I am trying to learn particle scattering. In the textbook, it says that we can derive
d(V1,V2)/d(Vg,V) = 1
from Center of Mass Coordinate formula:
V1 = Vg + m1/(m1+m2) * V
V2 = Vg - m1/(m1+m2) * V
Vg := d(Rg)/dt := Rg
V := dR/dt := R
Rg := (m1R1 + m2R2) / m1+m2
R := R1-R2
R should be small r with an arrow on top but I don't know how to put it here
I don't understand how can we get the first equation equal to 1. Does anyone know what it means?
Thank in advance
I'm guessing the complete set of equations in your book is the following: \begin{align} \vec{v}_1&=\frac{d\vec{r}_1}{dt}\\ \vec{v}_2&=\frac{d\vec{r}_2}{dt}\\ \vec{r}&=\vec{r}_1-\vec{r}_2\\ \vec{v}&=\frac{d\vec{r}}{dt}\\ \vec{r}_g&=\frac{m_1\vec{r}_1+m_2\vec{r}_2}{m_1+m_2}\\ \vec{v}_g&=\frac{d\vec{r}_g}{dt}\\ \end{align} It would follow from these that \begin{align} \vec{v}_1&=\vec{v}_g+\frac{\color{red}{m_2}}{m_1+m_2}\vec{v}\\ &\left(\textit{ not }\ \vec{v}_g+\frac{\color{red}{m_1}}{m_1+m_2}\vec{v}\ \text{ as you have given it}\right)\\ \vec{v}_2&=\vec{v}_g-\frac{m_1}{m_1+m_2}\vec{v}\ . \end{align} The quantity you want to evaluate is almost certainly the Jacobian \begin{align} \frac{\partial(\vec{v}_1,\vec{v}_2)}{\partial(\vec{v}_g,\vec{v})}&=\det\pmatrix{\frac{\partial v_{11}}{\partial v_{g1}}&\frac{\partial v_{11}}{\partial v_{g2}}&\frac{\partial v_{11}}{\partial v_1}&\frac{\partial v_{11}}{\partial v_2}\\ \frac{\partial v_{12}}{\partial v_{g1}}&\frac{\partial v_{12}}{\partial v_{g2}}&\frac{\partial v_{12}}{\partial v_1}&\frac{\partial v_{12}}{\partial v_2}\\ \frac{\partial v_{21}}{\partial v_{g1}}&\frac{\partial v_{21}}{\partial v_{g2}}&\frac{\partial v_{21}}{\partial v_1}&\frac{\partial v_{21}}{\partial v_2}\\ \frac{\partial v_{22}}{\partial v_{g1}}&\frac{\partial v_{22}}{\partial v_{g2}}&\frac{\partial v_{22}}{\partial v_1}&\frac{\partial v_{22}}{\partial v_2}}\\ &=\det\pmatrix{1&0&\frac{m_2}{m_1+m_2}&0\\ 0&1&0&\frac{m_2}{m_1+m_2}\\ 1&0&\frac{-m_1}{m_1+m_2}&0\\ 0&1&0&\frac{-m_1}{m_1+m_2}} \end{align} If you subtract the first row of this determinant from its third, and its second from its fourth (which doesn't change its value), you get $$ \det\pmatrix{1&0&\frac{m_2}{m_1+m_2}&0\\ 0&1&0&\frac{m_2}{m_1+m_2}\\ 0&0&-1&0\\ 0&0&0&-1}\ . $$ Then adding $\ \frac{m_2}{m_1+m_2}\ $ times the third row to the first and $\ \frac{m_2}{m_1+m_2}\ $ times the fourth row to the second (again, this will not change its value) you get $$ \det\pmatrix{1&0&0&0\\ 0&1&0&0\\ 0&0&-1&0\\ 0&0&0&-1}=1\ . $$