ec2-54-235-48-106.compute-1.amazonaws.com | ToothyWiki | RecentChanges | Login | Webcomic

A [lot of fuss] has been made about the LHC destroying the Earth.  One of [the major counter arguments] has been that cosmic rays are more energetic, and have not destroyed it yet.

DouglasReay asks: how valid is that argument?  I've not yet seen the calculations done anywhere, so here is my attempt at them:

It doesn't require that two cosmic rays collide.  If the moving cosmic ray particle has twice the energy of an LHC beam one then it can hit a stationary target (some piece of rock inside the Earth) and still be comparable.  So the two figures we need to compare are:

CALCULATION A: How many beam-beam high energy collisions will happen over the life-time of CERN ?

CALCULATION B: How many cosmic_ray-Earth_target high energy collisions have happened over the life-time of the Earth ?

CALCULATION A: How many beam-beam high energy collisions will happen over the life-time of CERN ?


Let's assume that CERN will run for 100 years, at its [highest projected luminosity].

AGE_OF_CERN = 3 x 10^9 seconds

CERN_COLLISIONS_PER_SECOND = 8 x 10^9  [source]

which gives

ANS (cern) = AGE_OF_CERN x CERN_COLLISIONS_PER_SECOND = 3 x 10^9 x 8 x 10^9 = 2.4 x 10^19

CALCULATION B: How many cosmic_ray-Earth_target high energy collisions have happened over the life-time of the Earth ?


AGE_OF_EARTH = 1.4 x 10^17 seconds [source]


The LHC will be producing collisions with up to 14 TeV?, which equals 99.9999991 % of speed of light for two protons each weighing .938 GeV?/c^2 at rest, or 2.24 x 10^-6 Joules  [source]

1.4 x 10^13 eV is well under the limit for the energy of Galactic Cosmic Rays, GCRs (those from outside the solar system) produced by supernovae, so we don't need the higher energy sort from [active galactic nuclei], quasars or radio galaxies.

The Earth's atmosphere gets hit by 100 cosmic rays per second per square meter, and the surface area of the Earth is 5 x 10^14 square meters which would mean:


However, not all the ones hitting the outer atmosphere are GCRs.  Higher energy ones are rarer.  At 10^16 eV, the rate is 2 per year per square meter. [source]

which would give:


However even at this lower figure, that still gives:

ANS (Earth) = AGE_OF_EARTH x EARTH_COLLISIONS_PER_SECOND = 1.4 x 10^17 x 3 x 10^7 = 4 x 10^24


During the life of the Earth so far, there have been more than 100,000 time as many collisions with higher energy than any the LHC will ever produce, even if it ran without pause at maximum capacity for the next 100 years.


Just looking at energy, the case for safety seems proven.  However, there is a complication.

It seems [some people] are saying that the resulting net momentum of the two colliding nuclei also matters.  Specifically, that are saying that if a cosmic ray hits a stationary target, whatever results will still have a high momentum, and therefore leave the Earth's gravitational field and not destroy it. [source] and [source]

It's not just that. Contrarotating beams in the LHC are not merely a way to get twice as much energy out of the same sized supercollider. It is also very important for the creation of the highest energy resonances that the net momentum must be zero. Hatchet-job proof: E for energy, p for momentum, m for rest mass, c for the speed of light. E squared equals (mc squared) squared plus (pc) squared, E and p are conserved and neither can ever be 0, m > 0. Therefore m (after) is proportional to 1 / |p (after)| for |p (after)| > 0. If you want gamma rays or neutrinos out of your machine, use a stationary target. If you want hadrons or heavier things (like black holes or the Higgs boson), use two colliding beams of equal momentum with as near to 180 degrees angle of incidence as humanly possible. --Requiem
ITYM four times as much energy. Also, doesn't the distinction in your last few sentences violate the principle that the universe should act the same whatever velocity it has? --Admiral

However is that valid?  It seems to make two assumptions:
A) that the resulting strange particle / black hole / whatever would never be significantly slowed down by further collisions as it passed through the entire thickness of the planet.
B) that the resulting collision would produce only a single cohesive mass, rather than multiple things shooting off in all directions at varying speeds.

Additional Safety Factor

There is nothing special about the Earth that makes it a particularly nice target for cosmic rays.  We'd notice if any stars within 1000 light years had been turned into black holes.  There are 1 x 10^7 of those, each having a surface area 1 x 10^4 times that of the Earth, so that gives us another 11 orders of magnitude safety factor.

But Physicists can be wrong

The latest kerfuffle is [a paper showing that 1 in 1000 physics papers have to be withdrawn due to mistakes].  From this [some people are saying that we therefore can't rely on the chance of the LHC destroying the world being less than 1 in 1000].

I have several thoughts on this.

1. So if 10 papers all independently say that the LHC won't destroy the world, would that mean the limiting probability fell to 1 in 10^30 ?
For a strict enough value of "independently", yes. --PT

2. Surely by the same reasoning the probability that NOT turning on the LHC will destroy the world is also at least 1 in 1000, so turning it on is no more risky than leaving it turned off.
Have you seen a peer-reviewed paper which computes the probability that not turning on the LHC will destroy the world? Anyway, it's irrelevant because it's not saying that the probability of the LHC destroying the world is 1 in 1000 but that the best estimate we can make for it is 1 in however many million plus or minus 1 in 1000. It's actually overly pessimistic there. --PT

3. Failure modes matter.  If a physics paper was withdrawn for being wildly innacurate (eg off by an order of magnitude) that is different from it being 100% incorrect.  If a paper says the chances of the LHC destroying the world are 1 in 10^16, and it is wrong by an order of magnitide, that still gives us at worst a chance of 1 in 10^15.
Hence "overly pessimistic" above. It's taking a worse-case best estimate. --PT

4. This is actually an interesting issue if you try to work it out correctly.  Suppose you have 10 papers each estimating the probability of some particular disasterous event (such as an oil spill in a particular location wiping out a particular species):
Paper 1 gives the chances as 9 in 10
Paper 2 gives the chances as 1 in 10^9
Paper 3 gives the chances as 1 in 10^9
Paper 4 gives the chances as 1 in 10^9
Paper 5 gives the chances as 1 in 10^12
Paper 6 gives the chances as 1 in 10^13
Paper 7 gives the chances as 1 in 10^14
Paper 8 gives the chances as 1 in 10^15
Paper 9 gives the chances as 1 in 10^16
Paper 10 gives the chances as 1 in 10^1,000,000

For simplicity's sake, let's assume that the authors are equally reliable, with a track record (from their institution) of:
being within 2 orders of magnitude of the correct answer, 90% of the time
being out by less than a factor of 2 on the orders of magnitude, 99% of the time
producing an invalid answer (eg wrongly typed number) which says nothing either way, 1% of the time

Do you go for the mode (1 in 10^9) ?
Do you go for the geometric mean (1 in 10^100,000) ?
Do you discard just two outliers (10^13) ?
Do go for the largest grouping within 2 orders of magnitude (10^14) ?

What if the authors are not equally reliable?  What if the papers are not independent?
Suppose the true probability is p. Then consider the function f(p) which is the probability that the papers would give the results above. I think the best estimator of p would be the value (or range of values) which maximises f(p). --PT

CategoryPhysics CategorySerious

ec2-54-235-48-106.compute-1.amazonaws.com | ToothyWiki | RecentChanges | Login | Webcomic
Edit this page | View other revisions | Recently used referrers
Last edited January 29, 2009 12:48 pm (viewing revision 11, which is the newest) (diff)