# The physical interpretation of the Minkowski spacetime diagram

Discussion in 'Physics & Math' started by geordief, Dec 21, 2017.

1. ### Neddy BateValued Senior Member

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1,570
Yes, but time and distance can easily be considered equivalent without having to use a Minkowski diagram. Consider the term "light year" which is a unit of distance, but which also tells you exactly how much time it would take for light to travel that distance.

Or consider the thought experiment commonly used in relativity, the "light clock," which is a hypothetical clock made of two parallel mirrors with a photon endlessly bouncing back and fourth between them. All I need to know is the distance between the mirrors, and I can tell you how much time it takes for each "tick" of the clock, at least in the reference frame of the light clock itself.

Both of the above concepts work fairly seamlessly with the concept of a three dimensional space, with x y and z as the three spacial axes. But a Minkowski diagram requires us to neglect one of those spacial axes (usually not a problem) and replace it with a ct axis which, as you say, is where that darn time thing shows up.

In that respect, Minkowski diagrams are not really all that different than any other type of graph which happens to have a time axis. But because relativity has some weird concepts such as relativity-of-simultaneity, which is well outside our everyday experience, some folks look for deeper answers in the Minkowski diagram. Some interpret them as showing that all time, past present and future, is one static "thing" and we are moving through it, as through a fourth spacial dimension. I find that to be going a little deeper than is necessary, but maybe that is just me.

3. ### geordiefRegistered Senior Member

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695
But acceleration does cause the time dilation doesn't it?

No I didn't claim that would break the (c) speed limit. All the same an instantaneous deceleration from .say .9c to -.9c has to be modeled somehow and we all surely know that it is a ridiculous scenario.

But,no I don't claim to be qualified to comment on this subject .I will give you the last word as I have probably misunderstood the discussion.

5. ### Neddy BateValued Senior Member

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1,570
No, as far as special relativity is concerned, time dilation is strictly the result of relative motion. By relative motion, I mean the perfectly reciprocal nature of motion, where you can lean on a street lamp by the side of a street and claim the cars are whooshing past where you stand, OR, you can drive in your car and claim that some person leaning against a street lamp is whooshing past where you sit in your driver's seat.

Since that relative motion is perfectly reciprocal, and since time dilation is the result of that relative motion, then, time dilation must also be perfectly reciprocal. But of course that leads to problems if one never looks deeply for where that perfect symmetry might break.

In the outward-and-return-journey model that we were discussing, it is what happens at the turn-around point which breaks the symmetry. The traveling-twin's turn-around requires considering a third reference frame, different than the two considered before.

Prior to the turnaround we had to use v=+0.866c in the calculations, and then after the turnaround, we had to use v=-0.866c where the negative sign represents the velocity in the opposite direction. That changes the traveling-twin's notion of simultaneity, and that is why, as far as he is concerned, his brother's wristwatch changes from displaying t=5 to displaying t=35 in the scenario we were using.

Yes, instantaneous acceleration/deceleration is a totally ridiculous scenario, but that was not meant to deceive in any way. It was just to make the calculations easier, and to showcase that all that is required to break the symmetry in that scenario is for the traveling-twin to stop, or reverse direction and head back home. The issue is not how gradually or abruptly the traveling-twin might decelerate or accelerate back toward their twin. The issue is simply that they did so, and that the stay-at-home-twin did not do so.

Last edited: Dec 28, 2017

7. ### Mike_FontenotRegistered Member

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23
OK ... I can see what was confusing you. The "t = 20-" DOESN'T mean "t = -20". In my terminology and variable-naming strategy, the variable "t" always refers to the traveling twin's age (in years), and it is usually the independent variable in the analysis. In the instantaneous turnaround scenario on my webpage, at the turnaround point, the traveling twin's age is t = 20 years, and that is the instant when he does his instantaneous turnaround. Immediately before the turnaround, the relative velocity between the twins is v = 0.866 ly/y (they are moving apart), and immediately after the turnaround, their relative velocity is v = -0.866 ly/y (they are moving toward each other). Since it makes no sense to assign two different velocities to a single instant in time, we need to label TWO times for those two different velocities, the first an infinitesimal time BEFORE the instant "t = 20", and the other an infinitesimal AFTER the instant "t = 20". The first such time is denoted "t = 20-", and the other such time is denoted "t = 20+".

