Do tachyons move faster than light?

I am trying to understand whether or not tachyons travel faster than light. The linked Wikipedia page shows some seemingly contradictory statements, and they are confusing.

For instance, the first sentence states that tachyons "always travel faster than the speed of light" whereas, in a later section, it is claimed that they are actually propagating subluminally. Is it true that tachyons represent faster-than-light particles, or not?

• – Paul Feb 27 '15 at 13:30

A tachyon is a particle with an imaginary rest mass. This however does not mean it "travels" faster than light, nor that there's any conflict between their existence and the special theory of relativity.

The main idea here is that the typical intuition we have about particles -- them being billiard ball-like objects -- utterly fails in the quantum world. It turns out that the correct classical limit for quantum fields in many situations is classical fields rather than point particles, and so you must solve the field equations for a field with imaginary mass and see what happens rather than just naively assume the velocity will turn out to be faster than light.

The mathematical details are a bit technical so I'll just refer to Baez's excellent page if you're interested ( http://math.ucr.edu/home/baez/physics/ParticleAndNuclear/tachyons.html ), but the conclusion can be stated very simply. There's two types of "disturbances" you can make in a tachyon field:

1) Nonlocal disturbances which can be poetically termed "faster than light" but which do not really represent faster than light propagation since they are nonlocal in the first place. In other words, you can't make a nonlocal disturbance in a finite sized laboratory, send it off to your friend in the andromeda galaxy and have them read the message in less time than it would take for light to get there. No, you could at best make a nonlocal disturbance that is as big as your laboratory, and to set that up you need to send a bunch of slower than light signals first. It's akin to telling all your friends all over the solar system to jump at exactly 12:00 am tomorrow: you'll see a nonlocal "disturbance" which cannot be used to send any information because you had to set it up beforehand.

2) Localized disturbances which travel slower than light. These are the only types of disturbances that could be used to send a message using the tachyon field, and they respect special relativity.

In particle physics the term "tachyon" is used to talk about unstable vacuum states. If you find a tachyon in the spectrum of your theory it means you're not sitting on the true vacuum, and that the theory is trying to "roll off" to a state of lower energy. This actual physical process is termed tachyon condensation and likely happened in the early universe when the electroweak theory was trying to find its ground state before the Higgs field acquired its present day value.

A good way to think about tachyons is to imagine hanging several pendulums on a clothesline, one after the other. If you disturb one of them, some amount of force will be transmitted from one pendulum to the next and you'll see a traveling disturbance on the clothesline. You'll be able to identify a "speed of light" for this system (which will really be the speed of sound in the string). Now you can make a "tachyon" in this system by flipping all the pendulums upside down: they'll be in a very unstable position, but that's precisely what a tachyon represents. Nevertheless, there's absolutely no way that you could send a signal down the clothesline faster than the "speed of light" in the system, even with this instability.

tl;dr: Careful consideration of tachyons makes them considerably different from science fiction expectations.

EDIT: As per jdlugosz's suggestion, I've included the link to Lenny Susskind's explanation.

http://youtu.be/gCyImLu0HSI?t=58m51s

• As far as I understand it, your points 1 and 2 are perfectly valid for any elementary particle. You can create entangled e.g. photons, send each in a different corner of your lab; the measurement of one of them changes their other counterpart (or at least your future measurement outcomes from the interaction with the counterpart). So is there nothing special about tachyons in this sense? – dominecf Oct 12 '20 at 15:20

In this answer we'll basically repeat Leandro M.'s good answer for a tachyonic field using formulas. (In contrast, note that the current version of the Wikipedia page mostly discusses the hypothetical notion of a tachyonic point particle, which by the very definition moves faster than the speed of light, which is widely believed to be irrelevant for modern physics, and which we will not discuss further here.)

Let us for simplicity use units where $$c=1=\hbar$$. Consider a spinless relativistic complex scalar field

$$\left(\partial_t^2-\partial_x^2+m_0^2\right)\phi(x,t)~=~0 \tag{1}$$

in 1+1 spacetime dimensions. The real and imaginary components, $${\rm Re}(\phi)$$ and $${\rm Im}(\phi)$$, are independent fields, since the eq. of motion (1) is linear.

The Lagrangian density for a spinless relativistic complex scalar field (1) is

\begin{align} {\cal L}~=~&|\partial_t\phi|^2 - {\cal V}, \cr {\cal V}~=~&|\partial_x\phi|^2+ m_0^2 |\phi|^2.\end{align}\tag{2}

A tachyonic mass $$m_0^2<0$$ corresponds to a potential density $${\cal V}$$ that is unbounded from below, which leads to an instability $$|\phi|\to \infty$$.

