Stuck with Kaon Decay: Finding Ek of Pions

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Homework Help Overview

The discussion revolves around the decay of a neutral kaon at rest into two pions, specifically focusing on determining the kinetic energies of the resulting pions given their rest masses and the kaon's rest mass. Participants express uncertainty about how to approach the problem without knowing the velocities of the pions.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants suggest using conservation of energy and momentum to analyze the decay process. There are discussions about the relationships between momentum and kinetic energy in relativistic dynamics, as well as the need to write down conservation equations.

Discussion Status

The conversation is ongoing, with participants exploring various equations and concepts related to relativistic energy and momentum. Some guidance has been provided regarding the conservation equations, and there is recognition of the need to consider both energy and momentum in the analysis.

Contextual Notes

Participants note the initial conditions of the kaon being at rest, which implies that the initial momentum is zero. There is also discussion about the implications of the pions having equal masses and how that affects their kinetic energies and gamma factors.

fredrick08
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Homework Statement


a neutral kaon at rest decays to pion- and pion+ , if kaon rest mass is 497.7MeV/c^2, and the pions rest mass is 139.6MeV/c^2, what are the kinetic energies of the resulting pions?

Im very stuck with this, and not sure where to start, how can i find Ek of pions, if i don't know their velocities? I've searched thorugh my textbook but can find any similar examples, please someone help.
 
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Use conservation of energy and momentum.
 
and conservation of momentum.

What are the relation between momentum and kinetic energy in relativistic dynamics?
 
i don't need to use invarient mass E^2=(pc)^2+(mc^2)^2?
 
Just write down the conservation equations in terms of [itex]E_0=mc^2, \; E=\gamma mc^2, \; T=(\gamma-1)mc^2[/itex].
 
maybe, there are several ways to do this, i would just to what Cyosis suggested.
 
so Ef=E1+E2=497.7MeV=.5mv^2+.5mv^2? i can't do anything from there...
 
You will have to use special relativity here. The total energy on one side equals the total energy on the other side. Total energy includes rest mass.

Edit: While your equation is wrong I don't see how you can say that you can't do anything from there. You already assumed that both speeds are equal (something you should show) so just solving for 1/2 m v^2 would give you the answer, albeit the wrong answer, because you're working classically.
 
so 497.7MeV=mv^2+2mc^2? I am sorry i completely don't understand... could you tell me the formula I am supposed to use, I am unfamiliar with this relativistic part
 
  • #10
Where does this mv^2 come from. Relativistic kinetic energy is not 1/2 mv^2. In a collision we have three types of energies. Total energy, E, relativistic kinetic energy, T and rest energy E_0. Therefore we know that [itex]E=T+E_0[/itex]. So what is the relativistic expression for kinetic energy?
 
Last edited:
  • #11
so Ek=T-Eo? my book says Ek=mc^2(gamma-1)
 
  • #12
I defined T as the kinetic energy so no. If [itex]E=T+E_0[/itex] then [itex]T=E-E_0[/itex]. You can call T E_k if you prefer that it does not matter as long as you don't mix them up.
 
  • #13
ok then T=gamma*mc^2-mc^2? but how do i find the gamma value, sine i don't know u?
 
  • #14
You are asked to find the kinetic energy not the velocity.

Can you please just write down the momentum conservation equation first, then the energy conservation equation?
 
  • #15
Pf=Pi? E/c=E/c+E/c and Ef=Ei
 
  • #16
A little bit more specific to this problem perhaps? What is the value of p-initial?
 
  • #17
p initial = 0, since the particle is at rest, therefore p final also has to equal 0?
 
  • #18
so are you saying that for the system total energy is m(k)c^2=2m(pi)c^2+2*gamma*m*u^2?
 
  • #19
You have way more information than that. You need to make use of it. I will list you all the variables you need to put in your equations.

[itex] E_{kaon},T_{\pi^-},T_{\pi^+},E_{0,\pi^-},E_{0,\pi^+},p_{kaon},p_{\pi^-},p_{\pi^+}[/itex]

The first thing you do when solving a problem is writing down all the variables that are relevant to the problem. Now use these variables to write down the conservation of momentum equation and the conservation of energy equation.
 
  • #20
Where does the u come from? And secondly how do you know that both pions have the same kinetic energy (it's true)?
 
  • #21
if this is true, don't i need the gamma value?? but i need the speed of the pions to calculate that? but if they have a speed then they have a momentum. which doesn't make sense, since initial momentum is 0?
 
  • #22
Momentum is a vector, both pions do have non-zero momentum. From this you can derive that the kinetic energy of both pions is the same. Write it down in equation form using the variables I supplied.
 
  • #23
ok so the whole thing is...

Ekaon=Eo(pi-)+T(pi-)+Eo(pi+)+T(pi+)

and momentum pkaon=p(pi+)+p(pi-)
 
  • #24
Yep that is correct. You know that pkaon=0 so what is the relation between p(pi+) and p(pi-)?
 
  • #25
they must be equal... and opposite?
 
  • #26
That is correct, since both pions have the same mass what does that mean for their gamma factor? What does that mean for their kinetic energy?
 
  • #27
their gamma values and kinetics energies will be equal?
 
  • #28
Yes and since they have the same mass what does this mean for their rest energies? Write down the equation for energy conservation again with the new information you have obtained.
 
  • #29
Ekaon=2Eo(pi)+2T(pi)
 
  • #30
That is correct, so what is the kinetic energy of a pion?
 

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