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Problem of the week #24 - September 10th, 2012

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Jameson

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Jan 26, 2012
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The Lorentz factor is important in many calculations for special relativity. If we consider \(\displaystyle m = f(v) = \frac{m_0}{\sqrt{1-v^2/c^2}}\) what is the inverse of $f$?

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Jameson

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Jan 26, 2012
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Congratulations to the following members for their correct solutions:

1) Reckoner

Solution (from Reckoner):

Note that \(f: [0, c)\to[m_0, \infty)\) is a bijection (it is one-to-one and onto), so an inverse exists. To find the inverse \(f^{-1}: \left[m_0, \infty\right)\to[0, c)\) we may solve for \(v\) in terms of the dependent variable \(m\):

\[\begin{eqnarray}m = \frac{m_0}{\sqrt{1 - v^2/c^2}} & \Rightarrow & \sqrt{1 - \frac{v^2}{c^2}} = \frac{m_0}m\\&\Rightarrow& 1-\frac{v^2}{c^2} = \frac{m_0^2}{m^2}\\&\Rightarrow& v^2 = c^2\left(1 - \frac{m_0^2}{m^2}\right)\\&\Rightarrow& v = c\sqrt{1 - \frac{m_0^2}{m^2}}\\&\Rightarrow& f^{-1}(m) = c\sqrt{1 - \frac{m_0^2}{m^2}}\end{eqnarray}\]

We may verify that \(f\) and \(f^{-1}\) are inverses by observing that

\[\left(f\circ f^{-1}\right)(m) = \frac{m_0}{\sqrt{1 - \left(c\sqrt{1 - m_0^2/m^2}\right)^2/c^2}} = m\]

and

\[\left(f^{-1}\circ f\right)(v) = c\sqrt{1 - \frac{m_0^2}{\left(m_0/\sqrt{1 - v^2/c^2}\right)^2}} = v.\]
 
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