I have a random paradox I seem to have found with three colliding observers, though to explain it properly I'm going to use a bit of visual aid and a bit of abstracting it from the math (or setting the math so that it works according to my example). For anyone who wants the TLDR, read this paragraph, for anyone who wants the precise details, read the next paragraph. We have 3 observers, A B and C that are some distance apart and moving towards each other. From B's point of view they all collide at the same time but from A's point of view they collide with B first, then C and then C hits B. It starts with 3 observers, A, B and C, which are each in a line and 6 units away (86542628m according to B, if you care about the math). (A->B = B->C = 6 units) I will represent each unit as a "-"; So to start with our scene looks like this: (A------B------C). From B's frame of reference, A and B are traveling towards B at c/sqrt(3) or 173085256 m/s. This number is special because when you put it through relativistic velocity addition the final velocity is 1.5 the original. This example would work with any speed but is easiest to demonstrate like this. Any observer can provide the time steps as long as they are the same time steps used across each instance, so each time step is going to be how long it takes for the observed distance from A->B to change by 2 units (1 second according to either A or B). These time steps don't effect any math or events, they are just freeze-frames for us to observe the system at and compare them to each other. Alright, so that's the groundwork. It's a little less complicated than the second paragraph makes it sound, a little more so than the first one does. So now I'm going to go through the example from the observer B's point of view. This is pretty straight forward. 0(A------B------C) 1(--A----B----C--) 2(----A--B--C----) 3(------ABC------) It is perfectly symmetrical, A and C hone in on B and collide at the same time. That is what we should expect to happen from all frames of reference. So now let's look at the point of view from observer A. Observer A will see the space between them all as contracted and will think that time is going slower than observer B did, but Lorentz contraction is homogeneous and I have chosen my phrasing and units of measurement carefully so that different clocks won't effect the problem. When adding velocities relativistically we use a different system of adding them. (Final V= (V1+V2)/(1+(V1*V2/c^2))) I'm sure everyone knows, but to be thorough; relativistic velocity addition will be less than V1+V2, especially for larger numbers. In my example relativistically adding the 2 velocities is 1.5 times the original, so C will travel 3 units of distance rather than 2 or the Galilean 4. Now let's see what happens. 0 (A------B------C) 1 (A----B-----C---) 2 (A--B----C------) 3 (AB---C---------) 4(B--AC------------) First A and B collide, then Observers A and C collide and then observers B and C eventually will collide on time step 6. Observer C will see the order in reverse. In special relativity it is not necessary for all observers to see the same order of events as other observers, but that only applies to things that are outside the cone if influence. We can find circumstances quite easily where a paradox occurs here, such as if they were 3 particles with known energy states, which one's scatter? Because of this it is clear that my proposed paradox does not have an effect in reality. I am looking for a potential resolution as to why. I have had this problem for a while now and have realized some things and ruled some others out. On principle this problem feels similar to the twins paradox, but as no acceleration ever occurs so general relativity never comes into play. The most promising sounding solution I've been offered as of yet is that observer A on creation (whenever that was) should see them self as being in the past (or seeing the others as being in their future) so should actually have their position shifted backwards by 3 seconds/time-steps. Unfortunately this falls apart because if you double the distance you need 6 time steps, and double it again and you need 12...etc which would imply that space time knows prophetically when it's next interaction will be and holds the particle in place for that long (infinity for light), which we have not observed. This reveals that instead the speed or distance traveled by observer C in comparison to observer B (according to observer A) must be altered.