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The relativistic energy-momentum equation is: $$E^2 = (pc)^2 + (mc^2)^2.$$ Also, we have $pc = Ev/c$, so we get: $$E = mc^2/(1-v^2/c^2)^{1/2}.$$

Now, accelerating a proton to near the speed of light, I get the following results for the energy of proton:

0.990000000000000   c =>    0.0000000011    J =      0.01    TeV 
0.999000000000000   c =>    0.0000000034    J =      0.02    TeV 
0.999900000000000   c =>    0.0000000106    J =      0.07    TeV 
0.999990000000000   c =>    0.0000000336    J =      0.21    TeV 
0.999999000000000   c =>    0.0000001063    J =      0.66    TeV 
0.999999900000000   c =>    0.0000003361    J =      2.10    TeV 
0.999999990000000   c =>    0.0000010630    J =      6.64    TeV
0.999999999000000   c =>    0.0000033614    J =      20.98   TeV 
0.999999999900000   c =>    0.0000106298    J =      66.35   TeV 
0.999999999990000   c =>    0.0000336143    J =      209.83      TeV 
0.999999999999000   c =>    0.0001062989    J =      663.54      TeV 
0.999999999999900   c =>    0.0003360908    J =      2,097.94    TeV 
0.999999999999990   c =>    0.0010634026    J =      6,637.97    TeV 
0.999999999999999   c =>    0.0033627744    J =      20,991.10   TeV

If the LHC is accelerating protons to $7 TeV$ it means they're traveling with a speed of $0.99999999c$.

Is everything above correct?

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  • $\begingroup$ You've just triggered a pet peeve of mine. That first column of equals signs is wrong. velocities are not energies. That's not an equality. $\endgroup$
    – dmckee --- ex-moderator kitten
    Commented Aug 12, 2018 at 8:59
  • $\begingroup$ The idea is that a proton that is traveling with the speed on colon one has the energy specified in columns 2/3. You're right that equal sign it's not appropriate...it should be kind of an arrow probably $\endgroup$
    – Albert
    Commented Aug 29, 2018 at 12:44

1 Answer 1

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Yes you are correct.

If the rest mass of a particle is $m$ and the total energy is $E$, then

$$ E = \gamma mc^2 = \frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}, $$

thus

$$ \frac vc = \sqrt{ 1 - \left( \frac{mc^2}E \right)^2 } \approx 1 - \frac12 \left( \frac{mc^2}E \right)^2 $$

The proton rest mass is 938 MeV, so at 7 TeV, the proton's speed is

$$ 1 - \frac vc = \frac12 \left( \frac{938\times10^6}{7\times10^{12}} \right)^2 = 9 \times 10^{-9} $$

meaning v ~ 0.999 999 991 c

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  • $\begingroup$ This answer is not quite correct. In the first equation, $E$ should be $\gamma mc^2 - mc^2$ rather than $\gamma mc^2$. Then $E/mc^2 = \gamma - 1 = \frac{1}{\sqrt{1-v^2/c^2}} - 1 \Rightarrow \left(\frac{1}{E/mc^2 + 1}\right)^2 = 1 - v^2/c^2 \Rightarrow v/c = \sqrt{1-\left(\frac{1}{E/mc^2 + 1}\right)^2}$ and we find that the proton's speed at 7 TeV is... still $0.999999991 c$. Although the result is essentially the same, I thought I'd point this out in case someone tries to use the same formula for something else. $\endgroup$
    – Thorondor
    Commented Jan 21, 2019 at 7:29

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