How Does Relativistic Deceleration Affect Velocity Calculations in Space Travel?

[Al68]

I was assuming a purely inertial situation, like a ship that moves inertially past the Earth rather than one that takes off from Earth after originally being at rest with respect to it. If the ship accelerates, the rod can't stay rigid, so in the Earth frame the back end of the rod will not begin to accelerate with the ship until well after the ship and the front end begin to accelerate--the back end will only start to accelerate once a signal moving at the speed of sound in the rod has reached it from the front end, starting at the moment the front end began to accelerate. Only after all parts of the rod have stopped accelerating and the rod is moving inertially again will its length be constant in the Earth's frame, and its new length will be shorter than its length before acceleration, because of Lorentz contraction. But like I said, it's easier just to think in terms of a ship that moves past the Earth inertially without ever accelerating.

It almost looks like, with an accelerating ship, you were right the first time, when you said that distance isn't like the length of a rigid object. Would this be correct?

Thanks,
Alan

 

[JesseM]

It almost looks like, with an accelerating ship, you were right the first time, when you said that distance isn't like the length of a rigid object. Would this be correct?

Once again, there is no such thing as "the" distance for an accelerating ship, because there is no single standard choice of coordinate systems for accelerating objects. If you use the specific type of coordinate system that I labeled "NI" in that long post from page 1, where the accelerating observer's definition of distance at any moment is defined to match that of their instantaneous inertial rest frame, then the argument I made in that post shows why we can figure out distance at any moment in the NI coordinate system by considering the distance in the rest frame of a passing inertial observer:

Now, this conclusion shouldn't be altered if we are talking about the instantaneous co-moving inertial rest frame of an accelerating observer rather than the constant inertial rest frame of an inertial ship. After all, we could imagine that we have both an inertial ship and an accelerating ship moving away from the earth, and that they cross each other's paths at the moment that they are at a distance d in the Earth's frame (they must have departed Earth at different times for this to work); at the point in spacetime where they cross, they both have the same instantaneous inertial rest frame, so they would both measure the same distance to the Earth in this instantaneous frame.

Of course, the distance in the NI system would not match that of a physical ruler carried by the accelerating observer, because that ruler would be constantly deformed due to the acceleration and thus wouldn't be a good measure of distance at all. But this is why I keep trying to emphasize that there is no standard way for an accelerating observer to define things like distance, because there's no single obvious physical way to measure them like for the inertial observer, so your choice of which non-inertial coordinate system to use is fairly arbitrary with no reason to say one is a better representation of the accelerating observer's "point of view" than any other.

 

[pervect]

But this is why I keep trying to emphasize that there is no standard way for an accelerating observer to define things like distance, because there's no single obvious physical way to measure them like for the inertial observer, so your choice of which non-inertial coordinate system to use is fairly arbitrary with no reason to say one is a better representation of the accelerating observer's "point of view" than any other.

I'm not sure how much this will help the discussion, but I can talk a little bit about the reasons the coordinate system commonly used for an accelerated observer is the coordinate system of an instantaneously co-moving observer.

Let's start with an object at a point. Through that point, there are an infinite number of vectors in 4-d space-time that could be consider the direction of "time".

When we settle on the velocity of an observer at that point, though, that picks out a unique vector that represents that particular's observer's concept of "time".

There will be a 3-dimensional space orthogonal to that time vector. This constitutes that particular observer's notion of "space".

The basis vectors that represent "space" at that point can be picked arbitrarily if one doesn't mind a rotating coordinate system. Rotating coordinage systems are sometimes annoying. To avoid this annoyance, there are procedures for picking out a "non-rotating" space vector. I won't go into the details at this point, I'll just mention that it's something that can be done with enough effort. Of course our choice is not really unique, we have a rotational degree of freedom in picking out our coordinate axes.

We've now set up a purely local coordinate system at that point. We now have to discusss how we extend that coordinate system to cover significant distances, rather than infinitesimal distances.

Relativity has a concept that's roughly similar to a "straight line" - it is called a geodesic.

The previous efforts have given us the information needed to construct the various time-like and space-like geodesics ("straight lines") passing through our initial point. These geodesics form the coordinate axes of our coordinate system. While we don't have to use relativity's notion of a "straight line" (i.e. a geodesic) to define how these coordinates propagate, it is a convenient and naturally choice, to make our axes the best possible equivalent to a "straight line".

The next issue comes up - how do we make "tic marks" on these coordinate axes, to represent equal intervals? We can use the concept of the Lorentz interval along the geodesic path to define where our "tic-marks" should be, such that distance between tic marks, along our geodesic paths, is uniform. This is also a rather convenient choice. (The technically minded might appreciate that this particular choice of tic-mark spacing makes the Christoffel symbol \Gamma^x_{xx} zero.

If we carry out this procedure in flat space-time, the above procedure will necessarily define as our coordinate system the coordinate system of an inertial observer at that point.

The above procedure is a bit more general than this, though, in that it can be carried out in curved space-times (where inertial frames do not really exist) as well as flat space-times (where inertial observers do exist).

So we can see that using the instantaneous inertial coordinates for the coordinate system of an accelerated observer is a rather "natural" choice. But it does have a limitation. Eventually, the different geodesic lines cross. In Euclidean geoemtry, parallel lines never cross, but unfortunately in the more general case, they do, and in particular, for any accelerated observer, a simple diagram shows that the timelike geodesics through different points in the trajectory of the accelerated observer (which we which to specify as the "origin" of our coordinate system) must intersect.

When this happens, when the geodesics ("straight lines") intersect, the coordinate system is no longer a 1:1 mapping. The result is that this general procedure for creating a coordinate system of an accelerated observer has a limited "range" - it is only a "local" coordinate system. The range limit is equal to c^2/a, where a is the accleration of the observer.

 

[JesseM]

I realize I was simplifying things a bit for the sake of the argument, in practice there may be many cases in general relativity where one particular coordinate system may be seen as the most "elegant" choice, or the easiest to work with mathematically (aside from the case of an accelerating observer, physicists might also say that the most elegant coordinate system for dealing with the spacetime outside a black hole is Schwarzschild coordinates, or that the most elegant coordinate system for dealing with cosmology is the one where the distribution of matter and energy appears most uniform in space). But the theory of special relativity was originally designed around the idea of inertial reference frames, and they have a very simple physical interpretation in terms of local readings on a network of inertial rulers and clocks; in general relativity there isn't anything like this, GR was actually designed around the idea that you should be able to use any coordinate system you want (this can be true in SR too if you formulate the laws in tensor form, but this wasn't how they were originally formulated or how they'd be described in undergraduate texts). It would actually be incorrect to apply the non-tensor form of SR to anything but an inertial frame, while the laws of GR could be applied in any non-inertial coordinate system you like. And even if the type of coordinate system you describe for an accelerating observer may be the most "natural" in some sense, I don't think there's any simple interpretation of the coordinates assigned to distant events in terms of local readings on measuring-devices like the inertial rulers and clocks of an inertial coordinate system.

