A few years ago, in a note in Nature, Gamow [1946] suggested that the whole universe might be in a state of uniform rotation and that this rotation might explain the observed rotation of the galactic systems. Indeed, if the primordial matter out of which the galaxies were formed by condensation was in a state of rotation, the galaxies themselves will possess a much faster rotation. For in consequence of the law of conservation of angular momentum, their angular velocity will increase as the square of the ratio of contraction. Therefore they will exhibit a rotation even in the coordinate system in which the primordial matter was at rest, that is to say, a rotation in the frame of reference defined by the totality of galaxies. In exactly the same way, the rotation of the galaxies in its turn can be used to explain the rotation of the fixed stars and planetary systems and therewith essentially all rotations occurring in astronomy. Of course, there arises at once the objection that by this theory the axes of rotation of all astronomical systems would have to be parallel with each other, which seems to contradict observation. But this difficulty is perhaps not insurmountable. For a closer statistical examination of the material furnished by observation might show that a certain degree of regularity in the directions of the axes of rotation is actually there, and the lack of complete regularity might be explainable by various disturbing influences. In particular, if the axes of rotation of matter in different places in space are not parallel with each other, the irregular motion of the galaxies will mix up the different directions of the axes.
Let us now consider the question of a rotating universe from the standpoint of theory and first of all from the standpoint of Newtonian physics. It is well-known that Newtonian physics gives a surprisingly good approximation for the expanding universes; so one might expect that the same will be true here; and this expectation is actually verified. Of course, if you want to apply the Newtonian law of attraction to cosmology, you have to formulate it as a local action principle
(because the integrals to which the distant action principle leads would in general not converge). According to this law, is not uniquely determined by, because we can require no boundary conditions. But one may make the tentative assumption that any values of and which satisfy this equation represent a physically possible state of affairs. Under this assumption,
is a possible gravitational potential for a homogeneous distribution of. It evidently has rotational symmetry around the-axis. But this potential (or rather the field of force produced by it) is exactly in equilibrium with the centrifugal force of a uniform rotation around the-axis with an angular velocity of. This Newtonian universe is the exact analogue of the relativistic one I am going to define shortly. The degree of approximation again is surprisingly good. The angular velocity for the relativistic solution differs only by a factor of from this one here. The privileged role which the-axis in this Newtonian world plays is only apparent. Actually, there would exist no experiment by which you could distinguish one place in the world from another. For if you describe this universe in a coordinate system which rotates together with matter, the only observable apparent forces would be the Coriolis forces, because the centrifugal forces are exactly annulled by the gravitational forces. But the Coriolis forces are independent of the place. So what is observable of the axis of rotation is only its direction, but not the place where it is located. This circumstance makes the analogy with the corresponding relativistic solution very close.
If we now consider the problem from the standpoint of relativity theory,the first question which arises is in what sense we can speak of a rotation of the whole universe in relativity theory, where we have no absolute space to which we could refer it. The answer, of course, is that in relativity theory, as a substitute for absolute space, we have a certain inertial field which determines the motion of bodies upon which no forces act. In particular, this inertial field determines the behaviour of the axis of a completely free gyroscope or of the plane of a pendulum, and it is with respect to the spatial directions defined in this way (by a free gyroscope or the plane of a pendulum) that matter will have to rotate. In the usual terminology, this means that it will rotate relative to the compass of inertia. You see that this kind of rotation does not involve the idea of an axis around which the whole world rotates. The world may be perfectly homogeneous and still rotate locally in every place, as is actually the case in the example I am going to give. Nevertheless, the world may be said to rotate as a whole (like a rigid body) because the mutual distances of any two material particles (measured by the orthogonal distance of their world lines) remain constant during all times. Of course, it is also possible and even more suggestive to think of this world as a rigid body at rest and of the compass of inertia as rotating everywhere relative to this body. Evidently this state of affairs shows that the inertial field is to a large extent independent of the state of motion of matter. This contradicts Mach’s principle but it does not contradict relativity theory.
