# Velocity of a proton

1. May 6, 2013

### Abigale

Hello,

I regard a particle in an accelerator. The particle has the kinetic energy of 7TeV.

I have callculated the momentum $$E=pc+mc^2\\ \Rightarrow p=\frac{1}{c} \sqrt{E^2 -(mc^2)^2} =7,00094~ TeV/c$$

After that I want to callculate the "velocity" and the "$\gamma$-factor".
But I am irritated and don't know which equations are allowed for this relativistic callculations.

For example I have found the equation $$\vec{p}=m\gamma\vec{v}$$
and
$$\gamma=\frac{1}{\sqrt{(1-\beta)}}~~~~~~~;\beta=\frac{v}{c}=\frac{pc}{E}$$

THX
Abby

2. May 6, 2013

### Bill_K

If it's the LHC you're talking about, 7 TeV represents the total energy of the particle, not just the kinetic energy.

Try E = γmc2.

3. May 15, 2013

### DEvens

That should be
$$E^2=(pc)^2+(mc^2)^2\\ \Rightarrow p=\frac{1}{c} \sqrt{E^2 -(mc^2)^2} \approx 7 TeV/c$$

The last assumes that the mass energy is very small compared to 7 TeV. If it's a proton then the mass is 0.938 GeV, which is quite small compared to 7 TeV.

You've got either gamma or beta wrong. I don't know your convention.
$$\gamma=\frac{1}{\sqrt{(1-\frac{v^2}{c^2})}}$$

But there's an easier way. The kinetic energy is $(\gamma -1) m c^2 = 7 TeV$ and so you can work out $\gamma= (7 TeV - m c^2) / m c^2$, and to three digits that's 7460. And then you can work out v. (Assuming I did the arithmetic correctly.)

$$v = c \sqrt{1-\frac{1}{\gamma^2}} \approx c (1 - \frac{1}{2(7460)^2}) \approx c(1-8.89 \times 10^{-9})$$