Particle Accelerators - Revision history

Lwcamp: /* Magnetic fields */

2026-10-24T06:07:32Z

Magnetic fields

← Older revision		Revision as of 22:07, 23 October 2026
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	It can be convenient to split up a particle's velocity into a part parallel to the magnetic field (the <b>longitudinal velocity</b>) and a part perpendicular to the field (the <b>transverse velocity</b>). The longitudinal velocity describes how fast the particle drifts along the magnetic field line. The transverse velocity determines how tightly and how quickly the particle circles around the field lines.		It can be convenient to split up a particle's velocity into a part parallel to the magnetic field (the <b>longitudinal velocity</b>) and a part perpendicular to the field (the <b>transverse velocity</b>). The longitudinal velocity describes how fast the particle drifts along the magnetic field line. The transverse velocity determines how tightly and how quickly the particle circles around the field lines.

	If the gyroradius is much smaller than the extent of the magnetic field, you get a case where the particle approximately cannot cross magnetic field lines. It just spirals around them. On the other hand, if the magnetic field ends before the particle can make a significant portion of a full revolution, then the particle ends up being somewhat deflected from its original trajectory. If the ~~gydroradius~~ is enough to deflect the particle into a region where the magnetic field has a different strength or direction, then things get complicated!		If the gyroradius is much smaller than the extent of the magnetic field, you get a case where the particle approximately cannot cross magnetic field lines. It just spirals around them. On the other hand, if the magnetic field ends before the particle can make a significant portion of a full revolution, then the particle ends up being somewhat deflected from its original trajectory. If the gyroradius is enough to deflect the particle into a region where the magnetic field has a different strength or direction, then things get complicated!

	When a particle is spiraling around the magnetic field while drifting down the field lines into regions where the magnetic field is stronger, it will circle around the field lines in a faster, smaller circle. This increases the kinetic energy in the spiraling motion; this kinetic energy must come from the kinetic energy of the longitudinal motion. If the field becomes too strong, the particle will slow down all the way and come to a stop. Then it will be reflected back, pushed the other way down the field lines toward the region where the field is not as strong. This phenomenon is called magnetic mirroring, and is responsible for the way radiation gets trapped in the Van Allen belts around Earth with the particles bouncing back and forth along the Earth's field lines and getting reflected back and forth at the poles where the field is strongest.		When a particle is spiraling around the magnetic field while drifting down the field lines into regions where the magnetic field is stronger, it will circle around the field lines in a faster, smaller circle. This increases the kinetic energy in the spiraling motion; this kinetic energy must come from the kinetic energy of the longitudinal motion. If the field becomes too strong, the particle will slow down all the way and come to a stop. Then it will be reflected back, pushed the other way down the field lines toward the region where the field is not as strong. This phenomenon is called magnetic mirroring, and is responsible for the way radiation gets trapped in the Van Allen belts around Earth with the particles bouncing back and forth along the Earth's field lines and getting reflected back and forth at the poles where the field is strongest.

Lwcamp: /* Nuclear collisions */

2026-10-22T02:57:06Z

Nuclear collisions

← Older revision		Revision as of 18:57, 21 October 2026
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	The simplest thing that can happen is an <b>elastic collision</b>, where no energy goes in to any internal nuclear state or into breaking up the nucleus. An elastic collision just ends up giving you displacement (or possibly phonons, but if you hit the nucleus you usually hit it hard enough to send it flying).		The simplest thing that can happen is an <b>elastic collision</b>, where no energy goes in to any internal nuclear state or into breaking up the nucleus. An elastic collision just ends up giving you displacement (or possibly phonons, but if you hit the nucleus you usually hit it hard enough to send it flying).

