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Relativistic mechanics Mass is nothing but energy. The body’s inertia depends on its speed Let’s consider a body at rest which absorbs two electromagnetic impulses According to the momentum conservation law: After the absorption, the body will remain at rest. Where is the energy gone? Maybe it has increased the temperature of the body, but … what can we say if the body is an electron? Let’s consider the situation by the point of view of an observer who is moving at speed V towards left. If the body will remain at rest in the previous frame, then it will still move at speed V also with respect to the new observer. • But, according to the momentum conservation law, we expect that the body’s momentum will be increased: mafterVafter E' mbeforeVbefore cos c V must be the same (Vafter = Vbefore)so, the only possibility is that the mass m must change! E’ = mc2 ; the energy E’ is transformed in mass. Let’s go now to check the fundamental law of dynamics, that is: F ma or, in a more universal form: dp F dt F = ma doesn’t work well! If we apply a constant force – say an electric force – to an electron: uniform magnetic field uniform electric field – + then, accordingly to F = ma , dv eE dt m eE v t m the electron would reach the speed of light! We only need to wait for time T: mc T eE Our reference frame Another frame as appears to us In special relativity the fundamental law of dynamics is dp F dt where the momentum p is defined by the product mv So we get: p = F t and, finally: v F t 2 c 1 F t A massive body can’t reach the speed c Why must be m = m0 in p = mv Let’s go back for a while to the relativistic kinematics. We saw that x and t aren’t invariant quantities as regards Lorentz’s Transformation. If we think to L-T as transformation of coordinates in a 4-D space, then we have to find what is the geometrical characteristic (invariant) in such a 4-D space – the so-called Minkowsky’s space-time – . In Euclidean geometry the distance between two points is invariant: 2 = (xA – xB)2 + (yA – yB)2 + (zA – zB)2 = = (x’A – x’B)2 + (y’A – y’B)2 + (z’A – z’B)2 = ’2 according to T-L, the following expression is invariant: s2 = c2t2 – x2 = c2t’2 – x’2 = s’2 Where (ct ; x ; y ; z) are the coordinates of the point (event) – ct is the time component and the other are the ordinary space coordinates – This expression can be taught as a way to calculate the magnitude of a quadrivector defined in a 4_D geometry. According to this geometrical interpretation we need to find a dynamics quadrivector whose components are the “classical” momentum : mv : mvx ; mvy ; mvz Because the conservation of momentum is a consequence of the homogeneity of space, the component of mv will appear in the “spatial places” of the dynamical quadrivector. ( ? ; mvx ; mvy ; mvz ) The first component, the time placed one, has to be related with energy, because energy conservation is a consequence of the homogeneity of time. So the quadrivector is: E ; mv x ; mv y ; mv z c This can be called energy-momentum quadrivector The magnitude of the energymomentum must be invariant: 2 2 0 2 E E 2 2 m v 2 c c where in the right member we have considered a reference frame in which a body is at rest. Let m0 be the (rest) mass of the body. Then: E0 = m0c2 and m = m0 2 4 m02 2 c 4 m 2 2 2 0c m0 v 2 2 c c 2(c2 – And, finally: v2) = c2 2 c 2 2 2 c v 1 1 2 !