41
$\begingroup$

It is possible to accelerate a charged particle in an electric field, how is it possible to accelerate a neutron? How can we control its velocity?

$\endgroup$
  • 10
    $\begingroup$ A practical beamed neutron source is usually a proton accelerator hitting a fixed target. The US has a facility called Spallation Neutron Source neutrons.ornl.gov/about. Here is more about the neutron spectrum of this facility than you ever may want to know... trace.tennessee.edu/cgi/… $\endgroup$ – CuriousOne Sep 18 '14 at 0:51
  • $\begingroup$ Can a gravitational field not be used in principle? $\endgroup$ – Bryson S. Sep 18 '14 at 14:02
  • 3
    $\begingroup$ And for sake of variety, here is the UK's version. $\endgroup$ – Holloway Sep 19 '14 at 8:56
40
$\begingroup$

Basically, the answer is no, it's not possible.

When we produce neutrons for research purposes, we have to produce them using nuclear reactions. They come out of the nuclear reactions with energies that are determined by the reaction, are not otherwise under our control, and that are on the MeV energy scale of nuclear physics. Examples of a neutron source would be a nuclear reactor or a plutonium-beryllium source.

If the nuclear reaction happens in a particle accelerator, then the center of mass frame can be moving at high velocities. If the beam is heavier than the target and the beam's velocity is high, then this can result in the production of neutrons that are kinematically focused and produced at a high velocity.

Once the neutrons are produced, it is possible to thermalize them, e.g., in water or paraffin. This produces a set of neutrons whose energy spectrum is Maxwellian and corresponds to the temperature of the substance used to thermalize them.

One can also filter neutrons by velocity. So, e.g., once a population of thermal neutrons has been produced, one could select a subset of them with some small range of velocities. Two techniques for doing this are measuring their time of flight (and simply ignoring those that have the wrong time of flight) and refracting them through a prism -- which, surprisingly, works very much like an optical prism.

A classic paper that uses and describes a bunch of these techniques is Zeilinger et al., "Single- and Double-slit diffraction of neutrons," Rev Mod Phys 60 (1988) 1067, in which they measured double-slit diffraction of neutrons. The paper can be found online (possibly illegally if your country's copyright laws are as repressive as mine).

$\endgroup$
  • 5
    $\begingroup$ Of course you can generate a very modest force in the Stern-Gerlach approach, but it's not worth anything for making a beam. $\endgroup$ – dmckee Sep 18 '14 at 0:46
  • 2
    $\begingroup$ It all depends on the magnetic field gradient... I am pretty sure that neutron stars produce quite remarkable neutron energies that way. That, of course, is not a useful technique here on Earth. $\endgroup$ – CuriousOne Sep 18 '14 at 0:53
  • 1
    $\begingroup$ ah Stern-Gerlach is a good approach. Even a tiny gradient over a long waveguide could be good enough for focusing the neutron beam, even if not producing much longitudinal acceleration. You should write a paper about it @dmckee $\endgroup$ – lurscher Sep 18 '14 at 2:05
  • 3
    $\begingroup$ Professional opinion too long for comment box, but Stern-Gerlach steering is not a practical way to separate neutrons by spin (though it is a possible concern if you're doing a parity experiment and sensitive to part-per-billion beam steering). $\endgroup$ – rob Sep 18 '14 at 2:16
  • 4
    $\begingroup$ Paraffin wax appears as a neutron moderator in lots of historical experiments, but its purpose (a solid, machinable moderator/absorber which doesn't handle much power) is nowadays handled by high-density polyethylene plastic. "Thermal" (room-temperature) neutrons typically come off of water moderators; "cold" neutrons typically are moderated by liquid hydrogen. There are also "ultra-cold" neutrons which are probably off-topic for this question. $\endgroup$ – rob Sep 18 '14 at 2:21
22
$\begingroup$

Although a neutron is electrically neutral, it has a non-zero magnetic dipole moment. It interacts with a magnetic field to give a potential

$$ U = \vec{\mu} \cdot \vec{B} $$

A gradient of magnetic field strength will give a force

$$ \vec{F} = \nabla|\vec{\mu} \cdot \vec{B} | $$

It's not possible to produce large, sustained field gradients, nor is it possible to rapidly modulate them in the way that charged particles are accelerated in radio frequency cavities. This means that it's only good for pretty small accelerations and low energies. Quantitatively, the neutron has $\mu \approx 50\,\mathrm{neV/T}$, and typical magnetic fields in laboratories are a few Tesla, so only neutrons with energies $\lt few \times 100\,\mathrm{neV}$ are appreciably affected.

