# 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. Commented Jul 10, 2020 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
Commented Jul 10, 2020 at 7:33
• If a object gains mass, $\Delta x \Delta p$ = $\Delta x m\Delta v$ = $\Delta x \Delta v = h/4\pi m$ Commented Jul 10, 2020 at 7:33
• @yu-v thanks for pointing, Commented Jul 10, 2020 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 $m\gamma$). Commented Jul 10, 2020 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 $\Delta x \approx 0$ Commented Jul 10, 2020 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
Commented Jul 10, 2020 at 13:09
• 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 :( Commented Jul 10, 2020 at 13:30
• @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
Commented Jul 10, 2020 at 14:11
• Ty for giving me the correct idea Commented Jul 10, 2020 at 14:29