Calculating Energy Loss of Muon Moving Through a Medium

[Andreas S-H]

Hello everyone. I have just complete an experiment calculating the speed of a muon. I got it to 2.6E8 m/s, however I know that they are created at close to speed of light to be able to get down to Earth's surface in their short lifespan. This speed could not have been its initial speed, as it with it's current speed could not get down to Earth from 15km. It most have lost some energy and therefore speed going though the air. I would love to calculate the amount of energy it has lost, is there a way to do that ?

In this experiment we had to show how naturally created muons in a height of 15km could get down to Earth and get detected when they only have a lifetime of 2.2μs

The data and how i calculated the speed is below.

We used two scintillator detectors.

We first calibrated the TDC module on our computer. Here we got the linear relationship to be

$y=0.45443ns*x + 13.471$

In the first measurement on of our detectors were raised $d=1.91m$ above the ground, with an uncertainty of $\sigma _d = 0.01m$. We detected 600 events. We got the mean to be $M_1 = 61.8704$ and the uncertainty of $\sigma _1 = 11.377$.

In the second measurement we put the two detectors on top of each other as to get as many events as possible. We got 901 events. We calculated the mean to be around $M_2=77.7572$ and uncertainty to be $\sigma _2 = 8.14527$.

With this information we calculated the average speed of the muons. $$v= \frac {d} {(M_2 - M_1)*0.45443E-9s} \approx 2.65E8 \frac {m}{s} $$

We then calculated the uncertainty of the speed. $$\sigma_{v}={\frac {1}{ 0.45443\,E-6\,s}\sqrt {{\frac {{\sigma_{d}}^{2}
}{ \left( M_{2}-M_{1} \right) ^{2}}}+{\frac {{d}^{2}}{ \left( M_{2}-M_
{1} \right) ^{4}} \left( {\frac {{\sigma_{1}}^{2}}{n_{1}}}+{\frac {{
\sigma_{2}}^{2}}{n_{2}}} \right) }}} \approx 9.06E6
$$

We calculated the Lorenz factor to be around 2.13

We then calculated the average total energy of the muons with $E=\gamma*m_0$

$$E_total=2.13*105MeV = 0.22GeV$$

We then calculated the time dilation and length contraction it would experience.
$$\delta t' = 2.13*2.2\mu s = 4.7 \mu s$$

$$L = 15km * \frac {1}{2.13} = 7.04km$$

So the total distance the muon should be able to travel in 4.7μs with a speed of $2.65E8 \frac {m}{s}$ should be $$d_muon = 2.65E8 \frac {m}{s} * 4.7E-6s = 1.25km $$

This is no where close to the required distance to get down to the earth. So we concluded that the muons had lost some energy and therefore speed on the way down. We theorized that it is because it passed through a medium, here the air. However we have no idea if we can calculate the energy lost.

Kindly Andreas

 

[Nugatory]

I have just complete an experiment calculating the speed of a muon.

You will get better and more helpful answers if you can tell us more?
Calculated from what data? How were these muons created and what did you measure to come up with a speed? And (most often overlooked) what are the uncertainties in your data?

 

[Andreas S-H]

You will get better and more helpful answers if you can tell us more?
Calculated from what data? How were these muons created and what did you measure to come up with a speed? And (most often overlooked) what are the uncertainties in your data?

Good point i have added the data

 

[Sagittarius A-Star]

$d=1.91m$
...
$M_1 = 61.8704$
...
$M_2=77.7572$
$$v= \frac {d} {(M_2 - M_1)*0.45443E-9s} \approx 2.65E6 \frac {m}{s} $$

My electronic calculator says, that the exponent is 8.

 

[Andreas S-H]

My electronic calculator says, that the exponent is 8.

oh my bad type error. However all the calculation were done in another document where it was 8. So everything should check out

 

[Sagittarius A-Star]

oh my bad type error. However all the calculation were done in another document where it was 8. So everything should check out

Yes. If you take into account the uncertainties for $M_1$ and $M_2$, what is the minimal absolute value for $M_2 - M_1$?

 

[Andreas S-H]

My electronic calculator says, that the exponent is 8.

Yes. If you take into account the uncertainties for $M_1$ and $M_2$, what is the minimal absolute value for $M_2 - M_1$?

Can you rephrase it? (im from Denmark and can't really make sense of the translation )

 

[Sagittarius A-Star]

Can you rephrase it? (im from Denmark and can't really make sense of the translation )

Ja. Hvis du tager hensyn til usikkerheden for $M_1$ og $M_2$, hvad er den minimale absolutte værdi for $M_2 - M_1$?

