Timeline for Why collide a moving particle with a particle at rest, rather than two moving particles?
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| Apr 10, 2019 at 7:05 | comment | added | Mateen Ulhaq | For the experiment with a particle at rest in the lab frame, the accelerated particle $b$ has energy $E_b \approx 2E$, right? | |
| S Mar 31, 2015 at 10:39 | history | suggested | andybuckley | CC BY-SA 3.0 |
Correct 'square' to 'square root' and correct the statements about LHC energy
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| Mar 31, 2015 at 10:37 | comment | added | PhotonBoom | @andybuckley, thanks for the additional info. I recognise your name, your material on Rivet definitely made my life easier for my thesis :p | |
| Mar 31, 2015 at 10:33 | review | Suggested edits | |||
| S Mar 31, 2015 at 10:39 | |||||
| Mar 31, 2015 at 10:31 | comment | added | andybuckley | For completeness, the LHC Run 1 actually finished (and took most data) with CoM energy of 8 TeV, not 7. The original question also asked about number of particles in a high-energy collision and the accuracy of the beam control: the LHC had roughly 10^11 protons in each of 2800 colliding bunches, each proton with 4 TeV of energy. The beams are 16 micrometres wide: colliding them is a technical feat! In each typical bunch crossing there are up to 60 pp interactions, to increase the luminosity; the downside is that overlaid uninteresting collisions (called pile-up) make event reconstruction hard. | |
| Mar 25, 2015 at 0:17 | history | edited | PhotonBoom | CC BY-SA 3.0 |
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| Mar 24, 2015 at 21:15 | comment | added | zeldredge | @easymoden00b : no, velocities do not add this way in special relativity | |
| Mar 24, 2015 at 20:49 | comment | added | easymoden00b | if each particle was moving at 0.6 the speed of light wouldn't the other particle be going 1.2 the speed of light from ones reference frame? | |
| Mar 24, 2015 at 19:54 | comment | added | PhotonBoom | @Jimnosperm, I sensed some confusion about what the CoM frame means so I edited the question to address this. The answer is to your question is always btw. | |
| Mar 24, 2015 at 19:52 | history | edited | PhotonBoom | CC BY-SA 3.0 |
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| Mar 24, 2015 at 19:19 | comment | added | Jim | It occurs to me. You assume at the end that masses are negligible, or at least that $m\ll2E_b$ (assuming both are protons, the rest mass can be called equal between them). Actually, the approximation is better stated as $m\ll p$ since $E$ is composed of $m$ and $p$ and $m\not\ll m$. Anyway, in the first case, we have $s\sim4(m^2+p^2)\sim4p^2$, but in the second case, you have $s\sim2m(m^2+p_b^2)^{1/2}$, which also must approximate as $s\sim2mp_b$. Looking at that, how often would we say it's really the case that $4p^2<2mp_b$? It seems like the second case has low energies | |
| Mar 24, 2015 at 17:09 | vote | accept | Ixrec | ||
| Mar 24, 2015 at 14:57 | history | edited | PhotonBoom | CC BY-SA 3.0 |
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| Mar 24, 2015 at 12:32 | history | edited | PhotonBoom | CC BY-SA 3.0 |
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| Mar 24, 2015 at 12:28 | history | edited | Robin Ekman | CC BY-SA 3.0 |
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| Mar 24, 2015 at 12:25 | history | answered | PhotonBoom | CC BY-SA 3.0 |