The Heisenberg Uncertainty Ray - A Hyper-Scanning Doomsday Weapon











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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?










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  • 31




    Sci-fi super-villain super weapons don't need to be reasonable, they need to have cool names and do cool technobabble things.
    – RonJohn
    yesterday






  • 12




    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 ;)
    – G0BLiN
    yesterday






  • 2




    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.
    – Joe Bloggs
    yesterday






  • 12




    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
    – Chronocidal
    yesterday






  • 4




    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"
    – user535733
    yesterday

















up vote
27
down vote

favorite
4












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?










share|improve this question


















  • 31




    Sci-fi super-villain super weapons don't need to be reasonable, they need to have cool names and do cool technobabble things.
    – RonJohn
    yesterday






  • 12




    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 ;)
    – G0BLiN
    yesterday






  • 2




    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.
    – Joe Bloggs
    yesterday






  • 12




    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
    – Chronocidal
    yesterday






  • 4




    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"
    – user535733
    yesterday















up vote
27
down vote

favorite
4









up vote
27
down vote

favorite
4






4





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?










share|improve this question













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?







science-fiction physics science






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asked yesterday









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  • 31




    Sci-fi super-villain super weapons don't need to be reasonable, they need to have cool names and do cool technobabble things.
    – RonJohn
    yesterday






  • 12




    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 ;)
    – G0BLiN
    yesterday






  • 2




    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.
    – Joe Bloggs
    yesterday






  • 12




    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
    – Chronocidal
    yesterday






  • 4




    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"
    – user535733
    yesterday
















  • 31




    Sci-fi super-villain super weapons don't need to be reasonable, they need to have cool names and do cool technobabble things.
    – RonJohn
    yesterday






  • 12




    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 ;)
    – G0BLiN
    yesterday






  • 2




    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.
    – Joe Bloggs
    yesterday






  • 12




    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
    – Chronocidal
    yesterday






  • 4




    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"
    – user535733
    yesterday










31




31




Sci-fi super-villain super weapons don't need to be reasonable, they need to have cool names and do cool technobabble things.
– RonJohn
yesterday




Sci-fi super-villain super weapons don't need to be reasonable, they need to have cool names and do cool technobabble things.
– RonJohn
yesterday




12




12




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 ;)
– G0BLiN
yesterday




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 ;)
– G0BLiN
yesterday




2




2




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.
– Joe Bloggs
yesterday




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.
– Joe Bloggs
yesterday




12




12




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
– Chronocidal
yesterday




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
– Chronocidal
yesterday




4




4




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"
– user535733
yesterday






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"
– user535733
yesterday












4 Answers
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up vote
58
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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?






share|improve this answer

















  • 24




    Dude - The last paragraph is golden.
    – WhatRoughBeast
    yesterday










  • 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.
    – Holger
    18 hours ago






  • 3




    @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.
    – Suthek
    15 hours ago










  • @Suthek you can not reconstruct the target, as you only know the exact positions of these objects, but not their momentum. That’s the whole point of the Uncertainty Principle. Perhaps, you’re fine with an imprecise reproduction, but then, you’re far away from the original question.
    – Holger
    12 hours ago










  • 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.
    – user3389672
    11 hours ago


















up vote
41
down vote













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.






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  • 7




    I love it when a hypothetical "could X exist/work?" question has an answer of "yes, it already exists"
    – Baldrickk
    16 hours ago










  • @Baldrickk Especially when the application is peaceful, rather than military :-)
    – cmaster
    7 hours ago


















up vote
8
down vote













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 .







share|improve this answer



















  • 7




    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.
    – cmaster
    yesterday










  • 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.
    – Jay
    yesterday










  • 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.
    – GretchenV
    13 hours ago


















up vote
4
down vote













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.






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    4 Answers
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    4 Answers
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    up vote
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    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?






    share|improve this answer

















    • 24




      Dude - The last paragraph is golden.
      – WhatRoughBeast
      yesterday










    • 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.
      – Holger
      18 hours ago






    • 3




      @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.
      – Suthek
      15 hours ago










    • @Suthek you can not reconstruct the target, as you only know the exact positions of these objects, but not their momentum. That’s the whole point of the Uncertainty Principle. Perhaps, you’re fine with an imprecise reproduction, but then, you’re far away from the original question.
      – Holger
      12 hours ago










    • 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.
      – user3389672
      11 hours ago















    up vote
    58
    down vote














    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?






    share|improve this answer

















    • 24




      Dude - The last paragraph is golden.
      – WhatRoughBeast
      yesterday










    • 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.
      – Holger
      18 hours ago






    • 3




      @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.
      – Suthek
      15 hours ago










    • @Suthek you can not reconstruct the target, as you only know the exact positions of these objects, but not their momentum. That’s the whole point of the Uncertainty Principle. Perhaps, you’re fine with an imprecise reproduction, but then, you’re far away from the original question.
      – Holger
      12 hours ago










    • 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.
      – user3389672
      11 hours ago













    up vote
    58
    down vote










    up vote
    58
    down vote










    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?






    share|improve this answer













    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?







    share|improve this answer












    share|improve this answer



    share|improve this answer










    answered yesterday









    Dave Sherohman

    3,6451118




    3,6451118








    • 24




      Dude - The last paragraph is golden.
      – WhatRoughBeast
      yesterday










    • 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.
      – Holger
      18 hours ago