8. ### Mike_FontenotRegistered Member

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23
You definitely DID do it. It is quite clear that your understanding of the twin "paradox" is very good.

An even easier way for the home twin (she) to get t' is to use the famous time-dilation result: t' = t / gamma.

Again, an even easier way for the home twin (she) to get t' is to use the famous time-dilation result: t' = t / gamma.

You are not being clear exactly what you mean above by the word "turnaround" here. A more precise way to phrase it would be "from the start to immediately before the turnaround", so that the answer does NOT include any ageing that happens to the home twin "during" the turnaround.

An even easier way for the traveling twin (he) to get this t is to use the famous time-dilation result: t = t' / gamma.

As above, you are not being clear exactly what you mean above by the word "turnaround". A more precise way to phrase WHAT YOU MEANT would be "from immediately before the turnaround to the end", so that the answer DOES include any ageing that happens to the home twin "during" the turnaround. And your answer to that is indeed 35 years. But it would be much better to treat the ageing by the home twin during the turnaround as a separate item, since that is the crux of the resolution of the twin "paradox". I.e., the ageing of the home twin, according to the traveling twin, is 5 years from the start to immediately before the turnaround, plus 30 years during the turnaround, plus 5 years from immediately after the turnaround until the end, for a total of 40 years. The original supposed paradox is that if the traveler's analysis implicitly assumes that the home twin doesn't age any during the turnaround, then the home twin should be the younger one when the twins are reunited. But he finds instead that she is the older one, and that seems to be a paradox to him. But there actually is no paradox, once it is realized that the home twin ages by a large amount during the turnaround.

Last edited: Dec 29, 2017
9. ### Neddy BateValued Senior Member

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1,570
Oh, okay, that makes more sense. The way I do it, I also change the velocity from v = +0.866 (before the instantaneous turnaround point) to v = -0.866 (after the instantaneous turnaround point). I just call it the turnaround point, without designating any particular notation for it.

Unlike you, I do not eliminate c entirely from the equations, I just choose units such that c=1. The only place c appears in the Lorentz transformation equations is as c² as the denominator of a fraction, so it is really not a problem to have it there. I just evaluate the numerator of the fraction, and do nothing with the simple denominator of 1.

When I posted my method, I had intended to use the exact same scenario that you had used, but I was doing it from memory, and I mistakenly thought that you had the final time of the traveling twin being 20, rather than it being 20 at the turnaround point. So all my results were basically half of your results. Sorry about that.

In my next post I will re-do my method to make it as much like yours as possible.

10. ### Neddy BateValued Senior Member

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1,570
I use the following more traditional variables, and I just re-use them as needed without any subscripts or anything. I separate all of the sections with explanatory text. Hopefully it will make sense to you:
t = time as measured by the Stay-At-Home-Twin
x = location of the Traveling-Twin as measured by the Stay-At-Home-Twin
t' = time as measured by the Traveling-Twin
x' = location of the Stay-At-Home-Twin as measured by the Traveling-Twin
v = relative velocity between the two twins
γ = gamma = Lorentz factor = 1 / √(1 - (v²/c²))

FIRST LET'S SEE HOW IT IS DONE FROM THE EASIEST REFERENCE FRAME:

From Start to Turnaround, (from the reference frame of the Stay-At-Home-Twin):
v = 0.866c
t = 40.00
x = vt = 34.64
Substituting the above values and γ=2 into the equation below gives:
t' = γ(t - (vx / c²)) = 20.00

From Turnaround to End, (from the reference frame of the Stay-At-Home-Twin):
v = -0.866c
t = 40.00
x = vt = -34.64
Substituting the above values and γ=2 into the equation below gives:
t' = γ(t - (vx / c²)) = 20.00

Totals, (from the reference frame of the Stay-At-Home-Twin):
t = 40.00 + 40.00 = 80.00 = Final Age of Stay-At-Home-Twin
t' = 20.00 + 20.00 = 40.00 = Final Age of Traveling-Twin
x = 34.64 + (-34.64) = 0.00 = Final Position of Traveling-Twin