 

[Al68]

Jesse,

OK, I almost forgot the reason I started this thread. I wanted to figure out the coordinate position of the Earth as a function of time in system NI. It looks like you've done that. But I also wanted to figure out how to do this for the ship's deceleration and coming to rest relative to earth. Is there a way to figure out the coordinate position of the Earth as a function of time in a decelerating system NI? That is in motion relative to Earth and starts decelerating at a specified distance D from Earth at a specified time T in the ship's frame, and eventually comes to rest relative to earth. In other words, similar to the ship coming to rest at the star system in the Twins Paradox.

Thanks,
Alan

 

[JesseM]

Jesse,

OK, I almost forgot the reason I started this thread. I wanted to figure out the coordinate position of the Earth as a function of time in system NI. It looks like you've done that. But I also wanted to figure out how to do this for the ship's deceleration and coming to rest relative to earth. Is there a way to figure out the coordinate position of the Earth as a function of time in a decelerating system NI? That is in motion relative to Earth and starts decelerating at a specified distance D from Earth at a specified time T in the ship's frame, and eventually comes to rest relative to earth. In other words, similar to the ship coming to rest at the star system in the Twins Paradox.

Thanks,
Alan

If you extend the relativistic rocket equations back through -t, I think it should give you the equation for a ship whose velocity is negative and decreasing until it comes to a stop at t=0 and d=0, then afterwards it turns around and its velocity is positive and increasing, with constant acceleration "a" in a single direction throughout (even if your acceleration is positive, if your current velocity is negative than your speed will be decreasing, so that is 'deceleration'). If this is correct then the equations I derived would be unchanged, you just have to keep in mind that for 'deceleration' you want T to be negative, with the ship reaching the position of the Earth at T=0, and D positive but decreasing until that point. Or, you could pick Tstop to be the moment the ship reaches earth, and substitute (T-Tstop) in for T in those equations.

 

[Al68]

Jesse,

I was looking for equations for a ship that, after leaving Earth and traveling away from earth, decelerates to come to rest relative to earth. Like the ship coming to rest at the distant star system in the Twins Paradox. And the coordinate position of Earth as a function of time in the ship's frame during this deceleration.

Or, alternatively, the coordinate position of a star system (at rest with earth, that the ship is traveling to) as a function of time in the ship's frame during it's acceleration away from earth.

Thanks,
Alan

 

[pervect]

If you extend the relativistic rocket equations back through -t, I think it should give you the equation for a ship whose velocity is negative and decreasing until it comes to a stop at t=0 and d=0, then afterwards it turns around and its velocity is positive and increasing, with constant acceleration "a" in a single direction throughout (even if your acceleration is positive, if your current velocity is negative than your speed will be decreasing, so that is 'deceleration'). If this is correct then the equations I derived would be unchanged, you just have to keep in mind that for 'deceleration' you want T to be negative, with the ship reaching the position of the Earth at T=0, and D positive but decreasing until that point. Or, you could pick Tstop to be the moment the ship reaches earth, and substitute (T-Tstop) in for T in those equations.

That's the approach I took. Perhaps you could double check the solutions I gave to this at the start of the thread in post #3 and post #6? The only thing I would add is that I've assumed that c=1 in these solutions for simplicity.

i.e. my posted solution

v(t) = -a(t-tstop)/sqrt(1+(a^2(t-tstop)^2)

becomes

v(t) = -a(t-tstop)/sqrt(1+(a^2/c^2)*(t-tstop)^2)

when c != 1

I've also given some solutions for x(t) via a different method (in #6), they should also be able to be obtained by integrating the above expressions. They also assumed that c=1.

As we've discussed, the x(t) I've given is for the Earth coordinate x(t), the ships coordiante x(t) is x_ship(t) = x(t)*sqrt(1-(v/c)^2).

I'm pretty sure Meir Achu'z suggestion is wrong: the velocity around the turnaround point should be an even function of t:

ie. v(t-t_turnaround) = -v(v-t_turnaround)

 

[JesseM]

Jesse,

I was looking for equations for a ship that, after leaving Earth and traveling away from earth, decelerates to come to rest relative to earth. Like the ship coming to rest at the distant star system in the Twins Paradox. And the coordinate position of Earth as a function of time in the ship's frame during this deceleration.

Or, alternatively, the coordinate position of a star system (at rest with earth, that the ship is traveling to) as a function of time in the ship's frame during it's acceleration away from earth.

Thanks,
Alan

It should also be possible to use the trick of extending the relativistic rocket equations into negative t and T coordinates to answer this. We can use these equations to figure out the rocket's distance from the Earth if it started out from a distant star and accelerated in the direction of the earth; extending this into -t tells us the rocket's distance from the Earth as it travels along the path from the Earth to the star, decelerating in the direction of the star (which is the same as accelerating in the direction of the earth).

So, the relativistic rocket equation for distance from the star as a function of time, in the star's reference frame, would just be the usual one:

d = (c^2/a)*(\sqrt{1 - (at/c)^2} - 1)

If the Earth is located at a distance d_0 from the star, the function for the ship's distance from Earth d_e as a function of t in the star's frame would just be:

d_e = d_0 - d = d_0 - (c^2/a)*(\sqrt{1 - (at/c)^2} - 1)

...and this would also be the formula for the ship's distance from Earth in the -t when it is moving in the direction of the star and decelerating (Also, if you wanted the ship to accelerate in the direction of the star for the first half of the trip and then decelerate for the second half, then d_e could be taken as the distance from the midpoint, with the distance from Earth just being 2*d_e.)

