相关链接
关于旋转宇宙的讲座
(1949b)
哥德尔
几年前，在《自然》杂志的一篇文章中，伽莫夫（Gamow）[1946] 提出，整个宇宙可能处于一种统一旋转的状态，并且这种旋转可能解释星系系统的观测旋转。确实，如果由凝聚形成星系的原始物质处于旋转状态，那么这些星系本身将具有更快的旋转速度。因为根据角动量守恒定律，它们的角速度将随着收缩比的平方而增加。因此，即使在原始物质静止的坐标系中，这些星系也会表现出旋转，即在由所有星系定义的参考系中表现出旋转。完全相同地，星系的旋转可以用来解释恒星和行星系统的旋转，从而基本涵盖了天文学中所有的旋转现象。当然，这一理论立刻引发了反对意见，即根据该理论，所有天文系统的旋转轴必须彼此平行，而这似乎与观测结果相矛盾。不过，这一难题或许并非不可克服。对观测材料进行更详细的统计检验可能会显示旋转轴方向存在一定程度的规律性，而不完全规律性可能可以通过各种干扰因素来解释。特别是，如果不同空间位置的物质旋转轴并不平行，星系的非规则运动将混合这些轴的不同方向。
现在让我们从理论的角度，首先从牛顿物理学的角度来考虑旋转宇宙的问题。众所周知，牛顿物理学对膨胀宇宙提供了令人惊讶的良好近似；因此可以预计，这里也会是相同的结果；这一预期实际上得到了验证。当然，如果要将牛顿引力定律应用于宇宙学，就需要将其表述为一个局部作用原理：
（因为远程作用原理涉及的积分通常不收敛）。根据这一定律， 并未由 唯一决定，因为我们不要求边界条件。但可以进行一个试探性的假设：任何满足此方程的 和 的值表示物理上可能的状态。在这一假设下，
是一个可能的均匀分布 的引力势。显然，它在-轴周围具有旋转对称性。但这一势（或者由此产生的力场）与围绕-轴的均匀旋转的离心力完全平衡，其角速度为。这个牛顿宇宙正是我要稍后定义的相对论宇宙的精确类比。近似的程度再次令人惊讶地好。相对论解的角速度仅与此处的解相差一个 的因子。在这个牛顿宇宙中，-轴的特殊角色仅仅是表面的。实际上，不会存在任何实验能够区分宇宙中的一个位置与另一个位置。因为如果在一个与物质一起旋转的坐标系中描述这个宇宙，唯一可观测的表观力将是科里奥利力，因为离心力完全被引力抵消。但科里奥利力与位置无关。因此，可观测的旋转轴只有其方向，而不是其位置。这一情况使其与相应的相对论解的类比非常接近。
如果我们现在从相对论的角度来考虑这个问题，第一个出现的问题是，在相对论理论中，我们如何讨论整个宇宙的旋转，因为在这里我们没有绝对空间可以作为参考点。答案当然是，在相对论理论中，绝对空间被某种惯性场所替代，该惯性场决定了不受力作用的物体的运动。特别是，这种惯性场决定了完全自由的陀螺仪轴或摆平面的行为，并且正是相对于这种方式定义的空间方向（通过自由陀螺仪或摆平面），物质将不得不旋转。按照通常的术语，这意味着它将相对于惯性罗盘旋转。可以看到，这种旋转并不涉及一个围绕整个宇宙旋转的轴的概念。即使这个世界可能是完全均匀的，它仍然可以在每个地方局部旋转，就像我要给出的例子那样。然而，可以说，这个世界作为一个整体（类似于刚体）旋转，因为任何两个物质粒子的相互距离（通过其世界线的正交距离测量）在所有时间内保持不变。当然，也可以更具象地将这个世界视为静止的刚体，并将惯性罗盘视为相对于该刚体到处旋转的现象。显然，这种情况表明，惯性场在很大程度上独立于物质的运动状态。这与马赫原理相矛盾，但并不与相对论理论相矛盾。
现在让我们定量地研究相对于惯性罗盘的物质旋转在 和物质的速度矢量 中的意义。为此，最好使用局部惯性系统的概念。众所周知，在狭义相对论理论中，惯性系统是 取如下值的系统：
这意味着在这些坐标中，自由粒子的运动是均匀且直线的。当然，在广义相对论理论中，整个空间的这样的坐标系并不存在，但局部上是存在的，也就是说，可以这样选择坐标，使得在所考虑的点， 具有这些值，并且所有的导数 都为零，因此在所考虑点的邻域内， 的值在二阶大小范围内保持不变。微分几何的一个众所周知的定理表明，在黎曼空间的任何点，都存在满足这些要求的坐标系统。在接下来的讨论中，我有时不得不考虑在所考虑点仅满足第一个条件的坐标系统。我将称它们为“正交归一”（orthonormal），据我所知，这是正交函数系理论中用于描述相应概念的一个术语。