Let us perform a spatial Fourier transformation

\begin{align} \tilde{\phi}(p,t)~=~&\int_{\mathbb{R}} \!dx~e^{-ipx} \phi(x,t),\cr \phi(x,t)~=~&\int_{\mathbb{R}} \!\frac{dp}{2\pi}~e^{ipx}\tilde{\phi}(p,t).\end{align}\tag{3}

Then the wave equation (1) becomes a second-order linear ODE

$$\left(\partial_t^2+E_p^2\right)\tilde{\phi}(p,t)~=~0, \tag{4}$$

where

$$E_p~:=~\sqrt{p^2+m_0^2}.\tag{5}$$

The complete solution to the second-order linear ODE (4) is$$^1$$

\begin{align} \tilde{\phi}(p,t)~=~&\sum_{\pm} C_{\pm}(p)e^{\pm iE_pt}\cr ~=~& A(p)\cos(E_pt)+B(p)t~{\rm sinc}(E_pt),\end{align}\tag{6}

where

$$C_{\pm}(p)~=~\frac{1}{2}\left(A(p)\mp i \frac{B(p)}{E_p}\right)\tag{7}$$

are two integration constants. We next analyze various cases.

1. Waves localized in $$p$$-space (and hence non-local in $$x$$-space). Here we assume that the wave packet is almost monochromatic, so that the coefficient functions $$p \mapsto C_{\pm}(p)$$ are sharply peaked around a central momentum. Such a wave packet is hence non-local in $$x$$-space, cf. the Heisenberg uncertainty principle.

1a) Oscillatory case $$p^2>-m_0^2$$. The phase velocity is

$$v_p ~:=~\frac{E_p}{|p|} ~\left\{ \begin{array}{c} > \cr = \cr <\end{array}\right\}~ 1\quad\text{for}\quad m_0^2 ~\left\{ \begin{array}{c} > \cr = \cr <\end{array}\right\}~0. \tag{8}$$

The group velocity is

$$v_g ~:=~\frac{dE_p}{d|p|} ~\stackrel{(5)}{=}~\frac{|p|}{E_p}~=~\frac{1}{v_p} ~\left\{ \begin{array}{c} < \cr = \cr >\end{array}\right\}~ 1\quad\text{for}\quad m_0^2 ~\left\{ \begin{array}{c} > \cr = \cr <\end{array}\right\}~0. \tag{9}$$

The group velocity formula (9) is derived under the assumption that we may linearize the dispersion relation, i.e. the wave packet is assumed to be localized in $$p$$-space. In the tachyonic case $$m_0^2<0$$, the group velocity is faster than the speed of light.

1b) Exponentially growing/decaying case $$p^2<-m_0^2$$. Such non-travelling solutions (6) are only possible for tachyons $$m_0^2<0$$.

1. Waves localized in $$x$$-space. Assume that for each constant time slice $$t$$, the wave has compact support

\begin{align} {\rm supp}(\phi(\cdot,t))~:=~&\overline{\{ x\in \mathbb{R}|\phi(x,t) \neq 0\}}\cr ~=~&[a_-(t),a_+(t)]~\subset ~\mathbb{R}, \cr a_+(t)~:=~&\sup {\rm supp}(\phi(\cdot,t))~<~\infty,\cr a_-(t)~:=~&\inf {\rm supp}(\phi(\cdot,t))~>~-\infty , \end{align}\tag{10}

of the form of an interval with endpoints $$-\infty. Let us define for later convenience the midpoint and the half-length

\begin{align} c(t)~:=&~\frac{a_+(t)+a_-(t)}{2},\cr b(t)~:=~&\frac{a_+(t)-a_-(t)}{2}~\geq~0, \end{align}\tag{11}

respectively.

   ^ phi
|            _____
|           /     \_______________
|          /    b           b     \
--|---------|-----------|-----------|--------------> x
a-          c           a+


$$\uparrow$$ Fig. 1. A wave $$\phi(x)$$ with compact support $$[a_-,a_+]$$ along the $$x$$-axis. Time $$t$$ is suppressed from the notation.

Up until now the Fourier variable $$p$$ has been real. However, the second-order linear ODE (4) and its solution (6) make sense for complex momentum $$p\in\mathbb{C}$$. We may hence take advantage of complex function theory. The square root (5) has an asymptotic behaviour

$$E_p~\sim~\pm p\quad\text{for}\quad |p|\to\infty.\tag{12}$$

If the compactly supported function $$\phi(\cdot,t)\in {\cal L}^1(\mathbb{R})$$ is integrable, then the corresponding spatial Fourier transform $$\tilde{\phi}(\cdot,t)$$ is an entire function by Lebesgue's majorant theorem. Comparing eqs. (3a) and (10), the ultra-relativistic asymptotic behaviour is heuristically given as

$$\tilde{\phi}(p,t)~\sim~e^{-ia_{\pm}(t)p}\quad\text{for}\quad {\rm Im}(p)\to \pm \infty. \tag{13}$$

A rigorous mathematical characterization$$^2$$ of this spatial Fourier transform is provided by the Paley-Wiener (PW) theorem.

