What was the reaction for pion production during the observation of antiprotons?

[Physteo]

Hello guys.
I am studying the experiment conducted by Chamberlain, Segrè, Wiegand and Ypsilantis in 1955 at Berkeley (using the Bevatron) to observe antiprotons.
Here is the article: http://escholarship.org/uc/item/46p0z8w7#page-3

I have some questions.
1) While the desired reaction of proton to copper-proton target was
p p → p p p p'
there is a strong background of Pion production. Why? And what is the reaction for the pion production?

2) You can read at the bottom of page 3, that negative particles scattered at 21° from the bevatron beam have a momentum of 1.19 Bev/c.
Why?

I can attempt to answer just question 1: Pions are lighter than proton, so the threshold is lower. The reaction I figured out was
p n → π- p p


Thanks!

 

[mfb]

1) While the desired reaction of proton to copper-proton target was
p p → p p p p'
there is a strong background of Pion production. Why? And what is the reaction for the pion production?

p p -> p p + multiple pions (1, 2, 3, ...)
p p -> p λ K + multiple pions and similar reactions are possible, too.
Pions are a frequent product in all hadronic high-energy collisions, as they are very light.

2) You can read at the bottom of page 3, that negative particles scattered at 21° from the bevatron beam have a momentum of 1.19 Bev/c.
Why?

If they have a lower momentum, they are bent more, if they have a higher momentum, they are bent less. The numbers depend on the geometry of the setup - mainly the field strength and its size.

 

[Physteo]

Great :) Thanks for the quick reply!
But I have a further question. Why do they choose exactly 1.19 GeV/c ? I guess this corresponds to the threshold momentum expected for p'. But how can I compute this theoretically?

I computed the threshold kinetic energy of the incident proton, and this is K = 6*m (where m is the mass of the proton, and I am setting c = 1). So the threshold energy is
E = K + m = 7m = 6568.1 MeV
the momentum is then
<br /> p = (E^2 - m^2)^{\frac{1}{2}} = 6500.7 MeV/c<br />
for the conservation law of the momentum, if the final protons and the antiprotons are threshold-produced, their momentum should be
<br /> p/4 = 1625 MeV/c<br />
But this is different from 1.19 Gev/c.
Where's the problem?

 

[mfb]

In the paper, they get 5.6 GeV kinetic energy for the threshold. There is your difference.

Apart from that, 1.18 GeV could be an arbitrary value, it does not have to be the antiproton momentum at threshold. Usually, energies a bit higher than that give a better cross-section (=>more antiprotons).

 

[Salman2]

For my clarification. Would it be correct that we know experimentally that the antimatter proton has opposite electric charge, AND opposite spin for magnetic moment (i.e., negative for both), compared to the matter proton ? Are there any other differences ?

Edit: Of course one other difference within nucleus is that the antiproton \bar{p} has (\bar{u}\bar{u}\bar{d}) quark structure, whereas {p} has (uud).

So, the antiproton differs from proton in at least three different fundamental ways, two related to EM force and a second to STRONG force.

Question ? Does the antiproton show opposite beta decay WEAK force behavior to proton ? We know that matter {p} beta plus decays to a neutron, β+, and matter neutrino v. Does antiproton \bar{p} show this same WEAK force beta plus decay pattern ?

 

[mfb]

What do you mean with "negative spin"? Its isospin (with a meaningful sign) is reversed, but that is just a fancy name for the quark content.

Are there any other differences ?

All other charges and quantum numbers are reversed, too.

 

[Bill_K]

For my clarification. Would it be correct that we know experimentally that the antimatter proton has opposite electric charge, AND opposite spin for magnetic moment (i.e., negative for both), compared to the matter proton ? Are there any other differences ?

Don't forget the most important difference -- the antiproton travels backwards in time! (Just kidding)

 

[Salman2]

What do you mean with "negative spin"?

I mean what Gabrielse from CERN had to say about the spin related to the magnetic moment of the antiproton, as copy below from this link:

www.physicsworld.com/cws/article/news/2013/apr/08/atrap-nails-down-the-antiprotons-magnetic-moment

""Nevertheless, the team found that the magnetic moments of the antiproton and proton are "exactly opposite" – that is, equal in strength but opposite in direction with respect to the particle spins. "Here, exactly opposite means that the direction that the [magnetic moment] points with respect to the particle's internal spin is opposite to that of the proton, but the size of the [magnetic moment] is the same," says Gabrielse.""

 

[mfb]

The relative orientation between spin and magnetic moment is different, right.

 

[Salman2]

Don't forget the most important difference -- the antiproton travels backwards in time! (Just kidding)

Suppose two moments A and B, and they are connected by arrow of space-time relative to both, A <----> B. Suppose a matter proton with positive energy moving in space-time, the diagram would be: A ---{p}---> B. Now, suppose an antimatter proton with negative energy, would not the diagram be: A <---\bar{p}--- B ? Is this not predicted by Einstein SR equation ? Please note, I'm not suggesting negative energy exists, only that it is predicted iff it is a property of antimatter.

 

[Bill_K]

Sorry Salman2, please note my comment was intended as a joke. Negative energy states do not exist, and no particle of any kind travels backwards in time. We've discussed this many times here, so let's not do it all over again. Suggest you search for earlier threads on this.