How big are Collider Beams ?

[Widdekind]

How big are Collider Beams, used in High Energy collisions ? For example, in Deep Inelastic Scattering experiments, beaming Leptons (electrons, neutrinos) at nucleons, how physically big & broad are the Leptonic Wave Functions ?

[Widdekind]

We wish to produce a mono-chromatic beam, with momentum p = 20 GeV [[math]\lambda_C \equiv h / m c = 6 \times 10^{-17} m[/math]], and momentum spread [math]\Delta p / p[/math] of 1% [[math]\Delta x \equiv \hbar / \Delta p = 10^{-15}m[/math]]. The beam is 2 mm wide.

 

Alessandro Bettini. Introduction to Elementary Particle Physics, pg. 82.

 

 

 

LEP was a circular machine, in a tunnel 27 km long. This is testament to the problems with lightweight electrons & positrons, traveling around in circles, that such a distance is needed to enable them to reach 100 GeV without wasting too much energy in [synchrotron] radiation. To reach energies of several hundred GeV, in circular orbits, would require distances of hundreds of kilometers, which are out of the question. This is why linear colliders are planned for the longer-term future.

 

The idea is to have one linear accelerator of electrons, and another accelerating positrons... To have a decent chance of a collision in a [head-on] linear accelerator, where the beams meet only once, requires high intensity beams that are less than a micron (10^-6m) across.

 

Frank Close. [VSI] Particle Physics, pg. 57.

 

 

When the LEP began running in the 1990s [100 GeV], needle-like bunches of electrons & positrons would pass through each other, at the heart of the detector, every 22 microseconds (22 millionths of a second) [[math]22 \mu s \times c = 7 km[/math]]. Even though there were some million million particles in each bunch, the particles were thinly dispersed, so interactions between them were rare. An interesting collision, or 'event', only occurred about once every 40 times or so the bunches crossed.

 

ibid., pg. 59.

 

 

At the Large Hadron Collider (LHC)... the ATLAS detector will be five stories high (20 m), and yet able to measure particle tracks to a precision of 0.01 mm [10^-5m].

 

ibid., pg. 77.

 

 

 

We want to procure a source of protons, and accelerate them to... about 99.999999% the speed of light. Then we need to make a beam of these protons, no wider than the width of a human hair [~100 microns], and guide the beam in a circular path of a day or so, during which time a proton will travel 26 billion km...

 

 

The LHC beam... is comprised of bunches of protons, with each bunch separated by no less than 7.6 m... This is the distance between adjacent waves in the accelerating electric field [40 MHz]. Of course, it is possible for adjacent bunches to be farther apart than this. If not every wave is filled with protons, adjacent bunches could be multiples of this distance, depending on how many waves are skipped...

 

An individual bunch [of protons] in the LHC includes about 100 billion (10¹¹) protons. The actual shape of each bunch has a passing resemblance to a stick of uncooked spaghetti, although it is about 0.3 m (1 foot) long, and the width is less than a millimeter. There will be 2,808 bunches of protons orbiting in each direction, and aimed & focused to collide at four points around the LHC's perimeter. So, except for the actual beam width being about a hundred times smaller than a piece of spaghetti [1% of 1 mm], you can get a pretty good visualization, of the LHC's beam, as [being] about 3,000 pieces of uncooked spaghetti, each separated by about 7.6 m (25 feet), orbiting at 99.999999% the speed of light, in an orbit that is 27 km (17 miles) around. if you do the math, you find that about 3,800 lengths are needed to fill up the entire orbit. So, if there are 2,808 bunches, the entire accelerator is not filled. You have a concentrated group of bunches, each separated by the 7.6 m (25 feet), followed by a relatively long gap. This gap has many uses. Note that "long" is a relative term, and in the ballpark of a millionth of a second (microsec). This exceedingly brief time, during which there are no protons in a detector, is used for the detector to recover & reset itself.

 

Don Lincoln. The Quantum Frontier: The Large Hadron Collider, pp. 67,93.

 

 

 

The method proposed by Fermi is based on the assumption, that in the collision of high-energy nucleons, the energy appears simultaneously in a small volume, and that both nucleons come to rest, in the center-of-mass system.

