PHYS
From Special Relativity to Feynman Diagrams.pdf

# We shall first show that if we insist in defining the

• 549

This preview shows pages 51–54. Sign up to view the full content.

We shall first show that, if we insist in defining the mass as independent of the velocity, then the conservation of momentum can not hold in any reference frame , thus violating the principle of relativity. 2 Let us consider a simple process in which a mass m explodes into two fragments of masses m 1 = m 2 = m / 2 (or equivalently a particle of mass m decays into two particles of equal masses, see Fig. 2.2 ). We shall assume the conservation of linear momentum to hold in the frame S in which the exploding mass is at rest: v m = 0 = m 2 v 1 + m 2 v 2 v 1 = − v 2 . For the sake of simplicity we take the x -axis along the common direction of motion of the particles after the collision, so that v 1 ( y , z ) = v 2 ( y , z ) = 0. Let us now check whether the conservation of linear momentum also holds in a different frame S . We choose S to be the rest frame of fragment 1, which moves along the positive x -direction at a constant speed V = v 1 ( x ) v 1 relative to S , and let the explosion 2 Here and in the rest of this chapter, when referring to the conservation of the total linear momentum of an isolated system of particles, we shall often omit to specify that we consider the total momentum and that the system is isolated , regarding this as understood.

This preview has intentionally blurred sections. Sign up to view the full version.

40 2 Relativistic Dynamics Fig.2.2 Decay of a particle into two particles of equal masses Fig.2.3 Same decay in the rest frame of particle 1 occur at the instant t = t = 0 , see Fig. 2.3 . In the frame S , the velocity of the mass m before the explosion is obtained by applying the relativistic composition law for velocities ( 1.75 ): v ( m ) x = 0 v 1 1 0 · v 1 c 2 = − v 1 = − V , v ( m ) y , z = 0 . Analogously, after the explosion, the velocities of the fragments in S are given by: v 1 v 1 x = 0 , v 2 v 2 x = v 1 v 1 1 + v 2 1 c 2 , v 2 y = v 2 z = 0 . Having computed the velocities in S we may readily check whether the con- servation of linear momentum holds in this frame. It is sufficient to consider the components of the linear momenta along the common axis x = x ; before and after the explosion the total momenta in S are given respectively by: P in = m v ( m ) = − m v 1 , (2.2) P f in = m 2 v 1 + m 2 v 2 x = m 2 v 2 = − m v 1 1 + v 2 1 c 2 . (2.3)
2.1 Relativistic Energy and Momentum 41 Since m v 1 = m v 1 1 + v 2 1 c 2 , (2.4) we conclude that, in S the total momentum is not conserved . Or better, the principle of conservation of linear momentum (as defined in classical mechanics) is not covariant under Lorentz transformations thereby violating the principle of relativity. As such it can not be taken as a founding principle of the new mechanics. It is clear that, just as the principle of relativity can not be avoided in any physical theory, it would also be extremely unsatisfactory to give up the conservation of linear momentum; in the absence of it we would indeed be deprived of an important guiding principle for building up a theory of mechanics. To solve this apparent puzzle, it is important to trace back, in the above example, the origin of the non-conservation of the total momentum.

This preview has intentionally blurred sections. Sign up to view the full version.

This is the end of the preview. Sign up to access the rest of the document.
• Fall '17
• Chris Odonovan

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern