# Lorentz transformation lab to CM system

1. Feb 17, 2008

### WarnK

Which Lorentz transformation takes the lab system to the CM system?

Lab system: $$p_a = (E^{lab}_a, \vec{p}_a)$$ and $$p_b = (m_b, \vec{0})$$
CM system: $$p_a = (E^{CM}_a, \vec{p})$$ and $$p_b = (E^{CM}_b, -\vec{p})$$

For a binary reaction a+b->c+d, the textbooks I have say quite a lot about the kinematics of such reactions (Mandelstam variables and all that) but how do I go about doing a Lorentz transformation between the lab and CM system? That must be possible to do.

2. Feb 18, 2008

### WarnK

Anyone?

3. Feb 18, 2008

### Rainbow Child

The velocity of CM in the scattering of two particles with rest mass $m_1$ and $m_2$ is given by

$$\vec{u}_{CM}=\frac{\vec{p}_1+\vec{p}_2}{E_1+E_2}\,c^2$$

from which you can find the Lorentz transformation.

4. Feb 19, 2008

### WarnK

So a lorentz transformation from the lab frame to the CM frame would be
$$\left[ \begin{array}{cccc} \gamma & -\beta \gamma & 0 & 0 \\ -\beta \gamma & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right]$$
with $$\beta = ||\vec{u_{CM}}||$$ and $$\gamma = (1-\beta^2)^{-1/2}$$?
It doesn't make any sense to me.

5. Feb 19, 2008

### Rainbow Child

The crusial thing is to make sense in nature, not to any of us.
First system of center of mass for two particles is defined by the requirement, that $(\vec{p}_1+\vec{p}_2)_{CM}=0$. The general Lorentz transormation $(c=1)$ is

$$\Lambda=\begin{pmatrix} \gamma& \gamma\,\vec{\beta} \cr \gamma\,\vec{\beta} & \delta_{\alpha\beta}+\frac{\gamma-1}{\beta^2}\,\beta_\alpha\,\beta_\beta \end{pmatrix}$$

Take now the 4-vector $p^\alpha_1=(E_1,\vec{p}_1)$ and apply $\Lambda$ with $\beta=\frac{\vec{p}_1+\vec{p}_2}{E_1+E_2}$, to get for the 3-momentum $\vec{P}_1,\,\vec{P}_2$ to the CM system:

$$\vec{P}_1=\Lambda\,\vec{p}_1\Rightarrow \vec{P}_1=\gamma\,\left(E_1+\frac{\gamma}{\gamma+1}\,\vec{\beta}\,\vec{p}_1\right)\,\vec{\beta}+\vec{p}_1[/itex] [tex]\vec{P}_2=\Lambda\,\vec{p}_2\Rightarrow \vec{P}_2=\gamma\,\left(E_2+\frac{\gamma}{\gamma+1}\,\vec{\beta}\,\vec{p}_2\right)\,\vec{\beta}+\vec{p}_2[/itex] Adding now the two equations we get [tex]\vec{P}_1+\vec{P}_2=\gamma\,\left(E_1+E_2+\frac{\gamma}{\gamma+1}\,\vec{\beta}\,(\vec{p}_1+\vec{p}_2)\right)\,\vec{\beta}+\left(\vec{p}_1+\vec{p}_2\right)\Rightarrow \vec{P}_1+\vec{P}_2=\gamma\,\left(E_1+E_2+\frac{\gamma}{\gamma+1}\,\vec{\beta}\,(E_1+E_2)\,\vec{\beta}\right)\,\vec{\beta}+\left(E_1+E_2\right)\,\vec{\beta}\Rightarrow$$
$$\vec{P}_1+\vec{P}_2=\gamma\,(E_1+E_2)\,\left(1+\frac{\gamma}{\gamma+1}\,\beta^2\right)\,\vec{\beta}+\left(E_1+E_2\right)\,\vec{\beta}\Rightarrow \vec{P}_1+\vec{P}_2=0$$

Now does it makes sense?