	But beyond that you can cause any number of nuclear reactions. You can leave the nucleus in an excited nuclear state, from which it usually relaxes back down to its original state by emitting gamma rays. You can knock out nuclear particles, such as neutrons, protons, deuterons (a proton stuck to a neutron), tritons (a proton stuck to two neutrons), helions (a neutron stuck to two protons), alpha particles (two protons stuck to two neutrons), or potentially even larger nuclear fragments. ~~If you hit~~ a ~~heavy~~ nucleus like ~~uranium, you can cause it~~ to ~~split apart by fission~~. After the nucleus has been hit and any particles knocked out, it will be very hot and will cool off by evaporating off nuclear particles, which are usually neutrons but can occasionally be other nuclear particles as well. To get any of these <b>inelastic collisions</b>, you need to hit the nucleus hard enough. If your beam particle doesn't have enough energy to excite the nucleus or knock particles, these things simply can't happen. As it gets enough energy to excite a particular interaction channel (as they are called), that process becomes possible and as the energy increases the process becomes more and more likely.		But beyond that you can cause any number of nuclear reactions. You can leave the nucleus in an excited nuclear state, from which it usually relaxes back down to its original state by emitting gamma rays. You can knock out nuclear particles, such as neutrons, protons, deuterons (a proton stuck to a neutron), tritons (a proton stuck to two neutrons), helions (a neutron stuck to two protons), alpha particles (two protons stuck to two neutrons), or potentially even larger nuclear fragments. Particularly high energy impacts can knock a quark off of a proton or neutron in the nucleus, resulting in a spray of baryons (like protons and neutrons but also more exotic particles) and mesons (like pions but also more exotic ones) called a <i>hadronic jet</i> as the quarks tear apart and boil new quarks out of the vacuum to make all the newly created particles. After the nucleus has been hit and any particles knocked out, it will be very hot and will cool off by evaporating off nuclear particles, which are usually neutrons but can occasionally be other nuclear particles as well. If the nucleus was very heavy like uranium, it can cool off by splitting apart by fission instead. To get any of these <b>inelastic collisions</b>, you need to hit the nucleus hard enough. If your beam particle doesn't have enough energy to excite the nucleus or knock particles, these things simply can't happen. As it gets enough energy to excite a particular interaction channel (as they are called), that process becomes possible and as the energy increases the process becomes more and more likely.

	A nucleus is only about a femtometer across (1 × 10<sup>-15</sup> m). Which makes it only about 1 × 10<sup>-30</sup> m² in cross sectional area. In the meantime, chemical bonds between atoms are usually about 1/10th of a nanometer (1 × 10<sup>-10</sup> m) long. In condensed matter (liquids and solids), this means that you have somewhere about 1 × 10<sup>30</sup> atoms per cubic meter. A meter thickness of atoms thus presents just about about enough atoms that their nuclei can cover the projected area – in other words, you can expect a particle to go through (very roughly) about a meter of condensed matter before it whacks into a nucleus. More detailed estimates usually have the particles penetrating several tens of centimeters into solid or liquid targets between nuclear collisions.		A nucleus is only about a femtometer across (1 × 10<sup>-15</sup> m). Which makes it only about 1 × 10<sup>-30</sup> m² in cross sectional area. In the meantime, chemical bonds between atoms are usually about 1/10th of a nanometer (1 × 10<sup>-10</sup> m) long. In condensed matter (liquids and solids), this means that you have somewhere about 1 × 10<sup>30</sup> atoms per cubic meter. A meter thickness of atoms thus presents just about about enough atoms that their nuclei can cover the projected area – in other words, you can expect a particle to go through (very roughly) about a meter of condensed matter before it whacks into a nucleus. More detailed estimates usually have the particles penetrating several tens of centimeters into solid or liquid targets between nuclear collisions.

Lwcamp: /* High energy physics */

2026-10-22T02:51:35Z

High energy physics

← Older revision		Revision as of 18:51, 21 October 2026
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	A nuclear strike is a discrete event that will remove the impacting particle from the beam and will (except for elastic collisions) probably produce a spray of radiation.		A nuclear strike is a discrete event that will remove the impacting particle from the beam and will (except for elastic collisions) probably produce a spray of radiation.

