collisionparticle-physics
What would happen if an accelerated particle (like they create in the LHC) hit a person standing in its path?
Would the person die? Would the particle rip a hole? Would the particle leave such a tiny wound that it would heal right away? Something else?
Both the quora answers are incorrect. The idea that "nothing happens" is incorrect for reasons I explain in great detail below. The idea that somehow Jupiter spreads itself across the surface of the Sun or directly influences the luminosity of the Sun by doing so is wrong on many levels as pointed out by Victor Toth on the quora page and by Rob and Chris as answers here.
Instead I put forward a couple of scenarios where the large amount of accreted energy and/or angular momentum certainly do have an effect on the Sun and/or the radiation the Earth receives from the Sun.
Scenario 1: The scenario where Jupiter just drops into the Sun from its current position would certainly have short-term effects. But short-term here means compared with the lifetime of the Sun, not hundreds of years.
The kinetic energy of Jupiter at the Sun's surface would be of order $GM_{\odot}M_\mathrm{Jup}/R_{\odot} \sim 4\times 10^{38}$ joules.
The solar luminosity is $3.83 \times 10^{26}\ \mathrm{J/s}$.
The addition of this much energy (if it is allowed to thermalise) would potentially affect the luminosity of the Sun for timescales of tens of thousands of years. The exact effects will depend on where the energy is deposited. Compared with the binding energy of the star, the additional energy is negligible, but if the energy is dissipated in the convection zone then kinetic energy would do work and lift the convective envelope of the Sun. In other words, the Sun would both increase in luminosity and in radius. If the effects were just limited to the convective envelope, then this has a mass of around $0.02 M_{\odot}$ and so could be "lifted" by $\sim 4\times 10^{38} R_{\odot}^2/GM_{\odot}M_{\rm conv} \sim 0.05 R_{\odot}$.
So in this scenario, the Sun would both expand and become more luminous. The relevant timescale is the Kelvin-Helmholtz timescale of the convective envelope, which is of order $GM_{\odot}M_{\rm conv}/R_{\odot} L_{\odot} \sim $few $10^5$ years.
If the planet somehow survived and punched its way to the centre of the Sun, then much less energy would be deposited in the convection zone and the effects would be lessened.
On longer timescales the Sun would settle back down to the main sequence, with a radius and luminosity only slightly bigger than it was before.
This all assumes that Jupiter can remain intact as it falls. It certainly wouldn't "evaporate" in this direct infall scenario, but would it get tidally shredded before it can disappear below the surface? The Roche limit is of order $R_{\odot} (\rho_{\odot}/\rho_{\rm Jup})^{1/3}$. But the average densities of the Sun and Jupiter are almost identical. So it seems likely that Jupiter would be starting to be tidally ripped apart, but as it is travelling towards the Sun at a few hundred km/s at this point, tidal breakup could not be achieved before it had disappeared below the surface.
So my conclusion is that dropping Jupiter into the Sun in this scenario would be like dropping a depth charge, with a lag of order $10^{5}$ years before the full effects became apparent.
Scenario 2: Jupiter arrives at Roche limit (just above the solar surface) having mysteriously lost a large amount of angular momentum. In this case the effects may be experienced on human timescales.
In this case what will happen is Jupiter will be (quickly) shredded by the tidal field, possibly leaving a substantial core. At an orbital radius of $2 R_{\odot}$, the orbital period will be about 8 hours, the orbital speed about $300\ \mathrm{km/s}$ and the orbital angular momentum about $10^{42}\ \mathrm{kg\ m^2\ s^{-1}}$. Assuming total destruction, much of the material will form an accretion disc around the Sun, since it must lose some of its angular momentum before it can be accreted.
How much of the Sun's light is blocked is uncertain. It mainly depends on how the material is distributed in the disk, especially the disk scale height. This in turn depends on the balance of the heating and cooling mechanisms and hence the temperature of the disk.
Some sort of minimal estimate could be to assume the disk is planar and spread evenly between the solar surface and $2R_{\odot}$ and that it gets close to the solar photospheric temperature at $\sim 5000\ \mathrm K$. In which case the disk area is $3 \pi R_{\odot}^2$, with an "areal density" of $\sigma \sim M_{\rm Jup}/3\pi R_{\odot}^2$.
