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Dimensions of Spacetime

  1. Sep 21, 2010 #1
    I know of the gravitational analogy. The bending of spacetime due to a mass is analogous to a ball placed on a sheet, other balls in the region will be "attracted" towards eachother. My question is, if we have to simplify our 3 spacial dimensions to 2 dimensions for the analogy, does that insinuate 4 spatial dimensions, making a total of 5 spacetime dimensions?

    The only alternative is that objects bend spacetime along the time axis, (in other words that extra dimension in the analogy is time itself) This however is extremely strange and if it is the case then my mind is blown and I would require some explaining.
  2. jcsd
  3. Sep 21, 2010 #2
    5th dimension may be sometimes helpful, but it is not necessary in the formulation of the theory and in calculations. Visual imagination is not absolutely necessary. What is necessary is a precise understanding of the mathematics involved. In classical Lagrangian mechanics of planets we use multidimensional spaces - we program our computers to solve equations using many auxiliary variables. Only at the end we translate the results into the language of 3D space.
  4. Sep 21, 2010 #3
    Arguably, the simplest spacetimes, the FLRW and the Schwarzschild solution, can be embedded in no less than resp. 5 and 6 dimensions. More complicated spacetimes need more dimensions I think, if I am not mistaken, any spacetime can be embedded in 10 dimensions.

    Edited: Here is a paper that shows the Schwarzschild metric can be done with 5, but apparently with a 'twist': http://www2.hu-berlin.de/leibniz-sozietaet/archiv%20sb/061/11_burghardt.pdf [Broken]
    Last edited by a moderator: May 4, 2017
  5. Sep 21, 2010 #4


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    The isometric embedding of a Riemann manifold in a higher dimensional flat space is described by the Nash embedding theorems. The details of the embedding and especially the dimension of the target space depend on the required properties of the embedding function, whether the embedding function is smooth etc.

    There are other embeddings, e.g. the Whitney embedding theorem which drops the "Riemannian-requirement".

    Caution: The term "Riemann" means "Riemann in the strict mathematical sense" whereas we physicists often deal with pseudo-Riemann manifolds w/o stating this explicitly. A Riemann manifold always has a metric g which is a positive definite symmetric tensor. This does not applay to our spacetime metric with signature (-+++).

    I do not know if the details regarding the dimension of the target space depend on this positive definiteness and whether they can be translated to pseudo-Riemann manifolds as well.

    What I know for sure is that this embedding is absolutely useless when dealing with general relativity. Nobody needs this embedding and nobody ever cares about the higher dimensional embedding space.
  6. Sep 21, 2010 #5
    That nobody ever cares is simply not true. There are many papers and books about embedding of spacetimes in higher dimensions.
  7. Sep 21, 2010 #6


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    I said "... when dealing with general relativity"
  8. Sep 21, 2010 #7
    Yes, and?
  9. Sep 21, 2010 #8


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  10. Sep 21, 2010 #9
    So now you admit it is interesting mathematically?

    Indeed it is not required. But considering alternative mathematical models for a theory can give new insights.

    I have a feeling that anything I say will not convince you. I suspect you simply do not like to think of GR embedded in higher dimensions, that is your prerogative. But that does not mean that nobody finds it interesting, mathematically or for whatever reason.

    Much research has been done on it (mostly a few decades ago), and not much new seem to have come out of that, but all those people who worked on it deserve recognition and to describe that as "nobody cares" is just not correct to say the least.
    Last edited: Sep 21, 2010
  11. Sep 21, 2010 #10


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    Under no circumstances I wanted to insult anybody here. If you think I did - sorry for that!

    I just wanted to express what you have just written: "... and not much new seem to have come out of that". So embedding seems to be w/o any physical relevance.

    In the last decades a lot has been achieved regarding "extra dimensions"; nevertheless there seems to be no relation between KK extra dimensions and extra dimensions resulting from an embedding. Finding indications for such a relation would be of great interest.

    Another thing: One should mention that visualizing curvature via embeddings of 2-manifols in 3-space is both demonstrative and misleading. It's demonstrative b/c everybody becomes a feeling what curvature means. But at the same time it is misleading b/c one could get the impression that we are dealing with curvature in terms of embeddings into higher dimensional flat space. The latter one is simply not true, we never use this embedding in GR, we always stick to the intrinsic description of manifolds.
  12. Sep 21, 2010 #11
    Yes, I think you are right about that.

