John Cuthber Posted March 14, 2016 Posted March 14, 2016 Neutron stars are not typically thought of as "small" so they may be a little off- topic but neutrons have been seen to fall under gravity. https://en.wikipedia.org/wiki/Ultracold_neutrons#Observation_of_the_gravitational_interactions_of_the_neutron
Robittybob1 Posted March 14, 2016 Author Posted March 14, 2016 (edited) Neutron stars are not typically thought of as "small" so they may be a little off- topic but neutrons have been seen to fall under gravity. https://en.wikipedia.org/wiki/Ultracold_neutrons#Observation_of_the_gravitational_interactions_of_the_neutron I found the abstract only at this stage but it sounds really important, I wonder if it has gained acceptance? http://www.nature.com/nature/journal/v415/n6869/full/415297a.html The discrete quantum properties of matter are manifest in a variety of phenomena. Any particle that is trapped in a sufficiently deep and wide potential well is settled in quantum bound states. For example, the existence of quantum states of electrons in an electromagnetic field is responsible for the structure of atoms16, and quantum states of nucleons in a strong nuclear field give rise to the structure of atomic nuclei17. In an analogous way, the gravitational field should lead to the formation of quantum states. But the gravitational force is extremely weak compared to the electromagnetic and nuclear force, so the observation of quantum states of matter in a gravitational field is extremely challenging. Because of their charge neutrality and long lifetime, neutrons are promising candidates with which to observe such an effect. Here we report experimental evidence for gravitational quantum bound states of neutrons. The particles are allowed to fall towards a horizontal mirror which, together with the Earth's gravitational field, provides the necessary confining potential well. Under such conditions, the falling neutrons do not move continuously along the vertical direction, but rather jump from one height to another, as predicted by quantum theory The number of neutrons and mass would be highly correlated wouldn't it? So in the above abstract was the neutron falling from one height to the other the action of a quanta? A quanta of what? Energy? Is that a change in its mass? Nesvizhevsky is on YT. found a paper: http://arxiv.org/pdf/hep-ph/0306198.pdf "Measurement of quantum states of neutrons in the Earth’s gravitational field" @John - Do you understand what that paper is about? Edited March 14, 2016 by Robittybob1
swansont Posted March 14, 2016 Posted March 14, 2016 Individual Neutrons. it seems like you can have a very dense object quite close to another F = G m^2 / ( r^2 ). R^2 would be a very small number so can the force be measured? You quoted something in the OP about a measurement of neutrons being affected by the gravity of the earth. That's cutting-edge — something that defines the limit of what we can measure. You should be able to extrapolate from that. Neutrons are something like 1050 times less massive (10-27 kg vs 6 x 1024 kg). They would have to be 1025 times closer just to have the same tiny effect. (That would be ~10^-18 m) ——— Nesvizhevsky's paper describes neutron behavior in a quantum well that's defined by gravity.
Robittybob1 Posted March 14, 2016 Author Posted March 14, 2016 (edited) You quoted something in the OP about a measurement of neutrons being affected by the gravity of the earth. That's cutting-edge — something that defines the limit of what we can measure. You should be able to extrapolate from that. Neutrons are something like 1050 times less massive (10-27 kg vs 6 x 1024 kg). They would have to be 1025 times closer just to have the same tiny effect. (That would be ~10^-18 m) ——— Nesvizhevsky's paper describes neutron behavior in a quantum well that's defined by gravity. I looked up Neutron radius yesterday and they gave it as 8.00E-16 meters so that is quite close to "~10^-18 m". The experiment is then in the lab rather than in space. From that and the mass I worked out the G force between two protons was 2.93E-34 Newtons [Edit: not Joules.] Edited March 14, 2016 by Robittybob1
imatfaal Posted March 14, 2016 Posted March 14, 2016 I looked up Neutron radius yesterday and they gave it as 8.00E-16 so that is quite close to "~10^-18 m". The experiment is then in the lab rather than in space. From that and the mass I worked out the G force between two protons was 2.93E-34 Joules. "force...was...joules" - really?
Robittybob1 Posted March 14, 2016 Author Posted March 14, 2016 "force...was...joules" - really? Newtons sorry
swansont Posted March 14, 2016 Posted March 14, 2016 I looked up Neutron radius yesterday and they gave it as 8.00E-16 meters so that is quite close to "~10^-18 m". The experiment is then in the lab rather than in space. From that and the mass I worked out the G force between two protons was 2.93E-34 Newtons [Edit: not Joules.] No, it's not that close. It's 10x bigger. And that only gets you to this minuscule effect that was measured in the lab. The nuclear interaction between them is going to be much, much, much larger. Nuclear binding energies of order MeV. The energy mentioned in the paper were peV. 12 orders of magnitude different.
