The Virtuosi (Posts by Nic Eggert)https://thephysicsvirtuosi.com/enContents © 2019 <a href="mailto:thephysicsvirtuosi@gmail.com">The Virtuosi</a> Thu, 24 Jan 2019 15:05:00 GMTNikola (getnikola.com)http://blogs.law.harvard.edu/tech/rss- Report from the Trenches: A CMS Grad Student's Take on the Higgshttps://thephysicsvirtuosi.com/posts/old/report-from-the-trenches-a-cms-grad-student-s-take-on-the-higgs/Nic Eggert<p><img alt="Mmmm run172822 evt2554393033
3d" src="http://lh6.ggpht.com/-hPHBh1UVJic/TuvD-aa1MYI/AAAAAAAAAX4/BQgsVZulLkw/mmmm-run172822-evt2554393033-3d.jpg?imgmax=800" title="mmmm-run172822-evt2554393033-3d.jpg">
Hi folks. It's been an embarrassingly long time since I last posted, but
today's news on the Higgs boson has brought me out of hiding. I want to
share my thoughts on today's announcement from the CMS and ATLAS
collaborations on their searches for the Higgs boson. I'm a member of
the CMS collaboration, but these are my views and don't represent those
of the collaboration. The upshot is that ATLAS sees a 2.3 sigma signal
for a Higgs boson at 126 GeV. CMS sees a 1.9 sigma excess around 124
GeV. CERN is being wishy-washy about whether or not this is actually a
discovery. After all the media hype leading up to the announcement, this
is somewhat disappointing, but maybe not too surprising. First of all,
what does a 2 sigma signal mean? The significance corresponds to the
probability of seeing a signal as large or larger than the observed one
given only background events. That is, what's the chance of seeing what
we saw if there is no Higgs boson? You can think of the significance in
terms of a Normal distribution. The probability of the observation
corresponds to the integral of the tails of the Normal distribution from
the significance to infinity. For those of you in the know, this is just
1 minus the CDF evaluated at the significance. For a 2 sigma
observation, this corresponds to about 5%. For both experiments, there
was a 5% chance of observing the signal they observed or bigger if the
Higgs boson doesn't exist. In medicine, this would be considered an
unqualified success. So why is CERN being so cagey? In particle physics
we require at least 3 sigma before we even consider something
interesting, and 5 sigma to consider it an unambiguous discovery. The
reasons why the burden of proof is so much higher in particle physics
than in other fields aren't entirely clear to me. I suspect is has to do
with the relative ease of running the collider a little longer compared
to recruiting more human test subjects, to use medicine as an example.
Given what I've just told you that we need a 3 sigma significance in
particle physics, why is everyone so excited about a couple of 2 sigma
results? Well, the first reason is that both results show bumps at
approximately the same Higgs mass. Although it's not rigorous, you can
get a rough idea of what the significance of the combined results are by
adding the significances in quadrature. This gives us about 2.8 sigma.
Higher, but still not up to the magic number of 3. The explanation for
the excitement that is most compelling brings us to Bayesian statistics.
The paradigm of Bayesian statistics says that our belief in something
given new information is the product of our prior beliefs and a term
which updates them based on the new information. Physicists have long
expected to find a Higgs boson with a mass around 120 GeV. So our prior
degree of belief is pretty high. Thus, it doesn't take as much to
convince us (or me anyway) that we have observed the Higgs boson. In
contrast, consider the OPERA collaboration's measurement of neutrinos
going faster than the speed of light. This claims to be a 6 sigma
result, but no one expected to find superluminal neutrinos, so our (or
at least my) prior for this is much lower. (Aside: If the OPERA result
is wrong, it is likely due to a systematic effect rather than a
statistical one. Nevertheless, I stand by my point.) The final thing
that excites me about this observation is that what we've seen is
completely consistent with what we would expect to see from the Standard
Model. Forgetting about significances for the moment, when the CMS
experiment fits for the Higgs boson mass, they find a cross section that
agrees very well with that predicted by the Standard Model. In the plot
below, you're interested in the masses where the black line is near 1.