Let us now investigate quantitatively what a rotation of matter relative to the compass of inertia means in terms of the and the velocity vector of matter. For this purpose, it is best to use the concept of a local inertial system. An inertial system in special relativity theory, as you know, is one in which the have values:
which implies that the motion of a free particle is uniform and rectilinear in these coordinates. In general relativity theory, of course, such a coordinate system for the whole space does not exist, but it exists locally, i.e., the coordinates can be so chosen that at the point considered, the have these values, and moreover all derivatives vanish, so that also in the neighbourhood of the point considered, the will have these values up to magnitudes of the second order. It is a well-known theorem of differential geometry that at any point of a Riemannian space there exist coordinate systems satisfying these requirements. In the sequel I shall sometimes have to consider coordinate systems which at the point considered only satisfy the first condition. I shall call them “orthonormal,” which I am told is a term in use for the corresponding concept in the theory of orthogonal systems of functions.
Now the angular velocity we are interested in is the angular velocity relative to the directions of space defined by axes of gyroscopes moving along with matter. For this is the angular velocity which the astronomer who, with his measuring apparatus, moves along with matter will observe. Therefore we have to introduce an inertial system which moves along with matter, i.e., one at whose origin matter is at rest at the moment considered, or, in other terms, one at whose origin the velocity vector
has values
Then, because directions defined by gyroscopes remain fixed in an inertial system, the angular velocity we are looking for will be equal to the angular velocity at the origin of this coordinate system computed as in classical mechanics. Therefore we obtain the formulae:
(This equality evidently holds in consequence of the values of at the origin.) Of course, instead of a coordinate system in which the compass of inertia is at rest and matter rotates, you may for the computation as well take a coordinate system in which matter is at rest and the compass of inertia rotates. I mean, of course, a coordinate system in which matter is at rest everywhere (not only at the origin), so that vanish identically. Then, under the further assumption that the coordinate system is orthonormal at the origin, you obtain the formulae:
for the angular velocity at which the compass of inertia rotates in this system, which is the same as the angular velocity at which it rotates relative to matter. These formulae follow immediately from the fact that the motion of the axis of a free gyroscope in four-space is a parallel displacement.But a parallel displacement of a vector along the-axis is expressed by the formulae:
So the matrix expresses the apparent deformation and rotation which the vector space undergoes by parallel displacement, and it is known from elasticity theory how the rotation is separated from the deformation. Since the coordinate system is orthonormal, the difference between the two kinds of Christoffel symbols disappears (or rather is reduced to a difference in sign), and we obtain the formulae displayed.
But these two formulae only apply in the very special coordinate systems defined, and it is of course very desirable to have a covariant expression for the angular velocity, i.e., one which applies in every coordinate system. This covariant expression is:
As you can verify immediately, is a skew-symmetric tensor; is the corresponding vector obtained by inner multiplication with the-tensor. Using the skew symmetry of the-tensor, you can verify immediately that is always orthogonal to. Moreover, you can verify by direct computation in a few lines that, for these two special coordinate systems, for becomes equal to the expression for obtained before, and vanishes. Of course, strictly speaking, only these relations here justify the preceding definitions, i.e., show that they are independent of the particular inertial [coordinate] system chosen, and similarly for the second definition.
