5
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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 Aug 12 '18 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 Aug 29 '18 at 12:44
6
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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 Jan 21 at 7:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.