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I understand that there is no need for a fourth-dimensional space to bend into, but why do physicists seem to be against the idea? Is this simply because there is no proof of a fourth dimension, or is there some sort of evidence against a fourth dimension?

Wouldn't it be simpler to assume there is a larger dimension that spacetime is embedded in? It seems to me that it could simplify a lot of things if we assume there is a fourth dimension tangential to the three dimensions of space, so it feels like physicists must have some good reason to be so against the idea.

Also, of course, there is the simple intuition that if something is on a curved surface (like a ball or saddle) there must be "something" inside the "ball" or between the sides of the "saddle."

Can a physicist explain why not being embedded in a higher 4D space is simpler or more accurately describes observations?

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    $\begingroup$ Do you mean fifth dimension? Spacetime, which is the structure which is typically said to possess curvature, is already four-dimensional (unless you're modeling a universe with $d\neq 3$ spatial dimensions). $\endgroup$ – J. Murray Jun 21 at 20:12
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    $\begingroup$ Time is usually considered to be our fourth dimension plus the x y and z spatial dimensions. $\endgroup$ – Adrian Howard Jun 21 at 20:19
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    $\begingroup$ @Zaaikort Comments are meant for requesting clarification or suggesting improvements to the question. Other comments - such as those involving discussions, those which attempt to (at least partially) answer the question, and those which are no longer useful - are routinely removed by moderators. I don’t know what your previous comment was, but I assume that’s what happened. $\endgroup$ – J. Murray Jun 22 at 11:37
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    $\begingroup$ "It seems to me that it could simplify a lot of things" what things? $\endgroup$ – fqq Jun 22 at 11:55
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    $\begingroup$ I'm not an expert but perhaps Bertrand's Theorem is helpful? In particular, check out this post where they note that 4 or more spatial dimensions largely go against empirical observations $\endgroup$ – perpetuallyconfused Jun 22 at 16:14

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You can always embed a (spacetime) manifold in a sufficiently high-dimensional space (if you have a $d$ dimensional manifold it can be embedded in a space of $2d$ dimensions). But that doesn't specify which space it is - it could be any sufficiently high dimensional space. So assuming it is embedded doesn't tell you anything at all. Hence it is simpler to not invoke any embedding in the first place.

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – ACuriousMind Jun 26 at 10:33
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I would say the answer is just the scientific principle of parsimony: if an empirically inconsequential entity can be dropped from a theory, it is preferable to drop it. As you have pointed out, the embedding space is not needed mathematically, and thus it also does not impact the empirical claims of the theory. Hence, parsimony tells us to drop it. If you like, you could say that such is indeed because there "is no proof".

But I would also say that I personally feel parsimony, like many other philosophical guidelines in science, can be taken too dogmatically. While we should perhaps "default" to the most parsimonious theory, thinking about potential extensions thereof could at the very least yield new ways of thinking and solving problems and, perhaps even later, novel theories that actually do make claims that are of empirical consequence and can then be tested.

On the third hand, though, I'm not going to claim there is any good fruit from this specific entity. In fact, a lot of analogies based around it can be quite misleading in terms of pedagogy, because we can't really visualize the embedding spaces as well as we'd need to even to make them useful. For example, that ever-brought-up-in-pop-sci "rubber sheet" picture of how gravity works.

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    $\begingroup$ The idea of parsimony goes back a long way in physics. "We are to admit no more causes of natural things than such as are both true and sufficient to explain their appearances." — Isaac Newton $\endgroup$ – Michael Seifert Jun 22 at 11:44
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    $\begingroup$ @MichaelSeifert, it's a fair bit older than that: "We may assume the superiority, other things being equal, of the demonstration which derives from fewer postulates or hypotheses." -- Aristotle $\endgroup$ – Mark Jun 22 at 20:45
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    $\begingroup$ @Mark Good point! I might argue that Aristotle was doing "natural philosophy" rather than "physics" as we understand it, but certainly Newton wasn't the first person to articulate the principle. $\endgroup$ – Michael Seifert Jun 23 at 13:54
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Even if spacetime is embedded in something bigger, we don't have access to it or any way of making observations of it. This tells you that our theories should only be formulated using quantities that can be computed "intrinsically", without reference to any embedding. So Riemann curvature is in, but mean curvature or second fundamental form are out. Now you can work with space-time embedded in some random ambient space, as long as you are careful to check that the calculations you're doing are independent of the embedding. Or you can work intrinsically on the spacetime itself and not worry about checking anything.