	===High energy physics===		====High energy physics====

	As you get to higher and higher energies, you unlock all sorts of new and weird and interesting processes that can happen. Boiling exotic particles out of the vacuum, unifying forces that had been divorced since the big bang, and producing bizarre unstable resonances in the quantum fields. This is what physics build accelerators to measure. From the point of view of getting a beam to its target, or what happens to the target, however, the main effect is mostly that the interacting particle is removed from the beam and a bunch of additional radiation is produced.		As you get to higher and higher energies, you unlock all sorts of new and weird and interesting processes that can happen. Boiling exotic particles out of the vacuum, unifying forces that had been divorced since the big bang, and producing bizarre unstable resonances in the quantum fields. This is what physics build accelerators to measure. From the point of view of getting a beam to its target, or what happens to the target, however, the main effect is mostly that the interacting particle is removed from the beam and a bunch of additional radiation is produced.

Lwcamp: /* Nuclear collisions */

2026-10-22T02:50:36Z

Nuclear collisions

← Older revision		Revision as of 18:50, 21 October 2026
Line 617:		Line 617:
	The simplest thing that can happen is an <b>elastic collision</b>, where no energy goes in to any internal nuclear state or into breaking up the nucleus. An elastic collision just ends up giving you displacement (or possibly phonons, but if you hit the nucleus you usually hit it hard enough to send it flying).		The simplest thing that can happen is an <b>elastic collision</b>, where no energy goes in to any internal nuclear state or into breaking up the nucleus. An elastic collision just ends up giving you displacement (or possibly phonons, but if you hit the nucleus you usually hit it hard enough to send it flying).

	But beyond that you can cause any number of nuclear reactions. You can leave the nucleus in an excited nuclear state, from which it usually relaxes back down to its original state by emitting gamma rays. You can knock out nuclear particles, such as neutrons, protons, deuterons (a proton stuck to a neutron), tritons (a proton stuck to two neutrons), helions (a neutron stuck to two protons), alpha particles (two protons stuck to two neutrons), or potentially even larger nuclear fragments. If you hit a heavy nucleus like uranium, you can cause it to split apart by fission. ~~And if you hit~~ the nucleus ~~really hard you might just shatter~~ it ~~into many tiny fragments of those protons~~, neutrons~~, deuterons, tritons, helions, alpha~~ particles~~, and heavier nuclear fragments mentioned earlier~~. To get any of these <b>inelastic collisions</b>, you need to hit the nucleus hard enough. If your beam particle doesn't have enough energy to excite the nucleus or knock particles, these things simply can't happen. As it gets enough energy to excite a particular interaction channel (as they are called), that process becomes possible and as the energy increases the process becomes more and more likely.		But beyond that you can cause any number of nuclear reactions. You can leave the nucleus in an excited nuclear state, from which it usually relaxes back down to its original state by emitting gamma rays. You can knock out nuclear particles, such as neutrons, protons, deuterons (a proton stuck to a neutron), tritons (a proton stuck to two neutrons), helions (a neutron stuck to two protons), alpha particles (two protons stuck to two neutrons), or potentially even larger nuclear fragments. If you hit a heavy nucleus like uranium, you can cause it to split apart by fission. After the nucleus has been hit and any particles knocked out, it will be very hot and will cool off by evaporating off nuclear particles, which are usually neutrons but can occasionally be other nuclear particles as well. To get any of these <b>inelastic collisions</b>, you need to hit the nucleus hard enough. If your beam particle doesn't have enough energy to excite the nucleus or knock particles, these things simply can't happen. As it gets enough energy to excite a particular interaction channel (as they are called), that process becomes possible and as the energy increases the process becomes more and more likely.

