Revisiting the Flaws of the Light Clock in Special and General Relativity

[yuiop]

This is false, the only reason that there is such a disparity between the values is that you have chosen a very long "cruising" time and a very short period of acceleration. In reality, if you had chosen comparable values to the "cruise" time T_c vs. the acceleration time T_a you would have found that the difference in proper time between the two twins is equally spread between the two effects and, that both depend on acceleration a:

d \tau=2T_c/ \sqrt{1+(aT_a/c)^2}+4\frac{c}{a} asinh(aT_a/c)

for the accelerating twin

I know perfectly well how the proper time us calculated, so please, explain the term in \frac{c}{a} asinh(aT_a/c). Is this a function of v only?

The correct equation for the proper time d \tau of the traveling twin with cruising time T_c and acceleration time T_a (both as measured in the inertial stay at home twins rest frame) can be stated as a function of v only like this:

d \tau=\frac{T_c}{\gamma}+\frac{c T_a}{v \gamma} \, \, asinh(v \gamma/c)

Where v is the cruising velocity and \gamma is 1/\sqrt{1-v^2/c^2}.

For a numerical example let us say that the total cruising time T_c is 10 years and the total acceleration time T_a is also 10 years, (both measured in the rest frame of the stay at home twin) so that the inertial twin ages by 20 years during the round trip of the travelling/accelerating twin. The cruising velocity of the traveling twin is 0.8c relative to the stay at home twin and the gamma factor is 1/0.6. Just to be absolutely clear, the traveling twin goes away from home at 0.8c for 5 yers, then spends 5 years slowing down to a stop, another 5 years accelerating to 0.8c and finally the last 5 years cruising at 0.8c for a round trip time of 20 years, all as measured by the inertial twin.

The proper time that elapses on the traveling twins clock in years is:

d \tau=10*0.6+10*0.6/0.8* \,asinh(0.8/0.6) = 6+8.23959 = 14.2396

Note that most of the time dilation happens during the cruise phase even though equal times are spent cruising and accelerating according to the inertial twin. This is because on average the instantaneous relative velocity during the acceleration phase is less than the cruising relative velocity and the acceleration has not increased the time dilation.

Your equation is wrong and predicts the proper time of the traveling twin to be longer than the stay at home twins' proper time. Basically you screwed up the factors when hacking your derivation. You didn't get it from any textbook did you?

 

[JesseM]

Proper acceleration, far from being "just a constant" is a measurable physical entity.

Of course it's measurable, but it's a constant in the equation for elapsed proper time of an observer with constant proper acceleration, which is where the equation you wrote down came from (that equation wouldn't be used for calculating the elapsed proper time for an observer whose proper acceleration was varying). And if the proper acceleration is varying, you can still calculate v(t) and then derive the elapsed proper time from that.

True. As true as the fact that , when expressing v as a function of a, the final result is not a useful function of v but a function of a.

OK, but a is not a variable in the equation v(t) = at / sqrt[1 + (at/c)^2]) (or in the equation for proper time as a function of coordinate time which you wrote down), the only variable that v is a function of is the coordinate time t. Regardless of the rocket's motion (even if it has a varying proper acceleration), you can always express v as function where the only variable is t, though the function may of course include some numerical constants. And I understand the "clock hypothesis" to mean, first of all, that in the limit as we look at arbitrarily short time intervals dt, the amount of time elapsed on a clock during that interval will be equal to \sqrt{1 - v^2/c^2} \, dt, so even if two different clocks are at different positions or have different instantaneous accelerations, as long as they have the same instantaneous velocity they will have the same instantaneous rate of ticking in a given inertial frame. And second, as a corollary of the above, I understand the clock hypothesis to mean that if you know v(t) for a given clock between t₀ and t₁, the integral \int_{t_0}^{t_1} \sqrt{1 - v(t)^2/c^2} \, dt will give the elapsed proper time for this clock between t₀ and t₁. Both of these are nontrivial statements about how the laws of physics work in our universe, for example we could write down an alternate set of hypothetical laws where the instantaneous time dilation did depend on the instantaneous acceleration, so two clocks with the same instantaneous velocity could have different instantaneous rates of ticking. Hopefully you don't disagree with any of the above? Assuming you don't, my point was just that talking about the time dilation as a function of acceleration could potentially confuse someone who doesn't fully understand the clock hypothesis, so from a pedagogical point of view it's better not to talk in that way, even though nothing you said was incorrect on a technical level.

