# Communication between two observers moving at the same relative velocity

Let's say we have two ships moving at, for example, .8c. Let's put them 1 light-year apart and give them exactly the same velocity. I want to point a laser from one ship and hit the other ship.

My initial thought is that since the speed of light is constant I'll have to aim the light ahead, so that the beam is still moving forward at .8c, and toward the other ship at .6c so they add up to $.6^2 + .8^2 = 1^2$, meaning that the maximum speed of communication between these two ships is .6c.

Someone has pointed out to me that this is wrong, since one of the core principles of relativity is that velocities are relative. Because of this we should not be able to tell that we are going .8c and communication between the two ships should be completely normal.

How does this work out? I imagine that I am correct from the perspective of a stationary observer, but time dilation will make it appear that .6c is actually 1c onboard either of the ships.

To clarify, my main question is: which direction the will the light beam have to be directed from the transmitter to the receiver. Can I point the beam directly at the receiving ship or will I have to lead it? If it can be pointed directly at the receiving ship, how can you reconcile the frame of reference of a stationary observer with the frame of reference moving with the two ships?

• Whenever you talked about speed, you must also say w.r.t whom? In your question this part is missing. If you are clear on this, then, by applying [link] (physics.stackexchange.com/questions/103907/…) special relativity formula for relative speed, you will get to know that information would flow at the same speed for all the ships, i.e. c. Commented Dec 28, 2017 at 10:01
• Obviously you have to aim your laser pointer straight into the target - receiver. The beam will move at straight line towards the receiver in your frame. However , in the frame of the "stationary" observer this beam will have the same x- velocity as you. It's x- coordinate will always be the same as yours, so it will go through the laser's tube perfectly well. Look at episode 1 in this video youtube.com/watch?v=FQKp3FU8vR8.
– user139020
Commented Dec 28, 2017 at 16:12