	A nucleus is only about a femtometer across (1 × 10<sup>-15</sup> m). Which makes it only about 1 × 10<sup>-30</sup> m² in cross sectional area. In the meantime, chemical bonds between atoms are usually about 1/10th of a nanometer (1 × 10<sup>-10</sup> m) long. In condensed matter (liquids and solids), this means that you have somewhere about 1 × 10<sup>30</sup> atoms per cubic meter. A meter thickness of atoms thus presents just about about enough atoms that their nuclei can cover the projected area – in other words, you can expect a particle to go through (very roughly) about a meter of condensed matter before it whacks into a nucleus. More detailed estimates usually have the particles penetrating several tens of centimeters into solid or liquid targets between nuclear collisions.		A nucleus is only about a femtometer across (1 × 10<sup>-15</sup> m). Which makes it only about 1 × 10<sup>-30</sup> m² in cross sectional area. In the meantime, chemical bonds between atoms are usually about 1/10th of a nanometer (1 × 10<sup>-10</sup> m) long. In condensed matter (liquids and solids), this means that you have somewhere about 1 × 10<sup>30</sup> atoms per cubic meter. A meter thickness of atoms thus presents just about about enough atoms that their nuclei can cover the projected area – in other words, you can expect a particle to go through (very roughly) about a meter of condensed matter before it whacks into a nucleus. More detailed estimates usually have the particles penetrating several tens of centimeters into solid or liquid targets between nuclear collisions.

Lwcamp at 19:05, 7 March 2026

2026-03-07T19:05:02Z

← Older revision		Revision as of 11:05, 7 March 2026
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	[[Category:Beams]][[Category:Physics & Engineering‏‎]][[Category:Engineering‏‎]]		[[Category:Beams]][[Category:Physics & Engineering‏‎]][[Category:Physics & Math & Engineering]][[Category:Engineering‏‎]]

Lwcamp: /* Nuclear collisions */

2025-04-09T01:15:36Z

Nuclear collisions

← Older revision		Revision as of 17:15, 8 April 2025
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	But beyond that you can cause any number of nuclear reactions. You can leave the nucleus in an excited nuclear state, from which it usually relaxes back down to its original state by emitting gamma rays. You can knock out nuclear particles, such as neutrons, protons, deuterons (a proton stuck to a neutron), tritons (a proton stuck to two neutrons), helions (a neutron stuck to two protons), alpha particles (two protons stuck to two neutrons), or potentially even larger nuclear fragments. If you hit a heavy nucleus like uranium, you can cause it to split apart by fission. And if you hit the nucleus really hard you might just shatter it into many tiny fragments of those protons, neutrons, deuterons, tritons, helions, alpha particles, and heavier nuclear fragments mentioned earlier. To get any of these <b>inelastic collisions</b>, you need to hit the nucleus hard enough. If your beam particle doesn't have enough energy to excite the nucleus or knock particles, these things simply can't happen. As it gets enough energy to excite a particular interaction channel (as they are called), that process becomes possible and as the energy increases the process becomes more and more likely.		But beyond that you can cause any number of nuclear reactions. You can leave the nucleus in an excited nuclear state, from which it usually relaxes back down to its original state by emitting gamma rays. You can knock out nuclear particles, such as neutrons, protons, deuterons (a proton stuck to a neutron), tritons (a proton stuck to two neutrons), helions (a neutron stuck to two protons), alpha particles (two protons stuck to two neutrons), or potentially even larger nuclear fragments. If you hit a heavy nucleus like uranium, you can cause it to split apart by fission. And if you hit the nucleus really hard you might just shatter it into many tiny fragments of those protons, neutrons, deuterons, tritons, helions, alpha particles, and heavier nuclear fragments mentioned earlier. To get any of these <b>inelastic collisions</b>, you need to hit the nucleus hard enough. If your beam particle doesn't have enough energy to excite the nucleus or knock particles, these things simply can't happen. As it gets enough energy to excite a particular interaction channel (as they are called), that process becomes possible and as the energy increases the process becomes more and more likely.