Only if such people are unaware of basics physics.

Well, one of the points of the forum is to teach about basic physics to people who aren't necessarily aware of it. You brought up the clock hypothesis (or were you responding to someone else having brought it up earlier), so I just wanted to clarify for people reading what it means to say in the context of the clock hypothesis that time dilation is not a function of acceleration, which might confuse someone who just saw you saying that the elapsed time was a function of a.

In order to be precise, one always needs to take into account the accelerated period.

True, but before you said:

The other point that I was making to kev (that you also missed) is that he has inadvertenly ignored the difference due to the acceleration by making the acceleration period very small wrt the cruising period. If the two periods are comparable, the acceleration effect cannot be neglected. Do you disagree with that?

I took this to mean you were speaking about what was needed to get an approximately correct figure for the elapsed time, since even if the two periods are not comparable and the acceleration period is very small, any finite period of acceleration needs to be taken into account if you want a perfectly precise figure for the elapsed time (say, in a theoretical word-problem where the length of the acceleration period is known with total precision).

 

[Passionflower]

Well, one of the points of the forum is to teach about basic physics to people who aren't necessarily aware of it.

Then the first thing to teach should be that the \gamma factor between two observers is determined by their acceleration both proper and inertial. If we know that we can calculate the time dilation as \gamma is a factor in this. And so of course the total dilation depends on acceleration, as \gamma is determined by acceleration. Or that if we know the time dilation we can calculate the acceleration. These are useful things.

However if one starts with "oh, they are moving but we don't know how this started" then all bets are off since one cannot calculate time dilation from that. All one can do is dwell on completely inconsequential observed time dilation and length contraction using completely unphysical planes of simultaneity. "They keep going forever with a relative speed x, both observers measure the other observers clock to go slower and their length to be shorter So what? You think that should be the basics?

 

[yuiop]

All these "acceleration is not relevant" arguments are rather silly because if we do not take proper (or inertial) acceleration into account we cannot possibly measure which twin ages more!

Watch how ridiculous the "paradox" becomes when we consider acceleration to be irrelevant:

Have two twins co-located with a relative velocity of zero. They separate with an instant velocity of v, when they are a distance x removed from each other, their relative velocity instantly becomes -v. As soon as they meet their instant velocity becomes 0 again. Which twin ages more?

The "paradox" is only a paradox for those who consider acceleration to be irrelevant!

Hi Passionflower. You have a sort of point and it is just as confusing to say that relative proper times is only a function of relative velocities. It is better to say the twins paradox is resolved by the asymmetry of path lengths in spacetime as analysed from any inertial reference frame. Mostly the arguments here are semantic and philosophical as others have mentioned. The paradox can be analysed either in terms of velocity and spacetime path lengths or in terms of acceleration. Both methods are equally valid mathematically, but the acceleration explanation can be confusing to newcomers. You can calculate the time dilation of an accelerating particle in terms of the sum of its instantaneous velocities, but then you might ask how much additional time dilation should we factor in for the acceleration and the answer is none! On the other hand you can calculate the the time dilation of the same particle in terms of its acceleration and distance away at any instant and arrive at the same answer and in this case no additional time dilation is attributed to instantaneous velocity. They are two sides of the same coin. Basically v\gamma=aT_a and you can either use v\gamma or aT_a in the calculations (but not both).

I do agree that any differential ageing involves acceleration, but in some situations where both twins experience equal proper acceleration, you also have to factor in where and when in spacetime the acceleration occurred and this basically boils down to comparing path lengths in space time.

Starthaus gives an equation where the proper time of the constant velocity leg of the journey is a function of the acceleration at the turn around (which is mathematically correct when he gets his equation right), but philosophically it is not ideal, because it implies that the proper time that elapses during the constant velocity leg is applied retrospectively once we find out what the acceleration is at the turn around. Let us say that the twin launches away from Earth with constant velocity and passes Alpha Centauri at constant velocity. What is the proper time that elapses on the traveling twins clock between leaving Earth and passing Alpha Centauri? The acceleration explanation can not give an answer, because it does not know where,when or how quickly the twin is going to turn around, unless the clock has some sort of predictive ability. Note that there is a definitive proper time elapsed on the traveling clock between the events "passing Earth" and "passing Alpha Centauri" even if the traveling clock was always moving at constant velocity and no acceleration was involved.

 

[Passionflower]

The paradox can be analysed either in terms of velocity and spacetime path lengths or in terms of acceleration. Both methods are equally valid mathematically, but the acceleration explanation can be confusing to newcomers.

