# Can energy momentum four-vectors be used to solve a relativistic kinematics problems when a centre of mass energy $\sqrt s$ is given? [closed]

This is a follow-up to this previous question. We are encouraged to use 4-vectors to solve these types of problems:

Decay kinematics using 4-vectors. Here we look at the kinematics of particle decays and how 4-vectors can simplify this type of problem quite considerably. The point here is to use energy-momentum 4-vectors.

I am trying to solve the same problem using the exact same recipe that was used to solve this problem using four-vectors, namely,

A particular centre-of-mass energy is needed to create a new particle. We will do the calculation in a so-called fixed-target configuration (a) and in a collider configuration (b).

(a) A particular centre-of-mass energy is needed to create a new particle. We will do the calculation in a so-called fixed-target configuration.

A particle of mass $$m_1$$ and total energy $$E_1$$ in the lab frame hits a stationary particle of mass $$m_2$$. Show that the required particle energy for a given $$s$$ is:

$$E_1=\frac{s-{m_1}^2c^4-{m_2}^2c^4}{2m_2c^2}$$ where $$s$$ is the square of the centre-of-mass energy. This is often called a ‘fixed target’ configuration as experiments were historically often done by colliding a beam of particles with a stationary target material.

Let $$P_1$$ be the four-momentum of incident particle 1.

Let $$P_2$$ be the four-momentum of stationary particle 2.

Let $$P_T$$ be the four-momentum of the centre of mass energy $$\sqrt{s}$$.

$$P_1=\left(E_1, {\bf p_1}c\right)=\left(E_1,p_1^{x}c,0,0 \right)\qquad\text{(as these problems are all 1D)}$$

$$P_2=\left(E_2, {\bf 0}\right)=\left(m_2c^2,0,0,0 \right)\qquad\text{(as particle 2 is the fixed-target)}$$

$$P_T=\left(\sqrt s,{\ p_Tc}\right)=\left(m_Tc^2, p_T^xc, 0, 0 \right)$$

In the center of momentum frame, the relevant 4-momentum conservation is

\begin{align}P_T &=P_1+P_2 \\&\implies {P_1}^2+2P_1\cdot P_2+{P_2}^2={{P_T}}^2\\&\implies \color{blue}{{m_1}^2c^2}-2E_1m_2c^2+\color{blue}{{m_2}^2c^2}=\color{blue}{{m_T}^2c^2}=\frac{s}{c^2}\end{align}

Where in the last step I have replaced the 4-vector norms by the invariant masses (marked in blue) and noting that $$s={m_T}^2c^4$$.

On rearrangement I get:

$$E_1=\frac{s-{m_1}^2c^4-{m_2}^2c^4}{2m_2c^4}$$

But this is not the correct result:

$$E_1=\frac{s-{m_1}^2c^4-{m_2}^2c^4}{2m_2c^2}$$

Am I making a mistake here? Or can this relation simply not be derived using 4-vectors?

If it can't be derived using 4-vectors then please explain why.

Many thanks.

EDIT:

I've just realised something: While $$P_1$$ & $$P_2$$ are four-vectors representing a physical particle, however, above I write that "Let $$P_T$$ be the four-momentum of the centre of mass energy $$\sqrt{s}$$." Is it really plausible to define a four-vector for a system of particles?

• It is actually only a dimension problem: The 4-vectors $P_1$ and $P_2$ have the dimension of energy whereas the should have the dimension of momentum. Upon squaring $P_1$ and $P_2$ they change the dimension. Upon computing $P_T^2$ a $c^2$ is forgotten. One has to be careful to correctly bookkeeping the $c^2$ are simply setting $c=1$. Commented Oct 1, 2020 at 15:49

You define the components of $$P_i$$ to be energies $$E_i$$ or $$\mathbf{p}_i c$$. But when squaring you get the unit of momentum squared, $$P_i^2 = m_i^2 c^2$$.

With other words: Either define your 4-vectors such that the components are momenta, or make sure that your $$P^2$$ is an energy squared. (Later you won't care anymore anyway, since you will be using the convention $$c = 1$$ and everything is energy.)

• Thanks for your answer, I will try this tomorrow, will it lead the result: $E_1=\frac{s-{m_1}^2c^4-{m_2}^2c^4}{2m_2c^2}$? Commented Sep 28, 2020 at 21:04
• Yes, it gives the correct result. Just use $P_i = (E_i/c, \mathbf{p}_i)$, i.e. units of momentum for the components of 4-momentum. I would in general suggest to try to always check if all terms in an equation have the same unit. It's a fast way to quickly check ones computations for easy mistakes.
– drfk
Commented Sep 28, 2020 at 21:45
• Thanks for explaining, the strange thing is that my relativity lecturer told me that it doesn't matter whether $P_i = (E_i/c, \mathbf{p}_i)$ or $P_i = (E_i, \mathbf{p}_ic)$ is used, just as long as I'm consistent throughout a calculation. Why is that not the case here? Commented Sep 28, 2020 at 22:58
• Yes, in that question you are being inconsistent as soon as you use $P_i^2 = m_i^2 c^2$ in your "Update"-section. I would suggest you use $P = (E/c, \mathbf{p})$ as the 4-momentum (thus having unit momentum). Now, per definition of the scalar product for 4-vectors $P^2 = P_0^2 - P_1^2 - P_2^2 - P_3^2$. Surely, $P^2$ must then have the same dimension as the squared components of $P$. Thus, if the components have unit momentum, then $P^2$ must have unit momentum-squared (i.e. $m^2c^2$). (If you use components of unit energy, then $P^2$ must have unit energy-squared (i.e. $m^2 c^4$.)
– drfk
Commented Sep 28, 2020 at 23:57
• What is it supposed to mean to be the "four-momentum of the center-of-mass energy"? Makes no sense to me. $s$ is simply defined as $s = (P_1 + P_2)^2 c^2$, i.e. the sum of the 4-momenta of all particles in the scattering process. The components you gave for $P_T$ are not correct, but $P_T^2 = s/c^2$ is correct.
– drfk
Commented Sep 30, 2020 at 0:33