Moreover, this covariant representation of the angular velocity is of interest for the following reason: If is an arbitrary vector field in an-dimensional space, then the identical vanishing of this skew-symmetric tensor is the necessary and sufficient condition for the existence of a one-parametric system of-spaces which are everywhere orthogonal to the vectors of the field. This is a straightforward generalization of the well-known condition which a vector field in Euclidean three-space must satisfy if there is to exist a system of surfaces everywhere orthogonal to the vectors of the field. This condition requires that
The true reason why this is the necessary and sufficient condition in both cases is that (1) this expression vanishes if the rot of [i.e., the curl of ]vanishes; (2) if it vanishes for, it also vanishes for, where is an arbitrary scalar field, because this expression is simply multiplied by if is multiplied by. The terms containing derivatives of cancel out, as you can verify immediately. This, of course, must be so in view of the fact that the existence (or nonexistence) of surfaces orthogonal to the vectors depends only on the directions, not on the absolute values, of the vectors. So in view of the just explained geometrical meaning of the vanishing of the skew-symmetric tensor, we arrive at the conclusion that the angular velocity of matter relative to the compass of inertia is everywhere 0 then and only then if there exists a one-parametric system of three-spaces everywhere orthogonal to the world lines of matter. Now the existence of such a system of three-spaces means, in a certain sense, the existence of an absolute time coordinate. For these three-spaces yield a very natural definition of simultaneity independent of the coordinate system. So there is a very simple connection between the existence of a rotation of matter and the existence of an objectively distinguished world time.
This incidentally also was the way in which I happened to arrive at these rotating solutions. I was working on the relationship between Kant and relativity theory and, more particularly, on the similarity which subsists between Kant and relativistic physics insofar as in both theories the objective existence of a time in the Newtonian sense is denied. On this occasion one is led to observe that in the cosmological solutions known at present there does exist something like an absolute time. This has been pointed out by epistemologists, and it has even been said by the physicist Jeans that this circumstance justifies the retention of the old intuitive concept of an absolute time. So one is led to investigate whether or not this is a necessary property of all possible cosmological solutions.
The rotating cosmological solution I have found on this occasion has certain shortcomings from the physical viewpoint. In particular, it yields no red-shift for distant objects, and it contains closed time-like lines. It has, however, a certain interest in principle insofar as it is doubtless the simplest rotating solution and, moreover, the only homogeneous solution which rotates like a rigid body. The linear element is very simple indeed. It is given by the expression
where is a constant. Written as a sum of squares it reads like this:
The-lines in this coordinate system are the world lines of matter. This linear element is obtained from the underlined three-dimensional linear ￥element simply by adding the term. As you see, does not occur in the preceding terms. So the four-space with this consists of a one-parametric system of copies | of the three-space with the underlined metric. If we introduce cylindrical coordinates in these three-spaces, with the old-axis as rotational axis of the coordinate system, we obtain the expression
for the linear element, which gives a clearer picture of the structure of this three-space. It shows that it has rotational symmetry around the-axis. Moreover, the non-vanishing of the term shows that the-lines (i.e., the world lines of matter) are not orthogonal to the plane; they are orthogonal to the radius vector from the origin (since vanishes), but they are inclined to the plane and, more specifically, the inclination is always in the direction of increasing. So the world lines of matter are helices of a sort, as they must be in a rotating universe.
I shall give the geometrical meaning for this linear element later, but first I would like to show how the field equations can be directly verified very quickly, or at | least describe in a few words the way in which it is most easily done. You can first show that this four-space is homogeneous. Namely, the four transformations
(where to are real numbers) carry this linear element into itself. But by suitable choice of the, you can evidently take the origin of the coordinate system into any given point. So it is sufficient to verify the field equations at one point, say the origin. Now at one point you can always introduce orthonormal coordinates by a linear transformation. If you do this by the linear transformation
You can, without much calculation, obtain the components of the Riemann tensor by means of the formula
Since almost all first and second derivatives of the vanish, the computation is very easy. The result is that only the with two different indices [other than 3] do not vanish, and they all have the same value:
Hence, for the components of the contracted Riemann tensor, you obtain that only is not 0, and it is equal to 1. So, if you denote by the unit vector in the direction of the-axis, you have the equality
Or rather, this holds in case the constant is 1. In general, therefore, you have the equality
But whenever can be represented in this form and, in addition, is a constant (as it is in our case, because the space is homogeneous), the relativistic field equations can be satisfied in a very simple manner. We only have to set the cosmological constant equal to and then multiply this equality by and subtract the result from the equality above:
The resulting equation is identical with the field equations
if we determine the constant by this equation:
If we compute the angular velocity by means of the formulae given before, we obtain the value
For the density observed in our world, i.e. g/cm, this leads to a period of rotation of years. This way of verifying the field equations by direct computation is of course not very interesting, but it [is] surprisingly short. In less than one printed page you can give in this way a complete proof including all details for the existence of rotating solutions.
