# Thread: Protons in the LHC

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## Protons in the LHC

I read on one of the websites that the protons in the LHC will get up to about 99.9999% the speed of light (I don't remember which site).

1. Is that speed accurate?

2. Given that mass increases as one approaches the speed of light, how heavy will these actually get, or am I misunderstanding how this part of relativity works?

Pete  Reply With Quote

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The protons have an energy of about 7TeV, which is a lot for a proton but still only about 11.2E-7 joules. Divide by c^2 to get 1.24E-23 kilograms.

That's still a lot for a proton, which has a rest mass of 1.67E-27 kg.  Reply With Quote

3. So that's 7,425 times as massive then. Whoa.  Reply With Quote

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Considering mass basically follows a 1/x function, I thought it would be much heavier than that. I know the formula for mass in relativity (m/sqrt(1-v^2/c^2) but how does that relate to eV? (it's been a very long time since that and I don't teach eV at the highschool level)

Pete  Reply With Quote

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6,058 Originally Posted by peter eldergill Considering mass basically follows a 1/x function, I thought it would be much heavier than that. I know the formula for mass in relativity (m/sqrt(1-v^2/c^2) but how does that relate to eV? (it's been a very long time since that and I don't teach eV at the highschool level)

Pete
mass of a proton is 931 MeV IIRC.  Reply With Quote  Reply With Quote

7. Originally Posted by peter eldergill Considering mass basically follows a 1/x function, I thought it would be much heavier than that. I know the formula for mass in relativity (m/sqrt(1-v^2/c^2) but how does that relate to eV? (it's been a very long time since that and I don't teach eV at the highschool level)

Pete
eV is a very handy unit for the energy of charged particles. A particle with charge -or + e (the elemantary charge) is accelerated by a voltage difference of 1 Volt, it has the energy of 1eV. So the unit makes a lot of sense if you deal with accelerated particles.

So all you need is how to convert eV to Joule.

From What is said above it follows that

1 eV = 1.6022x10^-19 C x 1V = 1.6022 x 10^-19 A x s x V= 1.6022 x 10^-19 W x s = 1.6022 x 10^-19 J

where:
C = Coulomb SI unit of charge
(1.6022 x 10^-19 is the elementary charge e)
A = Ampere SI unit for current
s = second
W = 1 Watt SI unit of Power
J = Joule SI Unit for work/energy

ETA:
What did I do? Just playing around with formula for electrical power and current an the units.

the electrical power P = UxI written in units means W = VxA
Now the current is defined as "charge per time" A = Q/s which reveals Axs =Q
Now add that W=J/s or Wxs = J

This is all you need.

ETA: My High school teacher as well as my first lecturer in physics on university always insisted on "Look at the equations for the units!". When we did calculations we always wrote down the units and checked if the result made sense. It's like v=s/t which is written in units [m/s] = [m]/[s]. Obviously correct. This always helped as a double check. If you are supposed to calculat a Energy and the unit for the result is [kg]/[s] than something went wrong.
Last edited by AndreH; 2008-Sep-14 at 01:50 PM.  Reply With Quote

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