# Why is the space-time interval squared?

The space-time interval equation is this:

$$\Delta s^2=\Delta x^2+\Delta y^2+\Delta z^2-(c\Delta t)^2$$

Where, $\Delta x, \Delta y, \Delta z$ and $\Delta t$ represent the distances along various coordinates according to an observer, and $\Delta s$ is the space-time interval. All observers agree on the space-time interval, it is constant. My question is why is it squared? If we had in equation like this:

$$\Delta s'=\Delta x^2+\Delta y^2+\Delta z^2-(c\Delta t)^2$$

$\Delta s'$ would be constant as well. It would also never be imaginary. It would have units of $[length]^2$ instead of $[length]$ though.

Is there a theoretical or practical reason that we define the space-time interval based on squaring, or is it just to make it look similar to Pythagoras' theorem/give it simpler units or something else entirely?

• Well it's a generalized Pythagorean theorem, it's natural to define it like that, as in any (pseudo)metric space: it's a "distance" in the correct units. However, if you want to define the square as another quantity, you are welcome to use that notation. – orion May 30 '14 at 20:41

You are correct when you point out that any function of $\Delta x^2 + \Delta y^2 + \Delta z^2 - \Delta t^2$ will be constant and agreed on by all observers. So we could define $\Delta s$ to be its cosine...if all we were interested in was getting an invariant.

You are also right when you point out the dimensional issue. Measure time in light-centimeters, and distance along the x,y,z axes in centimetres. Then length is measured in centimetres, and so is time.... Then the right hand side has units cm$^2$, and hence, so does the left hand side. Using cosine or other, similar functions like the identity function you suggest, would produce a quantity that did not even have the units of length (and so, could not be proper time).

Now, definitions are arbitrary, so you could define Ps to be equal to $\Delta x^2 + \Delta y^2 + \Delta z^2 - \Delta t^2$ if you want, and you could give it any name you want. But would you be able to express the fundamental laws of Physics in terms of that quantity? It is a requirement of the principle of relativity that it be an invariant, and either Ps or cos(Ps) would satisfy that, but it is desirable that it make life easy for us in our formulas, since doing Physics is already hard enough. There are important reasons why we want to use the square root function instead of cosine or instead of the identity function which one of the other answers insists on.

There is more to it than just to make it look like Pythagoras' Theorem or make it look like pre-relativistic physics. These reasons do not become apparent until you get to General Relativity, or at least to Differential Geometry. This is your question, rephrased: Why do we want to study an invariant quantity with dimensions of length? (Which is the same as time).

The answer is that we want to be able to define $s$, the proper time, or, as I am expressing it, "the length of a path". It will be given by a line integral $s = \int ds$ along the path, and will be invariant for all observers. To an observer who is moving along that path, it will appear to be the elapsed time. Now it is quite basic that if first 2 cm of time elapse, and then another 3, the total elapsed time is 5 cm. So we need an additive quantity. Neither cosine nor Ps are additive, as simple examples show, but if we define $ds^2 = dx^2+ dz^2+dy^2-dt^2$, then it will be additive, by the higher-dimensional non-Euclidean analogue to Pythagoras' Theorem. that is why the squaring occurs, and it is indeed squaring a quantity $ds$, and when finite intervals are involved along straight lines, it is indeed the square of a quantity $\Delta s$ defined as $$\Delta s = \sqrt {\Delta x^2 + \Delta y^2 + \Delta z^2 - \Delta t^2}.$$

SHORT ANSWER We square $\Delta s$ so that we get an additive quantity along world-lines.

• the advanced answer is, you want it to be a tensor, so it has to be linear, so you have to take the square root of those squares for it to be linear. – joseph f. johnson May 30 '14 at 3:40
• I thought it would have to do with general relativity. – PyRulez May 30 '14 at 10:55
• Did Einstein know all this when he was defining the interval, that it would be necessary? Or did he have hunch that happened to work out? – PyRulez May 31 '14 at 23:40
• I suppose he wanted a space-time interval to be measured in dimensions of length: that is simple physical intuition. Later, special relativity was taught with differential geometry concepts because, in fact, Einstein was thinking about General Relativity from the very start of relativity, even during special relativity. And a tensor has to be linear, so it has to have dimensions of length, not length squared. – joseph f. johnson Jun 1 '14 at 2:25
• Well, length squared could of turned out to be additive, but it didn't luckily. – PyRulez Jun 1 '14 at 11:47

and Δs is the space-time interval.