Before entering into a discussion of the properties of this solution, I would like to say something about the heuristic argument by which this [solution] is most easily obtained and about the geometrical meaning of this linear element. This will also show the true reason why it satisfies the field equations. If we want to find the relativistic analogue for the Newtonian universe mentioned before, we must of course look for a stationary solution. Moreover, since the Newtonian universe rotates like a rigid body, we have to require that the mutual distances of any two material particles should remain constant in time, i.e., that any two world lines of matter should be equidistant—that is to say that a perpendicular drawn from a point of one line to another line must have the same length no matter what point of the first line you choose. So the world lines of matter in the four-space we are looking for must be a system of pairwise equidistant geodesic lines, and for each point of the space there must exist exactly one line of the system passing through it. However, this system of lines must differ from a system of parallels in Euclidean space insofar as a translation | of the whole space along these lines must not be a parallel displacement of the space. For if it were, the compass of inertia (whose behavior is described by a parallel displacement) would not rotate relative to matter.
Now this description of the system of world lines reminds one at once of a system of Clifford parallels in a three-space of constant positive curvature. It is a well-known fact (for which, moreover, I am going to give a very simple proof later on) that in a three-space of constant positive curvature, there exist in every direction systems of straight lines satisfying all requirements I enumerated. The difference from what we are looking for is only (1) that we are looking for such a system of lines in a four-space, [and] (2) that the metric of relativity theory is not positive definite but has the signature. But these two defects can easily be remedied. There exist, of course, also three-spaces of constant curvature and signature—and, as I am going to prove shortly, these three-spaces (or to be more exact those of positive curvature) also contain systems of Clifford parallels. Now from such a three-space we can very easily obtain a four-space and a system of lines in it which has the desired properties. Namely, we simply add to the linear element a term [assuming we have started with coordinates to], that is to say, we construct in the simplest possible way a four-space out of a one-parametric system of congruent three-spaces of the kind described. Then we define in one of these three-spaces a system of Clifford parallels and enlarge it by adding the lines obtained by the translation:. It can easily be verified that then also any two lines of the enlarged system will be equidistant. The only thing which remains to be seen is how we can ensure that the field equations of relativity theory will be satisfied. For this purpose, let us see what the contracted Riemann tensor will look like in the four-space we just constructed. Of course, this space is homogeneous. So it is sufficient to compute the Riemann tensor at one point. We may furthermore assume that the coordinate system is orthonormal at this point, [and] that, moreover, the first three basis vectors lie in one of the three-spaces of the one-parametric system of three-spaces used. The fourth basis vector will then lie in the direction of an-line. Moreover, the first basis vector can be so chosen as to lie in the direction of a line of the chosen system of Clifford parallels.
Now, about the values of the in such a coordinate system, we can say the following: All components must vanish, because the do not depend on and, moreover, the are constants. From these properties of the, you can derive immediately that all vanish if at least one of the three indices is 3. Next, you can derive the same thing for the, namely, that they vanish if at least one index is 3, and finally, the same holds for the, owing to the formula [above expressing in terms of and] and therefore also for the. So it follows that the matrix of the has only zeros in the last row and the last column. The numerical value for the remaining three-dimensional matrix follows from this formula for three-spaces of constant curvature, where is the curvature:
Since the coordinate system was assumed to be orthonormal, the have the values [in the matrix ()], so that we obtain for the matrix of the this form:
It is easily seen that an of this form can never satisfy the field equations, whatever the values of and and the unit vector may be. For if we write the field equations in the form
we see that they mean that by subtracting a multiple of from, we must obtain a tensor which is the product of a vector with itself, and this vector moreover must be time-like. But a diagonal tensor which is the product of a vector with itself can only have one component different from zero, and because the vector is to be time-like, this component must be the first one; but it is evidently impossible to make the last three vanish in this manner.