Robittybob1 Posted March 14, 2016 Author Posted March 14, 2016 (edited) No, it's not that close. It's 10x bigger. And that only gets you to this minuscule effect that was measured in the lab. The nuclear interaction between them is going to be much, much, much larger. Nuclear binding energies of order MeV. The energy mentioned in the paper were peV. 12 orders of magnitude different. Are you working with different radii than in Wikipedia? The free neutron has a mass of about 1.675×10−27 kg (equivalent to 939.6 MeV/c2, or 1.0087 u).[3] The neutron has a mean square radius of about 0.8×10−15 m, or 0.8 fm, 0.8 X 10-15 m is the same as 8.0E-16 isn't it? So that force was when the two neutons were theoretically touching. Is this not possible? What radius are you using please? Edited March 14, 2016 by Robittybob1
swansont Posted March 14, 2016 Posted March 14, 2016 0.8 X 10-15 m is the same as 8.0E-16 isn't it? So that force was when the two neutons were theoretically touching. Is this not possible? What radius are you using please? Yes. Still much bigger than 10-18. (Sorry, I had a typo before, and it propagated) 10-16 is 100x bigger than 10-18. That just makes your case 10x worse. They need to be at 10-18m to have peV of gravitational potential energy, and they can't get that close, by a fairly large factor. At this scale the details are unimportant. At least 12 frikkin' orders of magnitude smaller than the nuclear binding energy. It's not a factor. You can't measure the effect.
Robittybob1 Posted March 14, 2016 Author Posted March 14, 2016 (edited) size of the units. It is commonly used with the metric prefixes milli-, kilo-, mega-, giga-, tera-, peta- or exa- (meV, keV, MeV, GeV, TeV, PeV and EeV respectively). Thus meV stands for milli-electron volt. so that is 10 to the power of -3, 3,6,9,12,15, and 18 respectively. They are 8.0E-16 m but they need to be in the order of 8.0E-18 m apart to have a PeV of gravitational potential. OK I might have a better understanding of the experiment now. 2.93E-34 Joules is something else. absolute gravitational potential = Gm/(r^2) for a neutron = 1.75E-07 J/kg "At least 12 frikkin' orders of magnitude smaller than the nuclear binding energy." Edited March 14, 2016 by Robittybob1
imatfaal Posted March 14, 2016 Posted March 14, 2016 Do we see gravitational lensing of neutrinos? The answer at moment is no - but only because we are so rubbish at detecting neutrinos You would need a super nova behind (or close to behind) the sun or a nearby galaxy. But we can sometimes detect Supernovae behind things because the rays of light that (if they travelled in a straight line) should miss the earth are bent back by gravity and we can detect them (overly simplistic explanation but good enough). If Neutrinos from this Supernova are not affected by gravity then we should measure a single burst of neutrinos at the time needed to travel straight line distance (not light path geodesic) from Nova to earth. No need to worry about the fact that there is a galaxy in the way - the neutrinos will go through as if it wasn't there. If Neutrinos from the Supernova are affected by gravity then we should see a spread over time of neutrinos - from those that travelled direct, those that have been mildly bent, though to those which have been gravitationally lensed the most and followed the trajectory outside the lensing galaxy on the same path as the light.
swansont Posted March 14, 2016 Posted March 14, 2016 size of the units. so that is 10 to the power of -3, 3,6,9,12,15, and 18 respectively. They are 8.0E-16 m but they need to be in the order of 8.0E-18 m apart to have a PeV of gravitational potential. OK I might have a better understanding of the experiment now. 2.93E-34 Joules is something else. absolute gravitational potential = Gm/(r^2) for a neutron = 1.75E-07 J/kg "At least 12 frikkin' orders of magnitude smaller than the nuclear binding energy." Not PeV. peV. Peta vs pico. The difference between the two is 24 orders of magnitude.