Actually, many (most?) will say that the spacetime interval is $\Delta s^2$. In other words, $\Delta s^2$ is not the squared interval; it is the symbol for the interval.

Since this has been questioned in a comment, I provide some references below:

Here is the definition of the spacetime-interval. Suppose, as measured by a certain experimenter, two events are separated by a time $t$ and a spatial distance $x$. Then in terms of these numbers the spacetime-interval between the two events is the quantity $$s^2=x^2-c^2 t^2.\tag{17.1}$$ Notice that this is written as the square of a number $s$. The pacetime-interval is the quantity $s^2$, not $s$. In fact, we will not often deal with $s$ itself. The reason is that $s^2$ is not always positive, unlike distance in space. If $ct$ is larger than $x$ in Equation 17.1 then $s^2$ will be negative. In order to avoid taking the square-root of a negative number, physicists usually just calculate $s^2$ and leave it at that. You should just regard $s^2$ as a single symbol, rather than as the square of something.

What immediate information does the spacetime interval gives us? If the spacetime interval between events A and B is negative, then either $t_1$ or $t_2$ is negative. It follows that events A and B are timelike related, as illustrated in figure 12$a$. In this case it is possible for an inertial observer to be present at both events A and B. The elapsed time such an observer would measure between A and B is simply the square root of minus the spacetime interval, $\Delta t=\sqrt{-\text(interval)}$.

Also, from Space-time intervals:

The interval is defined by

$$\Delta s^2 = \Delta x^2+\Delta y^2+\Delta z^2-(c\Delta t)^2$$

Note that the symbol $\Delta s^2$ is usually taken as a fundamental quantity and not the square of some other quantity $\Delta s$.

And Sean Carroll writes in "Lecture Notes on General Relativity":

The interval is defined to be $s^2$, not the square root of this quantity.

Is than an theoretical or practical reason that we define the space-time interval based on squaring

Theoretically, the interval is the Minkowski dot (inner) product of a displacement four-vector with itself

$$\Delta s^2 = x^{\mu}x_{\mu}$$

which is invariant under the Lorentz transformation. This is analogous to the length squared of the 3-D displacement vector

$$l^2 = \mathbf x \cdot \mathbf x$$

However, the Minkowski inner product is not positive definite; the inner product can be positive or negative.

Practically, the sign of the interval determines whether the four-displacement is time-like or space-like (the interval is light-like if the interval is zero).

If the interval is time-like then the proper time is

$$\tau = \sqrt{\frac{|\Delta s^2|}{c^2}}$$

If the interval is space-like, the proper distance is

$$\sigma = \sqrt{|\Delta s^2|}$$

• So it is to look like Pythagoras' theorem, and they get to avoid imaginary numbers. Tricky. – PyRulez May 30 '14 at 0:11
• @PyRulez, who's they? – Alfred Centauri May 30 '14 at 0:46
• One can easily quote a mixed-up textbook to support your point, but you are quite wrong. In relativity, dimensions of length and dimensions of time are the same. Therefore, if the lengths are squared on one side, they have to be length squared on the other, too. And, it is so (delta x)^2, as you can see when you try integrating along a geodesic to get s instead of ds. – joseph f. johnson May 30 '14 at 1:18
• @AlfredCentauri I suppose Einstein and the text book companies. CONSPIRACY jk. – PyRulez May 30 '14 at 10:58
• @josephf.johnson, I've added additional references that show that you are quite wrong; my statement that "many will say that the spacetime interval is $\Delta s^2$" is a fact. Whether you agree with those authors or consider them "mixed up" not does not make my opening statement incorrect. Indeed, it appears likely that you don't quite understand what the authors are saying given your 2nd sentence above. – Alfred Centauri May 30 '14 at 12:00

or is it just to make it look similar to Pythagoras' theorem ...?

If you take a look at Einstein's "Relativity: The Special and General Theory", you will see in the Appendix I (just before equation (10)) that in order to derive the interval equation, Einstein actually did begin with the Pythagorean Theorem in 3D, which he put like this:

$$r = \sqrt{x^2 + y^2 + z^2} = ct$$

This way he showed the vector of light traveling in a three dimensional space.

He then transformed the equation in a number of ways, but the Pythagorean Theorem was the source of the whole equation.

