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So this is from Griffiths particle physics book about pion decaying into neutrino and muon.

The part I have problem with is that it states ``` \begin{equation} p_{\pi}^2=m_{\pi}^2c^2 , p_{\mu}^2=m_{\mu}^2c^2 \end{equation} as a true statement for the reaction. I am not sure if p means energy-momentum four vector or just momentum. But in any case I do not know how this is true?

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Using natural units $c=1$ what that relation implies is that the particles are on-shell. Whenever a particle is on-shell you have that, given the particle 4-momentum $p_\mu$, $$p_\mu p^\mu \equiv p^2 = m^2$$ where $m$ is the particle's mass.

If you put back the velocity of light, you'll get $$p_\mu p^\mu \equiv p^2 = m^2c^4$$

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An on-shell particle of mass $m$ and velocity ${\bf v}$ has a four-velocity $u_\mu = \gamma(c,{\bf v})$, with $\gamma=(1-{\bf v}^2/c^2)^{-1/2}$. Its relativistic momentum $p_\mu$ is then given by the expression $p_\mu = mu_\mu$. The Lorentz invariant magnitude of $p_\mu$ can then be calculated to be $$ p^2 \equiv \eta^{\mu\nu} p_\mu p_\nu = (p_0)^2 - (p_1)^2 - (p_2)^2 - (p_3)^2 = m^2c^2, $$ assuming the mostly-negative convention $\eta^{\mu\nu}={\rm diag}(1,-1,-1,-1)$.

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$p^2$ means $p^\mu p_\mu=p_0^2-p_x^2-p_y^2-p_z^2$.

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Usually, one denotes $p$ for the 4-vector, $\vec{p}$ for the 3-vector and $p_i$ or $p^\mu$ for the components.

So this is just the basic $p^2 = (mc)^2 = \frac{E^2}{c^2}-\vec{p}^2$ relation that one gets from $p^\mu = (\frac{E}{c}, \vec{p})$ and basically defines mass.

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