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*Velocity 4-vector* 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* =m<strong>v</strong> 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 <strong>P</strong>=m<strong>v</strong> (newton) because it is the 3-component part of the 4-vector ==> <strong>m= </strong> *m_0 * gamma* <strong> (this says that the total mass of the moving particle is bigger than its rest mass m_0)</strong> the temporal component of the momentum 4-vector as we defined above is P_t= m_0 * u_t * c = *m_0 * gamma ** c = <strong>m*c </strong> 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, <strong>P) ==> </strong> *P_j^2* <strong>= P_t^2- P^2 = (E_tot/c)^2- P^2 ==> </strong> *(m_0 * u_j* c)^2* <strong>= (E_tot/c)^2- P^2 ==> (u_j^2=1 from the 4-velocity)</strong> *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
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Topic revision: r1 - 2011-01-21
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smaria
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