# Essential physics 1 by Firk F.W.K.

By Firk F.W.K.

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Integrating gives x(t) = ∫F(t)dt + Ct + C´. 13) The constants of integration can be determined if the velocity and the position are known at a given time. 2) If a = g(x) = v(dv/dx) then vdv = g(x)dx. 14) 36 KINEMATICS: THE GEOMETRY OF MOTION v 2 = 2∫g(x)dx + D, therefore v 2 = G(x) + D so that v = (dx/dt) = ±√(G(x) + D). 15) Integrating this equation leads to ±∫dx/{√(G(x) + D)} = t + D´. 16) Alternatively, if a = d2x/dt2 = g(x) then, multiplying throughout by 2(dx/dt)gives 2(dx/dt)(d 2x/dt2) = 2(dx/dt)g(x).

Consider the result of two successive Lorentz transformations L1 and L2 that transform a 4-vector x as follows x → x´ → x´´ where x´ = L1x , and x´´ = L2x´. The resultant vector x´´ is given by x´´ = L2(L1x) = L2L1x = Lcx where Lc = L2L1 (L1 followed by L2). If the combined operation Lc is always a Lorentz transformation then it must satisfy LcT Lc = We must therefore have . 22) CLASSICAL AND SPECIAL RELATIVITY 53 (L2L1)T (L2L1) = or L1T(L2T L2)L1 = so that L1T L1 = , (L1, L2 ∈ L) therefore Lc = L2L1 ∈ L .

If we multiply t and t´ by the constants k and k´, respectively, where k and k´have dimensions of velocity then all terms have dimensions of length. In space-space, we have the Pythagorean form x2 + y 2 = r 2 (an invariant under rotations). We are therefore led to ask the question: is (kt)2 + x 2 an invariant under G in space-time? Direct calculation gives (kt)2 + x 2 = (k´t´) 2 + x´ 2 + 2Vx´t´ + V2t´2 = (k´t´)2 + x´ 2 only if V = 0 ! We see, therefore, that Galilean space-time does not leave the sum of squares invariant.