Is the gamma-factor for Tachyons real?

[Sagittarius A-Star]

TL;DR

Is the Gamma-factor for Tachyons real, as stated in the paper?

I found in a paper:

4.1.4 Tachyons
If $v>1$, that is, the particle is moving faster than the speed of light, then the norm of the 4-velocity (165) will necessarily be$|\dot x|^2>0$ and thus the path must be spacelike. In that case we can normalize to$|\dot x|^2=1$, which gives (50)
$\gamma = \frac {1}{\sqrt{v^2-1}}$
(50) Note that the Lorentz factor is real, as it should be. Naively, if one used the definition of $\gamma$ which was derived by assuming $v<1$, then $\gamma$ will be imaginary for $v>1$; but that is simply not the right normalization factor to use.
Furthermore, since $\gamma$ is real, so is the energy $E=\gamma m$. A tachyon need not have imaginary energy or mass, as is sometimes naively claimed.

Source (see page 35 and also chapter 4.1.2):
https://scipost.org/SciPostPhysLectNotes.10/pdf

Is this correct, and if yes, why? They use the (-+++) convention.

 

[PeterDonis]

Is this correct, and if yes, why?

IMO it's an abuse of terminology and notation, since $\gamma$ is supposed to be the factor that appears in the Lorentz transformation that brings you to the particle's rest frame (the paper uses the term "Lorentz factor"), but there is no Lorentz transformation that will turn a spacelike vector into a timelike vector, or, to put it another way, there is no such thing as a "rest frame" for a tachyon since a tachyon has a spacelike 4-velocity.

 

[PeterDonis]

there is no such thing as a "rest frame" for a tachyon since a tachyon has a spacelike 4-velocity

Note that the paper also abuses the term "at rest" by saying that a tachyon is "at rest" as its $v \rightarrow \infty$.

 

[Sagittarius A-Star]

since a tachyon has a spacelike 4-velocity.

They claim to "normalize" the norm of the 4-velocity. I don't understand, what this means.

 

[PeterDonis]

They claim to "normalize" the norm of the 4-velocity. I don't understand, what this means.

They're not normalizing the norm itself; that's always $\pm 1$ (in units where $c = 1$).

What they're claiming to "normalize" is the frame-dependent quantity $dx / dt$, i.e., the ordinary velocity in a given frame. But "normalize" is really another abuse of terminology in the paper; what they are actually doing for a tachyon is taking the limit of the product $\gamma v$ (using their non-standard definition of $\gamma$ for a tachyon) as $v \rightarrow \infty$. But $\gamma v$ is just one component of the 4-velocity; it's not either a 3-vector or a 4-vector by itself. So they're not "normalizing" anything.

 

[PeterDonis]

Is this correct

Another issue I see is the paper's definition of a tachyon's energy as $E = m \gamma$, with their definition of $\gamma$. That means that a tachyon "at rest" by their definition (i.e., in a frame in which the tachyon's 4-velocity is $(0, 1, 0, 0)$) has zero energy--not just zero kinetic energy, zero energy period, even though it has, according to the paper, positive invariant mass. Also, the tachyon will still have nonzero momentum in this frame--the obvious definition of momentum to use in the paper's treatment would be $p = m \gamma v$, which means that a tachyon "at rest" has $p = m$.

This seems to me to imply that it is impossible for a tachyon to exchange momentum or energy with anything other than (possibly) another tachyon. Consider, for example, what absorption of a tachyon by an ordinary particle would have to look like (the paper implies that this must be possible in its discussion of tachyons and time travel). Pick the frame in which the tachyon is "at rest", and suppose, to start with the simplest possible case, that the ordinary particle also starts out at rest (really at rest, $v = 0$) in this frame. Then it is impossible for the ordinary particle to absorb the tachyon--it would have to absorb zero energy, but also nonzero momentum, which would have to change its velocity in the chosen frame, but that would have to also change its energy. Similar remarks would apply to emission of a tachyon, and since a collision with a tachyon is equivalent to absorption and immediate emission, this would seem to me to imply that it is impossible for a tachyon to interact with a non-tachyon.