AND NOW LET'S SEE HOW IT IS DONE IT FROM THE OTHER REFERENCE FRAME:

From Start to Turnaround, (from the reference frame of the Traveling-Twin):
v = 0.866c
t' = 20.00
x' = -vt' = -17.32
Substituting the above values and γ=2 into the equation below gives:
t = γ(t' + (vx' / c²)) = 10.00

From Turnaround to End, (from the reference frame of the Traveling-Twin):
v = -0.866c
t' = 20.00
x' = vt' = -17.32
Substituting the above values and γ=2 into the equation below gives:
t = γ(t' + (vx' / c²)) = 70.00

Totals, (from the reference frame of the Traveling-Twin):
t = 10.00 + 70.00 = 80.00 = Final Age of Stay-At-Home-Twin
t' = 20.00 + 20.00 = 40.00 = Final Age of Traveling-Twin
x' = -17.32 - (-17.32) = 0.00 = Final Position of Stay-At-Home-Twin

11. ### Neddy BateValued Senior Member

Messages:
1,570
Sorry, I did not see your red post post until after I had posted my previous two posts. I think we were both posting at the same time.

Thank you, yours is also.

Yes, but I prefer to use the entire Lorentz transformation equations.

Okay, but since I am assuming the turnaround itself is instantaneous, it has no duration, and so no ageing can happen to the home twin "during" the turnaround. The way I am doing it, the velocity is either +0.866 or -0.866 and the stay-at-home twin is suddenly older when the velocity becomes -0.866.

The way I do it, there is no 5+30+5, because the first 5 is essentially lost when it suddenly becomes 35. So I have 35+5 only. For simplicity, I was taking advantage of the symmetry of the two legs of the journey, but I really should have shown the calculation of 5 again for the return leg.

(Or, now that I have re-posted it with the same numbers as you...)

The way I do it, there is no 10+60+10, because the first 10 is essentially lost when it suddenly becomes 70. So I have 70+10 only. For simplicity, I was taking advantage of the symmetry of the two legs of the journey, but I really should have shown the calculation of 10 again for the return leg.

Last edited: Dec 29, 2017
12. ### Neddy BateValued Senior Member

Messages:
1,570
Considering the details in the above series of posts, I think the second half of my calculations should be changed to the following:

AND NOW LET'S SEE HOW IT IS DONE IT FROM THE OTHER REFERENCE FRAME:

From Start to the Instant Before Turnaround, (from the reference frame of the Traveling-Twin):
v = 0.866c
t' = 20.00
x' = -vt' = -17.32
Substituting the above values and γ=2 into the equation below gives:
t = γ(t' + (vx' / c²)) = 10.00

In the Instant After Turnaround, the Previous Calculations are Replaced, (from the reference frame of the Traveling-Twin):
v = -0.866c
t' = 20.00
x' = vt' = -17.32
Substituting the above values and γ=2 into the equation below gives:
t = γ(t' + (vx' / c²)) = 70.00

From the Instant After Turnaround to End, (from the reference frame of the Traveling-Twin):
v = -0.866c
t' = 20.00
x' = -vt' = 17.32
Substituting the above values and γ=2 into the equation below gives:
t = γ(t' + (vx' / c²)) = 10.00

Totals, (from the reference frame of the Traveling-Twin):
t = 70.00 + 10.00 = 80.00 = Final Age of Stay-At-Home-Twin
t' = 20.00 + 20.00 = 40.00 = Final Age of Traveling-Twin
x' = -17.32 + 17.32 = 0.00 = Final Position of Stay-At-Home-Twin

