Tau leptonic decay - Lifetimes and modes

[unscientific]

Homework Statement

[/B]
(a) Explain lepton universality.
(b) Explain why decay mode is forbidden and find hadronic branching ratios.
(c) Find the lifetime of tau lepton.
(d) What tau decay mode would be suitable?
(e) Find the precision.
(f) How do you improve the results?
(g) Why is it much harder to measure lifetime of muon?

Homework Equations

The Attempt at a Solution

Part(a)[/B]
For weak interactions, the coupling of leptons to gauge bosons are the same for all leptons. Thus different decay modes have the same vertex factor.

Part(b)
It is forbidden because for weak charged interactions, a tau neutrino must be produced? By universality, branching ratio to electronic decay is 18%. Thus branching ratio to hadronic decay is 64%.

Part(c)
Given that $\Gamma \propto G_F^2 m_\tau ^5$, let us consider
$$\frac{\Gamma_{\left( \tau^- \rightarrow e^- + \bar {\nu_e} + \nu_\tau \right)}}{\Gamma_{\left( \mu^- \rightarrow e^- \bar{\nu_e} + \nu_\mu \right)}} = \left(\frac{m_\tau}{m_\mu}\right)^5$$
$$\frac{t_\mu}{t_\tau} = \left(\frac{m_\tau}{m_\mu}\right)^5$$
$$t_\tau \approx 2 \times 10^{-12} s$$

For lepton decay of tau lepton, only $18%$ decays to electronic mode.
For lepton decay of muon, it only electronic mode is possible.

Factoring in branching ratio, $t_\tau \approx 3 \times 10^{-13} s$.

I have a feeling that this is a roundabout way to calculate the tau lepton's lifetime. Is there a more straightforward way to calculate this?

Part(d)
To achieve better tracks in a cloud chamber, the muonic decay is preferred over the electronic decay, since mass of the electron is much smaller than that of a muon. Is this right?

Part(e)
Not sure how to find the uncertainty, as we are not given the speed. Can we assume a speed of like $\approx 100m s^{-1}$? Then the uncertainty would simply be $10^{-8}s$.

I'm not sure on the last 2 parts..

 

[mfb]

To achieve better tracks in a cloud chamber, the muonic decay is preferred over the electronic decay, since mass of the electron is much smaller than that of a muon. Is this right?

How is the different mass an argument?
Both decays have a serious issue: how do you find the point where the tau decayed?

You can calculate the speed of the tau - the Z is produced at rest in the detector frame.
Where does 100m/s come from? In general, all particles in high-energy physics move at speeds that are significant fractions of the speed of light.

Cloud chambers mainly have historic value - there are much better detectors available now.

 

[unscientific]

How is the different mass an argument?
Both decays have a serious issue: how do you find the point where the tau decayed?

You can calculate the speed of the tau - the Z is produced at rest in the detector frame.
Where does 100m/s come from? In general, all particles in high-energy physics move at speeds that are significant fractions of the speed of light.

Cloud chambers mainly have historic value - there are much better detectors available now.

Part(d)
Since the decay position vertex is reconstructed from trajectories of charged particles, an electron is stable compared to a muon (lifetime of $2\mu s$) so we should use electronic decay?

Part(e)
Ok, then how can I find the uncertainty without knowing the speed. Can I assume that the speed is roughly $c$? Then uncertainty would be $\frac{10 \times 10^{-6}}{c} \approx 10^{-14}$.

Part (f)
I looked it up - there's something called hadronic calorimeters that can detect such processes?

Part (g)
I suppose the pp process undergoes strong interaction while the electronic process undergoes weak interaction? And for strong interactions to be detectable, deep scattering is required which makes it significantly harder.

 

[mfb]

Part(d)
Since the decay position vertex is reconstructed from trajectories of charged particles, an electron is stable compared to a muon (lifetime of $2\mu s$) so we should use electronic decay?

Muons are stable enough to neglect their decays, but how do you reconstruct a vertex with a single charged particle? How do you find out where the particle was produced?

Part(e)
Ok, then how can I find the uncertainty without knowing the speed. Can I assume that the speed is roughly $c$? Then uncertainty would be $\frac{10 \times 10^{-6}}{c} \approx 10^{-14}$.

The speed is roughly c but you have to consider time dilation. You'll need the energy (or velocity, or momentum) of the tau. Think of the Z decay and energy/momentum conservation.

Part (f)
I looked it up - there's something called hadronic calorimeters that can detect such processes?

Calorimeters don't do tracking. There are other detector types for this purpose.

Part (g)
I suppose the pp process undergoes strong interaction while the electronic process undergoes weak interaction? And for strong interactions to be detectable, deep scattering is required which makes it significantly harder.

The strong interaction cannot produce taus, but the particles created from the strong interaction are important.

 

[unscientific]

Muons are stable enough to neglect their decays, but how do you reconstruct a vertex with a single charged particle? How do you find out where the particle was produced?

I suppose there is some kind of experimental signature involved? Do we use muons or electrons then?

The speed is roughly c but you have to consider time dilation. You'll need the energy (or velocity, or momentum) of the tau. Think of the Z decay and energy/momentum conservation.

Since energy and momentum must be conserved at every vertex, the energy is spread to the muon/electron, its neutrino and radiation? To include the effect of time dilation, i multiply by $\gamma$?

Calorimeters don't do tracking. There are other detector types for this purpose.

The strong interaction cannot produce taus, but the particles created from the strong interaction are important.

If both the pp and e+e- are weak interactions, why is it harder to measure anyone over the other?

 

[mfb]

I suppose there is some kind of experimental signature involved? Do we use muons or electrons then?

For lifetime measurements: no, not at all. You use decays with at least two charged particles (which means at least three due to charge conservation). That allows to calculate a vertex.

