1
$\begingroup$

Consider the following reaction between a moving proton and a stationary proton

$$p + p \rightarrow p + p + \pi^0 + \pi^0 $$

Find the lowest kinetic energy (in the labsystem of reference) for which the reaction is possible.

My attempt:

My initial though was to use the expression for invariant mass together with the center of mass reference frame (LS=labsystem reference frame, CM= center of mass reference frame):

$E_{Tot, LS}^2 - p_{Tot, LS}^2c^2 = E_{Tot, CM}^2 - p_{Tot, CM}^2c^2 = E_{Tot, CM}^2$

But here comes my question. What values should I put into $E_{Tot, CM}^2$? Is this the total eneryg in the center of mass reference frame before the reaction (i.e. $E_{Tot, CM}=2m_pc^2$) or is it the total energy in the center of mass reference after the reaction (i.e. $E_{Tot, CM}=2m_pc^2 + 2m_\pi c^2$). According to the solution it should be the $2m_pc^2 + 2m_\pi c^2$, but I don't understand why $2m_pc^2$ is not correct?

$\endgroup$
14
  • $\begingroup$ Your expression for the total CM energy before the reaction is, of course, very wrong: there is lots of unaccounted momentum there. This energy is the same before and after the reaction: this is the point of the calculation. $\endgroup$ Jan 7 at 17:37
  • $\begingroup$ @CosmasZachos Just to clarify, with "your expression for the total CM energy", do you mean $2me_c^2$ or something else? How would the momentum enter into the total energy? As kinetic energy of the protons in the CM reference frame? $\endgroup$ Jan 7 at 19:21
  • $\begingroup$ $E_{TOT}^2=4( m_p^2c^4 +p^2 c^2)$ Surely you know the formula for the relativistic energy. $\endgroup$ Jan 7 at 19:28
  • $\begingroup$ ... but you never need to actually use this particular formula, do you? $\endgroup$ Jan 7 at 21:47
  • $\begingroup$ You probably meant to use $m_p$ (not $m_e$). $\endgroup$
    – robphy
    Jan 7 at 23:26

1 Answer 1

0
$\begingroup$

Some geometric thinking with an energy-momentum diagram might help,
where the 4-momentum vector has

  • a timelike-component (the relativistic-energy in the frame of the diagram) and
  • a spacelike-component (the relativistic momentum in that frame),
    and
  • where the rapidity-angle $\theta$ between the 4-momentum and the diagram's time-direction (the unit 4-velocity of the frame) encodes:
    --the velocity of the particle ($(v/c)=\tanh\theta$)
    --the relativistic momentum of the particle ($p/c=m\sinh\theta$)
    --the relativistic energy of the particle ($E/c^2=m\cosh\theta$)
    --the relativistic kinetic energy of the particle ($T/c^2=m(\cosh\theta-1)$)

So, for this process...

  1. Since there are two initial particles,
    the two initial 4-momenta will sum to the CM-4-momentum.
    This appears a triangle with future-timelike 4-momenta.
    Since these two particles are protons, this is actually a "Minkowski-isosceles triangle". The rapidity-angle in this triangle (wrt to the CM 4-momentum) has a simple relation to the rapidity-angle of the moving proton (wrt to the stationary proton).
    If you wish, you can work in the CM-frame, where the CM-4-momentum is purely timelike (i.e. has no spatial component). This may suggest the analogous Euclidean problem, whose strategy you could use with hyperbolic-trigonometry instead of ordinary circular-trigonometry.
  2. Since there are four final particles,
    the four final 4-momenta will sum to the CM-4-momentum (by total 4-momentum conservation).
    This appears as a five-sided figure.
    But with the two protons and two pions, this five-sided figure has two pairs of Minkowski-congruent sides.
  3. So, altogether, you will form a six-sided figure (with the CM-4-momentum eliminated) with the above sets of congruent sides.

Intuitively, for threshold production, the final particles will be at rest in the CM-frame.

If you draw the above,
you can solve a hyperbolic-trig problem for the "rapidity-angle with respect to the CM-4-momentum", where you know the lengths of the sides of this hexagon (which, at threshold, degenerates into a triangle [here, a Minkowski-isosceles one]).

This "rapidity-angle with respect to the CM-4-momentum"
can be expressed in terms of the "rapidity of the moving proton in the lab frame",
which can be used to calculate the kinetic energy of the moving proton in the lab frame.

(Given the proton mass and the pion mass, and the use of cosh and arccosh, the required proton kinetic energy can be neatly written that encodes the geometrical conditions used.)

(For more examples, see my answer to Momentum diagram for two colliding Particles )


To encourage the use of this approach (but without giving the complete answer), I updated my answer with an energy-momentum diagram that is not at threshold...
but could be modified to be "at threshold."

"New Particles" represents the total 4-momentum of the newly produced particles. robphy-energyMomentumDiagram

When appropriately set up, the method of solution is not unlike working with a free-body diagram.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.