Last edited: Dec 29, 2017
13. ### phytiRegistered Senior Member

Messages:
271

A graph with time in seconds and distance in meters, would be useless since the space unit is 3(10)8 times larger than the time unit.
If 1 cm = 1 sec, a signal to the moon and back would be represented as two horizontal lines spaced 2.5 cm. A scaling of the time is required for a useful graph.
By generalizing the time coordinate mathematically to a dimension ct, Minkowski provided the scaling and eliminated the issue of the nature of time vs space.
The history of time has always used an astronomical motion of an object. The light clock in its simplest form, is nothing more than light in motion. If the oscillations could be unfolded to form a straight path, it would be a light distance.
The Minkowski graph, with ct and x axis, represents the ratio of simultaneous object motion to light motion, vt/ct, v/c, or speed plots for an interval of time.
The example shows a 'twin' scenario. The blue lines are light signals.
At left, A leaves B, moving at a constant .6c, and returns at Bt = 10.
At right, B leaves A, moving at a constant .6c, and returns at At = 8.
The accelerations are instantaneous and discontinuous, thus unreal and misleading.
The popular response for a solution is, the one who returns ages less. A would experience an acceleration and B would not. As depicted, if the direction change requires zero time, there is no force applied, and A is not aware of any change. Even if a realistic curve for acceleration was used, the same could be applied to both A and B, and acceleration would not be the deciding factor. In addition the gamma factor contains the speed ratio v/c, and no terms for acceleration. If A moves simultaneously on a longer closed path than B, A must move faster and accumulates less time.
Notice A's world is smaller than B's world.

14. ### Mike_FontenotRegistered Member

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23
It's true that a human couldn't survive an "instantaneous" turnaround (between velocities that are a large fraction of the speed of light), or even a turnaround that lasts days or even months. But I give an example in my webpage that shows that a 1g turnaround that lasts for two years of the traveler's life (at a distance of about 40 ly at the start of the turnaround, according to the home twin) is actually QUALITATIVELY quite similar to the instantaneous turnaround scenario. In particular, most of the home twin's ageing (according to the traveler) occurs during the turnaround. She ages 17 years during the (mostly inertial) outbound leg, then about 64 years during the turnaround, and finally another 17 years on the (mostly inertial) inbound leg.

Actually, the above inbound ageing of the home twin is what WOULD have happened if the traveler HAD decided to go home after the turnaround. But in the scenario on the webpage, the traveler decides instead to do another turnaround in the opposite direction, so that at the end of the second turnaround, he ends up heading away from her again. The result is that during this second turnaround, according to the traveler, the home twin gets about 59 years YOUNGER. This result is not unexpected, because it is easy to see that if a second instantaneous turnaround is immediately done in the instantaneous turnaround scenario, the home twin's age is just restored to what it was immediately before the first instantaneous turnaround ... i.e., the two instantaneous turnarounds in opposite directions (with no time lapse in between them) just cancel out, and it is as if no instantaneous turnarounds had occurred at all.

15. ### Neddy BateValued Senior Member

Messages:
1,570
I would think an infinite amount of force would be required. Maybe A would not be aware of any change, but that would be because he would be dead from the force smashing his body to a bloody pulp.

16. ### Neddy BateValued Senior Member

Messages:
1,570
Yes, that ^ makes sense.

Did you see my new approach to the traveling-twin calculations in post #49? I took your advice and separated the"Instant Before Turnaround" from the "Instant After Turnaround". I think it makes more sense that way.

17. ### Mike_FontenotRegistered Member

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23
No, you're still not there. The home twin's ageing during the turnaround is absolutely critical for the resolution of the twin "paradox", and it should be clearly and separately identified and emphasized. The supposed "paradox" arises from many peoples' erroneous assumption that the home twin doesn't age any during the instantaneous turnaround.

The sequence should be (from the traveler's perspective):

First, compute the ageing of the home twin from the beginning of the trip until immediately before the instantaneous turnaround.

Second, compute the ageing of the home twin from immediately before the instantaneous turnaround until immediately after the instantaneous turnaround.

Third, compute the ageing of the home twin from immediately after the instantaneous turnaround until the twins are reunited.

Then add those three components up to get the home twin's age at the reunion.

18. ### Neddy BateValued Senior Member

Messages:
1,570
My method does show that the time on the home-twin's clock changes from t=10 just before the turnaround to t=70 just after the turnaround. That is quite a bit different than an "erroneous assumption that the home twin doesn't age any during the instantaneous turnaround."

Yes, and that is what I did here...

From Start to the Instant Before Turnaround, (from the reference frame of the Traveling-Twin):
v = 0.866c
t' = 20.00
x' = -vt' = -17.32
Substituting the above values and γ=2 into the equation below gives:
t = γ(t' + (vx' / c²)) = 10.00

Yes, and that is what I did here...

