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User: Athanasius

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  1. Re:Remote quantum surveillance on Can Quantum Entanglement Create Faster-Than-Light Communication? (mit.edu) · · Score: 1

    Technically you would only know, in an FTL sense, that the particle at the other location had the opposite value. Just because you agreed that a certain state of that certain particle would mean a certain action was taken/not taken doesn't mean that the other person didn't change their mind, or wasn't prevented from carrying out the agreed-upon course of action.

    You'd still only know if using some light-speed limited communication means to verify the outcome.

  2. Re:Why do you need to know the state? on Can Quantum Entanglement Create Faster-Than-Light Communication? (mit.edu) · · Score: 2

    What makes you think that the second particles wave function has collapsed, so far as you're concerned locally, before you force it to by attempting measurement? Entanglement only means that once one of the particles is measured the other when measured will have the complementary value.

    Please, honestly, give me a citation from somewhere/one trustworthy about this detection of wave function collapse without the particle interacting such that the entanglement has been destroyed. An repeatable, verifiable experimental result would be ideal.

  3. Re:Remote quantum surveillance on Can Quantum Entanglement Create Faster-Than-Light Communication? (mit.edu) · · Score: 1

    No.

    Let's assume you have a whole bunch of entangled pairs set up, so that you can consult one per day, month, year, whatever. That still doesn't help. When Twin A checks his particle 1 and sees it has (spin, polarisation, whatever) value +1 all you know is that when Twin B measure their matching particle they'll get value -1. That's it. A did not, and cannot choose that his particle measures as +1 rather than -1. All entanglement means is that the pair of particles will have complementary values measured.

    And, no, you can't assign life event, decisions, or any other information to which particle you measure out of the set. Then Twin B would need to measure them all to see... what? Which has changed? No, that doesn't work because the moment you measure any of the entangled particles you lose the entanglement. Measure them all, record the state, measure them all later... oops, now the measurements are no longer correlated with the state of the particles with Twin A. You literally have no idea if Twin A measured any of them without using a (maximally) light-speed conventional connection to ask.

  4. Re:Quantum science is in it's infancy on Can Quantum Entanglement Create Faster-Than-Light Communication? (mit.edu) · · Score: 2

    There's a difference between "we have no idea how to do this" (heavier than air flight 500 years ago) and "we have a lot of experimental evidence demonstrating that this works in a manner that will absolutely not allow for that" (the current knowledge about quantum mechanics and what it means for entangled pairs of particles).

  5. Re:Why do you need to know the state? on Can Quantum Entanglement Create Faster-Than-Light Communication? (mit.edu) · · Score: 3, Informative

    You can't tell if the state has changed without measuring it. The first of the entangled pair of particles (one at home, one on your spaceship) to be measured will mean the other will be measured (when it is) in a complementary state. That's all that happens. We're not talking about some particle giving off a photon of light when its partner is measured or anything like that. Also measuring breaks the entanglement. Purposefully changing the state of one of them also breaks the entanglement. So you can't have a bunch of them that you keep on measuring, waiting for one of them to change state. It just doesn't work that way.

  6. The problem is that:

    1. You don't get to choose which particle of the pair gets which value. It's random as to if it's -1 here and +1 there or vice versa.
    2. That's pretty much it. Look up Bell's Theorem https://en.wikipedia.org/wiki/... for a proof that the future-measured state of a given entangled particle isn't pre-determined (no "hidden values"). The only thing you know from a pair being entangled is that whatever value you measure on one particle for the entangled property, you will get the opposite/complementary value on the other particle when it is also measured.

    Thus all you can actually transmit using such a system is a random stream of data, with the knowledge that the matching data at the other end is complementary (and thus can be used to derive what you have). Also, this assumes no-one trying to intercept the transmitted particles in the middle. If you don't get your implementation correct it's possible to do so without detection, or for that to be detectable precisely because it broke the entanglement and thus you don't even have a complementary data set.

  7. SSL Certificate Pinning on Millions Of Waze Users Can Have Their Movements Tracked By Hackers (fusion.net) · · Score: 1

    So, Waze need to have the app properly implement SSL Certificate Pinning (in order to prevent a MITM SSL proxy that works via an additional Certificate Authority). Of course then it's likely still vulnerable to some reverse engineering of the app to get around that.

  8. Re:Why Bomb? on US Begins Dropping 'Cyberbombs' On ISIS (nytimes.com) · · Score: 1

    I don't know, perhaps these 'cyberbombs' have a good chance of causing collateral damage, just like the "boom" ones.

  9. Assuming you're prepared to believe this guy had a good enough grasp on all the physics involved, this was debunked ~8 months ago: https://www.reddit.com/r/EmDri...

  10. When I looked into that some time recently I believe for a visible red light laser you'd see significant dispersion after less than 10km

    Actually we bounce laser beams off of the Moon to measure Earth/Moon distance on a regular basis.

    From your link:

    At the Moon's surface, the beam is about 6.5 kilometers wide

    That's just emphasises my point, although some of that will be atmosphere-caused diffusion of course. Also the "significant dispersion" in the calculation I made was on the order of 10x the radius compared to when the beam exited the laser apparatus as the context was using lasers as weapons.

