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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?

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  • $\begingroup$ 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, 2014 at 10:19

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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.

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  • $\begingroup$ 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, 2014 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, 2014 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, 2014 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, 2014 at 12:49
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    $\begingroup$ @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, 2014 at 12:53
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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.

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