Some experiments trap ultra-cold neutrons in a magnetic bottle -- in a trap, there's always a restoring force, and this is the method that they use.

To make a high energy neutron beam, accelerators produce a high energy beam of charged particles (usually protons), and use a target to convert them to neutrons at the last possible instant.

$\endgroup$
  • $\begingroup$ This is true, but it's worth being quantitative: the neutron has $\mu\approx 50\,\mathrm{neV/T}$, so only ultra-cold neutrons with energies $E \lesssim \text{few}\times100\,\mathrm{neV}$ energies will experience easily-measurable accelerations due to laboratory fields. $\endgroup$ – rob Sep 30 '14 at 5:43
5
$\begingroup$

Yes! Neutrons are electrically neutral, but they have a magnetic moment. You can accelerate a bar magnet with a magnetic field, so you can also accelerate a neutron with a magnetic field. For most beams, the change in energy is pretty negligible, but there's a major exception for ultra-cold neutrons (UCN), which just so happen to be my specialty.

UCN have less than 300 neV of kinetic energy (read: slow enough bounce off a polished Ni wall), and they gain 60neV/T, so a 5T magnet (imagine a very strong MRI magnet) can more than double the energy of a neutron, provided its going slow enough (and is polarized correctly).

UCN experiments often have a region where the UCN are "converted" (read: down scattered from higher energies) that has to support gas pressure, and a region that works best under high vacuum. Those to regions are separated by a thin piece of metal (often aluminum) called a foil. You'd lose a lot of your neutrons in the foil if you just let the fast ones through, so you typically set up a strong magnet around the foil, which accelerates the neutrons so they have more energy than the foil's barrier. When the neutrons exit the magnet, they lose that energy again, and return to the <300neV population you want to study.

$\endgroup$
  • $\begingroup$ In terms of feasability, would it be possible to get a neutron accelerated to multi-MeV - GeV range using an AC oscilating electro-magnet? $\endgroup$ – Hydro Guy Sep 23 '14 at 21:48
  • 1
    $\begingroup$ This answer would be more useful if it specified that getting a linear acceleration (rather than just a rotation) requires a non-uniform magnetic field. $\endgroup$ – dmckee Sep 23 '14 at 23:00
  • 1
    $\begingroup$ @HydroGuy, that would be really hard. In fact, we would love the opposite effect: neutrons are "made" (liberated, really) in the MeV range, (from reactors and spallation sources), but we need them in the neV scale. That's a magnetic field of something like $10^{14}$ T, which is somewhere between one thousand and one million times more intense than what we think magnetars put out. Our technology isn't quite that advanced yet! $\endgroup$ – Menger Sponge Apr 19 '16 at 18:20
3
$\begingroup$

The neutron can be attached to a proton via the Strong Force by colliding a high-energy proton with the neutron, and then the proton-neutron atom can be accelerated with a regular electric field. Gravity can also accelerate a neutron.

$\endgroup$
  • 3
    $\begingroup$ By "proton-neutron atom," do you mean a deuteron? $\endgroup$ – Ben Crowell Sep 19 '14 at 5:24
3
$\begingroup$

How is it possible to accelerate a neutron?

Neutrons have a dipole moment, so they may be 'accelerated' insofar as they will turn in a magnetic field – that is their primary interaction with the electromagnetic field.

It is possible to accelerate a charged particle in an electric field, how is it possible to accelerate a neutron?

Neutrons also interact by the gravity, strong and weak forces, so in typical (Earth based) experiments, we wait for neutrons to get kicked out of nuclei in radioactive isotopes. One could also aim the neutrons downward to accelerate them with the gravity force, but that won't really change the speed much here on Earth. Especially since they have a 10 minute half-life.

How can we control its velocity?

Magnetic fields do no work, so their speed will remain constant, but as I said, you can get them to turn during their brief lives.

If I were being pedantic (and I do physics, so that's a thing), I would say the easiest way is to accelerate a charged particle that has neutrons.

