1
$\begingroup$

We consider 2D metric: $ds^2={d\theta}^2+\sin^2(\theta)d{\phi}^2$ (1)

Is it possible to transform the above metric to the form: $ds^2=dx^2+dy^2$ (2)

Let's check: Initially we write equation (2) in the form: $ds'^2=dx^2+dy^2$ (3)

For relations (1) and (3) we use the following transformations:

$\theta=f_1(p,q)$

$\phi=f_2(p,q)$

$x=f_3(p,q)$

$y=f_4(p,q)$

$d(\theta)=[ \partial{f_1} / \partial{p} ]dp + [ \partial{f_1}/ \partial{q}] dq$

$d(\phi)=[\partial{f_2}/\partial{p} ]dp + [\partial{f_2}/\partial{q}] dq$

$dx=[\partial{f_3}/\partial{p} ]dp + [\partial{f_3}/\partial{q}] dq$

$dy=[\partial{f_4}/\partial{p} ]dp + [\partial{f_4}/\partial{q}] dq$

Using the above transformations in (1) and (3) we have: $ds'^2=[(\partial{f_1}/\partial{p})^2+\sin^2(f_1)(\partial{f_2}/\partial{p})^2]dp^2+[(\partial{f_1}/ \partial{q})^2+\sin^2(f_1)(\partial{f_2}/\partial{q})^2]dq^2$ $+2[(\partial{f_1}/\partial{p} )( \partial{f_1}/\partial{q})+\sin^2(f_1) (\partial{f_2}/\partial{p} )( \partial{f_2}/\partial{q})]dpdq \qquad$ (4)

And

$ds^2=[(\partial{f_3}/\partial{p})^2+(\partial{f_4}/\partial {p})^2]dp^2+[(\partial{f_3}/\partial{q})^2+ (\partial{f_4}/\partial{q})^2]dq^2$ $+2[(\partial{f_3}/\partial{p} ) (\partial{f_3}/\partial{q})+ (\partial{f_4}/\partial{p}) (\partial{f_4}/\partial{q})]dpdq \qquad$ (5)

To make $ds^2=ds'^2$ we may consider the following equations: (denoted by SET A): $(\partial{f_1}/\partial{p})^2+\sin^2(f_1)(\partial{f_2}/\partial{p})^2= (\partial{f_3}/\partial{p})^2+(\partial{f_4}/\partial{p})^2$ (A1)

$(\partial{f_1}/\partial{q})^2+\sin^2(f_1)(\partial{f_2}/\partial{q})^2= (\partial{f_3}/\partial{q})^2+(\partial{f_4}/\partial{q})^2$ (A2)

$\frac{\partial{f_1}}{\partial{p}}\frac{\partial{f_1}}{\partial{q}}+ \sin^2(f_1) \frac{\partial{f_2}}{\partial{p}} \frac{\partial{f_2}}{\partial{q}} =\frac{ \partial{f_3}}{\partial{p}}\frac{\partial{f_3}}{\partial{q}}+\frac{ \partial{f_4}}{\partial{p} } \frac{\partial{f_4}}{\partial{q}}$ (A3)

If SET A has solutions for the functions $f_1, f_2, f_3$ and $f_4$ we are passing from relation (1) to relation (2) by coordinate transformation ($ds^2$ is preserved and it is being carried from one manifold to another). Is it really possible, that is, can a sphere be flattened?

What are the conditions[from the theory of differential equations] for the existence/non-existence of solutions for SET A?

[You may consider the following calculations to assess the nature of the problem Separate numbering starts from here.