The ATLAS experiment actually sees more signal than one would expect.
This is likely just a statistical fluctuation, and explains why the
ATLAS result has a higher significance. <img alt="GUIDO HIGGS CERN SEMINAR pdf
page 43 of 60
1" src="http://lh6.ggpht.com/-tHKpXH_FDfM/TuvD_jbM8wI/AAAAAAAAAYA/-LWjE0AqDog/GUIDO_HIGGS_CERN_SEMINAR.pdf%252520%252528page%25252043%252520of%25252060%252529-1.png?imgmax=800" title="GUIDO_HIGGS_CERN_SEMINAR.pdf (page 43 of 60)-1.png">
<img alt="ATLAS Higgs pdf page 34 of
68" src="http://lh6.ggpht.com/-dLvDz4KoVuU/TuvEBK3mg3I/AAAAAAAAAYI/4x-6m2b-g0M/ATLAS-Higgs.pdf%252520%252528page%25252034%252520of%25252068%252529.jpg?imgmax=800" title="ATLAS-Higgs.pdf (page 34 of 68).jpg">
In conclusion, while CERN is being non-committal, in my opinion, we have
seen the first hints of the Higgs boson. This is mostly due to my high
personal prior that there the Higgs boson exists around the observed
mass. Unfortunately, Bayesian priors are for the most part a qualitative
thing. Thus, ATLAS and CMS are sticking to the hard numbers, which say
that what we have looks promising, but is not yet anything to get
excited about. I'll close by reminding you all to take this all with a
grain of salt. There is every possibility that this is just a
fluctuation. I'll remind you that at the end of last summer, CMS and
ATLAS both showed a <a href="http://resonaances.blogspot.com/2011/07/higgs-wont-come-out-of-closet.html">3 sigma
excess</a>
around 140 GeV, which went away just a month later at the next
conference. So let's cross our fingers that next year's data will give
us a definitive answer on this question. By the way, if anyone wants to
know more, fire away in the comments. I'll do my best.</p>Higgsparticle physicsStatshttps://thephysicsvirtuosi.com/posts/old/report-from-the-trenches-a-cms-grad-student-s-take-on-the-higgs/Tue, 13 Dec 2011 17:36:00 GMT
- Why Black Holes from the Large Hadron Collider Won't Destroy the Worldhttps://thephysicsvirtuosi.com/posts/old/why-black-holes-from-the-large-hadron-collider-won-t-destroy-the-world/Nic Eggert<p>Hi everyone. As this is my first post, I thought I'd introduce myself.
Like the rest of the Virtuosi, I'm a graduate student in physics at
Cornell University. I work in experimental particle physics, in
particular on the Compact Muon Solenoid, one of the detectors at the
Large Hadron Collider. I'll post more on what I actually do at some
point in the future, but I thought I'd start with a post in the spirit
of some of the other fun calculations that we've done. My goal is to
convince you that black holes created by the LHC cannot possibly destroy
the world.
To start with, the main reason no one working on the LHC is too
concerned about black holes is because of <a href="http://en.wikipedia.org/wiki/Hawking_radiation">Hawking
radiation</a>. While we
usually think of black holes as objects that nothing can escape from,
Stephen Hawking predicted that black holes actually do emit some light,
losing energy (and mass) in the process. In the case of the little bitty
black holes that the LHC could produce, they should just evaporate in a
shower of Hawking radiation.
That's great you say, but Hawking radiation has never actually been
observed. What if Hawking is wrong and the black holes won't evaporate?