But if we now remember why we started the construction with three-spaces of constant curvature, we see that we did it only because they contained a system of equidistant geodesics. So if we were able to modify the metric in our three-spaces in such a way that the chosen system of Clifford parallels remained a system of equidistant geodesics, the new three-spaces would be as good for our purposes as the old ones, and we might be able in this way to introduce a new arbitrary parameter which could make it possible to satisfy the field equations. Now this is actually possible and in a very simple manner. We only have to stretch the metric in an arbitrary ratio in the direction of the lines of the chosen system of Clifford parallels. By stretching the metric of a space (let’s say the Euclidean plane) in a certain direction, I mean the following thing: We decompose the linear element into a component parallel to the direction and a component orthogonal to it and multiply the first component with the factor, while we leave the second one unchanged, so that we obtain the expression
for the new distance. Owing to the equalities here, this operation means the same as adding to the linear element a certain multiple of its projection on. The same operation can be performed in any Riemannian space if, instead of the direction, there is given a system of lines simply covering the whole space. If we denote by the vector field which consists of all tangent vectors of unit length to the lines of the given system, we obtain the expression
for the new linear element. For is the orthogonal projection of the linear element on the lines of the system, and we have to add that to the old linear element multiplied by a factor. It is easily seen that if the system of lines along which we stretch the metric is a system of equidistant lines, it will be a system of equidistant lines also in the new metric (simply because distances orthogonal to the lines of the system are not changed at all by the operation of stretching). Moreover, if the system of lines consisted of geodesics, the lines will be geodesics also in the new metric. This is particularly easy to see in our case, because the lines of a system of Clifford parallels are axes of rotational symmetry both for the space and for the system of lines. Hence they will be axes of rotational symmetry also for the new metric; but any axis of rotational symmetry is a geodesic line.
The rotational symmetry of the new metric also allows one to determine very easily the form of the contracted Riemann tensor of the new metric in the coordinate system defined before. A symmetric tensor in a three-space determines an ellipsoid (or some other surface of the second degree) in the vector space of each point. But if an ellipsoid in a three-space is to have rotational symmetry, it must be an ellipsoid of revolution, one of whose principal axes has the direction of the axis of rotational symmetry and the other two [of which] are orthogonal to it; and the same holds for other surfaces of the second degree. Hence in the coordinate system chosen before (whose first axis is the axis of rotational symmetry), the must have diagonal form, and the second and third entries on the diagonal must be equal to each other. So the whole four-dimensional tensor must have the form
where and evidently will depend on the ratio of stretching. Hence we may hope that by a suitable choice of we may be able to make. But in that [case] will be equal to if is the first unit vector of the coordinate system. If, however, the once have this form, I have shown before how the field equations can be satisfied, provided is positive. For it must be equal to a multiple of the density of matter. Now it turns out that, and only this value, actually gives and, and this is the space I defined before.
