# Constraints on interactions of tachyons

1. Mar 12, 2012

### bcrowell

Staff Emeritus
Now that the superluminal neutrino fiasco is winding down, I'm interested in seeing if I can consolidate what I know about tachyons. One of the things I learned from following the OPERA debacle is that you can have tachyons without Lorentz violation, or you can have FTL particles (still called tachyons?) with Lorentz violation.

I guess it's been known for a really long time that tachyons can't be charged.

This paper http://arxiv.org/abs/1109.5682 by Giudice, Sibiryakov, and Strumia talks about Lorentz-violating Lagrangians for neutrinos, and says that they would cause the superluminality to extend to electrons and muons, since they couple to neutrinos through the weak interaction.

The Cohen-Glashow model, http://arxiv.org/abs/1109.6562 , also says you'd get Cherenkov-like radiation from superluminal particles coupling to the weak force. This paper also seems to assume Lorentz violation.

Do these results extend to tachyonic particles in models without Lorentz violation? In other words, is there some kind of general theorem that says that tachyons not only can't be charged, but can't participate in weak interactions without running into these other problems?

What happens with tachyons that couple to the strong force? Are these also ruled out because they would make protons superluminal via the Giudice mechanism?

Re gravity, I guess the equivalence principle requires that tachyons, if they exist, *do* interact gravitationally.

Baez has this nice discussion of tachyons: http://math.ucr.edu/home/baez/physics/ParticleAndNuclear/tachyons.html

He claims that the tachyonic telephone can't exist, based on the behavior of the Klein-Gordon equation. This confuses me, because when the error in the OPERA result hadn't been found yet, plenty of theorists were running around building theories in which tachyonic neutrinos were propagating information faster than c from CERN to Gran Sasso. How did they avoid the constraint referred to by Baez?

Last edited: Mar 12, 2012
2. Mar 13, 2012

### Haelfix

Baez is correct.

Also, it does not make sense to talk about 'Tachyons' unless Lorentz invariance is an exact symmetry of the world. They are only strictly speaking well defined within that context. So for instance, the hypothetical existence of superluminal neutrinos should never be confused with Tachyons.

Now assuming that SR is correct, the modern interpretation of 'Tachyons' comes from the theory of spontaneous symmetry breaking and critical phenomena.

It is always the case that the presence of a Tachyon in a physical particle spectrum, arises b/c the physicist in question has stupidly expanded around a bad point in the potential. The vacuum will spontaneously condense and roll to a new 'safe' spot.

The canonical example is expanding the Higgs field around an unstable maximum, and important examples involve the dynamics of cosmic inflation.
http://en.wikipedia.org/wiki/Tachyonic_field

3. Mar 13, 2012

### Ilmrak

Interesting, I never thought about tachyons this way although I knew about negative $m^2$ for fields around an unstable maximum xD

Does this mean tachyons can only arise as solution for an hamiltonian convex in the fields?
Is there some meaningful way to eventually treat such an (unbound from below) hamiltonian?

Ilm

4. Mar 14, 2012

### Haelfix

It only looks like its unbounded from below in a very localized neighborhood of the potential. So I mean the classic example is the potential

V(phi) = lambda/4 ( phi^2 - v^2)^2

It has a term that looks like -m^2/2 phi^2. Near the global maximum at phi = 0 (the symmetric phase), the field modes with wave number less than m have an exponential character consistent with what the dispersion for tachyonic fields look like as first studied in this paper

http://relativitycalculator.com/images/superluminal_velocities/possibility_faster_than_light.pdf

Of course physically these modes eventually settle into a minima around <phi> = +/- v, after a phase transition from the symmetric phase to the broken phase, and once they are there all the effective masses are strictly positive as they must be.

I like the mechanical analogy the Wiki page gives, which is similar to the Ferromagnetism example that is often taught:

"There is a simple mechanical analogy that illustrates that tachyonic fields do not propagate faster than light, why they represent instabilities, and helps explain the meaning of imaginary mass (negative mass squared).
Consider a long line of pendulums, all pointing straight down. The mass on the end of each pendulum is connected to the masses of its two neighbors by springs. Wiggling one of the pendulums will create two ripples that propagate in both directions down the line. As the ripple passes, each pendulum in its turn oscillates a few times about the straight down position. The speed of propagation of these ripples is determined in a simple way by the tension of the springs and the inertial mass of the pendulum weights. Formally, these parameters can be chosen so that the propagation speed is the speed of light. In the limit of an infinite density of closely spaced pendulums, this model becomes identical to a relativistic field theory, where the ripples are the analog of particles. Displacing the pendulums from pointing straight down requires positive energy, which indicates that the squared mass of those particles is positive.
Now consider an initial condition where at time t=0, all the pendulums are pointing straight up. Clearly this is unstable, but at least in classical physics one can imagine that they are so carefully balanced they will remain pointing straight up indefinitely so long as they are not perturbed. Wiggling one of the upside-down pendulums will have a very different effect from before. The speed of propagation of the effects of the wiggle is identical to what it was before, since neither the spring tension nor the inertial mass have changed. However, the effects on the pendulums affected by the perturbation are dramatically different. Those pendulums that feel the effects of the perturbation will begin to topple over, and will pick up speed exponentially. Indeed, it is easy to show that any localized perturbation kicks off an exponentially growing instability that affects everything within its future "ripple cone" (a region of size equal to time multiplied by the ripple propagation speed). In the limit of infinite pendulum density, this model is a tachyonic field theory."

5. Mar 14, 2012

### kurros

Haelfix sort of says it, but to be explicit, barring some exotic circumstances I am forgetting or never learned about, the answer is no. If your potential really is unbound from below (not just "locally" unbound from below), you have no ground state, so your theory is unstable and you are screwed.

6. Mar 15, 2012

### Ilmrak

This was the situation I was thinking about.
So tachyons, in every sensate QFT, are simply a consequence of the choice of an unstable vacuum state, that then decay in a stable vacuum that contains no tachyons.

I'll exploit your willingness with another (maybe a bit stupid) question

Could tachyons be used in some approximation of a highly exited state?
After all we are full of unstable particles, maybe we could use tachyons as a good approximation of some short life state?

Ilm

7. Mar 15, 2012

### kurros

That I have no idea about. People do talk about about vacuum transitions though, which I assume must have something to do with, say, kicking the higgs field hard enough that you reach all the way back up to the pre-electroweak symmetry breaking vacuum and have it roll back down in a different direction, the intermediate stage of which I guess is some tachyon-like situation, but I haven't really read about such things in any meaningful detail.