Source:
https://translate.google.com/?hl=de...bsolute value for $M_2 - M_1$?&op=translate

 

[Andreas S-H]

Ja. Hvis du tager hensyn til usikkerheden for $M_1$ og $M_2$, hvad er den minimale absolutte værdi for $M_2 - M_1$?

Source:
https://translate.google.com/?hl=de&sl=de&tl=da&text=Yes. If you take into account the uncertainties for $M_1$ and $M_2$, what is the minimal absolute value for $M_2 - M_1$?&op=translate

I don't think i have learned that yet on college.

Is it this ?

$\frac {\sigma _1}{\sqrt {n_1} = \pm 0.46447}$ and $\frac {\sigma _2}{\sqrt {n_2} = \pm 0.27137}$

So is the answer $77.48584-61.40583 = 16$ ?

 

[PeroK]

Hello everyone. I have just complete an experiment calculating the speed of a muon. I got it to 2.6E8 m/s, however I know that they are created at close to speed of light to be able to get down to Earth's surface in their short lifespan. This speed could not have been its initial speed, as it with it's current speed could not get down to Earth from 15km. It most have lost some energy and therefore speed going though the air. I would love to calculate the amount of energy it has lost, is there a way to do that ?

In this experiment we had to show how naturally created muons in a height of 15km could get down to Earth and get detected when they only have a lifetime of 2.2μs

The data and how i calculated the speed is below.

We used two scintillator detectors.

We first calibrated the TDC module on our computer. Here we got the linear relationship to be

$y=0.45443ns*x + 13.471$

In the first measurement on of our detectors were raised $d=1.91m$ above the ground, with an uncertainty of $\sigma _d = 0.01m$. We detected 600 events. We got the mean to be $M_1 = 61.8704$ and the uncertainty of $\sigma _1 = 11.377$.

In the second measurement we put the two detectors on top of each other as to get as many events as possible. We got 901 events. We calculated the mean to be around $M_2=77.7572$ and uncertainty to be $\sigma _2 = 8.14527$.

With this information we calculated the average speed of the muons. $$v= \frac {d} {(M_2 - M_1)*0.45443E-9s} \approx 2.65E8 \frac {m}{s} $$

We then calculated the uncertainty of the speed. $$\sigma_{v}={\frac {1}{ 0.45443\,E-6\,s}\sqrt {{\frac {{\sigma_{d}}^{2}
}{ \left( M_{2}-M_{1} \right) ^{2}}}+{\frac {{d}^{2}}{ \left( M_{2}-M_
{1} \right) ^{4}} \left( {\frac {{\sigma_{1}}^{2}}{n_{1}}}+{\frac {{
\sigma_{2}}^{2}}{n_{2}}} \right) }}} \approx 9.06E6
$$

We calculated the Lorenz factor to be around 2.13

We then calculated the average total energy of the muons with $E=\gamma*m_0$

$$E_total=2.13*105MeV = 0.22GeV$$

We then calculated the time dilation and length contraction it would experience.
$$\delta t' = 2.13*2.2\mu s = 4.7 \mu s$$

$$L = 15km * \frac {1}{2.13} = 7.04km$$

So the total distance the muon should be able to travel in 4.7μs with a speed of $2.65E8 \frac {m}{s}$ should be $$d_muon = 2.65E8 \frac {m}{s} * 4.7E-6s = 1.25km $$

This is no where close to the required distance to get down to the earth. So we concluded that the muons had lost some energy and therefore speed on the way down. We theorized that it is because it passed through a medium, here the air. However we have no idea if we can calculate the energy lost.

Kindly Andreas

I can't follow what you're doing. Are you using the muon decay rate and how many muons decay between the two detectors and calculate the muons' proper time from that and hence the speed?

 

[PeroK]

PS I don't understand the why you are using the uncertainty is the speed. You want an estimate of the actual speed, surely?

 

[Sagittarius A-Star]

I mean this:

We got the mean to be $M_1 = 61.8704$ and the uncertainty of $\sigma _1 = 11.377$.
...
We calculated the mean to be around $M_2=77.7572$ and uncertainty to be $\sigma _2 = 8.14527$.

Your formula for $v$ depends on the difference $M_2 - M_1$, so $v$ is also uncertain.

 

[Andreas S-H]

I can't follow what you're doing. Are you using the muon decay rate to estimate how many muons decay between the two detectors and calculate the muons' proper time from that and hence the speed?

We are using the time difference between detectors going off. This is our setup. So as far as i know we aren't using their decay at all

 

[Andreas S-H]

PS I don't understand the why you are using the uncertainty is the speed. You want an estimate of the actual speed, surely?

I do, but he told us we have to do it this way

 

[Andreas S-H]

O

I mean this:

Your formula for $v$ depends on the difference $M_2 - M_1$, so $v$ is also uncertain.

It definitely is, but is $\sigma _v$ no the uncertainty for the speed ?