As to the derivation of the two forms of the linear element from the geometrical interpretation just given, there is not much to be said about the second form. In order to obtain it, you only have to introduce cylindrical coordinates with the lines of the system of Clifford parallels as-lines of the coordinate system. In order to obtain the first form, a closer examination of the group of transformations of our space is necessary. For this purpose let us first consider the simplest representation of the group of an ordinary three-space of constant positive curvature. We can think of such a space as the surface of a sphere, or rather, hypersphere, around the origin in a Euclidean four-space. Such a sphere is given by the equation
Now we can associate with each point of our space, i.e., the sphere, a quaternion
which, owing to the preceding equation, will have the norm 1. For the norm by definition is the product of with its complex conjugate, which gives the sum of squares
Then, owing to the fact that the norm of a product is the product of the norms, a multiplication of with a quaternion of norm 1 will carry this sphere into itself. This will be true no matter whether you multiply from the left or from the right, hence also if you multiply simultaneously from both sides. So the equation
represents a motion of the sphere into itself, and, vice versa, every motion of the sphere into itself can be represented in this way. Now it is easily proved that if you apply the powers of one quaternion, where is a real number, to one fixed point of the sphere, the resulting one-dimensional system of points will form a geodesic on this sphere. In order to prove this, it is sufficient to remark that,, lie on a geodesic. But this follows because (a), (b) there subsists a linear relation among,, and 1, namely, (here is a real number), and (c) the geodesics on the sphere are given by linear equations. If you now perform this operation with the same quaternion on all points of the sphere, you will obtain a system of geodesics covering the whole space, and it is clear that if any two lines of the system have a point in common they will be identical with each other, i.e., any two different lines of the system will be non-intersecting. Moreover, there exist transformations of the sphere which carry each line of the system into itself, i.e., which translate the whole sphere along these lines; namely, the transformation
for an arbitrary number, has that property. But from the existence of such transformations, it evidently follows that any two lines of the system are equidistant. So any such system of lines is a system of Clifford parallels.
Now literally the same considerations can be made if, instead of a four-dimensional Euclidean space, you start out with a space in which the square of the distance between two points is defined by the expression
(with two negative squares) and if you consider, instead of a sphere, the hypersurface of second degree defined by
only instead of the quaternions you now have to take another four-parametric algebra in which the norm [of the element]
is given by
Such an algebra can best be defined in the following way: You take the complex quaternions, i.e., those for which the coordinates are arbitrary complex numbers, and then consider that subalgebra of the complex quaternions in which the first two coordinates are real and the last two coordinates are pure imaginary. So the unities of this algebra are
If you denote [the latter three] by to, you obtain the rules of multiplication
which show that this number system (with real coefficients to) actually is an algebra, i.e., the product of any two numbers in the system is again a number of the system, and, moreover, the associative law holds, because it holds for the complex quaternions. This algebra evidently should be called the algebra of hyperbolic quaternions. Strangely enough, that name occurs in the literature in a different meaning, but anyhow I shall use this name in the sequel. (Or does there perhaps exist a name in use for this algebra?)
Now you can repeat with the hyperbolic quaternions all considerations made before. The fact that the norm of a hyperbolic quaternion may vanish also for non-zero quaternions is of no consequence, since we have to do only with quaternions of norm 1. There is, however, still another difference between the ordinary and the hyperbolic quaternions, namely, that the latter contain three-parametric subalgebras. For example, the subsystem defined by is such a subalgebra. As the unities of this subsystem you evidently can take the three quaternions
But they reproduce each other by multiplication according to the rules
Now it is to be expected that you will obtain a particularly simple form for the linear element if you take the points corresponding to this subalgebra as one of the coordinate planes. This is actually so—namely, you then obtain the first form of the linear element
[By means of the hyperbolic quaternions, you also obtain a very simple representation of the group of this space (or rather of the group of the underlined three-space; the extension to the four-space is then trivial). Namely, the group of this space will evidently consist of all those transformations of the space of constant curvature (from which we started) which also carry the given system of Clifford parallels into itself (not each line separately, of course, but the whole system). But these transformations are evidently those of the form
where is the quaternion defining the system of Clifford parallels in the manner explained above. As you see, this group has four real parameters, namely and the three parameters in (since is a hyperbolic quaternion of absolute value 1). Therefore the group of transformations of this four-space will be five-parametric, because also the transformations evidently carry the space into itself. Of course, this group of transformations must be five-parametric because (1) the space is homogeneous, which requires four parameters, and (2) it has rotational symmetry at every point, which gives you one more parameter.]
