# Special Relativity Paradox at low speed

Admittingly this seems rather simple to be a real problem with the theory so there is problably something I’m not understanding. However I’m very interested on how to overcome this paradox. So here goes.

It involves 4 observers A1, A2 (not moving wrt each other) in inertial reference frame A; B1, B2 (not moving wrt each other) in ref B; and a light flash source in space. A1, A2, B1, B2 have synchronized clocks.

A1 is situated at the light source and is not moving wrt the light source, A2 is a far distance away. The flash goes off at exactly t=0 On A1’s clock. The light reaches A2 at exactly t=6000 hours on his clock. Simply enough.

B1 and B2 are always located on the same line that A1 and A2 are on. B2 is moving toward the light source. B1, B2 are moving 50km/h wrt to A1,A2 such that B1 is coinciding with A1(and the source) at t=0, and B2 is coinciding with A2 when the light reaches A2.

So from the info above: when the flash goes off at t=0, B2 is 300,000km further away from the source than A2 is. The question is what time is it on B2’s clock when the flash hits him???

Just looking at ref frame B’s perspective the light should take 6000 hours and 1 second to reach B2 since the distance is 300,000km longer than A1 to A2.

However the light hits A2 and B2 in the same spot and same instant. B2’s clock is synchronized with A2’s. So how can they have a different reading???

I am only considering invariant speed of light here. Time dialation and length contraction are negligible at these speeds. And I don’t see a problem with synchronizing clocks at these speeds.

Thank You

• When you say "A1, A2, B1, B2 have synchronized clocks", what exactly do you mean? Whose clocks are synchronized with whose? For observers moving relative to each other, at what time are they synchronized? And for observers not in the same location, how are they synchronized? – David Z Mar 13 '18 at 1:26
• All the clocks are synchronized. All of them could be together beforehand and then synchronize. Then they all could move out slowly to get into the positions described. In the problem described the speed difference is only 50 km/h so I don’t see how this could be a problem (or could it?) – Alex777 Mar 13 '18 at 1:56
• It is a problem. Whenever relativity is involved, you can't have clocks in relative motion that stay synchronized. – David Z Mar 13 '18 at 3:31

## 3 Answers

The problem comes from assuming that the clocks in frame A and the clocks in frame B can be synchronized. This would mean that all four would agree on some event occurring simultaneously ($t = t' = 0$), but this is impossible.

Let's look at the Lorentz transform for the starting time of ship $B_2$ (the receiving ship) at the point where the light pulse is released ($t_0 = 0$): $$t'_0 = \gamma\left(t_0 - \frac{vx_0}{c^2}\right) = -\frac{\gamma vx_0}{c^2}$$ where $x_0$ is the distance between the B ships in the A frame, $v$ is the velocity of the $B$ ships in the A frame, and $\gamma$ is the Lorentz factor of the B ships' velocity. Notice that this is not zero, whereas the clock on $B_1$ would be observed to read zero according to the A ships. So, the A ships observe that the B ships are not starting their clocks at the same time. In fact, plugging in the numbers from your setup, $$v = 13.9\ \textrm{m/s},$$ and $$x_0 = 6000\ \textrm{hr}\times 3600\ \textrm{sec/hr}\times 3\cdot10^8\ \textrm{m/s} = 6.5\cdot10^{15}\ \textrm{m},$$ results in a difference of $+1\$second ($v$ and $x_0$ have opposite signs since $B_2$ is headed towards the origin). This means that, according to the A-frame ships, ship $B_2$ would have started its clock one second before $B_1$ would have seen the light pulse--that is, according to the calculation above, ship $B_2$ already has one second on its clock when the light pulse is released).

Spatially separated clocks with different velocities cannot be synchronized. The A ships will say the B ships' clocks did not start at the same time, and vice versa. The A clocks can be synchronized with each other and will keep the same time forever after because they are at rest with respect to each other. The same is true of the B clocks. Clocks $A_1$ and $B_1$, because they come to the same location, can synchronize their clocks. Afterwards, they will observe the counterpart clock to run at a different speed given by a constant factor ($1/\gamma$). The same is true of $A_2$ and $B_2$. However, it is impossible to get all four clocks to agree on a time to call $t = 0$ since simultaneity is frame-dependent.

• Thank you so much for your response. I briefly reacquainted myself with the Lorenz T. I meant for my question to be more of a though experiment. – Alex777 Mar 13 '18 at 13:29
• @Alex777 For a quick summary: time dilation and length contraction are negligible in this setup (contributing only nanoseconds to the time difference). It is the inability of the clocks in different frames to be started at the same time that causes the difference in time measurements. – Mark H Mar 13 '18 at 13:44
• Sorry posted the last comment prenmature. I was going to say that the Lorenz T is not needed here since you would come up with the same answer if the Lorenz factor =1. And the premises that my question is contending is the invariant speed of light which the Lorenz Transformation uses as one of its basic assumptions. After reading your answer, I definitely agree that it comes down to synchronizing the clocks. But why can’t you synchronize the clocks within acceptable limits? – Alex777 Mar 13 '18 at 14:06
• Theoretically Suppose all the observers start at the source and synchronize there, then they move out from the source at 0.1 m/s into the necessary positions (would take a long time) but the total time dialation would be 0.001 seconds (quick unreliable calculation). Then there is the acceleration to 50km/h, but it’s ONLY 50km/h. Couldn’t you know the clocks are in sync to some degree of accuracy? – Alex777 Mar 13 '18 at 14:21
• I may have got you thinking? ;) or not :( – Alex777 Mar 13 '18 at 14:36

Your mistake is here :

the light should take 6000 hours and 1 second to reach B2 since the distance is 300,000km longer than A1 to A2.

This is wrong because you constructed the problem to make sure B2 was at A2 when the pulse arrived at A2.

The "extra time" ( 1 sec ) would take it to B2's starting position, not it's position when the pulse reaches it.

If you do the math, it turns out that the time T it takes for light to go from A1 to B2 in the B frame is:

where t is the time in the A frame. Using 6000 hours for t and 50 km per hour for v we get T~6000 hours and 1 second.

Trust the Lorentz transformation and the math.