1. Relativistic mechanics
Mass is nothing but energy.
The body’s inertia depends on its speed

2. Let’s consider a body at rest which absorbs two electromagnetic impulses

3. 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?

4. Let’s consider the situation by the point of view of an observer who is moving at speed V towards left.

5. 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:

6. V must be the same (Vafter = Vbefore)so,
the only possibility is that the mass m must change!
 E’ = mc2 ;
the energy E’ is transformed in mass.

7. Let’s go now to check the fundamental law of dynamics,
that is: or, in a more universal form:

8. F = ma doesn’t work well!
If we apply a constant force – say an electric force – to an electron: uniform magnetic field uniform electric field – +

9. then, accordingly to F = ma ,
the electron would reach the speed of light! We only need to wait for time T:

10. Our reference frame

11. Another frame as appears to us

12. In special relativity the fundamental law of dynamics is
where the momentum p is defined by the
product mv So we get: p = Ft
and, finally:

13. A massive body can’t reach the speed c

14. Why must be m = m0 in p = mv
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 – .

15. 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

16. according to T-L, the following expression is invariant:
s2 = c2t2 – x2 = c2t’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.

17. According to this geometrical interpretation
we need to find a dynamics quadrivector whose components are the “classical” momentum : Because the conservation of momentum is a consequence of the homogeneity of space, the component of will appear in the “spatial places” of the dynamical quadrivector.

18. ( ? ; 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: This can be called energy-momentum quadrivector

19. The magnitude of the energy-momentum must be invariant:
where in the right member we have considered a reference frame in which a body is at rest.

20. Let m0 be the (rest) mass of the body.
Then: E0 = m0c2 and m = m0  2(c2 – v2) = c2  And, finally: !