# How can the spacetime interval be negative?

It’s my understand that the spacetime interval is analogous to the distance squared between two points in euclidean space. If the spacetime interval is the distance squared between two events in spacetime, how could that distance squared ever be negative? That would mean that the distance is an imaginary number, and i would expect it to only be a positive number or zero. How is this possible if spacetime is an actual surface? Sorry if i’m thinking too physically here.

• "It’s my understand that the spacetime interval is analogous to the distance squared between two points in euclidean space" it isn't, it isn't the sum of squares. Jun 20 '19 at 8:34
• @gented I’m not sure what you mean. Do you mean it IS the sum of squares? Jun 20 '19 at 8:57
• @gented Oh wait, i think i get what you’re saying now. It isn’t the same because of the signature? Jun 20 '19 at 9:05
• Yes, exactly: it isn't the sum of positive things. Some of them are summed, some other are subtracted :) Jun 20 '19 at 9:31
• @gented That makes sense. I’m just struggling with the ontology of it since in most things i see people start off explaining this stuff using euclidean space. i suppose it doesn’t really make sense to “demand” that intervals be positive definite since the notion of “distance” breaks down in a manifold that has time as a dimension? Jun 20 '19 at 10:13

The answer is quite simple, the reason is because you are measuring distances differently. In an Euclidean space your metric is diagonal with signature $$(+,+,+,+)$$ which means if you have vectors in such space $${\bf x} = (x_1,x_2,x_3,x_4)$$ and $${\bf y}=(y_1,y_2,y_3,y_4)$$, the distances square between them is $$({\bf x-y})\cdot({\bf x-y})_{\rm{Euclidean}} =({\bf x-y})^2 = (x_1-y_1)^2+(x_2-y_2)^2+(x_3-y_3)^2+(x_4-y_4)^2\ge 0$$ However Minkowski metric is diagonal and has signature $$(+,-,-,-)$$ or $$(-,+,+,+)$$, so time is a bit special and is labeled usually with the index 0, so for 4-vectors, $${\bf x} = (x_0,x_1,x_2,x_3)$$ and $${\bf y}=(y_0,y_1,y_2,y_3)$$, you compute distances in the following way $$({\bf x-y})\cdot({\bf x-y})_{\rm{Minkowski}} =({\bf x-y})^2 = (x_0-y_0)^2-(x_1-y_1)^2-(x_2-y_2)^2-(x_3-y_3)^2$$ as you can see in this case, the norm of a vector is not always positive, because the metric is Lorentzian meaning not positive-definite.

• Thank you for your answer. I think my question though deals more with ontology which i suppose is a mistake i usually make. I think what i’m really curious about, is if the metric is not positive definite, and is still used to calculate distances on the manifold, how can the distance ever be imaginary? Is my mistake that the meaning of “distance” breaks down once you introduce time as a dimension? i’m struggling to understand how spacetime, if it physically exists, could have a negative interval. It doesn’t seem to make sense Jun 20 '19 at 9:17
• @Thatpotatoisaspy Just a note: the cause of the negative interval is not "time as a dimension". You can easily construct 4-dimensional spacetime with signature (++++), so you have time there, but the metric is positive definite. Such spacetime of course is not the one we live in, but mathematically it is perfectly possible. The main issue is that our spacetime has hyperbolic geometry with signature (+++-). This is very different from the good old Euclidean spacetime, so in our case the "distance" really cannot (and does not) look like the nice simple Euclidean distance.
– mpv
Jun 20 '19 at 10:14
• @mpv Thanks, and yeah i know i was just trying to sort of understand ontologically how something like an interval cannot be positive definite. It’s a bit confusing to think about Jun 20 '19 at 10:17
• It is just confusing if you want to stick to the word "distance" in my opinion, that is why "space-time interval" is probably a better term. At the end of the day what is important to understand are the implications on causality and so on Jun 21 '19 at 14:02

Concerning the spacetime intervals, there are 4 contradicting concepts. Depending on the convention you chose you could adopt the signature (-, +, +, +) or (+, -, -, -), and you could chose to extract the root or not. But that is not enough. We might presume that for lightlike phenomena it should be clear that the corresponding spacetime interval is zero, but even here there is no unanimous opinion, in particular with respect to the question if we can say that the proper time of photons in vacuum is zero or not.