Comparing eqs. (6), (12), and (13), we deduce that the front velocity is generically$$^3$$ the speed of light,

$$\frac{da_{\pm}(t)}{dt}~=~\pm 1, \tag{14}$$

i.e. the endpoints $$a_{\pm}(t)$$ of the compact support move with the speed of light, independently of the mass square $$m_0^2$$. This is because the mass is not important in the ultra-relativistic limit (12). In particular, the support (10) of a position-localized wave packet does not expand faster than the speed of light, not even in the tachyonic case $$m_0^2<0$$.

References:

--

Footnotes:

$$^1$$ The latter form of eq. (6) is manifestly free of the square root ambiguity (5) by using even functions, i.e. the cosine and sinc function. The Fourier-transformed wave $$\tilde{\phi}(\cdot,t)$$ is holomorphic iff the two coefficient functions $$A(\cdot)$$ and $$B(\cdot)$$ are. If the wave $$\phi\in\mathbb{R}$$ is real, then the Fourier-transformed wave satisfied

$$\tilde{\phi}(p,t)^{\ast}~=~\tilde{\phi}(-p^{\ast},t),\tag{15}$$

iff

$$A(p)^{\ast}~=~A(-p^{\ast}), \qquad B(p)^{\ast}~=~B(-p^{\ast}).\tag{16}$$

$$^2$$ Here is a rigorous proof of eq. (14). Assume that $$\phi(\cdot,t\!=\!0)\in {\cal L}^2(\mathbb{R})$$ is (i) square-integrable, and (ii) has compact support

$$-\infty~<~a_-(t\!=\!0) ~\leq ~a_+(t\!=\!0)~<~\infty.\tag{17}$$

[The square-integrability (i) is a technicality to get inside the realm of the Paley-Wiener (PW) theorem. Then by Cauchy-Schwarz's inequality, the function $$\phi(\cdot,t\!=\!0)\in {\cal L}^1(\mathbb{R})$$ is integrable.]

By shifting the $$x$$-axis if necessary, we may assume that the initial support midpoint $$c(t\!=\!0)=0$$ is zero, i.e.

$$\infty~>~a_0~:=~a_+(t\!=\!0)~=~-a_-(t\!=\!0)~\geq~0. \tag{18}$$

In this way we get an initial globally defined holomorphic Fourier transform of exponential type $$a_0$$

$$\forall p\in \mathbb{C}: ~~ |A(p)|~\stackrel{(6)}{=}~|\tilde{\phi}(p,t\!=\!0)| ~\stackrel{(3a)}{\leq}~ Ke^{a_0|p|}, \tag{19}$$

where

\begin{align} K~:=~&\int_{\mathbb{R}}\!dx~ |\phi(x,t\!=\!0)|\cr ~=~&\int_{[-a_0,a_0]}\!dx~ |\phi(x,t\!=\!0)|~<~\infty.\end{align}\tag{20}

[Conversely, the ineq. (19) together with the Paley-Wiener (PW) theorem guarantees that the support

$${\rm supp}(\phi(\cdot,t\!=\!0))~\subseteq~[-a_0,a_0] \tag{21}$$

is inside the interval $$[-a_0,a_0]$$. The proof of eq. (21) is a straightforward exercise in closing an integration contour in the upper or lower half-plane of the complex $$p$$-plane.]

Assuming that the support $${\rm supp}(\phi(\cdot,t\!=\!t_0))$$ remains compact for at least one other time slice $$t_0\neq 0$$, it is then necessary that the coefficient function $$B(\cdot)$$ is an entire function of exponential type

$$\exists L,b_0>0~ \forall p\in \mathbb{C}: ~~ |B(p)|~\leq~ Le^{b_0|p|}.\tag{22}$$

It must be possible to choose $$b_0\leq a_0$$, because else the front velocity would be infinite, which is physically unacceptable.

