# Tachyons are bosons?

• B

## Main Question or Discussion Point

Are tachyons force Particles/messenger particles ? Is so do they act messenger between two entangled particles and allow faster than light information exchange? Thank for the answer.

## Answers and Replies

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PeterDonis
Mentor
2019 Award
Tachyons don't exist, to the best of our current knowledge. They do not appear in any of our current theories.

vanhees71
But "hypothetically" since they have mass -1 maybe?

Drakkith
Staff Emeritus
Are tachyons force Particles/messenger particles ? Is so do they act messenger between two entangled particles and allow faster than light information exchange? Thank for the answer.
But "hypothetically" since they have mass -1 maybe?

No, there is no evidence for any of the above and no reason to think they do.

DrDu
But "hypothetically" since they have mass -1 maybe?
I thougt they had ##m^2=-1##.

Debaa
I thougt they had ##m^2=-1##.
My bad their m=√-1

Demystifier
Gold Member
Tachyons don't exist, to the best of our current knowledge. They do not appear in any of our current theories.
Or do they? Higgs field before symmetry breaking can be thought of as a tachyon field with ##m^2<0##. Nevertheless, it does not propagate faster than ##c##. It has been discussed in more detail in

martinbn
Or do they? Higgs field before symmetry breaking can be thought of as a tachyon field with ##m^2<0##. Nevertheless, it does not propagate faster than ##c##. It has been discussed in more detail in
What is the precise definition of a tachyon? (This is a B thread so I can ask clarifying questions, right? )

Demystifier
Gold Member
What is the precise definition of a tachyon? (This is a B thread so I can ask clarifying questions, right? )
Tachyons are objects with ##m^2<0##, but the meaning of the parameter ##m## depends on the context. It may be the "mass" of the particle or the "mass" of the field.

In the particle case, ##m## defines the relation between energy ##E## and 3-momentum ##{\bf p}## through
$$E^2-{\bf p}^2=m^2$$

In the field case, one considers a field ##\phi(t,{\bf x})## which can be Fourier transformed in terms of plane waves ##e^{-i(\omega t- {\bf k}\cdot{\bf x})}##. Here ##m## defines the relation between frequency ##\omega## and wave 3-vector ##{\bf k}## through
$$\omega^2-{\bf k}^2=m^2$$
Is that precise enough?

PeterDonis
Mentor
2019 Award
Higgs field before symmetry breaking can be thought of as a tachyon field with ##m^2<0##
Can you clarify what you are referring to here?

PeterDonis
Mentor
2019 Award
I thougt they had ##m^2=-1##.
They don't have to have ##m^2 = -1##. They just have ##m^2 < 0##.

DrDu
martinbn
Tachyons are objects with ##m^2<0##, but the meaning of the parameter ##m## depends on the context. It may be the "mass" of the particle or the "mass" of the field.

In the particle case, ##m## defines the relation between energy ##E## and 3-momentum ##{\bf p}## through
$$E^2-{\bf p}^2=m^2$$

In the field case, one considers a field ##\phi(t,{\bf x})## which can be Fourier transformed in terms of plane waves ##e^{-i(\omega t- {\bf k}\cdot{\bf x})}##. Here ##m## defines the relation between frequency ##\omega## and wave 3-vector ##{\bf k}## through
$$\omega^2-{\bf k}^2=m^2$$
Is that precise enough?
The particle case is clear to me because it connects with the causal structure. It's what I thought tachyons are. In the field case it is not clear to me why that should be called tachyons (or anything at all), and how does the definition go in a general space-time?

Demystifier
Gold Member
Can you clarify what you are referring to here?
I am referring to the Higgs potential
$$V(\phi)=-\frac{\mu^2}{2}\phi^2+\frac{\lambda}{4}\phi^4$$
where ##\mu^2>0## and ##\lambda>0##. For small ##\phi## you can ignore the ##\lambda##-term, so what remains is a "mass" term with a wrong sign. Does it help?

vanhees71
Gold Member
2019 Award
But the point is that for this potential pertubation theory around ##\phi=0## doesn't make sense, because it's a maximum of the potential rather than a minimum. That's why you expand around the minimum,
$$V'=\phi (-\mu^2+\lambda \phi^2)=0,$$
i.e., around ##\phi_0=\mu/\sqrt{\lambda}##.

You can, to a certain extent, define QFTs of tachyons. See, e.g.,

J. Dhar, E.C.G. Sudarshan, Quantum Field Theory of interacting tachyons, Phys. Rev. 174, 174 (1968)

Demystifier
Gold Member
In the field case it is not clear to me why that should be called tachyons (or anything at all), and how does the definition go in a general space-time?
I guess you know that quantization of fields leads to quantum states that can be interpreted as quantum particles. If they are states with definite energy and momentum, then their energy and momentum satisfies the same relation as that for the corresponding classical particles. That explains why such fields are called tachyon fields.

Concerning general spacetime, it's much easier to write down the partial differential equation which the fields satisfy. This is the Klein-Gordon equation
$$(\nabla^{\mu}\nabla_{\mu}+m^2)\phi(x)=0$$
in general spacetime with metric signature ##(+,-,-,-)##, where ##m^2<0## for tachyon fields.

vanhees71
Demystifier
Gold Member
But the point is that for this potential pertubation theory around ##\phi=0## doesn't make sense, because it's a maximum of the potential rather than a minimum.
Well, it depends on what do you mean by "doesn't make sense". Mathematically, it makes sense if you are studying a regime in which ##\phi## is close to zero. It is certainly not easy to satisfy this condition in an LHC experiment, but in principle it is not impossible. Initial conditions are, in principle, arbitrary, so there is no physical principle which would forbid ##\phi(t=0)=0##. For a short time after such initial condition, the system would behave as a tachyon field.

vanhees71
Gold Member
2019 Award
That's an interesting gedanken experiment. However, I've no clue, how you'd experimentally make ##\phi=0## at some time ##t##.

DrDu
Terms luke "soft modes" and " glass transition" come to my mind.

PeterDonis
Mentor
2019 Award
I am referring to the Higgs potential
Ah, ok. @vanhees71 has already raised the points I would make.

Demystifier
Gold Member
Terms luke "soft modes" and " glass transition" come to my mind.
Can you be more explicit?

DrDu
I think there is a better chance to observe these phenomena in solid state systems where you can rapildy sweep the parameter mu^2. If you do this slowly, the excitations will become "soft" near the point where mu vanishes. If you do it rapidly, long range collective modes may freeze out and you get a glass transition.

Demystifier
Tachyons don't exist, to the best of our current knowledge. They do not appear in any of our current theories.
They do, though. As Demystifier pointed out, they show up whenever you are perturbing around an unstable vacuum. What is true is that no particles which travel faster than light (for a reasonable definition of "travel") can exist in any reasonable quantum field theory. But a particle can be a "tachyon" (an imaginary mass solution of the linearized equations of motion around an unstable vacuum) and still travel no faster than light, respecting causality. See e.g. http://math.ucr.edu/home/baez/physics/ParticleAndNuclear/tachyons.html and http://physics.stackexchange.com/questions/166095/do-tachyons-move-faster-than-light . Look especially at Qmechanic's excellent answer.

Demystifier
Drakkith
Staff Emeritus