Heisenberg principle, little question

[fluidistic]

\Delta x \Delta p \geq \frac{\hbar}{2}.
Say I want to measure the best I can the position of an electron, in detriment of its momentum (i.e. velocity since I assume that I know its mass quite well).
When \Delta x \to 0, \Delta p should tend to +\infty but there's the c limit so that I can't make \Delta x \to 0. Unless I should consider the relativistic mass of the electron and not the rest mass in the \Delta p =mv part of the inequality? So m would tend to +\infty and I'm not really limited by a maximum limit of velocity and I can get a very precise measure for \Delta x.

 

[maxverywell]

\Delta p =mv

\Delta p is not a particle's momentum but it's uncertainty in its momentum and it can be huge.

 

[alxm]

You can't assert p=mv, which is the non-relativistic momentum, and then assert that v \leq c (and hence p \leq mc) because of relativity.

Relativistic momentum is p = \frac{m_0 v}{\sqrt{1 - \frac{v^2}{c^2}}}, so v \rightarrow c as p \rightarrow \infty.

 

[fluidistic]

\Delta p is not a particle's momentum but it's uncertainty in its momentum and it can be huge.

I know.

You can't assert p=mv, which is the non-relativistic momentum, and then assert that v \leq c (and hence p \leq mc) because of relativity.

Relativistic momentum is p = \frac{m_0 v}{\sqrt{1 - \frac{v^2}{c^2}}}, so v \rightarrow c as p \rightarrow \infty.

Ah ok. My m standed for the relativistic mass. That's what I meant in

Unless I should consider the relativistic mass of the electron and not the rest mass in the p =mv part of the inequality? So m would tend to +\infty and I'm not really limited by a maximum limit of velocity and I can get a very precise measure for \Delta x .

Though now I don't understand what I meant by "I'm not really limited by a maximum limit of velocity".
Anyway I get the idea. And the \Delta p is the relativistic momentum, which is what it seems I was doubting on.
Thanks guys, question solved.