Large Hardon Collider

[Max™]

Nothing less than one of these:
http://telescopereviewsuk.files.word...ritage130p.jpg
http://telescopereviewsuk.wordpress....ian-telescope/

http://www.telescope.com/control/tel...ian-telescopes

A good dobsonian scope will generally give the best bang for the buck for a starting setup, from what I've seen.

[Max™]

http://www.newsdaily.com/stories/tre...s-cern-budget/

GENEVA, Sep. 17, 2010 (Reuters) — Europe's particle research center CERN unveiled budget cuts Friday that will force it to temporarily close its accelerators for a year in 2012, but said its flagship "Big Bang" machine will mainly be unaffected.

Announcing the trimmed-down budget, in which governments will provide 135 million Swiss francs ($133.4 million) less over a five-year period to 2015, CERN said its high-profile drive to study the origins of the cosmos would continue as planned.
It said it would delay upgrades to the Large Hadron Collider's beam intensity by one year, achieving this in 2016 instead of 2015, meaning scientists will have to wait longer for experiments to gather data at a faster rate.
A particle accelerator is a machine that propels a beam of sub-atomic particles at high speed. Physicists use the machines to create high-energy collisions so they can study the properties of the fundamental building-blocks of matter.

CERN operates a network of accelerators, including the world's biggest, the Large Hadron Collider (LHC), which opened two years ago to test predictions of high-energy physics.

CERN had previously announced that the LHC would not run in 2012 "for purely technical reasons." It said it would now also shut down all of its other accelerators in 2012 as it focuses its resources on the most critical research.
"The whole CERN accelerator complex will now join the LHC in a year-long shutdown," the institute said in a statement. "CERN management considers this a good result for the laboratory given the current financial environment."

Scientists and technical staff staged a protest outside CERN's main building on the French-Swiss border near Geneva last month over the possibility of budget cuts.

The reduced government contributions come as European governments seek to slash non-essential spending in the wake of a global financial crisis. Scientists say cuts to research budgets will reduce innovation and job creation, thus damaging economic recovery in the long-term.

[Kuya]

They should just cancel the project and let the free market fund these experiments. It would be more efficient that way.

[Max™]

http://www.wired.com/wiredscience/20...hysics-at-lhc/

http://www.wired.com/images_blogs/wi...collisions.png

After nearly 6 months of smashing particles, the Large Hadron Collider has seen signs of something entirely new. Pairs of charged particles produced when two beams of protons collide seem to be associated with each other even after they fly apart.
“It is a small effect, but it is very interesting in itself,” said physicist Guido Tonelli, spokesperson for the LHC’s CMS experiment. Tonelli and colleagues announced the results in a seminar at CERN September 21 and in a paper submitted to the Journal of High Energy Physics.
The LHC finally got up and running in March after more than a year of false starts. Beams of protons were smashed together in the 17-mile-long ring at energies of 7 teraelectronvolts (TeV) — three times the energy that had been achieved before.

When two protons collide, they produce a flurry of smaller, short-lived charged particles that fly away from each other at certain angles and speeds. The CMS (Compact Muon Solenoid) experiment at the LHC detects the path each of these particles takes. Physicists can then use those tracks to reconstruct what happened at the heart of the collision, like reassembling shards of glass from a broken window.
In the new experiment, the CMS team took data on the charged particles produced in hundreds of thousands of collisions. The team observed the angles the particles’ paths took with respect to each other, and calculated something called a “correlation function” to determine how intimately the particles are linked after they separate. The plot of the data ends up looking like a topographical map of a mountain surrounded by lowlands and a long ridge behind it.
In the most basic case (below, left), the data looked exactly like the physicists expected it to. But in cases where at least 110 charged particles were produced, the team saw a funny ridge-like structure extending away from the mountain peak (below, right).
That ridge essentially means that particles in some pairs are flying away from each other at close to the speed of light along one axis, but are oriented along the same angle in the other axis.
It’s as if two particles somehow talked to each other when they were produced, the physicists said. This phenomenon has never been seen before in proton-proton collisions, though it resembles something seen at RHIC (the Relativistic Heavy Ion Collider) at Brookhaven National Laboratory in New York. That effect was interpreted to be from the creation of hot dense matter shortly after the collisions.
The CMS team collected the data in mid-July, and spent the rest of the summer trying to blame it on an error or artifact of the data.
“We are here today because we didn’t succeed to kill it,” Tonelli said. As far as the team can tell, the effect is real.
But where it comes from, nobody knows. There are a lot of possible explanations, and the team is not ready to choose one yet.
“This is a subtle effect, and careful work is required to establish its physical origin,” said MIT physicist Gunther Roland at the seminar at CERN. “So fire away.”


Read More http://www.wired.com/wiredscience/20...#ixzz10C8Qmig1

http://www.wired.com/images_blogs/wi...sion-plots.png

[tyven]

WHAT DOES IT ALL MEAN?

[Max™]

They're still not sure, the RHIC reported finding some instances of particles appearing to exhibit correlation over long distances.

The LHC experiments were expected to show it was just signal noise, basically, but instead they found the same effect.

[Eliseos]

WHAT DOES IT ALL MEAN?

Blow shit up, fuck bitches, get money.

[Mizango]

Blow shit up, fuck bitches, get money.

Yep.

[Kaylia]

Solar flare changing the decay rate, fine-structure constant varying in the universe, particles showing new behaviors.....Theoretical physicist are going to have fun in the next few years. Can't wait to see next revolution....it's kinda exciting.

WHAT DOES IT ALL MEAN?

God exists. Only Him could have designed beautiful symmetric curved pattern .

