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.