# Heisenberg's Uncertainty Principle Question

stephen8686
Homework Statement:
Show that the smallest possible uncertainty in the position of an whose speed is given by $\beta=\frac{v}{c}$ is: $$\Delta x_{min}=\frac{h}{4\pi m_o c}\sqrt{1-\beta^2}$$
Relevant Equations:
$$\Delta x \Delta p=\frac{h}{4\pi}$$
So with the $\gamma=\frac{1}{\sqrt{1-\beta^2}}$ it seems obvious that relativistic momentum, $p=\gamma m_o v$ is supposed to be used.
Then $$\frac{ dp}{dv}=m_o(1-\beta^2)^{-1/2}+m_o v (\frac{-1}{2}(1-\beta^2)^{-3/2}(\frac{-2v}{c^2}))=m_o(\frac{1}{\sqrt{1-B^2}}+\frac{\beta}{(1-\beta^2)^{3/2}})=\frac{m_o}{(1-B^2)^{3/2}}$$
so $\Delta p=\frac{m_o\Delta v}{(1-B^2)^{3/2}}$

But this doesn't exactly fit the expression that I'm supposed to show. I don't know what do do with the $\Delta v$. In addition to this, I don't think I conceptually understand the question. Why should there be a $\Delta x_{min}$? Why can't the uncertainty in momentum be huge and the uncertainty in x be reduced to a dirac delta function?