Combining eqs. (19) and (22) with eq. (6), then for an arbitrary time slice $$t$$, we get a globally defined holomorphic Fourier transform of exponential type $$a_0+|t|$$,

$$\exists M>0~\forall p\in \mathbb{C}: ~~|\tilde{\phi}(p,t)|~\leq~ M e^{|m_0||t|}e^{(a_0+|t|)|p|} .\tag{23}$$

In eq. (23) we have used the triangle inequality

$$|E_p|~\stackrel{(5)}{\leq}~ \sqrt{|p|^2+|m_0|^2}~\leq~|p|+|m_0|. \tag{24}$$

Conversely, the ineq. (23) together with the PW theorem now guarantees that the support (10) is inside the interval

$$[-a_0-|t|,a_0+|t|]~\subset~\mathbb{R},\tag{25}$$

i.e. the front velocity is less or equal to the speed of light, as we wanted to show.

$$^3$$ We assume a generic situation, where the coefficient functions $$C_{\pm}(p)$$ do not vanish for $$|{\rm Im}(p)|\to\infty.$$

• Qmechanic: "basically repeat Leandro M.'s good answer using formulas. [...] spinless relativistic complex scalar field [...]" -- 1: Would you please relate the formulas and results of your answer to @Leandro M.'s remark that "people typically just calculate the effective potential and see if it has an imaginary part, which sidesteps the issue of talking about "particle" states." 2: "1a) [...] The phase velocity is [...] The group velocity is [...]" -- What about calculating signal front velocity? And what about cases "1b" and "2"? – user12262 Feb 27 '15 at 18:21
• The cases 1a and 1b have non-compact support, so the front velocity is not defined in these cases. The front velocity in case 2 was already calculated in the first version (v1) of the answer. – Qmechanic Feb 27 '15 at 20:53
• user12262, if you ask that as question I can give you some more mathematical detail on how to actually compute decay rates in theories with tachyons. The formulas given in this answer justify the statements I asserted without proof: even if fields have imaginary masses superluminal propagation cannot occur. They also show an instability -- if exponentially growing/decaying solutions are allowed, then initial conditions must be carefully tuned to avoid an exponentially growing energy density. When quantum effects are included, you can't do that balancing perfectly and the vacuum decays. – Leandro M. Feb 28 '15 at 5:13
• @Leandro M.: "if you ask that as question I can give you some more mathematical detail" -- Well, the one specific self-contained question which I asked already in commenting on your answer is: "How to determine the momentum of a tachyon, at least in principle?". (Perhaps that's already addressed at PSE; otherwise I may get around asking directly.) Apart from that, Qmechanic's response here makes me wonder whether in these cases it is justified at all to treat signal front velocity as a definite real quantity, "$c = 1$". – user12262 Feb 28 '15 at 18:17
• The fact causality is not violated no matter the sign of $m^2$ is trivial: Causality properties of solutions of linear PDEs are always described by the principal part of the equation. It is $g^{\mu\nu}\nabla_\mu\nabla_\nu$ here, in all cases. The sign of $m^2$ is irrelevant. – Valter Moretti Sep 26 '15 at 21:32

Here is another viewpoint- that tachyons are photons in their own frame of reference. The electrons that we know and love always seem to have the same characteristics, such as spin. Is this because only one spin value is allowed, or do the electrons simply not interact with photons put out by electrons with a different spin to themselves? Notice that when you are listening to one wavelength, you are excluding all the other possible wavelengths, even though they exist. If we could speed up the spin values of an atom, it might then seem to disappear from our known world! If we doubled the spin rate, it might be emitting photons that travel twice as fast as any photon we can intercept! But in its' own frame of reference, tempo would be twice as fast, so it would only record the photons as going at the speed of light. The only way to utilise such physics would be to plan for a trip to a star with antimatter rockets, just as we would in normal space, and add some tempo-boosting machinery. The rocket would seem to disappear to us on Earth, and take just a few years to reach the star, though the astronauts would have aged more than us. So Hyperspace is simply the same space with particles and photons that ignore us.

• -1. This answer bears little resemblance to mainstream physics. – Chris Feb 9 '18 at 7:20
• And is mainstream physics always right? Besides, I thought the principle of particle discrimination was part of mainstream physics- I'm just asking if spin could be another factor. Maybe dark matter is composed of electrons and protons with a different spin value. I also imagine that in some environments, like in black holes, some electrons might get squeezed into becoming one super-electron. Such an electron should have spin higher than 'ordinary' electrons. The maths works out. – Nicholas Gray Feb 12 '18 at 0:37
• Mainstream physics can be wrong, but this site is explicitly for questions and answers regarding mainstream physics. See the meta post on this policy. tl;dr personal theories are not allowed- only theories that have been published in a reputable journal. As for your answer, I very much doubt the maths work out, but if they do you should write them out and publish them, not post them on a Q&A site. – Chris Feb 12 '18 at 1:05
• Tachyons are not mainstream, but they are being discussed here. – Nicholas Gray Feb 12 '18 at 6:50
• The general rule of thumb seems to be that if something is not forbidden, then it exists. So tachyons are (probably) real particles. – Nicholas Gray Feb 12 '18 at 6:51

if tachyon is faster than speed of light it means that the theory that was given by the albert einstien is wrong .because he mentioned that the substance travelling at a speed of light should have infinite amount of energy