Bible Thumper said:
So you're saying that at 0.99c (relative to Earth) there will be situations when objects in relative motion to me can appear going only 0.1c? Since it's a well-known fact that Maxwell's constancy of the velocity of light law is maintained for any frame of reference that's going that speed
Any frame of reference that's going what speed? Again, there is no valid inertial frame that's moving at exactly light speed, but it's true that the speed of light is always measured to be c in the frame of any observer that's moving at any speed less than c.
Bible Thumper said:
then what you're saying is that:
- At 0.999c relative to Earth, there can be some objects that, relative to me, appear to go 0.1c.
Sure. Using the
formula for the addition of relativistic velocities, if you are moving at 0.999c
away from the Earth as measured in the Earth's frame, and you measure an object moving at 0.1c
away from the Earth in your frame, in the Earth's frame of reference the object will be moving at (0.999c + 0.1c)/(1 + 0.999*0.1) = 1.099c/1.0999 = 0.999181744c, slightly faster than you. On the other hand, if you are moving away from the Earth but the object is moving
towards the Earth at 0.1c in your frame, then in the Earth's frame the object will be moving away from Earth at (0.999c - 0.1c)/(1 - 0.999*0.1) = 0.899c/0.9001 = 0.998777914c, slightly slower than you.
Bible Thumper said:
2. Once my speed goes from 0.999c to 1.0c (relative to Earth or any reference, for that matter), that object that was going 0.1c relative to me now takes an awesome jump in velocity to 1.0c?
No, it's impossible for you to move from 0.999c to 1.0c, and even if you could do that, you would no longer have your own inertial rest frame at 1.0c so it would be meaningless to ask how fast any object is moving "relative to you" (since that phrase is normally used as shorthand for 'how fast the object is moving in your rest frame'). But if you like we can look at that object that was moving at 0.1c towards the Earth in your frame when you were moving at 0.999c in the Earth's frame, and ask how fast it will be moving in your frame if you jump up to moving at 0.999999999c in the Earth's frame. I showed above that the object would be moving at 0.998777914c in the Earth's frame, and after your change in speed the Earth will be moving at 0.999999999c in your frame, so we can apply the velocity addition formula again to conclude the object will now be moving at (0.999999999c - 0.998777914c)/(1 - 0.999999999c*0.998777914) = 0.001222085c/0.001222087 = 0.999998c, very close to light speed.
Bible Thumper said:
Is this correct? I can't see how something can appear to go from 0.1c relative to me to 1.0c relative to me just because I made an incremental increase in my velocity, to 1.0c.
The incremental increase in your velocity from 0.999c to 0.999999999c in my example is accompanied by very large changes in the time dilation factor on your clocks and the Lorentz contraction factor on your ruler, as measured in the Earth frame (the clocks are slowed down by, and the rulers shrunk by, a factor of \sqrt{1 - v^2/c^2}). And each observer measures the "speed" of any object in their own rest frame using rulers and clocks at rest in their frame, in order to calculated (change in position)/(change in time) for two points on the object's path.