Really? So then please explain to me mathematically how without any acceleration (proper or inertial) you can say which twin ages more. Use any example you like.

Mostly the arguments here are semantic and philosophical as others have mentioned.

They become philosophical, and frankly irrelevant, as soon as one completely ignores acceleration as there is absolutely nothing to calculate in that case.

 

[JesseM]

The correct equation for the proper time d \tau of the traveling twin with cruising time T_c and acceleration time T_a (both as measured in the inertial stay at home twins rest frame) can be stated as a function of v only like this:

d \tau=\frac{T_c}{\gamma}+\frac{c T_a}{v \gamma} \, \, asinh(v \gamma/c)

Where v is the cruising velocity and \gamma is 1/\sqrt{1-v^2/c^2}.

It depends on how you choose to express things, your method avoids having to know the value of the proper acceleration while starthaus' equation assumed we did know it. If we do know the value of the proper acceleration a, we also know the rocket will be decelerating from v to 0 for T_a/2 and accelerating from 0 to v for T_a/2, and according to the http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html , the proper time for a rocket that accelerates with constant proper acceleration a starting from a speed of 0 would be given by tau = (c/a)*arcsinh(aT/c), so for T=T_a/2 we get tau = (c/a)*arcsinh(aT_a/2c), and by the symmetry we can double that to get the total proper time elapsed from the beginning of the acceleration to the end, giving tau = (2c/a)*arcsinh(aT_a/2c). So, I get a total elapsed time of:

d \tau=\frac{T_c}{\gamma}+\frac{2c}{a} \, \, arcsinh(a T_a / 2c)

If we want to express it without reference to a, another equation on the relativistic rocket page says that velocity v as a function of coordinate time T would be v(T) = aT / sqrt[1 + (aT/c)^2], so if we know it takes a time of T_a/2 to go from 0 to the cruising speed v, that means that v = aT_a / 2 * sqrt[1 + (aT_a/2c)^2], which means:

v^2 = a^2 T_a^2 / 4*[1 + (a^2*T_a^2 / 4c^2)]
4v^2 + 4*v^2*a^2*T_a^2 / 4c^2 = a^2 * T_a^2
4v^2 + (v^2/c^2)*a^2*T_a^2 = a^2*T_a^2
4v^2 = a^2*T_a^2*(1 - v^2/c^2)
a^2 = 4v^2 / (T_a^2 * (1 - v^2/c^2))
a = 2v*gamma/T_a

Substituting that into my equation above:

d \tau=\frac{T_c}{\gamma}+\frac{c T_a}{v \gamma} \, \, arcsinh(v\gamma /c)

...which is the same as the equation you wrote down.

edit: whoops, I read your post hastily and thought you were objecting to the part of starthaus' post where he wrote the equation \frac{c}{a} arcsinh(aT_a/c) for the proper time during the acceleration phase, but now I think you probably objecting to this other equation starthaus posted:

d \tau=2T_c/ \sqrt{1+(aT_a/c)^2}+4\frac{c}{a} asinh(aT_a/c)

Maybe in this equation starthaus is actually letting T_a represent half the acceleration period, which would explain why he uses asinh(aT_a/c) rather than asinh(aT_a/2c) as I did, but why is there a 4 out in front? I don't know where the first term comes from either...

 

[starthaus]

The correct equation for the proper time d \tau of the traveling twin with cruising time T_c and acceleration time T_a (both as measured in the inertial stay at home twins rest frame) can be stated as a function of v only like this:

d \tau=\frac{T_c}{\gamma}+\frac{c T_a}{v \gamma} \, \, asinh(v \gamma/c)

Where v is the cruising velocity and \gamma is 1/\sqrt{1-v^2/c^2}.

If you do the proper calculations, you'll find out that v\gamma=aT_a

Your equation is wrong

Err, no.

and predicts the proper time of the traveling twin to be longer than the stay at home twins' proper time.

Err, also no.

Basically you screwed up the factors when hacking your derivation. You didn't get it from any textbook did you?

No, I didn't hack it and I didn't get it from a textbook.

The proper time that elapses on the traveling twins clock in years is:

d \tau=10*0.6+10*0.6/0.8* \,asinh(0.8/0.6) = 6+8.23959 = 14.2396

...which clearly contradicts your earlier claim that the contribution of the acceleration period is negligible.

