The Student Room Group

Relativistic Decay Problem

Could someone help me solve this problem, or at least give me some hints as to how to tackle it? (Before anyone asks, this is beyond A-Level).
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The K-meson can decay at rest into a pair of charged pi-mesons. Given that the rest masses of the K and pi are 498 and 140 MeV/c^2 respectively, show that the speeds of the pions are 0.83c in the rest frame of the K.
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Thanks :smile:
Reply 1
the energy of both pi mesons are equal to conserve momentum.

498/2 = 249 MeV each

so 249 MeV = gamma * 140 MeV/c^2 * c^2 = gamma 140

249/140 = 1/root(1-v^2/c^2) or (140/249)^2 = 1-v^2/c^2
therefore:

v^2 = 1-(140/249)^2 c^2

so v = root((1-140/249)^2) * c
= 0.827c
Reply 2
Thanks ever so much for your help. Just one question, what formula are you using in this line. It appears to be E = gamma x rest mass x c^2, but I can't recall ever seeing that (probably me as don't grasp relativity well yet!):

Willa
so 249 MeV = gamma * 140 MeV/c^2 * c^2 = gamma 140


Thanks :smile:
Reply 3
jus a guess E=m* csquared :confused:
Reply 4
Jack Schitt
jus a guess E=m* csquared :confused:


Nope there's a gamma in there. Just quickly browsed through next week's lecture notes and I've spotted a similar formula (so I haven't missed it, we just haven't done it yet!) but want to check I'm right!
Reply 5
Hoofbeat
Nope there's a gamma in there. Just quickly browsed through next week's lecture notes and I've spotted a similar formula (so I haven't missed it, we just haven't done it yet!) but want to check I'm right!

rofl, i was just jking, i am only doing A2s, so i know nothing!
Reply 6
Jack Schitt
rofl, i was just jking, i am only doing A2s, so i know nothing!


Lol! Thanks anyways for taking the time to read my post :tongue: :biggrin:
Reply 7
Gamma is the symbol representing 1/sqrt(1-v^2/c^2)
Reply 8
Hoofbeat
Thanks ever so much for your help. Just one question, what formula are you using in this line. It appears to be E = gamma x rest mass x c^2, but I can't recall ever seeing that (probably me as don't grasp relativity well yet!)

that's the relatavistic formula for kinetic energy

(you will see that, as v -> 0, gamma.mc² -> mv²/2)
Reply 9
(you will see that, as v -> 0, gamma.mc² -> mv²/2)

what the hell! that's wrong. You mean that taking the bionomial expansion, Then gamma.mc^2 - mc^2 = mv^2/2
Reply 10
Willa
what the hell! that's wrong. You mean that taking the bionomial expansion, Then gamma.mc^2 - mc^2 = mv^2/2

what the hell!!!!! yes

gamma = 1/root(1-v²/c²) = (1-v²/c²)^(-1/2)

expanding:

1 + (-1/2)(v²/c²) + (-1/2)(-3/2)/2 (v²/c²)² + ...

as v-> 0, gamma -> 1 - v²/2c²

=> E -> (1 - v²/2c²)mc² = mc² + mv²/2 = RE + KE_classical
Reply 11
elpaw
what the hell!!!!! yes

gamma = 1/root(1-v²/c²) = (1-v²/c²)^(-1/2)

expanding:

1 + (-1/2)(v²/c²) + (-1/2)(-3/2)/2 (v²/c²)² + ...

as v-> 0, gamma -> 1 - v²/2c²

=> E -> (1 - v²/2c²)mc² = mc² + mv²/2 = RE + KE_classical



You expanded the series wrongly.

gamma = 1/root(1-v²/c²) = (1-v²/c²)^(-1/2)

expanding:

1 - (-1/2)(v²/c²) + (-1/2)(-3/2)/2 (v²/c²)² + ...
Reply 12
1 + (-1/2)(v²/c²) + (-1/2)(-3/2)/2 (v²/c²)² + ...

as v-> 0, gamma -> 1 - v²/2c²


if you put in v=0 into that you get....1!!!!!!!

so you're saying that gamma-> 1

which means E-> mc^2 as v->0.....which is obvious!

just think what you're saying!!!!!!! you want to say "for small v, all further terms than the first two are negligible"...hence:

gamma = 1 + .5(v^2/c^2)

note the (CORRECT) plus sign

hence kinetic E = (1+0.5(v^2/c^2))mc^2 - mc^2 (from gamma.mc^2 - mc^2)
= 0.5mv^2 = KINETIC ENERGY

you even got the negative sign in your last line wrong!!!! think what you're doing man!

total E = mc^2 + 0.5mv^2