Mathematical Physics Final Honours and Diploma
General Relativity
Some Two Dimensional Surfaces

Brian Dolan
This material is available on the web at: www.thphys.may.ie/staff/bdolan/teaching.html

The equation of a sphere of radius $a$ centred at the origin of 3-dimensional Euclidean space is

$$x^2+y^2+z^2=a^2.$$

Using cylindrical polars $(\rho,\phi,z)$ with $0\le\rho <\infty$, $0\le\phi <2\pi$ and $x=\rho\cos\phi$, $y=\rho\sin\phi$ so that $x^2+y^2=\rho^2$, this can be written as

$$\rho^2+z^2=a^2\qquad\Leftrightarrow\qquad z^2=a^2-\rho^2.$$

Note that this constrains the ranges of $\rho$ and $z$ to $0\le\rho\le a$ and $-a\le z\le a$. Thus

$$z=\pm\sqrt{(a^2-\rho^2)}\qquad\Rightarrow\qquad dz=\mp{\rho d\rho\over\sqrt{(a^2-\rho^2)}}.\eqno(1)$$

In 3-dimensions an infinitesimal line element is given by Pythogoras' theorem as

$$(ds)^2 = (dx)^2+(dy)^2+(dz)^2$$

or, in cylindrical polar co-ordinates

$$(ds)^2 = (d\rho)^2+\rho^2(d\phi)^2+(dz)^2.$$

Now imposing the constraint (1) gives

$$(ds)^2=(d\rho)^2+\rho^2(d\phi)^2+ \left({\rho^2\over a^2-\rho... ...2 =\left(a^2\over a^2 - \rho^2\right)(d\rho)^2+\rho^2(d\phi)^2.$$

Hence the metric is

$$g_{\alpha\beta}=\left(\matrix{{a^2\over a^2-\rho^2}&0\cr 0&\rho^2\cr}\right)$$

in these co-ordinates. Note that the component $g_{\rho\rho}$ is singular (i.e. diverges) at $\rho=a$ thought thes sphere is perfectly smooth and regular there. This singularity is an artefact of the co-ordinate system.

This form of the metric is related to the more usual one in terms of polar co-ordinates in 3-dimensions, $(r,\theta,\phi)$ with $r=a$ = constant, by $\rho=a\cos\theta$ so that $a^2-\rho^2=a^2\sin^2\theta$ and $d\rho=-a\sin\theta$ giving

$$(ds^2)=a^2\bigl\{(d\theta)^2+\sin^2\theta(d\phi)^2\bigr\}.\eqno(2)$$

Example 2: The paraboloid of revolution.

Consider the following equation in 3-dimensional Euclidean space

$$z^2=4a(\rho -a) \qquad \rho\ge a, \eqno(3)$$

where $\rho^2=x^2+y^2$ and $a$ is a real positive constant. In the $x-z$ plane ($y=0$) this reduces to $z^2=4a(x-a)$, which is the equation of a parabola. The surface generated by equation (3) is called a paraboloid of revolution.

Imposing the constraint (3) $z=\pm 2\sqrt {a(\rho-a)}$, which implies that

$$dz=\pm d\rho\sqrt{a\over\rho-a},$$

and putting this in the 3-dimensional Euclidean line element in cylindrical polars

$$(ds)^2 = (d\rho)^2+\rho^2(d\phi)^2+(dz)^2$$

we get

$$(ds)^2=(d\rho)^2+\rho^2(d\phi)^2+\left({a\over\rho-a}\right)(... ...left({\rho\over\rho-a}\right)(d\rho)^2+\rho^2(d\phi)^2.\eqno(4)$$

Thus the metric is

$$g_{\alpha\beta}=\left(\matrix{{\rho\over\rho-a}&0\cr 0&\rho^... ...left(\matrix{1-{a\over\rho}&0\cr 0&{1\over \rho^2}\cr}\right).$$

Note that $g_{\alpha\beta}$ has a singularity when $\rho=a$, but it is clear from the figure that this is not a true singularity of the surface - it is a co-ordinate artifact.

We can now use the line element (4) to determine the distance between two points on the surface. Consider, for example, two points $A$ and $B$ with the same value of $\phi$, $\phi_A=\phi_B$ and different values of $\rho$, $\rho_A>\rho_B>b$. The distance along the curve between them with $\phi=\hbox{const}\Rightarrow d\phi=0$ is

$$ds=d\rho\sqrt{\rho\over\rho-a}.$$

Integrating this from $\rho_B$ to $\rho_A$ gives the distance between $A$ and $B$, $S_{AB}$, as

$$s_{AB}=\int^{\rho_A}_{\rho_B} \sqrt{\rho\over\rho-a}\; d\rho ... ...\sqrt{(\rho-a)}+ \sqrt\rho\right]^{\rho=\rho_A}_{\rho =\rho_B}.$$

Thus if $\rho_B=a$ and $\rho_A=2a$, for example, then

$$s_{AB}=a(\sqrt 2 + \ln(1+\sqrt 2)).$$

Example 3: The 2-sheeted hyperboloid of revolution.