	A nucleus is only about a femtometer across (~~<math>~~1 \times 10^{-15}</~~math~~> m). Which makes it only about ~~<math>~~1 \times 10^{-30}</~~math~~> m² in cross sectional area. In the meantime, chemical bonds between atoms are usually about 1/10th of a nanometer (~~<math>~~1 \times 10^{-10}</~~math~~> m) long. In condensed matter (liquids and solids), this means that you have somewhere about ~~<math>~~1 \times 10^{30}</~~math~~> atoms per cubic meter. A meter thickness of atoms thus presents just about about enough atoms that their nuclei can cover the projected area – in other words, you can expect a particle to go through (very roughly) about a meter of condensed matter before it whacks into a nucleus.		A nucleus is only about a femtometer across (1 × 10<sup>-15</sup> m). Which makes it only about 1 × 10<sup>-30</sup> m² in cross sectional area. In the meantime, chemical bonds between atoms are usually about 1/10th of a nanometer (1 × 10<sup>-10</sup> m) long. In condensed matter (liquids and solids), this means that you have somewhere about 1 × 10<sup>30</sup> atoms per cubic meter. A meter thickness of atoms thus presents just about about enough atoms that their nuclei can cover the projected area – in other words, you can expect a particle to go through (very roughly) about a meter of condensed matter before it whacks into a nucleus. More detailed estimates usually have the particles penetrating several tens of centimeters into solid or liquid targets between nuclear collisions.

	This assumes that a particle that goes through a nucleus interacts with it. This is usually a good assumption for particles that interact by the strong nuclear force, such as protons, neutrons, mesons, and the atomic nuclei of ions. But many particles cannot interact by the strong nuclear force, such as electrons or muons. These are much less likely to interact with an atomic nucleus even if they pass through it, so they will end up going considerably farther between hitting nuclei than the above assumption. But nuclei can interact electromagnetically (they are made up of charged protons, after all, and at a smaller level of charged quarks), so electrons and muons can produce some nuclear interactions; just far less than actual nuclear particles like protons and ions.		This assumes that a particle that goes through a nucleus interacts with it. This is usually a good assumption for particles that interact by the strong nuclear force, such as protons, neutrons, mesons, and the atomic nuclei of ions. But many particles cannot interact by the strong nuclear force, such as electrons or muons. These are much less likely to interact with an atomic nucleus even if they pass through it, so they will end up going considerably farther between hitting nuclei than the above assumption. But nuclei can interact electromagnetically (they are made up of charged protons, after all, and at a smaller level of charged quarks), so electrons and muons can produce some nuclear interactions; just far less than actual nuclear particles like protons and ions.

Lwcamp: /* Beam propagation in vacuum with self-forces */

2025-04-09T01:09:16Z

Beam propagation in vacuum with self-forces

← Older revision		Revision as of 17:09, 8 April 2025
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	<blockquote>		<blockquote>
	<i>Example</i>: Let's use our 1 MW, 10 MeV electron accelerator from before, with its 2 cm aperture and ~~<math>~~K = 1.37 \times 10^{-9}</~~math~~>. Suppose we are using this as an electron cannon, and shooting at a target 2 km away. We want to know how small of a spot we can direct on to our target. We use <~~math~~>~~R_m~~ = 0.01~~</math>~~ m (the beam radius at the aperture) and ~~<math>~~z = 2,000~~</math>~~ m for the distance to the target. This gives us ~~<math>~~F(~~\chi~~) = 10.46~~</math>~~. Looking at the graph, ~~<math>~~F(~~\chi~~)~~</math>~~ is about 10 where ~~<math>\chi</math>~~ is about 12. So we know that at the target the beam spot at the target is about 12 times larger than at the aperture, or about 12 cm across.		<i>Example</i>: Let's use our 1 MW, 10 MeV electron accelerator from before, with its 2 cm aperture and K = 1.37 × 10<sup>-9</sup>. Suppose we are using this as an electron cannon, and shooting at a target 2 km away. We want to know how small of a spot we can direct on to our target. We use R<sub>m</sub> = 0.01 m (the beam radius at the aperture) and z = 2,000 m for the distance to the target. This gives us F(χ) = 10.46. Looking at the graph, F(χ) is about 10 where χ is about 12. So we know that at the target the beam spot at the target is about 12 times larger than at the aperture, or about 12 cm across.
	</blockquote>		</blockquote>

