Question

# A certain particle of mass m has momentum of magnitude mc. What is $\beta$?

Solution

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Step 1
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$\textbf{(a)}$ In a relativistic limit the momentum for a particle is given by:

\begin{align} p=\gamma mv\end{align}

the Lorentz factor in terms of the speed parameter is given by:

$\gamma=\dfrac{1}{\sqrt{1-\beta^2}}$

substitute into (1) to get (note that $v=\beta c$):

$p=\dfrac{m\beta c}{\sqrt{1-\beta^2}}$

for a particle with momentum of $p=mc$, we get:

$\dfrac{m\beta c}{\sqrt{1-\beta^2}}=mc$

$\beta=\sqrt{1-\beta^2}$

$\beta=\dfrac{1}{\sqrt{2}}=0.707$

$\boxed{\beta=0.707}$

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