Would a high energy bottom quark 'decay' to a top quark? 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?
 A: 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.
A: 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.
