Available energy in particle collision derivation

[Piyu]

Hello, I am trying to derive the formula given in by

.

I understand we need to move it to the center of momentum frame to solve. Using the relativistic energy equations. I find that :

E = E₁ + E₂
E₁ = sqrt((Mc)²+(pc)²)
E₂ = sqrt((mc)²+(pc)²)

where both have equal magnitudes momentum p since its the center of momentum frame.

However, this is where i get stuck. no matter how i add them up i can't seem to get rid of the square root signs and solve the equation.

 

[gasperkm]

Hi,

I'm assuming you got the equation from Wikipedia - it's correct, however, Wikipedia is rather vague on where they actually got the equation from.

The starting equation you are looking for is,
[tex]E_a^2 = \left(\sum_{i=1}^n E_i \right)^2 - \left(\sum_{i=1}^n p_i c \right)^2[/tex]
where the sums are over each particle included in the system. In your case, these would be the moving particle and the stationary particle. I must admit I don't completely understand the equation (we mentioned it in the particle physics course), but I guess Wikipedia is correct about the net momentum not being able to convert the whole kinetic energy into mass (represented as the subtraction in the equation). The squares are probably there, just to ensure that the available energy is constant before and after the collision.

E = E1 + E2
E1 = sqrt((Mc)2+(pc)2)
E2 = sqrt((mc)2+(pc)2)

Also be sure to take the correct equation for energy (you are missing a square on [itex]c[/itex] at [itex](m c)^2[/itex]).

Hope this has been helpful.