Now I would like to say a few words about the physical properties of this solution. First of all, you can verify immediately that there exists no red shift for distant objects in this solution. This is true for every solution in which the world lines of matter are equidistant. For any such solution admits a one-parametric group of translations of the space along the world lines of matter (or, to be more exact, transformations expressible as
if the coordinate system is so chosen that the-lines are the world lines of matter and the-coordinate is the "Eigenzeit" measured along these world lines. Now if you apply this transformation to a light ray leading from one particle of matter to another one, you will again obtain a light ray between the same two particles, but one which leaves the first particle seconds later and also arrives at the second particle seconds later. But this means that light signals sent from one particle of matter to another arrive with the same time intervals in which they are being sent. Hence also, the time intervals between two succeeding crests of a light wave will be the same at the source and at the reception of light, i.e., there will be no red shift. This consideration shows quite generally not only that an expansion implies a red shift but also vice versa, [that] a red shift implies an expansion, no matter whether the universe rotates or not, i.e., a rotation alone can never produce a red shift.
The next question to be considered is the value of the angular velocity of the galaxies to be expected in a rotating world of this type. We have obtained before a value for the angular velocity of the universe. In order to derive from it the angular velocity of the galaxies, we have to divide it by the square of the ratio of contraction by which the galaxies were formed from the homogeneous primordial matter, provided, of course, we assume that the galaxies are homogeneous spheres and rotate like rigid bodies. But for a rough estimate of the order of magnitude, such an assumption may be made. (Moreover, it turns out that a concentration of mass toward the center of the galaxies changes surprisingly little the value obtained, provided in the rotation the centrifugal and gravitational forces are in equilibrium, which is assumed by astronomers to be true.) Now the ratio of contraction can be estimated from the observed average ratio between distance and diameter of galaxies. This ratio, according to Hubble, is about 200:1. So we have to divide this period of rotation by, which leads to a value of about years. As I understand, the periods of rotation of galaxies have been determined only in a few cases. For these few cases the observed value is somewhat larger, but not very much larger, than this number. It is, on an average, about years. But of course this agreement or disagreement is of very little significance, because in consequence of the lack of an expansion, this solution can hardly be the real universe anyway.
The next property of our universe which is of physical interest is the aforementioned existence of closed time-like lines, which I would like to discuss now. If you examine the second form of the linear element,
you will find that the coefficient of changes its sign for sufficiently great values of. Namely, for,, i.e., the coefficient of is less than zero; for, however, we get the opposite inequality, and therefore the coefficient of is greater than zero. If you choose so that, the coefficient of becomes equal to zero, i.e., the circle with radius around the origin in the plane is a null line. Now a null line is the path of a possible light signal. It is not necessarily the path of a light ray in vacuum because it need not be a geodesic line, and actually it isn’t a geodesic line in our case. But by means of a sufficient number of mirrors, you can force a light ray to go along any null line. To be more exact, you can force it to go along any polygon of geodesic null lines approximating the given null line. If, in particular, you send a light signal along a null line running back into itself, this means that the light signal will come back at exactly the same moment at which it is sent.
Now the reason for this circle being a null line, of course, is that, exactly as the world lines of matter, so also the light cones surrounding them are getting more and more inclined to the plane as you move away from the origin of the coordinate system. Hence finally their inclination is getting so great that they touch the plane. Therefore if you increase still further, the light cones will even intersect the plane, i.e., they will in part be situated below the plane. But this implies that there also exist null lines which, starting from some point in the plane, circle around the origin but at the same time are getting farther and farther below the plane as you go along them, so that they will return to the world line of matter from which they started at some point below the plane; that is to say, a light signal sent along such a null line will return earlier than it was sent. So it is possible to send light signals into the past and moreover, since a light path can be approximated as closely as you wish by a path of a material particle, you can even travel into the past on a rocket ship of sufficiently high velocity. Of course, these considerations seem to prove the physical impossibility of these worlds. But since these closed lines of particles and of light have an immense radius (it is of the same order of magnitude as the world radius of the Einstein static universe), these experiments are many orders of magnitude above what could actually be done, and therefore it is questionable whether anything follows from these considerations. [For there may be some impossibility in principle involved which is not known at present.] At any rate, however, the problem of whether our universe rotates is not necessarily connected with this question, because there certainly also exist rotating solutions without closed time-like lines.
