16
$\begingroup$

In all examples that I know, tachyons are described by scalar fields. I was wondering why you can't have a tachyon with spin 1. If this spinning tachyon were to condense to a vacuum, the vacuum wouldn't be Lorentz invariant---seems exotic but not a-priori inconsistent. Is there some stronger consistency requirement which rules out spinning tachyons?

Here's another confusion: I was reading Wikipedia, which claims that tachyons should be spinless and obey Fermi-Dirac statistics(?). (They reference an original paper by G. Feinberg which unfortunately I am not wealthy enough to download). The claim about Fermi-Dirac statistics is baffling---isn't the Higgs field a boson? Does anyone understand what they're talking about?

$\endgroup$
7
  • 4
    $\begingroup$ Higgs boson is not a tachyon, so what are you talking about? Also, spin-statistics theorem holds only for states with non-negative mass and also under other assumptions. As soon as any of those is violated more exotic possibilities arise. See the list of assumptions here: en.wikipedia.org/wiki/Spin-statistics_theorem#Proof $\endgroup$
    – Marek
    Commented Feb 21, 2011 at 11:17
  • 1
    $\begingroup$ The Higgs is a boson, but I'm not sure how that's related to your question. The Higgs is not a faster-than-light particle. $\endgroup$ Commented Feb 21, 2011 at 11:18
  • 3
    $\begingroup$ The Higgs particle is not a tachyon, but appears when expanding the Higgs field around the vacuum of spontaneously broken symmetry. Around the vacuum where the Higgs field has vanishing expectation value, the full gauge symmetry is restored but the Higgs field has negative mass^2. $\endgroup$ Commented Feb 21, 2011 at 11:57
  • 4
    $\begingroup$ The Higgs particle is a tachyon in the only sense that modern field theory accepts--- a particle which makes an unstable vacuum for the zero-charge state. $\endgroup$
    – Ron Maimon
    Commented Dec 31, 2011 at 11:57
  • $\begingroup$ related: physics.stackexchange.com/q/30944 $\endgroup$
    – user4552
    Commented May 5, 2013 at 2:13

4 Answers 4

9
$\begingroup$

It's not completely self-evident but it is true that in consistent theories, tachyons have to be scalar particles - much like the Higgs boson when expanded around the maximum of the potential (zero vev) - which obey, of course, Bose-Einstein statistics. (The claim about Fermi-Dirac is just wrong, or was meant to apply to Faddeev-Popov ghosts or similar fields, not physical tachyons.)

In non-interacting string theory, one sees that this conclusion is true because the ground state energy of a single-string Hilbert space is equal to $L_0=-1$ for the scalar tachyon, so any addition of spin - through the string oscillators - raises $L_0$ at least by one, bringing us to the massless or massive level (non-negative $m^2$).

$L_0=-1$ was true for the bosonic string ground state; in the case of the superstring, using the RNS formalism, the ground state has $L_0=-1/2$ but we also have antiperiodic fermions that only raise $L_0$ by $1/2$: that's still enough to show that any addition of spin - which only comes via internal oscillators - brings us to the massless or massive level, above the tachyonic interval.

The scalar character of the tachyon may also be seen in the effective field theory. Dirac or Weyl tachyons are impossible because the Dirac mass term $$-m \bar \psi \psi$$ has to be Hermitian. It implies that $m$ has to be real, which means that the particle is positively massive. A tachyonic fermion would need an imaginary $m$ but that would produce a non-Hermitian action.

The same is true for spin-one particles. Spin-one particles may only get their mass consistently by the Higgs mechanism: the relevant term arises from the covariant version of the kinetic terms for the Higgs fields: $$D_\mu \phi^\dagger D^\mu \phi$$ Again, this has to be Hermitean, and if it is Hermitean, $\phi^\dagger \phi$ that is left if the Higgs field has a nonzero vev, is automatically positively definite, which produces the usual mass term $$m^2 A_\mu A^\mu/2$$ with a positive coefficient.