Very little effort is done by theoretical physics to resolve these differences of conventions. It is thought that this question is of no particular importance.

However, our current spacetime concept has a big problem with quantum mechanics. Up to now, all attempts of quantization of spacetime failed, our spacetime concept seems not to be compatible with quantum mechanics.

The following information may lead to a solution of these problems:

1. First we should decide for one of the above-mentioned contradicting concepts. The solution of a signature (-, +, +, +) and with extraction of the root seems to correspond to the physical circumstances. ds = $$d\tau$$. Proper time is a physical object-related notion for distances (the time according to a clock following a given object).

2. After having decided for the spacetime interval concept of proper time, we realize that only timelike intervals are defined, while spacelike intervals are not defined, they are negative squares or imaginary numbers. At first sight, this seems to be inadmissible, but if we check we will realize that the continuity of spacetime in space direction is a mere assumption only, that means, we assume that spacetime is a manifold without ever having proved it, and even though continuity of spacetime is not compatible with quantum mechanics. At my knowledge, nobody ever tried to prove that spacelike spacetime intervals exist.

3. Historically, the mathematics of spacetime were first described by Minkowski in 1908. It was a fascinating idea that by Lorentz symmetry, space and time were linked by the equation of the spacetime interval, and rapidly ideas of a fourdimensional manifold emerged. A fourdimensional manifold would have been possible in Newton's spacetime (and there were some theories in this direction), but not in a Lorentzian spacetime where spacelike spacetime intervals are imaginary (according to the explanations above). After the famous lecture of Minkowski, the spacetime manifold was generally adopted as mathematical model for spacetime, a simple equation was considered to be a manifold. 20 years later, when the question of quantum gravity came up, nobody mistrusted the manifold character of spacetime, even if such manifold is not compatible with quantum mechanics.

4. Physics today: Please be aware that today the concept of a continuous spacetime manifold is still mainstream. But according to the information above 1-3, it is impossible that you get a coherent answer on the basis of the mainstream model, because it is an intrinsic contradiction of the model, in the core of quantum gravity.

• This answer is, to say the least, wrong in many aspects: first of all the eventual observables do not depend on which signature you choose, so that isn't a problem at all. Second of all, failure of quantising general relativity has nothing to do with the concept of space-time. Third of all space-time manifolds are mathematically well defined and equipped with all properties you need, so the theories we have are indeed coherent. Jun 20 '19 at 12:04
• @gented, thank you for your comment! 1. The question is not about observables but about negative squares in spacetime intervals 2. It is exactly spacetime (not GR which has been experimentally proved) which is not compatible with quantum mechanics, and quantization of the existing spacetime manifold concept is tried without success 3. If so, why are there negative square/ imaginary spacetime intervals? Jun 20 '19 at 12:13
• There is no "negative square" in space-time interval because they aren't squares of anything: they are norms of vectors in a non-Euclidean manifold and they may be negative and still be well-defined. 2. No, it isn't - gravity isn't (perturbatively) renormalisable because it doesn't have the right natural units, which is a consequence of the fact that inertial mass and gravitational mass are the same. 3. Negative space-time intervals are perfectly fine, why would that be a problem? Jun 20 '19 at 12:30
• @gented, Squares or not: in physics there are both opinions. And if you don't see a problem, you did not get the point of Thatpotatoisaspy's question. "Shut up and calculate" is not a sufficient answer for his question. Jun 20 '19 at 12:49
• I am just stating that your answer is partially wrong as you are claiming that quantum gravity has problems because of reasons that aren't really a problem - the real reasons being different ones. Then if you want to discuss the philosophical implications of non-Euclidean geometry please go ahead, but don't claim that it enters quantum gravity because it doesn't. Jun 20 '19 at 12:55