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

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?

Answer

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.

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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.