ok....at first thanks - you gave me quite homework to do, it will take a few days until I manage to read that book but I just can't sit like that after what you wrote. firstly I want to remark that I don't know what that damn word manifold mean - I cant find it anywhere - I can just suppose. But this isn't at most important right now. Boud, I have a few remarks to what wrote. Let me quote:
possible, so we came to the conclusion that the 2-torus in three dimensional Euclidean space is not an good example of two dimensional closed space of zero curvature. But if we introduced
Agreed. Unless we redefine the metric. For example: Let's put the 2-torus in the X-Y plane, centred on (0,0,0), with radius R.
Then the metric of E^3 (Euclidean 3-space, i.e. R^3 with flat metric) is
ds^2 = dx^2 + dy^2 + dz^2 = (dr^2 + r^2 d\theta^2) + dz^2 where r=\sqrt{x^2+y+2}
Under this metric, the 2-torus does not, in general, have zero curvature, and it is inhomogeneous (curvature different on different points).
I assume that theta is an angle between x and y axies and (r^2 = x^2+y^2), right ? If so then this is just an infinitesimal length in E^3 space in cylindrical coordinates. ok, I understand this point. (btw. but even here I have an uncertainty, look at the end, because it's a little different topic)
So let's just make a new metric on the 2-torus:
ds^2 = (dr^2 + R^2 d\theta^2) + dz^2 where r=\sqrt{x^2+y+2}
only difference: "r^2" -> "R^2".
( but I think dr should be = 0 in this metric )
(Exercise: Is this a metric on R^3?)
as you wrote this is a metric on flat torus. but in fact it's not true. if we use cylindrical sthight-linear coordinates than the equation r=R is an equation of tube, thus this is a metric on a surface of tube not torus. If we go to the coordinates like: r' = r \phi' = \phi z' = \rho arcsin(z/\rho)
where \rho is an radius of torus (the one from it's center) then the same equation r' = R describes torus. For the first equation r=R I checked Pythagoras's theorem integrating for each side of chosen triangle, and finally came to the conclusion that 1=1 as expected (meaning it's true). now I wanted to do the same thing with an equation r'=R which would be for torus. In fact in that prime coordinates, tube becomes torus so I thought since ds is invariant, all that should be showed is to check if that transformation is an isometry. Well it is, under condition that \rho = infinity or z=0. And this is good for nothing. I'm still gonna spend some time trying to find such function of the complementary - third side of an triangle on a surface of torus so after integrating along that side and putting it into the Pythagoras's theorem I will find equality. but I am sceptic because...
the fourth spital dimension than it becomes possible. Like for
possible: yes; necessary: no.
But the problem now is how to put this idea into the real world. All these considerations about the topology of space were made from "the higher ground" - I mean from the point of view of a space with at lest two dimensions more that the space (earlier a transparency or just a piece of paper) we were thinking of. The problem appears if we talk about the topology of the Universe where there are only three spital dimensions. To make it close like 3-torus it is essential to introduce 3 more dimensions (it's six) to make it possible. Then I say that, that all topoloy thing
Only if you *assume* that the only spaces which are physically possible are n-dimensional Euclidean spaces.
... that's just the point - as a basis the first thing was that we were looking among spaces of zero curvature which are Euclidean spaces, so I do assume only Euclidean spaces. The metric tensor g in cylindrical coordinates of torus is (I guess)
| R^2 0 | g= | | | 0 1/[1-(z-\rho)^2] |
( and for tube is with g_22 = 1.) It depends on point, and in general det(g) might be negative so it can't be an Euclidean space. Can it ? (and now I am not in some 3-dimensional space but just two). So I bring back my implication about topology science and 3+n - dimensional space. Despite of all what was said, don't we assume k almost zero these days ?
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this is bizzare: Let's imagine tube. From mathematical point of view we can define it's curvature in each of it's points along two perpendicular directions (along the axis of symmetry and the other one). the first radius of curvature will be infinity, while the other one will be R (radius of the tube). Living on a surface of tube I could also measure is's curvature by measuring it's circumference and dividing by 2pi. Now I can imagine a 2-space of shape of this tube. Although I computed the Pythagoras's theorem is satisfied (so the curvature must be zero) how can I deal with the fact, that it is flat in the circumstances I mentioned above.
thanks for references. when I finally get satisfied with this problems because if this going to be like this (I working on it for at least week) I won't start with my master thesis - this is quite off my topic ;) nevertheless very interesting, although maybe for you - boring.
pozdrawiam, bartek
On Fri, 2002-11-15 at 05:26, szajtan odwieczny wrote:
ok....at first thanks - you gave me quite homework to do, it will take a few days until I manage to read that book but I just can't sit like that after what you wrote. firstly I want to remark that I don't know what that damn word manifold mean - I cant find it anywhere - I can just suppose.
a manifold is: rozmaitosc rozniczkowa. let me put some words here which should clear something:
1. manifold is only a "imaginary" structure, you can think of it as of something smooth without peak, with or without edges 2. manifold itself is not very useful in GR (OTW) 3. riemann manifold is a manifold with given metric. only with the metric you get the recipe for calculating distances. 4. the shape of the metric is not important at all - the condition is that there are mappings (for n-dim manifold) which LOCALLY map M^n into R^n with given coordinate system (there is no coordinate system on the manifold itself). set of mappings should cover the whole manifold. 5. in fact the metric is also not described on the manifold, but on R^n, into whih the manifold can be at every its point mapped.
keeping this in mind, a 2-torus can be mapped EVERYWHERE (but locally) into R^2, and assuming metric ds^2 = dx^2 + dy^2 its flat. thats all. simple and easy.
how to get curvature from the metric? 1. calculate connection coeficients (wspolczynniki koneksji) \Gamma^\mu_{\nu\rho} 2. calculate Riemann tensor
if all the terms in the Riemann tensor =0, space is flat. for ds^2 = dx^2 + dy^2 is flat for sure, because even connection =0.
regards - michal
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Michal Frackowiak -
szajtan odwieczny