Heisenberg indeterminacy principle

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The forum discussion centers on the Heisenberg indeterminacy principle, specifically calculating the minimum uncertainty in the position of an electron given its speed is less than the speed of light. Using the formula (Δx)(Δp) = h/4π, where h is Planck's constant, the user calculated Δx to be approximately 3.049 x 10^-59 meters. However, the discussion introduces a relativistic version of the uncertainty principle, suggesting that as velocity approaches the speed of light, the minimum uncertainty in position could theoretically be zero. This highlights the complexities of applying quantum mechanics to relativistic scenarios.

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penmanta0711
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No object can travel faster than the speed of light, so it
would appear evident that the uncertainty in the speed of
any object is at most 3 * 10^8 m s
(a) What is the minimum uncertainty in the position
of an electron, given that we know nothing about
its speed except that it is slower than the speed of
light?

Known
(Δx)(Δp)=h/4∏ hence Δx=h/4∏(mass of e-)(velocity)
Mass of e-=9.10x10^-31 kg
Velocity is equal to 3x10^8 m s
I plug in my values and came up with 3.049 x10^-59meters
My answer seems a bit wonky..could you double check..and if wrong give me hints to solve properly...thank you
 
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There is a relativistic version of the uncertainty principle, ΔxΔv >= m2/(2* E3) where we are using units such that c = h-bar = 1. E = √p2+m2. where m is the rest mass and p is the relativistic momentum p = λmv. p can rise unbounded as v → c and thus so can E. Δv = c = 1, so Δx >= 0. The minimum uncertainty is 0. Or so I'd like to think.
 

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