Prove e(charge of proton)=1.6E-19 | Mev.fm

[chikou24i]

How to prove that e(the charge of proton)=1.6E-19 Mev.fm

 

[ChrisVer]

That is obviously wrong, where did you find this conversion?
It's 1.6 \times 10^{-19} ~C

 

[vanhees71]

In SI units you have
$$\alpha=\frac{e^2}{4 \pi \epsilon_0 \hbar c} \simeq 1/137,$$
which is dimensionless. This is the Sommerfeld fine structure constant describing, in natural units, the interaction strength of charged particles due to their electromagnetic interaction ($e\simeq 1.6 \cdot 10^{-19} \mathrm{C}$ is the charge of a proton, $\hbar$ is Planck's modified constant, and $c$ the speed of light, $\epsilon_0$ is an artifact of the "ugly" SI units, which are made for practical purposes rather than to expose the beauties of a consistent relativistic formulation of classical and quantum electrodynamics).

In modern high-energy physics, rationalized Gaussian (also called Heaviside-Lorentz) units are used, which brings the full symmetry of space and time and the correct relativistic dimensions of the electromagnetic field forward (electric and magnetic field components having the same unit). In addition, theoreticans have a hard time with their math anyway, and thus reduce the effort further by choosing "natural units", measuring angular momenta, actions, etc. in units of $\hbar$. Thus effectively they set $\hbar=1$. Then, as space and time are very much more symmetric in relativistic than in Newtonian physics, it nearly doesn't make sense to measure them in different units. This is the more true since also in the SI units the speed of light is just a conversion factor with a once-and-for all fixed value. So you can as well set $c=1$ and measure velocities in units of $c$. This implies that energies and momenta have the same unit, usually using MeV or GeV, which is convenient when working in high-energy physics as at the LHC. Also lengths and times have the same unit. Here a convenient scale is fermi (or femtometer, which is $10^{-15} \;\mathrm{m}$).

To convert back to usual units, one simply has to do some dimensional analysis of the quantity in question and use the relation $\hbar c \simeq 0.197 \;\text{MeV} \; \text{fm}$.

Then you need to remember just one more thing: Cross sections are usually given in milli-barns. Then you just have to remember that $10 \; \mathrm{mb} = 1 \; \mathrm{fm}^2=10^{-30} \; \mathrm{mb}$.

 

[ChrisVer]

In SI units you have

\alpha=\frac{e^2}{4 \pi \epsilon_0 \hbar c} \simeq 1/137,​

which is dimensionless.

Just to point out something (which I always disliked as an undergrad), the last equality is true at a given energy scale.

 

[vanhees71]

It's valid at low energy scales. The running of the em. coupling leads to $\alpha(\Lambda=m_{Z}) \simeq 1/128$.

 

[Vanadium 50]

Do you two think this is helping the OP?

 

[ChrisVer]

Do you two think this is helping the OP?

How are you supposed to help someone who asks a false question and then doesn't reply back ? - where did he find this "conversion". I don't have the charisma of a psychic..

 

[vanhees71]

I indeed hope, it helps to tell the matters as they are.

 

[chikou24i]

First of all I'm sorry for the late of replying, Vanhees71 Thank you for the answer you really helped me and ChrisVer I asked a question for you to help me or to correct me if I was wrong, and thank you for you too for the notice.

 

[ChrisVer]

ChrisVer I asked a question for you to help me or to correct me if I was wrong,

Sorry then, the way your question was written, you made me believe you had seen this conversion somewhere (and asked for a "proof"), and that's why I asked for the source ("where did you find it?").