The upsilon is an unstable particle that decays into a tau and an antitau according to the reaction: enter image description

The upsilon is at rest relative to the laboratory when it decays. The momemtum of the tau relative to the laboratory is $4.40GeV c^{-1}$. The rest mass of the tau and of the anti tau is $1.78GeVc^{-2}$

Determine the rest mass of the upsilon.

First, why are masses of the leptons represented in terms of $GeVc^{-2}$? Lets say I want to express the mass of a proton in terms of $GeVc^{-2}$, how can I do this conversion?

According to the answer:

enter image description here

Honestly this is very confusing to me, why $Mc^2 = 2E$? and why is $E$ the square root of the mass squared and the momentum squared, how do you even formulate this?


1 Answer 1


Particle physicists like to express the masses, momenta, energies of particles in some form of electron volts and a power of c. (Yes, that metric propaganda in school lied to you..it is not always a perfect 10).

It makes a lot of sense, since the one place where it is clear that rest mass and energy are interchangeable is the interaction point / target end station of a particle accelerator.

The conversion from SI to practicing physicist is as follows:

$$ \frac{M}{[1 \rm GeV/c^2]} = \frac 1{10^9}\times \frac{Mc^2}{|e|\times 1\,{\rm V}} $$

where $e$ is the charge of the electron. Note that many theorists consider the speed of light to be a dimensionless constant:

$$ c = 1$$

so that one rarely says "G eV over c squared", preferring just, "GeV", pronounced "gee eee vee" in the States, and perhaps "jev" elsewhere.

Regarding your problem, this is solvable by conservation of 4-momentum. Setting $c=1$, the initial 4-momentum is:

$$ p_i^{\mu}= (M_{\Upsilon}, \vec 0)$$

While the final state is:

$$ p_f^{\mu}= p_{\tau}^{\mu} + p_{\bar{\tau}}^{\mu} = p_i^{\mu}$$

If we call the tau's 3-momenta $\vec p$, and enforce the conservation of 3-momenta, then:

$$p_{\tau}^{\mu} =(\sqrt{|\vec p|^2+m_{\tau}^2 }, \vec p)$$

$$p_{\bar \tau}^{\mu} =(\sqrt{|\vec p|^2+m_{\tau}^2 }, -\vec p)$$

From here, the answer should be straightforward.

  • $\begingroup$ Hi JEB thanks for the explanation! Really helpful ;) $\endgroup$
    – CountDOOKU
    Commented Sep 28, 2020 at 0:43

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