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I know that the spacetime interval is the analog of the (square of the) Euclidean distance in spacetime. Also, I understand that it is an invariant quantity and determines spacetime's causal structure. Whether timeline intervals are positive and spacelike intervals negative (or vice versa) is a matter of convention. However, the light cone structure of spacetime is not a matter of convention. It is a real thing. And the interval measures something in relation to that real thing. However, it is still not exactly clear to me what it measures and how.

For illustration, assuming the (+, −, −, −) convention, consider the following cases:

  • Case 1: spacetime events A and B are spacelike separated, and the spacetime interval between them is −7.
  • Case 2: spacetime events C and D are spacelike separated, and the spacetime interval between them is −5.
  • Case 3: spacetime events E and F are timelike separated, and the spacetime interval between them is 5.
  • Case 4: spacetime events G and H are lightlike separated, and the spacetime interval between them is 0.

I have a few interconnected clarificatory question about the above:

  1. A and B (Case 1) are further apart than C and D (Case 2). Does that mean that A and B are somehow more causally disconnected than C and D? Or is causal disconnection an all-or-nothing value (meaning two events are either causally disconnected or not)?
  2. Are C and D (Case 2) equally close as E and F (Case 3), given that the absolute value of both intervals is equal? More specifically, is there any way to compare the magnitudes of intervals of different types, or are they incommensurable?
  3. Are G and H (Case 4) maximally close, and thus closer together than any of the other pairs of events mentioned? If an interval of 0 does not represent being maximally close, then what makes it 0?
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    $\begingroup$ The spacetime square-interval is the analogue of the square-of-the-Euclidean-distance. $\endgroup$
    – robphy
    Jul 6 at 1:41
  • $\begingroup$ Thanks, that is right. I edited the post accordingly. $\endgroup$
    – Maverick
    Jul 6 at 11:25

6 Answers 6

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With your metric, the spacetime interval is given by

\begin{equation} \Delta s^{2} = c^{2}\Delta t^{2} - \Delta x^{2} - \Delta y^{2} - \Delta z^{2} \end{equation}

which, as you've already mentioned, is Lorentz invariant and its value is therefore independent of frame of reference.

If $\Delta s^{2} > 0$ then the events are timelike, meaning two observers could communicate with one another. If $\Delta s^{2} < 0$ then the events are spacelike and can never communicate with one another - the distance between them is greater than the distance light can travel in the time interval between the two events, $\Delta s^{2} = c^{2}t^{2} - d^{2} < 0 \implies |d| > c\Delta t$. This is what you have referred to as causal disconnection. There aren't various degrees of causal disconnection, you either can causally effect an event or you can't. This is why the invariance of the spacetime interval is necessary - you don't want to boost into a new frame to find that two events are now causally connected!

The special case of $\Delta s^{2} = 0$ is when $|d| = c\Delta t$, i.e. the separation of events is such that they can only communicate using light. They are causally connected.

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  • $\begingroup$ Many thanks! That is very helpful in clarifying causal disconnection. But it does not answer my full question. Could you please also say something about my other concerns: namely, incommensurability and closeness? $\endgroup$
    – Maverick
    Jul 5 at 10:33
  • $\begingroup$ The 'maximally close' question you have is what I've answer in the final two sentences, namely that $\Delta s^{2} = 0$ implies that the distance between two events is such that they can only communicate using light speeds. As for the question on 'incommensurability' I would generally avoid comparing the magnitudes of spacetime intervals if they are of different signs. This isn't may area of research, so someone may correct me, but I don't know of a situation where we'd be enormously interested in comparing two spacetime intervals of opposite sign but equal magnitude $\endgroup$
    – Niall
    Jul 5 at 10:50
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The following spacetime diagram might help. It's drawn on "rotated graph paper" so that it's easier to see the tickmarks along segments and to visualize the square-interval between events. (The grid is based on the spacetime-paths of the light-signals in an inertial light-clock at rest in this diagram. It turns out that we are working with light-cone coordinates, using the eigenbasis of the Lorentz boost transformations in $(1+1)$-Minkowski spacetime.)

RRGP-robphy-causalDiamonds

For concreteness, consider the following events with $(t,x)$-coordinates:
$A=(0,0)$, $B=(3,4)$, $D=(2,3)$, $F=(3,2)$, $H=(2,2)$, and $F_2= (5.25,-4.75)=(\frac{21}{4}, -\frac{19}{4})$, which can be verified by counting up diamond diagonals vertex-to-opposite-vertex along the vertical and horizontal directions.

Observe, for instance, that the square-interval of $AF$ is $$\Delta s_{AF}^2=(F_t-A_t)^2-(F_x-A_x)^2=(3-0)^2-(2-0)^2= (9)-(4)=5.$$ Now construct the "causal diamond of $AF$" for the timelike-segment $AF$ by drawing the light-cones of $A$ and $F$ and finding the intersection of the causal future of $A$ and the causal past of $F$ (the blue-shaded diamond with timelike-diagonal AF).
Physically, these are the events that can be influenced by $A$ and then can influence $F$.

