# Help with relativistic momentum question

1. Jan 5, 2013

### DunWorry

1. The problem statement, all variables and given/known data
a Pion travelling at 0.93c decays into a muon which travels in the same direction and a neutrino which travels in the opposite direction. Use conservation of energy and momentum to find the energy of the muon as determined in the rest frame of the original pion. You should assume the muon has a mass of 106MeV/c^2 and the pion has a mass of 104 MeV/C^2, the neutrino is massless.

2. Relevant equations

E$^{2}$ = C$^{2}$P$^{2}$ + M$^{2}$C$^{4}$

E = $\gamma$M$_{0}$C$^{2}$
P = $\gamma$mv

3. The attempt at a solution
Hmmmm well first I tried the momentum and said ($\gamma$mv)$_{pion}$ = ($\gamma$mv)$_{muon}$ - P$_{neutrino}$

Energy conservation Energy$_{muon}$ = Energy$_{pion}$ - Energy$_{neutrino}$.

The Energy$_{neutrino}$ = CP$_{neutrino}$ as it is massless
and Energy$_{pion}$ = $\sqrt{C^{2}P^{2} + M^{2}C^{4}}$ Since we know the Pions velocity we can work out the energy of the Pion, however I'm not sure how to work out the momentum of the neutrino.

I thought since it said in the rest frame of the Pion, well relative to itself then it is at rest initally, so the momentum = 0. That means in the frame of the Pion ($\gamma$mv)$_{muon}$ = P$_{neutrino}$ and I thought well we should just sub it, except we do not know the speed of the muon either so we can't work out its momentum.

The answer just gives it as E = $\frac{m_{pion}^{2}+m_{muon}^{2}}{2m_{pion}}$ = 110 MeV I'm not sure how they got this can someone help please?

Last edited by a moderator: Jan 7, 2013
2. Jan 5, 2013

### oli4

Hi DunWorry BiHappy.
The result is wrong, there is a missing c² factor.
Other than that, you can get to the same formula, I think you are on the right track.
Here are the steps.
1) don't worry about the initial values, they are here to confuse you, indeed, in the rest frame of the particle, the momentum is 0 and is conserved, so you know that the momentum of the neutrino and of the muon is the same (in magnitude, which is all you care about)
2) conservation of energy, has you said, the initial energy is just the mass of the pion times c². let's call it E0
so you have E0=E1+E2, and you want E2 (E1 for the energy of the neutrino, and E2 for the energy of the muon)
Threfore, as you already found, E2=E0-E1
3) square this value and use your other references equation, in just two lines, you should find a value for PC (P is the momentum of either the neutrino or the muon)
but E2=E0-E1 with E1=PC, so you just apply the found value of PC and you will get to the final answer, but with the c² factors which must be there unless you put C=1

Cheers

3. Jan 5, 2013

### DunWorry

Sorry what do you mean by 'other references' equation?

4. Jan 5, 2013

### oli4

Ah, sorry, I meant your list of other relevant equations.
You have everything you need there
Once you have E2=E0-E1, you square it and use your relevant equations to find a redundant PC² that you eliminate. that will give you PC in term of the masses of the involved particles.
You then come back to E2=E0-E1=E0-PC and substitute for what you found.
It really is not easy but maybe I am doing a poor job at explaining. if you give up tell me so and I will be much more explicit but I think you would be disappointed because you can certainly get it yourself easily which you would realise and it would be too late.

Cheers...

5. Jan 5, 2013

### DunWorry

Hmmm ok I'm not sure if this is what you mean but this is what I did.

E2 = E0 - E1, I guess you want to square so you can use E$^{2}$ = c$^{2}$p$^{2}$ + m$^{2}$c$^{4}$

So if I square this I get E2$^{2}$ = E0$^{2}$ + E1$^{2}$ - 2E1E0

Writing out E2$^{2}$ in full and subbing E1 = cp I get

C$^{2}$P$^{2}$ + m2$^{2}$C$^{4}$ = E0$^{2}$ - 2E0(CP) + C$^{2}$P$^{2}$

So the c^2p^2 cancel, and I rearrange for CP = $\frac{E0^{2} - m2^{2}c^{4}}{2E0}$

Then I sub this into E2 = E0 -cp

to get E2 = E0 - $\frac{E0^{2} - m2^{2}c^{4}}{2E0}$

Putting in the numbers I get an answer of

E2 = $\sqrt{8x10^{16}}$ - $\frac{8x10^{16} - (106x10^{6})^{2}}{2\sqrt{8x10^{16}}}$

I get 161 MeV, which is off the answer, also I can't see how you get the expression with the masses, also what is wrong with what I just done to give me wrong answer? =/

6. Jan 5, 2013

### oli4

Hi, I don't know about the numerical values, I didn't check them, but the given answer was wrong anyway since it was missing the c² factors.
What you did looks correct, in order to get to the equation you want, express CP not in terms of E0 but in terms of the value of E0 and that should be it, you just expand E0 in what it is and some simplification will happen leading to the final equation.
(you can also do that after instead of before with your own final equation, look at it closely and try to put this E0 outside of the fraction inside it and you are there already)
Cheers.

7. Jan 5, 2013

### TSny

The mass of a charged pion is about 139.6 Mev/c2.

8. Jan 5, 2013

### DunWorry

hmmm ok thanks for your help. To get the expression,
I got E1 = $\frac{E0^{2} - m2^{2}c^{4}}{2E0}$
Writing in full and pulling out some C from top and bottom

=$\frac{c^{2}(p^{2} + m0^{2}c^{2} - m2^{2}c^{2}}{2c \sqrt{p^{2}+m0^{2}c^{2}}}$

Now I'm a bit stuck because I got some p and some square root on the bottom. I noticed if I let P = 0 then I get

E1 = $\frac{c^{2} (m0^{2} - m2^{2})}{2m0}$

I got 3 questions:
1. is this what you mean its missing a factor of c^2?
2. in the answers it says m0^2 + m2^2 on top, instead of minus, is it incorrect?
3. Just thinking why did we let p=0? Since its in the particles own frame, I can see its momentum from its frame of reference is zero. However since it is travelling at 0.93c relative to the lab then in the lab frame it will have some momentum. I thought the equation E$^{2}$ = c$^{2}$p$^{2}$ + m$^{2}$c$^{4}$ was invariant and thus the same in all inertial frames, how can p be zero, and have some value at the same time?

Thanks

9. Jan 5, 2013

### oli4

Hi DunWorry,
remember that you are in the rest frame, E0 is just the mass of the pion times c², it really is that simple, P is completely gone, we got rid of it in the previous steps.
Cheers...

10. Jan 7, 2013

### vela

Staff Emeritus
It's pretty common to use units where c=1 and omit it from the equations.

The equation isn't invariant; the quantity $E^2-(pc)^2$ is invariant. That means if a particle has energy E and momentum p in one inertial frame and energy E' and momentum p' in another inertial frame, and you calculate $E^2-(pc)^2$ and $E'^2-(p'c)^2$, you'll they will be equal to each other. Specifically, you'll find they are both equal to $(mc^2)^2$.