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I have been taught that in Classical Mechanics, the total energy of a system of two particles in the Centre of Mass Frame is given by

$$ E_\mathrm{total} = \frac{1}{2}MV^2 + \frac{1}{2}\mu v_r^2 $$

where $M$ is the total mass of the system, $\mathbf{V}$ is the velocity of the centre of mass, $\mu$ is the reduced mass and $\mathbf{v_r}$ is the separation velocity of the two particles.

Furthermore, I have been taught that the first term on the right hand side, $\frac{1}{2}MV^2$ represents kinetic energy of the centre of mass and the second term $\frac{1}{2}\mu v_r^2$ is the "energy to do Physics with" - that is, the energy which can be transfered to any particles that would exit a collision of the original particles.

My question is this: Is it possible to find analogous expressions for each of these terms in the relativistic case? According to wikipedia the total energy of the COM frame is given by

$$ E_\mathrm{tot} = Mc^2 $$

but is it possible to find an equivalent to this "energy to do Physics with?"

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  • $\begingroup$ Is $v_r$ the separation velocity with the lab velocities of the particles? If so, the KE in the center of mass frame does not include the $\frac{1}{2}MV^2$. $\endgroup$
    – Bill N
    Aug 8, 2015 at 23:08
  • $\begingroup$ Yes I have made a small edit to clarify this, also $v_\mathrm{r}$ is independent of the frame in the classical case. I think you will find that if you expand the first equation, that $E_\mathrm{total}$ as I have stated it is equal to the sum of the kinetic energies of the particles in the LAB frame. I think you are correct in saying that the KE in the COM frame is only $\frac{1}{2}mv_r^2$ but the total energy includes the $\frac{1}{2}MV^2$ corresponding to the energy carried by the motion of the centre of mass. $\endgroup$
    – mcgarc
    Aug 8, 2015 at 23:33
  • $\begingroup$ Yes, what you have is the lab frame KE. $\endgroup$
    – Bill N
    Aug 8, 2015 at 23:45

2 Answers 2

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Particle 1 and particle 2 each have relativistic momentum 4-vectors: $$\pmatrix{\frac{E_1}{c}\\p_{1x}\\p_{1y}\\p_{1z}}\text{ and } \pmatrix{\frac{E_2}{c}\\ p_{2x}\\p_{2y}\\p_{2z}}$$, so the total momentum 4-vector is $$\pmatrix{\frac{E_1+E_2}{c}\\p_{1x}+p_{2x}\\p_{1y}+p_{2y}\\p_{1z}+p_{2z}},$$ where $E_1=m_1c^2+K_1$ and $E_2=m_2c^2+K_2$.

These have squared magnitudes of $E_j^2/c^2-(p_{jx}^2+p_{jy}^2+p_{jz}^2)=(m_jc)^2$, where $j$ represents 1, 2, or the totals.

The squared magnitude of the total is $$(m_{CM}c)^2 = E_1^2/c^2+E_2^2/c^2+2E_1E_2/c^2-[(p_{1x}+p_{2x})^2+(p_{1y}+p_{2y})^2+(p_{1z}+p_{2z})^2]$$

Doing the algebra and substituting the $(m_1c)^2$ and $(m_2c)^2$ terms we get $$m_{CM}^2c^2 = m_1^2c^2 + m_2^2c^2 + 2(\frac{E_1E_2}{c^2}-\vec{p}_1\cdot\vec{p}_2)$$

Notice the mass in the center of mass is not simply the sum of the individual masses. You might think of this mass as "what you can do physics with" although one might interpret it differently.

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  • $\begingroup$ what would be the Lorentz transformation to the CM frame? $\endgroup$ Jul 25, 2022 at 14:19
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Say in the Lab frame the total 4-momentum of the system is p = (E/c, p), where E is the total energy and p is the total 3-momentum. Also let M be the total mass of the system. Like in non-relativistic mechanics, in Special Relativity the COM frame is the frame where the total 3-momentum is null, ${\it p_C} = (E_C/c, {\bf 0})$, ${\bf p}_C = {\bf 0}$. With this definition, $E_C$ is the system's energy in the COM frame and the equivalent of your "energy to do physics with", while the Lab frame total 3-momentum p is the COM 3-momentum, just as in the non-relativistic case. The relationship between $E_C$, $E$, and p follows from Lorentz invariance, which requires

$$ {\it p}^µ {\it p}_µ = {\it p}_C^µ {\it p}_{C,µ} = M^2 c^2 $$

Therefore $(E_C/c)^2 = (E/c)^2 - {\bf p}^2 = M^2 c^2$ or $ E = \sqrt{({\bf p}c)^2 + E_C^2}$ . Now define the kinetic energy $K_C$ of the COM as that part of the total energy in the Lab frame that is in excess of the system's energy in the COM frame: $$ K_C = E - E_C = \sqrt{({\bf p}c)^2 + E_C^2} - E_C $$

Conversely, this allows us to write $E = K_C + E_C$, which is the exact equivalent of the non-relativistic counterpart $E = MV^2/2 + E_C$. When $({\bf p}c)^2 << E_C^2 = (Mc^2)^2$, we get $K_C \approx [E_C + ({\bf p}c)^2/2E_C] - E_C = {\bf p}^2/2M$ and we retrieve $E = {\bf p}^2/2M + E_C$. Note that from a relativistic point of view, we have the mass-energy relationship $E_C = Mc^2$, but $E_C$ is actually the "energy to do physics with" in the COM frame. The last thing we need in order to describe the COM completely is the boost β from the COM frame to the Lab frame. We can find it if we write the Lab total 4-momentum p as the Lorentz transform by boost β of the COM 4-momentum, like this: ${\it p} = (E/c, {\bf p}) = (𝛾E_C/c, 𝛾{\bf β} E_C/c)$, where $𝛾 = 1/\sqrt{1-{\bf β}^2}$ as usual. This gives ${\bf β} = {\bf p}c/E$ and $E = 𝛾E_C$. Hope this helps.

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