Question

4) Prove the Lorentz transformation relations for energy and momentum.

Answer 1

sol": Let us Consider two frames s and swhere s' is moving with speed u with respect to the s. frame. Let the coordinate of an event in s frame is (t, x, y, z) and the coordinate of that event in S' frame is (t's x's y', z'). We will take the boost is along x-direction ie, the frame &' is moving with velocity v along x-dirn with respect to the strome. we know the position and time co-ordinates of the two frames s and s' are related by Larendz transform- ť = (t - vxr X'= (*- it)r ation, given by - - and Z' =Z where, y = I VI-Vyz Now let let p = 0%. B - Then can be written as a ct' = (ct - fx) X = (X- ßc tr the above transformations . The and Z above = z equations can be written as in metrix form; ot' / -rf vool yil Z') x TN

ar we can write - X 0 000 1=1-rß r 1 0 oollx' 1 0 1 0 X X where, x° = ct, x'=x, x=y and y² =z and yor = ct'. X = X', y'ay' and y?! Z" Nowe o we can see that the Lorentz tranformation matrix Tv -rf 0 o1 I -vp rool so ation - (or enorgy-momentum), the transton- four momentum i the following i po = -rer 0 0 0 Il pe 0 1 2 1 Od (Ek) Y b3 / -rß o ol /EC) 1-rß rool Px where. po = Elc ..o o o l \ P² / so the energy. momentum transformations are E%= r Efe - YB PX I p² =P I t p ² = Pž 1 = E=(E-vPx) op and pé = Pz B = -rf 86 + BE Px = v(Px - EVE) 3)
so we see that the energy transforms like the position coordinate and the momentum transforms like the time coordinates.We have calculated here the Lorentz transformation for energy and momentum when there is a boost along x-axis that is why the y and z-components of momentum are unchanged. Similarly we can find the transformation relation for boost along y and z-axis.