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Determining Momentum of a Particle from Radius, Distance, and Angle?

by Anonymous2258
Tags: angle, determining, distance, momentum, particle, radius
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Anonymous2258
#1
Jan21-13, 12:35 PM
P: 9
I didn't know whether to put this in homework or in here because it is not homework but it is a question, so sorry in advance! I like to learn about quantum physics as a hobby, and I found this great link about working with actual particle accelerator data by an organization called Quarknet. Here is the link:
http://quarknet.fnal.gov/projects/su...nt/index.shtml

I thought I would try and use this data(after clicking on "project context") to do what the site says and determine the mass and lifetime of the Z boson. I understand I need to use the
E2 = m2c4 + pc2
formula, and I've found the energies, but the only problem is I don't know how to determine momentum. They give you the XY radius as well as the Z component of the particle's path, as I think it travels in a helical path because of the electromagnetic field. They also give you the cosine of the x, y, and z angles. I am totally lost on how to solve for momentum and I'd really like to understand how one would go about doing this. I know that momentum for relativistic particles is not simply mass times velocity, so could someone please help? Thank you!
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mfb
#2
Jan22-13, 11:33 AM
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P: 11,589
##r=\frac{p_t}{qB}## with the transversal momentum pt (see here, for example). Geometry allows to get the total momentum p just as in the nonrelativistic case.
Anonymous2258
#3
Jan22-13, 02:32 PM
P: 9
Quote Quote by mfb View Post
##r=\frac{p_t}{qB}## with the transversal momentum pt (see here, for example). Geometry allows to get the total momentum p just as in the nonrelativistic case.
Thank you very much. I've seen that you can set m(v2/r) equal to the force from the magnetic field q(vb) and get that, but since it is the cross product, am I going to have to use sinθ to get the magnitude? Or is the magnetic field always going to be perfectly perpendicular to the particle's velocity. I'm thinking I might have to do something along those lines because they did give me the cosine of the X, Y, and Z angles. Here's a sample collision image so you can more easily visualize it
http://quarknet.fnal.gov/projects/su...t/theta2.shtml
And also, these are electron-positron collisions, so am I going to have to worry about charge, or do I just use 1.6*10-19 which simplifies the equation when put into electron-volts.

mfb
#4
Jan22-13, 04:37 PM
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P: 11,589
Determining Momentum of a Particle from Radius, Distance, and Angle?

I don't know which θ you mean.
Or is the magnetic field always going to be perfectly perpendicular to the particle's velocity.
No, but you can ignore the motion along the magnetic field (=z-component) for pt.

Quote Quote by Anonymous2258 View Post
And also, these are electron-positron collisions, so am I going to have to worry about charge, or do I just use 1.6*10-19 which simplifies the equation when put into electron-volts.
Well, you can only measure the ratio of momentum to charge. Luckily, all particles which reach the detector are uncharged (not relevant here) or with +- 1 elementary charge, so you can use this value.
Anonymous2258
#5
Jan22-13, 09:19 PM
P: 9
Quote Quote by mfb View Post
I don't know which θ you mean.

No, but you can ignore the motion along the magnetic field (=z-component) for pt.


Well, you can only measure the ratio of momentum to charge. Luckily, all particles which reach the detector are uncharged (not relevant here) or with +- 1 elementary charge, so you can use this value.
Oh okay gotcha...so just as a final check can you tell me if I am doing the calculations right?
I'm using the total energy for a single event as E, in GeV (I already adjusted for the calibration). For pc I'm using .3Br where B is the magnetic field in Teslas and r is the XY radius in meters.
I'm putting it into
E2 = m2c4 + (pc)2
and solving for m, does that seem right? I'm not getting what the Z boson mass should be. I'm basically just subtracting the momentum term from the energy squared term and square rooting it to get the mass in GeV/c2.
mfb
#6
Jan23-13, 09:40 AM
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P: 11,589
Wait... the Z boson is not measured in the detector at all.

