4 ms·

Huh. I wonder if you had, like, an entangled system of X and Y, if something comes under the gravitational influence of that system X-Y, what directions would t

Huh. I wonder if you had, like, an entangled system of X and Y, if something comes under the gravitational influence of that system X-Y, what directions would the gravitation acceleration go? Would it go towards the middle of the two? Would the "detection" of the gravitation acceleration make it collapse towards X/Y? If you didn't detect the gravity acceleration, would the third body enter into the entangled system via gravity? I am about as far from an expert as it's possible to be, but I always wondered what would happen if an entangled system exerted its gravity on something else.

yarg 4y ago
Not necessarily the middle. It's gravity (approximately an inverse square law) operating across the probabilistic distribution of the locations of the system.
- yarg 4y ago
  Just to expand on this, the solution is the problem is a probabilistic integral across infinitely many three-body problems. https://en.wikipedia.org/wiki/Three-body_problem https://en.wikipedia.org/wiki/Three-body_problem Actually, the gradient function is the probabilistic integral across the gradient functions of infinitely many three-body problems. The actual solution is the integral of that.
jerf 4y ago
While not commonly expressed this way as it is usually phrased in terms of much flashier black holes, this is one of the questions that we need a (correct) theory of quantum gravitation to deal with. I think the best answer is nobody knows. I think you could make a decent handwavy case that this probably produces some sort of gravity entanglement that will collapse (or whatever, insert your favorite theory here) later, as it would ultimately be very strange if gravity is not itself "quantum" w.r.t. also ending up with imaginary probabilities and the quantum effects that naturally follow from that, but this has neither been successfully formalized into math nor experimentally verified.
- MilStdJunkie 4y ago
  Oh interesting! That makes sense! I guess measuring the gravitational attraction of entangled systems is probably a few dozen orders of magnitude beyond any sensor that exists, unless it's entangled black hole stuff. Even if you cranked up a proton to 90% C, that's still just adding maybe an order of magnitude, then you have to entangle it with another relativistic proton, and then shoot another particle so it can maybe enter the two entangled relativistic protons' gravitational field(s?). This sounds like a great experimental design to prove a whole lotta nothin'.
- zmgsabst 4y ago
  > as it would ultimately be very strange if gravity is not itself "quantum" w.r.t. also ending up with imaginary probabilities and the quantum effects that naturally follow from that Is there an argument for this rather than the other way? I’m probably misremembering, but I thought I’d seen something about spacetime being continuous and particles being braids in that aether — but did that blow up when string theory didn’t happen at the LHC?
  - kadoban 4y ago
    > but did that blow up when string theory didn’t happen at the LHC? String theory isn't really possible to rule out via any experiment I've ever seen. At best you can rule out some versions of it, but even that is hard AFAIK.
  - gpderetta 4y ago
    AFAIK non-stringy quantum gravity theories have been proposed.
- riskable 4y ago
  Couldn't you do an experiment where you entangle two particles at different altitudes and measure to see which one drops faster? If they're entangled and influenced by their local gravity then you'd expect both to fall at some half-way point (their average) in terms of gravitational influence since they'd each be influenced by the other's gravity (equally?). If they both drop at the same exact speed (and it's not what it should be from a locality standpoint) it would confirm that local gravity is having an impact on both particles at the same time. In other words, if you drop a particle from 10 meters at sea level it would fall at a certain rate. However, if you entangle it with a particle that's 6,000 meters up that particle at sea level might drop at a rate equivalent to what it would at around 3,000 meters (assuming gravity influences both particles equally). If that's how it worked then in theory you could entangle a large mass of particles here on earth with some out in space and you'd only need a fraction of the amount of energy for the particles here on earth to reach escape velocity.
whatshisface 4y ago
Nonrelativistic quantum mechanics offers an answer to your question. You can write a Hamiltonian involving a Newtonian gravitational potential and the answer you will get out is that your third particle, Z, will enter into a superposition of two states, one being attracted to point X and the other to point Y.
- monktastic1 4y ago
  Indeed. And one result of _that_ is that we can get decoherence even if the objects are otherwise entirely shielded: https://physics.stackexchange.com/questions/34993/reversing-gravitational-decoherence https://physics.stackexchange.com/questions/34993/reversing-...
  - cwillu 4y ago
    Worth noting that this is a question by Scott Aaronson

yarg 4y ago

Not necessarily the middle. It's gravity (approximately an inverse square law) operating across the probabilistic distribution of the locations of the system.