Another way of saying this is that it's easier to figure out which components of the metric/curvature tensor have which physical meaning when you're working intrinsically.

Moreover, starting with an embedding and figuring out formulas for the metric etc can be a pain. Conversely, if you know the metric, it can be a pain to find an embedding which induces it. As other people have mentioned, there are abstract existence theorems that tell you any Riemannian manifold can be embedded isometrically in some higher dimensional space but it can be difficult to find the embedding explicitly. As a fun example of this, a flat torus is very easy to visualise intrinsically (just draw a parallelogram and identify opposite sides in pairs) and can be embedded isometrically in 4d without too much difficulty. It can also be embedded isometrically in 3d if you allow your embedding to be only once differentiable (by the Nash-Kuiper embedding theorem) and a team of mathematicians produced a computer generated image of what that would look like:

http://hevea-project.fr/ENPageToreDossierDePresse.html

i.e. like an insanely complicated fractal.

Of course this example is irrelevant to spacetime because you want second derivatives to make sense of curvature, but it illustrates nicely the fact that by assuming you have an embedding you're adding a lot of complication to the theory which you never actually need or use.

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  • $\begingroup$ Here's a nice article about Nash embedding: quantamagazine.org/… $\endgroup$ – Zaaikort Jun 22 at 9:07
  • $\begingroup$ @Evans, I'm sure this is a very rookie follow up question, but when you say, "Even if spacetime is embedded in something bigger, we don't have access to it or any way of making observations of it." couldn't we argue that the four-dimensional spacetime in General Relativity describes gravity better with Einstein's four-velocity and is therefor evidence of a real fourth-dimension (time) with space-like qualities? $\endgroup$ – JDUdall Jun 24 at 20:41
  • $\begingroup$ @Evans, or are you saying that even though spacetime is four-dimensional masses don't necessarily need to be thought of as three-dimensional objects in a four-dimensional embedded space? We can just call it four-dimensional space and not worry about any extrinsic curving of three-dimensional objects into a four-dimensional space? $\endgroup$ – JDUdall Jun 24 at 20:41
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    $\begingroup$ @JDUdall: I was talking about 4d space-time being embedded in something higher dimensional (like 5 or more). You can certainly think of 3d space as being embedded in 4d space-time, but that doesn't help understand curvature: it's the curvature of 4d space-time itself (not of a spacelike slice) which is responsible for gravity. $\endgroup$ – Evans Jun 24 at 20:51
  • $\begingroup$ @Evans, I am with you on the 4D curvature of spacetime being responsible for gravity. I was conceptualizing matter as 3D objects (with 4D inertial vector/4-velocity) embedded in a 4D spacetime. In this way matter distorts spacetime "into" the higher (4th) dimension and then mass moves/accelerates along that path. It seems to simplify how Special/General Relativity works to visualize it geometrically like that. There must be something I'm still missing if physicists don't find this helpful/representational. $\endgroup$ – JDUdall Jun 24 at 21:25
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I have a lot of sympathy with this question, because I think it is true that physicists have been a little too strong on the idea that embedding is "wrong". Embedding is a well-defined mathematical idea that gives the same predictions for observable phenomena as does a treatment of the same manifold without embedding. So it is not ruled out by observation. It is ruled in or out merely by human assessments of what is the least misleading or most elegant formulation.

What is against the embedding idea is, I suppose, first that it may require more than one extra dimension, and second that it may suggest to the non-expert that the rest of the higher-dimensional space is available to be explored or measured. What is for the embedding idea is the help it gives to our intuition about curvature, and the fact that it sometimes offers nice ways to treat manifolds and find geodesics. For example, Flamm's paraboloid gives a good insight into proper distances around a black hole. And who has ever thought about a wormhole without picturing one of those tubes in their mind?

Overall, then, I think the message from experts to the general public here should not be that to think of spacetime as embedded in a higher-dimensional space is wrong, but rather that this notion is a useful mathematical method and it has limited uses. The embedding picture is in some respects misleading, because it seems to suggest that the rest of the higher-dimensional space is "there" in the same sense that the 4-dimensional spacetime and its matter content is "there", but this is not so, because the higher-dimensional space was simply conjured into existence by a mathematician waving a pen.

