Question

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

Solution

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Answered 6 months ago
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(a)\textbf{(a)} In a relativistic limit the momentum for a particle is given by:

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

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

γ=11β2\gamma=\dfrac{1}{\sqrt{1-\beta^2}}

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

p=mβc1β2p=\dfrac{m\beta c}{\sqrt{1-\beta^2}}

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

mβc1β2=mc\dfrac{m\beta c}{\sqrt{1-\beta^2}}=mc

β=1β2\beta=\sqrt{1-\beta^2}

β=12=0.707\beta=\dfrac{1}{\sqrt{2}}=0.707

β=0.707\boxed{\beta=0.707}

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