 

[starthaus]

d \tau=2T_c/ \sqrt{1+(aT_a/c)^2}+4\frac{c}{a} asinh(aT_a/c)

I guess in this equation starthaus is actually letting T_a represent half the acceleration period, so the second term is correct. Not sure where he got the first term though...

...from knowing math and physics

 

[JesseM]

Then the first thing to teach should be that the \gamma factor between two observers is determined by their acceleration both proper and inertial.

Not necessarily, what about two observers who pass one another moving inertially in space and note their ages as they pass? And even if you assume they started at rest relative to one another, if you assume acceleration was quasi-instantaneous (as is commonly assumed in these kinds of problems to make the math simpler), then the question of which one accelerated initially is irrelevant to figuring out their elapsed time, all that matters is which one accelerated to turn around once they were a significant distance apart. Then all you have to do is pick a frame, find the velocity of the one that turned around both before and after the turnaround (as well as the time between departure and turnaround in the frame you chose), as well as the velocity of the one that moved inertially between departure and reunion, and from that you can figure out how much elapsed time both have experienced between departure and reunion.

 

[yuiop]

Substituting that into my equation above:

d \tau=\frac{T_c}{\gamma}+\frac{c T_a}{v \gamma} \, \, arcsinh(v\gamma /c)

...which is the same as the equation you wrote down.

edit: whoops, I read your post hastily and thought you were objecting to the part of starthaus' post where he wrote the equation \frac{c}{a} arcsinh(aT_a/c) for the proper time during the acceleration phase, but now I think you probably objecting to this other equation starthaus posted:

d \tau=2T_c/ \sqrt{1+(aT_a/c)^2}+4\frac{c}{a} asinh(aT_a/c)

I guess in this equation starthaus is actually letting T_a represent half the acceleration period, so the second term is correct. Not sure where he got the first term though...

If you start with my equation:

d \tau=\frac{T_c}{\gamma}+\frac{c T_a}{v \gamma} \, \, arcsinh(v\gamma /c)

and substitute aT_a for v\gamma and \sqrt{1+(aT_a/c)^2} for \gamma you end up with the equation Starhaus should have ended up with if hadn't blundered:

d \tau=T_c/ \sqrt{1+(aT_a/c)^2}+\frac{c}{a} arcsinh(aT_a/c)

Even if you allow for the fact that Starhaus might of meant T_c and T_a were the the cruising time and accelerating time of the the outwrad leg only, rather than the round trip time, the equation he should have ended up with is:

d \tau= 2T_c/ \sqrt{1+(aT_a/c)^2}+2\frac{c}{a} arcsinh(aT_a/c)

Either way, Starthaus messed up.

However, he is right that it does produce the mathematical oddity that the proper time of the outward leg is apparently determined by the acceleration at the end of the leg, which just shows you have to be careful in how you physically interpret the equations.

 

[JesseM]

...from knowing math and physics

I edited my post, the second term appears to be incorrect even if you let T_a be half the acceleration period. Were you indeed making that assumption? In that case the total proper time elapsed during the acceleration period should be 2\frac{c}{a} \, arcsinh (aT_a/c), not 4\frac{c}{a} \, arcsinh (aT_a/c).

 

[starthaus]

the equation he should have ended up with is:

d \tau= 2T_c/ \sqrt{1+(aT_a/c)^2}+2\frac{c}{a} arcsinh(aT_a/c)

Either way, Starthaus messed up.

Err, you are wrong again. The equation is correct, read here.

However, he is right that it does produce the mathematical oddity that the proper time of the outward leg is apparently determined by the acceleration at the end of the leg,

I never claimed such nonsense. You need to understand my post.

which just shows you have to be careful in how you physically interpret the equations.

Yes, you also need to be careful in getting your math right.

 

[starthaus]

I edited my post, the second term appears to be incorrect even if you let T_a be half the acceleration period. Were you indeed making that assumption? In that case the total proper time elapsed during the acceleration period should be 2\frac{c}{a} \, arcsinh (aT_a/c), not 4\frac{c}{a} \, arcsinh (aT_a/c).

You are making the same error(s) as kev. My formula is correct.

 

[JesseM]

If you start with my equation:

d \tau=\frac{T_c}{\gamma}+\frac{c T_a}{v \gamma} \, \, arcsinh(v\gamma /c)

and substitute aT_a for v\gamma and \sqrt{1+(aT_a/c)^2} for \gamma

Why would you do that, though? According to my calculations (which ended up with the same equation you got), a = 2v*gamma/Ta, so that implies v*gamma = a*T_a/2, not aT_a. Did I make an error?