Concerning a tachyonic vev, well, any vev of a tachyonic field that solves the equations of motion has to be Lorentz-breaking - because it is a non-constant function of spacetime. The spin of a tachyon would just add one more aspect to this story. But it is intuitively natural that the tachyons have to be scalars - the value of the tachyon away from the maximum of the potential measures "how much the instability has already advanced", and this quantity is naturally a scalar.

$\endgroup$
20
  • 1
    $\begingroup$ Thank you for your answer Lubos! You claim that massive vectors can only be made to consistently interact via the Higgs mechanism (or maybe stueckelburg.) I have heard this before, but I am wondering if there is a reference I can visit to understand this myself. The claim in wikipedia about tachyons being fermions seems unbelievable.. but I was curious about where this strange idea came from. $\endgroup$ Commented Feb 21, 2011 at 14:49
  • 1
    $\begingroup$ Dear @truebeliever1234, a tachyon can't have a constant (in spacetime) vev because a constant vev doesn't satisfy the equations of motion, $(\nabla_\mu \nabla^\mu + m^2)T=0$. The first term with the derivatives drops for a constant $T$ and the second one doesn't, so the equation is violated. That's the point of tachyons that instead of constants, you must either have waves moving in spacelike directions, or exponentially growing/decreasing functions of time. Constants are not allowed. $\endgroup$ Commented Feb 21, 2011 at 15:15
  • 3
    $\begingroup$ @truebeliever Have a look at the following paper and papers that cite it. As I recall, some of what they found in this paper was later reinterpreted by Sen&Zwiebach in terms of decay of unstable D-branes. The stuff on Lorentz violation probably involves an uncontrolled approximation and as far as I know has not been backed up by later studies. Still, the idea is intersting. Spontaneous Breaking of Lorentz Symmetry in String Theory. V.Alan Kostelecky, (Indiana U.) , Stuart Samuel, (City Coll., N.Y.) . IUHET-139, CCNY-HEP-88/4, May 1988. 8pp. Published in Phys.Rev.D39:683,1989. $\endgroup$
    – pho
    Commented Feb 21, 2011 at 21:27
  • 1
    $\begingroup$ I agree with Lubos too. ;) $\endgroup$
    – pho
    Commented Feb 22, 2011 at 13:47
  • 1
    $\begingroup$ No, higher spin particles still have the normal spin, not rapidity, even if they are tachyons and vectors or tensors because eg spin in the z direction is the dependence of the wave function on the angle in the xy plane and x, y are totally unaffected by the particle movement in t, z. In an environment, a spinning particle may become a tachyon and the normal spin must clearly still be conserved. Also, nonzero rapidity would not allow normalized wave functions by the usual rules. $\endgroup$ Commented Jan 8, 2023 at 4:17
2
$\begingroup$

If the polarizations of a tachyonic vector boson are Lorentz covariant, then its norm has to be indefinite, not positive definite, or positive semidefinite as in the case of massless vector bosons. This is true despite the fact that the polarizations have to be transverse to the 4-momentum because the 4-momentum is spacelike. Negative probabilities don't make any sense.

Tachyonic scalars have to be bosons, not fermions!

$\endgroup$
2
  • $\begingroup$ I have seen this argument, but I don't believe it holds water since the ghost polarizations only appear if you assume that the spin-1 tachyon must have a particle interpretation---specifically that the field can be expanded in Fourier modes with spacelike 4-momenta. However, in all physical applications of tachyons the 4-momentum is actually timelike, and then the mass-shell condition implies that the field grows exponentially in time. This is a signature of instability of the vacuum. Under this assumption you find that the free spin-1 tachyon does not propagate ghosts. $\endgroup$ Commented Feb 21, 2011 at 15:42
  • 1
    $\begingroup$ If you perform the Fourier transform of the spatial wave vectors, then if $m^2 = - k^2$, when the magnitude of the spatial wave vector exceeds $k$, we still have spacelike 4-momenta. $\endgroup$
    – QGR
    Commented Feb 21, 2011 at 16:00
2
$\begingroup$

Your question was solved by Eugene Wigner when he classified all possible representations of the Poincare group, i.e. all possible groups that can be translated in space or time, and boosted by changes of velocity in relativity.