yarg 4y ago
Just to expand on this, the solution is the problem is a probabilistic integral across infinitely many three-body problems. https://en.wikipedia.org/wiki/Three-body_problem https://en.wikipedia.org/wiki/Three-body_problem Actually, the gradient function is the probabilistic integral across the gradient functions of infinitely many three-body problems. The actual solution is the integral of that.

jerf 4y ago

While not commonly expressed this way as it is usually phrased in terms of much flashier black holes, this is one of the questions that we need a (correct) theory of quantum gravitation to deal with. I think the best answer is nobody knows. I think you could make a decent handwavy case that this probably produces some sort of gravity entanglement that will collapse (or whatever, insert your favorite theory here) later, as it would ultimately be very strange if gravity is not itself "quantum" w.r.t. also ending up with imaginary probabilities and the quantum effects that naturally follow from that, but this has neither been successfully formalized into math nor experimentally verified.

MilStdJunkie 4y ago
Oh interesting! That makes sense! I guess measuring the gravitational attraction of entangled systems is probably a few dozen orders of magnitude beyond any sensor that exists, unless it's entangled black hole stuff. Even if you cranked up a proton to 90% C, that's still just adding maybe an order of magnitude, then you have to entangle it with another relativistic proton, and then shoot another particle so it can maybe enter the two entangled relativistic protons' gravitational field(s?). This sounds like a great experimental design to prove a whole lotta nothin'.
zmgsabst 4y ago
> as it would ultimately be very strange if gravity is not itself "quantum" w.r.t. also ending up with imaginary probabilities and the quantum effects that naturally follow from that Is there an argument for this rather than the other way? I’m probably misremembering, but I thought I’d seen something about spacetime being continuous and particles being braids in that aether — but did that blow up when string theory didn’t happen at the LHC?
- kadoban 4y ago
  > but did that blow up when string theory didn’t happen at the LHC? String theory isn't really possible to rule out via any experiment I've ever seen. At best you can rule out some versions of it, but even that is hard AFAIK.
- gpderetta 4y ago
  AFAIK non-stringy quantum gravity theories have been proposed.
riskable 4y ago
Couldn't you do an experiment where you entangle two particles at different altitudes and measure to see which one drops faster? If they're entangled and influenced by their local gravity then you'd expect both to fall at some half-way point (their average) in terms of gravitational influence since they'd each be influenced by the other's gravity (equally?). If they both drop at the same exact speed (and it's not what it should be from a locality standpoint) it would confirm that local gravity is having an impact on both particles at the same time. In other words, if you drop a particle from 10 meters at sea level it would fall at a certain rate. However, if you entangle it with a particle that's 6,000 meters up that particle at sea level might drop at a rate equivalent to what it would at around 3,000 meters (assuming gravity influences both particles equally). If that's how it worked then in theory you could entangle a large mass of particles here on earth with some out in space and you'd only need a fraction of the amount of energy for the particles here on earth to reach escape velocity.

whatshisface 4y ago

Nonrelativistic quantum mechanics offers an answer to your question. You can write a Hamiltonian involving a Newtonian gravitational potential and the answer you will get out is that your third particle, Z, will enter into a superposition of two states, one being attracted to point X and the other to point Y.

monktastic1 4y ago
Indeed. And one result of _that_ is that we can get decoherence even if the objects are otherwise entirely shielded: https://physics.stackexchange.com/questions/34993/reversing-gravitational-decoherence https://physics.stackexchange.com/questions/34993/reversing-...
- cwillu 4y ago
  Worth noting that this is a question by Scott Aaronson