# B Tachyons are bosons?

1. Mar 7, 2017

### Debaa

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.

2. Mar 7, 2017

### Staff: Mentor

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

3. Mar 7, 2017

### Debaa

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

4. Mar 7, 2017

### Drakkith

Staff Emeritus

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

5. Mar 8, 2017

### DrDu

I thougt they had $m^2=-1$.

6. Mar 8, 2017

7. Mar 8, 2017

### Demystifier

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

8. Mar 8, 2017

### martinbn

What is the precise definition of a tachyon? (This is a B thread so I can ask clarifying questions, right? )

9. Mar 8, 2017

### Demystifier

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?

10. Mar 8, 2017

### Staff: Mentor

Can you clarify what you are referring to here?

11. Mar 8, 2017

### Staff: Mentor

They don't have to have $m^2 = -1$. They just have $m^2 < 0$.

12. Mar 8, 2017

### Debaa

13. Mar 8, 2017

### martinbn

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?

14. Mar 8, 2017

### Demystifier

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?

15. Mar 8, 2017

### vanhees71

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)

16. Mar 8, 2017

### Demystifier

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.

17. Mar 8, 2017

### Demystifier

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.

18. Mar 8, 2017

### vanhees71

That's an interesting gedanken experiment. However, I've no clue, how you'd experimentally make $\phi=0$ at some time $t$.

19. Mar 8, 2017

### Demystifier

20. Mar 8, 2017

### DrDu

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