# Relativistic speed/energy relation. Is this correct?

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 if 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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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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