    • 3




      @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.
      – Suthek
      15 hours ago










    • @Suthek you can not reconstruct the target, as you only know the exact positions of these objects, but not their momentum. That’s the whole point of the Uncertainty Principle. Perhaps, you’re fine with an imprecise reproduction, but then, you’re far away from the original question.
      – Holger
      12 hours ago










    • 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.
      – user3389672
      11 hours ago














    • 24




      Dude - The last paragraph is golden.
      – WhatRoughBeast
      yesterday










    • 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.
      – Holger
      18 hours ago






    • 3




      @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.
      – Suthek
      15 hours ago










    • @Suthek you can not reconstruct the target, as you only know the exact positions of these objects, but not their momentum. That’s the whole point of the Uncertainty Principle. Perhaps, you’re fine with an imprecise reproduction, but then, you’re far away from the original question.
      – Holger
      12 hours ago










    • 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.
      – user3389672
      11 hours ago








    24




    24




    Dude - The last paragraph is golden.
    – WhatRoughBeast
    yesterday




    Dude - The last paragraph is golden.
    – WhatRoughBeast
    yesterday












    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.
    – Holger
    18 hours ago




    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.
    – Holger
    18 hours ago




    3




    3




    @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.
    – Suthek
    15 hours ago




    @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.
    – Suthek
    15 hours ago












    @Suthek you can not reconstruct the target, as you only know the exact positions of these objects, but not their momentum. That’s the whole point of the Uncertainty Principle. Perhaps, you’re fine with an imprecise reproduction, but then, you’re far away from the original question.
    – Holger
    12 hours ago




    @Suthek you can not reconstruct the target, as you only know the exact positions of these objects, but not their momentum. That’s the whole point of the Uncertainty Principle. Perhaps, you’re fine with an imprecise reproduction, but then, you’re far away from the original question.
    – Holger
    12 hours ago












    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.
    – user3389672
    11 hours ago




    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.
    – user3389672
    11 hours ago










    up vote
    41
    down vote













    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.






    share|improve this answer

















    • 7




      I love it when a hypothetical "could X exist/work?" question has an answer of "yes, it already exists"
      – Baldrickk
      16 hours ago










    • @Baldrickk Especially when the application is peaceful, rather than military :-)
      – cmaster
      7 hours ago















    up vote
    41
    down vote













    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.






    share|improve this answer

















    • 7




      I love it when a hypothetical "could X exist/work?" question has an answer of "yes, it already exists"
      – Baldrickk
      16 hours ago










    • @Baldrickk Especially when the application is peaceful, rather than military :-)
      – cmaster
      7 hours ago













    up vote
    41
    down vote










    up vote
    41
    down vote









    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.






    share|improve this answer












    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.







    share|improve this answer












    share|improve this answer



    share|improve this answer










    answered yesterday









    cmaster

    2,739514




    2,739514








    • 7




      I love it when a hypothetical "could X exist/work?" question has an answer of "yes, it already exists"
      – Baldrickk
      16 hours ago










    • @Baldrickk Especially when the application is peaceful, rather than military :-)
      – cmaster
      7 hours ago














    • 7




      I love it when a hypothetical "could X exist/work?" question has an answer of "yes, it already exists"
      – Baldrickk
      16 hours ago










    • @Baldrickk Especially when the application is peaceful, rather than military :-)
      – cmaster
      7 hours ago








    7




    7




    I love it when a hypothetical "could X exist/work?" question has an answer of "yes, it already exists"
    – Baldrickk
    16 hours ago




    I love it when a hypothetical "could X exist/work?" question has an answer of "yes, it already exists"
    – Baldrickk
    16 hours ago












    @Baldrickk Especially when the application is peaceful, rather than military :-)
    – cmaster
    7 hours ago




    @Baldrickk Especially when the application is peaceful, rather than military :-)
    – cmaster
    7 hours ago










    up vote
    8
    down vote













    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 .







    share|improve this answer



















    • 7




      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.
      – cmaster
      yesterday










    • 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.
      – Jay
      yesterday










    • 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.
      – GretchenV
      13 hours ago















    up vote
    8
    down vote













    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 .







    share|improve this answer



















    • 7




      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.
      – cmaster
      yesterday










    • 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.
      – Jay
      yesterday










    • 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.
      – GretchenV
      13 hours ago













    up vote
    8
    down vote










    up vote
    8
    down vote









    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 .







    share|improve this answer














    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 .








    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited yesterday

























    answered yesterday









    L.Dutch

    72k22173347




    72k22173347








    • 7




      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.
      – cmaster
      yesterday










    • 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.
      – Jay
      yesterday










    • 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.
      – GretchenV
      13 hours ago














    • 7




      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.
      – cmaster
      yesterday










    • 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.
      – Jay
      yesterday










    • 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.
      – GretchenV
      13 hours ago








    7




    7




    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.
    – cmaster
    yesterday




    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.
    – cmaster
    yesterday












    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.
    – Jay
    yesterday




    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.
    – Jay
    yesterday












    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.
    – GretchenV
    13 hours ago




    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.
    – GretchenV
    13 hours ago










    up vote
    4
    down vote













    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.






    share|improve this answer

























      up vote
      4
      down vote













      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.






      share|improve this answer























        up vote
        4
        down vote










        up vote
        4
        down vote









        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.






        share|improve this answer












        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.







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered 9 hours ago









        UIDAlexD

        3,206519




        3,206519






























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