So now we can do the same sort of substitutions that I did originally to translate this into D and T for the NI system. As before, D = d*\sqrt{1 - v^2/c^2}, and likewise the distance to the Earth D_e in the NI system would be D_e = d_e * \sqrt{1 - v^2/c^2}. So, plugging in:

D_e = \sqrt{1 - v^2/c^2} * [d_0 - (c^2/a)*(\sqrt{1 - (at/c)^2} - 1)]

And as before, we can plug in t = (c/a)*sh(aT/c) to get:

D_e = \sqrt{1 - v^2/c^2}*[d_0 - (c^2/a)*(ch(aT/c) - 1)]

I realize in retrospect that there's a problem with both this equation and the ones I derived earlier for accelerating away from the earth--they still include v, which is in the original inertial coordinates rather than in the NI system's coordinates. So we should really plug in v(T) from the relativistic rocket page, which is v = c * th(aT/c), giving:

D_e = \sqrt{1 - (th(aT/c))^2}*[d_0 - (c^2/a)*(ch(aT/c) - 1)]

And the page on hyperbolic trig identities says 1 = (sech(x))^2 + (tanh(x))^2, so 1 - (tanh(x))^2 = (sech(x))^2, so this simplifies to:

D_e = sech(aT/c) * [d_0 - (c^2/a)*(cosh(aT/c) - 1)]

Now just plug in negative values for T (or let Tstop be the time the ship reaches the star and substitute T-Tstop in for T in the above equation), and this should be the distance from the Earth in the NI system as the ship decelerates towards the star. As for the velocity and acceleration of the Earth in the NI system, I just realized that because I failed to take into account that v was a function of T, my earlier derivations of dD/dT and dD^2/dT^2 were incorrect. To differentiate the expression above you'd use the http://www.math.hmc.edu/calculus/tutorials/prodrule/ of calculus as well as the chain rule I mentioned earlier, and you'd have to know that the derivative of sech(x) is -tanh(x)*sech(x) while the derivative of tanh(x) is 1 - tan^2(x) (from this page). I don't know how interested you are in the velocity and acceleration of the Earth in the NI system though.

 

[Al68]

Jesse,

Thanks for your response, I will have to take some time to look at it. But I do have another question.

If I'm on a rocket accelerating away from earth, my coordinate velocity relative to Earth after acceleration will be v = at/\gamma.
Can this be explained by saying that as my velocity relative to Earth increases due to acceleration, \gamma increases, resulting in an increasing amount of length contraction of my distance from earth, which partially counteracts the increase in velocity relative to Earth due to proper acceleration. This causes my coordinate acceleration relative to Earth to be equal to a/\gamma, and therefore my coordinate velocity relative to Earth to be equal to at/\gamma. And my coordinate velocity relative to Earth will be v = a_c t, where a_c refers to coordinate acceleration.

Is this correct?

Thanks,
Alan

 

[JesseM]

That's the approach I took. Perhaps you could double check the solutions I gave to this at the start of the thread in post #3 and post #6?

The only thing I noticed was you had the formula v(t) = -a(t-tstop)/sqrt(1+(a^2(t-tstop)^2), whereas the way I was approaching the problem I don't think that first "a" would be negative, it should be the same as the formula on the relativistic rocket page but with (t-tstop) substituted for t, giving v(t) = a(t-tstop) / sqrt[1 + (a/c)^2*(t-tstop)^2]. Did you change the definition of which direction was positive acceleration? If so, would this also change the definition of positive vs. negative velocity? Other than that minor issue everything in those posts looked good to me, although I couldn't check the equations in #6 that you found using maple. I hadn't known about x^2 - c^2*t^2 being constant for constant acceleration, that's a nice result...sometime I need to go back and study those relativistic rocket equations were derived, I don't remember my college SR course dealing much with accelerations.

I'm pretty sure Meir Achu'z suggestion is wrong: the velocity around the turnaround point should be an even function of t:

ie. v(t-t_turnaround) = -v(v-t_turnaround)

If it's an even function, and we set t=0 as the turnaround, then wouldn't that mean v(t) = v(-t)? But shouldn't its velocity a given amount of time before the turnaround have the opposite sign as its velocity a given amount of time after the turnaround, since it changed direction when it turned around? It seems to me like the position as a function of time should be an even function of t, but not the velocity (same as for a ball tossed upwards, where t=0 is the maximum height where the ball switches from rising to falling).

 

[JesseM]

Jesse,

Thanks for your response, I will have to take some time to look at it. But I do have another question.

If I'm on a rocket accelerating away from earth, my coordinate velocity relative to Earth after acceleration will be v = at/\gamma.

I think you have the wrong formula there--the relativistic rocket page says that v(t) is at / \sqrt{1 + (at/c)^2}, and anyway it wouldn't make sense to have v be a function of gamma since gamma is itself a function of v.

 

[pervect]

I think the formula is actually technically correct, because gamma = sqrt(1+(at/c)^2).

It's not particularly useful, unfortunately: if you don't know v, you don't know gamma either as Jesse points out.

I don't see offhand how the formula can have any physical significance. Letting c=1

1) at can be greater than the speed of light, so it can't be any sort of physical velocity.

2) at is not the rapidity either. (aT is the rapidity).

Perhaps there is some deeper significance to the formula that I'm missing, but I dont' see it.

 

[Al68]

Jesse,

As pervect pointed out, \gamma = \sqrt{1 + (at/c)^2} = 1/\sqrt{1 - (v/c)^2}.
It’s obviously more convenient use \gamma = \sqrt{1 + (at/c)^2} when you don’t already know the coordinate velocity. One could also use v = at * \sqrt{1 - (v/c)^2} and solve for v to get v =at/ \sqrt{1 + (at/c)^2}, but this is less convenient.

I have also seen the term “proper velocity” used to describe at. And coordinate velocity = proper velocity divided by \gamma. And I’ve seen proper velocity defined as momentum divided by mass. And of course “proper velocity” can exceed c, but coordinate velocity cannot exceed c. And the velocity restriction in SR applies to coordinate velocity. But that’s not really relevant here.

It just looks to me like the increasing effect of length contraction (during acceleration) is the reason that coordinate velocity (after acceleration) is equal to v = at/\gamma instead of just at. Were it not for this length contraction, velocity would be equal to at, just like it does in classical physics, or at low speeds. Or maybe we could say that the increasing effect of length contraction (during acceleration) is the physical reason that a ship’s coordinate velocity can never exceed c, no matter how long it accelerates.

Is there a better way to physically describe why coordinate velocity does not equal at (at relativistic speeds)?

Thanks,
Alan

 

[JesseM]

It just looks to me like the increasing effect of length contraction (during acceleration) is the reason that coordinate velocity (after acceleration) is equal to v = at/\gamma instead of just at.

Length contraction of what, in which coordinate system? Remember, we're not dealing with coordinates in the non-inertial NI system here, we're just talking about v and t in the inertial coordinate system of the earth. The Earth will see the length of the rocket itself shrink of course, but for the purposes of this formula the rocket is being treated as a point. And in the Earth's frame, the Earth's own rulers, which are used to measure how the rocket's distance increases, are not shrinking.

