# Kinetic Energy in the Center of Mass Frame

Given two bodies moving along a line, I can find the velocity of the center of mass frame. Taking the time derivative of the Galilean transformation I get

$$v'=v-v_{cm}$$

By definition, the total momentum of the center of mass frame is $$0$$. From this, I can then find the velocity of the center of mass frame.

\begin{align*} 0 &= \sum m_iv'\\ &= \sum m_i(v_i-v_{cm}) \\ &= \sum (m_iv_i-m_iv_{cm}) \\ &= \sum (m_iv_i)-v_{cm}\sum m_i \\ v_{cm} &= \frac{\sum m_iv_i}{\sum m_i} \end{align*}

Now, I am now trying to find the relationship between the total kinetic energy of the particles in the original frame given by $$T$$ and the total kinetic energy in the center of mass frame $$T_{cm}$$. The original kinetic energy is

$$T = \sum\frac{1}{2}m_iv_i^2$$

and I believe the kinetic energy in the center of mass frame is given by

\begin{align*} T_{cm} &= \sum\frac{1}{2}m_i(v_i-v_{cm})^2\\ &= \sum\frac{1}{2}m_i(v_i^2-2v_iv_{cm}+v_{cm}^2)\\ &= \sum\frac{1}{2}m_iv_i^2 - \sum m_iv_iv_{cm} + \sum\frac{1}{2}m_iv_{cm}^2 \end{align*}

and substituting back in (and rearranging) gives

$$T = T_{cm} + \sum m_iv_iv_{cm} - \sum\frac{1}{2}m_iv_{cm}^2$$

However, this does not match the correct solution of

$$T = T_{cm} + \sum\frac{1}{2}m_iv_{cm}^2$$

I believe I am conceptually missing something when I am setting up the total kinetic energy in the center of mass frame, but I can't figure it out for the life of me. Does anybody have a hint as to what I am doing incorrectly?

Starting with

$$T = T_{cm} + \sum m_iv_iv_{cm} - \sum\frac{1}{2}m_iv_{cm}^2$$

use

$$v_{cm} = \frac{\sum m_iv_i}{\sum m_i}$$

which can also be written as

$$\sum m_iv_i = v_{cm}\sum m_i$$

and substitute back in to get

$$T = T_{cm} + \sum\frac{1}{2}m_iv_{cm}^2$$