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 fluidistic Sep1-10 11:57 AM

Heisenberg principle, little question

$$\Delta x \Delta p \geq \frac{\hbar}{2}$$.
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 Sep1-10 12:44 PM

Re: Heisenberg principle, little question

Quote:
 Quote by fluidistic (Post 2862167) $$\Delta p =mv$$
$$\Delta p$$ is not a particle's momentum but it's uncertainty in its momentum and it can be huge.

 alxm Sep1-10 01:25 PM

Re: Heisenberg principle, little question

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 Sep1-10 04:49 PM

Re: Heisenberg principle, little question

Quote:
 Quote by maxverywell (Post 2862223) $$\Delta p$$ is not a particle's momentum but it's uncertainty in its momentum and it can be huge.
I know.
Quote:
 Quote by alxm (Post 2862283) 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
Quote:
 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.

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