Were it not for this length contraction, velocity would be equal to at, just like it does in classical physics, or at low speeds.

I would say the reason velocity is not equal to at is because "a" does not refer to the coordinate acceleration in the Earth's frame, where v and t are being measured; the actual coordinate acceleration is constantly decreasing as the ship approaches the speed of light.

Can you phrase your argument in a way that only refers to what's happening in the inertial frame of the earth, without considering any non-inertial coordinate systems like the NI system?

 

[Al68]

OK, in the inertial frame of the earth, a ship’s velocity after acceleration will be v = at/\gamma due to the increasing effect of lorentz contraction of the distance of the ship from Earth as velocity increases. This increasing effect of lorentz contraction of the distance to the ship increasingly counteracts the increasing distance of the ship due to acceleration. And if a is the proper acceleration of the ship, and a_c is the coordinate acceleration of the ship, a_c = a/\gamma due to the fact that the effect of lorentz contraction is increasing with time during acceleration. And v = a_c*t, or at/\gamma, instead of v = at as would be the case in classical physics or at low speeds. And the ship’s coordinate velocity will never reach c no matter how long it accelerates due to the increasing effect of lorentz contraction of the ship’s distance from earth. All of this is in earth’s frame.

(edit)And the coordinate acceleration is constantly decreasing as the ship approaches the speed of light because \gamma is constantly increasing.

In other words, were it not for lorentz contraction of the distance between Earth and the ship, v would equal at just like in classical physics.

Is this better?

Thanks,
Alan

 

[JesseM]

OK, in the inertial frame of the earth, a ship’s velocity after acceleration will be v = at/\gamma due to the increasing effect of lorentz contraction of the distance of the ship from Earth as velocity increases.

I still don't understand what you mean by this. If distance is measured by physical rulers, what ruler is lorentz-contracting here? The ship's increasing velocity won't cause the Earth's own ruler to contract in the Earth's frame, and it is this ruler that is being used to measure distance in the Earth's frame. The only thing I can think of is that you would be talking about a ruler carried by the ship, but we've already discussed the problems with an acclerating observer trying to measure distances with accelerating rulers, and in any case the length of rulers carried by the ship plays absolutely no role in the Earth's calculation of the distance and velocity as a function of time in the coordinates of its own inertial rest frame.

And if a is the proper acceleration of the ship, and a_c is the coordinate acceleration of the ship, a_c = a/\gamma due to the fact that the effect of lorentz contraction is increasing with time during acceleration.

Are you sure the coordinate acceleration in the Earth's frame would be a/\gamma? Did you actually take the derivative of v(t) with respect to t? If not, this would probably be a worthwhile exercise, and if you know enough calculus to use the chain rule and the product rule, it shouldn't be too hard to differentiate the expression for v(t) on the relativistic rocket page (if not I can show you how).

 

[Al68]

If we say that, after acceleration, the distance between the Earth and the ship is lorentz contracted in the ship's frame, doesn't the distance between the ship and the Earth have to be lorentz contracted in Earth's frame?
And if two ship's pass each other, and each ship's crew measures the relative coordinate velocity between them, won't this coordinate velocity be based on length contracted measurements of distance, which are reciprocal?
Without lorentz contraction \frac{d^2 x}{dt^2} could be constant, and \frac{dx}{dt} would have no limit. Obviously the variable t does not change, since it is always read on the same clock at rest with the observer.
But with lorentz contraction, \frac{d^2 x_c}{dt^2} decreases at the same rate that 1/\gamma decreases with time, where x_c is the length contracted distance.

As far as a_c = a/\gamma, this is obvious from the relativistic rocket equation, since coordinate velocity equals coordinate acceleration times time. And v = at/\gamma. This is same relativistic rocket equation that is used everywhere for coordinate velocity.

It's my understanding that the relativistic rocket equation is derived by applying the lorentz transformation to acceleration to get a_c = a/\gamma

I can derive the relativistic rocket equation if I assume a_c = a/\gamma, and v refers to coordinate velocity.

v = a_c t
v = at/\gamma
v = at \sqrt{1 - (v/c)^2}, just solve for v, and
v^2 = (at)^2 (1 - (v/c)^2)
v^2 = (at)^2 - (at)^2 (v/c)^2
v^2 + (at)^2 (v/c)^2 = (at)^2
v^2(1 + (at/c)^2) = (at)^2
v^2 = (at)^2/(1 + (at/c)^2)
v = at/\sqrt{1 + (at/c)^2}

And it's obvious that \gamma = 1/\sqrt{1 - (v/c)^2} = \sqrt{1 + (at/c)^2}.
The second form is more convenient to use when you don't already know v.

Now I just have to hope I didn't make a mistake here.

Thanks,
Alan

 

[JesseM]

If we say that, after acceleration, the distance between the Earth and the ship is lorentz contracted in the ship's frame, doesn't the distance between the ship and the Earth have to be lorentz contracted in Earth's frame?

But what does it mean to say "the distance contracted"? Usually when you use a phrase like that you're comparing one frame to another--for example, in the purely inertial case, we found that if the distance in the Earth's frame at the moment of an event on the ship's worldline was d, then the distance in the ship's frame at the moment of that same event was contracted by d*\sqrt{1 - v^2/c^2}. Notice that if we're talking about the distance at the moment of that particular event, this isn't reciprocal, the distance is definitely bigger in the Earth's frame than the ship's frame, not vice versa.

Anyway, when you talk about distance contracting "after acceleration" in the ship's frame (again, I would prefer to call it 'the NI coordinate system' to make clear that it isn't the only choice of coordinate system for the ship), it seems like you're not comparing two coordinate systems but talking about the distance at one time vs. another in the ship's coordinate system. But since D(T) actually grows with T, this can't be what you mean by "contracting"--maybe you just mean that a ruler at rest in the Earth's frame will be shrinking in length in the NI system? And although we can't really talk about a ruler "at rest" in the NI system since the acceleration would deform it, I suppose we could imagine two small rockets moving in such a way that the coordinate distance between them in the NI system was constant. In this case I think, though I'm not sure, that the distance between them would shrink in the Earth frame too...I don't think it'd just be by the gamma-factor of the ship at that moment in the Earth's frame, though, because the distance between these two rockets would be constantly changing in the Earth's frame.

