Velocity addition for tachyons How does the velocity of a tachyon transform under a Lorentz boost? Suppose we only consider motion along the $x$ direction for simplicity. If the velocity of the tachyon is $u$ in the lab frame, what is the velocity $u'$ of the tachyon in a frame moving with velocity $v$ (slower than $c$) relative to the lab frame? Can we still use the formula
$$
u' = \frac{u-v}{1-\frac{uv}{c^2}}
$$
 A: Yes, the velocity addition formula
$$u' = \frac{u-v}{1-\frac{uv}{c^2}}$$
is valid also for tachyons.
See the article "On Tachyon Lorentz Transformation"
(American Jornal of Physics 37, 1281 (1969)).
Notice also that it correctly predicts $|u'| > c$ if $|u| > c$.
A: The short answer is that it is valid, but that misses a lot of subtleties.
Using three-vectors for velocity in special relativity is fairly unnatural even for ordinary speeds, and it's worse for tachyons, since their three-velocity can be "infinite" and, in a certain sense, "time reversed" (see below), and those situations can't be described properly with three-vectors.
It's better to use a four-vector to represent speed. The four-velocity at a point is just a tangent vector to the worldline at that point. A three-velocity $(u_x,u_y,u_z)$ is equivalent to an unnormalized four-velocity of $(1,u_x,u_y,u_z)$, or any positive scalar multiple of that.
All four-vectors transform in the same way. Under a Lorentz boost in the $x$ direction by $v, |v|<1$ (using a three-velocity $v$ in anticipation of recovering the formula quoted in the question), that four-velocity becomes $(γ(1 - vu_x), γ(u_x - v1), u_y, u_z)$. You can renormalize that to make the time component equal to $1$ again, getting
$$\left( 1,\; \frac{u_x-v}{1-vu_x},\; \frac{u_y}{γ(1-vu_x)},\; \frac{u_z}{γ(1-vu_x)} \right)$$
and if you drop the $1$, and add explicit factors of $c$, you get a generalization of the formula from the question (which is the special case where $u_y=u_z=0$).
Nothing in that derivation depends on the speed $u$ being light speed or less as such. However, when you renormalize the boosted four-vector, you have to divide by the time component, and when the speed is tachyonic (and only when it's tachyonic), that time component may be zero. In that case, you can represent the speed by a "formally infinite" three-velocity that has an infinite magnitude and a direction like that of any nonzero vector. However, this is mathematically suspect, and really the only reason it makes sense is because it's equivalent to the four-vector form. There is also no natural reason for the velocity to be treated as "infinite" in the first place. Whether a velocity is "infinite" in this sense is frame-dependent and hence not really a property of the velocity.
The time component can also be negative, and because of that, when you divide by it, you lose sign information. (This is when "the tachyon appears to travel backwards" as you observed in a comment.) Whether the sign information matters depends on the nature of the hypothesized tachyons. If tachyons have an arrow of time, and the four-velocity points in the direction of increasing proper time (as one normally imagines it does for sublight speeds), then the direction of the arrow of time is lost in the conversion. However, if tachyons have an arrow of time, then you can use them to send signals into your own past (using a "tachyonic antitelephone"), so perhaps they can't have one and the loss of sign information doesn't matter.
Ultimately, though, the three-velocity is just a poor way of thinking about the velocity of tachyons, even though it can be adapted to that purpose. It's better to stick with the four-velocity for both subluminal and superluminal speeds.
A: The question is generic, and the issue of tachyons is a red herring (and, in fact, the issue of Relativity itself is a bit of a red herring - as you're about to see). What you're actually asking is what the transform of a line in space-time is.
Suppose, without loss of generality, the line is given in parametric form by $(x,t) = \left(x_0 + λ a, t_0 + λ b\right)$, for $λ ∈ ℝ$, where we assume $(a,b) ≠ (0,0)$. This line is viewed, in temporal terms, as a motion along the $x$ axis at a speed $u$, where $a = u b$. If $b = 0$, then the speed is infinite: $u = ∞$, and the line is viewed, in temporal terms, not as a motion at all, but as an actual line in space that exists at an instant - namely: at time $t_0$.
Since, we're only asking about speeds (and slopes), then without any real loss of generality, assume also that the line passes through the origin $(0,0)$, with $\left(x_0, t_0\right) = (0,0)$.
The action of the coordinates under a boost by a velocity $v$ in a direction parallel to the $x$ axis is given by:
$$(x, t) → \left(\frac{x - v t}{\sqrt{1 - α v^2}}, \frac{t - α v x}{\sqrt{1 - α v^2}}\right).$$
This is generic to a large variety of cases, as follows:

*

*$α = 0$: Geometries where infinity is an absolute speed; i.e. the geometry of non-relativistic theory.