 

[JesseM]

You are making the same error(s) as kev. My formula is correct.

Care to point out the error? Do you agree that if we look at the second half of the acceleration period, the rocket starts from a speed of 0 and accelerates for a coordinate time of T_a in your notation? If so, the http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html page says that if a rocket starts from 0 and accelerates for a coordinate time of t, its elapsed proper time (which they denote with T) is:

T = (c/a) sh^-1(at/c)

Do you disagree with that equation? If not, you should presumably agree that during the second half of the acceleration the rocket experiences a proper time of (c/a)*arcsinh(aT_a/c), so by symmetry the total elapsed proper time during the acceleration phase is just double that.

 

[starthaus]

Why would you do that, though? According to my calculations (which ended up with the same equation you got), a = 2v*gamma/Ta, so that implies v*gamma = a*T_a/2, not aT_a. Did I make an error?

Yes, you did, you need to think about the definition of "cruising" speed.

 

[starthaus]

Care to point out the error?

Please read my answer to kev's alleged finding an error in my formula. You are both missing two periods of acceleration.
The formulas in post #80 are correct. Yours and kev's are not.

 

[Passionflower]

Not necessarily, what about two observers who pass one another moving inertially in space and note their ages as they pass? And even if you assume they started at rest relative to one another, if you assume acceleration was quasi-instantaneous (as is commonly assumed in these kinds of problems to make the math simpler), then the question of which one accelerated initially is irrelevant to figuring out their elapsed time.

Obviously because you measure the elapsed time between the event when they pass with a relative speed and when they return. What happens before is irrelevant.

all that matters is which one accelerated to turn around once they were a significant distance apart.

Exactly, acceleration again!

 

[JesseM]

Please read my answer to kev's alleged finding an error in my formula. You are both missing two periods of acceleration.

OK, so you're saying that the rocket also initially accelerated for a time T_a to achieve the cruising speed of v, and then at the end decelerated for a time of T_a as well. Still, I trust you have no objection to my and kev's equations in a different scenario where the rocket just passed by the "stationary" observer (i.e the one who remains inertial and has a velocity of zero in the frame we're considering) already moving at speed v, continues to cruise for time T_c/2, then accelerates to turn around for time T_a, then cruises back towards the "stationary" observer, passing him after another period of T_c/2 (and they compare clocks as they pass).

Anyway, given the scenario you are considering with initial acceleration and final deceleration, I have no objection to the second term, let me think about the first term...but before I do that, can you tell me if T_c represents the total cruising time in both directions, or just the cruising time in one direction?

 

[starthaus]

Anyway, given the scenario you are considering with initial acceleration and final deceleration, I have no objection to the second term, let me think about the first term...but before I do that, can you tell me if T_c represents the total cruising time in both directions, or just the cruising time in one direction?

one direction, this is why it gets doubled in the final formula

 

[yuiop]

Why would you do that, though? According to my calculations (which ended up with the same equation you got), a = 2v*gamma/Ta, so that implies v*gamma = a*T_a/2, not aT_a. Did I make an error?

I think it only because you are considering a,v and gamma for half the acceleration phase while I am considering the total acceleration phase. We end up with the same results.

 

[yuiop]

...which clearly contradicts your earalier claim that the contribution of the acceleration period is negligible.

I was talking about a different scenario where the acceleration phase was much more extreme and took place in seconds. With very extreme acceleration with the acceleration phase period tending to zero, the time dilation due to acceleration becomes negligable. The greater the acceleration is, the more you can ignore it.

 

[JesseM]

one direction, this is why it gets doubled in the final formula

OK, thanks. Then the starting equation in your scenario and with your definitions would be:

d \tau=\frac{2T_c}{\gamma}+\frac{4c}{a} \, \, arcsinh(a T_a / c)

Another equation on the relativistic rocket page says that velocity v as a function of coordinate time T would be v(T) = aT / sqrt[1 + (aT/c)²], so if we know it takes a time of T_a to go from 0 to the cruising speed v, that means that v = aT_a / sqrt[1 + (aT_a/c)²], so:

v² = a²*T_a² / (1 + a²*T_a²/c²)

v² = c² * a² * T_a² / (c² + a²*T_a²)

v²/c² = a² * T_a² / (c² + a²*T_a²)

(1 - v²/c²) = [(c² + a²*T_a²) - (a² * T_a²)] / (c² + a²*T_a²) = c² / (c² + a²*T_a²) = 1 / (1 + a²*T_a²/c² )

gamma = 1/sqrt(1 - v²/c²) = sqrt(1 + (aT_a/c)²)

So plugging that into the top equation,

d \tau= \frac{2T_c}{\sqrt{1 + (aT/c)^2}} +\frac{4c}{a} \, \, arcsinh(a T_a / c)

...which agrees with what you had.

