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Electron,protons ,quarks masses?

2006-07-21 04:40:09 · 7 answers · asked by goring 6 in Science & Mathematics Mathematics

and how long would it take to do so.

2006-07-21 04:48:51 · update #1

and how much energy does it take ?

2006-07-21 07:02:41 · update #2

7 answers

The Super Proton Synchrotron (an older ring collider) is used to accelerate electrons and positrons to nearly the speed of light. These are then injected into the ring. As in all ring colliders, the LEP's ring consists of many magnets which force the charged particles into a circular trajectory (so that they stay inside the ring), RF accelerators which accelerate the particles with radio frequency (RF) waves and quadrupoles that focus the particle beam (i.e. keep the particles together). (Note that 'accelerating' here does not really mean that the particles get much faster, as they already are very close to the speed of light in the beginning. But they gain a lot of kinetic energy and so become more massive because mass and energy are equal according to the theory of special relativity.) When the particles are accelerated to maximum energy (and focused to so-called bunches), an electron and a positron bunch is made to collide with each other at one of the collision points of the detector. When an electron and a positron collide, they annihilate to a virtual particle, either a photon or a Z boson. The virtual particle almost immediately decays into other elementary particles, which are then detected by huge particle detectors.

2006-07-21 04:44:23 · answer #1 · answered by Justsyd 7 · 1 0

The Super Proton Synchrotron (an older ring collider) is used to accelerate electrons and positrons to nearly the speed of light. These are then injected into the ring. As in all ring colliders, the LEP's ring consists of many magnets which force the charged particles into a circular trajectory (so that they stay inside the ring), RF accelerators which accelerate the particles with radio frequency (RF) waves and quadrupoles that focus the particle beam (i.e. keep the particles together). (Note that 'accelerating' here does not really mean that the particles get much faster, as they already are very close to the speed of light in the beginning. But they gain a lot of kinetic energy and so become more massive because mass and energy are equal according to the theory of special relativity.) When the particles are accelerated to maximum energy (and focused to so-called bunches), an electron and a positron bunch is made to collide with each other at one of the collision points of the detector. When an electron and a positron collide, they annihilate to a virtual particle, either a photon or a Z boson. The virtual particle almost immediately decays into other elementary particles, which are then detected by huge particle detectors.

Source(s):

http://en.wikipedia.org/wiki/large_elect...

2006-07-21 04:45:49 · answer #2 · answered by rod 5 · 0 0

There are "component" which could commute a the linked fee of sunshine: gentle and gravity you're precise, gentle can't be interpreted as some thing that has no mass. yet what's it then. Einstein in call for equation "E=mc^2" let us know that capability and mass are suitable, categories of a similar component. Mass is, in accordance to Einstein is an fairly great concentration of capability which would be converted to a greater "direct" style of capability. once you commute to speeds close to the linked fee of sunshine, your mass starts turning out to be gentle, the main direct style of capability (lower back, in accordance to Einstein), and once you eventually commute on the linked fee of sunshine, all your mass turns into gentle and there isn't any coming lower back, gentle can't decrease velocity. as a results of fact of this gentle velocity is the "familiar" velocity decrease, as a results of fact gentle is the main direct source of capability and travels at that velocity!!!!! :O

2016-12-10 11:39:43 · answer #3 · answered by Anonymous · 0 0

Yes. Since the particles themselves have small mass, it doesn't even take that long (on the order of microseconds to get relativistic speeds).

2006-07-21 05:15:59 · answer #4 · answered by mathematician 7 · 0 0

Why go almost, Get on to tachyon speeds ..Various multiples of "C"

2006-07-21 04:46:31 · answer #5 · answered by chopper 007 2 · 0 0

yes

2006-07-21 04:42:28 · answer #6 · answered by shyam 2 · 0 0

maybe.

2006-07-21 04:43:45 · answer #7 · answered by Timothy C 1 · 0 0

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