  • Note that the number of "light-clock diamonds" in the "causal diamond of $AF$" (the "area") is $5$.
  • Note that for another event $F_2$ on the same hyperbola as $F$ with center at $A$, the area is still $5$, as expected, since $F_2$ and $F$ are related by a boost about event $A$.

By similar constructions, the areas of the causal diamonds can be counted off.

  • For the spacelike-segment $AB$, the diamond has area 7. Since $AB$ is spacelike, we can assign the square-interval to be minus-the-area of the causal diamond: $-7$. (For a physical interpretation, it appears one has to determine the timelike-diagonal of the causal diamond where $AB$ is the spacelike-diagonal, then use the interpretation of the causal diamond of the timelike-diagonal (akin to $AF$).)

  • For the lightlike-segment $AH$, the diamond has area 0.
    Since $AH$ is a causal relation from $A$ to $H$, the set of events $E$ that can be influenced by $A$ and then influence $H$ is non-empty... but the set has measure zero.

(The details and motivation of this approach are in my paper
"Relativity on Rotated Graph Paper"
Am. J. Phys. 84, 344 (2016) https://doi.org/10.1119/1.4943251 ;
see also an early draft at https://arxiv.org/abs/1111.7254 .)


So, my answer to your question,

What does the spacetime interval measure?

is that, in $(1+1)$-Minkowski spacetime, the square-interval for a displacement $AF$ is a measure of the area of the causal diamond associated with $AF$. When $AF$ is causal, the diamond can be interpreted as the set of events that can be influenced by $A$ and then influence $F$.

(In $(3+1)$-Minkowski spacetime, the square-interval is associated with the spacetime-volume of the causal diamond, together with some combinatorial factors. One can find a $(1+1)$-slice of the causal diamond to return to a $(1+1)$-case.)

Since the terms "closeness" and "causal disconnectedness" are not well defined, I can't say more about those notions.


UPDATE: (to address the comment)

Some details (taken from my paper referenced above):

Instead of rectangular coordinates $(t,x)$,
we can use light-cone coordinates $[u,v]$, where $u$ increases along $\nearrow$ and $v$ increases along $\nwarrow$, with $$u=t+x\qquad v=t-x$$

So, $F=(3,2)$ can be expressed as $F=[5,1]$.
Note that $$uv=(t+x)(t-x)=t^2-x^2$$ which shows the connection between signed-area and square-interval.

Thus, $A=(0,0)$, $B=(3,4)$, $D=(2,3)$, $F=(3,2)$, $H=(2,2)$, and $F_2= (5.25,-4.75)=(\frac{21}{4}, -\frac{19}{4})$,
can be expressed as
$A=[0,0]$, $B=[7,-1]$, $D=[5,-1]$, $F=[5,1]$, $H=[4,0]$, and $F_2= [0.5,10]$, which can be seen from the diagram.

So, Diamond(A,H) has area $(4-0)(0-0)=0$, and Diamond(A,D) has area $(5-0)(-1-0)=-5$... verified by counting clock-diamonds (grid boxes).

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  • $\begingroup$ Thank you, that is very helpful! Regarding "causal disconnectedness": in my usage, for any two spacetime events A and B, A and B are causally connected iff A causally influences B. Causal disconnection is the negation of that. Regarding "closeness": my question was an aspect of my question about commensurability. I still don't understand whether the different kinds of spacetime intervals can be meaningfully compared in terms of magnitude. For example, just looking at the graph you posted indicates that (A, H) and (A, D) are the same size, but I have a feeling this can't be right. $\endgroup$
    – Maverick
    Jul 6 at 7:33
  • $\begingroup$ @Maverick Diamond(A,H) has zero area, whereas Diamond(A,D) has signed-area -5. The diamond-edges are lightlike $\endgroup$
    – robphy
    Jul 6 at 12:42
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  1. A and B (Case 1) are further apart than C and D (Case 2). Does that mean that A and B are somehow more causally disconnected than C and D? Or is causal disconnection an all-or-nothing value (meaning two events are either causally disconnected or not)?

Its all-or-nothing value

  1. Are C and D (Case 2) equally close as E and F (Case 3), given that the absolute value of both intervals is equal? More specifically, is there any way to compare the magnitudes of intervals of different types, or are they incommensurable?
  2. Are G and H (Case 4) maximally close, and thus closer together than any of the other pairs of events mentioned? If an interval of 0 does not represent being maximally close, then what makes it 0?

Depends.

You can easily compare two vectors that are colinear as one is multiple of the other. You can then look at absolute value of this multiple and check whether it is bigger or smaller than one. I.e. $$u=\lambda v \land |\lambda| >1\Rightarrow u>v.$$ In ordinary space, you have rotation which is length preserving operation. You can then rotate any general vector $v'$ so that it is colinear with $u$ and compare them.

In spacetime however, things get a little complicated. Metric preserving operations are lorentz transformations and there are no lorentz transformation that would mix vectors from null/time-like/space-like categories.

