define the u_4 vector note the differentiation with respect to ds.

u=(u_t, u_x, u_y, u_z) = (cdt/ds, dx/ds, dy/ds, dz/ds) note that by its definition the 4-velocity is dimensionless like beta=v/c

take the u^2 = u_t^2-u_x^2-u_y^2-u_z^2 = [ (cdt)^2- dx^2-dy^2-dz^2]/ ds^2 = ds^2/ds^2 = 1

express ds^2=gamma^(-2) c^2 dt^2 > (c^2dt^2)/(ds^2) gamma^(2)

then the u_t component above is cdt/ds= sqrt(c^2dt^2/ds^2) = sqrt (gamma^2) => u_t gamma

everybody should reproduce the above

and then u_4 =( gamma, v/c gamma)

Momentum 4-vector

newtonian P =mv where v=3-velocity

Define 4-momentum as follows

P_j= m_0 * u_j * c where u_j is the the famous 4-velocity (i.e. j runs 0(time) , 1(x), 2(y), 3(z))

then the 3-momentum using the 3-velocity (v_a) can be expressed as

P_a=m_0 * u_a * c = m_0 * (gamma v_a/c ) * c (taking the 3-velocity from above)

= m_0 * gamma * v_a this should be equal to P=mv (newton) because it is the 3-component part of the 4-vector => m m_0 * gamma (this says that the total mass of the moving particle is bigger than its rest mass m_0)

the temporal component of the momentum 4-vector as we defined above is

P_t= m_0 * u_t * c = m_0 * gamma * c = m*c

from the Taylor expanded mass we had m*c^2=m_0*c^2+KE and we said the total energy of the moving particle is

E_tot=m*c^2 = m c* * c = P_t *c > P_t E_tot/c ==>

the temporal component of the momentum is the total energy divided by c

So now we look for the direct relation between E_tot, and P_j

P_j = (P_t, P) => P_j^2 P_t^2- P^2 = (E_tot/c)^2- P^2 ==>

(m_0 * u_j c)^2* = (E_tot/c)^2- P^2 ==> (u_j^2=1 from the 4-velocity)

m_0^2 *c^2 = E_tot^2/c^2 - P^2 ==>

E_tot= +- sqrt(m0^2*c^4+P^2c^2)

-- Main.smaria - 2011-01-21