我们感兴趣的角速度是相对于由与物质一起运动的陀螺仪轴定义的空间方向的角速度。这是天文学家通过其测量设备观察到的角速度，他与物质一起移动。因此，我们需要引入一个与物质一起运动的惯性系统，即在所考虑的时刻其原点处物质静止的系统，或者换句话说，其原点处速度矢量
取值为
然后，由于惯性系统中陀螺仪定义的方向保持固定，我们寻找的角速度将等于在经典力学中计算出的该坐标系统原点处的角速度。因此，我们得到公式：
（这一等式显然是由于在原点处 的取值而成立的。）当然，可以不用一个惯性罗盘静止而物质旋转的坐标系统，而是选用一个物质静止而惯性罗盘旋转的坐标系统。我指的是一个物质在所有地方（不仅仅是原点）都静止的坐标系统，因此 恒为零。然后，在进一步假设坐标系统在原点是正交规范的情况下，可以得到公式：
这描述了惯性罗盘在该系统中的旋转角速度，这与其相对于物质的旋转角速度相同。这些公式直接来源于自由陀螺仪轴在四维空间中的运动是平行位移的事实。但是沿着-轴的向量 的平行位移由以下公式表示：
因此，矩阵 表示通过平行位移向量空间所经历的表观变形和旋转，并且根据弹性理论，旋转如何与变形分离是已知的。由于坐标系是正交的，两种 Christoffel 符号之间的差异消失（或者更确切地说，减少为符号上的差异），我们得到了所展示的公式。
但这两个公式仅适用于定义的非常特殊的坐标系，而显然，非常希望能有一个协变表达式用于角速度，即一个适用于每个坐标系的公式。这个协变表达式为：
你可以立即验证， 是一个反对称张量； 是通过与-张量的内积得到的相应向量。利用-张量的反对称性，你可以立即验证 总是与 正交的。此外，通过直接计算可以在几行内验证，对于这两个特殊的坐标系， （当 时）等于之前得到的 表达式，并且 为零。当然，严格来说，只有这些关系在此处证明了先前定义的合理性，即证明它们与选择的特定惯性[坐标]系无关，第二个定义同样如此。
此外，这种角速度的协变表示在以下方面也很有意义：如果 是 维空间中的一个任意向量场，则这个反对称张量 的恒等消失是存在一个-维空间单参数系统的必要且充分条件，这些空间处处正交于该场的向量。这是欧几里得三维空间中一个向量场 必须满足的众所周知的条件的直接推广，如果要存在一组处处正交于该场向量的曲面，则该条件要求：
真正的原因是，这在两种情况下都是必要且充分的条件，其原因在于：(1) 如果 的旋度（即 的 curl）为零，那么这个表达式将消失；（2）如果对 消失，那么对 也会消失，其中 是一个任意标量场，因为当 被 乘以时，该表达式仅仅乘以。包含 导数的项会相互抵消，你可以立刻验证这一点。当然，这必须是如此，因为与向量 正交的曲面的存在（或不存在）仅取决于方向，而不取决于向量的绝对值。因此，鉴于刚刚解释的关于反对称张量 消失的几何意义，我们得出结论：物质相对于惯性罗盘的角速度处处为 0，当且仅当存在一个单参数的三维空间系统，该系统处处与物质的世界线正交。现在，存在这样的三维空间系统意味着，在某种意义上，存在一个绝对时间坐标。因为这些三维空间提供了一种独立于坐标系的同时性定义。因此，物质旋转的存在与一个客观上区分的世界时间的存在之间有着非常简单的联系。
顺便说一下，这也是我得出这些旋转解的方式。我正在研究康德与相对论之间的关系，特别是研究康德与相对论物理学之间的相似性，因为在这两种理论中，都否认了牛顿意义上时间的客观存在。在这种情况下，人们会观察到，在目前已知的宇宙学解中，确实存在类似绝对时间的东西。这一点已被认识论学者指出，甚至物理学家 Jeans 也曾表示，这种情况为保留旧的直觉绝对时间概念提供了理由。因此，人们会去探讨这种性质是否是所有可能的宇宙学解的必要属性。
