# Available energy (particle collision)

For a discussion about the meaning of the term in classical thermodynamics, see Exergy.

In particle physics, the available energy is the energy in a particle collision available to produce new matter from the kinetic energy of the colliding particles. Since the conservation of momentum must be held, a system of two particles with a net momentum may not convert all their kinetic energy into mass - and thus the available energy is always less than or equal to the kinetic energy of the colliding particles. The available energy for a system of one stationary particle and one moving particle is defined as:

$E_a = \sqrt{2 E_t E_k + (m_t c^2)^2 + (m_k c^2)^2}$

where

$E_t$ is the total energy of the target particle,
$E_k$ is the total energy of the moving particle,
$m_t$ is the mass of the stationary target particle,
$m_k$ is the mass of the moving particle, and
$c$ is the speed of light.

## Derivation

This derivation will use the fact that:

$(mc^2)^2 = E^2-P^2c^2$

From the principle of the conservation of linear momentum:

$P_a = P_k$

Where $P_a$ and $P_k$ are the momentums of the created and the initially moving particle respectively. From the conservation of energy:

$E_T= E_t+E_k$

Where $E_T$ is the total energy of the created particle. We know that after the collision:

$(E_a)^2=(E_T)^2-(P_a)^2 c^2$
$(E_a)^2=(E_t+E_k)^2-(P_k)^2 c^2$
$(E_a)^2=(E_t)^2 +(E_k)^2 + 2 E_t E_k-(P_k)^2 c^2$

Donating this last equation (1). But

$(m_k)^2 c^4=(E_k)^2-(P_k)^2 c^2$

and since the stationary particle has no momentum

$(m_t)^2 c^4=(E_t)^2$

Therefore from (1) we have

$(E_a)^2=(m_k)^2 c^4+(m_t)^2 c^4 + 2 E_t E_k$

Square rooting both sides and we get

$E_a = \sqrt{ (m_t c^2)^2 + (m_k c^2)^2+2 E_t E_k}$