 

[PeroK]

I do, but he told us we have to do it this way

What's the significance of the measurements of 901 and 600 events?

 

[Sagittarius A-Star]

We then calculated the uncertainty of the speed. $$\sigma_{v}={\frac {1}{ 0.45443\,E-6\,s}\sqrt {{\frac {{\sigma_{d}}^{2}
}{ \left( M_{2}-M_{1} \right) ^{2}}}+{\frac {{d}^{2}}{ \left( M_{2}-M_
{1} \right) ^{4}} \left( {\frac {{\sigma_{1}}^{2}}{n_{1}}}+{\frac {{
\sigma_{2}}^{2}}{n_{2}}} \right) }}} \approx 9.06E6
$$

From what is the term before the square-root calculated? Is there the exponent correct?

 

[Andreas S-H]

What's the significance of the measurements of 901 and 600 events?

He didn't really tell us. If i had to guess. It would be that fewer muons would get registered because they would have to almost travel in a vertical line to hit both detectors and we can therefore use it for a difference.

Here is what the data looks like

 

[PeroK]

Given that I can't even figure out what $M_1$ and $M_2$ represent, I don't think I can help!

 

[Andreas S-H]

Given that I can't even figure out what $M_1$ and $M_2$ represent, I don't think I can help!

that's okay. Our teacher have just said that everything was correct. And confirmed that the muons did in fact loose a large amount of their energy in the air. he also gave us the amount of energy it lost, however he still won't say how he calculated it

 

[Sagittarius A-Star]

So we concluded that the muons had lost some energy and therefore speed on the way down.

I don't think the energy loss is significant for the speed, because in the following document they measured $\beta \approx 1$, that means $v \approx c$:
https://web.mit.edu/jgross/Public/mit-classes/8.13/muons-paper.pdf

I assume, that in the second measurement most of the muons were not received vertically because the detectors were too close together, see in the link FIG. 10 on page 5.

In the second measurement, what was the distance between the sensors, when you put the detectors on top of each other?

 

[Andreas S-H]

I don't think the energy loss is significant for the speed, because in the following document they measured $\beta \approx 1$, that means $v \approx c$:
https://web.mit.edu/jgross/Public/mit-classes/8.13/muons-paper.pdf

I assume, that in the second measurement most of the muons were not received vertically because the detectors were too close together, see in the link FIG. 10 on page 5.

In the second measurement, what was the distance between the sensors, when you put the detectors on top of each other?

they were lying on each other so i would say 0m

 

[Sagittarius A-Star]

they were lying on each other so i would say 0m

What time interval shows the x-axis at the second measurement?

 

[PeroK]

What time interval shows the x-axis at the second measurement? By which signal is the oscilloscope triggered?

Are $M_1$ and $M_2$ times in some unspecified units?

 

[Sagittarius A-Star]

Are $M_1$ and $M_2$ times in some unspecified units?

To my understanding, this are the x-axis values (without unit) of the maximums of the shown curves. According to the OP, one x-axis unit relates to $0.45 ns$.

 

[Sagittarius A-Star]

I would love to calculate the amount of energy it has lost, is there a way to do that ?

According to the following site, it looses 2 GeV:
https://cosmic.lbl.gov/SKliewer/Cosmic_Rays/Muons.htm

We first calibrated the TDC module on our computer. Here we got the linear relationship to be

$y=0.45443ns*x + 13.471$

How does this calibration ensure in detail, that one unit on the x-axis relates to $0.45443 ns$?

they were lying on each other so i would say 0m

What are approximately width, length and hight of the detectors?

 

[Andreas S-H]

According to the following site, it looses 2 GeV:
https://cosmic.lbl.gov/SKliewer/Cosmic_Rays/Muons.htmHow does this calibration ensure in detail, that one unit on the x-axis relates to $0.45443 ns$?What are approximately width, length and hight of the detectors?

We, used different tick times for the TDC when we got the times we tried to use linear regression on it. here we got a $R^2 = 0.9991$.

The height is approx 10cm width 50cm and length is about 1.2m

 

[Sagittarius A-Star]

We, used different tick times for the TDC when we got the times we tried to use linear regression on it. here we got a $R^2 = 0.9991$.

Which signal does the y-axis show while the calibration?

The height is approx 10cm width 50cm and length is about 1.2m

Does that mean, that in the second measurement some (not vertically received) muons travel 130cm between detection on the left/rear side of the top detector and the detection on the right/front side by the bottom detector?

Is 1.2 m also the length of the physically active detection device in the detector box?

If yes and if they move with c, this would contribute up to 4.3 ns to $.45ns * (M_2 - M_1)$ in the formula for v.

Then it would be better to have in both measurements several meters distance between the detectors.