In connection with the temporal structure of this universe, I would like to point out one more fact. It may seem that in consequence of the possibility of travelling into the past it becomes impossible to distinguish consistently between the positive and negative direction of time. But this is not at all true. It does remain possible to divide the light cones into two classes, the positive and negative ones, in such a way that a limit of positive light cones is again a positive light cone—this continuity of course is a necessary requirement if the division is to make sense physically—since a division can evidently be effected by defining the positive light cones to be those which contain the positive direction of the-lines in this coordinate system. This has two consequences: (1) that after such a definition has been made, each possible world line has a well-defined orientation which determines for any two neighbouring points on it which is earlier and which is later; and (2) that in whatever way you may travel around in space-time, your own positive direction of time will always agree with the positive direction of time subsisting in the places where you arrive. In particular, a closed time-like line will always look like that in figure 1,
(Figure 1)
and never like that in figure 2,
(Figure 2)
which would mean that on returning you would find your surroundings look like a film running in the wrong direction.
But in spite of the existence for the whole world of a uniform positive direction of time, there exists, on the other hand, no uniform time for the whole world. This assertion can be given a quite precise meaning. Namely, one may say that a coordinate is a time coordinate if always increases if one moves along any possible world line of matter in its positive direction. In special relativity theory and in the non-rotating universes, such time coordinates evidently exist for the whole four-dimensional space. But the existence of closed time-like lines implies that such a time coordinate for the whole four-dimensional space cannot exist. For if you move along a closed time-like line in its positive direction, the time coordinate for the initial point must be the same as for the end point (because initial and end point coincide), but then cannot have been increasing along the whole line. But the existence of a time coordinate in this sense is not absolutely necessary in order that the solution may have a physical meaning. The existence of a well-defined positive direction of time, however, is necessary. For only the direction of time determines, for example, what it means to send a light signal, since the light is to be found only in the positive half of the double light cone issuing from the space-time point from which the signal is sent. The positive direction of time also allows you to formulate the second law of thermodynamics, at least for closed systems; whether it could be applied to the whole universe I do not know.
I would like to mention one more physical property of our universe—namely, that the light rays in empty space orthogonal to the axis of rotation are circles, i.e., they revert to their origin, arriving, however, always at a later moment than they were sent. This effect has the consequence that looking in a direction orthogonal to the axis of rotation you can see only up to a certain finite distance, and this again has the consequence that the apparent density of galaxies would have to decrease more and more if you look in such a direction. But for the distances covered by the telescopes of today, this effect would be pretty small. That the light rays orthogonal to the axis of rotation are circles again shows the close agreement with the Newtonian approximation. For if you introduce in the Newtonian approximation a coordinate system in which matter is at rest (i.e., a rotating coordinate system), then also the paths of free particles will be circles in consequence of the Coriolis forces.
There are two more facts of some interest concerning this rotating universe. Namely, (1) by making certain identifications of the points, you can construct out of it finite rotating universes. But they also have a time of finite extension. (2) This universe is essentially the only rotating universe which is spatially homogeneous and has equidistant world lines of matter, where “essentially” means up to the question of connectivity in the large. So all other rotating solutions have either a contraction or an expansion or both at the same time, namely, a contraction in one direction of space and an expansion in another direction. And there actually exist rotating and expanding solutions, but I have not yet represented them in analytic form.
