# Properly accelerating observer in special relativity and the synchronicity of their own clocks

I am a little confused about what a properly accelerating observer will see in special relativity. When using Rindler coordinates, we always have two observers, so I think I am getting confused as to which observer will see what, and whether that observer is looking at the original lattice or a new lattice.

Suppose we have two frames $$K$$ and $$K'$$ at rest relative to one another. By frame I mean a lattice of rulers indicating unit lengths and clocks which register events at a certain rate (light clocks). Both $$K$$ and $$K'$$ initially share the same lattice and origin $$(t,x) = (0,0)$$. At this origin event, $$K'$$ experiences a proper acceleration $$\alpha$$.

1. Does $$K'$$ continue to see their lattice of clocks tick at the original rate during the course of the acceleration? At any instant, they can be thought to occupy an instantaneous coincident inertial frame, and in inertial frames all clocks tick at the same rate, to any observer, either co-moving with the frame or not. So at any instant are they measuring events against this instantaneous inertial frame, where all clocks are in sync? Will they then notice the clocks slowing down, as they progress through new inertial reference frames?

2. How does $$K$$ see the rate of the clocks in $$K'$$ tick spatially? If $$K'$$ is a distance $$d$$ away, do they see the clock at $$d$$ tick at the same original rate as any one of their clocks in $$K$$?

3. This video shows how an accelerating observer in a rocket ship would see some lattice of lengths: https://www.youtube.com/watch?v=Otp-eBHLJzQ. Is this lattice of lengths observations of the original lattice, now belonging to $$K$$? The lattice of lengths belonging to $$K'$$ appears not to change during acceleration, correct?

Thanks.

• If that is supposed to be a "first person" view, how come you can see the spaceship? There are some good elements to the visualization, but it seems to be more for entertainment than education (you can tell from the comments that most people don't get it at all). It is more akin to the "unphysical" view in my videos here: youtube.com/playlist?list=PLvGnzGhIWTGR-O332xj0sToA0Yk1X7IuI - my others are all genuinely first person. Commented Oct 2, 2023 at 17:43
• it would help if you included the world line of the $K'$ origin in the coordinates of $K$.
– JEB
Commented Oct 2, 2023 at 21:47

Suppose we have two frames $$K$$ and $$K'$$ at rest relative to one another. By frame I mean a lattice of rulers indicating unit lengths and clocks which register events at a certain rate (light clocks). Both $$K$$ and $$K'$$ initially share the same lattice and origin $$(t,x) = (0,0)$$. At this origin event, $$K'$$ experiences a proper acceleration $$\alpha$$.

Although you expressed your opening paragraph in terms of Rindler coordinates, what you have described here is different. In Rindler coordinates the acceleration, $$\alpha$$, is constant (as a function of time), whereas here you have $$\alpha$$ change at $$t'=0$$. That is fine, but we will need to use a slightly more general metric than the Rindler metric.

One thing that you did not specify, but which I infer from your question, is that you wish the lattice $$K'$$ to remain rigid. This is possible in principle, and the resulting motion is called Born rigid motion. The motion of the entire non-rotating lattice is uniquely determined by the motion of the origin and the requirement of Born rigid motion.

1. Does $$K'$$ continue to see their lattice of clocks tick at the original rate during the course of the acceleration?

No. See below.

At any instant, they can be thought to occupy an instantaneous coincident inertial frame,

This is not correct. The metric in the accelerated frame is not the standard Minkowski metric, but rather it is given by $$ds^2 = -\left(1+\frac{\alpha_x x}{c^2}+\frac{\alpha_y y}{c^2}+\frac{\alpha_z z}{c^2} \right) c^2 dt^2 + dx^2 + dy^2 + dz^2$$ where $$\alpha_x$$, $$\alpha_y$$, and $$\alpha_z$$ are functions of $$t$$ describing the acceleration of the origin. For $$\alpha_y = \alpha_z =0$$ and $$\alpha_x = \alpha$$, a constant, this reduces to the usual Rindler metric in Kottler–Møller coordinates, which is not the metric of an inertial frame.

and in inertial frames all clocks tick at the same rate, to any observer, either co-moving with the frame or not. So at any instant are they measuring events against this instantaneous inertial frame, where all clocks are in sync?

In this frame there is a pseudo-gravitational time dilation of the lattice clocks given by $$\gamma = \left(1+\frac{\alpha_x x}{c^2}+\frac{\alpha_y y}{c^2}+\frac{\alpha_z z}{c^2} \right)^{-1/2}$$ therefore the clocks do not remain in synch. The clocks lower in the pseudo-gravitational potential tick more slowly and thus progressively lose synchronization with higher ones.

1. How does $$K$$ see the rate of the clocks in $$K'$$ tick spatially?

They are time dilated according to the ordinary time dilation of moving clocks in inertial frames.

I cannot comment on the video.