# Lets try solving this

#### panthera

hi,

I have a problem and I am trying it out and would like to have suggestions as well. I would post the solution after I finish and will have opinions .Meanwhile I would like others to try this....I shall be obliged to get any ans...

An unstable particle of rest energy 1000MeV decays into a meu-meson (Mo csqr=100MeV) and a neutrino with a mean life, when at rest, of 10^-8 sec.
a) calculate the mean decay distance when the particle has a momentum of 1000MeV/c.
b) What is the energy of the meu-meson if it is emitted at an angle 15 degrees.

ps- i guess its the spelling of meu as i can't write the symbol,correct if it's wrong...
and ^ stands for "raised to the power"
Thanks

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#### OlderDan

Homework Helper
panthera said:
hi,

I have a problem and I am trying it out and would like to have suggestions as well. I would post the solution after I finish and will have opinions .Meanwhile I would like others to try this....I shall be obliged to get any ans...

An unstable particle of rest energy 1000MeV decays into a meu-meson (Mo csqr=100MeV) and a neutrino with a mean life, when at rest, of 10^-8 sec.
a) calculate the mean decay distance when the particle has a momentum of 1000MeV/c.
b) What is the energy of the meu-meson if it is emitted at an angle 15 degrees.

ps- i guess its the spelling of meu as i can't write the symbol,correct if it's wrong...
and ^ stands for "raised to the power"
Thanks
I have an answer for a) as the product of the particle velocity times its time-dilated mean life. The velocity is found from the initial momentum and rest energy. Have you found this?

I'm working on b)

#### panthera

OlderDan said:
I have an answer for a) as the product of the particle velocity times its time-dilated mean life. The velocity is found from the initial momentum and rest energy. Have you found this?

I'm working on b)
thanks...yes i have found a) in the same way...and have got the ans 3m...

for 2nd one i am using four vectors......i always face difficulty in solving these problems using other methods...its almost solved but i hav not yet put the values to get the ans...

#### panthera

four vectors----

Px =( Px, jEx)

P μ =(Pμ ,jEμ)

P ν =( P ν , jE ν)

[I tried to put arrow head but couldn't, its getting displaced]

Now,

P ν = Px - P μ

Squaring we get

P ν 2 = (Px - P μ) 2

=>mx2 + mμ2 – 2Ex E μ + 2Px P μ cos θ = 0, θ = 15 degrees

by putting values and neglecting mμ2 we get

E μ =1115.36 MeV [here, E μ is equivalent to P μ ]

is this correct?

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#### OlderDan

Homework Helper
Deleted erroneous calculations and related comments

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#### panthera

OlderDan said:
I agree with your mean distance of 3m. The total energy I got is (my subscript p is your subscript x)

$$E_p = 1000 \sqrt{2}Mev$$

I think you have a sign error in your last equation, but it may only be the term you neglected. If you neglect the rest mass of the muon and look in the rest frame of the particle you would have equal and opposite momentum and equal energies for the decay products. It does not seem reasonable that the energies would be very different in the lab frame, but your result gives the muon nearly 80% of the energy.

You are correct to assume the muon rest mass is a small effect, but I see no need to neglect it. Once the hard algebra is completed, it is just as easy to include it. My final result looks like this

$$E_\mu = \frac{{2E_p p_p c\cos \theta + m_{p0} ^2 c^4 - m_{\mu 0} ^2 c^4 }}{{2\left( {E_p + p_p c\cos \theta } \right)}}$$

$$E_\mu = \frac{{2\left( {1000\sqrt 2 } \right)\left( {1000} \right)\cos 15^{\rm{o}} + 1000^2 - 100^2 }}{{2\left( {1000\sqrt 2 + 1000\cos 15^{\rm{o}} } \right)}}Mev = 781.9Mev$$

$$E_\nu = E_p - E_\mu = 1000\sqrt 2 Mev - 781.9Mev = 632.3Mev$$

If you neglect the muon rest mass, the muon energy is 784Mev. Don't take my word for it, but I think this is correct.
thanks for the solution ...well i am not finding any sign error in my equation; its by solving 4 vectors... only i have multiplied the original eqn with -1 to write it in simplified way[i have not written all the steps here]....your equations are correct but i think we have a sign mismatch.. ...i generally avoid the term mμ sqr in such problems as in case of high energy muons, its negligible....

thanks

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#### OlderDan

Homework Helper
panthera said:
thanks for the solution ...well i am not finding any sign error in my equation; its by solving 4 vectors... only i have multiplied the original eqn with -1 to write it in simplified way[i have not written all the steps here]....your equations are correct but i think we have a sign mismatch.. ...i generally avoid the term mμ sqr in such problems as in case of high energy muons, its negligible....

thanks
I did make an error, but it was worse than just a sign mistake. I now agree with your original equation, but my earlier result is way off. I have reproduced your result neglecting the rest mass of the muon. Without the approximation, it's a messy thing to be sure.

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#### panthera

thanks for your suggestions and cooperation..okay...so i can infer that my ans is correct

thanks & regards

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#### OlderDan

Homework Helper
panthera said:
thanks for your suggestions and cooperation..okay...so i can infer that my ans is correct

thanks & regards
Yep.. looks good to me now. I even managed to get through it without neglecting the muon rest mass. I thought it might be useful to have the general result in hand I got 1116.84Mev. Obviously not a big dfference for this problem.

#### panthera

by NOT neglecting, i got somewhat similar results as yours...

cheers

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