Lwcamp: /* Beam propagation in vacuum with self-forces */

2025-04-09T01:06:56Z

Beam propagation in vacuum with self-forces

← Older revision		Revision as of 17:06, 8 April 2025
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	</blockquote>		</blockquote>

	To get actual numbers, take a look at the picture below. It shows the shape of a beam emitted from an accelerator. The beam reaches a minimum width of <~~math~~>~~R_m~~</~~math~~> at a position ~~<math>~~z=0~~</math>~~. At any distance ~~<math>~~z~~</math>~~ along the beam from this minimum, the beam width will be ~~<math>~~R(z)~~</math>~~.		To get actual numbers, take a look at the picture below. It shows the shape of a beam emitted from an accelerator. The beam reaches a minimum width of R<sub>m</sub> at a position z = 0. At any distance z along the beam from this minimum, the beam width will be R(z).
	<table class="center" style="width: auto; margin-left: auto; margin-right: auto;">		<table class="center" style="width: auto; margin-left: auto; margin-right: auto;">
	<tr>		<tr>
	<td width=600>[[File:Self-charge_shaped_beam.png\|600 px\|frameless]]		<td width=600>[[File:Self-charge_shaped_beam.png\|600 px\|frameless]]
	</table>		</table>
	It is useful to define ~~<math>\chi</math>~~ as the fractional amount by which the beam expands		It is useful to define χ as the fractional amount by which the beam expands
	<div class="center" style="width: auto; margin-left: auto; margin-right: auto;"~~><math~~>		<div class="center" style="width: auto; margin-left: auto; margin-right: auto;">
	~~\chi~~ = ~~\frac{~~R(z)~~}{R_m}.~~		χ = R(z) / R<sub>m</sub>.
	</~~math~~></div>		</div>
	So if the beam doubles in radius, then ~~<math>\chi~~ = 2~~</math>~~.		So if the beam doubles in radius, then χ = 2.

	We also need to introduce a function ~~<math>~~F(~~\chi~~)~~</math>~~, which doesn't have any simple expression (it is defined as an integral over the reciprocal of a logarithm), so it is easiest just to give it in tables and figures.		We also need to introduce a function F(χ), which doesn't have any simple expression (it is defined as an integral over the reciprocal of a logarithm), so it is easiest just to give it in tables and figures.
	<table class="center" style="width: auto; margin-left: auto; margin-right: auto;">		<table class="center" style="width: auto; margin-left: auto; margin-right: auto;">
	<tr>		<tr>
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	</table>		</table>
	To use these to get the beam size at any distance from the beam minimum, we can use<ref>Stanley Humphries, Jr., "Charged Particle Beams", Originally published in 1990 by John Wiley and Sons (QC786.H86 1990, ISBN 0-471-60014-8)</ref>		To use these to get the beam size at any distance from the beam minimum, we can use<ref>Stanley Humphries, Jr., "Charged Particle Beams", Originally published in 1990 by John Wiley and Sons (QC786.H86 1990, ISBN 0-471-60014-8)</ref>
	<div class="center" style="width: auto; margin-left: auto; margin-right: auto;"~~><math~~>		<div class="center" style="width: auto; margin-left: auto; margin-right: auto;">
	F(~~\chi~~) = ~~\sqrt{~~2 K~~} \, \frac{~~z~~}{R_m}~~		F(χ) = √[2 K] z / R<sub>m</sub>
	</~~math~~></div>		</div>
	and		and
	<div class="center" style="width: auto; margin-left: auto; margin-right: auto;"~~><math~~>		<div class="center" style="width: auto; margin-left: auto; margin-right: auto;">
	~~\frac{~~F(~~\chi~~)~~}{\chi}~~ = ~~\sqrt{~~2 K~~} \, \frac{~~z}{R(z)}.		F(χ) / χ = √[2 K] z / R(z).
	~~</math>~~</div>		</div>