我在此情况下发现的旋转宇宙学解从物理观点来看存在某些缺陷。特别是，它对遥远天体没有红移，并且包含闭合类时间线。然而，就原理而言，它具有一定的兴趣，因为它无疑是最简单的旋转解，而且是唯一一个像刚体一样旋转的均匀解。线元非常简单，表达式如下：
其中 是一个常数。写成平方和的形式如下：
在此坐标系中，-线是物质的世界线。此线元是从下划线标记的三维线性元中通过添加项 来简单地扩展线元。如你所见， 未出现在前面的项中。因此，带有该 的四维空间由一个三维空间的单参数系统副本组成，其度量下划线。如果我们在这些三维空间中引入柱坐标，并将旧的-轴作为坐标系的旋转轴，我们可以得到表达式
这个线元素清晰地展示了这个三维空间的结构。它显示了围绕-轴的旋转对称性。此外，非零项 表明-线（即物质的世界线）并不正交于 平面；它们与从原点发出的半径向量正交（因为 消失），但它们倾斜于 平面，更具体地说，倾斜方向总是指向 增大的方向。因此，物质的世界线是一种螺旋线，正如在旋转宇宙中它们必须如此。
稍后我会给出该线性元素的几何意义，但首先，我想展示场方程如何可以快速直接验证，或者至少用几句话描述最容易完成的方式。你可以首先证明这个四维空间是齐次的。即，以下四个变换
（其中 到 是实数）使得此线性元素保持自身不变。但通过适当选择，你显然可以将坐标系的原点移动到任意给定点。因此，只需在一个点（例如原点）验证场方程即可。现在，在一个点处，你总是可以通过线性变换引入正交规范坐标。如果通过以下线性变换实现：
通过以下公式，你可以无需过多计算就得到黎曼张量的分量：
由于 的几乎所有一阶和二阶导数都为零，计算非常简单。结果是，仅当 有两个不同的指标 [不包括 3] 时，它们才不为零，且它们都具有相同的值：
因此，对于收缩黎曼张量的分量，你会得到仅有 不为 0，且其值为 1。因此，如果用 表示沿-轴方向的单位向量，你可以得到等式
更确切地说，这在常数 的情况下成立。一般而言，你可以得到以下等式：
然而，只要 能以这种形式表示，并且 是常数（在我们的例子中确实如此，因为空间是齐次的），相对论场方程就可以通过非常简单的方式满足。我们只需要将宇宙学常数 设置为，然后用 乘以上述等式，并从上述等式中减去结果：
由此得到的方程与场方程相同：
如果通过以下方程确定常数：
并使用之前给出的公式计算角速度，我们可以得到值
对于我们世界中观测到的密度，即 g/cm，这导致一个旋转周期为 年。这种通过直接计算验证场方程的方式当然不太有趣，但它[确实]出奇地简短。通过这种方式，你可以在不到一页的篇幅中完成包括所有细节的完整证明，以证实旋转解的存在。
在讨论该解的性质之前，我想先谈谈获得该解的启发性论证以及该线性元素的几何意义。这也将展示该解为何满足场方程的真正原因。如果我们想找到之前提到的牛顿宇宙的相对论类比，我们当然必须寻找一个静态解。此外，由于牛顿宇宙像刚体一样旋转，我们必须要求任何两个物质粒子之间的相互距离在时间上保持恒定，即，任何两条物质的世界线必须是等距的——也就是说，从一条线上某一点到另一条线的垂线，无论选择第一条线的哪个点，其长度必须相同。因此，在我们寻找的四维空间中，物质的世界线必须是成对等距的测地线系统，并且对于空间的每个点，系统中必须恰好有一条线通过它。然而，这种线系统必须不同于欧几里得空间中的平行线系统，因为沿这些线对整个空间进行平移，不能是空间的平行位移。否则，惯性罗盘（其行为由平行位移描述）将不会相对于物质旋转。