*$α > 0$: Geometries that possess a finite non-zero absolute speed $c = 1/\sqrt{α}$; i.e. the geometry that Special Relativity is subject to.

*$α < 0$: Euclidean geometry, where $t$ is actually a spatial dimension and not a time dimension at all, and speeds are just slopes.

The transform is subject to the following limitation:
$$1 - α v^2 > 0,$$
which only has effect in the cases $α > 0$, restricting the speed of the boost transform to $|v| < c$.
So, $u$ has no restriction, but $v$ does - when in the case $α > 0$ of Relativity.
For Euclidean geometry $α < 0$, the transforms - when written in this form - describe only rotations between -90 degrees and +90 degrees, and are only a subset of the full range, since you can flip over by 180 degrees, but we won't worry too much about that; since this case is not central to our question here.
Under transform, the equation of the line becomes
$$(x,t) = \left(λ a, λ b\right) → \left(λ \frac{a - v b}{\sqrt{1 - α v^2}}, λ \frac{b - α v a}{\sqrt{1 - α v^2}}\right) = \left(Λ (a - v b), Λ (b - α v a)\right),$$
where $Λ = λ/\sqrt{1 - α v^2}$. This describes a line with velocity
$$u' = \frac{a - v b}{b - α v a}.$$
In the case $b ≠ 0$, where $u = a/b$ is finite and non-zero, then upon division of both top and bottom by $b$, this becomes:
$$u' = \frac{u - v}{1 - α u v}.$$
In the case $b = 0$, where $u = ∞$, it becomes:
$$u' = \frac{1}{-αv}.$$
For the non-relativistic case, $α = 0$, it stays infinite: $u' = ∞$, otherwise, for Relativity, where $α = 1/c^2$, it becomes $u' = -c^2/v$.
Finally, in the case where $α u v = 1$, we have $u' = ∞$. That includes the above-mentioned case for non-relativistic theory. For Relativity, with $u v = c^2$, since $|v| < c$, then this may only apply if $|u| > c$ - i.e. to tachyons.
So, tachyons are not "things moving through space in time" at all, but exist as just lines in space that exist (in at least one frame of reference) at an instant.
The non-relativistic version of this corresponds to the instantaneous form of what was informally termed as a "line of force", as it conveys an instantaneous transfer of impulse across space.
You can generalize this further to include the "Carrollian" and "static" universes, by adding in a second deformation coefficient $β$ and generalizing the transform to:
$$(x, t) → \left(\frac{x - β v t}{\sqrt{1 - αβ v^2}}, \frac{t - α v x}{\sqrt{1 - αβ v^2}}\right).$$
Here, light speed is $c = \sqrt{β/α}$, when $αβ > 0$, with the cases already considered corresponding to $β = 1$. The corresponding transform then, instead, becomes
$$u' = \frac{a - β v b}{b - α v a}.$$
For the Carrollian and static universes, $β = 0$, one has:
$$u' = \frac{a}{b - α v a}.$$
It's actually better, then, consider inverse speeds, instead, and write
$$\frac{1}{u'} = \frac{1}{u} - α v.$$
Zero speed $u = 0$ (i.e. $a = 0$) transforms into zero speed: $u' = 0$: in both the static and Carrollian universes, i.e. 0 is an absolute speed. In the static universe, one also has $α = 0$, and the transform is just $u' = a/b = u$, so that all speeds are absolute in the static universe. Otherwise, for the Carrollian universe (where $α ≠ 0$), infinite speed $u = ∞$ and $b = 0$, would transform as $u' = 1/(-αv)$, as in the case of Relativity.