Anyway, if this is something like what you were getting at, how does the fact that the distance between these two rockets shrinks in the Earth's frame tell us anything about why the velocity in the Earth's frame is at/gamma? I still can't see any rhyme or reason in your argument, the velocity in the Earth's frame doesn't have anything to do with whatever the distance between those rockets may be. Remember, once again, the fact that the NI system isn't fundamentally any more physically valid than some other non-inertial coordinate system you could come up with--you could use some different coordinate system for the ship such that two rockets at constant coordinate distance in this system were actually increasing in distance in the Earth's frame, it wouldn't change your conclusions about v(t) in the Earth's frame one iota.

And if two ship's pass each other, and each ship's crew measures the relative coordinate velocity between them, won't this coordinate velocity be based on length contracted measurements of distance, which are reciprocal?

I'm not sure what you mean here either--maybe it would be better to focus on an inertial case like this since it's a lot less complicated than the accelerating case, and would probably help me understand what you're getting at in the accelerating case. The thing is, if two ships pass each other, each ship uses its own rulers and clocks to measure the speed of the other ship, and of course those rulers and clocks aren't contracted or slowed in the ship's own frame! I suppose if each ship wanted to try to figure out how fast the other ship was measuring it to move, then it could look at how fast it was moving past marks and how fast clocks are ticking on the other ship's measuring system...but I'm still not getting what is meant by the phrase "each ship's crew measures the relative coordinate velocity between them, won't this coordinate velocity be based on length contracted measurements of distance".

Without lorentz contraction \frac{d^2 x}{dt^2} could be constant, and \frac{dx}{dt} would have no limit. Obviously the variable t does not change, since it is always read on the same clock at rest with the observer.

It's not just lorentz contraction though, if you removed lorentz contraction but kept time dilation and the Einstein clock synchronization convention (which would mean there was a preferred frame, but never mind), then "constant a" in the sense that the coordinate acceleration in the ship's instantaneous inertial frame stays the same throughout the journey would still mean non-constant acceleration as seen in a single inertial frame.

But with lorentz contraction, \frac{d^2 x_c}{dt^2} decreases at the same rate that 1/\gamma decreases with time, where x_c is the length contracted distance.

Your argument about the acceleration is wrong--see below.

As far as a_c = a/\gamma, this is obvious from the relativistic rocket equation, since coordinate velocity equals coordinate acceleration times time.

Velocity = coordinate acceleration times time only works when coordinate acceleration is constant, not when it's varying over time, in which case velocity is found by integrating a(t) (note that integrating a constant 'a' with respect to 't' gives a*t). You might think that if v = at/gamma, then differentiating with respect to t would give a/gamma...but this is the problem with defining v(t) as a function of both t and gamma, where gamma is itself a function of t. You can't treat gamma as a constant for the purposes of differenting v(t) with respect to t, you have to take into account that it's a function of t as well. What you really need to do to find the coordinate acceleration in the Earth's frame is differentiate this expression with respect to t:

v(t) = at / \sqrt{1 + (at/c)^2}

Only when you've differentiated the whole thing will you know the coordinate acceleration as a function of t and the proper acceleration a. Try this and see what you get...again, I can help with the math if you need it.

 

[Al68]

Jesse,

I should not have said that a_c = a/\gamma, where a_c is coordinate acceleration. I should have said that \overline{a_c} = a/\gamma, where \overline{a_c} is average coordinate acceleration. And that coordinate velocity equals average coordinate acceleration times time.

(edit)And if v = \overline{a_c} t, then:

\frac{dv}{dt} = \overline{a_c}, since at any t, \overline{a_c} will be a constant.

And the equation \Delta v = at/\gamma remains true, according to every source I have seen.

And we do not have time dilation without lorentz contraction in SR. We have both. But if we measure time on a clock at rest with us, it will be proper time. If we measure the length of an object in motion relative to us, it will not be the object's proper length, it will be length contracted according to SR, if we use a ruler at rest with us. So, if we (on earth) were to measure a ship's coordinate velocity by measuring the length of the ship and the time difference between the front of the ship and the rear of the ship to pass a marker (in Earth's frame), we know from SR that the proper length of the ship is greater than the coordinate length of the ship we measured. We know that the distance we measured was length contracted, because we are familiar with SR, not because we ever measured the proper length of the ship. We would have to be at rest with the ship to measure that. But we could have someone on the ship measure it's length for us, with a ruler that was identical to ours (before the ship was in motion relative to us), and tell us the proper length of the ship, and it would match up with our calculation. Or maybe we measured the length of the ship with the same ruler while it was at rest with earth. Either way, the length of the ship will be length contracted when we measure it in motion relative to us. But all of Earth's calculations of velocity, and the total distance of the ship from Earth at any specified time (by Earth's clock), will be based on the Earth's measurements of the length of the ship.

This leads me to believe that \Delta v = at/\gamma instead of \Delta v = at (like in classical physics) because of length contraction, not because of time dilation. Because when we measure the coordinate velocity of an object in motion relative to us, we use a clock at rest with us, and a ruler at rest with us. But the ship, who's length we measured, is not at rest with us. And the faster the ship is going, the shorter we will measure it's length to be with our ruler. We will measure it's length to be Proper Length/\gamma. This is basic SR. So, in Earth's frame, if the ship is accelerating, it's coordinate length is getting shorter with time. And if we use this coordinate length of the ship, suitable markers, and our clocks, to measure velocity, velocity will increase with time according to the relativistic rocket equation, \Delta v = at/\gamma, or \Delta v/t = a/\gamma.

I know I don't always word things properly, so I hope this makes sense this time.

Thanks,
Alan

 

[JesseM]

Jesse,

I should not have said that a_c = a/\gamma, where a_c is coordinate acceleration. I should have said that \overline{a_c} = a/\gamma, where \overline{a_c} is average coordinate acceleration. And that coordinate velocity equals average coordinate acceleration times time.

When you say "coordinate velocity equals average coordinate acceleration times time", do you mean the change in coordinate velocity between the two ends of the time-interval you're averaging the acceleration, ie \overline{a_c} = \Delta v / \Delta t? I think it is true that the average value of a function's derivative over some interval is just the difference in values of the function at the two ends of the interval divided by the size of the interval...this is another way of saying that the average value of a function's derivative between two points is the same as the slope of the line connecting the function at those points, and we know that the derivative at any point is like the "instantaneous slope" at that point. So this seems right to me, although my calculus is a little rusty so I'm not sure how you'd prove it. And since v can be written as at/\gamma, that means \Delta v between two times t1 and t2 would be a(t_2 - t_1)/\gamma, or a \Delta t /\gamma for that interval, which means if \overline{a_c} = \Delta v / \Delta t is true, it must also be true that \overline{a_c} = a/\gamma.

