A rocket of rest length L starts at rest on Earth, then accelerates.
(all measurements will be in Earth's reference frame because the rocket doesn't have an inertial frame. If you use other frames in your answer, please define them clearly)
The fast-moving rocket is Lorentz-contracted.
This means at any particular time, the back of the rocket has traveled further than the front of the rocket. As the rocket accelerates, this difference in distance-traveled gets shorter.
That means the back of the rocket is moving faster than the front.
First question: how could I calculate the Lorentz-contracted length of the rocket when it doesn't have a single speed? Knowing the length would make the next part much easier.
Is there just one answer to how the speed/acceleration of the front and back of the rocket must be related?
Suppose the front of the rocket accelerates at uniform rate a. Then at time t it has covered a distance of (1/2 a t^2) and has a speed of (at).
If the whole rocket were moving the same speed, it would have an apparent length of L√(1-(at)^2). Since the back is moving even faster than the front, the rocket is contracted even more.
The back of the ship has thus traveled MORE than the distance 1/2 at^2 + (L - L√(1-(at)^2).
The front of the ship has had an average speed of at/2. The back of the ship has had an average speed of more than:
at/2 + L (1 - √(1-(at)^2) )/t
There's no limit to how long the rocket can be.
For any given a and t (other than 0), I could find a value of L that makes this average speed faster than the speed of light.
Is there a flaw in this argument?
Or does the front of an object have a speed limit that is less than c and dependent on the object's total length?
This scenario was based on a thought experiment in my textbook in which instead of a fixed acceleration the rocket had reached a "final speed" such that its length was L/2 (this speed should be √3/2). Does it make any difference to analyze the rocket after both ends have (at different times) finished accelerating?