How Does Relativistic Deceleration Affect Velocity Calculations in Space Travel?

[Al68]

I don't think it makes sense to explain things like $$v(t) = at/ \gamma (t)$$ and $$a_c (t) = a / \gamma (t)$$ solely in terms of Lorentz contraction

Jesse, I guess I have to agree, since length contraction and time dilation are so intertwined, that we really can't separate them.

I think you may have misunderstood what Einstein was arguing. I am sure he would have agreed that 1) the standard non-tensor laws of special relativity only work in inertial frames, and 2) as long as you stick to an inertial frame, you will always get the same prediction about the ages of the twins when they reunite. He probably did not see this as satisfactory because the fundamental distinction between inertial and non-inertial frames in 1) went against his "Machian" view of physics (are you familiar with principle[/url] and the influence it had on Einstein's thinking?), so although he wouldn't disagree that 1) and 2) would "resolve" the twin paradox in SR, I think he saw SR itself as problematic for this reason, and wanted a new theory that would apply the same laws to non-inertial observers as inertial ones. He was ultimately successful in this, since GR has the same laws in all coordinate systems, inertial and non-inertial alike. And with GR you can understand the twin paradox in terms of a coordinate system where the traveling twin is at rest throughout the journey, it just means that the traveling twin will see a gravitational field during the period where inertial observers see him accelerating--see this section of the Twin Paradox page.

Yes, I agree that Einstein saw SR as problematic because inertial frames were considered different from non-inertial frames. But according to several sources on the net, including this one, http://en.wikipedia.org/wiki/Twin_paradox Einstein's GR resolution of the Twins Paradox is considered to be faulty today. So I would disagree that he was ultimately successful. At least in this respect. It's my understanding that he never considered any of the SR resolutions of the Twins Paradox to be satisfactory.

Are you saying that the logical fallacy of proof by authority can occur only when someone uses the words "proof" and "authority"?

I've gone back and reread the passage of yours that I quoted, and, even if you did not intend it that way, I don't think it's too big a leap to intrepret what you wrote as

[(Einstein understood SR as well as anyone) /\ (Einstein viewed the twin paradox as a problem for SR] => (the twin paradox is a problem for SR)

That is not what I intended, although I can understand why it could be interpreted that way, given all the internet discussions on this issue. I am trying to understand the Twins Paradox better. I consider SR to be correct, so even if I were to ultimately choose to disagree with the accepted Twins Paradox resolution, it will not cause me to view the Twins Paradox as a problem for SR. It would cause me to view SR as a problem for the accepted Twins Paradox resolutions.

And I certainly don't believe in proof by authority, if I did, I wouldn't have so many questions.

Why questions tend to be very deep and difficult, but this doen't mean that we shouldn't ask them.

Thanks, I agree.

No, we do not assume this. It is a consequence of our definition of an inertial observer as one who remains at rest (or moving at constant velocity) in an inertial reference frame. You can test it by calculation on some simple examples.

For example, consider a simple twin paradox scenario in which the traveling twin goes out in a straight line for 5 years at a speed of 0.8c, traveling a distance of 4 light-years, turns around "instantaneously" and returns at the same speed. In the Earth's reference frame (assumed to be inertial), let us calculate the total spacetime path length for the stay-at-home twin (A) and the traveling twin (B).

Event #1 is the traveling twin's departure, at x = 0 and t = 0.

Event #2 is the traveling twin's turnaround, at x = 4 ly and t = 5 yr.

Event #3 is the traveling twin's return, at x = 0 and t = 10 yr.

The spacetime path of twin A has just one straight-line segment, with length

[tex]\Delta s_A = \sqrt{(t_3 - t_1)^2 - (x_3 - x_1)^2} = \sqrt{10 - 0)^2 - (0 - 0)^2} = 10[/itex]

The spacetime path of twin B has two straight-line segments, with total length

[tex]\Delta s_B = \sqrt{(t_2 - t_1)^2 - (x_2 - x_1)^2} + \sqrt{(t_3 - t_2)^2 - (x_3 - x_2)^2} = \sqrt{(5 - 0)^2 - (4 - 0)^2} + \sqrt{(10 - 5)^2 - (0 - 4)^2} = 6[/itex]

This calculation is easiest in the inertial reference frame in which twin A is at rest throughout, but we can use any other inertial reference frame, and get the same values for $\Delta s_A$ and $\Delta s_B$.

I think this last statement is the key. You will only get these values if you use the numbers from an inertial frame for the calculation. If we were to assume that the ship could be viewed the same way, we would get different values.

Like this: We consider the ship's frame (B) to be preferred instead of the inertial frame (A). And we say that the distance is given as 4 ly in the ship's frame. And:

Event #1 is the traveling twin's departure, at x' = 0 and t' = 0.

Event #2 is the traveling twin's turnaround, at x' = 4 ly and t' = 5 yr.

Event #3 is the traveling twin's return, at x' = 0 and t' = 10 yr.

Now the spacetime path of twin B has just one straight-line segment, with length

$$\Delta s_B = \sqrt{(t'_3 - t'_1)^2 - (x'_3 - x'_1)^2} = \sqrt{10 - 0)^2 - (0 - 0)^2} = 10$$

And the spacetime path of twin A has two straight-line segments, with total length

$$\Delta s_A = \sqrt{(t'_2 - t'_1)^2 - (x'_2 - x'_1)^2} + \sqrt{(t'_3 - t'_2)^2 - (x'_3 - x'_2)^2} = \sqrt{(5 - 0)^2 - (4 - 0)^2} + \sqrt{(10 - 5)^2 - (0 - 4)^2} = 6$$.

And of course this would add up, since the distance traveled would now be $$4/\gamma$$ or 2.4 light years each way in Earth's frame.

The only difference here is, I considered the non-inertial frame as the "preferred frame" and said that the Earth turned around relative to the ship's frame at a distance specified in the ship's frame.

I'm not saying this is correct, just that the math can be done either way. Of course, I know that I did not use an inertial frame for my numbers. And by definition, the inertial frame is the one with the longest spacetime interval. And this will always result in the most time passing for the inertial frame (between any two events). I have to repeat that I do not consider my above calculations to be correct. I just wanted to illustrate that mathematics doesn't provide the answer, it only describes the answer. And it is very easy to use mathematics to describe a wrong answer, as I think I have shown above.

It seems to me that you're asking basically, "why are inertial reference frames special?" or "why is inertial motion special?"

Yes, that is my question.