    But the two approaches are identical in that they give the same results, so one is not more right than the other. Since we do not need the embedding one could apply Occam's razor.

    But I think one cannot state that physically spacetime is embedded or not, we simply do not know, in fact we do not even know if spacetime itself is physical or simply a mathematical model. For the physics it does not matter, if it computes right it is right, whether we model it within higher dimension, a curved surface or with turtles upon turtles.
  13. Sep 21, 2010 #12


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    It boils down to "what is physically relevant".

    If you have a model A that does everything you need, then why should you introduce a model B? Usually I like Ockham's razor, but somethimes I think it's hard to say which approach is simpler. Does Ockham's razor tell us that we should either drop the canonical or the path integral formalism? I don't thinl so.

    In our case here I wouldn't say that we should drop the embedding simply b/c it introduces extra dimensions. I don't think that this automatically means that it becomes more difficult. The fact that it's equivalent isn't a better reason as well. If it's equivalent then we should keep the embedding approach and abandon the standard one.

    But there is no "embedding approach". I do not see that it is better, explains more, provides powerful calculational tools, opens up a new perspective, ... nothing. Soit's fair to say that it's irrelevant.

    That has nothing to do with the question if spacetime is embedded or not. I am absolutely with you that we currently do not know what spacetime really IS. Therefore in principle every new idea might be helpful ...
  14. Sep 21, 2010 #13
    If I am understanding this correctly, it does not really matter how many dimensions we consider there are. However, then what is the significance in saying spacetime is 4-dimensional (3 spatial and 1 time)?
    Does that mean it is equally true to say spacetime is 5 dimensional or 10 dimensional?
  15. Sep 21, 2010 #14
    No you need a minimum of 4 dimensions. These dimensions are curved in GR.

    By analogy: consider the planet Earth. The earth is a curved two dimensional surface, of course it is embedded in 3 dimensions (and four if we include relativity) but for calculations we can just do fine thinking it is a two dimensional curved surface. It is the same with GR, we use a four dimensional curved surface (sometimes called hypersurface).
  16. Sep 21, 2010 #15
    That depends on what you want to do with these dimensions. With what structure you want to fill them. How do you want to relate your "spacetime" to our 3D technology.
  17. Sep 21, 2010 #16
    If you used a minimum of 4, then in what dimension does spacetime "bend" due to gravity?

    I don't understand using earth as 2 dimensional surface, I know of it as a 3 dimensional mass.

    I just want to understand how many dimensions there are in spacetime.
  18. Sep 21, 2010 #17
    Let's take a ball, can you understand that it has a 2 dimensional surface that is curved?

    If you consider it a 4 dimensional curved surface that question becomes irrelevant, all you need to know is that the surface is curved.

    Lookup Riemann, he was a mathematician who developed the necessary math to work on curved surfaces.
  19. Sep 21, 2010 #18
    Yeah, but that is just imaginary, it is still a collection of particles in 3-space. What I mean is that I don't see how the 2 dimensional surface is used in physical calculations.
  20. Sep 21, 2010 #19


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    Don't take this analogy too seriously.

    But yes, gravity, in the weak field limit, can be seen as the bending of "time".
  21. Sep 21, 2010 #20
    "If you use a minimum of 4, in what dimension does spacetime "bend" due to gravity? (johnnyb42)

    I'm not the pro, so I'll be more prosaic. First, spacetime does not bend due to gravity. Gravity IS the curvature ("shape") of spacetime.

    You can kind of think of gravity/spacetime curvature as directional, in that if you apply no force to resist it (you're in free-fall), it will carry you along it's internal stream. In the vicinity of Earth, free-fall is "down" at an increasing rate. So the most relevaqnt dimensions would be "down" and "time". If you are self-propelling in an orbit around Earth, all three spatial dimensions and the time dimension are relevant, as you continuously accelerate "through and against" earth's downward curved spacetime to maintain your own course.

    Well, this is kind of cartoonish, but anyway...
  22. Sep 21, 2010 #21
    Atyy, what did you mean by "in the weak field limit" ?
  23. Sep 21, 2010 #22
    It is not imaginary at all. When you want to calculate the area of a circle drawn on a ball you use real calculations on a curved surface. We do not have to take into account into which dimension it is curved to know the answers, it is similar with a curved 4-dimensional surface.