$ds^2=d\theta^2+Sin^2(\theta)d\phi^2$ ------------------- (1)

$ds’^2=dx^2+dy^2$ --------------------------- (2)

Transformations:

$\theta=\theta(x,y)$

$\phi=\phi(x,y)$

$d\theta=\frac{\partial \theta}{\partial x}{dx}+\frac{\partial \theta}{\partial y}{dy}$

$d\phi=\frac{\partial \phi}{\partial x}{dx}+\frac{\partial \phi}{\partial y}{dy}$

Using the above differentials in (1) we have:

$ds^2= [(\frac{\partial \theta}{\partial x})^2+Sin^2(\theta)(\frac{\partial \phi}{\partial x})^2]dx^2$ $+[(\frac{\partial \theta}{\partial y})^2+Sin^2(\theta)(\frac{\partial \phi}{\partial y})^2]dy^2$ $+[\frac{\partial \theta}{\partial x} \frac{\partial \theta}{\partial y}+ Sin^2(\theta)\frac{\partial \phi}{\partial x} \frac{\partial \phi}{\partial y}] dx dy$

-------------------- (3)

Relations (1) and (2) would become identical $[ds^2=ds’^2]$ if the following equations [SET A] are satisfied if:

$(\frac{\partial \theta}{\partial x})^2+Sin^2(\theta)(\frac{\partial \phi}{\partial x})^2=1$ -------- A1

$(\frac{\partial \theta}{\partial y})^2+Sin^2(\theta)(\frac{\partial \phi}{\partial y})^2=1$ ------- A2

$[\frac{\partial \theta}{\partial x} \frac{\partial \theta}{\partial y}+ Sin^2(\theta)\frac{\partial \phi}{\partial x} \frac{\partial \phi}{\partial y}]=0$ ---------------- A3

We will look for a solution for which the following hold true:

SET B

$\frac{\partial \theta}{\partial x}=\psi$ ------------ B1

$\frac{\partial \theta}{\partial y}=\chi$ ------------ B2

$\frac{\partial \phi}{\partial x}=\frac{1}{Sin \theta}\chi$ -------------- B3

$\frac{\partial \phi}{\partial y}= - \frac{1}{Sin \theta}\psi$ ------------- B4

$\psi^2+\chi^2=1$ ----------- B5

Exact differential conditions:

[SET C]

$\frac{\partial \psi}{\partial y}=\frac {\partial \chi }{\partial x}$ ----------- C1

$\frac{\partial}{\partial y}\frac{1}{Sin \theta}{\chi}=-\frac{\partial}{\partial x}\frac{1}{Sin \theta}{\psi}$ -------------------- C2

Differentiating C2 we have:

$-\frac{Cos^2 \theta}{Sin \theta}\frac{\partial \theta}{\partial y}\chi+\frac{1}{Sin\theta}\frac{\partial \chi}{\partial y}= \frac{Cos^2 \theta}{Sin \theta}\frac{\partial \theta}{\partial x}\psi+\frac{1}{Sin\theta}\frac{\partial \psi}{\partial x}$

Using B1 and B2 in the above step we have,

$\frac{Cos^2 \theta}{Sin \theta}(\chi^2+\psi^2)=\frac{1}{Sin \theta}(\frac{\partial \psi}{\partial x}+\frac{\partial \chi}{\partial y})$

Since

$(\psi^2 + \chi^2)=1$

We may write:

$Cot \theta =\frac{\partial \psi}{\partial x}+\frac{\partial \chi}{\partial y}$

Therefore our final equation is:

$\frac{\partial^2 \theta}{\partial x^2}+\frac{\partial^2 \theta}{\partial y^2}=Cot \theta$ -------------- D

If the above equation is solvable we can flatten a sphere.

Theorema Egregium:The Gaussian curvature of a surface is invariant under local isometry.

Incidentally an isometry comes under diffeomorphisms which involve invertible functions which are differentiable 'r' times [or infinite times]. This is suggestive of linear functions[or bijections]. But the functions representing $\theta $and $\phi$ could be of non-linear nature]

[It is important to note that tensor equations are preserved in coordinate transformations where the value of the line element,ds^2, does not change. But in General Relativity the same tensor equations[ex: the Geodesic equation,Maxwell's Rquations in covariant form] are considered applicable in all manifolds--flat spacetime and curved spacetime. The stated equations are considered invariant in all manifolds]

  • 2
    Seeing as the sphere isn't homeomorphic to a disk, no.2011-11-14
  • 3
    Any metric can locally be flattened, but it won't work globally.2011-11-14
  • 0
    If you want to leave the world of smooth manifolds with smooth metics, you can flatten the sphere almost everywhere. Think of a cone on a standard disk. I imagine that this has a curvature tensor that is distribution valued.2013-07-16

3 Answers 3

8

This is not possible, not even locally.