However, you can take absolute value of spacetime interval. This will define equivalence relation between vectors that have the same absolute value of spacetime interval. You can then compare two position 4-vectors $u$ and $v$ by choosing one as a reference and using the other vector to pick vector colinear with the first which is of the same equivalence class as the second vector.

Still bigger complication are null vectors. These form a three dimensional vector space, so you can have two noncolinear null vectors $u_0, v_0$. To compare them, you would need to transform one of them to be colinear with other, but there is no structure in spacetime that would let you prefer one over other.

So while the closeness of position 4-vectors (two events) can be extended to include both time-like and space-like categories, the null vectors are simply uncomparable, unless they happen to be colinear.

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This will not answer all your questions but hopefully it will provide some intuition. The contour given by $\Delta s^2=1$ for a timelike interval can be parametrized using $(ct,x)=\gamma\,(1,\beta)$, i.e. it is the Lorentz transform of $(ct,x)=(1,0)$ for general $\beta$. This gives the following interpretation of this curve. Consider all possible intertial observers which pass through the origins. So this is like a bunch of spaceships all passing through the origin at $t=0$, each spaceship having a different velocity. The position of these spaceships at their percieved $ct=1$ is the curve given by $\Delta s^2=1$. We can also frame it as follows: it is the locus of all points which are a proper time $c\tau=1$ from the origin but generalized for all possible observers. Written like this it can be generalized to Galilean relativity. For Galilean relativity the 'proper time' is always the same and so the spacetime interval becomes just the time interval between two events.

enter image description here

Now we can repeat this for a spacelike interval $\Delta s^2=-1$, as shown below. This time the curve given by all points which are a proper distance $-1$ away, generalized for all possible observers. Again, for Galilean relativity the 'proper distance' is always the same and so the spacetime interval becomes just the distance.

enter image description here

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This is another answer for the questions below your cases. To answer your questions directly, 1) if two events are spacelike-separated, it doesn't matter how far apart they are. They are just spacelike-separated. They can't communicate or causally affect each other. You can't be more unable to do something. You're just unable to. 2) and 3) are a little tricky to think about. In more mathematical terms, the distance between two points in a given space is determined by what's called a metric function. This function, call it $g$, satisfies some very natural properties: 1- the distance between a point and itself is zero ($g(x,x)=0$)

2- the distance between any two distinct points x and y is greater than zero ($g(x,y)>0$)

3- Symmetry: distance between x and y is the same as distance between y and x ($g(x,y)=g(y,x)$)

4- Triangle inequality: for any three points x, y, and z, $g(x,z)< g(x,y)+g(y,z)$

Those are all natural and intuitive properties of the concept of distance. But notice how spacetime immediately manifests itself differently from that concept. You immediately see that distance in spacetime is allowed to be negative, and that two distinct points can be separated by a zero-length curve. This is why distance in spacetime is very different from distance in Euclidean space. Spacetime is what's called a "Pseudo-Riemmanian manifold" which is a kind of space with these bizzare properties. That's why it's not straightforward to understand what it means when the interval is zero or negative. But this is where the physics comes in. If you draw a light cone and take the distance between any two points on it, it will be zero, even though there are points that are clearly farther apart than others. But they're only farther apart according to our simple notion of distance, not spacetime's distance. (It truly is fascinating). So the physics tells you that this just means that these points can only communicate by light rays. They are not "maximally close" because this notion of closeness no longer applies here. The same goes for events separated by negative intervals. They're just events that can't communicate at all and can't causally affect each other, even if they "look" the same distance apart on paper. This is only because paper has natural distance properties but spacetime doesn't.

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I'll start from "the light cone is a real thing. And the interval measures something in relation to that real thing". You're right. Special relativity postulates that all inertial observers agree on the speed of light. This is why the light cone the same for all observers and is "a real thing". Since nothing can travel faster than light, then events that are causally connected are determined by the light cone. Points inside the future light cone of point A are points that can be influenced by A. Points in the past light cone of A are points that can influence A. Points ON the light cone of A are points which can send or receive light signals to or from A. And finally, points outside the light cone of A are points that can neither influence nor be influenced by A and cannot communicate with A. And all observers agree on that. This is what the spacetime interval measures. It tells you whether two events are causally related ($\Delta s^2 >0$) or not causally related ($\Delta s^2 <0$) or whether they can communicate by light signals only ($\Delta s^2 =0$). That's what the sign of the interval tells you. The magnitude of the interval isn't as easy to interpret. You can have two events separated in time but not in space, and two other events separated in space but not in time such that $\Delta s^2$ for the two events is the same. You wouldn't be able to distinguish which is which based only on the magnitude of the interval. All we can say that if $\Delta s^2$ between events A and B is less than $\Delta s^2$ for events C and D, then that means that A and B are closer in spacetime to each other than C and D. It's helpful to think about distance in spacetime as a whole, not just spatial or temporal separation because they can't be inferred from the magnitude of the interval. I think that adresses your cases. Please leave a comment if you have any further questions or things you want to further clarify.

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