    I want to help you but you have to be open to learning something new, perhaps it would be better if you opened a new topic and ask about what it means when we say curved spacetime and how the question as to in what dimension it is curved into is irrelevant for GR.
  24. Sep 21, 2010 #23
    So is it simply, 4 dimensions of spacetime, 3 spatial and a time dimension, and these dimensions are not cleanly differentiable, but rather form a 4 dimensional "mesh" we refer to as spacetime?

    What I mean is, is it wrong to thing of space time as literally, 3 dimensions of space, and 1 dimension of time? If it is not wrong, then what I am further saying is, the 3 spatial dimensions stretched into another dimension, it is either the time dimension or yet another spatial dimension. My confusion is, if we must use at least 4 dimensions in calculations, that means that the bending is in the time direction...

    For example, (PassionFlower) your analogy to the area of a circle on a sphere, there is no time dimension considered here. So the 2 dimensional spherical surface is curved into another spatialdimension, I am just recognizing the other dimension, and it must require at least 3 dimensions to calculate this area. If we go to reality, our 3 dimensions are bent, but they must be bent into another dimension, I only ask if it is time they are bending into or if they bend into another spatial dimension.
  25. Sep 21, 2010 #24
    Well that question is actually a bit more complicated, and many people often make a mistake here. The time an observer measures with an ideal clock is the distance between two points in spacetime, so from that perspective time is not a dimension at all!

    The confusion comes because the coordinate chart to map spacetime is usually chosen in such a way that the time the observer measures overlaps one dimension of in this chart. And then this dimension is conveniently called time, but this is not a physical statement about spacetime but an arbitrary, albeit very convenient, choice that places the observer in a preferred position in the chart of spacetime.

    First of all you have to understand something about curvature, basically there are two kinds of curvature, extrinsic and intrinsic. Let's take an example with a sheet of paper. When we bend a sheet of paper we have extrinsic curvature, the sheet is curved into one dimension, however the intrinsic curvature is zero. All drawings on this sheet continue to give the same areas for closed loops regardless of how we bend it. But if we take that same sheet of paper, make it very wet and then let it dry, the sheet is both extrinsically and intrinsically curved. If we verify the areas we will discover they are in fact changed! For the calculations the extrinsic curvature is irrelevant but the intrinsic curvature is not.

    So, with that information, the answer is no, unless we have completely flat spacetime as in SR all dimensions are curved. Can you think why if only one dimension is curved the spacetime is not really curved at all?

    But with coordinate charts (continuous lines we draw on the curved surface) we have a lot of flexibility (we basically can draw them any way we like) and in some instances we can 'push' all the curvature into only one dimension, but that is only due to the flexibility of using charts.

    But the point is that for the calculations it does not matter where it is bent into. Riemann discovered we can express curvature in terms of intrinsic curvature and for GR that is all we need. The question as to in what dimension things are curved into has, so far at least, not given any physical interpretation. Thus Ockham's razor applies!

    Let me answer it this way, if we consider spacetime as 3+1 dimensions, then no the 3 dimensions are not curved into the other 1 dimension.

    Hope this helps, and hopefully my explanation is not too far off the money for the experts.
    Last edited: Sep 22, 2010
  26. Sep 21, 2010 #25


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    To be more precise, I should have said "Newtonian limit".

    In general relativity, the gravitational field consisting of 10 components that change over spacetime. Typically, one describes spacetime by dividing it into space and time. When one does that, one of the 10 components is a "time" component.

    In Newtonian gravity, the gravitational field consists of only 1 component that changes over space.

    Since we know that Newtonian gravity is a very good theory when the gravitational field is weak and when velocities are small (these two conditions are called the "Newtonian limit"), then the general relativistic description with 10 components should reduce to the Newtonian description in the Newtonian limit. When one takes the Newtonian limit in general relativity, one does indeed recover equations that look like Newton's equations. The 1 component of the Newtonian gravitational field turns out to be related to the "time" component of the 10 components of the general relativistic gravitational field.

    You can hear an explanation of it in lecture 5 "Einstein's Field Equations" http://ocw.mit.edu/courses/physics/8-224-exploring-black-holes-general-relativity-astrophysics-spring-2003/lecture-notes/ [Broken] . He talks about the Newtonian limit and warped time at about 1 hour and 5 minutes into the lecture.
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