 

[Passionflower]

The greater the acceleration is, the more you can ignore it.

Sorry but this is just absurd, you cannot ignore acceleration, great or small, in the twin experiment. Without acceleration there is no absolute time dilation.

I am still waiting for an example of the twin experiment without acceleration but "in terms of velocity and spacetime path lengths", which, if I understand correctly, in your view does not require acceleration.

 

[yuiop]

It is obvious that the equations given by Starthaus are symbol for symbol copied from wikipedia, so it would have have been helpful if he credited wikipedia and gave a link to where the wikipedia scenario is described less vaguely than Starthaus described it.

In the wikipedia scenario there are 4 acceleration periods and 2 cruising periods so that explains why our equations differ, because the scenarios and definitions of the periods differ.

I think you will find if you use my equations with T_a and T_c defined as total times for the whole journey, you can use the same equation for any of number of acceleration and cruising phases, without having to have different equations for different twins scenarios, as long as the acceleration is the same for all the acceleration phases.

 

[yuiop]

Sorry but this is just absurd, you cannot ignore acceleration, great or small, in the twin experiment. Without acceleration there is no absolute time dilation.

I am still waiting for an example of the twin experiment without acceleration but "in terms of velocity and spacetime path lengths", which, if I understand correctly, in your view does not require acceleration.

In an earlier reply to you I acknowledged that in any situation where differential ageing occurs then acceleration is involved, but you can calculate the time dilation without using the acceleration. That is not the same as saying acceleration is not required.

 

[Passionflower]

In the wikipedia scenario there are 4 acceleration periods and 2 cruising periods so that explains why our equations differ, because the scenarios and definitions of the periods differ.

I have no desire to be squeezed between Starthaus and the others in this argument but that is the situation in the standard twin "paradox", you need 4 accelerations and optionally two cruising periods..

 

[JesseM]

Sorry but this is just absurd, you cannot ignore acceleration, great or small, in the twin experiment. Without acceleration there is no absolute time dilation.

But if the acceleration is instantaneous, you don't actually need to consider the acceleration phase when calculating the elapsed proper time. If you know the ship cruised at speed v₁ for a coordinate time of t₁, then instantaneously accelerated to turn around so it had a speed of v₂ in the other direction and took an additional coordinate time of t₂ to reunite with the inertial observer, then the elapsed time for the observer that turned around is just t₁*sqrt(1 - v₁²/c²) + t₂*sqrt(1 - v₂²/c²). So in this sense, the acceleration can be ignored in your calculations.

I have no desire to be squeezed between Starthaus and the others in this argument but that is the situation in the standard twin "paradox", you need 4 accelerations and optionally two cruising periods..

Most textbook discussions of the twin paradox assume instantaneous accelerations.

 

[Passionflower]

So in this sense, the acceleration can be ignored in your calculations.

Acceleration is essential in the twin experiment. Without acceleration there is no absolute time dilation.

Show me a calculation where you completely ignore acceleration (proper or inertial) that shows an absolute time dilation and I show you where you made a mistake!

 

[JesseM]

Acceleration is an essential in the twin experiment. Without acceleration there is no absolute time dilation.

Show me a calculation where you completely ignore acceleration (proper or inertial) that shows an absolute time dilation and I show you where you made a mistake!

I don't know what you mean by "completely ignore", but by "ignore" I only mean that you don't have to consider the contribution that the accelerating phase makes to the total elapsed proper time, since you are treating the acceleration as instantaneous (I think my meaning was fairly clear from the context, especially given my comment 'you don't actually need to consider the acceleration phase when calculating the elapsed proper time'). In that sense the calculation I already gave you in my last post ignores acceleration, though of course acceleration plays a role in that it explains why v₁ on the outbound leg may be different than v₂ on the inbound leg (and since inertial paths are geodesics and geodesics always maximize proper time, it plays a conceptual role in understanding why the inertial twin is always the one who ages more than the one who turns around, similar to the idea that a straight line between two points in Euclidean geometry always has a shorter length than any bent path between the same points).