1
$\begingroup$

High energy positrons and electrons can form $D$ mesons when they interact with each other and annihilate, but I was wondering what those energy levels would have to be.

Specifically, would a fixed target containing electrons, and a 6GeV positron beam be enough to start forming them? And, if so, roughly how many would you expect (like, 1 out of every 10000 interactions or 1 out of every 10 kinda thing)?

Edit: Also, I probably should have asked this in the original question, but does anyone know what kind of materials might be appropriate to use a target in this case?

$\endgroup$
1
$\begingroup$

The $D^{+-}$ has a mass of $m = 1869.65 +/- 0.05 MeV$

A pair has to be produced to conserve the quantum numbers of input with output, This means that the center of mass energy should be at least twice this number even for threshold production.

If you do the four vector algebra you will see that the energy in the center of mass system is too small. The momentum of the electron is to all intents zero, thus only the energy term is left in the invariant mass calculation. The energy of the electron is 0.5MeV to be multiplied with the 6 GeV of the positron beam and the square root has to be taken . The energy in the center of mass of a $6GeV$ positron beam hitting an electron at rest is way below the threshold of the two D .

That is why a collider was needed in order to first observe them.

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

The $D^0$ has a mass of 1.865 GeV. But that's not enough. To make your $D^0$ you need to conserve energy as well as momentum - so you can't just make a stationary $D^0$, you have to make one moving with the same momentum as the original positron beam.

The standard way to calculate the required energy for such problems is to evaluate the invariant quantity $s=(\sum E_i)^2 + (\sum \vec p_i^2)$, which befire the reaction, using lab frame quantities, is $(E+m_e c^2)^2 - \vec p^2 c^2=2 m_e c^2 E +m_e^2 c^4$ and after the collision, using centre of momentum system, is $1.865^2$ at threshold. Equating these and putting in the numbers gives $E=3478$ GeV. A lot more than 6.

And even then - the process $e^+e^- \to D^0$ is a second order weak process and the cross section is so small you would never see it. To get a reasonable rate you need, as Anna V says, the reaction $e^+e^- \to D^+D^-$. That would take twice as much energy in the cms, and so four times as much in the lab...

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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