Well, the usual next argument is that <a href="http://en.wikipedia.org/wiki/Cosmic_ray">cosmic
rays</a> from space bombard the
earth all the time, producing collisions many times more energetic than
what we'll be able to produce at the LHC. To me, this is a fairly
convincing argument. However, let's pretend we don't know about these
cosmic rays and that there's no Hawking radiation. We can calculate what
effect black holes produced by the LHC would have on the earth if they
do stick around.
To start out with, the most massive black hole the LHC could produce
would be around 10 Tera-electron-volts, or TeV. We're probably
overestimating here. The eventual goal is for the LHC collisions to be
14 TeV, but producing a particle with the entire collision energy is
incredibly unlikely (see <a href="http://www.scientificblogging.com/quantum_diaries_survivor/fascinating_new_higgs_boson_search_dzero_experiment">Tomasso Dorigo's
post</a>
for more details on why, along with more details than you probably
wanted to know about hadron colliders). However, we want to think about
the worst case scenario here, and we're just going to do an order of
magnitude calculation, so 10 TeV is a good number. Note that I'm using a
particle physics convention here of giving masses in terms of energies
using E=mc^2. For reference, 10 TeV is about 1000 times smaller than a
small virus.
Now from the mass of our black hole, we can get its size by calculating
something called the <a href="http://en.wikipedia.org/wiki/Schwarzschild_radius">Schwarzschild
radius</a>. The
Schwarzchild radius for a black hole of mass m is given by
$$r = \frac{2Gm}{c^2}\text{.}$$
Here G is Newton's gravitational constant and c is the speed of light.
Plugging our mass in gives us
$$r = 10^{-50} \text{meters.}$$
This is incredibly small! In fact as I was writing this, I realized that
it's actually smaller than the Planck length, which means our equation
for the Schwarzschild radius may be somewhat suspect. Nonetheless, let's
hope that if we ever figure out quantum gravity, it gives us a
correction of order one and proceed with our calculation, which is just
an order-of-magnitude affair anyway.
Now, anything that enters the Schwarzschild radius of the black hole is
absorbed by it. The lightest thing that we could imagine the black hole
swallowing is an electron. Let's figure out how long on average a black
hole would have to travel through material with the density of the earth
before it absorbs an electron. In the spirit of considering the worst
case scenario, we'll have the black hole travel at the speed of light,
and consider the earth to be the density of lead.
We could do a complicated cross-section calculation to find the rate at
which the black hole accumulates mass, but we can also get it right up
to factors of pi through unit analysis. We know that the answer should
involve the area of the black hole, the density of the earth, and the
speed of the black hole. We want our answer to have units of mass per
time to represent the mass accumulation rate of the black hole. The only
combination that gives the right units is
$$a=\frac{\text{mass}}{\text{time}} = \rho c
r^2=\frac{10,000\text{kg}}{\text{m}^3}\frac{3\times 10^8
\text{m}}{\text{s}}(10^{-50}\text{m})^2} =
10^{-88}\text{kg/s}\text{.}$$
Alright, now that we know how fast our black hole accumulates mass,
let's figure out how long it takes it to accumulate an electron. The
electron mass is
$$m_e = 10^{-30}\text{kg,}$$
so the time to accumulate an electron is
$$t = \frac{m_e}{a} = 10^{50}\text{s.}$$
Now, the current age of the universe is 10^17 seconds. The time it
takes our black hole to accumulate an electron is longer than the age of
the universe by many orders of magnitude! So, if the LHC produces black
holes, and if Hawking is wrong, the black holes will just fly straight
through the earth without interacting with anything. Even if we take the
size of the black hole to be the Planck length, our black hole
accumulates an electron in 10^25 seconds, which is still much longer
than the age of the universe.
So the moral of the story is that you should be excited about the new
discoveries that the LHC might produce, and you don't need to worry
about black holes.</p>end of the earthLHCphysicshttps://thephysicsvirtuosi.com/posts/old/why-black-holes-from-the-large-hadron-collider-won-t-destroy-the-world/Sun, 16 May 2010 19:53:00 GMT