For pc I'm using .3Br where B is the magnetic field in Teslas and r is the XY radius in meters.
This gives the transverse momentum of the electron/muon. It is sin(θ) of the total momentum where θ is the angle between z-axis and particle track.

Using an assumption for the particle mass, you can get the total energy of the particle. Calorimeter data could be used, too, but in real data the uncertainty there is problematic.

Once you have the 4-momenta of the Z decay products, you can add them to get the 4-momentum of the Z-boson candidate and calculate its mass via your formula.
Anonymous2258
#7
Jan23-13, 06:02 PM
P: 9
Quote Quote by mfb View Post
Wait... the Z boson is not measured in the detector at all.


This gives the transverse momentum of the electron/muon. It is sin(θ) of the total momentum where θ is the angle between z-axis and particle track.

Using an assumption for the particle mass, you can get the total energy of the particle. Calorimeter data could be used, too, but in real data the uncertainty there is problematic.

Once you have the 4-momenta of the Z decay products, you can add them to get the 4-momentum of the Z-boson candidate and calculate its mass via your formula.
So .3Br is the momentum perpendicular to the track and I need to find the x, y, and z components of the momentum? What 4-momenta am I adding, am I using the momenta from more than one event? Sorry, and I don't really understand how I'm supposed to use the 4-momentum in the energy mass momentum equation because it seems to be a vector in spacetime(sorry, I'm not really that clear on what 4-momentum is, I had never heard it before).
mfb
#8
Jan24-13, 11:25 AM
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P: 11,589
You are mixing two different steps here.

Starting from physics: A Z-boson is created and immediately decays into two muons (or electrons, I'll consider muons here). Those muons fly through your detector and get detected. You cannot see the Z-boson itself, so you have to work backwards:

You have to detect both muons in the detector. For each individual muon, calculate its 4-momentum (start with that).

If an event just contains a single muon, you might have missed one, or there was no Z boson. In any way, you can do nothing here.
If an event contains at least one positive and one negative muon, they might come from the decay of a single particle. If that is true, their added 4-momenta must be the 4-momentum of the decaying particle. So you add the calculated 4-momenta, and calculate the mass of that (possible) particle. If the muons really come from a Z boson, the calculated mass will be the Z-mass. If they come from a different particle, the calculation will give the mass of this other particle. If they were produced independently, the calculated mass is random. If you collect a lot of those muon pairs, you will see events with the Z-mass, plus contributions from other particles, and some background.

So .3Br is the momentum perpendicular to the track and I need to find the x, y, and z components of the momentum?
Right

Sorry, and I don't really understand how I'm supposed to use the 4-momentum in the energy mass momentum equation because it seems to be a vector in spacetime
p^2 is ##\vec{p}^2##, a scalar, and energy is a scalar as well. But first you need the momentum components.
Anonymous2258
#9
Jan24-13, 05:49 PM
P: 9
Quote Quote by mfb View Post
You are mixing two different steps here.

Starting from physics: A Z-boson is created and immediately decays into two muons (or electrons, I'll consider muons here). Those muons fly through your detector and get detected. You cannot see the Z-boson itself, so you have to work backwards:

You have to detect both muons in the detector. For each individual muon, calculate its 4-momentum (start with that).

If an event just contains a single muon, you might have missed one, or there was no Z boson. In any way, you can do nothing here.
If an event contains at least one positive and one negative muon, they might come from the decay of a single particle. If that is true, their added 4-momenta must be the 4-momentum of the decaying particle. So you add the calculated 4-momenta, and calculate the mass of that (possible) particle. If the muons really come from a Z boson, the calculated mass will be the Z-mass. If they come from a different particle, the calculation will give the mass of this other particle. If they were produced independently, the calculated mass is random. If you collect a lot of those muon pairs, you will see events with the Z-mass, plus contributions from other particles, and some background.


