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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 $E<pv$.

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  • $\begingroup$ Let us continue this discussion in chat. $\endgroup$
    – user87745
    Mar 7, 2016 at 20:38
  • $\begingroup$ It's not clear to me what this question is asking for. There is no explicit question in the body, and the things you calculate do not really make sense to me. Maybe have a look at the excellent answers to Do tachyons move faster than light?, where behaviour of tachyons is analysed in detail. $\endgroup$
    – ACuriousMind
    Mar 15, 2016 at 20:11
  • $\begingroup$ The explicit question is what is asked in the title. In the body, I have elaborated my point. The problem with the expressions for Energy and Momentum by Bulanuik and Sudarshan is my concern and that I have explained in the body. $\endgroup$
    – user87745
    Mar 16, 2016 at 13:10

1 Answer 1

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

You've already got a problem here in that ${\sqrt{1-v^2}}$ is derived from Pythagoras's theorem, such that the hypotenuse of a right-angle triangle represents the speed of light in natural units wherein c=1, the base represent speed v as a fraction of c, and the height is the Lorentz factor. See the simple inference of time dilation on Wikipedia:

enter image description herepublic domain image by Mdd4696

Think of the wave nature of matter, and of a helical spring viewed sideways, such that it looks like this: /\/\/\/\/\. Now mentally stretch it to emulate a particle moving at a faster speed. Divide each isosceles triangle in half vertically, then think of each right-angle triangle as per the illustration above. Keep stretching the spring until it's stretched out tight, like this _________. The height is now zero, because v=c and so ${\sqrt{1-v^2}}$ is zero. You can't stretch the spring further, and the Lorentz factor can't be for example ${\sqrt{1-1.414^2}}$. You end up with an imaginary speed and a non-real solution.

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

It isn't clear, because you have an undefined result. The Lorentz factor is based on Pythagoras's theorem, and there are no right-angle triangles where the base is longer than the hypotenuse.

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 $uv>1$.

I'm afraid it's the wrong conclusion. In seismology we have P-waves and S-waves. The former are longitudinal waves, and they propagate faster than the latter, which are transverse waves. Light is a transverse wave too. If there was some other wave that propagated faster than light, its energy would not be zero.

This negative energy is explained with the help of the accompanied reversal of chronological order

You should forget about that. See this answer about time dilation and the related answer about the nature of time. Clocks feature some kind of regular cyclical internal motion and give a cumulative called the time. There is no such thing as negative motion, so there's no such thing as negative time.

what I could not understand is that how the energy can be negative

It can't be.

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.

A frame is little more than a state of motion. If there was some tachyonic wave moving faster than light, the non-tachyonic wave nature of matter means you will never be able to move such that it appeared to be moving slower than light.

So one must assume that some different expression for the energy and momentum should be at work for tachyonic particles.

Correct.

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