TheScienceOrca said:
Lets say Alice is floating in the middle of the universe at 0,0
a ship at 5,0 is moving up the y-axis at .9c parallel to alice (x will always stay 5).
The x dimension is superfluous here, and it will make the math easier if we leave it out and only deal with the ##t## and ##y## coordinates. That's what I'll do below; everything I do will be valid regardless of what ##x## coordinates (or ##z## coordinates, for that matter) we assign to Alice and the ship and the bullet, as long as they are all constant.
TheScienceOrca said:
If this rocket also shoots a bullet in the same direction (up the y-axis at .9c relative to itself).
You stated the bullet would be moving .994c relative to alice and the rocket still .9c.
Yes. (More accurately, 0.9945c, which is the accuracy I'll use in calculations below.)
TheScienceOrca said:
In 1 second relative to Alice, according to your statements the bullet would be .994c LS away and the rocket .9c LS away.
Yes.
TheScienceOrca said:
Which means the bullet is only .094c away from the rocket
Relative to Alice; *not* relative to the rocket. Distances get transformed when you change frames, just like velocities do. Also there is relativity of simultaneity to consider. See below.
TheScienceOrca said:
even though the bullet is traveling .9c relative to the rocket.
But not relative to Alice. You have to be very careful not to switch frames in mid-stream, so to speak, which is what you did in the sentence I just quoted (in two parts so you can see exactly where you switched--in between the two parts I quoted).
TheScienceOrca said:
This is because as 1 second has passed for Alice, but not a full second for the bullet right?
No, it's more than that. Let's work out the coordinates that you asked for.
We have the following events, given with their coordinates in Alice's frame:
Event A: Alice and the ship start out co-located at (0, 0), and the ship fires the bullet at the same instant. The ship moves at 0.9c relative to Alice, and the bullet moves at 0.9945c relative to Alice.
Event B: The ship is located at (1, 0.9) after 1 second relative to Alice. (Note that we're using coordinates in which time is in seconds and distance is in light-seconds.)
Event C: The bullet is located at (1, 0.9945) after 1 second relative to Alice.
Now what we want are the coordinates of events B and C relative to the ship (note that event A has the same coordinates relative to the ship, since it's the origin of both frames). This is easily obtained via the Lorentz transformation; if ##t, y## are the coordinates relative to Alice, and ##t', y'## are the coordinates relative to the ship, then we have:
$$
t' = \gamma \left( t - \frac{v y}{c^2} \right)
$$
$$
y' = \gamma \left( y - v t \right)
$$
where ##\gamma = 1 / \sqrt{1 - v^2 / c^2}##. For ##v = 0.9c##, we have ##\gamma = 2.294##, and this gives the following event coordinates relative to the ship:
Event B: (0.4359, 0) (note that we expect ##y' = 0## here because the ship is at rest at ##y' = 0## in its own frame)
Event C: (0.2408, 0.2168)
Note carefully several things:
(1) In the ship's frame, event B happens *less* than 1 second after event A. This is an example of time dilation: only 0.4359 seconds elapse on the ship between two events that are 1 second apart for Alice.
(2) In the ship's frame, event C happens *before* event B (whereas in Alice's frame, they happen at the same time). This is an example of relativity of simultaneity: events that are simultaneous in one frame are not simultaneous in another frame. But it also means that, if we want to know how far away the bullet is from the ship at event B, in the ship's frame, event C is the *wrong* event to look at. Instead, we need to look at:
Event D: (0.4359, 0.3923) is the event where the bullet is, in the ship's frame, when the ship is at event B (note that the time of this event, in the ship's frame, is the same as the time of event B). The bullet is 0.3923 light seconds away from the ship, in the ship's frame, when 0.4359 seconds have elapsed, because the bullet is moving at 0.9c relative to the ship. (We can verify this, by the way, by taking the ratio of ##y'## to ##t'## for event C; ##0.2168 / 0.2408 = 0.9##, as expected.)
And just for completeness, we can transform the coordinates of event D back to Alice's frame, simply by inverting the sign of ##v## in the equations above; this gives
Event D: (1.81, 1.8) in Alice's frame. Notice that the ##y## coordinate here is 1.8 = 0.9 + 0.9; this is not a coincidence. It has to be that way because of how all the math combines: the way velocities add, and the way coordinates transform. It might be instructive for you to work out, from the various equations already given, how this comes about.