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Relativity and pi-meson decay

  • Thread starter Vuldoraq
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  • #1
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[SOLVED] Relativity and pi-meson decay

Homework Statement



A [tex]\pi[/tex] meson of mass [tex]m_{\pi}[/tex] at rest, decays into a muon of mass [tex]m_{\mu}[/tex] and a neutrino of zero rest mass, [tex]m_{v}=0[/tex]. Find the momentum of the muon.

Homework Equations



I think this is the right equation though I'm not sure if I should use four vectors.

[tex]E^{2}=(pc)^{2}+(mc^{2})^2[/tex]

The Attempt at a Solution



Total energy before the decay is the same as the total energy after the decay, so,

[tex]E_{\pi}=m_{\pi}c^{2}[/tex]

[tex]E_{\mu}=P_{\mu}c+m_{\mu}c^{2}[/tex]

[tex]E_{v}=P_{v}c[/tex]

Giving a total energy of,

[tex]E_{\pi}=E_{\mu}+E_{v}[/tex]

[tex]m_{\pi}c^{2}=P_{v}c+P_{\mu}c+m_{\mu}c^{2}[/tex]

So the momentum is,

[tex]P_{\mu}=m_{\pi}c-P_{v}-m_{\mu}c[/tex]

I have a horrid feeling this is all wrong as it seems too easy. Please could someone let me know if I'm wrong in my reasoning?
 
Last edited:

Answers and Replies

  • #2
alphysicist
Homework Helper
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Hi Vuldoraq,

Homework Statement



A [tex]\pi[/tex] meson of mass [tex]m_{\pi}[/tex] at rest, decays into a muon of mass [tex]m_{\mu}[/tex] and a neutrino of zero rest mass, [tex]m_{v}=0[/tex]. Find the momentum of the muon.

Homework Equations



I think this is the right equation though I'm not sure if I should use four vectors.

[tex]E^{2}=(pc)^{2}+(mc^{2})^2[/tex]

The Attempt at a Solution



Total energy before the decay is the same as the total energy after the decay, so,

[tex]E_{\pi}=m_{\pi}c^{2}[/tex]

[tex]E_{\mu}=P_{\mu}c+m_{\mu}c^{2}[/tex]
I think there is a math error in this last line; it doesn't seem to follow from the equation you have listed in the "Relevant equations" section. (Taking the square root doesn't just cancel the squares on the right hand side.)

Also, I think you should also use conservation of momentum in this problem.
 
  • #3
272
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Hey Alphysicist, thanks for the pointers

Correcting my error would give the muon energy to be,

[tex]E_{\mu}=\sqrt{(P_{\mu}c)^{2}+(m_{\mu}c^{2})^{2}}[/tex]

Which means the total energy is,

[tex]m_{\pi}c^{2}=\sqrt{(P_{\mu}c)^{2}+(m_{\mu}c^{2})^{2}}+P_{v}c[/tex]

Giving, after moving stuff around,

[tex]P_{\mu}=\sqrt{c^{2}(m_{\pi}^{2}-m_{\mu}^{2})+P_{v}^{2}+2m_{\pi}P_{v}c}[/tex]

Does this look better?

How would I apply cons of momentum in this problem?
 
  • #4
alphysicist
Homework Helper
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About conservation of momentum: looking at the right hand side of your final expression, you know the masses of the pion and the muon, but you do not yet know the momentum of the neutrino, so you cannot evaluate the muons momentum which is what they ask for.

So go ahead and write down the equation for momentum conservation. What does the fact that momentum is conserved indicate about the momenta of the muon and neutrino?
 
  • #5
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Does it mean they have equal and opposite momenta? So the equations then become,

[tex]P_{v}=-P_{\mu}[/tex]

Which makes the right hand side of my equation become,

[tex]P_{\mu}=\sqrt{c^{2}(m_{\pi}-m_{\mu})^{2}-P_{\mu}^{2}-2m_{\pi}P{\mu}c}[/tex]

Which after rearranging and applying the quadratic formula becomes,

[tex]P_{\mu}=(1/2)*(m_{\pi}c \ +/- \ c\sqrt{2m_{\mu}^{2}-m_{\pi}^2})[/tex]

Which seems to be a negative momentum? Which value gives the actual momentum? Assuming I've gone right so far.
 
Last edited:
  • #6
alphysicist
Homework Helper
2,238
1
They should have equal and opposite momenta. However, in terms of the energy equations you have written, I think you should substitute:

[tex]
p_{\nu} = p_{\mu}
[/tex]

because I believe we want the magnitudes here. (For example, you found for the neutrino that its energy is E = p c because it's massless. Even if the momentum is negative, the energy is positive so we just use the magnitude of p.)
 
  • #7
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I see what you mean, this would then give a momentum of,

[tex]P_{\mu}^2=c^{2}(m_{\pi}^2-m_{\mu}^2)+P_{\mu}^2+2m_{\pi}cP_{\mu}[/tex]

Which reduces to,

[tex]P_{\mu}=-\frac{c(m_{\pi}^2-m_{\mu}^2)}{2m_{\pi}}[/tex]

I don't understand why it's negative though?

Thanks again for your help! :smile:
 
  • #8
alphysicist
Homework Helper
2,238
1
Vuldoraq,

I'm having some trouble following the work because of the changes made along the way. However, if we look back at post #3:

Hey Alphysicist, thanks for the pointers

Correcting my error would give the muon energy to be,

[tex]E_{\mu}=\sqrt{(P_{\mu}c)^{2}+(m_{\mu}c^{2})^{2}}[/tex]

Which means the total energy is,

[tex]m_{\pi}c^{2}=\sqrt{(P_{\mu}c)^{2}+(m_{\mu}c^{2})^{2}}+P_{v}c[/tex]
This looks okay to me.

Giving, after moving stuff around,

[tex]P_{\mu}=\sqrt{c^{2}(m_{\pi}^{2}-m_{\mu}^{2})+P_{v}^{2}+2m_{\pi}P_{v}c}[/tex]
I think there is an error here. It should have a minus sign and should be:

[tex]P_{\mu}=\sqrt{c^{2}(m_{\pi}^{2}-m_{\mu}^{2})+P_{v}^{2}-2m_{\pi}P_{v}c}[/tex]

Now with this, plug in the momentum relation ([itex]p_{\mu}=p_{v}[/itex]) and solve for the momentum. I think it gives the same answer but without the minus sign. Do you get the answer?
 
  • #9
272
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Hey Alphysicist,

I plugged in the corrected value and my answer is the same, and no longer negative,

[tex]P_{\mu}=\frac{c(m_{\pi}^2-m_{\mu}^2)}{2m_{\pi}}[/tex]

Thanks for your help, I would never have got there without it.

Vuldoraq
 

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