I have a question regarding the usage of the centre-of-mass definition (which I thought required the total momentum, $p_T=0$) in the solution to the question given below:

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.

The solution says:

In the fixed target experiment, the total energy $E_T=E_1+m_2c^2$ and the total momentum magnitude is $p_Tc=\sqrt{E_1^2-m_1^2c^4}$. The square of the centre-of-mass energy is, therefore, $$s=m_T^2c^4=E_T^2-p_T^2c^2=E_1^2+2E_1m_2c^2+m_2^2c^4-E_1^2+m_1^2c^4$$ $$=2E_1m_2c^2+m_1^2c^4+m_2^2c^4$$ Rearranging, gives the required particle energy: $$E_1=\frac{s-m_1^2c^4-m_2^2c^4}{2m_2c^2}$$

But, in the centre-of-mass frame $p_T=0$, as shown in this image* below:


So therefore,

$$p_T=\sqrt{E_1^2-m_1^2c^4}=0$$ and as a result, $$E_1=m_1c^2$$

My question is, why is the author using the centre-of-mass definition (which I thought required zero total momentum) when the total momentum is actually non-zero (only one particle is stationary)?

*Image in body is from ICL dept. of Physics


The author explicitly states

We will do the calculation in a so-called fixed-target configuration.

So the calculation is done in what is usually called the laboratory frame, and calls it fixed-target in order to emphasize that the target is at rest in this frame.

Because of Lorenz transofrmations, any inertial frame can be used as long as energy and momentum are conserved in the given frame.

The center of mass is used only in order to define s

where s is the square of the centre-of-mass energy.

The calculation is in the lab. s is the "length" of the sum of the four vectors describing the particles involved, for any inertial frame, this is invariant for all inertial frames.

  • $\begingroup$ Thank you for your answer, I'm still quite confused. It was my understanding that the relation $s=m_T^2c^4$ can only be used in the centre of mass frame, this is what is alluded to in that image I posted in my question and in the link to Wikipedia under special relativity it states that in the relation $ m_0{}^2 =\left(\frac{E}{c^2}\right)^2-\left(\frac{p}{c}\right)^2$ the momentum part disappears. Is there some difference between centre of mass energy and centre of mass frame? $\endgroup$ – Electra May 9 '20 at 4:45
  • $\begingroup$ It is all in the four vectors. The image gives you the definition of invariant mass ( mass same in all inertial frames, the length of the four vector describinga single particle see the link I give) which is mathematically the same as the energy in the center of mass if you have more than one particle, and it defines S fpr a system of particles. It is all in the energy momentum four vectors and the algebra of four vectors $\endgroup$ – anna v May 9 '20 at 5:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.