This discussion focuses on the derivation of relativistic energy and momentum using rapidity as a more convenient variable than velocity. The key equations presented include the energy function ##E = m g(\theta) = m c^2 \cosh(\theta)## and the momentum function ##p = m f(\theta) = m c \sinh(\theta)##. The derivation emphasizes the importance of conservation laws and the transition from classical to relativistic physics, highlighting the differences in how energy and momentum are treated. The discussion also touches on the implications of the Lagrangian formalism in describing massive and massless particles.
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ergospherical said:I mean there is a radius vector ##\mathbf{X}## whose derivative ##\dot{\mathbf{X}} = \dfrac{\mathbf{P}}{E}## characterises the motion of the system as a whole, in that ##E = \gamma(\dot{X}^2)M## as @PAllen showed me yesterday.
Rindler: Relativity - Special General Cosmological - Exercise 6.5 said:By considering two equal particles traveling in opposite directions along parallel lines, show that the CM (center of mass) of a system in one IF does not necessarily coincide with its CM in another IF. Prove that, nevertheless, if the particles of the system suffer collision forces only, the CM in ervery IF moves with the velocity of the ZM frame.
Of course, to treat a complete closed system you have to take particles and fields as dynamical quantities (with the usual unsolved problems of interacting point particles) but also then the energy is not proportional to the particle mass.PAllen said:Is that really an exception? You must include the 4 momentum carried by fields in total 4 momentum of a system.
Yes, in this way you define the total mass of a closed system, but that's not the mass of the particle. It's mass defined by ##P_{\mu} P^{\mu}=M^2 c^2##, where ##P^{\mu}## is the (conserved!) total four-vector of the system.PAllen said:I should clarify that I was thinking of COM in the sense of center of momentum. Specifically, given a total 4 momentum of an arbitrary isolated system, it is trivially decomposed into its mass times a 4 velocity. The total energy is then mass times gamma, the time component of the 4 velocity. A boost by the corresponding velocity takes you to a frame where gamma is zero. We can talk about the velocity of this COM frame (in the original frame) without needing to discuss any notion of center of energy computed by analogy to center of mass (using radius vectors).
But note that the Rindler exercise only asks you to prove a fact for collisions only; it does not state the result must be false for more general systems.
Never, ever, did I say anything about particle mass. I explicitly said that’s not what I was referring to, from the very first post on this.vanhees71 said:Yes, in this way you define the total mass of a closed system, but that's not the mass of the particle. It's mass defined by ##P_{\mu} P^{\mu}=M^2 c^2##, where ##P^{\mu}## is the (conserved!) total four-vector of the system.