 

[Vanadium 50]

Things are complicated with tachyons. One can either assume a real mass, in which case the four-momentum is complex, or one can assume the four-momentum is real, in which case the mass is imaginary. Cohen and Glashow (PRL107, 181803) use δ (defined in that paper) as a real substitute for γ.

In the imaginary mass ansatz,
<br /> v = c \sqrt{1+\frac{(-im)^2}{E^2}}<br />

You can see that as E goes up, v goes down.

 

[Sagittarius A-Star]

They're not normalizing the norm itself; that's always $\pm 1$ (in units where $c = 1$).

They change it for tachyons from $-1$ (see equation 166) to $+1$ (see equation 169).

[edit: The norm will have, with the usual gamma, "automatically" a different sign for tachyons compared to tardyons (=objects with $v<c$). Reason: the "normal" $\gamma ^2$ in equation (166) will have a different sign accordingly. The "normalize" could then only be meant as preventing the norm from having different signs for tachyons and tardyons.]

 

[vanhees71]

IMO it's an abuse of terminology and notation, since $\gamma$ is supposed to be the factor that appears in the Lorentz transformation that brings you to the particle's rest frame (the paper uses the term "Lorentz factor"), but there is no Lorentz transformation that will turn a spacelike vector into a timelike vector, or, to put it another way, there is no such thing as a "rest frame" for a tachyon since a tachyon has a spacelike 4-velocity.

Indeed, a tachyon must be "faster than light" in any reference frame.

One should be aware that, as far as relativistic QFT is concerned, you can define to a certain extent a theory for free tachyons but there's trouble with the S-matrix concerning the linked-cluster principle and also a stable ground state. Usually tachyonic fields indicate an instability and leads to "condensation", i.e., spontaneous symmetry breaking of a global symmetry with the occurrence of massless Goldstone modes and massive particles.

 

[Vanadium 50]

I'm not sure I agree 100% with @vanhees71

I absolutely agree that if your QFT predicts tachyons, it's a sign your theory is sick. I also agree that these are the sicknesses that come out of the mathematics. However, I don't want to go as far as to say that if there were tachyons, they would behave in these troublesome ways. I would instead say we have not developed a consistent theory of tachyons.

I'd further say that the problem occurs much earlier in the chain than QFT. In QM, spacelike commutators are zero. With tachyons, that's not necessarily true. In QM, operators are Hermitian. With imaginary masses, this is obviously false. I would say if someone really wanted a QFT that worked for tachyons, they would need to rederive much of what we take for granted under different assumptions.

 

[vanhees71]

I think, relativistic QT should be formulated as a QFT, because you always deal with creation-destruction processes when you deal with relativistic energies. The trouble with tachyons, as far as I understand it, which brought me to my statements above, is that in the definition of the S-matrix you have the time-ordering and this only leads to an Poincare covariant S-matrix for the representations of the poincare group with $p \cdot p =m^2 \geq 0$ (see Weinberg, QT of Fields, vol. 1) and the validity of the linked-cluster principle. On the other hand, this is only a sufficient condition. I'm not sure whether it's also a necessary condition for the construction of a stable Poincare covariant QFT. In that sense it might indeed just be our lack of an idea how to describe tachyons with all the constraints, particularly those related with (Einstein) causality, but so far none has been found AFAIK. It's also the question, whether it's needed since there's not the slightest hint for the existence of tachyons in nature yet.

 

[PeterDonis]

They change it for tachyons from $-1$ (see equation 166) to $+1$ (see equation 169).

Calling that "normalization" is another abuse of terminology.

 

[Sagittarius A-Star]

Calling that "normalization" is another abuse of terminology.

What you cite from my posting #8 must be a wrong interpretation of me about what they are doing. I corrected my interpretation shortly afterwards, as describe in #8 below, in the "edit" section.

 

[PeterDonis]

I corrected my interpretation shortly afterwards

I wasn't saying that you are adopting the term "normalization" to describe what you were describing. I was saying that the paper does.