Wigner found the invariants of the Poincare group, first the mass, and second the spin. Furthermore he found the allowed spins which depend on the character of the mass.

For m^2>0, spin = 0,1/2,1,3/2,... and Polarization may be measured from p = -s,1-s ... s

For m^2=0, spin = 0,+/-1/2,+/-1,+/-3/2 which are also are the possible measureable polarizations

When m^2<0, the only allowed representation of spin are the trivial spin=0, and a set of infinite dimensional groups. So there a no finite spin groups that describe tachyons. What the infinite dimensional groups allow you to do, I do not know, the math get somewhat hard there.

$\endgroup$
1
  • 5
    $\begingroup$ This assumes that the result of tachyonic condensation is Lorentz invariant, so that Wigner's classification applies. $\endgroup$
    – Ron Maimon
    Commented Dec 31, 2011 at 11:59
0
$\begingroup$

No, there's no controversy. It's just something that comes directly out of the symplectic classification associated with the symmetry group. For Relativity, that's the Poincaré group, and in the non-relativistic limit, it's the Galilei group and its central extension: the Bargmann group. In order to have consistency with the "correspondence principle" and to have a consistent non-relativistic limit, you should actually also be doing the symplectic decomposition with the (trivial) central extension of the Poincaré group, rather than the Poincaré group itself.

This is the picture that emerges:

Tardyons have 3 pairs of symplectic coordinates - which is what underlies the Heisenberg relation - if spin 0, and a 4th pair if spin non-zero. The extra pair are associated with the coordinates of a sphere. It corresponds to the degree of freedom usually labelled by $m$ (for the axial component of angular momentum) in quantum theory.

Luxons have 3 pairs of symplectic coordinates, if they belong to the sub-class which I've termed "helical". In the helical subclass, angular momentum is helical; i.e. its axis is parallel to linear momentum; and the helicity (i.e. the component parallel to the momentum) is an invariant. As a special case, this includes "spin 0" - which actually means helicity 0. The photon is a helical luxon, and it has non-zero helicity.

The luxons that don't fall within the helical sub-class have 4 pairs of symplectic coordinates. Unlike tardyons, the extra pair does not describe spin but is associated with the coordinates of a cylinder. This larger, and more generic, sub-family of luxons, I've termed "cylindrical".

There is no consensus on what, if anything, a cylndrical luxon may correspond to.

Tachyons have 3 pairs of symplectic coordinates only if they are "spin 0". Otherwise, they have 4; and the 4th pair is associated with either a one-sheeted hyperboloid, a double cone, or two-sheeted hyperboloid.

So, there are three subfamilies in the tachyon class. Again, there is no consensus on what (if anything) they would correspond to.

All "spin 0" classes have a well-defined canonical position vector $𝐫$ and the angular momentum $𝐉$ and moving mass moment $𝐊$ have the same decomposition as in non-relativistic theory: $𝐉 = 𝐫×𝐏$ and $𝐊 = M𝐫 - 𝐏t$, where $𝐏$ and $M$ the ("moving") mass (in relativity, "total energy" $E = Mc^2$ is often used in place of $M$).

The parameter $t$ is arbitrary. For a free non-interacting system, all the canonical quantities are constant, and the constancy of $𝐊$ entails the constancy of $𝐯 = d𝐫/dt = 𝐏/M$, which is just a back-door way of expressing The Law Of Inertia.

All of what I've described applies both relativistically and non-relativistically, except that the tachyon and luxon classes - for the non-relativstic version - converge into an unnamed class I've termed the "synchron". (Its defining property is $M = 0$, but $𝐏 ≠ 𝟎$.) Like the luxons, they also have a helical sub-family - inherited as the non-relativistic limit of the helical luxon class.

