Interpreting the various tracks in a hydrogen bubble chamber

[songoku]

Homework Statement

Look at the figure, which shows tracks in a hydrogen bubble chamber. One anti-proton hits a proton in the chamber. The resulting tracks show the tracks of various pairs of particles ($\pi^{+}$ / $\pi^{-}$ , $e^{+}$ / $e^{-}$ ) which are created in the collision.

(a) Identify the incoming anti-proton track.

(b) How did you identify the incoming anti-proton? What does this tell you about its kinetic energy compared to the resulting particles?

(c) How does the picture illustrate the conservation of matter/energy?

(d) Identify any pair of particles (e.g. $e^{+}$ / $e^{-}$). How do you know that the tracks show you a pair of identical but opposite particles?

(e) How does the picture illustrate the conservation of charge?

(f) How does the picture illustrate the conservation of momentum in two dimensions?

Relevant Equations

None

(a) I don't know how to identify which one is the anti-proton track. The anti-proton hits proton at point in the figure where all the lines are coming together but how to identify which one is anti-proton?

Do I need to know a certain rule regarding the color of each particle?

Thanks

 

[BvU]

Hehe, the particles have no colour. The tracks have been given a colour afterwards, to help you.
A bubble chamber is usually placed in a strong magnetic field, perpendicular to the plane of the picture.

$\$

 

[songoku]

Hehe, the particles have no colour. The tracks have been given a colour afterwards, to help you.
A bubble chamber is usually placed in a strong magnetic field, perpendicular to the plane of the picture.

$\$

Okay so the magnetic field will cause magnetic force to act on the particle and the particle will move in a somewhat circular trajectory but how to identify which one is anti-proton (since all charged particle will be deflected)?

Thanks

 

[BvU]

There is one track that changes curvature; what does that mean for the momentum ?

 

[songoku]

There is one track that changes curvature; what does that mean for the momentum ?

Sorry I still don't know which track that changes curvature? Is it: A1 to A2 or B1 to B2 or C1 to C2 or D1 to D2 or none of them?

Momentum should be conserved in this case so changing curvature would mean there is collision between particles.

Or maybe you mean the momentum of certain particles would change (possibly decrease) due to the change in curvature?

Thanks

 

[Tom.G]

Hint:
Why did you not label one or two tracks?

 

[BvU]

somewhat circular trajectory

Yes, in particular: $r\propto mv$

Sorry I still don't know which track that changes curvature? Is it: A1 to A2 or B1 to B2 or C1 to C2 or D1 to D2 or none of them?

A1 is the straightest track; the exercise composer also hasn't given it a colour ...
So my money is on A1 as answer to part (a). It curves to the right, so green appears to indicate negative.

Now I humbly must admit my 'changes momentum' wasn't very well thought through
(even though very advanced in years, I only worked with near zero COM momentum in colliders ) :
If the $p^-$ and the $p^+$ annihilate, then A1 --> A2 for the $p^-$ is nonsense.

The hypothesis A1 --> A2 for the $p^-$ leaves us searching for a forward $p^+$ which is absent.
(D1 as outgoing $p^+$ has too much transverse momentum)

Reverting to the problem statement:
(b) We have no information on the energy of the incoming $p^-$. Momentum conservation for the center of mass requires a large fraction of that to go into the products forward motion. The annihilation adds almost 2 GeV distributed over at least 8 particles (zero charge particles are invisible in a cloud chamber). I suppose the expected answer is that the incoming $p^-$ has the highest kinetic energy of all particles shown. But I don't feel very convinced.

I'm sorry I can't offer much help on (c) and (d).

There are four outgoing green tracks of negative and four of positive particles. No neutral particles that decay within the visible region - let's assume all products are seen in the picture. At least we see charge conservation is satisfied (e).

For (f) we would need to add the momenta of all eight tracks. The sum should be the incoming $p^-$ momentum.

$\$

 

[songoku]

Hint:
Why did you not label one or two tracks?

No particular reason. I just thought what I did would be wrong because I labeled the tracks randomly. I mean the correct pair could be A1 and B1 or A1 and C1 instead of A1 and A2

A1 is the straightest track; the exercise composer also hasn't given it a colour ...
So my money is on A1 as answer to part (a).

Is this a general rule, the straightest track would be the initial particle? Or maybe this is an educated guess?

It curves to the right, so green appears to indicate negative.

By "curves to the right", do you mean curves clockwise? And do you assume the direction of magnetic field is into the plane of the picture?

If the $p^-$ and the $p^+$ annihilate, then A1 --> A2 for the $p^-$ is nonsense.

If they annihilate, two photons should be created and we would not be able to detect the photons in the bubble chamber?

The hypothesis A1 --> A2 for the $p^-$ leaves us searching for a forward $p^+$ which is absent.
(D1 as outgoing $p^+$ has too much transverse momentum)

When anti-proton hits proton, does the proton at rest initially?

Reverting to the problem statement:
(b) We have no information on the energy of the incoming $p^-$. Momentum conservation for the center of mass requires a large fraction of that to go into the products forward motion. The annihilation adds almost 2 GeV distributed over at least 8 particles (zero charge particles are invisible in a cloud chamber). I suppose the expected answer is that the incoming $p^-$ has the highest kinetic energy of all particles shown. But I don't feel very convinced.

I'm sorry I can't offer much help on (c) and (d).

Actually you have helped, a lot.

For (c), I am thinking something like the total kinetic energy of red and green tracks should be equal to the total KE of anti-proton and proton

For (d), I am thinking B1 and D2 because they have somewhat similar radius but opposite curvature

Thanks

 

[BvU]

Is this a general rule, the straightest track would be the initial particle?

The straightest track is the track with the highest momentum

And do you assume the direction of magnetic field is into the plane of the picture?

The problem statement says it's a $p^-$

When anti-proton hits proton, does the proton at rest initially?

For all practical purposes, yes. (It only has some minimal thermal kinetic energy)

 

[songoku]

The straightest track is the track with the highest momentum The problem statement says it's a $p^-$For all practical purposes, yes. (It only has some minimal thermal kinetic energy)

Ah ok ok

Now I have something to build my argument upon.

Thank you very much for the help and explanation BvU, Tom.G