	<blockquote>		<blockquote>

Lwcamp: /* Beam propagation in vacuum with self-forces */

2025-04-09T00:59:56Z

Beam propagation in vacuum with self-forces

← Older revision		Revision as of 16:59, 8 April 2025
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	<blockquote>		<blockquote>
	<i>Example</i>: Let's consider an electron beam with ~~<math>~~P = 1~~</math>~~ MW of power and ~~<math>~~V = 10~~</math>~~ GeV energy particles. It needs to have a current of ~~<math>~~I = 0.0001~~</math>~~ A to get this power with this particle energy. The beam will initially be 2 cm wide (~~<math>~~R = 0.01~~</math>~~ m). At this particle energy, ~~<math>\gamma~~ = 19600~~</math>~~, and ~~<math>\beta</math>~~ is so close to one as to make no difference. For a reasonable normalized emittance of <~~math~~>~~\epsilon_n~~ = 1\times 10^{-6}</~~math~~> m~~<math>\cdot</math>~~rad, we end up with a geometrical emittance of ~~<math>\epsilon~~ = 5.1 \times 10^{-11}</~~math~~> m~~<math>\cdot</math>~~rad.		<i>Example</i>: Let's consider an electron beam with P = 1 MW of power and V = 10 GeV energy particles. It needs to have a current of I = 0.0001 A to get this power with this particle energy. The beam will initially be 2 cm wide (R = 0.01 m). At this particle energy, γ = 19600, and β is so close to one as to make no difference. For a reasonable normalized emittance of ε<sub>n</sub> = 1 × 10<sup>-6</sup> m·rad, we end up with a geometrical emittance of ε = 5.1 × 10<sup>-11</sup> m·rad.

	Plugging these numbers in, we find that ~~<math>~~K = 1.56 \times 10^{-21}</~~math~~>. The quantity ~~<math>(\epsilon~~/R~~)^2~~ = 2.6 \times 10^{-17}</~~math~~>. So for this particle beam, emittance will dominate over self-forces and we only really need to worry about emittance. Only if you try to focus it down to about a tenth of a millimeter or less will ~~<math>(\epsilon~~/R~~)^2</math>~~ be comparable to ~~<math>~~K~~</math>~~ such that you would need to take self-forces into account.		Plugging these numbers in, we find that K = 1.56 × 10<sup>-21</sup>. The quantity ε²/R² = 2.6 × 10<sup>-17</sup>. So for this particle beam, emittance will dominate over self-forces and we only really need to worry about emittance. Only if you try to focus it down to about a tenth of a millimeter or less will ε²/R² be comparable to K such that you would need to take self-forces into account.
	</blockquote>		</blockquote>

	<blockquote>		<blockquote>
	<i>Example</i>: Now consider an electron beam with the same ~~<math>~~P = 1~~</math>~~ MW of power but particles with only ~~<math>~~V = 10~~</math>~~ MeV energy. This requires a current of ~~<math>~~I = 0.1~~</math>~~ A. we'll use the same 2 cm wide (~~<math>~~R = 0.01~~</math>~~ m) beam width. At this particle energy, ~~<math>\gamma~~ = 20.57~~</math>~~, and ~~<math>\beta~~ = 0.999~~</math>~~. For the same normalized emittance of <~~math~~>~~\epsilon_n~~ = 1\times 10^{-6}</~~math~~> m~~<math>\cdot</math>~~rad, we end up with a geometrical emittance of ~~<math>\epsilon~~ = 4.86 \times 10^{-8}</~~math~~> m~~<math>\cdot</math>~~rad.		<i>Example</i>: Now consider an electron beam with the same P = 1 MW of power but particles with only V = 10 MeV energy. This requires a current of I = 0.1 A. We'll use the same 2 cm wide (R = 0.01 m) beam width. At this particle energy, γ = 20.57, and β = 0.999. For the same normalized emittance of ε<sub>n</sub> = 1 × 10<sup>-6</sup> m·rad, we end up with a geometrical emittance of ε = 4.86 × 10<sup>-8</sup> m·rad.