(And I got the downvote because ... I reminded the history? Well, I guess it does not pay to study the sources ...)

EDIT: PyRulez commented below that: "This doesn't really explain why the space-time interval was squared (only the distances need to be)".

Well, $x$ ($\Delta x$) is a distance, $y$ is a distance, $z$ is a distance and $ct$ - as I showed above (or what simply follows from the fact that it is velocity multiplied by time) - is also a distance. Now, what do you call the result of adding and subtracting distances (squared)? Einstein called it a "line element" or "linear element" and he wrote in "The Foundation of the General Theory of Relativity" (p. 119):

"The magnitude of the linear element pertaining to points of the four-dimensional continuum in infinite proximity, we call ds".

If we have a continuum and we add/subtract what we call distances in this continuum, then we must obtain a distance as a result.

The shortest euclidean distance between three points namely 1,2,3 follows that $dist(1,3)=dist(1,2)+dist(2,3)$.

where $dist(x,y)$ is the vector between points x and y.

Now we from daily experience know that space in itself without time is euclidean.

Now this linearity relation should be carried over to space-time. Why ?

Because if there are three simultaneous events for an observer , then their space-time distance must equal the euclidean distance and thus will follow the linearity condition.

So we expect that The space-time interval between any events a,b and c must also follow the relation

$dist^* (a,c)=dist^* (a,b)+dist^* (b,c)$

where $dist^*$ is the space-time interval distance vector.

which will be followed only if the unit of space interval is length and not length$^2$.

This linearity relation also makes the math easier and lets us do things in special relativity similar to pre-relativity days like defining velocity,kinetic energy,momentum in a similar way newton did and they follow the same kind of vector/scalar addition respectively the way they did in pre-relativity days.

Now if you still did everything the same way like defining momentum to be $p=$ $m$ x (new metric) $/$ proper time and energy to be having units $p^2/2m$.

Where new metric denotes the metric to be $\Delta s'$ in the question.

You won't be having the relations like energy conservation,momentum conservation to hold true in the same mathematical form they used to do earlier in pre-relativity mechanics.

So either treat minkowski metric to be of the dimensions $length$ or completely change the way you defined momentum,energy and everything before relativity so that your theory remains consistent with the universe.

The latter seems a very daunting task to do than the former. To summarise : .

Our equations retain their old pre-relativity mathematical form is the core reason why we take space-time interval to be of the units of length. Also our distance still is a vector quantity.

• @PyRulez I have edited the answer. – Isomorphic May 29 '14 at 21:32
• -1; The premise of the answer is false: the correct statement is the triangle inequality, the rest does not seem to answer the question. – Robin Ekman May 30 '14 at 0:38
• The answer is poorly organised, but basically right on target. To have units of length, it has to be s^2, not s. To be additive along geodesics, it also has to be s^2, not s. – joseph f. johnson May 30 '14 at 1:15
• Triangle inequality is one more condition that is there along with linearity. – Isomorphic Jun 28 '14 at 8:13

One reason among many is that if a negative distance is just as good as a positive distance (ie, you can use a negative distance with one direction just the same as a positive distance with the opposite direction) then $$s^2$$ makes that fact clear because in the world where you're working with squares (such as the equation you posted), a Real distance with either signum is just as good because both come out the same.

• This is very unclear... – ZeroTheHero Apr 16 '19 at 20:13
• You might be attempting to make an important point. Or you might be wrong. I can't tell, because what you have written is barely coherent. Please fix it. – PM 2Ring Apr 17 '19 at 7:56
• indeed, I just re-read it. It sounded great as I wrote it. I'll fix – codeshot Apr 19 '19 at 7:56
• Ok, that's a lot clearer, but I don't think it actually answers the question. – PM 2Ring Apr 25 '19 at 4:24
• Well, answering this question completely is hard, there are many reasons why we communicate the way we do and many are lost to history. But the answer with the tick answers the question worse because the question was about why do we write $s^2$ and use $s$ instead of $s'$ and by extension use $\pm \sqrt{s'}$ - one of the main reasons is because $s^2$ communicates the relation between positive and negative $s$ while $s' = ... , s = \pm \sqrt{s'}$ doesn't communicate that anywhere near as well when $s^2$ is so commonly understood to imply that $s = \pm \sqrt{s^2}$ – codeshot Apr 27 '19 at 8:15