 

Because of the Lorentz Contraction, this volume is not spherical, but is compressed in the direction of motion of the nucleons prior to the collision. The amount of compression is given by the well-known factor:

 

[math]\sqrt{1 - \frac{v^2}{c^2}} = \frac{M c^2}{E'} = \gamma^{-1}[/math]

where M is the mass of the nucleon, and E' its energy, in the center-of-mass system... The volume of the region in which energy appears is (order of magnitude):

 

[math]\Omega \approx \frac{M c^2}{E'} \left( \frac{\hbar}{\mu c} \right)^3 = \frac{\lambda_{C,\pi}^3}{\gamma}[/math]

since [math]\hbar / \mu c[/math] ([math] \mu[/math] is the [pi-]meson mass) is of the order of the interaction range.

 

L.D.Landau & Ya.Smorodinsky. Lectures on Nuclear Theory, pg. 105.

Edited August 3, 2010 by Widdekind

[^_^]

This is the distance between adjacent waves in the accelerating electric field [40 MHz].

 

 

LHC RF is 400.789599 MHz

 

 

although it is about 0.3 m (1 foot) long

 

Bunch length is ~ 7.55mm (0.0755m)

 

 

http://lhc.web.cern....signReport.html

 

Chapter 2 is your friend.

 

This exceedingly brief time, during which there are no protons in a detector, is used for the detector to recover & reset itself.

It is more of a case of the abort gap for the extraction kicker to fire.

In fact, just ignore that reference entirely.

[Widdekind]

Thanks for the response. I found this reference, as well:

 

A wave packet is a wave [function] corresponding to each electron. The length of an electron wave packet, in a field-emission electron beam, is [math]1 \; \mu m[/math]... Interference only occurs, when two electrons come as close as [math]1 \; \mu m[/math] ['back-to-back'], but the average distance between electrons is 1 m [~3 ns between emissions], at the shortest, in our electron beam, even when we use the brightest field-emission electron source. [Thus] the probability that two electrons overlap is extremely small...

 

In a typical electron diffraction [Double Slit] experiment, [the distance between the two slits] [math]d = 3 \AA[/math] and [math]\lambda = 0.04 \AA[/math]. Therefore the divergence angle of the electron beam should be less than 7 x 10^-3 rad = 0.4 degree. It is not difficult to obtain such a degree of the divergence angle of the electron beam. But, when we want to get an interference pattern, from two slits separated by a macroscopic distance, say, [math]d = 10 \; \mu m[/math], then [math]\alpha < 2 \times 10^{-7} rad = 1.2 \times 10^{-5} degrees[/math]. We have to use a highly collimated electron beam...

 

The wavelength of electrons, accelerated up to 100 KeV, is only [math]0.04 \; \AA[/math].

 

Akira Tonomura. The Quantum World Unveiled by Electron Waves, pp. 19,24,59.

 

If the uncertainty, in the transverse momentums, of the electrons in the beam, is [math]\Delta p_{\bot} = p \; sin( \alpha ) \approx p \; \alpha[/math] (Small Angle Approximation), where [math]p = \hbar / \lambda[/math], then, from, the Heisenberg Uncertainty Principal [math]\Delta x \approx \hbar / \Delta p[/math], w.h.t. [math]\Delta x_{\bot} \approx \lambda / \alpha[/math]. Then, for the wide-angle electron beam [math]\left( \alpha = 7 \times 10^{-3} \; rad \right)[/math], w.h.t. [math]\Delta x_{\bot} \approx 6 \; \AA[/math]; whereas, for the more focused electron beam [math]\left( \alpha = 2 \times 10^{-7} \; rad \right)[/math], w.h.t. [math]\Delta x_{\bot} \approx 20 \; \mu m[/math]. Perhaps these transverse spatial extents, of the electrons' wave packets, can be compared, to their longitudinal lengths [math]\left( \approx 1 \; \mu m \right)[/math] ?

 

Note that, if p = m v (Classical limit), then [math]\beta \approx \lambda_C / \lambda[/math], where the Compton Wavelength of electrons is [math]\lambda_C = 0.0243 \AA[/math]. For 100 KeV electrons, with wavelengths [math]\lambda = 0.04 \AA[/math], w.h.t. [math]\beta = 0.6[/math]. Thus, if the electrons' wave packets are [math]1 \; \mu m[/math] long, then their emission must take about t = L / v = 6 x 10^-15 s.

Edited August 7, 2010 by Widdekind

[Widdekind]

In Neutron Interferometers, using thermal neutrons,

 

the cross-section of the beam is around 1 sq. cm, and we may think of each neutron as being represented by a wave-pulse around 10^-3 cm long... the wave-pulse is, thus, about the size & shape of a small postage stamp.

 

A.Whitaker. Einstein, Bohm, and the Quantum Dilemma, pg. 315.