(edit)And if v = \overline{a_c} t, then:

Are you saying that the instantaneous value of v(t) at some time t is equal to the average acceleration times t? That is definitely not right, although as I said above, I think it's true that \Delta v = \overline{a_c} \Delta t.

And we do not have time dilation without lorentz contraction in SR. We have both. But if we measure time on a clock at rest with us, it will be proper time. If we measure the length of an object in motion relative to us, it will not be the object's proper length, it will be length contracted according to SR, if we use a ruler at rest with us.

True, although strictly speaking you have to use both rulers and clocks at rest with respect to you to measure the length of a moving object--"length" is understood to mean "the position coordinate of one end minus the position coordinate of another end when the positions of each end are recorded at the same time", so how you define simultaneity matters here.

So, if we (on earth) were to measure a ship's coordinate velocity by measuring the length of the ship and the time difference between the front of the ship and the rear of the ship to pass a marker (in Earth's frame), we know from SR that the proper length of the ship is greater than the coordinate length of the ship we measured.

True, although you don't actually need to know the distance between the two ends of the object to measure its speed--an alternate method would just be to measure the position of the front end at one time and compare with the position of the front end at a later time.

We know that the distance we measured was length contracted, because we are familiar with SR, not because we ever measured the proper length of the ship. We would have to be at rest with the ship to measure that. But we could have someone on the ship measure it's length for us, with a ruler that was identical to ours (before the ship was in motion relative to us), and tell us the proper length of the ship, and it would match up with our calculation. Or maybe we measured the length of the ship with the same ruler while it was at rest with earth. Either way, the length of the ship will be length contracted when we measure it in motion relative to us. But all of Earth's calculations of velocity, and the total distance of the ship from Earth at any specified time (by Earth's clock), will be based on the Earth's measurements of the length of the ship.

No, again, it isn't necessary to know anything about the length of the ship, you could just measure the front end of the ship at different times. And in the case of an accelerating ship, at a given moment not every part of the ship is moving at the same speed, and botht the ship and any rulers on board are being distorted by acceleration, an effect separate from length contraction. In practice the relativistic rocket equations are either assuming the ship can be treated as pointlike, or they can be interpreted as only applying to a single point on the ship (the front, the middle, whatever) whose proper acceleration is constant.

If you try to extend the method of measuring speed by measuring the time between the front end and back ending passing a particular fixed point to the case of the relativistic rocket, you run into trouble because there is some finite time interval between the two measurements while the speed of the rocket is changing (and the speed of different parts of the rocket is not even the same at a single moment in your frame), and also because, again, the rocket is being physically distorted by the acceleration. You could try to eliminate the issue of distortion by the idea I suggested in my last post, where you have two tiny rockets programmed to accelerate in such a way that their separation is constant in the NI coordinate system, and they are treated like the "front" and "back" of an imaginary rocket which would appear to be at rest in the NI system. But the first issue won't go away here, you're still measuring the time between the front rocket and the back rocket passing you, but they are both constantly changing speed, and their speeds won't even be the same at a single moment in your inertial frame.

In contrast, when making two measurements of the front of the rocket, you could make the time-interval between the measurements as small as you wish, thus getting as close as you want to a measurement of the front's instantaneous velocity in your frame. So this is really a much better way to think of measuring the speed of an accelerating object.

This leads me to believe that \Delta v = at/\gamma instead of \Delta v = at (like in classical physics) because of length contraction, not because of time dilation.

OK, just to run through it again to check that I've got this right: you're saying that one way of measuring the speed of an inertial object would be based only on two items of knowledge, 1) the time between the front and back of an object passing a single clock at rest in our frame, and 2) knowledge of the proper length of the object, which we can use to figure out its length in our frame. You want to extend this procedure to an accelerating object in some way. An accelerating object does not really have a "proper length", but I suggested that we could at least add the condition that its length in the object's own instantaneous co-moving inertial frame is constant (which is the same thing as saying its length in the NI coordinate system is constant), perhaps just by creating an imaginary object whose front and back ends are mini-rockets programmed to accelerate in such a way that this works out.

As I said, there is the problem that the front and back ends/mini-rockets will not have the same instantaneous velocity in our frame, but I suppose that in the limit as the distance between the two ends in the co-moving frame goes to zero, the difference in instantaneous velocities in our frame would go to zero too. It could be that in this limit your argument makes sense, I'm not quite sure.

Here's another way I might make sense of the argument. Instead of having the rocket accelerating continuously, have its velocity increase in a series of discrete steps--ie have the rocket move at constant velocity, then instantaneously accelerate, then move at constant velocity again, etc. (with the time the rocket stays at constant velocity being constant in its co-moving inertial frame for that segment of the trip, and the amount that the velocity increases on each instantaneous acceleration also being constant in the co-moving inertial frame before that acceleration). In the limit as the time between accelerations goes to zero, I think this would become identical to the regular accelerating rocket. But with discrete accelerations, there would be no problems in using your method of measuring speed during each constant-velocity segment. Would it then make sense to say that the reason that the increase in velocity seen in our frame gets smaller with each jump, even though the increas in velocity is constant in the co-moving frame before each jump, is just because the length of the ship in our frame gets smaller after each jump? I'm not sure if I can see how that explains it, but if I could then I think the same explanation would make sense in the continuously-accelerating case, since that should just be the limit of this discrete-jump case. I'll have to think about it...can you think of a way of explaining yourself how the smaller change in velocity with each jump could be understood exclusively in terms of the ship's length in our frame getting smaller with each jump, and the fact that the change in velocity is constant in the ship's co-moving frame before each jump? It seems to me that explaining how the changes in velocity get smaller in our frame would also require knowing how each co-moving frame's clock speed and definition of simultaneity is different from ours, but I'm open to being convinced otherwise.

 

[Al68]

(edit)And if v = \overline{a_c} t, then:

Are you saying that the instantaneous value of v(t) at some time t is equal to the average acceleration times t? That is definitely not right, although as I said above, I think it's true that \Delta v = \overline{a_c} \Delta t.

I meant that the final velocity (v) after a period of acceleration (t) would equal the average coordinate acceleration times time. I was assuming that v = 0 at t = 0. Just like the relativistic rocket equation, v = at/\gamma is only true if we assume that v = 0 at t = 0. I know it is a little sloppy to use v and \Delta v interchangeably, and t and \Delta t interchangeably, even though they are not interchangeable. But everybody else does it.

