# How can spacetime be intrinsically curved? [closed]

Why it is assumed that spacetime is intrinsically curved, in contrast to extrinsically curved?

I can't see why this should be. For example, if you take a piece of 2-d paper, how can you curve it without a 3-d space.

Of course, the answer will be mathematical, but, as Feynman once told me:

If you can't put a theory in words, you haven't understood it at all.

So has anyone a non-mathematical answer?

• – John Rennie Jan 1 at 8:44
• Your question is confusing and a bit rude. You ask about intrinsic curvature, then give an example of extrinsic curvature and make a comment about embedding. Finally, you cast doubt on the understanding of anyone that can't explain the answer (to what?) in words. – m4r35n357 Jan 1 at 10:29
• "If you can't put a theory in words, you haven't understood it at all" – I couldn't disagree more. I would probably change this phrase to "if you can't put a theory in math, you haven't understood it at all" – Prof. Legolasov Jan 1 at 13:26
• It is not a matter of opinion. Maths is the language of science, period, and "agreeing with Feynman" will not give you any insights into your question. And you still don't seem to understand intrinsic and extrinsic curvature. – m4r35n357 Jan 1 at 14:10
• This is well explained in any standard GR textbook, but too large to reasonably fit in the scope of a SE answer. I recommend Sean Carroll’s lecture notes on GR. – Dale Jan 1 at 15:06

A lot of other answers here go into much more detail, but i feel like the simplest answer to your question is just that it doesn’t matter whether or not we model spacetime curvature intrinsically or extrinsically. The idea is, if we can model spacetime curvature intrinsically then there is basically no reason to assume the existence of any other space that the spacetime manifold lives in. The theory does not require it, and it doesn’t change the predictions of the theory even if it was embedded in some other space so there’s no point.

I think you may need to learn to accept separating the mathematics of a theory from its ontology. The reality is, we can make excellently accurate predictions, but it is ultimately very hard to pinpoint the “true” nature of reality so long as these things are experimentally unobservable. And at the end of the day you’d be hard-pressed to find any modern-day physicist who thinks general relativity is fully true, as most are seeking some sort of quantum theory of gravity that has general relativity as the classical limit.

I apologise for veering off into more philosophy than physics in my answer, but you did ask for a non-mathematical answer (although i am not entirely sure if this will meet your demands).

EDIT: To be clear, the intrinsic curvature of a manifold is a type of curvature observable even to “inhabitants” of the manifold itself, while extrinsic curvature is only observable from an embedding. Both can be seen in a manifold that’s embedded in a space, it’s just that Extrinsic curvature requires an embedding while intrinsic does not. I was not trying to imply that they are the same thing as a comment pointed out. Apologies if i make any errors in my answers, i am a bit rusty on Riemannian/Pseudo-Riemannian geometry.

• This answer seems to suggest that intrinsic and extrinsic curvatures are the same thing, which is wrong (for example, in the ADM formalism they are actually canonically conjugate variables, so far from being equal to each other). If you didn’t intend to give such an impression, maybe add a paragraph explaining that both can be derived from an embedding, but one depends on the embedding while the other doesn’t? – Prof. Legolasov Jan 1 at 15:49
• @Prof. Legolasov I wasn’t trying to suggest that they’re the same thing, i was just trying to use the same language as OP. But i guess i could have gone into more detail about what the terms actually mean, it’s just that it seems beyond the scope of OP is asking for, especially since they wanted a non-mathematical answer – Thatpotatoisaspy Jan 1 at 17:00

The answer lies in Riemannian geometry, but the question is, what is it trying to tell us? I'll talk about curved space (Riemannian geometry), and then curved spacetime (pseudo-Riemannian geometry).

I define space to be the set of locations (mathematically, all locations are conceptualized as points). Now the thing is, I have to specify relationships between these locations. For example, some locations are closer to each other than others, the set of locations is $$3$$-dimensional, etc. These are all real properties of the set.

This entire process of mathematically characterizing all the properties of what we know of as space comes in "three layers."

1. The first is the topological: for each location, this specifies a set of locations that are "near" the original location. This is basically the "closeness" relations between points. Actually, a more accurate word summarizing this layer is probably "continuity."
2. The second is differential: we need a way to discriminate between zig-zag lines and smooth curves (because we know objects in outer-space take smooth trajectories). This is the layer that makes it possible to do calculus (which physics is based on).
3. The third is Riemannian: even though we specified continuity and smoothness in the above layers, there is still something missing. We need to specify length scales and perpendicularity. Essentially, we need to define relationships between different directions at a point (which two directions are perpendicular, etc.). All of the relevant info is summarized in the metric tensor $$g_{\mu\nu}$$.

Now a change in the metric means a change in the relationships between points and directions. For example, beginning with Bernhard Riemann in the mid-19th century, mathematicians realized that we can modify the Pythagorean theorem in certain ways and still get self-consistent mathematics (for example, if $$dx$$ and $$dy$$ are the legs of a right triangle and $$ds$$ is the hypotenuse, then instead of writing $$dx^{2} + dy^{2} = ds^{2}$$ maybe we can write $$2dx^{2} + dx\,dy + 3dy^{2} = ds^{2}$$), but this is because all they're doing is changing the metric tensor on a manifold.

