Natural unit

January 21, 2011

GoLuckyRyan Basic, no math, operational , , Leave a comment

on Size, Energy and Unit, we know that the speed of light better be equal to 1, that simplify the equation of relativity.

c = 1

now, we impose 1 more things, the Reduced Planck constant,  \hbar also set to 1. that simplify all equations with angular momentum or spin.

\hbar = 1

the Angular momentum:

J^2 \left| l,m \right > = j(j+1) \hbar ^2 \left| l,m\right >

J_z \left|l,m \right> = m \hbar \left|l,m\right>

now becomes :

J^2 \left| l,m \right > = j(j+1) \left| l,m\right >

J_z \left|l,m \right> = m \left|l,m\right>

when we want to calculate the real value, we can recover the \hbar by considering the dimension. the reduced Planck constant has dimension

[\hbar] = [kg][m^2][s^{-1}] = [energy][second]

for example,

E = \omega \hbar

on the sum of 4 momentum and excited mass

January 17, 2011

GoLuckyRyan Basic, relativity , , Leave a comment

when we have a decay process, there are many fragments, we can measure their momentum and energy and construct the 4-momentum

\vec {P_i} = ( E_i , p_i )

we use the c = 1 unit as usual.

to find out the mass before the decay, we can use

\sum E_i^2 - \sum p_i^2 = m_{excite}^2

the reason for the term “excited mass”, we can see by the following illustration.

consider a head on collision of 2 particles in C.M. frame, with momentum p and energy E1 and E2.

the mass for each one is given by

m_1 = \sqrt {E_1^2 - p^2 }

m_2 =\sqrt {E_2^2 - p^2 }

but if we use the sum of the 4 momentum and calculate the mass,

\sqrt { (E_1 +E_2)^2 - (p - p)^2} = E_1+E_2

which is not equal to

\sqrt { E_1^2 - p^2} + \sqrt{E_2^2 - p^2 }

in fact, it is larger.

the reason for its larger is, when using the sum of 4 momentum, we actually assumed the produce of collision is just 1 particles, and the collision is inelastic. Thus, if we think about the time-reverse process, which is a decay, thus, some of the mass will convert to K.E. for the decay product.

Special Relativity I

December 17, 2010

GoLuckyRyan Basic, theory , , Leave a comment

i just state the formula and the usage of it.

the basic equation is

E^2 = (p c)^2 + (m c^2)^2

where E is total energy, p is momentum

here we can see the advantage of using MeV as unit of mass. the equation is now further simplified by using MeV/c as momentum unit.

E^2 = p^2 + m^2

which is Pythagorean theorem!

the speed of the particle is from the formula

\displaystyle \beta = \frac {v}{c} = \frac {p}{E}

For example, proton mass is 940 MeV/c2, if we say an proton is moving at 94MeV, or a 94 MeV proton. we mean, the KINETIC Energy (K.E.) of proton is 94MeV. The total energy is

E = m + K.E.

Thus, a 94 MeV proton is moving at 41.7% of light speed. by using a right-angle triangle of base 10, side 11, and the hight is Square-Root 21.

another way around is, a proton at 90% speed of light, how much K.E. it has? which is around 3000 MeV or 3 GeV [Giga eV]