3
$\begingroup$

The masses of the Z and W particle sum almost exactly to the mass of the Top quark,within the errors:

Z + W = 80.385±0.015 + 91.1876±0.0021 = 171.57 ±0.0171 GeV

Top quark 172.9± 1.5 GeV

A: Is this one of those simple coincidences?

B: The Z,W particles are decays of the T?

C: Someone has a not too cranky idea connecting them?

EDIT: After consideration of dmckee and Lubos posts.

How about instead of a decay from a t quark, collide a $W^\pm$ and a $Z$ to produce a red top and anti-red bottom.

$$W^+ + Z^0 \to t(r) + \bar{b}(\bar{r})$$

this conserves charge, spin, color, confinement and energy - provided the excess energy of the bottom quark comes from the kinetic term of the collision. It immediately decays as lubos and dmckee pointed out in an early question

EDIT 2: Also note decay time of t-quark is $4.2\ 10^{-25} s$, nearly matching the W,Z decays times of $3.0\ 10^{-25} s$ , although I'm yet to find an uncertainty for these.

And with incredible hubris I'm calling this the Metzgeer Momentary Meson $t\bar{b} $
:) joke

Improve this question
$\endgroup$
0

2 Answers 2

Reset to default
5
$\begingroup$

You run into several practical problems immediately.

  • The total angular momentum of a system of a $W^\pm$ and a $Z$ is an integer, but the top has spin $\frac{1}{2}$.

  • The charge of a system involving one $W$ and one $Z$ would be $\pm 1$. The top has charge $\frac{2}{3}$.

These two could be repaired by assuming that there is a spin $\frac{1}{2}$ charge $-\frac{1}{3}$ particle involved. However...

  • The decay modes would generate some really funny implications. Start with the predominate mode: $t \to W^+ + b$ means that we're suggesting that a $Z$ is related to a $b$ (and a anti-down or anti-strange?) somehow.

    Wait! what?

    We started with one quark related to two weak vector bosons because the masses came out close and now we have some other quark related to one weak vector boson even though the masses are totally different. And even if we believe that what are we going to do about the origin of the lighter quark generation? Or are we to believe that only the top and bottom are related to the weak bosons?

Any way, as you may have guessed, I'm going with coincidence all the way. They have to happen sometimes.

Improve this answer
$\endgroup$
6
  • 2
    $\begingroup$ One also fails to conserve the color in the original decay - a confined colorful particle can't decay to colorless ones such as W,Z. ... I am utterly unimpressed by the "coincidence" that the sum of two particles' masses is a third particle mass within 1%. I wouldn't even call it a coincidence. It's just a description of one of the normal situations. $\endgroup$
    – Luboš Motl
    Commented Jun 5, 2012 at 3:51
  • $\begingroup$ Wait up - there's another coincidence - the decay times of t,Z,W are all close to 10^-25 s, so the masses match up and the decay times. That hints at more than a coincidence. $\endgroup$
    – metzgeer
    Commented Jun 5, 2012 at 9:50
  • 1
    $\begingroup$ @metzgeer Ah...that is not a coincidence instead it is an example of "weak universailty". All three are a single weak vertex with roughly the same mass difference which means roughly the same phase space for the products, so roughly the same half-life follows. $\endgroup$
    – dmckee --- ex-moderator kitten
    Commented Jun 5, 2012 at 16:33
  • 1
    $\begingroup$ Also, $m_{\rm Higgs} \approx \sqrt{m_Z m_t}$. Just a coincidence. There are lots of them if you look. They mean nothing. $\endgroup$
    – Matt Reece
    Commented Jun 6, 2012 at 4:43
  • $\begingroup$ @Matt Reece: "Just a coincidence." Someone disagrees: arxiv.org/abs/1209.0474 ... In the longer run (which may not be so far away), I expect to find that there are quite a few similarly simple quantitative relations which do have explanations, but which have been dismissed as "coincidence". The difficulty is to identify which such relations do mean something, and which ones don't. $\endgroup$
    – Mitchell Porter
    Commented Sep 5, 2012 at 2:10
0
$\begingroup$

It could happen as long as conservation of momentum was possible which it could be with some changes. Also conservation of color charge needs to happen. Otherwise it makes sense.

$$W^+ + Z^0 \to t(r) + \bar{b}(\bar{r})$$ $$t(r) \to b(r) + {W^+}$$

If we add these two equation we get: $$W^+ + Z^0 \to W^+ $$

This is a perfectly normal reaction that can happen directly without any top quarks interfering. The top quark version is just the more rarer process.

Improve this answer
$\endgroup$
2
  • $\begingroup$ To solve this all you need to do is substitute the top quark in the top equation. The bottom and the antibottom can be treated as opposites and can be canceled. Now all you are left with is the W+ boson. Color is also conserved in the above reaction. $\endgroup$
    – Roghan Arun
    Commented May 8, 2020 at 2:31
  • $\begingroup$ The energy of the system can be provided by the kinetic energy of the Z boson and W boson. Also the rest mass could be added in. $\endgroup$
    – Roghan Arun
    Commented May 8, 2020 at 2:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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