$\endgroup$
  • 2
    $\begingroup$ I'm not sure that it's true that "magnetic fields do no work". This is true for the Lorentz q(v x B) force, but not for mu.Grad(B). $\endgroup$ – Gremlin Sep 19 '14 at 9:08
2
$\begingroup$

I think it may be possible, by making the neutrons precess in a controled way by exposing them with microwaves in a sufficiently strong magnetic field, as in a NMR device. Then instead of applying a magnetic gradient from a conventional magnet, one could use a focussed pulse from a laser to create a magnetic mean-field gradient in order to accelerate the neutron. The laser-pulse will need to be timed along with the orientation of the precessing neutron to obtain a coherent force in a desired direction. The important difference here is that the energy of the laser-pulse can be focussed on a volume of the order of lambda^3, so that for a given pulse energy the strength of the magnetic field scales as lambda^-3, and the gradient of this field will pick up another factor lamda^-1. The timescale at which the field gradient can be maintained does scale with lambda, so the resulting change in velocity can be expected to scale with lambda^-3.

$\endgroup$
  • $\begingroup$ Hi, welcome to Physics. You might like to know that the neutron's magnetic moment is about 50 nano-eV per tesla. A neutron undergoing a spin flip on the surface of a magnetar would have an energy change of only a few kilo-eV. I think your proposal is conceptually interesting but the arithmetic is not practical. $\endgroup$ – rob Jan 7 '18 at 0:19
  • $\begingroup$ Thanks for your comment. I find a value for the neutrons magnetic moment equal to -0.966 E-26 J/T on en.wikipedia.org/wiki/Neutron, which is equal to -600 nano eV. This is indeed a small number but significantly larger than the value you state. $\endgroup$ – Pedro Jan 8 '18 at 2:20
  • $\begingroup$ Please note that the energy change associated with a spin flip of the neutron in a 1 Tesla field is equal to 2 times 600 neV, which corresponds to about 0.12J/mole. Using that a mole of neutrons weighs about a gram this amount of energy is good for an acceleration from zero speed to about 15 meters per second. $\endgroup$ – Pedro Jan 8 '18 at 6:05
  • $\begingroup$ Your conversion is off by a factor of ten. There's a bunch of interesting physics that happens if you have neutrons with 100 neV energies, including trapping of one spin state by tesla-level fields. Look for "ultra cold neutrons." You have the speed about right, but I'm not sure that's what the asker had in mind. $\endgroup$ – rob Jan 8 '18 at 10:33
  • $\begingroup$ Rob, thanks for pointing out that I was indeed a factor 10 off with the magnetic momeny, ie, -0.966 E-26J/T equals about -60 nano eV. $\endgroup$ – Pedro Jan 9 '18 at 6:47
2
$\begingroup$

One could try to pair super-cold neutrons with equally cold free electrons through their magnetic dipole interaction, similar to the way electrons form Cooper-pairs inside a super-conductor. Note that the about 1000 times weaker magnetic moment of the neutron compared the electron will make an e-n pairing correspondingly less stable compared to e-e pairing, but on the other hand the absence of a repulsive electrostatic force between e and n will make a pairing easier. The pairing may be enhanced by the fact that an e-n pair obeys bosonic Fermi-statistics meaning that they can populate low energy states which are not accessible to individual electrons in the case where the electrons are degenerate. Depending on the strength of the e-n pairing one could accelerate the pairs up to some maximum level of acceleration using an electric field, and subsequently remove the electrons by applying a more powerful electric field.

$\endgroup$
1
$\begingroup$

One could use the fact that ultra-cold neutrons can scatter off a polished surface of a metrial such as diamond or nickel with 100% reflectivity as long as the perpendicular velocity is below a velocity threshold of the order of v=7m/s. This may sound like a small velocity -change, but since the neutron can be localized in an area of the order of its wavelength, which is of the order of its spatial wavelength d=50nm for a neutron travelling at 7m/s, the classical acceleration at the scattering surface can be estimated as v^2/(2*d) which is of the order of 5E8 m/s^2, which is a significant level of acceleration. This is of course a handwaving estimate of the upper bound of the level of acceleration at the surface of a neutron-mirror, but the scale of the result seems encouraging. One could then use a fast-spinning neutron-mirror where the ultra-cold neutrons start near the axis of rotation of the mirror and radially move outward due to the lack of a centripetal force, in order to accelerate the neutrons up to a speed equal to the maximum local velocity of the mirror.

$\endgroup$
  • $\begingroup$ Rather than posting several different answers, you can edit your old answer and combine all information into a single post. Also, note that this site supports mathjax, which is an engine that will typeset math formulas using latex-like code. See here for a tutorial. Cheers! $\endgroup$ – AccidentalFourierTransform Apr 8 '18 at 2:43

protected by AccidentalFourierTransform Apr 8 '18 at 2:48

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.