	Plugging these numbers in, we find that ~~<math>~~K = 1.37 \times 10^{-9}</~~math~~>. The quantity ~~<math>(\epsilon~~/R~~)^2~~ = 2.36 \times 10^{-11}</~~math~~>. For this particle beam, self-forces will have a much greater effect than emittance.		Plugging these numbers in, we find that K = 1.37 × 10<sup>-9</sup>. The quantity ε²/R² = 2.36 × 10<sup>-11</sup>. For this particle beam, self-forces will have a much greater effect than emittance.
	</blockquote>		</blockquote>

Lwcamp: /* Beam propagation in vacuum with self-forces */

2025-04-09T00:47:28Z

Beam propagation in vacuum with self-forces

← Older revision		Revision as of 16:47, 8 April 2025
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	where q is the charge per particle, I is the electric current in the beam, 1 / (4 π ε<sub>0</sub>) = 8.9875517923 × 10<sup>9</sup> N m²/C² is the Coulomb constant, m is the particle mass, β is the speed as a fraction of the speed of light, γ is the Lorentz factor, c is the speed of light, and R is the radius of the beam. If K is much larger than ε²/R² the beam will be dominated by self-force expansion; for K much smaller than ε²/R² it will be dominated by emittance (and if K ≈ ε²/R² you get a mix of both).		where q is the charge per particle, I is the electric current in the beam, 1 / (4 π ε<sub>0</sub>) = 8.9875517923 × 10<sup>9</sup> N m²/C² is the Coulomb constant, m is the particle mass, β is the speed as a fraction of the speed of light, γ is the Lorentz factor, c is the speed of light, and R is the radius of the beam. If K is much larger than ε²/R² the beam will be dominated by self-force expansion; for K much smaller than ε²/R² it will be dominated by emittance (and if K ≈ ε²/R² you get a mix of both).

	(Useful values: the charge of an electron is ~~<math>~~q = -1.602176634 \times 10^{-19}</~~math~~> C and that of the proton is ~~<math>~~q = +1.602176634 \times 10^{-19}</~~math~~> C. The electron mass ~~<math>~~m = 9.1093837015 \times 10^{-31}</~~math~~> kg and for a proton ~~<math>~~m = 1.67262192369 \times 10^{-27}</~~math~~> kg. Ions have a mass of ~~<math>~~m = 1.66053906660 \times 10^{-27}</~~math~~> kg times their standard atomic weight. The speed of light is ~~<math>~~c = 299792458~~</math>~~ m/s. If you know the total ~~Power <math>~~P~~</math>~~ in watts delivered by your beam and you know the energy of each particle ~~<math>~~V~~</math>~~ in eV, then the beam current in amperes is ~~<math>~~I = P/V~~</math>~~.)		(Useful values: the charge of an electron is q = -1.602176634 × 10<sup>-19</sup> C and that of the proton is q = +1.602176634 × 10<sup>-19</sup> C. The electron mass m = 9.1093837015 × 10<sup>-31</sup> kg and for a proton m = 1.67262192369 × 10<sup>-27</sup> kg. Ions have a mass of m = 1.66053906660 × 10<sup>-27</sup> kg times their standard atomic weight. The speed of light is c = 299792458 m/s. If you know the total power P in watts delivered by your beam and you know the energy of each particle V in eV, then the beam current in amperes is I = P/V.)

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