Irodov 1.148: Energy Splits into Internal and Bulk – Amit & Deepali Sinha Foundation

Problem Irodov 1.148

A system of particles has total mass $m$ . In the frame where its center of inertia is permanently at rest, the system’s total energy (the kinetic energy of the particles’ motion relative to that center, plus any internal potential or interaction energy) equals $\tilde E$ . That center-of-inertia frame translates, as a whole and at constant velocity $\vect V$ , relative to a second inertial frame $K$ . Find the total energy $E$ of the same system of particles as measured in frame $K$ .

Left: three point masses inside a schematic boundary, center of inertia at rest, total mass m and total energy E. Right: the same system in frame K, where the center of inertia moves with velocity V. — Left: three point masses inside a schematic boundary, center of inertia at rest, total mass $m$ and total energy $\tilde E$ . Right: the same system in frame $K$ , where the center of inertia moves with velocity $\vect V$ .

Given

$m$
$\tilde E$
$\vect V$

Find

E

, the total energy of the system in frame

K

Solution

Step 1Add the frames’ velocities and expand the kinetic energy in

K

Every particle’s velocity in $K$ is its velocity in the center-of-inertia frame plus the one common velocity $\vect V$ of that frame, by ordinary Galilean addition. Writing the system’s kinetic energy in $K$ as a sum over its particles and substituting this relation expands each term into three pieces.

\vect v_i = \vect v_i' + \vect V

\begin{aligned} T &= \sum_i \frac12 m_i\,\vect v_i\cdot\vect v_i\\ &= \sum_i \frac12 m_i\bigl(\vect v_i'+\vect V\bigr)\cdot\bigl(\vect v_i'+\vect V\bigr)\\ &= \sum_i \frac12 m_i\vect v_i'\cdot\vect v_i' \;+\; \vect V\cdot\sum_i m_i\vect v_i' \;+\; \frac12 V^2\sum_i m_i \end{aligned} \tag{1}

Step 2The cross term vanishes by the definition of the center-of-inertia frame

The primed frame is, by hypothesis, the frame in which the center of inertia sits still, so the system’s total momentum measured there is zero. Substituting this, and $\sum_i m_i=m$ , into the expansion (1) removes the middle term entirely.

\sum_i m_i\vect v_i' = \vect 0

\begin{align*}T &= \sum_i \frac12 m_i\vect v_i'\cdot\vect v_i' \;+\; \vect V\cdot\vect 0 \;+\; \frac12 mV^2\end{align*}

The surviving sum is exactly the kinetic energy $T'$ measured in the center-of-inertia frame, plus one extra term carrying $V$ .

T = T' + \dfrac{1}{2}mV^2

(2)

Step 3The internal energy is the same in both frames

Writing $U$ for whatever part of the total energy is not translational kinetic energy (interaction potential energy, internal binding or thermal energy), the total energy in each frame is kinetic plus this frame-independent $U$ .

\begin{aligned} E &= T+U\\ \tilde E &= T'+U \end{aligned} \tag{3}

Step 4Combine to get

E

Subtracting the second line of (3) from the first and inserting (2) isolates $E$ in terms of the given quantities.

\begin{align*}E-\tilde E &= T-T'\\ &= \frac12 mV^2 \qquad \text{by \text{(2)}}\end{align*}

E = \tilde E + \dfrac{1}{2}mV^2

(4)

Check

Dimensionally $[m][V^2]=(\text{kg})(\text{m/s})^2=\text{J}$ , an energy, matching $\tilde E$ and $E$ . As $V\to0$ the two frames coincide (the center of inertia is then at rest in $K$ too), and the formula correctly gives $E\to\tilde E$ with no leftover correction.

Remark

The derivation never used how many particles the system has or what forces bind them together, only Galilean velocity addition and the fact that momentum is zero in a system’s own center-of-inertia frame. So the same correction $mV^2/2$ , the kinetic energy the system would have if its whole mass sat at the center of inertia moving at $\vect V$ , carries any Newtonian system’s total energy between its center-of-inertia frame and any other inertial frame, whether that system is a handful of point particles, an extended rigid body, or a cloud of gas exchanging heat internally.