- #1

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- Homework Statement:
- Show that the smallest possible uncertainty in the position of an whose speed is given by [itex]\beta=\frac{v}{c}[/itex] is: [tex]\Delta x_{min}=\frac{h}{4\pi m_o c}\sqrt{1-\beta^2}[/tex]

- Relevant Equations:
- [tex]\Delta x \Delta p=\frac{h}{4\pi}[/tex]

So with the [itex]\gamma=\frac{1}{\sqrt{1-\beta^2}}[/itex] it seems obvious that relativistic momentum, [itex]p=\gamma m_o v[/itex] is supposed to be used.

Then [tex]\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}}[/tex]

so [itex] \Delta p=\frac{m_o\Delta v}{(1-B^2)^{3/2}}[/itex]

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

Then [tex]\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}}[/tex]

so [itex] \Delta p=\frac{m_o\Delta v}{(1-B^2)^{3/2}}[/itex]

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