INTERPRETATION OF BUBBLE CHAMBER IMAGES Bubble Chamber images

INTERPRETATION OF BUBBLE CHAMBER IMAGES Bubble Chamber images look very beautiful, but at the same time very complicated to try to interpret! The picture on the left shows the tracks left by charged particles and antiparticles Fermilab bubble chamber: 4. 6 m in diameter in a 3 T magnetic field

HOW ARE TRACKS FORMED? A moving charged particle interacts with the liquid along its path via the Coulomb force it exerts on atomic electrons. In this way it transfers enough energy to start the liquid boiling and thus leaves behind it a trail of small bubbles. However: A photon has no charge and therefore does not interact via the Coulomb force. Therefore it cannot leave a trail of bubbles Q. What other particles would not leave a track in a bubble chamber ?

With so many complicated tracks to look at, how can the paths of individual particles be interpreted? We can do one of the following - 1 Relate the movement of electrons produced by Compton Scattering with the direction of the magnetic field 2 Relate the movement of the deflected particles with their charge and the direction of the magnetic field 3 Relate the production of particle / antiparticle pairs (matter & antimatter) to the energy of the photon which produced them 4 Apply conservation of momentum to a head-on collision between an electron and a positron

1 Relate the movement of electrons produced by Compton Scattering with the direction of the magnetic field applied to the bubble chamber Notice the trajectory of the spiralling lone electron indicated by the arrow this electron was knocked out of the atom that originally held it by a high energy photon. Questions 1. Identify the other examples of this interaction in the picture above 2. What is the direction of the magnetic field? 3. Why doesn’t the photon leave a track ?

Answers 1. Other Compton electrons are indicated by the arrows (above) 2. From the rule for the force exerted by a magnetic field on a current - the field is going into the page 3. The photon’s path is not visible because it does not carry any charge and so does not cause the formation of bubbles in the chamber……

2 Relate the movement of deflected particles with their charge and the direction of the magnetic field y Remember that the direction of the force on a moving charged particle is found using the rule for the force exerted by a magnetic field on a current. x So the electrons all turn clockwise in the above diagram z

B A An electron ( e - ) and a positron (e + ) leave the tracks shown in the picture above Question: Which is which?

e- B e+ A Answer: The particle which deflects to the right (track B) is an electron, e. The particle which deflects to the left (track A) is a positron, e+

3 Relate the production of particle / antiparticle pairs (matter & antimatter) to the energy of the photon which produced them Two photons are produced at C where a positron annihilates with an electron. Just one travels to D where it interacts with a nucleus in the liquid and materializes into an electron/positron (e- e+) pair. ee+ D C To a good approximation, all of this photon’s energy is shared by the e-/e+ pair. Question Calculate the kinetic energy of the e-/e+ pair, knowing that the momentum of the photon is 265 31 Me. V/c. ( me = 0. 511 Me. V/c 2 )

Answer E = pc If the photon momentum is 265 Me. V/c then E = 265 Me. V ee+ Converting to Joules, D C 1 Me. V = 1. 6 x 10 -13 J E = 265 x 1. 6 x 10 -13 = 4. 24 x 10 -11 J = K(e-, e+)

4 Apply the Conservation of Momentum to a head-on collision between an electron and a positron E e+ e- At point E a rare event took place. For a snooker example of a head on collision, click below The positive track (e+) seems to change into a negative track (e- ). What happened was that the positron made a head-on collision with an electron. Question What is the linear momentum of the electron if the incoming positron’s linear momentum was 54 15 Me. V/c? . ( me = 0, 511 Mev/c 2 ) (melecton= mpositron)

E e+ e- Answer Visually, just before and just after E, the tracks seem to be more or less equally curved – thus having the same momentum. Since the track of the positron appears to have stopped at the point of collision, all of its momentum must have been transferred to the electron. Therefore the linear momentum of the electron must be about 54 15 Me. V/c since the mass of the electron is the same as that of the positron