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According to me, an object gains relativistic mass as it approaches the speed of light, and $$\Delta x \Delta p \ge\frac {\hbar}{2}$$ So objects with speeds close to $c$, should show less uncertainty in position because an object with a small de broglie wavelength is less likely to spread.

$$\lambda = \frac{h}{m_0v}\sqrt{1-\frac{v^2}{c^2}}$$ $$\sqrt{1-\frac{v^2}{c^2}} \rightarrow 0$$ $$\lambda \rightarrow 0$$

Shouldn't $\Delta x \rightarrow 0$ too?

In short does the uncertainty principle hold true if $\Delta p$ is relativistic? Or it only takes non-relativistic mass into the account but is still correct even at speeds close to $c $?

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    $\begingroup$ "So objects with speeds close to c, should show less uncertainty in position." Why does that follow? BTW, in special relativity there's no upper bound on momentum. $\endgroup$
    – PM 2Ring
    Commented Jul 10, 2020 at 7:24
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    $\begingroup$ also note that $\Delta x \Delta p \geq \hbar/2$, not equal. The uncertainty can be very large both momentum and position $\endgroup$
    – user245141
    Commented Jul 10, 2020 at 7:33
  • $\begingroup$ If a object gains mass, $\Delta x \Delta p$ = $\Delta x m\Delta v$ = $\Delta x \Delta v = h/4\pi m$ $\endgroup$
    – Tim Crosby
    Commented Jul 10, 2020 at 7:33
  • $\begingroup$ @yu-v thanks for pointing, $\endgroup$
    – Tim Crosby
    Commented Jul 10, 2020 at 7:35
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    $\begingroup$ You don't need to bring relativistic mass into it. Momentum in SR is $p=mv\gamma$, where $m$ is the rest mass and $\gamma$ is the Lorentz factor. (If you insist on using the deprecated concept of relativistic mass, that equals $m\gamma$). $\endgroup$
    – PM 2Ring
    Commented Jul 10, 2020 at 8:19

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The point is that the "$p$" in $\Delta p$ may not have the properties that you think it has because $$p= \frac{m_0 v}{\sqrt{1 - v^2/c^2}} $$ where $v$ is the coordinate velocity $dx/dt$ and $m_0$ the rest mass of the particle. Notice that when $v\to c$ then $p \to \infty$!!

This $p$ is what is conserved in collisions and thus has a meaning for dynamics, unlike the kinematic velocity $v$. In other words, if you do not know $p$ well, you do not know e.g. outcomes of collision experiments well, and that applies even if this corresponds to a very small uncertainty in velocity $\Delta v$.

Now of course, if you reduce $\Delta x$ greatly, the Heisenberg uncertainty principle tells you that $\Delta p > \hbar/(2\Delta x)$. Since $p$ can attain any value in $(-\infty,\infty)$ without violating relativity (see above), there is no conflict.

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  • $\begingroup$ I didn't say that both theories conflict each other, I I don' understand, if the De Broglie wavelength of an object $\approx 0$ does it imply that $\Delta x \approx 0$ $\endgroup$
    – Tim Crosby
    Commented Jul 10, 2020 at 12:35
  • $\begingroup$ @TimCrosby The De Broglie wavelength is the wavelength of a particle with sharp $p$ and thus completely delocalized position. The quasi-classical link between De Broglie wavelength and the Heisenberg uncertainty principle can be made by considering that you are "probing" the particle with a secondary particle-wave with momentum $P$. This causes a momentum disturbance to the primary particle $\Delta p \sim P$ and determines its position up to half the wavelength of the probing particle $\Delta x \sim \lambda_{\rm DB}/2 \sim \hbar/(2P)$. $\endgroup$
    – Void
    Commented Jul 10, 2020 at 13:09
  • $\begingroup$ Then why do macroscopic objects show negligible uncertainty in position and momentum?, I thought that their De Broglie wavelength being so small that it almost looked like one spike.was the reason :( $\endgroup$
    – Tim Crosby
    Commented Jul 10, 2020 at 13:30
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    $\begingroup$ @TimCrosby Macroscopic objects show small relative uncertainty of momentum compared to their total momentum $\delta p/p \ll 1$, and tiny relative uncertainty compared to their size $R$, $\delta x/R \ll 1$. This is absolutely no problem as long as $pR \gg \hbar$. The absolute uncertainties will actually always be way larger than the quantum limit in any real experimental context! $\endgroup$
    – Void
    Commented Jul 10, 2020 at 14:11
  • $\begingroup$ Ty for giving me the correct idea $\endgroup$
    – Tim Crosby
    Commented Jul 10, 2020 at 14:29

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