34
$\begingroup$

From what I know about the Heisenberg Uncertainty Principle, one of its consequences is that when you learn a great deal about the position of a particle, you greatly increase the uncertainty of the particle's momentum. This got me thinking about a super-accurate scanner that will measure the positions of all the particles in an object (for example a building) to an extremely high degree of certainty. This would cause the momentum of all of these particles to become extremely uncertain. I am not sure where the energy would come from for this, but if the numbers are extreme enough, most of the particles should have extremely high momentums in random directions, causing the object to be obliterated.

Is my understating of the uncertainty principle horribly flawed, or does this have enough reasonableness for a science-fiction super villain weapon?

$\endgroup$
  • 36
    $\begingroup$ Sci-fi super-villain super weapons don't need to be reasonable, they need to have cool names and do cool technobabble things. $\endgroup$ – RonJohn Dec 2 '18 at 9:35
  • 16
    $\begingroup$ A very accurate scanning of a large target (e.g. the sub-atomic level scan of a building as you suggested) must direct a lot of energy at the target - at that point, you don't really need the scanning aspect of the device anymore - the energy itself is what's weaponized, not how you interpret the results... You can still have the technology be based on a very accurate scanner, so it keeps the very cool name ;) $\endgroup$ – G0BLiN Dec 2 '18 at 11:43
  • 2
    $\begingroup$ I feel like learning the position of everything in the target will be a side effect of you blowing the target apart, not it’s cause. $\endgroup$ – Joe Bloggs Dec 2 '18 at 12:15
  • 12
    $\begingroup$ An easy-to-visualise but not-particularly-accurate way to think about the Heisenberg Uncertainty Principle: To "see" things, we bounce photons of them. Now, if you replace "photons" with "bouncy balls", and try to "see" an 18-wheeler truck, not much changes. But, try to "see" an empty plastic milk bottle, and by the time your ball returns the bottle has been sent flying off elsewhere by the very act of "seeing" it $\endgroup$ – Chronocidal Dec 2 '18 at 15:46
  • 4
    $\begingroup$ I find that dropping a hundred-meter-tall Schroedinger Box onto an unsuspecting city has a higher rate of return and has a lower electric bill. "Pay me the billion dollars...or I'll open the box. 50/50 chance the giant cat will run amok! Either way, the cleanup will be monumental! Muhahahahah" $\endgroup$ – user535733 Dec 3 '18 at 1:48
70
$\begingroup$

I am not sure where the energy would come from for this

It would come from the scanner itself. In order to make extremely-high-precision scans, the scanner itself has to direct considerable energy at the target of the scan. The more precision you want, the more energy you need to pump into it. Much of that energy is absorbed by the target, which is what would cause the particles to fly apart when scanned.

It should be obvious that, in the real world, it would be much easier to build a device (let's call it a "disintegration ray") which just hits the target with all that energy and scatters its constituent particles to the wind without bothering to also determine and record the original positions of the particles. Building a machine gun is easier than building a machine gun which also tracks where its bullets end up and uses that to determine what the target originally looked like.

So I'd call it possible, but extremely impractical in real-world terms.

As a supervillain super-weapon, however, I really like it. The concept makes me smile and it's based on technobabble that sounds reasonable if you don't think too hard about it. And aren't those really the most important factors when designing a doomsday weapon?

$\endgroup$
  • 28
    $\begingroup$ Dude - The last paragraph is golden. $\endgroup$ – WhatRoughBeast Dec 2 '18 at 13:59
  • 1
    $\begingroup$ Actually, the “disintegration ray” fulfills the specs of the scanner, as when it hits the target, you know its position precisely, if you cared to remember the exact starting position and direction of your ray. So you have measured the precise position and raised the uncertainty of the momentum. That’s what it is all about. Whether you actually do remember the position and direction of the ray, is irrelevant. $\endgroup$ – Holger Dec 3 '18 at 10:17
  • 3
    $\begingroup$ @Holger I think the argument is that you don't just know the "position" of the target, but the actual position and state of each atom/molecule the target consists of. The upside to this, as Dave stated, is that you could use that information to reconstruct what the target originally looked like on a molecular level. If you ask me, this sounds like a thing DC's Brainiac would have. $\endgroup$ – Suthek Dec 3 '18 at 13:13
  • 1
    $\begingroup$ I think this would be hilarious. All the superheroes scoff at the overly villainesque name of the superweapon and just call it a disintegration ray despite the creators protests to he contrary. Then to prove his point the villiain creates imperfect clones of "disintegrated" superheros by combinging the exact knowledge of the position of all of the atoms in their body with an algorithm to estimate the probable initial velocities. $\endgroup$ – user3389672 Dec 3 '18 at 16:48
  • 1
    $\begingroup$ @GretchenV But then surely the same principle would apply to the "frozen copy" mentioned in the comment I was replying to. You couldn't create a frozen copy any more than you could a perfect copy without having complete control of both the particle's position and its momentum. Otherwise, as you said, you would place the particle in the exact position you want, and then it would be left with a completely unpredictable momentum (as opposed to no momentum at all). What I was saying is that if we had the ability to create a "frozen" copy, we should have the ability to make a perfect copy as well. $\endgroup$ – Abion47 Dec 4 '18 at 1:31
54
$\begingroup$

This "weapon" has already been built. It's called the European X-Ray Free Electron Laser (XFEL), and it's used for super-accurate imaging of molecules.

