on the sum of 4 momentum and excited mass

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

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

Sqrt[ Sum[E_i]^2 - Sum[p_i]^2 ] = excited mass.

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

m1=Sqrt[ E1^2 - p^2 ]
m2=Sqrt[ E2^2 - p^2 ]

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

Sqrt[ (E1+E2) ^2 - (p - p)^2 ] = E1 + E2

which is not equal to Sqrt[ E1^2 - p^2 ] + Sqrt[ E2^2 - 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.

Building of special relativity graphically.

Special Relativity II

We are going to talk about coordinate transform from "center of momentum frame" to "Laboratory frame".

At the center of momentum [C.M.] frame, the total momentum is zero.

Refer here on Google Docs. for the Mathematica 7 code and calculation steps.

anyone want to have the .nb file, feel free to ask.

i am just discuss on the result.

1) In the C.M. frame, energy of each particle reserved, there is no exchange in energy and momentum. after collision, they just change the moving direction.

2) the energy is C.M frame is always smaller then Lab frame, or other fame. which is also from the face that, at the CM frame, total momentum is zero and the corresponding energy is the Proper Energy.

3) In Lab frame, the scatter angle is always smaller than 90 degree for incident particle's mass > target particle.

4) In Lab frame, The larger the scatter angle, the smaller the momentum and larger the momentum transfer.

Special Relativity I

i just state the formula and the usage of it.

the basic equation is

E² = (p c)² + (m c²)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.

E² = p² + m²

which is Pythagorean theorem!

the speed of the particle is from the formula

 β = v / c = p / E

For example, proton mass is 940MeV/ 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

Mass + K.E. = 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]