For all cases, $M^2 - α P^2$ is an invariant, with the distinction between the relativistic and non-relativistic cases being drawn, respectively, by $α = 1/c^2 > 0$ versus $α = 0$. The distinction between tardyon, luxon and tachyon rests entirely in the sign of this invariant.

If $M^2 - α P^2 > 0$, then they are tardyons and are distinguished by the property that they have both a center of mass and a rest frame in which $𝐏 ≠ 𝟎$. The value of $M$ in that frame is $m$ - the "rest mass" and $M^2 - α P^2 = m^2$. For the non-relativistic case $M$, itself, is the rest mass $m$ and is invariant.

If $M^2 = α P^2$ and $𝐏 ≠ 𝟎$, then it's a luxon (which, in the non-relativistic case, becomes a synchron with $M = 0$). They have no rest frame, and therefore the term "rest mass" makes no sense for them. But since $M^2 - α P^2 = 0^2$, it's common practice to wing it and just say that their "rest [sic] mass" is 0. However, referring to what follows, you could treat them like "tachons with zero impulse", just as well as "tardyons with zero rest mass". They're a limit case of both classes.

In the non-relativistic limit, $Π = |P|$ is an invariant and may be considered as an "impulse", with a synchron being that which conveys an instantaneous action-at-a-distance transfer of impulse; i.e. the instantaneous version of what is traditionally called a "line of force". They have infinite speed and exist only at an instant. They don't move in time at all, but exist in space as purely spatial objects. The "time of existence" is canonical for them, and is given by $t = -𝐊·𝐏/P^2$, and is an invariant for this class.

If $M^2 = α P^2$, trivially, with $𝐏 = 𝟎$ and $M = 0$, then that's the "homogeneous" class, which I'll describe further below.

If $M^2 < α P^2$, then it's a tachyon. This can only happen in the relativistic case $α > 0$. They have no rest frame, so the term "rest mass" makes no sense here, either. Instead, it's the impulse $Π^2 = P^2 - M^2/α$ that has meaning. That's the momentum square magnitude in any "synchron" frame of reference where the tachyon is instantaneous at infinite speed. In an infinite speed frame, for tachyons, $M = 0$, like the synchron.

What makes all this important is that it runs central to the issue of the spin-orbit decomposition.

For the spin non-zero classes, you need a spin-orbit decomposition in order to even be able to talk about spin. And that only happens with the tardyon class, with it being given by $$𝐉 = 𝐫×𝐏 + 𝐒, \hspace 1em 𝐊 = M𝐫 - 𝐏t + \frac{α𝐏×𝐒}{m + M}.$$ You can't write down anything meaningful, without $𝐫$ (and $m$).

If the latter of these relations is solved for $𝐫$, you get $𝐫 = (𝐊 + 𝐏t)/M$, plus the extra contribution $-α𝐏×𝐒/(M(m+M))$ that is entirely relativistic and spin-dependent - and together this gives you the classical version of what is called the Newton-Wigner position vector.

There's nothing like this for tachyons, and there is only marginally something like this for luxons ... only in the case of the helical subclass, where you can sorta write down a position operator (notwithstanding the "no position operator" theorem in the literature you may see mis-used as a folklore theorem on its "impossibility"). But there's no spin for the helical sub-class, and no 4th pair of symplectic coordinates at all. It's just 3 pairs (given by the components of $𝐫$ and $𝐏$ themselves), and they satisfy the same complementarity relation as in the spin 0 case.