现在，对世界线系统的这种描述让人立刻想起在具有常正曲率的三维空间中的 Clifford 平行线系统。这是一个众所周知的事实（此外，我稍后会给出一个非常简单的证明），在具有常正曲率的三维空间中，存在符合我列举的所有要求的直线系统。我们寻找的与之不同之处仅在于：(1) 我们寻找的是在四维空间中的这样一个线系统，[以及] (2) 相对论的度规不是正定的，而是具有 签名。但是，这两个问题可以很容易地解决。当然，也存在具有 签名的常曲率三维空间——正如我稍后将证明的，这些三维空间（或者更确切地说，那些正曲率的）也包含 Clifford 平行线系统。从这样的三维空间中，我们可以非常容易地构造一个四维空间及其具有期望性质的线系统。具体来说，我们仅需要在线性元素中添加一项 [假设我们从坐标 到 开始]，也就是说，我们以最简单的方式从所描述类型的同构三维空间的一参数系统中构造出一个四维空间。然后我们在其中一个三维空间中定义一个 Clifford 平行线系统，并通过平移生成的线扩展它：。可以很容易地验证，这样得到的任意两条线也将是等距的。剩下要解决的问题是如何确保相对论的场方程能够得到满足。为此，我们来看一下在我们刚刚构建的四维空间中，约化的黎曼张量的形式。当然，这个空间是均匀的。因此，只需在一点 计算黎曼张量即可。此外，我们可以假设该点的坐标系是正交规范的，[并且]前三个基矢量位于所用三维空间的单参数系统的一个三维空间中。第四个基矢量将位于-线的方向上。此外，第一个基矢量可以被选定为位于选择的 Clifford 平行线系统的一条线上。
现在，关于在这种坐标系中 的值，我们可以说如下几点：所有 的分量都必须为零，因为 不依赖于，而且 是常数。从这些 的性质可以立即推导出，如果三个指标中至少有一个是 3，则所有 都为零。接下来，你可以对 推导出相同的结论，即，如果至少有一个指标是 3，则它们为零。最后，对于，根据 [上面用 和 表示的公式]，它们也为零，因此 也同样如此。因此， 的矩阵在最后一行和最后一列上只有零值。其余三维矩阵的数值由关于常曲率三维空间的公式得出，其中 是曲率：
由于假设的坐标系是正交规范的， 的值 [如矩阵 () 中所示]，因此我们得到 的矩阵形式如下：
显然，这种形式的 无法满足场方程，无论、 和单位向量 的值如何。因为如果我们将场方程写为以下形式：
我们会发现，这意味着通过从 中减去 的倍数，我们必须得到一个由自身乘积构成的张量，并且该向量还必须是类时的。但一个对角张量如果是自身乘积，则只能有一个分量不为零，而由于该向量必须是类时的，这个分量必须是第一个分量；然而，显然不可能让其余三个分量以这种方式消失。
如果我们现在回想为什么最初使用具有常曲率的三维空间进行构造，就会发现，我们这样做只是因为它们包含一个等距测地线系统。因此，如果我们能够以某种方式修改三维空间中的度规，使得所选的 Clifford 平行线系统仍然是一个等距测地线系统，那么新的三维空间将和旧的同样适合我们的目的，并且我们或许可以通过这种方式引入一个新的任意参数，使得可能满足场方程。这实际上是可能的，并且非常简单。我们只需要按照 Clifford 平行线系统的线的方向，将度规按任意比率 拉伸即可。通过拉伸一个空间（例如欧几里得平面）的度规至某一方向，我的意思是以下操作：我们将线元素 分解为一个与方向 平行的分量 和一个与其正交的分量，然后将第一个分量乘以系数，而将第二个分量保持不变，这样我们就可以得到以下表达式：
用于新的距离。根据此处的等式，这一操作意味着在原线元素上添加其在方向 上投影的某个倍数。在任何黎曼空间中都可以执行相同的操作，如果给定的方向不是，而是覆盖整个空间的一组线。如果我们用 表示由系统中所有单位长度的切向量组成的矢量场，我们可以得到以下表达式：
用于新的线元素。其中 是线元素在系统线上的正交投影，我们需要将其乘以系数 并加到原线元素上。可以很容易地看出，如果沿着我们拉伸度规的线系统是等距线系统，那么在新度规下它仍然是等距线系统（因为系统线正交方向上的距离在拉伸操作中完全没有改变）。此外，如果线系统由测地线组成，则在新度规下这些线仍然是测地线。这在我们的情况中特别容易看出，因为 Clifford 平行线系统既是空间的旋转对称轴，也是线系统的旋转对称轴。因此，它们在新度规下也将是旋转对称轴；但任何旋转对称轴都是测地线。