So, in short, the idea of intrinsic curvature of space is just the idea that relationships between locations and directions are changed in some non-standard way.

Of course, the question is about spacetime. I define spacetime to be the set of events, and I define an event to be a location with a specified time. We go through the same process of defining relationships between different events, except the third layer is pseudo-Riemannian geometry (this is done to account for time).

Einstein's proposal was that gravity is the result of changes in relationships between events and directions.

Why these changes happen or how they happen is a question that is currently unanswered, and it is thought that quantum gravity might provide an answer.

What I want to emphasize is that at no point was space or spacetime conceptualized as a fabric or an aether. There is a lot of confusion regarding this point, and it is the reason why I was interested in this question in the first place.

Some people might think that if spacetime is not a fabric, then it is nothing. But that's not correct, because it is defined as the set of events, and we can identify real properties about this set. It is a concept, and concepts are neither material things nor are they nothing.

Spacetime is just an abstract set of events with relationships between these events, and curvature refers to how these relationships are altered.

• This is not an explanation in words (although your comment is useful: +1) – descheleschilder Jan 1 at 13:18

Ok, let's try putting intrinsic curvature in words like OP requested.

Take a space, and attach a little arrow to some point of this space.

Now drag this arrow around a loop, but be sure not to rotate it (relative to its local surroundings) at any point.

If the arrow still comes back rotated, that's due to the intrinsic curvature of space.

Example: a 2-d sphere. Attach the arrow to the north pole. Drag it parallel to the direction of the arrow until it touches the equator. The arrow should now be facing south. Now drag it orthogonal to its direction along one quarter of the equator. The arrow still is facing south. Now drag it to the north pole. If you do everything correctly, the arrow should come back rotated by $$90$$ degrees. That's how you know that a sphere has intrinsic curvature.

• That's how you show why (or how) the 2-d sphere is curved. But the sphere itself is curved in a 3-d space. – descheleschilder Jan 1 at 13:50
• @descheleschilder you are incorrect, the sphere has intrinsic curvature that doesn't depend on whether it is embedded into a higher dimensional space or not. Nothing in my answer even assumes such an embedding (besides the references to the "north pole", "equator", etc, but those are there only for your convenience) – Prof. Legolasov Jan 1 at 13:51
• Well, Legolas, then you are correct. Congratulation and a happy New Year!!!. – descheleschilder Jan 1 at 15:23
• @descheleschilder don’t be rude. Happy new year to you, too. – Prof. Legolasov Jan 1 at 15:34
• What do you mean rude/? I say that you are correct and wish you a happy New Year! – descheleschilder Jan 1 at 15:56

tl;dr: There is no reason why it should be so, it just is.

The foremost reason why, in physics, we assume 3+1 spacetime to be non-euclidean(and in fact curved when massive) is because experiments overwhelmingly vindicate it. Its the same with bounded interaction speed (light speed), relativistic invariance, least action principle(conservation of energy etc), bell's inequality(quantum nature) etc.

There is no compulsion on nature to have obliged to this idea-she just did.

For all we know, things could have existed in flat spacetime instead of a curved one and graivty,or whatever, would have done its own thing, without affecting space at all. Infact before GR, that's how physicists thought.

Anyways, your question,imho, seems to be simpler and perhaps, a deeper one.

is there a difference between space, and embedded space?

Mind you there may be mathematically, so we modify to

Is there a physically observable difference between space, and embedded space?

What you are asking is for curved space to exist, is the embedding space necessary?

Mathematicians are pretty comfortable with treating spaces without any embedding. A surface can just exist-no need for a volume for it to hover in. It does so with all its properties attached-curvature,roughness,holes etc. As a matter of fact, even a measly point can boast of its existence without any assist from "$$3$$Dness".

So why do you need embeddings? Well because we live in a $$3$$D world. So we take these self-contained surfaces, embed them into our world,and see what we get. That's the reason you can't ever see the real Klein bottle.
By a similar token, we live in a 3+1D world. There are no more dimensions. We don't need an embedding to experience this world. So this world just exists-no apriori requirement for higher dimmension.

How does this make any intuitive sense--that embedding dimensions aren't needed for a space to exist? Well, think of it in this trite way--say people live on this unembedded surface(curved or not). They can make measurements on that surface--from atomic scale to across oceans(the Reimannianness). They then can compare these measurements with their knowledge of Euclidean geometry to conclude if their space is Euclidean(or even curved). Nowhere in this process did they have to measure anything along the embedding dimension. From measurements entirely within this space, they could establish their space's curvature(and other props.);and thats the point-the space has all the machinery they need to characterise that space. Funnily enough, they may even ascertain their space has holes,arguably something must fill them, and yet they'll never see that hole or what fills it.

Remember that just because one may expect some extra dimmension is necessary to bend into, it need not actually be so. An analogy would be to assume you need something to propagate into.