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marcus
#40
Jan11-05, 03:31 PM
Astronomy
Sci Advisor
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P: 23,235
I just happened to see a 1998 webpage that mentions the LHC and says that it will give the pair of colliding protons an collision energy of 5.4 TeV

http://www.nupecc.org/nupecc/report9...nc/node16.html

Is this right? I dont know what to expect from the LHC when it begins to operate up to spec, say in 2007.

but anyway, suppose it is right. and each proton has a kinetic energy of about 2.7 TeV, when they meet. What is that in natural?

by serendipity it turns out right around 10-15

I know that because it turns out that if you were to use a quarter of a volt, to measure voltage, then your 'electron quartervolt' would come out
10-28 of the natural energy unit

so we can take 2.7 x 1012 eV and multiply by 4 and get
around 11 x 1012 "eQ", which is 11 x 10-16 of the natural energy unit. Given that it is several years off and we dont know what energy LHC will actually attain, I am willing to call that 10-15

Now we have this "avogadro-like" number 2.6E18 which is how many proton rest masses make a natural mass unit-----or how many proton rest energies make a natural energy unit (same ratio).
And that tells us that to go from proton rest energy to LHC energy is a factor of 2.6E5. Have I made some mistake with powers of ten? It looks like what the accelerator does is increase the energy of a proton by 260,000 (a quarter of a million) and then smack two together head-on.

the speed that the proton must be going is, as you might expect, so close to 1 that, if I try to find the speed naively, my calculator cannot calculate it---it just says exactly 1.
I would have to proceed indirectly or else get a calculator with more than 12 digit accuracy because

[tex]\beta ^2 = 1 - (\frac{1}{260,000})^2[/tex]

Does anyone know if this figure of 5.4 TeV agrees with current expectations? Or, if not, what the current target energy is at LHC?

If you had to write beta, the speed, out as a decimal number it would be
11 nines followed by a three

beta = 0.999999999993