How Does Particle Decay Relate to the Speed of Resulting Particles?

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Homework Help Overview

The discussion revolves around the decay of a particle at rest with mass M into n identical smaller particles, each with mass m. Participants are exploring the relationship between the speed of the resulting particles and the conservation of energy and momentum in the context of particle physics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants are attempting to relate the rest energy of the initial particle to the energies of the resulting particles, questioning the absence of kinetic energy in certain formulations. There are discussions about using relativistic equations and the implications of conservation laws.

Discussion Status

The conversation is ongoing, with some participants providing alternative formulations and questioning assumptions about energy and momentum conservation. There is no explicit consensus, but various interpretations and approaches are being explored.

Contextual Notes

Participants note that the problem becomes more complex with an increasing number of decay products, particularly when considering momentum in three dimensions. There is also a reminder that M must be greater than nm, which is a constraint in the problem setup.

fredrick08
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Homework Statement


A particle at rest with mass M, decays into n identical smaller particles with equal mass, m. Show that speed of the particles is given by

u=c*root(1-(((n^2)(m^2))/M^2))


The Attempt at a Solution


this one i don't really know where to start, M has a rest energy... m's have Et=Eo+Ek so,

Mc^2=(n*mc^2)+(n*(1/2)mv^2)?
 
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lol that doesn't even make sense...
 
no one has any ideaS?
 
is it Mc^2=ynmc^2=>y=M/nm=>1/root(1-B^2)=M/nm=>B=root(1-(nm)^2/M^2)=v/c=>v=c*root(1-(nm)^2/M^2) this works out but can someone tell me why there is no kinetic energy part in the equation?? does ymc^2= total energy, because i thought it was just relativistic rest mass?
 
Well M would be the rest mass of the initial particle, m is the rest mass of the smaller particles, and M > nm. The total energy is conserved, and so is momentum.

It should be straightforward for two particles (colinear) and three particles (coplanar). Four or more starts getting complicated because momentum is in three dimensions.

Try with 2 particles (products), then 3.

Since the products are identical (m), there is some symmetry which is the key.


Has one considered E2 = p2c2 + m2c4
 
Last edited:

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