8
$\begingroup$

The reason for the long life time of $B$-hadrons is that the CKM element $|V_{tb}| > 0.999$, meaning that the preferred decay of the $b$-quark is to a $t$-quark (and vice versa). However because $m_{top} >> m_{bottom}$ this `isn't allowed' for $b$-quarks.

My feeling is that at very high energies, i.e $E_{bottom} >> m_{top}$, that the decay $b \rightarrow t$ would be permitted. I would therefore conclude from this as well, that the lifetime of the $b$-quark would change dramatically as $E_{bottom} \rightarrow m_{top}$. Is that correct? If not, why not?

$\endgroup$
  • $\begingroup$ But static isolated $b$-quarks don't exist. In a quantum field theory I would think that boosting the entire $b$-quark field into a static frame of reference is impossible, as the gauge transformation would compensate for this by adding additional gauge bosons to compensate for the energy (which themselves can mediate the decay $b \rightarrow t$ $\endgroup$ – kd88 Jun 26 '14 at 10:19
6
$\begingroup$

It doesn't matter whether the $b$-quark is highly energetic, it can never decay to a top quark and a $W$-boson if it is on mass shell, by which I mean, $p^2=E^2 - \vec p^2 =m_b^2$. To see this, consider energy-momentum conservation, $$ b^\mu = W^\mu + t^\mu \Rightarrow m_b^2 = M_W^2 + m_t^2 + 2W\cdot t = M_W^2 + m_t^2 + 2 E_t M_W $$ However, since the energy $E_t$ is positive, energy-momentum cannot be conserved in the decay - the left-hand-side cannot equal the right-hand-side for the measured particle masses.

Now, if the $b$-quark is not on mass-shell, $p^2\neq m_b^2$, the decay is possible. An off-shell $b$-quark could be an internal line - a virtual particle - in a Feynman diagram. Decay widths (i.e. lifetimes) are different for particles that are not on mass shell. However, since particles off-mass shell are not propagating (they are internal in Feynman diagrams), when we talk of a life-time, we always mean an on-shell particle.

$\endgroup$
  • $\begingroup$ Just because a particle has $p^2 \neq m^2$, it doesn't mean that it is off-mass shell. For example you can accelerate a proton to 1 TeV ($>> m_p$), but it's mass is still $m_p$. $\endgroup$ – kd88 Jun 26 '14 at 12:32
  • $\begingroup$ @jk88 I mean $p^2 = g_{\mu\nu}p^\mu p^\nu = E^2 - \vec p^2 = m^2$ $\endgroup$ – innisfree Jun 26 '14 at 12:37
  • $\begingroup$ I see. However doesn't boosting the $b$-quark field into the rest frame of the $b$-quark introduce additional energy terms associated with gauge bosons? $\endgroup$ – kd88 Jun 26 '14 at 12:46
  • $\begingroup$ i.e. the term $\frac{1}{4}(\bf{W^{\mu\nu}W_{\mu\nu}} + B^{\mu\nu}B_{\mu\nu})$. Surely the emission of a $W$ boson according to this term would produce a top quark. $\endgroup$ – kd88 Jun 26 '14 at 12:49
  • 2
    $\begingroup$ @jk88 it is easiest to think about in the $b$'s rest frame, but not necessary. A Lorentz boost is a change of co-ordinate frame. It doesn't have any physical effects (physics is Lorentz boost invariant). It doesn't introduce any additional terms with gauge bosons. The terms you list are Lorentz scalars - things that don't change under boosts. $\endgroup$ – innisfree Jun 26 '14 at 12:53
0
$\begingroup$

If you consider the decay from the frame where the b-quark is at rest: then p=0 and E=m(bottom quark)*c2 and it can't decay to the more massive top quark. It would violate the energy energy-momentum conservation law.

$\endgroup$

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.