Gamma factor when doing four momentum problems

[bonbon22]

Homework Statement

Why does the gamma factor drop off when doing a four momentum problem?

Relevant Equations

ϒ = gamma factor = lorentz factor = 1-( v^2/c^2)

So if i had this problem where i am squaring a four momentum vector with itself which gives

P² = ($\gamma mc$ )² - $\gamma$²$m$²$\vec v$ *$\vec v$

I have been told that the gamma factor is not considered at all. why would the gamma factor drop off? Does this rule apply to any four momentum problem?

 

[Orodruin]

First of all, you have the wrong expression for the gamma factor, i.e., you have written $\gamma = 1 - v^2/c^2$ when it should be $\gamma = 1/\sqrt{1-v^2/c^2}$. That it falls out from the expression for $P^2$ is basic algebra from there.

 

[bonbon22]

First of all, you have the wrong expression for the gamma factor, i.e., you have written $\gamma = 1 - v^2/c^2$ when it should be $\gamma = 1/\sqrt{1-v^2/c^2}$. That it falls out from the expression for $P^2$ is basic algebra from there.

If i had this other scenario where one proton is smashing into another which is stationary. Which creates a new particle, also stationary. Then balancing the four momentum i get P1 + P2 = P3 squaring i get P1^2 + P2^2 +P1.P2 =P3^2 . Since particle 2 let's say in this case is stationary the v2 becomes zero so using the Lorentz factor equation plugging in i get gamma 2 = 1. So is it possible then for the P1.P2 term to lose the gamma1 factor?

 

[Orodruin]

If i had this other scenario where one proton is smashing into another which is stationary. Which creates a new particle, also stationary.

This would violate conservation of momentum as you would have non-zero momentum before the collision and zero momentum after.

P1^2 + P2^2 +P1.P2 =P3^2

It should be $p_1^2 + p_2^2 + 2p_1 \cdot p_2 = p_3^2$.

So is it possible then for the P1.P2 term to lose the gamma1 factor?

It is not clear to me exactly what you are asking here.