[[小猪：光程差从字面上就可以理解，就是光走过的路程的“长度”差异，这个“长度”并非几何长度，而是几何长度与折射率的乘积；所以光程的严格表达是∫n(x)dx， 结果等于ct，  

小猪很容易的得出结论，在地球上静止的光纤，两个方向有不同的光程，因此有光程差。所以光纤陀螺可以测到地球的加速度。按同样的逻辑，在实验室中静止的迈克尔逊干涉仪也应该有光程差，请小猪解释。  

【【【【JQS代劳回答：

折射率的变换公式很类似于速度叠加公式：

n'=(n+v/c)/(1+nv/c)，其中n为相对于介质静止的观测者观察到的折射率，如果介质以速度v运动的话，那么这个介质的折射率就是n'，n'=(n+v/c)/(1+nv/c)。这一公式可以用Lorentz变换得到。

对“在地球上静止的光纤，两个方向有不同的光程，因此有光程差”的数学诠释：

这个折射率的差别是：n'=(n+v/c)/(1+nv/c)与n'=(n-v/c)/(1-nv/c)的差别，其中v为地球速度或者光纤速度。

echelson实验没有以上因折射率不同导致的光程差，因为空气或者真空的折射率为n=1，无论对于：n'=(n+v/c)/(1+nv/c)还是n'=(n-v/c)/(1-nv/c)，这个n'都是1。

久光先生，以上问题的答案应该自己去解决，而不是去问询别人。】】】

一段直光纤在旋转系中适当放置时光在其中相反方向传递也有光程差，

所谓有没有光程差，当然是说光程差是否等于0 。小猪想彻底否定光速不变说吗？

您不要老用“光速是否相等”这样的误导自己的术语。在非惯性系中必须明确地区分“表观光速”和“固有光速”，象您这样容易混淆概念的人，在使用上最好不要省略“光速”二字前面的定语。

自己研究下面的论文摘录去吧。

In the ECEF frame used in the GPS, the unit of time is the SI second as realized by the clock ensemble of the U.S. Naval Observatory, and the unit of length is the SI meter. This is important in the GPS because it means that local observations using GPS are insensitive to effects on the scales of length and time measurements due to other solar system bodies, that are time-dependent.

Let us therefore consider the simplest instance of a transformation from an inertial frame, in which the space-time is Minkowskian, to a rotating frame of reference. Thus, ignoring gravitational potentials for the moment, the metric in an inertial frame in cylindrical coordinates is

and the transformation to a coordinate system rotating at the uniform angular rate is

This results in the following well-known metric (Langevin metric) in the rotating frame:

where the abbreviated expression for the square of the coordinate distance has been used.

The time transformation t=t' in Eqs. (3>t' is really determined in the underlying inertial frame. It is an example of coordinate time. A similar concept is used in the GPS.

Now consider a process in which observers in the rotating frame attempt to use Einstein synchronization (that is, the principle of the constancy of the speed of light) to establish a network of synchronized clocks. Light travels along a null worldline, so we may set in Eq. (4>. Then

and solving for (cdt') yields

The quantity is just the infinitesimal area in the rotating coordinate system swept out by a vector from the rotation axis to the light pulse, and projected onto a plane parallel to the equatorial plane. Thus, the total time required for light to traverse some path is

Observers fixed on the earth, who were unaware of earth rotation, would use just for synchronizing their clock network. Observers at rest in the underlying inertial frame would say that this leads to significant path-dependent inconsistencies, which are proportional to the projected area encompassed by the path. Consider, for example, a synchronization process that follows earth's equator in the eastwards direction. For earth, and the equatorial radius is 6,378,137 m, so the area is . Thus, the last term in Eq. (7>

From the underlying inertial frame, this can be regarded as the additional travel time required by light to catch up to the moving reference point.