新度规的旋转对称性还允许人们非常容易地确定新度规的约化黎曼张量的形式，在先前定义的坐标系中，三维空间中的对称张量 确定了每个点矢量空间中的一个椭球（或其他二次曲面）。但是，如果三维空间中的椭球具有旋转对称性，则它必须是一个旋转椭球，其主轴之一是旋转对称轴的方向，另外两个主轴与其正交；其他二次曲面同样遵循这一规则。因此，在先前选择的坐标系中（其第一主轴是旋转对称轴）， 必须是对角形式，且对角线上的第二和第三项必须相等。因此，整个四维张量 的形式必须是
其中 和 显然取决于拉伸比率。因此，我们可以希望通过合适的 值选择使得。但在这种情况下， 将等于，如果 是坐标系的第一个单位向量。然而，如果 一旦具有此形式，我之前已经展示过如何在满足场方程的情况下， 为正。因为它必须等于物质密度的某个倍数。现在发现，当且仅当 时，实际上会使 且，这就是我之前定义的空间。
至于从几何解释中推导线元素的两种形式，对于第二种形式，没有太多可说的。要得到它，只需引入以 Clifford 平行线系统为-线的柱坐标系即可。为了得到第一种形式，则需要更仔细地检查我们空间的变换群。为此，让我们首先考虑普通具有正曲率的三维空间群的最简单表示。我们可以将这样的空间看作是欧几里得四维空间中围绕原点的球面，或者更确切地说是超球面。这样一个球面由以下方程给出：
现在我们可以将空间中的每个点，即球面，与一个四元数关联起来：
由于前面的方程，该四元数的范数为 1。根据定义，范数是 与其复共轭的乘积，结果是平方和：
然后，由于积的范数是各自范数的乘积，将 与范数为 1 的四元数相乘将使该球面保持不变。这无论从左乘还是右乘都成立，因此即使同时从两侧相乘也成立。因此，方程
表示球面内的一种运动，反之，球面内的每一种运动都可以通过这种方式表示。现在很容易证明，如果你将一个四元数 的幂应用于球面上的某个固定点（其中 是实数），由此得到的单维点集将构成球面上的一条测地线。为了证明这一点，只需注意到、、 位于同一条测地线上。这是因为：(a)；(b)、 和 1 之间存在线性关系，即（其中 是一个实数）；(c) 球面上的测地线由线性方程给出。如果你现在对球面上的所有点执行操作，你将得到覆盖整个空间的测地线系统，并且很明显，如果该系统中的任意两条线有一个公共点，它们将相互重合，即，该系统中的任意两条不同的线将不会相交。此外，存在将该系统中的每条线映射到自身的球面变换，即将整个球面沿着这些线平移的变换；即，变换
对于任意数，具有此属性。但从此类变换的存在可以显然得出，该系统中的任意两条线是等距的。因此，任何这样的线系统都是一个 Clifford 平行线系统。
现在，如果不是从四维欧几里得空间开始，而是从一个点之间的距离平方定义为
（有两个负平方）的空间出发，并且如果考虑一个二次超曲面定义为
而不是球面；那么不仅要使用四元数，还需要使用另一种四参数代数，其中元素的范数为
由下式给出：
这种代数可以通过以下方式最好地定义：取复数四元数，即，其中的坐标 是任意复数，然后考虑复数四元数的一个子代数，其中前两个坐标是实数，后两个坐标 是纯虚数。因此，该代数的单位元素是：
如果将 表示为，则可以得到以下乘法规则：
这些规则表明，该数系（带有实系数 到）实际上是一个代数，即，系统中任何两个数的乘积仍是系统中的一个数，并且，由于其适用于复数四元数，结合律也成立。这种代数显然应该被称为双曲四元数代数。有趣的是，该名称在文献中以不同的含义出现，但无论如何，我将在接下来的讨论中使用此名称。（或者，这种代数是否已经有一个通用的名称？）
现在可以将之前对普通四元数的所有考虑扩展到双曲四元数。双曲四元数的范数可能为零，即使对于非零四元数，这一点无关紧要，因为我们只处理范数为 1 的四元数。然而，普通四元数和双曲四元数之间还有另一个区别，即后者包含三参数子代数。例如，由 定义的子系统就是这样一个子代数。显然，该子系统的单位元素可以是以下三个四元数：
但它们通过以下乘法规则彼此再现：
现在可以预期，如果将对应于此子代数的点作为一个坐标平面，则可以获得线元的特别简单的形式。这实际上确实如此，即：你随后获得了线元的第一种形式：
