# Universal reference frame based on the speed of light/special relativity

Everything I've read has said that there is no universal static reference frame, but based on my understanding of special relativity I don't understand why it can't be determined.

Here is an experiment I can think of that would seem to determine "universal static", what am I missing? Assuming 1 dimensional space for a second, our reference frame (a spaceship) is moving and an indeterminate speed $V_0$. We take two probes with atomic clocks and send one in each direction at some fraction of the speed of light ($V_0+V_p$ and $V_0-V_p$). After a period of time, the probes return at a slower speed than they left (taking advantage of the fact that time dilation effects increase exponentially with velocity). Upon returning, the clock in the probe sent in the positive direction ($V_0+V_p$) should show less time past than the clock on our spaceship, and the probe sent in the opposite direction ($V_0-V_p$) should show that more time has past than the clock on our spaceship, this tells us which direction we are moving with respect to "universal static", and with repeated experiments we can determine our exact velocity relative to it.

• How can you send probes out at $V_0+V_p$ and $V_0-V_p$ if you don't know what $V_0$ is? – Brian Moths Nov 30 '17 at 20:21
• You don't know what V0 is, that's what you are determining experimentally, but if you fire a probe in both directions, their speeds will be V0+Vp and V0−Vp – Steven Pick Nov 30 '17 at 20:38
• No, their speeds will only be $V_0+V_p$ and $V_0-V_p$ if you fire them with the appropriate velocities, which can't be determined unless you know $V_0$, so this experiment is impossible to conduct in the first place. – Brian Moths Nov 30 '17 at 20:41
• Vp is known, V0 is unknown, so the resulting velocities with respect to the to be determined universal reference frame is V0+Vp and V0−Vp. What is impossible about that? – Steven Pick Nov 30 '17 at 20:51
• That's not how velocities add. Also, time dilation is not exponential in velocity. You seem to assume that an absolute reference frame exist, and time dilation is wrt to this frame. This is not how relativity works. – fqq Nov 30 '17 at 21:10

It is hardly possible to detect “universal” frame this way. Non-inertial clock will dilate at the same magnitude when it meets inertial clock again despite of direction of its motion.

Your question looks like the twin paradox in the preferred frame theory (Lorentz Ether theory). It is often said that this theory is empirically equivalent to the SR. That means, this theory predicts the same outcome of your thought experiment (or any other experiment) as Special Relativity. In Lorentz theory, moving observer lives by his own “local time”, which is different from “absolute, universal” or “ether” time.

https://en.wikipedia.org/wiki/Lorentz_ether_theory

It should be noted, that there is no "paradox" related to the ether theory. In the ether theory, the paradox that has arisen in the depths of special relativity is resolved by means of elementary algebraic methods.

We can simulate the both cases that you propose in the framework of the ether theory, so as to see whether it is possible to detect absolute frame.

1) Let’s consider what would happen if one of two twins who are at rest in the ether at one point, flies at speed $v$ to a distant point and then after a while returns to twin $A$ remaining at rest.

If for the twin flying in the ether his “local time” characterizing the rate of physical processes in his body and the pace of the movement of his clock on both segments of his flight (there and back) slows down due to interaction with the ether, then the lapse of his “local time” will be $1/\sqrt {1-v^2/c^2}$ times less than for the twin at rest in the ether, and the “travelling” twin will get less “old”. The turn of the travelling twin, provided it is virtually instantaneous, has no practical effect on the ratio of times of both twins.

2) Now let's calculate, what will happen if the two twins are flying side by side in the ether at speed $v$ – with their “local time” passing slower – then one of them stops, staying at rest in the ether for some time, then catching up with the travelling twin. I think it is equivalent to your probe, that was sent in “opposite direction”. The twin who continued his flight in the ether with no information about the fact of his motion in the ether perceives this maneuver of his brother as a round trip to a distant point.

An obvious answer is that, since according to the ether theory after the twin’s stop in the ether his time will pass faster than the “local time” of his twin brother who continues his flight, and then when the twin stopping in the ether after some time catches up with the missing brother, he will age more than the latter. The “local” time of the twin catching up with his flying brother will actually flow slower than for the flying brother. This is due to the faster speed of the twin catching up with his brother. As a result, the brother making a stop in the ether will age not more, but less than his twin brother who has not interrupted his flight.

Let us demonstrate that if the proper times of the motion there and back of the non-inertial twin relative to the inertial twin are equal, then for the non-inertial twin it will take $1/\sqrt {1-v^2/c^2}$ times less time than for the moving inertial twin, and the non-inertial twin will age less.

Let at the time of stop of one of the twins in the ether the clocks of the parting twins show zeros. Suppose that after making a stop for some time, the twin who has lagged behind, at the moment $t_1$ of the ether time when his clock (because of the stop) was showing this time, left at speed $u$, such that $v<u<c$, following his brother flying away from him. The distance between the twins at the start of the twin who has left behind is equal to $vt_1$. Setting out, the twin left behind will catch up with the twin flying at a constant speed $v$ at the point in time $t_2$, having spent the time equal to $vt_1/(u-v)$. During this period, by the clock of the twin following the flying away brother at speed u, there will be a lapse of proper time, which is $1/\sqrt {1-v^2/c^2}$ times less than the ether time and equals $vt_1\sqrt{1-(u/c)^2}/(u-v)$. Let us assume the velocity $u$ such that the proper time $t’_2-t’_1$ of the catching up twin is numerically equal to the time $t_1$ of his stay at rest relative to the ether, i.e. $t’_2-t’_1=t_1$ or

$$t_1=vt_1\sqrt {1-(u/c)^2/(u-v)} (1)$$

This equation meets the condition under which the twin spends the same proper time on a trip to a distant point and back. By elementary transformations of the equation (1) we can obtain the value of velocity $u$, which is equal to $\frac {2v}{1+(v/c)^2}$ . Substituting this value in the expression for the time $vt_1/(u-v)$ required for the return of the twin, and summing the time $vt_1/(u-v)$ and the time $t_1$, we obtain the ether time spent by the lagging behind twin on the stop and return to the flying twin. This time is equal to $2t_1/(1-v^2/c^2)$. Since the clock of the inertial twin flying at a speed $v$ go $1/\sqrt {1-v^2/c^2}$ times slower than the clock at rest in the ether, the flying twin will determine the time spent by the lagging behind twin on the stop and return to the flying twin as a quantity meeting the equality:

$$t’_2=2t_1/\sqrt {1-(v/c)^2}$$

Since the time elapsed for the non-inertial twin by the moment of his return is numerically equal to $2t_1$, and the time of the inertial twin is numerically equal to $2t_1/\sqrt{1-(v/c)^2}$, then the lapse of time for the non-inertial twin is $1/\sqrt {1-v^2/c^2}$ times shorter, and he has aged less than the inertial twin has.

This way we can see, that non–inertial twin (or clock) will show $1/\sqrt {1-v^2/c^2}$ less time than inertial one, despite of direction of its motion in ether. We get absolutely the same result as in the Special Relativity, but resolution of the paradox is very simple.

https://arxiv.org/abs/1201.1828