# Super luminal speed paradox? [duplicate]

This question already has an answer here:

Chosing a reference frame in which the Earth is at rest and doesn't rotate is related, but different. Here I'm asking if there is a paradox and if my attempt at resolution has merit, and if not where exactly does it fail. Attempt at resolution poses a scenario in which motion of the lab frame is safely assumed to be zero and inertial.

The Earth rotates a full cycle every 24 hours. Alhpa Centauri is 4 light years away. The Newtonian formula for speed of an object moving in a circle is $$\vec{v}=\vec{\omega}\times \vec{r}$$

Where $$\vec{v}$$ is velocity along the circle, $$\vec{\omega}$$ is angular velocity and $$\vec{r}$$ is distance from axis of rotation.

Using the above numbers, we get $$v=(2 \pi/24hrs)\times 4lyrs=(\pi/3)(lyrs/hr)$$

$$\frac{1 lyr}{1hr}=\frac{1\times c \times 8766hrs}{1hr}=8766c$$

This means $$v=(\pi/3)(8766c)>c$$, an impossibility.

Does Alhpa Centauri hit superluminal speeds in any coordinate system ? I'm skeptical, based on the following.

Possible Answer

Consider a frame, spinning in the lab about its z-axis, that axis being oriented towards Alpha Centauri. Consider another Cartesian frame not rotating in the lab. Set up a "light clock" in the lab consisting of a light beam bouncing between two mirrors, one on the floor, one on the ceiling, separated by a distance $$L$$.

The motion of the light beam moves perpendicularly to the direction of relative motion. So the distances along the z axis are expected to be preserved. The same principle applies to distances along the r-axis. In addition, a light beam moving radially in ward or outward for one frame should have the same motion in the other. It could be detected and this probably null, not only by the above argument, but arguments from isotropy.

Let primes represent measurements in the rotating frame, unprimed, the non-spinning lab frame. By the above, Then for one half tick of the lab's clock $$r_0=r_0'$$, $$\Delta{z}=\Delta {z'}=L$$:

$$c\Delta t=L$$ $$c^2(\Delta t')^2=r_0^2 d\theta'^2+L^2$$

Swapping out $$L$$, and with some algebra:

$$\Delta t'^2=\frac{(\Delta{t})^2}{1-\frac{r_0^2\omega'^2}{c^2}}$$

Where $$w'=d\theta'/dt'$$. So there is a formula for time dilation.

Letting $$\Delta{t}=L/c:$$

$$\Delta t'^2=\frac{L^2}{c^2-r_0^2w'^2}$$

Divide both sides into $$L^2=\Delta{z'}^2=\Delta z^2$$.

$$\frac{L^2}{(\Delta t')^2}=c^2-r_0^2w'^2$$

$$r_0^2w'^2=c^2-\frac{L^2}{(\Delta{t'})^2}$$

So:

$$r_0^2w'^2+\frac{L^2}{(\Delta{t'})^2}=c^2$$

The first term on the left side of the equation represents angular motion. The second term represents motion along the z axis.

$$(\frac{\Delta{z'}}{\Delta{t'}})^2=c^2-r_0^2w'^2$$

The total squared distance traveled according to the rotating frame is $$c^2\Delta{t'}^2$$. From this we subtract the vertical squared distance traveled, then divide by $$\Delta{t'^2}$$ to get the sum of the square of the components of other contributors to velocity:

$$v_{alt}^2=\frac{w'^2r_0^2L^2}{c^2-r_0^2w'^2}\frac{1-w'^2r_0^2/c^2}{(L^2/c^2)}=w'^2r_0^2$$

So regardless of any object's position away from the common axis, the square of either component will be less than $$c$$. So Alpha Centauri is not traveling at the speed of light however far away it might be or however rapidly the earth rotates.

## marked as duplicate by John Rennie, Jon Custer, Kyle Kanos, user191954, Aaron StevensNov 3 '18 at 20:23

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

## 1 Answer

Short answer: A spinning frame of reference is no inertial frame.

Longer answer:
To measure any physical property directly, you need to measure from inside an inertial frame, i.e. a non-accelerating frame. (Or a free falling frame in general relativity.) If you want to use a non-inertial frame, you have to take extra care of accelerations within your frame.

Earth is spinning. A point on a spinning surface will constantly change it's velocity. Thereby, a spinning surface is no inertial frame. So, before you take any measurement from the surface of the earth, you have to take into account it's own movement.

Another way to think of this is the reverse: point a laser-pointer towards Alpha Centauri. Once the photons hit AC, they would whizz across it at much above the speed of light. However, no Alpha Centurians could exchange information with this process: The photons hitting one spot of AC are not the same photons hitting the next spot. Though the point is quite fast, it's not a physical object, but rather an interpretation.

Similarly, while photons from AC quickly change direction over the course of one day, that's not a physical motion of the source, but rather our interpretation of what we see on an accelerated frame of reference.

P.S.: Kudos for deriving special relativity.

• The surface of a cylinder is intrinsically flat. The light beam moving only vertically in the lab frame is moving along the surface vertically and laterally. The direction is changing, but not the speed. $r_0$ is expected to be constant. What about $w'$? The light beam is essentially traveling along a cylindrical surface in a path the shape of a helix whereas the geodesic between source point and AC would not make a full turn. This suggest a pseudoforce that can be added to the geodesic equation allowing for the motion. Perhaps it could help compensate for the non-inertia aspects. – R. Romero Nov 1 '18 at 17:35
• Do we need to factor in the actual earth frame in detail? In the thought experiment, the lab is assumed to be inertial. The motion of AC can be modeled in the lab frame as a point of constant position 4 ly away from the origin. From here, the question is, how fast does AC move according to the rotating frame. By my math above, it would be less than $c$ however far it is. – R. Romero Nov 1 '18 at 18:08
• You need to factor in (or rather, factor out) any acceleration. Acceleration in this case means rotation. So yes, you would have to account for any rotation before measuring speed. – GammaSQ Nov 1 '18 at 19:04