In the Instant After Turnaround, the Previous Calculations are Replaced, (from the reference frame of the Traveling-Twin):
v = -0.866c
t' = 20.00
x' = vt' = -17.32
Substituting the above values and γ=2 into the equation below gives:
t = γ(t' + (vx' / c²)) = 70.00

Demonstrating that the time on the home-twin's clock changes from t=10 just before the turnaround to t=70 just after the turnaround.

Yes, and that is what I did here...

From the Instant After Turnaround to End, (from the reference frame of the Traveling-Twin):
v = -0.866c
t' = 20.00
x' = -vt' = 17.32
Substituting the above values and γ=2 into the equation below gives:
t = γ(t' + (vx' / c²)) = 10.00

The reason there are only two components to add is because the home twin's clock changed from t=10 just before the turnaround to t=70 just after the turnaround. The first component which was 10 was replaced by the second component which was 70, which then needs to be added to third component which was 10. And that is what I did here...

Totals, (from the reference frame of the Traveling-Twin):
t = 70.00 + 10.00 = 80.00 = Final Age of Stay-At-Home-Twin
t' = 20.00 + 20.00 = 40.00 = Final Age of Traveling-Twin
x' = -17.32 + 17.32 = 0.00 = Final Position of Stay-At-Home-Twin

It seems to me that you think the Lorentz transformation equation I used in the second component is supposed to be 60, and then the first component of 10 needs to be added to that 60 to obtain 70. But the Lorentz transformation equation I used in the second component gave me 70 right out of the box.

I think this disagreement comes from the fact that you have been using your own method instead of the Lorentz transformation equations. I would like to learn your method, but I find the variable names and methodology confusing. Could you please post in this thread your solution to this scenario, using the same variable names that I used?

t = time as measured by the Stay-At-Home-Twin
x = location of the Traveling-Twin as measured by the Stay-At-Home-Twin
t' = time as measured by the Traveling-Twin
x' = location of the Stay-At-Home-Twin as measured by the Traveling-Twin
v = relative velocity between the two twins
γ = gamma = Lorentz factor = 1 / √(1 - (v²/c²))

I would like to see how you use those variables to calculate the 60 instead of 70. Thank you.

EDIT: From your website it appears you get 60 by multiplying -34.64 * (-1.732) what are those in terms of x, x', t, t', v, gamma, etc.?

Last edited: Dec 29, 2017
19. ### Confused2Registered Senior Member

Messages:
487
Methinks length contraction is the same regardless of travelling away or towards so the travelling twin will/should see the same time (yet to be determined) on the home clock (using a telescope) both (immediately) before and (immediately) after the instantaneous turn around.

Last edited: Dec 30, 2017
20. ### Neddy BateValued Senior Member

Messages:
1,570
Yes, that is a good point.

But remember, the time seen on the distant wrist watch (through the telescope) is not its current time, it is an earlier time. That is because we know it takes a certain amount of time for light to travel a certain amount of distance.

To calculate the current time from the apparent time, the question a relatively moving observer in such circumstance must ask himself is, "How can I calculate what the distance between myself and this distant wristwatch would have been at the earlier time which I am currently seeing?"

If you think about it, you should see that the answer to that question depends on whether he is moving away from the wristwatch (in which case he would say the earlier distance must have been smaller than it is currently), or whether he is moving toward the wristwatch (in which case he would say the earlier distance must have been greater than it is currently).

So the traveling twin, using the above logic, should conclude two different distances (and thus two different current times on the distant wrist watch), depending on whether he performs the calculation just before the turnaround or just after the turnaround.

21. ### Mike_FontenotRegistered Member

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23
You (following Einstein) assign the time variables t and t' to the two twins' ages. The disadvantage of those choices is that people often have trouble remembering which twin's age got which variable name. I wanted to use variable names which would be easier to remember and harder to confuse.

I use the variable "t" to always denote the age of the traveling twin ... that much, you do have to remember. And, as a first step, I use the pronounceable acronym "CADO" to denote the home twin's age. "CADO" stands for "Current Age of a Distant Object". The "Object" is a Person (specifically, the home twin), but I used "Object" because I wanted a pronounceable acronym, and CADP and CAHT aren't pronounceable. So you also DO have to remember that the "Object" always means the home twin. In addition, the variable "t" is usually considered to be the independent variable, which can be specified as part of the problem statement, and the variable "CADO" is usually considered to be the dependent variable. So when I want to be explicit, I write the home twin's age, when the traveling twin's age is "t", as "CADO(t)".