This time we start with 3-dimensional Minkowski space with line element

$$(ds^2)=-c^2(dt)^2+(dx)^2+(dy)^2$$

or, using 2-dimensional polar co-ordinates $x=\rho\cos\phi$ and $y=\rho\sin\phi$,

$$(ds^2)=-c^2(dt)^2+(d\rho)^2+\rho^2(d\phi)^2.$$

The constraint for the 2-sheeted hyperboloid is

$$c^2t^2-\rho^2=a^2\qquad\Rightarrow \qquad ct=\pm\sqrt{(\rho^2+a^2)},$$

where a is a positive real constant.

Differentiating the constraint gives

$$cdt=\pm d\rho\left({\rho\over\sqrt{\rho^2+a^2}}\right).$$

We need only consider one sign, e.g. the plus sign, so as to restrict ourselves to one sheet of the surface. Thus

$$(ds)^2=-\left({\rho^2\over\rho^2+a^2}\right)(d\rho)^2+(d\rho)... ...eft({a^2\over\rho^2+a^2}\right) d\rho^2 +\rho^2(d\phi)^2 \ge 0.$$

Thus the resulting 2-dimensional metric is positive definite (i.e. all distances are space like) and has components

$$g_{\alpha\beta}=\left(\matrix{{a^2\over\rho^2 +a^2}&0\cr 0&\rho^2\cr}\right).$$

An alternative co-ordinate system for this surface is to use $(\alpha,\phi)$ where $\rho=a\sinh\alpha$. Then $d\rho=a\cosh\alpha d\alpha$ and $\rho^2+a^2=a^2\cosh^2\alpha$ and

$$(ds^2)= a^2\bigl\{(d\alpha)^2+\sinh^2\alpha(d\phi)^2\bigr\}.\eqno(5)$$

Note the similarity (and the difference!) between this metric and the metric on the 2-dimensional sphere given in equation (2).

An important aspect of this surface which is obscured by the derivation here, but hinted at by the form in equation (5) and its similarity to the 2-dimensional sphere in equation (2), is that the surface is actually homogeneous - its intrinsic geometry is such that it would look the same at every point on the surface to a 2-dimensional bug living on it, just as a perfect sphere would look the same at every point to a 2-dimensional bug living on it. This is not apparent from the figure here, but that is because of the way we have embedded the surface in 3-dimensional Minkowski space in order to derive the 2-dimensional metric. Both the sphere and the 2-sheeted hyberboloid have a very high (indeed the highest possible, in a well defined sense) degree of symmetry.

Example 4: The 1-sheeted hyperboloid of revolution.

As for the 2-sheeted hyperboloid, we start with 3-dimensional Minkowski space using 2-dimensional polar co-ordinates

$$(ds^2)=-c^2(dt)^2+(d\rho)^2+\rho^2(d\phi)^2.$$

The constraint for the 1-sheeted hyperboloid is

$$c^2t^2-\rho^2=-a^2\qquad\Rightarrow \qquad ct=\pm\sqrt{(\rho^2-a^2)},$$

where a is a positive real constant. This time $\rho$ must be constrained so that $\rho\ge a$.

Taking infinitesimals gives

$$cdt=\pm d\rho\left({\rho\over\sqrt{\rho^2-a^2}}\right).$$

Thus

$$(ds)^2=-\left({\rho^2\over\rho^2-a^2}\right)(d\rho)^2+(d\rho)... ...2 =-\left({a^2\over\rho^2-a^2}\right) d\rho^2 +\rho^2(d\phi)^2.$$

Line elements can be either space-like or time-like on this surface. For example, curves of constant $\rho$ are space-like while curves of constant $\phi$ are time-like (since $\rho\ge a$). This is an example of a 2-dimensional curved space-time. The metric has components

$$g_{\alpha\beta}=\left(\matrix{-{a^2\over(\rho^2 -a^2)}&0\cr 0&\rho^2\cr}\right),$$

with $a\le\rho <\infty$. Again $g_{\rho\rho}$ is singular ar $\rho=a$ but, as is clear from the diagram, the surface is perfectly regular there. Note that the range $a\le\rho <\infty$, $0\le\phi <2\pi$ only covers have of the hyperboloid, e.g. the half with $t\ge 0$.

