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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?

Electra
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  • 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 $c=1$. – Frederic Thomas Oct 01 '20 at 15:49

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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.)

drfk
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  • Thanks for your answer, I will try this tomorrow, will it lead the result: $E_1=\frac{s-{m_1}^2c^4-{m_2}^2c^4}{2m_2c^2}$? – Electra Sep 28 '20 at 21:04
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    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 Sep 28 '20 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? – Electra Sep 28 '20 at 22:58
  • Well, you haven't been consistent, as I tried to explain in my answer. You used $P_i = (E_i, \mathbf{p}_i c)$ (which has the unit energy). But then in the derivation where you squared $P_i$ you had $P_i^2 = m_i^2 c^2$ which has the unit momentum-squared. Just look at your definition of $P_2$ and then what you wrote in the derivation (highlited in blue). – drfk Sep 28 '20 at 23:03
  • I thought that the $m_i^2c^2$ was the invariant mass? Sorry to ask, but which are the correct dimensions then? Does that mean they are not being consistent in this question also? I'm really confused now. – Electra Sep 28 '20 at 23:14
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    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 Sep 28 '20 at 23:57
  • Thank you so much for your help so far; I just have one more question for you: Is $s$ always equal to ${m_T}^2c^4$ and not ${m_T}^2c^2$, ie. not convention dependent? – Electra Sep 29 '20 at 20:53
  • $\sqrt{s}$ is the CM energy. As an energy it must have the dimensions of mass times velocity-squared. Again, I would strongly suggest that you take care that all quantities you use have the units you would expect and if not see how the quantities would have to be redefined to be consistent. See that a momentum has the c-factors it needs to have the units of momentum. For an energy, see that the c-factors result in an energy. As you are mixing conventions you need to check that yourself (it is also very common in textbooks having to do with relativity to set $c=1$, so be wary of that). – drfk Sep 29 '20 at 21:23
  • Sorry, but .. . . . . . . Is $s$ always equal to ${m_T}^2c^4$ and not ${m_T}^2c^2$? – Electra Sep 29 '20 at 21:30
  • The reason I'm desperate for an answer to that last comment is because of this question; if you look at the answers and comments to that question, you will see that no-one seems to know what $s$ is. – Electra Sep 29 '20 at 21:53
  • If $s$ is the center-of-mass energy-squared, then yes, it is $m^2 c^4$ as only this has the unit of an energy-squared. But everyone could define it differently (and your citations show that many do), as long as it is internally consistent. That is why I said, you have to check for yourself, that every quantity has the unit you expect it to have. You can not simply read different sources and combine parts of definitions from one source with calculations from other sources using different convention. There is simply no absolute rule, apart from it it having to be internally consistent. – drfk Sep 29 '20 at 22:11
  • Thanks for explaining that :) – Electra Sep 29 '20 at 22:33
  • Hi, please see my last edit, is it plausible? Many thanks ^^ – Electra Sep 29 '20 at 23:58
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    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 Sep 30 '20 at 00:33
  • So put simply, you are saying that the relation $E_1=\frac{s-{m_1}^2c^4-{m_2}^2c^4}{2m_2c^2}$ cannot be derived using four vectors (which is what this question was all about)? Is there a reason why you didn't tell me this sooner? – Electra Sep 30 '20 at 01:37
  • I am absolutely not saying that at all. Every kinematic relation in Minkowski space can be derived using 4-vectors. I feel I'm being trolled. If not, I'm sorry I couldn't help you. – drfk Sep 30 '20 at 15:09
  • You are no being trolled, I am very sorry if it seemed that way, please accept my sincerest apologies. Have you figured out a way to reach $E_1=\frac{s-{m_1}^2c^4-{m_2}^2c^4}{2m_2c^2}$ using 4-vectors? I will try to limit how often I ask questions to you. – Electra Sep 30 '20 at 20:22