Since energy and momentum must be conserved at every vertex, the energy is spread to the muon/electron, its neutrino and radiation?

I don't understand that question. The energy of the tau can be computed based on the Z decay, this has nothing to do with the tau decay.

To include the effect of time dilation, i multiply by $\gamma$?

You probably want to divide by that.

If both the pp and e+e- are weak interactions, why is it harder to measure anyone over the other?

That question does not make sense at all.

 

[unscientific]

For lifetime measurements: no, not at all. You use decays with at least two charged particles (which means at least three due to charge conservation). That allows to calculate a vertex.
I don't understand that question. The energy of the tau can be computed based on the Z decay, this has nothing to do with the tau decay.
You probably want to divide by that.
That question does not make sense at all.

Part (d)
I made some progress. The electronic mode is helicity suppressed due to the electron's mass. So the $\tau \rightarrow \nu_\tau + \mu + \nu_\mu$ is preferred.

Part (e)
Still not sure on this bit to find the uncertainty. Can I use the uncertainty priniple $\Delta E \Delta t \sim \hbar$ to give $E \sim \Delta E \sim 3 \times 10^{-22}$ somehow?

Final part (f)
For production of tau muon, this process requires pulling quark-antiquark pairs out of the vacuum during the pp collision process in the LHC. This involves dissolving the inner proton structure by pulling the quarks apart to increase the exchange gluon energy. Given that many other products are produced alongside, detecting the specific tau muon makes it significantly harder.

 

[mfb]

Part (d)
I made some progress. The electronic mode is helicity suppressed due to the electron's mass. So the $\tau \rightarrow \nu_\tau + \mu + \nu_\mu$ is preferred.

They have nearly the same branching fraction, but both are unsuitable. You cannot measure the point where the tau decayed with just a single charged track. You need at least two charged tracks to reconstruct a vertex - and due to charge conservation this means you need at least three charged particles.

Part (e)
Still not sure on this bit to find the uncertainty. Can I use the uncertainty priniple $\Delta E \Delta t \sim \hbar$ to give $E \sim \Delta E \sim 3 \times 10^{-22}$ somehow?

No, the uncertainty comes from the resolution of the flight distance, a purely classical effect.

If a Z at rest decays to two tau, what is the energy of each tau?
If a tau lives for one lifetime, how far does it travel before it decays? You'll need special relativity.
What is the relative uncertainty on this value if your measurement has 10µm uncertainty?

Final part (f)
For production of tau muon, this process requires pulling quark-antiquark pairs out of the vacuum during the pp collision process in the LHC. This involves dissolving the inner proton structure by pulling the quarks apart to increase the exchange gluon energy. Given that many other products are produced alongside, detecting the specific tau muon makes it significantly harder.

What is a tau muon?
The other collision products are an issue, yes. The unknown tau energy is another one (you have to measure it, with its own uncertainty).

 

[unscientific]

For lifetime measurements: no, not at all. You use decays with at least two charged particles (which means at least three due to charge conservation). That allows to calculate a vertex.

That's right, I was confusing pion decay with tau decay. I looked up wikipedia for tau decays, I think $\tau \rightarrow \nu_\tau + \pi^{+} + \pi^{-} + \pi^{-}$ is possible?

No, the uncertainty comes from the resolution of the flight distance, a purely classical effect.

If a Z at rest decays to two tau, what is the energy of each tau?
If a tau lives for one lifetime, how far does it travel before it decays? You'll need special relativity.
What is the relative uncertainty on this value if your measurement has 10µm uncertainty?

The energy of the tau would be half of Z's mass, $45 594 MeV/c^2$ which gives $\gamma = 25.7$. Due to length contraction, $l = \frac{10^{-5}}{\gamma} = 3.9 \times 10^{-7} m$. Tau is moving close to speed of light, so $\Delta t_\tau = \frac{l}{c} = 1.3 \times 10^{-15} s$, which is about 1% of the lifetime of tau.

What is a tau muon?
The other collision products are an issue, yes. The unknown tau energy is another one (you have to measure it, with its own uncertainty).

I meant tau sorry.

I think an entire summary of the $e+e-$ collider is

 

[mfb]

That's right, I was confusing pion decay with tau decay. I looked up wikipedia for tau decays, I think $\tau \rightarrow \nu_\tau + \pi^{+} + \pi^{-} + \pi^{-}$ is possible?

Yes.

The energy of the tau would be half of Z's mass, $45 594 MeV/c^2$ which gives $\gamma = 25.7$.

Right.

Due to length contraction, $l = \frac{10^{-5}}{\gamma} = 3.9 \times 10^{-7} m$.

That does not work. For the tau the detector is length contracted, for us the tau lifetime is dilated (longer). Not the other way round.

I think an entire summary of the $e+e-$ collider is
[ATTACH=full]178330[/ATTACH]

That is a possible process, right.

 

[unscientific]

That does not work. For the tau the detector is length contracted, for us the tau lifetime is dilated (longer). Not the other way round.

Ok so just working on timescales, we factor in time dilation once.
$$\delta t = \gamma \frac{\delta l}{c} = 9 \times 10^{-13} s$$
This is three times the lifetime!

 

[mfb]

Somehow you always put the gamma factor in the wrong place.

 

[unscientific]

Somehow you always put the gamma factor in the wrong place.

Ok. So the error in length is $\delta l = 10^{-5}m$. Speed of tau is $c$. Time in our frame is $\delta t = \frac{\delta l}{c}$. But we want to find the time in the tau's rest frame, so $\delta t_0 = \frac{1}{\gamma} \delta t = \frac{l}{\gamma c} = 1.3 \times 10^{-15}$. That's what I got before..

 

[mfb]

But now you got it with the right argument.