# General Lorentz boost of four-momentum in CM frame, particle physics

In particle physics, we observe a scattering of the type: $$a+b \rightarrow c+d$$

Known quantities in the LAB frame are a, b and c. I want to transform c into the CM frame of the initial state and then investigate the $$\cos(\theta)_{CM}$$ distribution. How would I apply the Lorentz transformation to energy-momentum four vectors in natural units, generally speaking?

Other posts that I've found only cover special cases where the transformation is along one of the axes. I would like to know what the general form of the corresponding Lorentz matrix is, considering none of the momentum components in the LAB frame are zero.

Also, other posts often argue with the velocity between reference frames - I'm not sure how I would apply this concept to energy-momentum four vectors, since not velocities are given, but momenta. How would I calculate $$\gamma$$ and $$\beta$$? I assume I would start with calculating the invariant mass of the initial state and setting it equal between LAB and CM frame.

I apologize if this is considered a redundant post, but what I've found so far is not helpful enough.

• This and this not helpful? Feb 7 at 17:49

In the future, please try to ask one question per post. This question has several parts, so I will address each one briefly.

How would I apply the Lorentz transformation to energy-momentum four vectors in natural units, generally speaking?

The Lorentz transform, $$\Lambda_\mu^{\mu'}$$, transforms energy-momentum four-vectors in the same way as it transforms spacetime coordinates: $$(E,\vec p) = P^\mu$$$$P^{\mu'}=\Lambda_{\mu}^{\mu'} P^\mu$$

I would like to know what the general form of the corresponding Lorentz matrix is, considering none of the momentum components in the LAB frame are zero.

The Lorentz matrix for a boost in an arbitrary direction is given by: $$\Lambda_\mu^{\mu'} = \begin{bmatrix} \gamma &-\gamma v_x/c &-\gamma v_y/c &-\gamma v_z/c \\ -\gamma v_x/c&1+(\gamma-1)\dfrac{v_x^2} {v^2}& (\gamma-1)\dfrac{v_x v_y}{v^2}& (\gamma-1)\dfrac{v_x v_z}{v^2} \\ -\gamma v_y/c& (\gamma-1)\dfrac{v_y v_x}{v^2}&1+(\gamma-1)\dfrac{v_y^2} {v^2}& (\gamma-1)\dfrac{v_y v_z}{v^2} \\ -\gamma v_z/c& (\gamma-1)\dfrac{v_z v_x}{v^2}& (\gamma-1)\dfrac{v_z v_y}{v^2}&1+(\gamma-1)\dfrac{v_z^2} {v^2} \end{bmatrix}$$

I'm not sure how I would apply this concept to energy-momentum four vectors, since not velocities are given, but momenta. How would I calculate γ and β?

With the energy-momentum four-vector $$P^\mu$$ it is straightforward to obtain velocity 3-vector, $$\vec v$$, as follows: $$P^\mu = (E,\vec p)$$$$\vec v = \frac{c^2 \vec p}{ E}$$ Once you have $$\vec v$$ then you can calculate $$\beta=v/c$$ and $$\gamma=1/\sqrt{1-\beta^2}$$ as usual.

So, to transform into the center of momentum frame, simply calculate $$a+b$$ and then use that to calculate $$\vec v$$ of the center of momentum. Use that $$v$$ to transform to the frame where the center of momentum is at rest.

• Excellent answer, many thanks! Feb 9 at 23:13