Is Heisenberg's uncertainty principle consistent with Special Relativity?

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 $$c$$?

• "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. – PM 2Ring Jul 10 '20 at 7:24
• also note that $\Delta x \Delta p \geq \hbar/2$, not equal. The uncertainty can be very large both momentum and position – user245141 Jul 10 '20 at 7:33
• If a object gains mass, $\Delta x \Delta p$ = $\Delta x m\Delta v$ = $\Delta x \Delta v = h/4\pi m$ – Tim Crosby Jul 10 '20 at 7:33
• @yu-v thanks for pointing, – Tim Crosby Jul 10 '20 at 7:35
• 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 $m\gamma$). – PM 2Ring Jul 10 '20 at 8:19

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.
• 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 $\Delta x \approx 0$ – Tim Crosby Jul 10 '20 at 12:35
• @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 $\Delta x \sim \lambda_{\rm DB}/2 \sim \hbar/(2P)$. – Void Jul 10 '20 at 13:09
• @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! – Void Jul 10 '20 at 14:11