What is the meaning of the negative sign in $\Delta s^2 = \Delta x^2 + \Delta y^2 + \Delta z^2 - (c\Delta t)^2$?

In the equation of the spacetime interval formula $$\Delta s^2 = \Delta x^2 + \Delta y^2 + \Delta z^2 - (c\Delta t)^2$$ is there meaning for the minus sign before the $$(c\Delta t)^2$$ or is it just a pure mathematical stuff?

Another question, sometimes I see the formula as $$\Delta s^2 = (c\Delta t)^2 - \Delta x^2 - \Delta y^2 - \Delta z^2$$ so why it have two different forms?

The relative minus sign between the $$x,y,z$$ and $$t$$ terms is fundamental. This minus sign is the reason we can't "turn around and go the opposite direction" in time, like we can in space.

The quantity $$\Delta \tau$$ defined by $$(c\Delta \tau)^2 = (c\Delta t)^2 - \big( (\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2 \big) \tag{1}$$ is the proper time experienced by an object moving in such a way that its spatial coordinates change by $$\Delta x,\,\Delta y,\,\Delta z$$ during the coordinate-time interval $$\Delta t$$, assuming that it is moving inertially throughout that interval. The proper time is defined only when the right-hand side of (1) is non-negative, as it must be for the motion of any physical object. A spacetime interval for which (1) is positive is called timelike, and a spacetime interval for which (1) is zero is called null. The null case corresponds to something moving at the speed of light.

The minus sign in (1) therefore imposes a speed limit: nothing can move faster than $$c$$. More accurately, nothing can pass through a locally inertial frame faster than $$c$$. (For clarification, see https://physics.stackexchange.com/q/400458.)

The quantity $$\Delta \ell$$ defined by $$(\Delta \ell)^2 = \big( (\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2 \big) - (c\Delta t)^2 \tag{2}$$ is the proper distance between two points. It is defined only when the right-hand side is non-negative. An interval for which the right-hand side of (2) is positive is called spacelike.

The metric of spacetime, which defines time and geometry, can be implicitly specified by writing down either the equation for proper time (as in equation (1)) or by writing down the equation for proper distance (as in equation (2)). The special cases (1) and (2) are for flat spacetime, the arena of special relativity. For curved spacetime, the expressions can be more complicated; but the right-hand sides of the proper-time and proper-distance equations are always each other's negatives, so the same spacetime metric may be specified either way.

Appendix

This appendix explains the opening statement about why the minus sign means that we can't turn around in time.

Equation (1) is a discrete version of $$(c\,d\tau)^2 = (c\,dt)^2 - \big( (dx)^2 + (dy)^2 + (dz)^2 \big) \tag{3}$$ where $$dt,dx,dy,dz$$ are infinitesimal coordinate-increments along a smooth worldline (curve in spacetime). To represent the history of a physical object, the right-hand side of (3) must be non-negative. The proper time increment $$d\tau$$ is defined only for such worldlines. The claim is that if we parameterize the worldline by expressing the coordinates $$t,x,y,z$$ as smooth functions of a single parameter $$u$$ such that distinct values of $$u$$ correspond to distinct points along the worldline, then $$dt/du$$ has the same sign everywhere along the worldline. (For a timelike worldline, we could use $$u=\tau$$, but using a generic $$u$$ accommodates lightlike worldlines, too.)

Here's a proof. The condition that the right-hand side of (3) must be non-negative is the same as the condition $$\left(c\,\frac{dt}{du}\right)^2 \geq \left(\frac{dx}{du}\right)^2 + \left(\frac{dy}{du}\right)^2 + \left(\frac{dz}{du}\right)^2. \tag{4}$$ Since the functions are smooth (because physical motions are smooth), the only way $$dt/du$$ can change sign is if it is equal to zero somewhere. According to equation (4), it cannot be zero except where $$dx/du$$, $$dy/du$$, and $$dz/du$$ are all zero. But we chose the parameterization $$u$$ to be such that distinct values of $$u$$ correspond to distinct points along the worldline. If the $$x,y,z$$ derivatives are all zero, then this requires $$dt/du\neq 0$$. Therefore, $$dt/du$$ cannot be zero anywhere, so it cannot change sign.

This is how the minus sign in equation (3) prevents a physical object from "turning around" in time, like it can in space.