In hydrostatic equilibrium, the scale height will be $\sim kT/g m_\mathrm H$, where $g$ is the gravitational field and $m_\mathrm H$ the mass of a hydrogen atom. The gravity (of a plane) will be $g \sim 4\pi G \sigma$. Putting in $T \sim 5000\ \mathrm K$, we get a scale height of $\sim 0.1 R_{\odot}$.
Given that Earth is in the ecliptic plane and this is where the disk will be, then a large fraction, $\gt 20\ \%$, of the sunlight reaching the Earth may be blocked. To work out if this is the case, we need to work out an optical depth of the material. For a scale height of $0.1 R_{\odot}$ and a planar geometry, then the density of the material is $\sim 3\ \mathrm{kg/m^3}$. Looking though this corresponds to a column density of $\sim 10^{10}\ \mathrm{kg/m^2}$.
For comparison, the solar photospheric density is of order $10^{-12}\ \mathrm{kg/m^3}$ and is only the upper $1000\ \mathrm{km}$ of the Sun. Given that the definition of the photosphere is where the material becomes optically thick, we can conclude that a tidally shredded Jupiter is optically thick to radiation and indeed the sunlight falling on the Earth would be very significantly reduced – whether or not the amount of radiation impacting the Earth is reduced or increased is a tricky radiative transfer problem, since if the disk were at $5000\ \mathrm K$ and optically thick it would be kicking off a lot of radiation!
How long the accretion disk would remain, I am unsure how to calculate. It depends on the assumed viscosity and temperature structure and how much mass is lost through evaporation/winds. The accreted material will have radiated away a large fraction of its gravitational potential energy, so the energetic effects will be much less severe than Scenario 1. However, the Sun will accrete $\sim 10^{42}\ \mathrm{kg\ m^2\ s^{-1}}$ of angular momentum, which is comparable to its current angular momentum. The accretion of Jupiter in this way is therefore sufficient to increase the angular momentum of the Sun by a significant amount. In the long term this will have a drastic effect on the magnetic activity of the Sun – increasing it by a factor of a few to an order of magnitude.
I am confused about how exactly this is even possible. As far as I understand one needs to create an ultra-high vacuum in a particle collider to let different particle beams collide.
The ultra high vacua in the present collider experiments are needed for two reasons
To make sure that the detectors are getting the debris of pure proton-proton hits, and not of random molecules ( which still exists because no vacuum is complete,and have to be included in a background for the Monte Carlo simulations of the interaction).
Keeping the beam intact , i.e not losing energy and direction through the scattering you describe, and becoming useless for accurate measurements.
In the early times when one was studying interactions of particles in bubble chambers and detectors , the beams were crerated in vaccuum, but arrived at the detectors through air, at least in bubble chambers I am sure it is so, because I worked on Kaon experiments at CERN. That is because the incoming particles were seen one by one parallel, any interactions would take them out of the beam and they were not so many because the intensity of the beam was kept so that a small number of particles arrived at the detector.
Picture from CERN 2-metre hydrogen bubble chamber exposed to a beam of positive kaons with energy 10GeV entering from the bottom of the picture.
If we lost the beam we used to joke that "a cat had entered the beam line".
For more intense beams used in electronic detectors the air beam would be dangerous, even if not intense enough to be visible.
Edit after finding this reference, page 343:
The first fast ejected beam from the PS was obtained on 12 May 1963. This photograph shows the beam lighting up blocks of plastic scintillator as it travels in air on its way into the South Hall
The photo is in order to show how the beam interacted with the scintilators, no light is seen in the air .The hall was well lighted all times, the photo could not have been taken with the hall lightings on.
So the physicist must have been checking a beam line, thinking the synchrotron was off, but it either was still on, or it came on without alarms going off.
Best Answer
Amazingly this actually happened to a Russian scientist called Anatoli Bugorski (WARNING: this is pretty gruesome).
The beam basically just killed all the tissue it passed through. The symptoms were the relatively mundane ones expected from tissue death.
The LHC has a much, much greater energy than the one that struck Bugorski, so it would cause a lot more heating and presumably burning of neighbouring tissue. How much extra damage there would be depends on how rapidly the beam is absorbed, and I must admit I don't know this. The total LHC beam energy is 362 MJ, which is enough to turn 150kg of water at body temperature to steam. If any significant fraction of this was absorbed by your head the resulting explosion would probably not leave much of your head behind.