There's a question on my homework about the process $e^{-} e^{+} \rightarrow \tau^{+} \tau^{-}$. Specifically, it is claimed that the minimum energy required of the colliding positron and electron beams is slightly less than twice the $\tau$ mass, and I am asked to explain this deviation and compute it. In the context of this course, we have only discussed the relativistic kinematics of such processes (which would predict a minimum energy of twice the $\tau$ mass), so I am not sure what might be responsible. I'm grasping at straws here, but might it have something to do with the Coulomb interaction between the opposite charges? Thanks in advance.

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    $\begingroup$ It may (or may not) be significant that particle physicists often use "energy" in this kind of context to mean the kinetic energy of the beam. Of course in the reaction in question the electrons will be in the ultra-relativistic regime so the difference is trivial, so that may not be what is intended. $\endgroup$ Commented Feb 12, 2014 at 20:20
  • $\begingroup$ Maybe the homework intends you not to count electron rest masses as required "energy" required. Perhaps, the idea here is that they're going for energy that an experimentalist would put into the two electrons to get the reaction to go, and that the energy from the electron's rest mass comes for free. $\endgroup$
    – QuantumDot
    Commented Feb 12, 2014 at 20:22
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    $\begingroup$ Another possibility. If the Tau-pair is produced almost at rest they could (in principle, I don't know if it is possible in practice) form a short-lived bound state at total energy just slightly less that $2m_\tau$. $\endgroup$ Commented Feb 12, 2014 at 20:28
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    $\begingroup$ I'm not sure, but I think maybe @mikefallopian may be on the right track with the Coloumb interaction. The Coloumb interaction between two electrons at the collision point is $\sim 2 \alpha m \sim 10keV$, not negligible. In the extreme case, if you for example released an electron positron from rest a short distance away from each other, I presume they could collide and produce new particles. $\endgroup$
    – JeffDror
    Commented Feb 12, 2014 at 20:30
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    $\begingroup$ I talked to my professor and what he is looking for is, in fact, the tau-antitau bound state having mass slightly less than $2m_\tau$ $\endgroup$ Commented Feb 13, 2014 at 14:39

1 Answer 1


Assuming that $\tau^-\tau^+$ can form a bound state similarly to positronium($e^-e^+$), all we need is the form of the ground state of positronium, specifically that it is proportional to the reduced mass of the pair: $\mu = \frac{m_1m_2}{m_1+m_2}$.

Knowing that positronium's ground state is $(-13.6/2)= -6.8$ eV and that the new reduced mass for the Tau particle bound state is simply $\frac{m_{Tau}}{2} = \frac{m_{Tau}}{m_e}\mu_{e^-e^+}$

It follows that the energy of the Tau particle bound state is this factor, $\frac{m_{Tau}}{m_e}$ multiplied by $-6.8$ eV. $(-6.8)(3477) = -2.36\times10^4$ eV


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