# 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 7TeV 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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