Upsilon Meson Decays and B Meson Decays: Feynman Diagrams and Distance Traveled

[unscientific]

Homework Statement

(a)Draw feynman diagrams of upsilon meson. Why is decay to $q\bar q$ states suppressed? Explain why width for 4s is much wider.
(b) How do B mesons decay? Why is no other type of interaction possible? Draw feynman diagrams.
(c) Find the distance traveled by 1s and 4s.
(d) Why is asymmetry useful here? Find $\beta$. Find $E_+, E_-, \beta$.

Homework Equations

The Attempt at a Solution

Part(a)[/B]
Feynman diagrams are given by

Decay to other $q \bar q$ is cabbibo suppressed by $\sin \theta_c$.
I'm not sure why the width at 4s is particularly wide.

Part(b)
Weak interactions. Only weak interactions are possible as the flavour is not conserved. Feynman diagrams are shown below:

Part(c)
How can I work out $c\tau$ if I don't even have their lifetimes?

 

[unscientific]

bumpp

 

[mfb]

Huh? Where did my post go.

Decay to other $q \bar q$ is cabbibo suppressed by $\sin \theta_c$.

Why should it?
There is no decay to $q \bar q$ present in your diagrams.

I'm not sure why the width at 4s is particularly wide.

Same comment as usual: check the particle masses to find possible decays. Which interactions are responsible for them?

Weak interactions. Only weak interactions are possible as the flavour is not conserved. Feynman diagrams are shown below:

Right (but the W sign is odd, one B+ should be B- and the D- label are missing).

How can I work out $c\tau$ if I don't even have their lifetimes?

You have their decay widths, that is nearly the same.

 

[unscientific]

Why should it?
There is no decay to $qq¯q \bar q$ present in your diagrams.

True. Sorry I was thinking of inter-generation mixing in weak interactions. I think it is suppressed because strong interactions are preferred over weak $(Z^0)$ and EM $(\gamma)$ interactions. If that's the case, then wouldn't EM/weak production of lepton/anti-lepton pairs be suppressed as well?

Same comment as usual: check the particle masses to find possible decays. Which interactions are responsible for them?

I think something happened between $3s$ and $4s$ state, something extra must have been produced between $10.355GeV$ and $10.580 GeV$. I think it's the production of $B^{+}B^{-}$ pair of bosons.

Right (but the W sign is odd, one B+ should be B- and the D- label are missing).

That's right. Thanks for pointing that out.

You have their decay widths, that is nearly the same.

Since $\Delta E \sim \Gamma$ and $\Delta E \Delta \tau \sim \hbar$,their lifetimes are $\tau \sim \frac{\hbar}{\Gamma}$. The values are $\tau_{1s} = 1.3 \times 10^{-20} s$ and $\tau_{4s} = 2.7 \times 10^{-23}s$. The distance traveled before decaying is given by $c\tau_{1s} = 3.8 \times 10^{-12}m$ and $c\tau_{4s} = 8.1 \times 10^{-15} m$. The hierarchy is B mesons travel furthest, followed by 1s and then 4s. Not sure why this is the case.

 

[mfb]

True. Sorry I was thinking of inter-generation mixing in weak interactions. I think it is suppressed because strong interactions are preferred over weak $(Z^0)$ and EM $(\gamma)$ interactions. If that's the case, then wouldn't EM/weak production of lepton/anti-lepton pairs be suppressed as well?

What is suppressed relative to what now?

I think something happened between $3s$ and $4s$ state, something extra must have been produced between $10.355GeV$ and $10.580 GeV$. I think it's the production of $B^{+}B^{-}$ pair of bosons.

And $B^0 \bar B^0$, right.

Since $\Delta E \sim \Gamma$ and $\Delta E \Delta \tau \sim \hbar$,their lifetimes are $\tau \sim \frac{\hbar}{\Gamma}$. The values are $\tau_{1s} = 1.3 \times 10^{-20} s$ and $\tau_{4s} = 2.7 \times 10^{-23}s$. The distance traveled before decaying is given by $c\tau_{1s} = 3.8 \times 10^{-12}m$ and $c\tau_{4s} = 8.1 \times 10^{-15} m$. The hierarchy is B mesons travel furthest, followed by 1s and then 4s. Not sure why this is the case.

Think of the available interactions for decays.

 

[unscientific]

What is suppressed relative to what now?

And $B^0 \bar B^0$, right.

Suppressed relative to strong interactions.

Think of the available interactions for decays.

