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Heisenberg principle, little question

  1. Sep 1, 2010 #1


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    [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].
  2. jcsd
  3. Sep 1, 2010 #2
    [tex]\Delta p[/tex] is not a particle's momentum but it's uncertainty in its momentum and it can be huge.
  4. Sep 1, 2010 #3


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    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].
  5. Sep 1, 2010 #4


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