The spacetime interval in special relativity, $ds$, is defined as $$ ds^2=c^2dt^2-dx^2-dy^2-dz^2 $$ with the $(+,-,-,-)$ Minkowski sign convention. The value of $ds^2$ can be positive, zero, or negative depending on whether the displacement is timelike, null/lightlike, or spacelike, respectively. However, now I am asking about the value of $ds$ itself, without the square over it. When is $ds$ greater than zero, and when is it negative? I have already know the formula $ds=\pm \sqrt{(ds)^2}$. But I don't know whether I should put $+$ or $-$ before the square root. I don't think both + and - can be OK simultaneously for ONE particle.

This is important because if we let the proper time be $\tau$, then from the invariance of the interval, $$ ds^2=c^2d\tau^2. $$ But I just cannot tell whether $ds=cd\tau$ or $ds=-cd\tau$. So I feel it is necessary to know the sign of $ds$.


In this answer, I'm using the $+---$ signature.

Most of the spacetimes we deal with in GR as possible realistic models for our universe have certain special properties. One of these properties is that the spacetime is time-orientable. Time-orientability means that: --

  1. It's possible to define an arrow of time at every point in the spacetime, i.e., at each point we have a definition of which light cone is the future light cone and which is the past light cone.

  2. This definition never changes discontinuously from point to point.

This is closely analogous to the notion of orientability in geometry with a Riemannian signature.

When a spacetime is time-orientable, then choosing an orientation allows us to define $ds>0$ for infinitesimal spacetime displacement vectors that lie in the future light cone, and $ds<0$ for those that lie in the past (in $+---$ signature).

In $-+++$ signature, we have similar ideas, but $ds$ is not the same as the proper time $d\tau$.

  • $\begingroup$ Could you give an example of such an orientation? $\endgroup$ – Ma Joad Feb 23 at 22:28
  • $\begingroup$ @MaJoad: For example, in Minkowski space, we can choose the future light cone to be the one lying in the positive $t$ direction. $\endgroup$ – Ben Crowell Feb 24 at 2:10

OP wrote (v3):

The value of $ds^2$ can be positive, zero, or negative depending on whether the displacement is timelike, null/lightlike, or spacelike, respectively.

Well, it follows by taking the square root that $ds=\pm \sqrt{(ds)^2}$ is real, zero, or imaginary, respectively. Whether to use the $+$ or $-$ branch of the square root depends on the context/specifics/conventions of the physical set-up.

  • $\begingroup$ I am afraid that that is not what I have asked. I have already written in my question that I know your formula $ds=\pm \sqrt{(ds)^2}$. But I don't know whether I should put $+$ or $-$ before the square root. I don't think both can be OK simultaneously in a particular situation. $\endgroup$ – Ma Joad Feb 23 at 13:29
  • $\begingroup$ @MaJoad The sign is conventional, so the answer depends on the conventions you are using. $\endgroup$ – Chris Feb 23 at 19:28
  • $\begingroup$ @Chris Could you give an example of such convention? My book is not clear about it. $\endgroup$ – Ma Joad Feb 23 at 22:33

$ds^2$ is really a shorthand for $d\vec s\cdot d\vec s$, which is shorthand for $g_{ab}ds^a ds^b$.

Thus, the real quantity to consider is not a scalar, but the displacement-vector $d\vec s$.
And vectors aren't "signed" (like scalars are) and can't be ordered (like scalars can).
Is the vector $(3\hat x - 4\hat y)$ positive or negative or zero? It's neither.
Is the vector $(3\hat x - 4\hat y)$ greater, less than, or equal to $(4\hat x - 3\hat y)$ ? It's neither.
(When one uses $-\vec v$, it means the vector you add to $\vec v$ to get $\vec 0$.)

What can be signed?

  • The magnitude $\|d\vec s\|=\sqrt{d\vec s\cdot d\vec s}$ is a nonnegative scalar, which is signed (here, zero or positive). [In some treatments in relativity, it could be pure-imaginary.]
  • A component along a direction (e.g. $ds_x=d\vec s\cdot \hat x$) is a scalar, which is signed.

So, what I am saying is that "$ds=\pm \sqrt{(ds)^2}$" doesn't really make sense (or is ambiguous)... unless you are asking about one of the above and maybe should have better notation.

The issue is really about geometry in general, not specific to relativity.

Suppose we are dealing with Euclidean space.
Given only "$ds^2=1$" (that is $d\vec s\cdot d\vec s=1$), then $d\vec s$ is a unit vector.
If you also know that $d\vec s$ has only an $x$-component, then either $d\vec s= \hat x$ or $d\vec s = -\hat x$.
If further $ds_x\equiv d\vec s\cdot \hat x \ >0$, then $d\vec s= \hat x$, otherwise since then $ds_x<0$, then $d\vec s= -\hat x$.

For your specific relativity problem,
the real question is therefore
Given $d\vec s\cdot d\vec s=c^2 d\tau^2$, is the spacetime-displacement vector $d\vec s=c d\tau \ \hat \tau$ or $d\vec s=-c d\tau \ \hat \tau$?
As a purely mathematical problem, you need more information. Both are acceptable "solutions".
Invoking the physical input that massive particles have future-timelike 4-velocities (which point into the future-lightcone) [that is $ds_{\tau}>0$ with the usual conventions], then you can conclude $d\vec s=c d\tau \ \hat \tau$.


Your question is a question of general geometry: If you have got a square (room) of 49 m2, what is the value of the length of the side of the square? The answer (and this is the answer to your question): in general, the value of a length is positive and intervals are specified in positive values. However, for special purposes, we can also imagine to work with positive and with negative lengths. But as long as there are no negative intervals defined, and in case of doubt, the value is positive. By the way, the result is equally positive for spacetime intervals within the future lightcone and within the past lightcone, there is no difference.


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