)f.\text{(CM.0.30)}$$

Incidentally, Equations CM.0.29 and CM.0.30 still hold if we replace ${\mathbf{\nabla }}_i$ with $-{\mathbf{\nabla }}_j$ or ${\mathbf{\nabla }}_{ij}$ (the gradient with respect to the coordinates ${𝐫}_{ij}$), which can be proved using the same argument. Note that this result implies that the potentials $V_{ij}$ and $V_{ji}$ are not completely independent, they must be related so that the forces resulting from their gradients are the same.

Now that we have the correct form of the potential function for the pairwise forces, we can try to integrate the pairwise forces' contribution to the work, from Equation CM.0.23. We cannot directly integrate each term separately, because

$$\int {𝐅}_{ji}·d{𝐫}_i=-\int {\mathbf{\nabla }}_i{V}_{ij}(|{𝐫}_i-{𝐫}_j|)·d{𝐫}_i$$

is not the integral of the derivative of a function of the single position ${𝐫}_i$. Instead, we should group pairs of terms. The ${𝐅}_{ij}$ term will give

$${𝐅}_{ij}=-{\mathbf{\nabla }}_j{V}_{ij}(|{𝐫}_i-{𝐫}_j|)=+{\mathbf{\nabla }}_i{V}_{ij}(|{𝐫}_i-{𝐫}_j|)=-{𝐅}_{ji},$$

just the negative of the force ${𝐅}_{ji}$. However, the force ${𝐅}_{ji}$ is dotted with the infinitesimal line element $d{𝐫}_i$, while the force ${𝐅}_{ij}$ is dotted with the line element $d{𝐫}_j$. So, the combined $i$ and $j$ terms of the sum in the work integral is

$$\underset{1}{\overset{2}{\int }}{𝐅}_{ji}·d{𝐫}_i+{𝐅}_{ij}·d{𝐫}_j=-\underset{1}{\overset{2}{\int }}{\mathbf{\nabla }}_i{V}_{ij}(r_{ij})·(d{𝐫}_i-d{𝐫}_j).\text{(CM.0.31)}$$

The differential $d$ is linear, so $(d{𝐫}_i-d{𝐫}_j)=d{𝐫}_{ij}$, and we already proved that ${\mathbf{\nabla }}_i{V}_{ij}={\mathbf{\nabla }}_{ij}{V}_{ij}$. We now have what we wanted, the integral of the derivative of one vector variable, ${𝐫}_{ij}$. Evaluating the integral for each pair and summing gives

$$\sum _{i,j}\underset{1}{\overset{2}{\int }}{𝐅}_{ji}·d{𝐫}_i=\sum _{i<j}{V}_{ij}(1)-V_{ij}(2),\text{(CM.0.32)}$$

where the sum bounds on the right-hand side make sure that we only count the pairs once. If we wanted to extend the sum to all values of $i$ and $j$, we can use the symmetry of $V_{ij}$ to see that $ij$ and $ji$ terms would be equal. Therefore, extending the sum overcounts by a factor of 2, which we can cancel by a factor of 1/2:

$$\sum _{i<j}{V}_{ij}(1)-V_{ij}(2)=\frac{1}{2}\sum _{i,j}{V}_{ij}(1)-V_{ij}(2).$$

Finally, we can see that defining the total potential energy as

$$V=\sum _i{V}_i+\frac{1}{2}\sum _{i,j}{V}_{ij}\text{(CM.0.33)}$$

allows us to conclude that the total energy $E=T+V$ is conserved, just as before.

Optional: Total kinetic energy in terms of the center of mass. As with the angular momentum, we can rewrite the total kinetic energy in terms of the center of mass's motion. To do so, we use the previously defined relationship for velocity relative to the center of mass, ${𝐯}_i={𝐰}_i+𝐕$, and just calculate $v_i^2$:

$$v_i^2=({𝐰}_i+𝐕)·({𝐰}_i+𝐕)=w_i^2+V^2+2{𝐰}_i·𝐕.$$

To find the total kinetic energy, we multiply by $m_i/2$ and sum over $i$:

$$T=\sum _i\frac{m_i{v}_i^2}{2}=\frac{1}{2}\sum _i{m}_i{w}_i^2+\frac{1}{2}\sum _i{m}_i{V}^2+\sum _i{m}_i{𝐰}_i·𝐕.$$

By commuting the mass-weighted sum through the time derivative and the dot product, we can see that the final term is zero by the same arguments we made earlier about the position of the center of mass in the $\mathbf{\rho }$ coordinates being zero, so we find that the total kinetic energy is

$$T=\frac{1}{2}M{V}^2+\frac{1}{2}\sum _i{m}_i{w}_i^2.\text{(CM.0.34)}$$

The total kinetic energy is the kinetic energy of the center of mass plus the sum of the kinetic energy of each particle about the center of mass. Again, if $V=0$, then the first term disappears. So, in each coordinate frame where the center of mass has no velocity, the kinetic energy is the same, just the kinetic energy measured in the frame where the center of mass is at the origin.

CM.0.3 References

1. Newton's laws of motion. https://www1.grc.nasa.gov/beginners-guide-to-aeronautics/newtons-laws-of-motion/.
2. H. Goldstein. Classical Mechanics. Addison-Wesley series in advanced physics. Addison-Wesley Press, 1950.

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