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    $\begingroup$ I don't think the extra dimensions (just look at string theory) are as big of a problem as the fact that embedding to higher dimensional space is far from unique. For example, you can embed 2D flat space in 3D flat space so that 2D space would be curved in 3D space, even though intrinsically it is flat, e.g. cylinder. And I guess, physicists have no wish to needlessly work with extrinsic curvature and additional arbitrary choices. $\endgroup$ – Umaxo Jun 22 at 9:46
  • $\begingroup$ Thanks; the point about non-uniqueness is a good one. $\endgroup$ – Andrew Steane Jun 22 at 11:22
  • $\begingroup$ @Umaxo but physicists seem to be OK with gauge freedom, which is a similar kind of non-uniqueness. $\endgroup$ – Ruslan Jun 22 at 14:03
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    $\begingroup$ Gauge freedom is a different kind of nonuniqueness: the bundle is fixed, the connection is the choice. Different bundles give measurably different physics. With embeddings, the analogue of picking a bundle is picking an ambient space (topology and geometry) which is a lot more choice than just picking a gauge group and some characteristic classes. If you fix an ambient space and look at embeddings into that, you get sigma models, which are studied (e.g. string theory). $\endgroup$ – Evans Jun 22 at 19:31
  • $\begingroup$ @Andrew Steane, Thank you for bringing up some of the standard issues physicist have with embedded space. I was thinking that some of the principals of General Relativity seemed to be more easily understood (to my mind) when considering an additional dimension that is perpendicular to the standard three spacial dimensions. Is orthogonality not a useful enough concept to go through the trouble of embedding in a higher space, or is it just superfluous and unnecessary? $\endgroup$ – JDUdall Jun 24 at 20:48
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To understand this you must first have a clear picture on the difference between intrinsic curvature and extrinsic curvature. Imagine an ant that lives on a sphere. To the ant the sphere looks flat because he is so tiny. But if he does certain experiments he might deduce that the sphere is actually curved. Let's imagine the ant carried a little gyroscope which will always point in the same direction. If the ant walks in what he considers a straight line the gyroscope will slowly tilt. This shouldn't happen for an actual straight line so we conclude the surface is actually curved.

enter image description here

Because the ant lives in a 3D universe but is constrained to a 2D surface we call this curvature extrinsic. The actual space isn't curved but because of the constraint of the sphere it appears like it is curved.

In contrast there is also intrinsic curvature. Imagine instead of an ant we have two-dimensional beings that live in a universe with the metric of a 2-sphere. Their cells are 2D. Their DNA is 2D. Their vision is based on light rays that propagate in 2D. These beings also experience curvature, but the curvature they experience isn't because of some extra dimension, it is purely because their universe is intrinsically curved. For example if these beings walk around a large enough triangle the angles will be larger than $180^\circ$. If they perform parallel transport along the triangle (like in the image below) he will not get back the same vector. Physicists believe our universe is intrinsically curved.

So which one is more natural? Extrinsic curvature might seem more natural to you because you are familiar with it. You are after all currently constrained to live on the surface of a 2-sphere. But from a physics perspective extrinsic curvature is a bit of a stretch. Both intrinsic curvature and extrinsic curvature could explain the curvature of spacetime but given two equal theories one should pick the simplest. Experimental measurements have ruled out a 'big' fifth dimension because that would be noticable in the strength of the fundamental forces$^\dagger$. There is still the possibility of small, curled up dimensions like proposed in string theory but that isn't what your question is about. So to include extrinsic curvature in your model of the universe you would first have to proof its existence and then also explain why the effects of this extra dimension aren't seen in experiments. Intrinsic curvature is the simpler, more natural option here even though it is counterintuitive.

$\dagger$ I don't have a source for this so if somebody has a source for it that would be great.

