Acceleration and special relativity?

[chroot]

We can then simply accept v(T0) = aT0

If anyone is interested, I figured out my error.... and it was such a simple one, I'm almost kicking myself for not realizing it. The error was in the sentence quoted above.

The problem was that I was thinking of the acceleration of the rocket as constant from the point of view of the stationary observer. This would require the rocket continually increase its throttle, as discussed in the thread http://www.sciforums.com/showthread.php?s=&threadid=15271

Instead, the derivation that others provided is that the acceleration is constant from the point of view of a person on the rocket. This means v(T₀) does not equal aT₀. v(T₀) actually equals aT₀ * 1/<font face=symbol>g</font>. This works out to

v = at / sqrt(1 + (at/c)²)

and leads to an arcsinh, rather than an arcsin, in the final answer.

I would like to add, though, that if the rocket's acceleration is constant from the point of view of a stationary observer, then my derivation is correct. It's the right answer, just the wrong problem.

- Warren

 

[§lîñk€¥™]

I'm a resonable guy. The bottom line, though, is that if (2) is correct, then (4) is correct too... without question. It just really got my goat when you and zanket were trying to tell me the integral of (3) is not (4) -- when textbooks and CA systems agree with me.

Well, for myself, I can only apologise for my ignorance of mathematics. As you said, if [2] was correct then [4] was correct, so I can understand why you felt aggrieved at my (and Zanket's) dismissal of your answer.

But at least the good side of this was that the reason for the disagreement was found by yourself, and in doing that, you've taught me two things rather than the one thing I asked for.

kind regards
Paul