They should just cancel the project and let the free market fund these experiments. It would be more efficient that way.

lol

[Eliseos]

How the hell do you do this integral without looking at an integral table or plugging into wolfram alpha?

http://latex.codecogs.com/gif.latex?...7B2%7D%7D%20dx

Wolfram won't give steps, and it lists the indefinite integral result as

http://latex.codecogs.com/gif.latex?...%7B2%7D%7Dx%29

and erf(x) is supposed to be http://functions.wolfram.com/GammaBetaErf/Erf/02/ which I have no idea what that is.

Woozie, if you do problem 2.11 in Griffiths, you'll have to do this integral. Expectation values can suck it hard.

EDIT: The answer of sqrt(pi)/2 from wolfram gives me the correct answer that I need, but yeah if this is given on an exam I'd fail it pretty fast.

EDIT2: I suck hard. The back cover of the book has the exact integral, it even includes the name of it (Gaussian) I looked up the derivation of it and I think I'll just use what the book gives. I hate math.

[Woozie]

Integrate by parts. And that's not the gaussian. It would be a gaussian if there was no x^2 term. The integral of the gaussian is sqrt(pi). That's something you should know from memory. If you're interested in seeing how it's derived, check out the wiki page. You should also know the guassian by heart for when there's an alpha constant in the exponential ( exp(-a*x^2) from negative infinity to positive infinity ...the answer is then sqrt(pi/a). )

To integrate this by parts, let u = x and du = x*exp(-x)

When you integrate by parts, you'll get

http://latex.codecogs.com/gif.latex?...fty}e^{-x^2}dx

The first term on the right is zero and the second is 1/2 times the gaussian, which you should memorize and know by heart. It evaluates to 1/2*sqrt(pi).

[Eliseos]

Hmm, in the back cover of Griffith's QM he calls it the Gaussian integrals, in just the e^(-x^2) integral where n is 0 for the x^(2n) term. The answer is the same since 0! is 1 and something raised to the 0th power is also 1. Idk he must have just generalized that integral and called it the gaussian. But yeah, your way of doing that integral is much simpler, and I hate integration by parts because I can never pick good substitutions for it I kept trying to have u be x^2 and dv be the e^(-x^2) term, and it was just a mess trying to combine everything.

EDIT: I'm dumb, the limits of integration on the back of the book gaussians are from 0 to infinity, not -infinity to infinity. The answers are still coming out the same though when I just multiply by two, since I'm checking whether it's an even or odd function before I do any of the integrals anyways.

[Kaylia]

Holy crap. I came to this thread to ask question about error functions as well...what was the odd.

Anyway, I'm trying to solve the integral of exp^-(ax+bx²) without mathematica, but can't figure out how. I'm not going to lie and pretend I spent more than 15 minutes trying, but does anyone has an idea how I could do it?

I know I should be able to solve this, but not doing any maths and physics for 6 months wasnt a good idea...I managed to forget everything in this short time.

[Eliseos]

Maybe try a u substitution? Let u = ax+bx^2, du = a + 2bx, do the easy integration with u and then plug back in u? Quantum mechanics has brought my math confidence way down so I am maybe probably wrong. If that doesn't work, could try splitting the exponentials up, so it'd be e^(-ax)*e^(-bx^2) and then integrate by parts, but yeah that might get messy also.

[Woozie]

Holy crap. I came to this thread to ask question about error functions as well...what was the odd.

Anyway, I'm trying to solve the integral of exp^-(ax+bx²) without mathematica, but can't figure out how. I'm not going to lie and pretend I spent more than 15 minutes trying, but does anyone has an idea how I could do it?

I know I should be able to solve this, but not doing any maths and physics for 6 months wasnt a good idea...I managed to forget everything in this short time.

I'm pretty sure an antiderivative doesn't exist. But if you're trying to evaluate this from negative infinity to positive infinity, you can just complete the square, and then use u substitution to make this into a gaussian.

Complete the square:
http://latex.codecogs.com/gif.latex?...sqrt{b}})^2}dx

u-substitution (u=sqrt(b)*x+a/(2*sqrt(b))
http://latex.codecogs.com/gif.latex?...fty}e^{-u^2}du

http://latex.codecogs.com/gif.latex?...frac{a^2}{4b}}

You might want to check my work. I think I may have misplaced a negative sign somewhere.

[Mizango]

I know I should be able to solve this, but not doing any maths and physics for 6 months wasnt a good idea...I managed to forget everything in this short time.

Yeah, I too learned that lesson long ago QQ

[Chun Liroy]

Since miz is making me post here....

"Two snow cats tow a housing unit to a new location.. shown by fig... The sum of the force F(A) and F(B) exerted on the unit by the horizontal cables is parallel to line L, F(B) is 30 degrees, F(A) is 50 degrees on either side of line L. F(A)= 4500N. Determine F(B) and the magn. of F(A)+F(B). I got that F=F(A)+F(B), so you do vector addition, and that theta=tan^-1(F(y)/F(x))... but I can't figure out how to do it with 2 variables missing -.- I bet it has something to do with substitution.. fux dat.

[Mizango]

lol, well that and I talking from my phone

[Chun Liroy]

Embarrassing... for me... A in calc 1 twice but can't do algebra based physics... -.-

[Mizango]

It happens. Last week I had an issue where I got stumped on a question from my daughters 5th grade homework, I too did well in Calc, but it turns out I was over complicating the problem. Needless to say I was not smarter than a 5th grader on that particular day.

Ill be back at my desk and in the office in the next 45-hour or so, I hope this isnt for your next class?