Robittybob1 Posted March 14, 2016 Author Posted March 14, 2016 Not PeV. peV. Peta vs pico. The difference between the two is 24 orders of magnitude. "1 joule is equal to 1.0E+12 picojoule" So pico is much smaller than Peta! I've never used these descriptions before. Thanks
Robittybob1 Posted March 15, 2016 Author Posted March 15, 2016 Why do you bring nuclear binding energy to a discussion on gravity? The free neutrons have more energy /mass than if they were bound but they don't just get attracted together by gravity and then lose this binding energy and fuse together. So how close can two neutrons be naturally without being forced together. In that experiment they seem to have neutrons going into their device and they are observed to fall in stages. Is that the Earth's gravity pulling them down in quanta? http://arxiv.org/pdf/hep-ph/0306198.pdf I'll reread the paper but I need a bit more understanding to start comprehending it.
swansont Posted March 15, 2016 Posted March 15, 2016 Why do you bring nuclear binding energy to a discussion on gravity? The free neutrons have more energy /mass than if they were bound but they don't just get attracted together by gravity and then lose this binding energy and fuse together. So how close can two neutrons be naturally without being forced together. In that experiment they seem to have neutrons going into their device and they are observed to fall in stages. Is that the Earth's gravity pulling them down in quanta? http://arxiv.org/pdf/hep-ph/0306198.pdf I'll reread the paper but I need a bit more understanding to start comprehending it. I bring it up because you can't get the neutrons that close together without the strong interaction coming into play. In that experiment it is individual neutrons being affected by the gravitational pull of the earth. They are isolated from any other interactions. You can't do this with two individual neutrons.
Robittybob1 Posted March 15, 2016 Author Posted March 15, 2016 I bring it up because you can't get the neutrons that close together without the strong interaction coming into play. In that experiment it is individual neutrons being affected by the gravitational pull of the earth. They are isolated from any other interactions. You can't do this with two individual neutrons. So is the "strong interaction" the same or connected to the "binding energy" in some way? OK I'll do some more homework.
Robittybob1 Posted March 24, 2016 Author Posted March 24, 2016 On another thread Strange linked to this article and it also should be linked here. ALPHA Probes Antimatter Gravity http://alpha.web.cern.ch/node/248 we can exclude the possibility that the gravitational mass of antihydrogen is more than 110 times its inertial mass, or that it falls upwards with a gravitational mass more than 65 times its inertial mass. So does that mean it could still fall upwards?
imatfaal Posted March 24, 2016 Posted March 24, 2016 On another thread Strange linked to this article and it also should be linked here. ALPHA Probes Antimatter Gravity http://alpha.web.cern.ch/node/248 So does that mean it could still fall upwards? Yes
Strange Posted March 24, 2016 Posted March 24, 2016 Is it possible to interpret these error bounds as probabilities; for example, that it is roughly twice as likely that they do fall down than they fall up?
swansont Posted March 24, 2016 Posted March 24, 2016 Is it possible to interpret these error bounds as probabilities; for example, that it is roughly twice as likely that they do fall down than they fall up? No, not really. Imagine a neutrino time-of-flight measurement (similar to the one a few years ago, but without the bias that gave the superluminal result). You determine that the speed of a neutrino is c ± 0.001 m/s. That doesn't mean there's a 50% chance that it travels faster than c — that would violate some well-established physics. To make that claim, you would need a result above c where c was statistically excluded.
Robittybob1 Posted March 24, 2016 Author Posted March 24, 2016 Yes What does fall upwards mean? Fall down seems to have a destination it will end up colliding with something, and we call that down from the perspective of an observer there but where does upward terminate? [i'm asking this is a general sense not specifically as occurred in their experiment.]
imatfaal Posted March 24, 2016 Posted March 24, 2016 What does fall upwards mean? Fall down seems to have a destination it will end up colliding with something, and we call that down from the perspective of an observer there but where does upward terminate? [i'm asking this is a general sense not specifically as occurred in their experiment.] Under the action of just gravity on earth the antimatter particle would go upwards - ie repulsion rather than attraction. NB This is very very unlikely - it is just that we do not have the tech to rule it out yet
swansont Posted March 24, 2016 Posted March 24, 2016 What does fall upwards mean? Fall down seems to have a destination it will end up colliding with something, and we call that down from the perspective of an observer there but where does upward terminate? [i'm asking this is a general sense not specifically as occurred in their experiment.] Up and down are indications of direction. They don't terminate, as such. The coordinate system doesn't care that the earth is there; the axes extend to infinity. That the earth gets in the way of the motion is a constraint of the specific problem.
Robittybob1 Posted March 24, 2016 Author Posted March 24, 2016 (edited) Up and down are indications of direction. They don't terminate, as such. The coordinate system doesn't care that the earth is there; the axes extend to infinity. That the earth gets in the way of the motion is a constraint of the specific problem. Do we then put a set of coordinates on an antimatter particle and one on a matter particle and see in which direction they move under the influence of their own gravity? If they move away from each other they fall upwards. That would work in deep space but all experiments are done under Earth's gravity which is going to define "down". So is "up" just the opposite to our down? http://www.nutriculamagazine.com/a-new-approach-of-gravity-effect-on-antimatter-cern/ article describing an actual experiment. Edited March 24, 2016 by Robittybob1
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