Right


p^2 is ##\vec{p}^2##, a scalar, and energy is a scalar as well. But first you need the momentum components.
Okay so I calculated momentum in the x, y, and z directions, and added them from two muons in the same event together. You said I need to calculate the 4-momentum. Does that require knowing the energy of the muon? And once I have the 4-momentum, how do I square it from there? The only vector multiplication I know that gives a scalar from 2 vectors is the dot product.
mfb
#10
Jan25-13, 10:34 AM
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P: 11,589
Does that require knowing the energy of the muon?
It is sufficient to know the momentum and its mass, this allows to calculate E via ##E^2=m^2 c^4+p^2c^2##

how do I square it from there
What do you want to square?

The only vector multiplication I know that gives a scalar from 2 vectors is the dot product.
And the dot product will give you p2.
Anonymous2258
#11
Jan28-13, 05:35 PM
P: 9
Quote Quote by mfb View Post
It is sufficient to know the momentum and its mass, this allows to calculate E via ##E^2=m^2 c^4+p^2c^2##


What do you want to square?


And the dot product will give you p2.
Alright so I calculated the momentum of the muons as well as the energy so that's the 4-momentum? And I want to square the combined 4-momentums of the two decay particles, right? So to do that do I just use this formula
http://en.wikipedia.org/wiki/Four-vector#Scalar_product
And then that'll give me the momentum to plug into the energy momentum equation? But what do I use for E in that equation, the combined energy of the muons?
mfb
#12
Jan29-13, 06:00 AM
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P: 11,589
Right

But what do I use for E in that equation, the combined energy of the muons?
Of course.

$$\vec{p_{\mu_1}}+\vec{p_{\mu_2}} = \vec{p_Z}$$
In components:

$$\begin{pmatrix} E_1 \\ p_{x1} \\ p_{y1} \\ p_{z1} \end{pmatrix} + \begin{pmatrix} E_2 \\ p_{x2} \\ p_{y2} \\ p_{z2} \end{pmatrix} = \begin{pmatrix} E_1+E_2 \\ p_{x1}+p_{x2} \\ p_{y1}+p_{y2} \\ p_{z1}+p_{z2} \end{pmatrix} = \begin{pmatrix} E_Z \\ p_{xZ} \\ p_{xZ} \\ p_{xZ} \end{pmatrix}$$
Anonymous2258
#13
Jan30-13, 09:31 PM
P: 9
Quote Quote by mfb View Post
Right

Of course.

$$\vec{p_{\mu_1}}+\vec{p_{\mu_2}} = \vec{p_Z}$$
In components:

$$\begin{pmatrix} E_1 \\ p_{x1} \\ p_{y1} \\ p_{z1} \end{pmatrix} + \begin{pmatrix} E_2 \\ p_{x2} \\ p_{y2} \\ p_{z2} \end{pmatrix} = \begin{pmatrix} E_1+E_2 \\ p_{x1}+p_{x2} \\ p_{y1}+p_{y2} \\ p_{z1}+p_{z2} \end{pmatrix} = \begin{pmatrix} E_Z \\ p_{xZ} \\ p_{xZ} \\ p_{xZ} \end{pmatrix}$$
Oh, okay! Then for calculating the 4-momentum of other particles besides the muon, what do I use for the energy component? Do i just the energy in the calorimeter? And one last thing, if I wanted to calculate branching ratios, is there any way to tell which kind of particles are in the detector or can I only divide them into hadrons, muons, and whichever fermions interact via the electromagnetic force?
mfb
#14
Feb1-13, 10:46 AM
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P: 11,589
For other particles, use the mass of other particles ;). Combining the energy measurement from the calorimeters with the momentum measurement of the tracking system would lead to large errors. If tracking does not give a reliable momentum measurement (for high-energetic particles, for example), you can use the measured energy and direction to reconstruct the momentum, again with the known mass of the particle.

If the particle type cannot be identified*, the usual approach is to test all hypotheses: Treat it as a possible pion (with the pion mass), look if it could fit to some interesting process with a pion. Treat it as a possible muon, look if it could fit to some interesting process with a muon, and the same for kaons (, electrons) and protons.

*if particle identification is crucial, the detectors have special parts for that.


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