And, you are right that my method of measuring velocity would not be accurate (except in the limit as we used a smaller and smaller ship), but it would be pretty close, since presumably the length of the rocket would be very small compared to the total distance involved. And the change in velocity during the time interval would be very small. And your idea of measuring velocity would be more accurate.

But my main question now is, if we always use clocks at rest with us to measure time, is there any other explanation for why v = at/\gamma is correct? Is there any other specific reason that (change in) coordinate velocity does not equal proper acceleration times (change in) time, like it does in Newtonian physics?

Thanks,
Alan

 

[JesseM]

I meant that the final velocity (v) after a period of acceleration (t) would equal the average coordinate acceleration times time. I was assuming that v = 0 at t = 0. Just like the relativistic rocket equation, v = at/\gamma is only true if we assume that v = 0 at t = 0. I know it is a little sloppy to use v and \Delta v interchangeably, and t and \Delta t interchangeably, even though they are not interchangeable. But everybody else does it.

OK, I see what you mean. v(t) for any time t is always equal to \Delta v for the interval between t=0 and that t.

However, I realize in retrospect there was a problem with the argument that \overline{a_c} = a/\gamma. I had said earlier that \Delta v between two times t1 and t2 would be equal to a(t_2 - t_1)/\gamma, but I forgot that gamma would be changing from one time to another, it would actually be a* t_2 /\gamma(t_2) - a* t_1 /\gamma(t_1). So if you plug that into \overline{a_c} = \Delta v / \Delta t, what you actually get is \overline{a_c} = a* t_2 /[\gamma(t_2) * (t_2 - t_1)] - a* t_1 /[\gamma(t_1) * (t_2 - t_1)], not \overline{a_c} = a/\gamma. Analogous to what you say above about v, if we let t_1 = 0 and t_2 = t in the above, then we get \overline{a_c}(t) = a * t /[\gamma(t) * t], or \overline{a_c}(t) = a / \gamma(t). This might be what you meant anyway, but the idea that \overline{a_c} is a changing function of t is important, as I'll explain below.

Although it's true that v(t) = \overline{a_c} t can be derived from \Delta v = \overline{a_c} \Delta t, that only works if you assume \overline{a_c} to be a constantly changing function of t, like my \overline{a_c}(t) = a / \gamma(t)--for any given t, you can look at the change in velocity from 0 to t and that's v(t), but the average of a_c also has to be taken from the time interval of 0 to the t you're looking at, so it won't be the same from one value of t to another. So the rest of your argument, where you said:

\frac{dv}{dt} = \overline{a_c}, since at any t, \overline{a_c} will be a constant.

...is incorrect, because \overline{a_c} is not constant.

But my main question now is, if we always use clocks at rest with us to measure time, is there any other explanation for why v = at/\gamma is correct? Is there any other specific reason that (change in) coordinate velocity does not equal proper acceleration times (change in) time, like it does in Newtonian physics?

Well, the way I think of it is based on the velocity addition formula. Like I said, constant acceleration can be thought of as a limiting case of making a series of instantaneous jumps in velocity, such that each time you make a jump, the magnitude of the increase is constant in your last co-moving rest frame, and the amount of time spent cruising inertially between jumps is also the same in each frame where you're at rest during the cruising period. But if you think about the way velocities add in relativity, it should be clear that an inertial observer will see the increase in velocity with each jump being smaller on successive jumps, Because if something is moving with respect to me with velocity v1 (in this case, the velocity before the jump), and then it increases in velocity by v2 in the frame where it was at rest when moving at v1, I will not see the velocity increase to v1+v2 but only to (v1+v2)/(1 + v1*v2/c^2), according to the velocity-addition formula. And the larger v1 is, the less this formula says the new velocity will have increased from v1, given constant v2. In classical mechanics, if the object was jumping according to this rule, then after 3 such jumps starting from rest I'd see the velocity as v2 + v2 + v2, but in relativity it'd only be v2 + (v2 + v2)/(1 + v2*v2/c^2) + (v2 + [(v2 + v2)/(1 + v2*v2/c^2)])/(1 + v2*[(v2 + v2)/(1 + v2*v2/c^2)]/c^2). And besides the fact that the increase in velocity is getting smaller with each jump as seen in my frame, there's also the fact that the time between jumps is getting longer and longer since the time is only supposed to be constant in the last co-moving rest frame before the jump, and each successive co-moving rest frame's clocks are ticking slower and slower in my frame. So this further decreases the rate that the velocity seems to increase in my frame, when compared with classical predictions. I think if you consider continuous acceleration as a limiting case of this sort of series of discrete jumps--and I think you could actually show rigorously that the accelerating case works out as the limit of the time between jumps and the size of each jump approaching zero, although I can't guarantee this--then you can see more easily why the acceleration in an inertial frame is continuously decreasing even if the proper acceleration "a" is constant.

 

[Al68]

Jesse,

Again, you are right, the correct form of the equation shoud be \overline{a_c}(t) = a / \gamma(t). And the correct form of the relativistic rocket equation should be v(t) = at/\gamma(t). I was being a little sloppy again. (edit) And you are correct that \overline{a_c} would not be constant with respect to t. That was a mistake on my part.

I don't think we could say that the velocity addition formula actually causes v = at/\gamma to be correct. One might even say that Lorentz contraction causes the velocity addition formula to be correct.

But I do have another question. In Einstein's own writings about the Twins Paradox it is clear that he did not agree with the resolutions commonly accepted today. He even said that the ship's twin should be able to consider himself at rest the entire time, and that the Twins Paradox could not be resolved in SR. This was supposedly one of his reasons for pursuing GR. He tried to resolve it in GR by considering the ship's twin to be at rest, but that resolution is considered faulty today. If one could argue that Einstein understood SR as well as anyone, why would he still view the Twins Paradox as a problem for SR?

Thanks,
Alan

 

[George Jones]

But I do have another question. In Einstein's own writings about the Twins Paradox it is clear that he did not agree with the resolutions commonly accepted today. He even said that the ship's twin should be able to consider himself at rest the entire time, and that the Twins Paradox could not be resolved in SR. This was supposedly one of his reasons for pursuing GR. He tried to resolve it in GR by considering the ship's twin to be at rest, but that resolution is considered faulty today. If one could argue that Einstein understood SR as well as anyone, why would he still view the Twins Paradox as a problem for SR?

This is an example of a logical fallacy - proof by authority. Einstein was a flesh-and-bllood human being who, like all flesh-and-blood human beings, made mistakes.

In any case, if Einstein really did hold these views, I would like to see some references.