【通过双曲四元数，你还可以得到该空间群（或者更准确地说，是下划线三维空间的群）的一个非常简单的表示；扩展到四维空间是显而易见的。也就是说，该空间的群显然由所有将恒曲率空间（从我们开始的空间）变换为自身的变换组成，这些变换同时也将给定的克利福德平行线系统变换为自身（当然不是每条线分别，而是整个系统）。但这些变换显然是以下形式的：
其中， 是定义克利福德平行线系统的四元数，如上所述方式。可以看到，这个群有四个实参数，即 和 中的三个参数（因为 是一个范数为 1 的双曲四元数）。因此，该四维空间的变换群将是五参数的，因为变换 也显然将空间变换为自身。当然，这个变换群必须是五参数的，因为（1）空间是齐次的，这需要四个参数；（2）它在每一点上具有旋转对称性，这为你提供了一个额外的参数。】
现在，我想对这个解的物理性质说几句话。首先，你可以立即验证，在这个解中不存在遥远物体的红移。这对于每个物质世界线等距的解都是正确的。对于任何这样的解，都可以接受一个沿物质世界线的空间平移的一参数群（或者，更准确地说，是可表示为以下形式的变换：
如果坐标系被如此选择，使得-线是物质的世界线，并且-坐标是沿这些世界线测量的“本征时间”。现在，如果你将这个变换应用于从一个物质粒子到另一个物质粒子的光线，你将再次获得连接相同两粒子的光线，但这条光线从第一个粒子出发的时间比之前晚 秒，同时也比之前晚 秒到达第二个粒子。但这意味着，从一个物质粒子发送到另一个物质粒子的光信号以相同的时间间隔到达。由此可知，光波的两个连续波峰之间的时间间隔在光的发射源和接收处是相同的，即，将会没有红移。这一考虑表明，不仅扩张意味着红移，反之亦然，[即]红移意味着扩张，无论宇宙是否旋转，即，仅靠旋转永远无法产生红移。
接下来需要考虑的问题是，在这种类型的旋转宇宙中，星系的角速度应具有怎样的值。我们之前已经得到了宇宙角速度的一个值。为了从中推导出星系的角速度，我们必须将其除以星系由均匀原始物质形成的收缩比的平方，当然，这是假设星系是均匀的球体并像刚体一样旋转的前提下。但对于数量级的粗略估计，这种假设是可以接受的。（此外，事实证明，如果质量向星系中心的集中程度发生变化，所得到的值变化惊人地小，前提是旋转中离心力和引力处于平衡状态，天文学家认为这是正确的。）现在，收缩比可以通过观测到的星系之间的平均距离与直径的比值来估算。根据哈勃的观察，这一比值大约为 200:1。因此，我们必须将这一旋转周期除以，这导致约 年的值。据我所知，星系的旋转周期仅在少数情况下被确定。对于这些少数情况，观测值略大于此数值，但并不太大。平均值大约是 年。但当然，这种一致性或分歧几乎没有意义，因为由于缺乏扩张，这个解几乎不可能是真正的宇宙。
我们的宇宙的下一个具有物理意义的性质是前面提到的闭合类时间线的存在，现在我想讨论这一点。如果你检查线元的第二种形式，
你会发现，当 足够大时， 的系数会改变符号。具体而言，当 时，，即 的系数小于零；而当 时，我们得到相反的不等式，因此 的系数大于零。如果你选择 使得，那么 的系数变为零，即，在 平面上以原点为中心的半径为 的圆是一个零线。现在，零线是光信号可能的路径。它不一定是光线在真空中的路径，因为它不需要是测地线，实际上在我们的情况下它也不是测地线。但是，通过足够数量的镜子，你可以迫使光线沿着任何零线前进。更准确地说，你可以迫使它沿着近似给定零线的测地零线多边形前进。特别是，如果你沿着一条返回自身的零线发送一个光信号，这意味着光信号将会恰好在发出时的同一时刻返回。
