We examine the possibility of tachyons under the criteria that they form the energy momentum four vector that transforms according to Lorentz Transformations.

Therefore,

$E$$'$ $=\displaystyle\frac{E-pv}{\sqrt{1-v^2}}$

Where $v$ is the relative velocity of frame $O'$ wrt frame $O$. If the velocity of the tachyon is $u$ in frame $O$ then it is clear that $E'<0$ if $E<pv$.

Now if we assume that $E = \displaystyle\frac{im_*}{\sqrt{1-u^2}}$ and $p = \displaystyle\frac{im_*u}{\sqrt{1-u^2}}$ ( Where $im_*$ is the 'rest mass' of tachyon and $i = \sqrt{-1}.$ ) then we arrive at the conclusion that $E'<0 $ for $uv>1$.

This negative energy is explained with the help of the accompanied reversal of chronological order and reinterpretation principle but what I could not understand is that how the energy can be negative if the expression for the energy is to remain $E' = \displaystyle\frac{im_*}{\sqrt{1-w^2}}$ where $w$ would be the velocity of the considered tachyon in frame $O'$. This can only happen if the $m_*$ itself alters its sign but then the question arises that for what frames $m_*$ is positive and for what frames it is negative and no conclusive or meaningful answer can be provided as our initial choice of frame $O$ was completely arbitrary.

So one must assume that some different expression for the energy and momentum should be at work for tachyonic particles. This results into unacceptability of the result that $E'<0$ for $uv>1$. All one should be able to say is $E'<0$ if $E<pv$.