The first metric you mention is that of a piece of a sphere, which has constant, positive curvature. The second one, that of the plane, has zero curvature. The two are not locally isometric, because of the Theorema Egregium.

  • 0
    Tensor equations preserve their form in coordinate transformations where the line element,where ds^2 is preserved.But in General Relativity tensor equations like the Geodesic equation,Maxwell's Equations[expressed in covariant form]are preserved in all manifolds--flat spacetime and curved spacetime.How does one explain this?2011-11-14
  • 0
    @AnamitraPalit, I have no idea what you mean!2011-11-14
  • 0
    @AnamitraPalit: The form is the same, but the values are different.2011-11-14
  • 0
    In a coordinate transformation the values of the individual variables[the components] may change,preserving the form of the equation.But in these transformations we are confined to the same manifold. The value of the line element is preserved.Preservation of ds^2 implies the preservation of the norm ,angles,dot product.But when we move from one manifold to another the metric itself is not supposed to be preserved.The form of the equation, the values of the individual variables should change--unless we devise transformations where the metric may be preserved.2011-11-14
  • 2
    Just for clarification, in my comment I was referring to diffeomorphism, not isometry. There certainly is a Riemanninan metric on the sphere which is flat in a neighborhood of some point. Clearly it is not isometric to the standard one!2011-11-14
  • 0
    In the theory of manifolds the original manifold is expressed as the union of subsets[on the manifold].Each point on a subset is has its image on Rn.These local charts are not necessarily infinitesimally small,so far as theory is concerned----we may cover a sphere with six large surfaces.The basic point is:What is the basic reason from the theory of differential equations that SET A in the question should not have solutionsie,be unsolvable.PDEs usually have a rich variety of solutions2011-11-15
  • 2
    @AnamitraPalit, as I wrote above, the reason is Gauss's Theorema Egregium. This is explained in essentially every textbook on differential geometry of surfaces.2011-11-15
0

Let’s solve the PDE:$$\frac{\partial^2 u}{\partial x^2}+ \frac{\partial^2 u}{\partial y^2}=\cot (u)$$ ------------(1) Subject to the constraint :$$(\frac{\partial u}{\partial x})^2+(\frac{\partial u}{\partial y})^2=1$$ ----------------- (2)

We write PDE (1) as: $$\frac{\partial^2 [(u+F(u))-F(u)]}{\partial x^2}+ \frac{\partial^2 [(u+F(u))-F(u)]}{\partial y^2}=\cot (u)$$ ----------- (3)

PDE (3) may be represented as the following two equations: $$\frac{\partial^2 (u+F(u))}{\partial x^2}+ \frac{\partial^2 (u+F(u))}{\partial y^2}=0$$--------- (4)

And

$$\frac{\partial^2 F(u)}{\partial x^2}+ \frac{\partial^2 F(u)}{\partial y^2}=-\cot (u)$$--------------- (5)

Now,

$$\frac{\partial F(u)}{\partial x}=\frac{dF}{du} \frac{\partial u}{\partial x}$$ $$\frac{\partial^2 F(u)}{\partial x^2}=\frac {d^2 F}{du^2} (\frac {\partial u}{\partial x})^2+\frac{dF}{du}\frac{\partial^2 u}{\partial x^2}$$

And

$$\frac{\partial F(u)}{\partial y}=\frac{dF}{du} \frac{\partial u}{\partial y}$$ $$\frac{\partial^2 F(u)}{\partial y^2}=\frac {d^2 F}{du^2} (\frac {\partial u}{\partial y})^2+\frac{dF}{du}\frac{\partial^2 u}{\partial y^2}$$ PDE (5) may be written as:

$$\frac {d^2 F}{du^2} [(\frac {\partial u}{\partial x})^2+(\frac {\partial u}{\partial y})^2]+\frac{dF}{du}[\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}]=-\cot(u)$$ Or, $$\frac {d^2 F}{du^2}\times {1}+\frac{dF}{du}\times \cot(u) =-\cot(u)$$ Or, $$\frac {d^2 F}{du^2}+\cot(u)\frac{dF}{du} =-\cot(u)$$