1s and 4s decays via strong interaction while the B mesons that only decay via weak interactions (as flavour is not conserved) and that corresponds to longer lifetimes. Stronger the interaction, the shorter the lifetime.

 

[mfb]

How does the 1s decay via strong interaction look like? That is exactly the qqbar decay that is missing here so far. And it is suppressed in the same way the charmonium you had a while ago rarely decays via the strong interaction.

while the B mesons that only decay via weak interactions (as flavour is not conserved) and that corresponds to longer lifetimes

Right.

 

[unscientific]

How does the 1s decay via strong interaction look like? That is exactly the qqbar decay that is missing here so far. And it is suppressed in the same way the charmonium you had a while ago rarely decays via the strong interaction.

Right.

OZI suppression. That explains a factor of 1000 out, which is typically the case for OZI suppression.

 

[unscientific]

Right.

Part (d)
Beam power asymmetry is useful as it avoids the centre of mass frame so new particles that are produced will move relative to the lab frame. This allows us to work out the production point of the decay.

Energy of 4s particle is given by $E = E_{-}(1+\delta)$. It's gamma factor is given by $\gamma = \frac{E_{-}(1+\delta)}{M_{4s}}$. It's speed in beta is given by $\beta = \left(1-\frac{1}{\gamma^2} \right)^{\frac{1}{2}}$. Distance it travels in lab frame is given by $L = \left(c\beta\right) \left( \gamma \tau \right)$ as we observe proper time to be dilated by a factor of $\gamma$.

Given that $E=10.58$ and $\delta=0.1$, we find that $E_{-} = 9.62~GeV$, $E_{+}=0.96~GeV$, $\gamma = 1.1$, $\beta = 0.42$ and $L = 3.74 \times 10^{15} ~m$.

 

[mfb]

Do you think 3.74*10¹⁵ m can be a reasonable answer?

Also, your value for E looks wrong.

 

[unscientific]

Do you think 3.74*10¹⁵ m can be a reasonable answer?

Also, your value for E looks wrong.

Sorry I meant $3.74 \times 10^{-15}m$.

 

[mfb]

For a B+, this is way too short.

 

[unscientific]

For a B+, this is way too short.

I'm not sure what went wrong. Consider two beams smashing together, with one moving at energy equal to mass of particle to be produced, the another moving towards it at 10% of that energy. Total energy is $1.1 \times M_{4s}$. The gamma factor of particle produced is correspondingly $\gamma =1.1$. That gives a speed of $0.46c$. Distance traveled is dilated time multiplied by speed, $v (\gamma \tau)$. Using the lifetime of 4s as $3 \times 10^{-23}~s$, I get distance traveled as $3 \times 10^{-15}~m$.

 

[mfb]

Consider two beams smashing together, with one moving at energy equal to mass of particle to be produced, the another moving towards it at 10% of that energy. Total energy is 1.1×M4s1.1 \times M_{4s}. The gamma factor of particle produced is correspondingly γ=1.1\gamma =1.1.

No, as the particle does not get produced at all. The center-of-mass energy is not sufficient. That is the wrong energy I mentioned.

For the flight distance of a B+ you should use the B+ lifetime, not the Y(4s) lifetime.

 

[unscientific]

No, as the particle does not get produced at all. The center-of-mass energy is not sufficient. That is the wrong energy I mentioned.

For the flight distance of a B+ you should use the B+ lifetime, not the Y(4s) lifetime.

How is the total energy insufficient? The energy is $1.1 \times M_{4s}$?

 

[mfb]

In the lab frame, yes, but producing an Y(4s) at gamma=1.1 would violate momentum conservation.

 

[unscientific]

In the lab frame, yes, but producing an Y(4s) at gamma=1.1 would violate momentum conservation.

So in the CM frame the total energy is $M_{4s}$ while in the lab frame one has $1.1 \times E_{-}$ energy. I would use invariance of the length of 4-vector to find the energies?

 

[mfb]

Correct.

 

[unscientific]

Correct.

$$M^2 = (1+\delta)E_{-}^2 - (p_1-p_2)^2$$
$$M^2 = (1+\delta)E_{-}^2 - p_1^2 - p_2^2 + 2p_1p_2$$
$$M^2 = \delta E^2(1-\delta) + 2m^2 + \sqrt{\left[ E_{-}^2-m^2\right]\left[ (\delta E_{-})^2-m^2 \right]}$$

Still can't solve for $E_{-}$..

 

[mfb]

E is the only unknown quantity in that equation.
You can neglect the electron mass - 0.0005 GeV are irrelevant. That simplifies the equations a lot.

 

[arivero]

Did you got the solution?