  11. Happy to help :).

  12. It doesn't. There's no way to slow it down, and given the size that's being talked about it wouldn't have enough power to have any hope of sending a detectable signal back (both because it wouldn't have enough stored energy and it also wouldn't have a big enough antenna to have any hope of aiming the signal back).

    I suspect this whole thing would be more aimed at developing technologies and inspiring others to solve the various problems to sending a useful probe in the future.

  13. Re:Obligatory Fermi on Hawking Backs $100 Million Interstellar Travel Project to Send 'Nano-Craft' To Nearest Star · · Score: 5, Interesting

    You can't collimate a laser beam that perfectly. When I looked into that some time recently I believe for a visible red light laser you'd see significant dispersion after less than 10km. Yes, in a vacuum. Even if you could align the internals perfectly you'd still get a small amount of diffraction where the beam leaves the apparatus.

    Over lightyears you're never going to maintain beam cohesion.

    This also both answers the GP's question for the period of time the such a probe is being accelerated and why it wouldn't be accelerated the whole distance. Indeed given the travel time, even if accelerated to very close to the speed of light, you'd not be aiming the laser at the destination system (it would move some by the time the probe got there).

  14. Re:Interesting, but.. on Hawking Backs $100 Million Interstellar Travel Project to Send 'Nano-Craft' To Nearest Star · · Score: 4, Informative

    No. Not being able to leverage quantum entanglement into actual FTL communications is a fundamental limit of how it works.

    To state it simply. If two particles have their state entangled for a property then measuring that property on one causes the same measured property on the other to have the opposite value but which way around these are is essentially random and impossible to control. The best you can use this for is to securely duplicate a sequence of random values, (and in the case of sending one half of each pair to another site, assuming your implementation doesn't have any problems, know if someone had at all intercepted those particles).

    This is why all current uses of the technology are used to send an encryption key which you then use to encrypt normal communications.

  15. Re:Interesting read about names on Names That Break Computers (bbc.com) · · Score: 3, Informative

    Someone already did: http://spaceninja.com/2015/12/...

  16. Re:Is 20 days wrong? on Company Behind Badlock Disclosure Says Pre-Patch Hype Is Good Marketing (csoonline.com) · · Score: 1

    This *appears* to be all about hitting the next Microsoft Patch Tuesday. I'm somewhat peeved that all the users of Samba are being made to wait on a fix until that day. I almost want someone else to figure out the vulnerability and publish it so as to get the patches released sooner.

  17. Re:Where's my IPv6 on The State of Slashdot: Https, Poll Changes, Auto-Refresh, Videos, and More · · Score: 3, Insightful

    You say that, but as soon as there's any code logging/recording IPs it'll need changing to handle IPv6 as well (is that field now so much longer that it has knock-on effects on other things?). And if there's any code that tries to do checks/manipulations on the addresses then you have a load more work to do as well.

  18. He was quick off the mark with this one, probably had it prepared: https://xkcd.com/1642/

  19. Re:.. visiting a web-site running as an onion ser. on Sensitive Information Can Be Revealed From Tor Hidden Services On Apache (dailydot.com) · · Score: 1

    Ah, I did wonder if this was a case of "exiting from a TOR tunnel means the traffic is coming from 127.0.0.1".

    To be honest that means TOR is breaking a lot of expectations about localhost. Really they ought to use some RFC1918 address space for that final hop.

    Still, as-is, you could reconfigure /server-status to only be allowed from your actual local IP (and any other safe IPs), and not include localhost in the list.

  20. Server Misconfiguration is news? on Sensitive Information Can Be Revealed From Tor Hidden Services On Apache (dailydot.com) · · Score: 2

    This is simply server misconfiguration. I can't comment on other distributions but Debian at least restricts /server-status URL access to localhost by default. You'd have to explicitly change this to allow from anywhere else.

  21. Re:Only incidentally similar to su on Systemd Absorbs "su" Command Functionality · · Score: 1

    Ah, there was an auth popup in one of the example screenshots when trying to use this to get to the local host.

  22. Re:Only incidentally similar to su on Systemd Absorbs "su" Command Functionality · · Score: 1

    So, wait, the reason I can't see any authentication going on in the examples is because it's actually creating a container that, presumably, has no access outside of itself? Which would mean it's actually useless as a replacement for su.

  23. Re:Drone It on Test Pilot: the F-35 Can't Dogfight · · Score: 1
    From TFA:

    ...consisted of traditional Basic Fighter Maneuvers in offensive, defensive and neutral setups at altitudes ranging from 10,000 to 30,000 feet.”

  24. Re:Drone It on Test Pilot: the F-35 Can't Dogfight · · Score: 1
    First let me apologise for RTFA'ing. But here's what was said about weight carried:

    The F-35 was flying “clean,” with no weapons in its bomb bay or under its wings and fuselage. The F-16, by contrast, was hauling two bulky underwing drop tanks, putting the older jet at an aerodynamic disadvantage. But the JSF’s advantage didn’t actually help in the end. The stealth fighter proved too sluggish to reliably defeat the F-16, even with the F-16 lugging extra fuel tanks. “Even with the limited F-16 target configuration, the F-35A remained at a distinct energy disadvantage for every engagement,” the pilot reported.