But the above name is ambiguous, because the home twin's age, when the traveling twin is "t" years old, doesn't have just one value ... it is generally different, depending on whether it's from the home twin's perspective or from the traveling twin's perspective (or even from someone else's perspective). So I add another letter to the end of the CADO acronym (separated by an underscore), to specify which twin's perspective I'm referring to: "CADO_H(t)" denotes the home twin's age when the traveling twin's age is "t", according to the home twin. And "CADO_T(t)" denotes the home twin's age when the traveling twin's age is "t", according to the traveling twin. The extra letters "H" and "T" denote "home twin's perspective" and "traveling twin's perspective", respectively.

Why do I use the "ages" of the two twins as variables, rather than the more general "time"? The reason is that "time" is sometimes viewed as kind of an abstract thing which may not necessarily have any "real" significance. In contrast, the "age" of a person has a very intuitive meaning ... it is a "real" quantity. And I think that is important in thinking about the twin "paradox".

The "CADO equation" is an equation that I derived long ago by staring at a Minkowski diagram. That's the easiest and clearest way to derive it ... it's actually quite easy to derive. It can also be derived in a purely analytical way from the Lorentz equations, but that way is more contorted and less clear ... I do it both ways in my Physics Essays paper:

"Accelerated Observers in Special Relativity", PHYSICS ESSAYS, December 1999, p629.

Given, though, that the Minkowski diagram is really just a graphical representation of the Lorentz equations, the CADO equation does clearly follow from the Lorentz equations.

The CADO equation has CADO_T(t) by itself on the left-hand-side, and three variables on the right-hand-side which are all from the perspective of the home twin ... one of those three variables on the right is CADO_H(t), and the other two are the relative velocity and the separation of the twins (which are both also considered as functions of "t"). This is very useful equation, because variables from the perspective of the home twin (who is inertial) are relatively easy to determine (and are well-known and non-controversial), whereas variables (particularly the time variable) from the (accelerating) traveling twin's perspective are not as widely known and are more controversial and difficult to obtain. The CADO equation lets us determine a more difficult quantity by evaluating a simple combination of three relatively easy quantities to determine.

22. ### Confused2Registered Senior Member

Messages:
487
Edit...@Neddy_Bates (or anyone)...
Sorry, can we take an instantaneous break at the turnaround point please? You offered before but I waited to see what anyone else thought. Assuming the turnaround point has a clock Einstein synced with the start point...
I'm really hoping the traveler clock reads 20 seconds and the turnaround clock reads 40 seconds. At which point the telescoped view of the home clock shows 40-17.32 =22.68 which (I think) is a new number in this thread. Immediately before stopping the traveler has a clock showing 20 seconds and thinks he is 17.32/2 light seconds from the start point so in his telescope view of the start clock (relative velocity 0.866c) he sees a time of 20-(17.32/2)=11.34 seconds. If the traveler imagined a virtual start point after the turnaround I think it would be (probably) 22.68 light seconds from the finish. As pointed out earlier - I'm a low paid clock-watcher - Lorenz (and yourself) are high fliers - I'm not saying I'm right - just where am I wrong?

23. ### Neddy BateValued Senior Member

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1,570
If you are familiar with the Lorentz transformation equations, then you should know that in the instant before the turnaround the traveling-twin gets t=10 for the time on the home-twin's clock, and, in the instant after the turnaround the traveling-twin gets t=70 for the time on the home-twin's clock.

The Lorentz transform equations do not produce a result of 60 which would then need to be added to the original result of 10. You can always subtract the original result of 10 from the 70, if you want to know the amount by which the the time on the home-twin's clock changes during the instantaneous turnaround.

Given that, can you please explain what your calculation -34.64*-1.732=60 represents? Is the -1.732 the velocity times two? Is it the velocity times gamma, which in this case happens to be two? That is what I am trying to figure out at this point. Thanks.