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?
Best Answer
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