现在，这个圆为零线的原因，当然是由于，与物质的世界线完全一样，围绕它们的光锥在远离坐标系原点时越来越倾斜于 平面。因此最终它们的倾斜度变得如此之大以至于它们接触到了 平面。因此，如果你进一步增大，光锥甚至会与 平面相交，即，它们的部分将位于 平面以下。但这意味着还存在一些零线，这些零线从 平面上的某个点出发，围绕原点旋转，但与此同时随着前进越来越远离 平面以下，因此它们会返回到它们从物质的世界线上某一点出发的地方的 平面以下；也就是说，沿着这样的零线发送的光信号将比其发送时更早返回。所以可以将光信号发送到过去，而且，由于光的路径可以任意接近地近似为物质粒子的路径，你甚至可以乘坐具有足够高速度的火箭飞船进入过去。当然，这些考虑似乎证明了这些世界在物理上的不可能性。但由于这些粒子和光的闭合线具有巨大的半径（其量级与爱因斯坦静态宇宙的世界半径相同），这些实验超出了实际可行的量级，因此是否可以从这些考虑中得出任何结论是值得怀疑的。[因为目前可能存在某种原则上的不可能性尚未被发现。]无论如何，关于我们的宇宙是否旋转的问题，与这一问题不一定相关，因为确实还存在没有闭合类时间线的旋转解。
关于这个宇宙的时间结构，我想指出另外一个事实。看起来，由于可以进入过去，变得无法一致地区分时间的正负方向。但事实并非如此。仍然可以将光锥分为两个类别，正的和负的，这样一来，正光锥的极限仍然是正光锥——这种连续性当然是物理意义上的划分需要满足的基本要求——显然可以通过定义正光锥为包含此坐标系中-线正方向的光锥来进行划分。这带来了两个结果：(1) 一旦做出这样的定义，每一条可能的世界线都有一个明确的方向性，用于确定它上任何两个相邻点之间哪个在前，哪个在后；(2) 无论你如何在时空中旅行，你自己的时间正方向将始终与到达地的时间正方向一致。尤其是，闭合的类时线的样子总是如图 1 所示，
(图 1 )
而不会像图 2 那样，
(图 2 )
这意味着当你返回时，你会发现周围的景象像一部倒放的电影一样。
尽管整个世界存在统一的时间正方向，但另一方面，却不存在统一的全世界的时间。这一主张可以被赋予一个相当精确的意义。也就是说，可以认为坐标 是一个时间坐标，当且仅当沿着任何可能的物质世界线在其正方向上移动时， 总是增加的。在狭义相对论和非旋转宇宙中，这种时间坐标显然适用于整个四维空间。但闭合类时间线的存在意味着这样的时间坐标对于整个四维空间来说是不可能存在的。因为如果沿着闭合的类时间线在其正方向上移动，那么时间坐标 在起点和终点处必须相同（因为起点和终点重合），但这就意味着 不可能在整个线路上都增加。然而，从这个意义上说，时间坐标的存在并不是为了使解具有物理意义所必须的。相反，一个明确定义的时间正方向的存在却是必要的。例如，只有时间的方向才能决定发送光信号意味着什么，因为只有在发出信号的空间时间点的双光锥的正半部分才能找到光。时间的正方向也允许你制定热力学第二定律，至少对于封闭系统是如此；至于是否可以应用于整个宇宙，我不得而知。
我想提到我们宇宙的另一个物理特性——即在空旷空间中与旋转轴正交的光线是圆形的，也就是说，它们会回到起点，但总是在比发出时更晚的时刻到达。这一效应的结果是，当你朝着与旋转轴正交的方向观察时，你只能看到有限的距离，这又导致了一个后果：如果你沿这样的方向观察，银河系的表观密度会越来越低。但是对于今天望远镜所覆盖的距离，这种效应将非常小。光线正交于旋转轴的轨迹为圆，这再一次表明了与牛顿近似的高度一致性。如果你在牛顿近似下引入一个物质静止的坐标系（即一个旋转坐标系），那么自由粒子的路径也将因为科里奥利力而是圆形的。
关于这个旋转宇宙，还有两个有趣的事实。其一，通过对某些点的特定标识，可以用它构造有限的旋转宇宙。但它们也有一个有限的时间范围。其二，这个宇宙本质上是唯一一个空间均匀且物质世界线等距的旋转宇宙，其中“本质上”是指在大范围连接性问题上。所有其他旋转解要么表现为空间的某一方向的收缩，要么是扩张，要么是同时存在的两者，即一个方向上收缩，而另一个方向上扩张。实际上确实存在旋转且扩张的解，但我尚未以解析形式表示它们。
Lecture on rotating universes (1949b)
Kurt Gödel
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.
