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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 $?

Tim Crosby
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1 Answers1

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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.

Void
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