# If the speed of causality changes, could you go FTL?

In the middle of some research, I reached a sort of confusion that I’d like to sort out. In flat space FTL is impossible, because in a Minkowski metric,

$$\mathrm{d}s^2=c^2 \mathrm{d}t^2-\mathrm{d}x^2-\mathrm{d}y^2-\mathrm{d}z^2,$$

any

$$v=\frac{\mathrm{d}\sigma}{\mathrm{d}t}>c$$

makes $$\mathrm{d}s^2<0$$, which isn’t allowed for anything (I use $$\mathrm{d}\sigma^2= \mathrm{d}x^2+ \mathrm{d}y^2+ \mathrm{d}z^2$$ for convenience). But in the metric

$$\mathrm{d}s^2=c^2 \mathrm{d}t^2-0.5\mathrm{d}x^2-\mathrm{d}y^2-\mathrm{d}z^2,$$

for example, it’s trivial to find two events which are separated by $$\mathrm{d}\sigma>c \mathrm{d}t$$ and where $$\mathrm{d}s^2>0$$, implying a causal connection between otherwise-distant points; the original bar that was placed on velocity due to too-high velocity making the trajectory curve spacelike then is lifted. This includes metrics like the exteriors of some black hole metrics, so it might seem.

But does that mean you could actually go FTL, or would special relativity still stop you? One way of saying it’s still impossible is by thinking about an onlooker stationary relative to a particular frame $$\mathcal{F}$$ while another tries to use this spacetime to their advantage to travel FTL in $$\mathcal{F}$$. Each observer would see the other’s velocity exceed $$c$$, and then their mass and whatnot become complex due to special relativity.

So in this case it seems like general relativity might allow distant points to be causally connected, and like it might allow FTL velocities, but special relativity seems to get in the way. Is there a resolution for this?

• What does FTL stand for?
– John
Commented Jun 5 at 19:51
• Faster Than Light (travel/communication) Commented Jun 5 at 20:09
• $\tanh^2(dx)$ is not defined. The best you could do is $\tanh^2(x)dx^2$. Commented Jun 5 at 20:11
• Good point. Updated it with something that actually makes mathematical sense. Commented Jun 5 at 20:53
• A real quadratic form is determined up to isomorphism by its signature. Commented Jun 6 at 1:21

What you are describing as a "causality change" is simply the concept of a metric in general relativity. Having a metric on a spacetime is equivalent if you want to the assignment of a light cone at each point in spacetime. Certainly it is possible to have different light cones at different points in spacetime. For Minkowski $$$$ds^{2} = c^{2}dt^{2} -dx^{2} -dy^{2} - dz^{2} = 0,$$$$ by focusing on planes with $$y,z = const.$$ or $$x,z = const.$$ or $$x,y = const.$$ $$$$\left(\frac{dx}{dt}\right)^{2} = c^{2} \quad \mathrm{or} \quad \left(\frac{dy}{dt}\right)^{2} = c^{2} \quad \mathrm{or} \quad \left(\frac{dz}{dt}\right)^{2} = c^{2}.$$$$ As you can see, in the Minkowski case, the opening of the light cone is the same at every point and in every direction. In the example you gave, the aperture of the light cone is the same at every point in spacetime but varies as the direction varies $$$$ds^{2} = c^{2}dt^{2} - \frac{1}{2}dx^{2} -dy^{2} - dz^{2} = 0,$$$$ by looking at the planes with $$y,z = const.$$ or $$x,z = const.$$ or $$x,y = const.$$ $$$$\left(\frac{dx}{dt}\right)^{2} = 2c^{2} \quad \mathrm{or} \quad \left(\frac{dy}{dt}\right)^{2} = c^{2} \quad \mathrm{or} \quad \left(\frac{dz}{dt}\right)^{2} = c^{2}.$$$$ Of course, the definition of FTL travel in General Relativity is more subtle, since you can always write a metric that has the causal structure you want. For example the Alcubierre's metric which is interpreted as a warp drive, does not violate special relativity. So to answer your question: no, special relativity is not sufficient to elminate such possibilities. However, there are problems with metrics exhibiting these behaviors: violation of energy conditions and horizons over which the stress-energy tensor of quantum field theories on curved-background cannot be renormalized leading to uncontrolled particle creation.
In special relativity, one cannot simply use the regular three-dimensional velocity $$v$$. Instead, one has to use the four-velocity $$u^\mu = \frac{dx^\mu}{d\tau}$$ where $$\tau = t\sqrt{1-\frac{v^2}{c^2}}$$ is the time measured by an observer moving with velocity $$v$$ and $$x^\mu = (ct,\vec x)$$ is the four-dimensional position in spacetime. Immediately we see that for $$v = c$$ things get funky. I am not entirely happy with your reasoning why the worldline of an observer moving with $$v>c$$ would be spacelike, you cannot just naively "divide by $$dt^2$$". The fact that something which can be given a correct physical interpretation comes out seems to me as pure chance. It also makes no sense to claim that $$d\sigma > c\;dt$$: These are differential forms which are used as basis (co)vectors. They are not ordered in any way, and their "sizes" cannot be compared as they are. To do this, you need the metric which then defines a quadratic form and assigns a scalar to a pair of vectors or covectors. Lastly, the metric you give is literally just the regular Minkowski metric with a rescaled $$x$$-coordinate (assuming that you meant to write $$dx^2$$). All the usual results of special relativity hold for this metric, including the fact that spacetime points with a spacelike separation cannot be causally connected.
• Okay, I think I should clarify given what you said in the edit: if $c_f=299792458$ is the speed of light in flat space, which is constant, and the speed of causality changed to be higher (like $2c_f$), would that allow your velocity to get higher than $c_f$ (but of course still not higher than the speed of light)? Of course the speed of light and causality are one in the same but if you can increase the speed of causality then it would seem as though you could raise the speed limit that you could reach, no? Commented Jun 6 at 0:36