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-   -   Heisenberg principle, little question (http://www.physicsforums.com/showthread.php?t=425779)

fluidistic Sep1-10 11:57 AM

Heisenberg principle, little question
 
[tex]\Delta x \Delta p \geq \frac{\hbar}{2}[/tex].
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 [tex]\Delta x \to 0[/tex], [tex]\Delta p[/tex] should tend to [tex]+\infty[/tex] but there's the c limit so that I can't make [tex]\Delta x \to 0[/tex]. Unless I should consider the relativistic mass of the electron and not the rest mass in the [tex]\Delta p =mv[/tex] part of the inequality? So m would tend to [tex]+\infty[/tex] and I'm not really limited by a maximum limit of velocity and I can get a very precise measure for [tex]\Delta x[/tex].

maxverywell Sep1-10 12:44 PM

Re: Heisenberg principle, little question
 
Quote:

Quote by fluidistic (Post 2862167)
[tex]\Delta p =mv[/tex]

[tex]\Delta p[/tex] 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 [tex]p=mv[/tex], which is the non-relativistic momentum, and then assert that [tex]v \leq c[/tex] (and hence [tex]p \leq mc[/tex]) because of relativity.

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

fluidistic Sep1-10 04:49 PM

Re: Heisenberg principle, little question
 
Quote:

Quote by maxverywell (Post 2862223)
[tex]\Delta p[/tex] 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 [tex]p=mv[/tex], which is the non-relativistic momentum, and then assert that [tex]v \leq c[/tex] (and hence [tex]p \leq mc[/tex]) because of relativity.

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

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 [tex] p =mv[/tex] part of the inequality? So m would tend to [tex]+\infty[/tex] and I'm not really limited by a maximum limit of velocity and I can get a very precise measure for [tex]\Delta x[/tex] .
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 [tex]\Delta p[/tex] is the relativistic momentum, which is what it seems I was doubting on.
Thanks guys, question solved.


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