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# .

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

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