Point is, whenever the XFEL hits a target molecule, all the electrons are simply blasted away from the molecule, and the more inert rest of atomic nuclei dissolves in a Coulomb-explosion. The scattered X-Rays are used to reconstruct the 3D positions of all the atoms in the instant that the molecule was hit.

This blasting away of the electrons is not by accident: The shorter the wavelength of the photons, the smaller features you can detect, because the change of the photon's momentum is greater. Of course, that change in momentum comes from somewhere, and it's the electron cloud that gets the kick in the opposite direction. So, the quality of the localization is directly related to the uncertainty of momentum after the measurement.

$\endgroup$
  • 11
    $\begingroup$ I love it when a hypothetical "could X exist/work?" question has an answer of "yes, it already exists" $\endgroup$ – Baldrickk Dec 3 '18 at 11:46
  • $\begingroup$ @Baldrickk Especially when the application is peaceful, rather than military :-) $\endgroup$ – cmaster Dec 3 '18 at 21:06
9
$\begingroup$

No, your conclusions are wrong.

Heisenberg uncertainty principles states that the uncertainty in the measurement of two conjugate variables, in this case position and momentum, is always higher than a certain value

$$\Delta x \cdot \Delta P_x > h/4 \pi$$

Mind that the uncertainty is on the measurement, not on the actual value (whatever "actual" might mean under the assumption of quantum physics).

Particles in macroscopic objects already move randomly in all direction due to thermal agitation, and that does not destroy the object. When you increase the agitation you are just increasing the object temperature, maybe melting it or evaporating it. But it has nothing to do with Heisenberg principle.

Also, citing this other answer:

$h/4 \pi = 1.054571726(47)×10^{−34} Js$

All it needs is some algebra to see that a kilogram ball moving at a micron per second and measurement accuracies of the order of a micron will still fulfill the HUP constraint as $h/4 \pi$ is a very small number. For classical dimensions $h/4 \pi$ is essentially zero and the HUP always holds.

When one goes to dimensions of less than nanometers and masses of the order of molecules then one is in the quantum mechanical regime and can start talking of uncertainties .

$\endgroup$
  • 8
    $\begingroup$ You are just not thinking about the correct energies, precisions, and masses here: a bullet may be well localized when it leaves the gun, but if you try to localize the electrons to subatomic precision, you'll find that the electrons are ricocheting off in all directions after the measurement. Also, the electrons don't really decide where they are, until you actually measure them. In quantum mechanics, you cannot measure without influencing the system, and that's what the uncertainty principle is all about. $\endgroup$ – cmaster Dec 2 '18 at 9:25
  • $\begingroup$ I would agree with the answer, the conclusions are wrong albeit not obviously due to the lack of understanding of the details therein. Thermal agitation is indeed not the only reason however again the words obliterated definitely do not apply at ALL in certain determined situations even given random input. $\endgroup$ – Jay Dec 2 '18 at 16:05
  • 1
    $\begingroup$ This answer is actually wrong because the premise of the doomsday weapon is that it attempts to "measure the position of all the particles in a macroscopic object" which I take to mean "each and every particle". Now, in one mole of protium (to keep this simple), there are 2 x 6.02214076×10<super>23</super> particles. The size of the atom is 120 pm. To locate the electron with respect to the proton with accuracy of 10 pm, would require bouncing a photon with 10 pm wavelength of it, but the protium dissociates at 91.2 pm or less incident photons. $\endgroup$ – GretchenV Dec 3 '18 at 14:42
8
$\begingroup$

You're misunderstanding the Heisenberg Uncertainty Principle, or rather what "measure" means. In order to measure something, we have to interact with it - on the nanoscale this commonly involves bouncing electrons off of the item we're imaging.

A scanning electron microscope (SEM) is a type of electron microscope that produces images of a sample by scanning the surface with a focused beam of electrons. - Wikipedia

Imagine "imaging" a pool table by analyzing the path the cue ball takes. It'd be trivial to figure out where a ball is - or more accurately was - at the moment of impact. The cost of this is that you've transferred some of the energy of your scanning cue ball into whatever ball you've measured, altering its momentum - you can no longer be certain the ball is at a standstill. The system you're proposing is one that increases uncertainty by deposing large amounts of energy into a target object. In other words, basically every projectile, energy, or nuclear weapon system known to man. The only difference is the word "measure." Attaching a telescope to your death-ray arguably fits the bill.

Rather amusing way to bypass a treaty, don't you think? It's not a planet killing Nicoll-Dyson beam, it's a Planetary Spectral Analyzer.

$\endgroup$
  • 3
    $\begingroup$ Haha, that last paragraph presents a particularly wonderful idea regarding this technology. $\endgroup$ – TheNewGuy Dec 4 '18 at 6:44
  • $\begingroup$ @TheNewGuy Makes me think of the Puppeteers from Ringworld. A species that abhors violence so greatly they don't possess any weapons of their own. That said, all of their peaceful equipment is dual purpose - flashlights can have their brightness and beam width tuned with a knob, all the way down to a useful tactical laser. Their digging beams are marvels of engineering that move material by selectively suppressing the positive or negative charge of atoms to turn them to dust - that also create near-nuclear explosions is you do both at the same time. $\endgroup$ – UIDAlexD Dec 5 '18 at 14:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.