• "This minus sign is the reason we can't "turn around and go the opposite direction" in time, like we can in space." Would you be kind enough to explain how this follows? – Philip Wood Nov 23 '18 at 22:44
• Dan, this is the clearest exposition of this I have seen to date, and reading it reminds me of why I started hanging around this site in the first place. Thanks for taking the time to post this answer and have a happy holiday. – niels nielsen Nov 23 '18 at 22:51
• @PhilipWood The plus sign in $x^2+y^2$ implies that this quantity is invariant under ordinary rotations: $x\mapsto x\cos\theta-y\sin\theta$, $y\mapsto x\sin\theta+y\cos\theta$. We can go from $\theta=0$ to $\theta=\pi$ to "turn around", reversing the signs of $x$ and $y$. But the minus sign in $(ct^2)-x^2$ means that this quantity is invariant under hyperbolic rotations (Lorentz transf's): $ct\mapsto ct\cosh\theta+x\sinh\theta$, $x\mapsto ct\sinh\theta+x\cosh\theta$, with $\cosh^2\theta-\sinh^2\theta=1$. No value of $\theta$ can change the sign of $t$, so we cannot "turn around" in time. – Chiral Anomaly Nov 23 '18 at 22:54
• @PhilipWood Thanks for the gracious feedback. Actually, I wasn't quite happy with the "proof" in my comment, because it doesn't explain how those abstract transformations relate to the motion of physical objects, and it's unnecessarily specific to flat spacetime. That was bugging my conscience, so I added an appendix to the main answer that addresses your question more directly and that can be generalized to curved spacetime (though I'm still only showing the flat case). – Chiral Anomaly Nov 23 '18 at 23:53
• Thanks again. The key step seems to be, "If the x,y,z derivatives are all zero, then this requires dt/du≠0." Are you saying that unless this were the case, then, locally, we wouldn't have different space-time points corresponding to different values of $u$? – Philip Wood Nov 24 '18 at 0:09

If we measured distance in light-seconds instead of meters, the constant c would be 1, and the metric distance element would simply become

$$Δs^2 = Δx^2 + Δy^2 + Δz^2 - Δt^2$$, or

$$Δs^2 = Δt^2 - Δx^2 - Δy^2 - Δz^2$$

(both forms are equivalent, because multiplying the vector by -1 does not change its squared length)

This metric distance element follows out of Maxwell's equations, which can be written in 4-space as one single equation (I reuse here c for the sake of clarity):

$$((1/c^2) (∂^2/∂t^2 ) - (∂^2/∂x^2) - (∂^2/∂y^2) - (∂^2/∂z^2) )A= μ_0 J$$

wherein $$A=(φ/c,(A_x,A_y,A_z ))$$ is the 4-potential composed of scalar and vector potential and $$J=(ρc,(J_x,J_y,J_z ))$$ is the 4-current density composed of charge and current

This equation is also called the Fundamental Equation of Electrodynamics, and is the 4-space equivalent of Poisson's equation in 3D space:

$$((∂^2/∂x^2) + (∂^2/∂y^2) + (∂^2/∂z^2))φ = -ρ$$

wherein φ is a potential and ρ a source density (or charge).

These equations follow from the general Stokes conservation (or accounting) law:

$$∫_Vdω = ∫_{dV} ω$$

stating that the change of inventory of a quantity dω inside a volume or hypervolume V equals the flow of said quantity ω through the surface or hypersurface dV of said volume.

In its differential form, it yields the law of Gauss:

$$div (ω(x)) = ρ(x)$$

wherein ω is the flowing quantity, and ρ the source density.

Expressing the flow ω as a gradient of a potential φ, one obtains:

$$ω(x)= -grad(φ(x))$$

This yields then Poisson's equation:

$$∆(φ(x)) = div(grad(φ(x))) = -ρ(x)$$

Poisson's equation is a flow-conservation equation in 3-dimensional space of metric signature (+,+,+). The metric distance element therein is

$$Δs^2 = Δx^2 + Δy^2 + Δz^2$$

The Fundamental Equation of Electrodynamics is a flow-conservation equation in 4-space of metric signature (+,-,-,-). The metric distance element therein is

$$Δs^2 = Δt^2 - Δx^2 - Δy^2 - Δz^2$$

Special Relativity is just about flow in 4-space, as is Electrodynamics.

• This doesn't really make much sense. It's true that Maxwell's equations are form-invariant under a Lorentz transformation, but if that is the point of this answer, then it's not being made very clearly. – Ben Crowell Mar 1 at 17:06
• The question was about the meaning of the negative sign in the metric distance element (or of the three negative signs, if one likes). A more profound answer is in my other post physics.stackexchange.com/questions/451979/… – Edgar Mueller Mar 3 at 18:11
• This may also help understanding: The fundamental equation of electrodynamics (FEE) has the metrics of 4-space (+,-,-,-). The simultaneous presence of positive and negative differential terms in a second-order differential operator means that its solutions are waves; in fact, the FEE is a wave equation, which has as well standing-wave solutions (eigensolutions); these represent the massive particles. – Edgar Mueller Mar 3 at 20:42