 

[jtbell]

But I do have another question. In Einstein's own writings about the Twins Paradox it is clear that he did not agree with the resolutions commonly accepted today.

Today? Einstein has been dead for over fifty years. Surely people have thought of new ways to look at the twin paradox since he died, and refined earlier solutions.

If one could argue that Einstein understood SR as well as anyone,

I am not familiar with the detailed history of the development of our understanding of SR, during Einstein's lifetime and afterwards. Nevertheless, I find it very plausible that there are many people today who understand SR better than Einstein ever did. Many people have been refininig our knowledge of the consequences and implications of SR, and testing its predictions experimentally.

 

[Al68]

This is an example of a logical fallacy - proof by authority. Einstein was a flesh-and-bllood human being who, like all flesh-and-blood human beings, made mistakes.

In any case, if Einstein really did hold these views, I would like to see some references.

What logical fallacy are you talking about? I never said anything about proof or authority. And if you assumed that I intended to challenge SR, you are not correct.

Mostly I was referring to Einstein's paper "Dialogue about objections to the theory of relativity". In this paper, he presents a resolution of the Twins Paradox. He says that special relativity is not suitable for resolving the issue. He then says the cause of the asymmetry between the twins is a pseudo-gravitational field created during the acceleration of the ship. He then uses the gravitational time dilation of GR to try to resolve the paradox.

The sources I have seen say that most physicists consider this resolution faulty. I would have to agree with that. I do in fact believe Einstein was wrong here.

This paper is available in libraries, but as far as I know, it is not available on the internet. I have looked. I only find other papers that reference it. And Wikipedia's article on the Twins Paradox references this and talks about it a little. http://en.wikipedia.org/wiki/Twin_paradox

(edit) there is also a link at the bottom of the Wikipedia article that discusses Einstein's resolution of the paradox.

And of course Einstein has been dead a long time, but the resolutions presented in textbooks today were around long before that.

And while there has been more experimental evidence to support SR since Einstein, he did not doubt SR, anyway, so this evidence would not be relevant to why Einstein believed what he did about the Twins Paradox.

And I believe most of the resolutions accepted today use the same version of SR found in Einstein's original 1905 paper. Although they make assumptions that are not explicitly stated in this 1905 paper, they normally reference Einstein's SR as their source.

(edit) It also seems strange that Einstein's thought's on the Twins Paradox are rarely mentioned in discussions of the Twins Paradox.

Thanks,
Alan

 

[jtbell]

Mostly I was referring to Einstein's paper "Dialogue about objections to the theory of relativity."

The Wikipedia article on the Twins Paradox gives the date of this article as 1918. This is almost ninety years ago, and rather early in the history of relativity theory. At this time, I wouldn't expect even Einstein to have grasped SR as it relates to this paradox, as deeply as many people do now.

I have the impression that much of the current thinking about the Twin Paradox dates from the 1950s or 1960s, and it wasn't until a bit later that the currently popular explanations started to appear in introductory textbooks. When I was an undergraduate in the early 1970s, my sophomore-level introductory modern physics book (whose first edition dated back to the mid 1950s, I think) still claimed that resolving the paradox requires GR. However, none of the similar textbooks that I've seen since I started teaching in the mid 1980s does this.

 

[Al68]

The Wikipedia article on the Twins Paradox gives the date of this article as 1918. This is almost ninety years ago, and rather early in the history of relativity theory. At this time, I wouldn't expect even Einstein to have grasped SR as it relates to this paradox, as deeply as many people do now.

I have the impression that much of the current thinking about the Twin Paradox dates from the 1950s or 1960s, and it wasn't until a bit later that the currently popular explanations started to appear in introductory textbooks. When I was an undergraduate in the early 1970s, my sophomore-level introductory modern physics book (whose first edition dated back to the mid 1950s, I think) still claimed that resolving the paradox requires GR. However, none of the similar textbooks that I've seen since I started teaching in the mid 1980s does this.

Well, you are probably right that most of the current thinking about the Twins Paradox has been after 1918. But Einstein certainly had a good grasp on SR, and obviously spent time thinking about the Twins Paradox. As had other physicists of the time.

But none of the resolutions I have seen even address Einstein's thoughts. Especially the question of why can't the ship's twin consider himself to be at rest the entire time. Note that if we did consider acceleration to be irrelevent, and the ship's twin to be at rest, then we would have to say that the Earth's twin changed reference frames relative to the ship. And the ship would never change reference frames. And I have heard that we cannot consider the ship to be at rest because the ship changes reference frames. And the ship changes reference frames because the ship accelerated (did not stay at rest). But this sounds like circular logic to me. Einstein thought that there was no reason (in SR) that we could not consider the ship to be at rest the entire time, resulting in the Earth twin aging less.

I am not saying Einstein was right or wrong about this part. I just haven't seen it proven either way. And all the explanations I have seen don't attempt to prove anything. They just show the math and diagrams that match up with the assumptions made.

Thanks,
Alan

 

[robphy]

Well, you are probably right that most of the current thinking about the Twins Paradox has been after 1918. But Einstein certainly had a good grasp on SR, and obviously spent time thinking about the Twins Paradox. As had other physicists of the time.

I think it's fair to say that, early on, Einstein did not make use of the geometric interpretations suggested by Minkowski. Formulated geometrically, the "paradox" is easily resolvable... in particular, by direct calculation (i.e. spacetime arc-length) without issues of "transformations" and "reference frames".

One should also note that modern relativists (who emphasize the geometric structure and not much on "issues of reference frames") interpret "SR" and "GR" differently from the early physicists and relativists and from many textbooks and pop-sci books that haven't caught up yet.

But none of the resolutions I have seen even address Einstein's thoughts. Especially the question of why can't the ship's twin consider himself to be at rest the entire time. Note that if we did consider acceleration to be irrelevent, and the ship's twin to be at rest, then we would have to say that the Earth's twin changed reference frames relative to the ship. And the ship would never change reference frames. And I have heard that we cannot consider the ship to be at rest because the ship changes reference frames. And the ship changes reference frames because the ship accelerated (did not stay at rest). But this sounds like circular logic to me. Einstein thought that there was no reason (in SR) that we could not consider the ship to be at rest the entire time, resulting in the Earth twin aging less.

I am not saying Einstein was right or wrong about this part. I just haven't seen it proven either way. And all the explanations I have seen don't attempt to prove anything. They just show the math and diagrams that match up with the assumptions made.

Thanks,
Alan

The key word missing in your post is "inertial" (which is not the same as "at rest"). The traveling twin can "regard himself at rest"... but he cannot regard himself as "inertial".