source of image

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  • $\begingroup$ Perhaps you should explain parallel transport, this would be a great answer. (Still upvoted for bringing up the real diffy geo stuff) $\endgroup$ – Buraian Jun 22 at 14:41
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    $\begingroup$ Actually, the procedure you described using parallel transport always only detects intrinsic curvature. Extrinsic curvature deals with how the tangent plane rotates in the ambient space as you move about (not necessarily around a loop). Consider a cilinder embedded in Euclidean space. This has no intrinsic curvature (parallel transporting a vector around a loop will always yield the same vector) but it has extrinsic curvature, as the tangent space changes as you move around the circumference. $\endgroup$ – Andrea Jun 22 at 20:40
  • $\begingroup$ @AccidentalTaylorExpansion, thank you for bringing up this point of intrinsic VS extrinsic curvature. Is it correct to say that a 2-torus is "extrinsically curved," but "intrinsically flat"? Wouldn't a torus be distorted in a fundamentally different "direction" if, for instance, a wave was moving across it's surface than if the wave was moving across a flat surface? In other words, why wouldn't the "extrinsically" of the curve be important? I guess from my ignorant perspective it seems like it is ironing out some important features of the universe to consider all curves intrinsic. $\endgroup$ – JDUdall Jun 24 at 20:59
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    $\begingroup$ @Andrea Good point. I'm not an expert on this stuff. I'll try to update my answer. $\endgroup$ – AccidentalTaylorExpansion Jun 28 at 12:19
  • $\begingroup$ @JDUdall I believe a torus has both positive and negative intrinsic curvature (don't quote me on that). A cylinder is nice example of a shape that has extrinsic, but not intrinsic curvature. A wave on a cylinder would behave as if it travelled on a flat plane as long as you ensure the wave there is no influence from the $r$ direction but only in the $z,\theta$ directions $\endgroup$ – AccidentalTaylorExpansion Jun 28 at 12:36
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The principle of Parsimony/Occam's razor imply that the simplest solution is the best. I.e. ignore any possibility of embedded space-time.

More to the point, it goes to the heart of what Science is. Scientific knowledge is based upon the idea that theories can be tied back to reality by observation - directly or indirectly; theories about the non-observable are the domain of religion and the like. I personally have a theory that the Universe was created and guided by giant orange dragon (he ate the turtles), but he stubbornly refuses to show himself. Unfortunately He is not observable so my theory is not scientific. Your proposed higher dimensional space is also non-observable and therefore not scientific.

Interestingly String Theory and the like have been criticised as being unscientific because their predictions are currently untestable; there are actually a number of grey areas but there is no argument that I'm aware of (doesn't mean much) that requires embedded space-time, and makes it or its presence observable.

Also, of course, there is the simple intuition that if something is on a curved surface (like a ball or saddle) there must be "something" inside the "ball" or between the sides of the "saddle."

I would suggest that intuition is not always a good tool whenever anyone wonders into abstract mathematics, at least to start with - another reason to get much more familiar with working with "bendy" 4D spacetime.

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You might be interested in the ADM, or the Hamiltonian formulation of GR. In that, you express the Einstein field equations as the evolution of a spacelike hypersurface in time. The full 4D manifold is then spanned by the history of this hypersurface. There you can really think of the hypersurface being bent in or out the 4th dimension of time. Indeed, this extrinsic curvature is directly related to the momentum of the hypersurface. There are many technicalities involved as you can imagine, including the fact that the choice of hypersurface, and the choice of the time function are highly arbitrary, but nonetheless it can give some neat examples.

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    $\begingroup$ +1 for ADM. The bending of the surface in the fourth dimension can be confusing. That's the extrinsic curvature, but the space-like slices have intrinsic curvature as well. $\endgroup$ – wdb Jun 24 at 22:24
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A good question. There are several good answers. Embedding four-dimensional spacetime in a higher dimension, but it does nothing to elucidate what the curvature of spacetime is. It is intrinsic, and it can be observed within the four dimensions. Consider the Earth revolving around the Sun. We observe the annual cycle of the variation of our weather and the angle of the Sun's path across the sky. We can attribute the Earth's path to a gravitational force, as we did before general relativity, or we can attribute it to the curvature of spacetime as predicted by general relativity. In the second case, the Earth is moving in a "straight line" in curved spacetime. This does not require a higher-dimensional space to embed spacetime in. Also, we can observe the curvature in minute variations between clocks at sea level and clocks at high altitudes. High altitude clocks run slightly slower than clocks at sea level. The time-like dimension of spacetime is stretched at high altitudes compared to at sea level. Again, there is no experimental or theoretical need to embed spacetime in a higher dimensional space.

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The answer to your question is intrinsic curvature. We happen to live in a universe where the curvature in our spacetime is intrinsic. We are able to tell this in our experiments. Like for example there is a beautiful example of the Shapiro delay, we who exist inside the 4 dimensional spacetime, are able to tell that this is caused by the intrinsic curvature of our spacetime.

enter image description here

The grid is supposed to show how the sheet has been stretched. The sheet is still 2-D, because we haven't stretched if upwards or downwards, but it's been deformed so that the spacing between grid lines changes. The key think (and the hardest to understand intuitively) is that for Flatlanders living on the sheet the grid lines still look straight. A Flatlander walking along the centre vertical grid line would think they were walking in a straight line, but would actually be moving in a curve. This type of curvature is what happens in general relativity. It's intrinsic not extrinsic. So to back to your question, you can't move behind the universe because there is no behind to move into. There are only the three spatial and one time dimensions - it's just that they are intrinsically curved.

