So, as I was browsing a bunch of the tachyon questions throughout the years in this forum, and an oddity of these hypothetical faster-than-light particles came to mind. Ordinary particles with mass always have an inertial frame of reference where they are at rest, thus their rest mass can be measured. Lightspeed particles, while never at rest, have an invariant velocity, so different frames of reference can agree on their energy. Not so for tachyons.

Tachyons not only are never at rest (like lightspeed particles), but their measured velocity is frame-dependent (like ordinary particles with mass). So, my question:

Question: Is there anything analogous to a rest mass that can be used as a common point of reference when working out the mass of tachyons, or is it completely frame-dependent? And, no, we cannot say the tachyon's frame of reference, because they cannot be treated as observers (same as with lightspeed particles, we cannot treat a photon as an observer).


1 Answer 1


The concept of an invariant mass still applies to tachyons, and (in units with $\hbar=c=1$) the relationship between the invariant mass $m$, tachyon energy $E$, and tachyon momentum $p$ would still be $$ E^2 = p^2 + m^2 $$ however, for tachyons, $m^2 < 0$. You are correct that you cannot interpret $m^2$ in terms of the rest frame of the tachyon, since it does not have one.

For a tachyon traveling at a constant speed $v>c$, you cannot boost to a frame where the tachyon is at rest ($v=0$), but you can boost to a frame where the velocity is infinite ($v=\infty$). The ida is that a tachyon will follow a spacelike path $x=vt$ for constant $v>c$, and you can always boost such a path so that it occurs entirely at one time. Graphically, for an ordinary particle you can make the particle's trajectory into the time axis in the rest frame. For a tachyon you can make the trajectory into the $x$ axis. Then you could interpret $m^2$ as the result you get taking the limit of $E^2-p^2$ as you approach this infinite velocity frame (this is a subtle limit since both $E^2$ and $p^2$ will diverge in that frame). Of course, $m^2=E^2-p^2$ in any frame, but if you want to make use of the special frame that is analogous to the rest frame of an ordinary particle, that is one way to do it.


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