The above ODE gives us the form for F to be used in PDE (4) which is Laplace’s equation in two dimensions

We obtain: $$ u+F(u)=Soln \;of \;Laplace’s\; eqn.\; in \; two \;dimensions$$

[Differentiating the above relation twice wrt to x and y and adding the results we have, $$F''(u)[(\frac{\partial u}{\partial x})^2+(\frac{\partial u}{\partial y})^2]+F'(u)[\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}]=-[\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}]$$ Again F(u) satisfies relation (5):$$F''(u)+\cot(u)F'(u)=-\cot(u)$$ The simplest choice would be that the PDE's (1) and (2) are getting satisfied simultaneously.This a reflection of the fact that the constraint $\psi^2+\chi^2=1$ has been used in the construction of relation D in the Question]

In the final step the particular solutions for F and that for Laplace's equation are chosen in a manner such that equation (2) is satisfied.

[u has been used instead of $\theta$ ,which has been used in the previous postings]

-5

A sphere may be flattened by a one to many mapping[strictly speaking these mappings are not functions because of their multiple valued nature]which preserves the value of $ds^2$ . But the metric coefficients are not preserved. The manifold itself changes. The transformed manifold contains multiple images of each point of the sphere.

In our example $\theta$ and $\phi$ [I am referring to the link in the question]are single-valued functions of the ordered pairs(x,y) but x and y may not be single-valued functions of ordered pairs $(\theta,\phi)$.

You may separate out different subsets from the original image set to get a bijective [one -one ,onto-mapping] for each subset . But for these subsets the curvature elements are not supposed to change[Theorema Ergregium]

So far as the differential equations in the question [or in the link provided in the question] are concerned $\theta$ and $\phi$ are single valued functions of the ordered pairs(x,y). So there is really nothing is wrong or unrigorous about the differential equations.

But we may start our calculations[referring to the ones in the link in the question] with

$x=x(\theta,\phi)$ And $y=y(\theta,\phi)$ The existence of solutions in either of the procedures will not indicate towards the violation of Theorema Egregium. But we can definitely flatten a sphere by suitable mappings for which the value of $ds^2$ is preserved.

  • 5
    To be honest, I do not understand what all this even means *at all*.2011-11-20
  • 0
    We do have certain mappings by which a sphere may be flattened with the preservation of $ds^2$. This does not violate Theorema Egregeim.We can use mappings which do not belong to the category of one to one correspondence In fact your answer in incomplete [and incorrect] in the fact that it completely denies the possibility of flattening a sphere2011-11-21
  • 0
    Well, I do not understand your comment at all either. I guess the simplest will be to simply drop the subject :)2011-11-21
  • 0
    Did you see the calculations on the link[in the question].The equations are not unsolvable ones.2011-11-21
  • 0
    Whether a set of PDE's should be solvable or not is the issue here,so far as the flattening of the sphere is concerned.Theorema Egreguim does not stand in the way of the existence such solutions2011-11-21
  • 0
    In view of its incompleteness[rather incorrectness] the answer given by Mariano Suarez-Alvarez should be voted down.The incorrectness of the answer given by M S-Alvarez consists in the fact that it asserts a complete impossibility of the flattening of a sphere, in complete disregard to certain types of mappings I have mentioned in my answer.2011-11-21
  • 2
    -1: This answer is wrong. Gauss's theorema egregium asserts that curvature is a local isometry invariant, so there is no map which preserves the value of $ds^2$ *and* makes it flat.2011-11-21
  • 0
    @Anamitra I'm no specialist in topology, but I'm certain one of the basis of topology is avoiding any creasing and tearing, which can't be avoided if one is to continuosly deform a sphere into a plane!2012-04-18