Universe being flat and why we can't see or access the space "behind" our universe plane?

Now to understand the difference between extrinsic and intrinsic curvature is to look at the grid in the example and understand that the grid is intrinsically curved, because it is not bending in a hypothetical higher dimension. The curvature does not need to extend to a higher dimension (like you say).

And this is the case with our universe, because in our 4 dimensional spacetime, we are able to tell from experiments, that there is curvature, and it is intrinsically curved.

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  • $\begingroup$ Right, but the FOURTH dimension (time) also distorts in a gravity field, not just the three spacial dimensions. Could we think of our three-dimensional spacial universe as embedded in a higher four dimensional spacetime, with time modeled as extrinsic curves into this higher fourth dimension? $\endgroup$ – JDUdall Jun 30 at 1:13
  • $\begingroup$ @JDUdall I thought of that too this way sometimes, like the curvature manifests in the temporal dimension. $\endgroup$ – Árpád Szendrei Jun 30 at 1:34
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Because it doesn’t bend into anything, by definition of Spacetime and the relation between mass and curvature (Einstein's field equations).

The graphical image of a part of an $N$-dimensional space bending into an ($N+1$)-dimensional space is a misleading metaphor from popular science where a massive object modifying the curvature of space-time is (misleadingly) represented by something like a bowling ball placed on a supple diaphragm (like a trampoline) and bending the surface of said trampoline from a quasi 2-dimenensional flat disk into a 3-dimensional funnel-like object.

  • Inaccurate (misleading/popular) metaphor:

enter image description here

  • better illustration:

enter image description here

Spacetime is a concept based on the use of four coordinates (x, y, z, t): a 4 dimensional space, as opposed to the description, based on the (classical, Newtonian) choice of 3 coordinates of space (x, y , z) and 1 (separate, independent) coordinate of time (t).

Using the relativistic description, a planet in free space describes a geodesic which is a straight line, whereas a planet orbiting around another one follows a geodesic which reflects the fact that spacetime is locally affected by gravity (described as curvature).

So the classical description :

« Planet B orbits on an elliptic trajectory in space around planet A, due to the gravitational force between A and B »

Is the equivalent to say:

« The masses of A and B affect spacetime locally around A and B, giving spacetime (not space) a given local curvature function of said masses, in such a way that the projection of the position of A in three dimensions, as a function of time results in an elliptical trajectory »

But in that 4 dimensional description of the phenomenon, at no step in the process does the 4 dimensional spacetime «bend into» anything of a higher order of dimension. The 4 dimensional space does not need a 5th dimension to describe the evolution of systems it is concerned with.

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    $\begingroup$ "Because it doesn’t." is a matter of philosophical debate. Until such a time when we can actually measure the presence or absence of such a higher dimensional ambient space, Occam's razor says we may as well work without it. But I don't think anyone currently can actually, honestly claim that it isn't there. $\endgroup$ – Arthur Jun 22 at 10:47
  • $\begingroup$ It’s not… from the point of vue of the (coherent and logically self-contained) of special relativity (…is my point). If said formalism gets in turn generalized, then it is entirely other (separate) story… $\endgroup$ – Serge Hulne Jun 22 at 10:51
  • $\begingroup$ The graphical image of a part of an N-dimensional space bending into an (N+1)-dimensional space is a misleading metaphor from popular science where a massive object modifying the curvature of space-time is (misleadingly) represented by something like a bowling ball placed on a supple diaphragm (like a trampoline) and bending the surface of said trampoline from a quasi 2-dimenensional flat disk into a 3-dimensional funnel-like object. $\endgroup$ – Serge Hulne Jun 22 at 16:25
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    $\begingroup$ I'm fully aware of that. But just because one popular, faulty embedding is stuck in peoples' minds, that doesn't mean a true, correct embedding doesn't exist. An ambient space isn't part of the current standard theory because, again, Occam. But declaring categorically that it doesn't exist is unscientific. It is more correct to stay agnostic about the possibility, but ignore it for the time being. $\endgroup$ – Arthur Jun 22 at 18:37
  • $\begingroup$ I'm not saying it does not exit. I'm saying it does not exist in the theory of relativity, because it's unnecessary. $\endgroup$ – Serge Hulne Jun 22 at 19:42

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