There is also the homogeneous class. They have no center of mass, and (in fact) are invariant with respect to spatial and temporal translations, but may have non-zero $𝐉$ and $𝐊$. For relativity, the vectors $𝐒_± = 𝐉 ± 𝐊/\sqrt{-α}$ are each analogous to spin, except that they are complex, because $α > 0$. So, there are 2 pairs of symplectic coordinates; unless both vectors vanish (i.e. if $𝐉 = 𝟎$ and $𝐊 = 𝟎$), in which case they are also isotropic and boost-invariant (i.e. invariant with respect to change in the moving frame of reference) and have 0 symplectic coordinates. An "isotropic, boost-invariant, homogeneous" system is just another way of describing a "vacuum", and may be taken as the definition of such. So, the general case may be thought of as a "vacuum with spin". For the non-relativistic case, there may be 0, 1 or 2 sets of symplectic coordinates; the case of 0 being (again) the vacuum, the case of 1 being where $𝐊 = 𝟎$ and $𝐉 = 𝐒 ≠ 𝟎$ (with $S^2$ then being an invariant), and the case of 2 being where $𝐊 ≠ 𝟎$, with $K^2$ and $𝐉·𝐊$ both being invariant. In the relativistic case, for the homogeneous class, $𝐉·𝐊$ and $K^2 - α J^2$ are the invariants, and one can distinguish 3 sub-classes based on the sign of the latter invariant, with the $K^2 - α J^2 = 0$ case being the relativistic version of the above-mentioned $𝐊 = 𝟎$, $𝐉 ≠ 𝟎$ non-relativistic sub-class.

If you were to work out the symplectic decomposition in detail, what you would end up finding out is that - like the Wigner classification - the vectors $𝐖 = M𝐉 + 𝐏×𝐊$ and $W_0 = 𝐏·𝐉$ actually play a central role in everything. They are what, together, comprise 4 components what is known as the Pauli-Lubanski vector. There's also actually good reason to also treat $W_4 = 𝐊·𝐉$ as a 5th component, but it is not really needed here. They have an invariant: $W^2 - α {W_0}^2$ and the vector satisfies the identity $𝐏·𝐖 = M W_0$, which limits its number of independent components to just 2 (plus $W_4$ as a third orphan component).

For the so-called "spin 0" family, this invariant is always 0, because $𝐖 = 𝟎$ and $W_0 = 0$, for this class. It's actually from there, that you get the trivial spin-orbit decomposition: you just define $𝐫 = (𝐊 + 𝐏t)/M$, where $t$ can be arbitrarily chosen, and substitute into $𝐖 = 𝟎$ to solve for $𝐉$ and get $𝐉 = -𝐏×𝐊/M = 𝐊×𝐏/M = 𝐫×𝐏$.

For the helical families, the invariant is still 0, but $W_0 ≠ 0$. This can only happen in the non-relativistic case with synchrons, or in the relativistic case with luxons. In all cases, $W_0$ is an invariant with $𝐖$ being parallel to $𝐏$. The "helicity" (which is defined as the component of angular momentum along an axis parallel to $𝐏$) is just $W_0/P$, itself and is invariant.

For the tardyons, the invariant $W^2 - α {W_0}^2$ is always 0 or positive, and if it is positive, then it is $W^2 - α {W_0}^2 = m^2 S^2$: the square of the product of rest-mass and spin. In that case, the independent components of $(𝐖, W_0)$ trace out a sphere of radius $m S$.

For luxons, the invariant is 0 (for the helical subfamily) or positive (for the larger cylindrical family). In the latter case, the independent components of $(𝐖, W_0)$ trace out a cylinder of radius $\sqrt{W^2 - α {W_0}^2}$.

For tachyons, the invariant $W^2 - α {W_0}^2$ may be positive, zero or negative; and the independent components of $(𝐖, W_0)$ trace out a one-sheeted hyperboloid (if positive), a double cone (if zero) or a two-sheeted hyperboloid (if negative).

So, what corresponds to "spin" for tardyons, morphs into "cylindrical" something or another when you pass through the boundary case of luxons, into something totally different, involving hyperboloids or double-cones, by the time you get to tachyons. I have no idea what it represents or corresponds to; and I don't think any consensus has ever been arrived at in the literature, either, on what it ought to represent.

I worked much of this out in more explicit detail here

What's the physical meaning of the statement that "photons don't have positions"?

in the process of addressing the issues of what corresponds to a "spin-orbit decomposition" and "position operator" for helical luxons. So, you might consider this an addendum to that earlier reply.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.