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They measured a bigger gravitational effect on the particle, because the superpositional pair of the particle flew closer to a mass than the actually measured p

by Arwill 5y ago

They measured a bigger gravitational effect on the particle, because the superpositional pair of the particle flew closer to a mass than the actually measured particle. Regarding the "never touch", gravity decreases with distance squared, so it diminishes quickly with distance. There is a big difference in being near the mass, as opposed to feeling the dimished effect of it from far.

pdonis 5y ago
> They measured a bigger gravitational effect on the particle No, they didn't. They measured a phase shift in the particle's wave function. There is no "gravitational force" in free fall, and the particles were in free fall. > gravity decreases with distance squared The Newtonian gravitational force does, but the Newtonian gravitational force is irrelevant for an experiment conducted in free fall, as this one was. The gravitational potential is the key thing being measured, and it's not the potential due to the Earth, it's the potential due to a 1-kg "source mass". > There is a big difference in being near the mass, as opposed to feeling the dimished effect of it from far. The particles were near the 1-kg source mass.
- hilbert42 5y ago
  "The gravitational potential is the key thing being measured, and it's not the potential due to the Earth, it's the potential due to a 1-kg "source mass"." Right, it seems that for many the 'potential' worldview is hard to grasp (it was for me too until it drawned on me that it's important). I blame this on poor training and poor textbooks, they don't emphasize the importance of potentials. Also, we seem to grow up with a 'fields' perspective, electric fields and so on. Maxwell was on the 'right' path with his original formulation of his equations where potentials were involved. However, when Heaviside reformulated them to the 'vector' view we quickly lost the 'potential' one. No doubt, Heaviside's formulation is incredibly useful in electrical engineering and eleconics and as you'd know that's how they usually appear in textbooks. Trouble is, outside advanced physics texts the 'potential' view is usually omitted. Educators really need to fix this. Another problem is that the description of a potential is not up to scratch. All too often we seem to be stuck with highschool physics descriptions - those that involve pith balls. The concept that we never measure absolute energy, but only differences often gets lost when describing potentials. You'd reckon that after Feynman's well documented whingeing about the fact he wasn't taught about potentials early enough that you'd think by now educators would have had sufficient time to have rewritten their notes but apparently they've not.
- Arwill 5y ago
  >Each of those two sets of atoms were split into superpositions, with one path traveling closer to the mass than the other, separated by about 25 centimeters One path of the particle in superposition was closer to the 1.25Kg mass than the other path, and they did measure a difference when doing that. I don't know if you are trying to be pedantic, or just want to contradict. I know what you are saying, but the the expression "not touching the field" makes perfect sense to me. Try plotting the 25cm distance difference for the 1.25Kg mass, and see if it makes a difference or not...
  - pdonis 5y ago
    > they did measure a difference They measured a phase shift in the wave function, as I said. They did not measure any direct difference in "gravitational effect" on the particles, as for example a difference in bending of their trajectories due to the source mass would be. > the expression "not touching the field" makes perfect sense to me The problem with it, as several commenters have pointed out, is that you can't shield anything from gravity. The "not touching the field" comes from electromagnetism, where you can shield things from the field. So the "not touching the field" interpretation, while it works for EM, does not work for gravity.
- martopix 5y ago
  > There is no "gravitational force" in free fall, and the particles were in free fall. what is free fall in a reference frame is a particle subject to a force in another reference frame.
  - pdonis 5y ago
    No, it isn't. Free fall is invariant: attach an accelerometer to the object and it reads zero. That is true regardless of your choice of reference frame.

by Arwill 5y ago

pdonis 5y ago

> They measured a bigger gravitational effect on the particle No, they didn't. They measured a phase shift in the particle's wave function. There is no "gravitational force" in free fall, and the particles were in free fall. > gravity decreases with distance squared The Newtonian gravitational force does, but the Newtonian gravitational force is irrelevant for an experiment conducted in free fall, as this one was. The gravitational potential is the key thing being measured, and it's not the potential due to the Earth, it's the potential due to a 1-kg "source mass". > There is a big difference in being near the mass, as opposed to feeling the dimished effect of it from far. The particles were near the 1-kg source mass.

hilbert42 5y ago
"The gravitational potential is the key thing being measured, and it's not the potential due to the Earth, it's the potential due to a 1-kg "source mass"." Right, it seems that for many the 'potential' worldview is hard to grasp (it was for me too until it drawned on me that it's important). I blame this on poor training and poor textbooks, they don't emphasize the importance of potentials. Also, we seem to grow up with a 'fields' perspective, electric fields and so on. Maxwell was on the 'right' path with his original formulation of his equations where potentials were involved. However, when Heaviside reformulated them to the 'vector' view we quickly lost the 'potential' one. No doubt, Heaviside's formulation is incredibly useful in electrical engineering and eleconics and as you'd know that's how they usually appear in textbooks. Trouble is, outside advanced physics texts the 'potential' view is usually omitted. Educators really need to fix this. Another problem is that the description of a potential is not up to scratch. All too often we seem to be stuck with highschool physics descriptions - those that involve pith balls. The concept that we never measure absolute energy, but only differences often gets lost when describing potentials. You'd reckon that after Feynman's well documented whingeing about the fact he wasn't taught about potentials early enough that you'd think by now educators would have had sufficient time to have rewritten their notes but apparently they've not.
Arwill 5y ago
>Each of those two sets of atoms were split into superpositions, with one path traveling closer to the mass than the other, separated by about 25 centimeters One path of the particle in superposition was closer to the 1.25Kg mass than the other path, and they did measure a difference when doing that. I don't know if you are trying to be pedantic, or just want to contradict. I know what you are saying, but the the expression "not touching the field" makes perfect sense to me. Try plotting the 25cm distance difference for the 1.25Kg mass, and see if it makes a difference or not...
- pdonis 5y ago
  > they did measure a difference They measured a phase shift in the wave function, as I said. They did not measure any direct difference in "gravitational effect" on the particles, as for example a difference in bending of their trajectories due to the source mass would be. > the expression "not touching the field" makes perfect sense to me The problem with it, as several commenters have pointed out, is that you can't shield anything from gravity. The "not touching the field" comes from electromagnetism, where you can shield things from the field. So the "not touching the field" interpretation, while it works for EM, does not work for gravity.
martopix 5y ago
> There is no "gravitational force" in free fall, and the particles were in free fall. what is free fall in a reference frame is a particle subject to a force in another reference frame.
- pdonis 5y ago
  No, it isn't. Free fall is invariant: attach an accelerometer to the object and it reads zero. That is true regardless of your choice of reference frame.