An alternative co-ordinate system for this surface is to use $(\beta,\phi)$ where $\rho=a\cosh\beta\ge a$. Then $d\rho=a\sinh\beta d\beta$ and $\rho^2-a^2=a^2\sinh^2\beta$ giving

$$(ds^2)=a^2\bigl\{-(d\beta)^2+\cosh^2\beta(d\phi)^2\bigr\}.$$

Three Dimensional Generalisations

Consider the metric on the 2-dimensional sphere

$$g_{\alpha\beta}=\left(\matrix{{1\over 1-\rho^2}&0\cr 0&\rho^2\cr}\right)$$

and the upper segment of the 2-sheeted hyperboloid

$$g_{\alpha\beta}=\left(\matrix{{1\over 1+\rho^2}&0\cr 0&\rho^2\cr}\right),$$

where we have set $a=1$ for simplicity. Including the metric for flat 2-dimensional Euclidean space in 2-dimensional polar co-ordinates $(\rho, \phi)$,

$$g_{\alpha\beta}=\left(\matrix{1&0\cr 0&\rho^2\cr}\right),$$

we see that all three of these 2-dimensional surfaces can be represented by the single formula,

$$g_{\alpha\beta}=\left(\matrix{{1\over 1-k\rho^2}&0\cr 0&\rho^2\cr}\right),$$

where $k=0,+1,-1$ for flat space, the sphere and the hyperboloid respectively.

It is clear that the flat plane and the sphere are homogeneous and isotropic about every point in that they have the same curvature at every point and look the same in all directions on the surface at every point, for example on a cloudy day at a point on the earth's surface where the sea is perfactly flat the horizon looks the same in all directions. Indeed, it is possible to prove, though we will not do it here, that any space which is isotropic about every point must inevitably be homogeneous - this is true in any number of dimensions, not just two. It is perhaps not quite so clear, but will assumed here anyway, that the hyperboloid is also isotropic about every point. Indeed it can be proven that these are the only three metrics which give rise to surfaces which are isotropic about every point.

These three metrics also have three dimensional versions. Using three dimensional polar co-ordinates $(r,\theta,\phi)$ where $0\le r<\infty$, $0\le\theta\le 2\pi$ and $0\le\phi <2\pi$ these are

$$(ds)^2=\left({1\over 1-kr^2}\right)(dr)^2 + r^2\bigl\{(d\theta)^2+\sin^2\theta(d\phi)^2\bigr\}.\eqno(6)$$

Flat 3-dimensional Euclidean space is the case $k=0$, The case $k=1$ is a 3-dimensional sphere. This is hard to visualise but can be obtained by imposing the constraint

$$w^2+x^2+y^2+z^2=1$$

in 4-dimensional Euclidian space with the line element

$$(ds)^2=(dw^2)+(dx^2)+(dy^2)+(dz^2),$$

where $(w,x,y,z)$ are 4-dimensional Cartesian co-ordinates. Similarly $k=-1$ is a 3-dimensional space, analogous to the hyperboloid. It can be obtained by imposing the constraint $c^2t^2-r^2=1$ in 4-dimensional Minkowski space, with line element

$$(ds^2)=-c^2(dt)^2+(dr)^2 +r^2\bigl\{(d\theta)^2+\sin^2\theta(d\phi)^2\bigr\}.$$

It can be shown that the three line elements given by equation (6) are the only three kinds that are isotropic about every point. They are therefore relevant to the study of cosmology, the study of the large scale structure and evolution of the universe, for the following reasons.

The distribution of galaxies in our universe is believed to be isotropic about us and the temperature of the universe is observed to be the same in all directions, i.e. isotropic, at $2.73^0$K to within a few parts per million. Since we do not expect that there is anything special about the point at which our planet sits we assume that the universe is similarly isotropic about every point and therfore the 3-dimensional space of our universe must be of the form of equation (6), though which of the three possible values of $k$ applies to our universe is not known. Just as the 2-dimensioanl sphere has a finite area and the flat plane and the hyperboloid have infinite area, so the 3-dimensional sphere has a finite volume and flat 3-dimensional space and the 3-dimensional hyperboloid have infinite volume. For this reason a universe with $k=+1$ is called closed and a universe with $k=0$ or $-1$ is called open. It is not known at the moment whether our universe is closed or open.

The paraboloid of revolution also has a three dimensional version

$$(ds)^2= \left({r\over r-a}\right)(dr)^2+ r^2\bigl\{(\theta)^2+\sin^2\theta(d\phi)^2\bigr\}.\eqno(4)$$

This describes three dimensional space outside a black hole of radius $a$.