Wednesday, 18 March 2015

The Forever War (by Joe Haldeman) and Albert Einstein’s special theory of relativity

(Originally posted on Friday, 11 November 2016; expanded on 14 November 2016)

I am currently re-reading (once again) an awesome novel by Joe Haldeman titled “The Forever War”. I am going to review it in detail, but here I would like to mention that time dilation plays an important role in the plot of this novel. It made me ponder on the main aspects of Albert Einstein’s special theory of relativity again. Some of the things I found on the net were truly jaw-dropping and I decided to analyse this topic in detail.

This time I realised that to fully understand Albert Einstein’s special theory of relativity you have to simultaneously analyse three different things that change when travelling at speeds close to the speed of light: time, length and mass. What’s worse these three things can be seen from two different perspectives and it seems that some internet sites switch the perspectives back and forth, which is utterly confusing.

First, I have to comment on the “change of mass”. It's a concept that is a kind of simplification, but mass is something we understand by heart and this is why I use it here. The point of special theory of relativity is that anything with non-zero mass can NEVER be accelerated to the speed of light – the more the speed of a spaceship approaches the speed of light the harder it is to keep the spaceship accelerating. On the Earth (with speeds much lower that the speed of light) the acceleration is harder when a particular thing (a car for example) is heavier and this is why we can imagine that when the speed of the spaceship approaches the speed of light the spaceship is much “heavier” than normal and we need a much more powerful engine to keep it accelerating in the same way. Finally, with the speed VERY close to the speed of light, the “mass” of the spaceship would go towards infinity and we would need and infinite engine power to keep it accelerating at all. We can also look at this problem from another point of view. When we have a particular space engine and we use it at its maximum power then the more the speed of the spaceship approaches the speed of light the slower the spaceship will be accelerating (the force coming from the engine stays the same, but the mass of the spaceship gets bigger and bigger). This is the only thing that is messed up in The Forever War, but ONLY at ONE moment.


Example I (“twins”)

The most famous example of how the special theory of relativity works is the example about twins – one twin stays on the Earth and the other one travels on a spaceship and comes back after some time. The twin who was on the Earth will be older then the twin who was on the spaceship. For the sake of the example below (for the sake of easy calculations) I assume that the spaceship accelerates and decelerates “almost instantly”, which in reality is not possible. I skip the acceleration/deceleration process because otherwise I would have to use Einstein’s general theory of relativity, which would be much more complicated.

Let’s assume that the spaceship (with the moving twin) travels to a star that is exactly 1 light-year away and then comes back, so the total distance is 2 light-years, and the speed of the spaceship is 0.9 times the speed of light (0.9c). The formula for relativistic factor (Lorentz factor), which calculates the changes of time, length and mass, is this:

The relativistic factor for the assumed speed of the spaceship is 2.294. Below there are two different versions of calculations that show what exactly happens from each perspective. Lengths and masses “are different” while moving, but they “come back” to normal at the end of the journey. The time on the other hand does NOT ”come back” to normal because the time had already passed.

Example I (“twins”) – version 1 of the calculations.
Observer: the stationary twin (the twin on the Earth).
Reference point (for the speed of the moving twin): the Earth.
I.1.a. Distance of the journey (length seen in the same way): 2 light years (unchanged distance/length).
I.1.b. Length of the spaceship (length seen in a different way): normal length of the spaceship / 2.294 = 0.4359 times the normal length of spaceship (irrelevant value).
I.1.c. Time of the journey: 2 light-years / 0.9c = 2.2222 years (time on the Earth).
1d. Mass of the spaceship: normal mass of the spaceship? (there’s no way to measure the mass of the spaceship from the Earth, but it's irrelevant – anyone outside the spaceship doesn't really care about its mass).

Example I (“twins”) – version 2 of the calculations.
Observer: the moving twin (the twin on the spaceship).
Reference point (for the speed of the moving twin): the Earth.
I.2.a. Distance of the journey (length seen in a different way): 2 light years / 2.294 = 0.8718 light-years.
I.2.b. Length of the spaceship (length seen in the same way): normal length of the spaceship (unchanged length).
I.2.c. Time of the journey: 0.8718 light-years / 0.9c = 0.9687 years (time on the spaceship).
I.2.d. Mass of the spaceship: normal mass of the spaceship * 2.294 = 2.294 times the normal mass of the spaceship (more than twice as “heavy”, but still able to accelerate further in a decent way).

Example I (“twins”) – time comparison.
I.3. One day/month/year on the spaceship was equal to 2.294 days/months/years on the Earth (2.2222 years on the Earth / 0.9687 years on the spaceship).

Example I (“twins”) – interesting things to notice and explain.
4a. The LENGTHS changed for BOTH observers by the same factor, but different things changed (temporarily) for different observers. To the stationary twin it was the length of the spaceship that changed (everything else was stationary to him) and to the twin on the spaceship it was everything other than spaceship that changed (only the spaceship was stationary to him). The reference point (the real stationary point in the example) was the Earth because it was the Earth that was NOT accelerating or decelerating (except for its journey around the sun, but the solar system as a whole can be considered as a stationary point, more or less).
4b. It took 0.9687 years on the spaceship to travel a distance that the light itself travels for 2 years. How is that possible?
The problem is that both these times are given by different observers AND for different reference points. The time of 2 years is given by the observer who is watching the LIGHT from the “outside” and the time of 0.9687 years on the spaceship is given by the observer who is watching the SPACESHIP from the “inside”. As we can see in the above example for the “outside” observer the time of the journey of the spaceship is 2.2222 years on Earth, which is MORE than 2 years (the time of the journey of the light), so for the single “outside” observer everything is OK. The funny question is what time would be given from the perspective of the light itself. The answer is jaw-dropping: ZERO!!! For example if a spaceship would reach 0.99999999999999 times the speed of light (there are 14 “9”) then the relativistic factor would be 7073895.381! It means that one day/month/year on the spaceship would be equal to 7073895.381 days/months/years on the Earth! Of course it will never be possible to reach such a high speed because it would also mean that the mass of the spaceship would be 7073895.381 times the normal mass of the spaceship – there would be no way to generate enough force to accelerate such a “heavy” monster.
4c. What was the ONLY problem with Forever War?
This one fragment is dubious:
    “It's not as though we'd be actually lost,” he said with a rather wicked expression. “We could zip up in the tanks, aim for Earth and blast away at full power. We'd get there in about three months, ship time.”
    “Sure,” I said. “But 150,000 years in the future.” At twenty-five gees, you get to nine-tenths the speed of light in less than a month. From then on, you're in the arms of Saint Albert.

The way it is written it is clear that a 3 month journey from the spaceship perspective would mean 150,000 Earth-years. The AVERAGE relativistic factor would have to be 600,000 (150,000 years on the Earth / 0.25 years on the spaceship), but it would mean that the mass of the spaceship would be 600,000 times its normal mass. No way I could imagine an engine technology that would succeed to speed up such a “heavy” spaceship. Thankfully in other parts of the book there were no such horrendously high relativistic factors.
By the way: the relativistic factor of 600,000 is “achieved” by the speed of 0.9999999999986111c (on one of the sites given below you can find an online calculator that allows you, among other things, to do such reverse calculations in a very easy way).
4d. The twins paradox is explained by a different HISTORY of acceleration/deceleration.
The twins paradox is an example of a wrong application of the special theory of relativity. Somebody argued that from the point of view of the spaceship it was the Earth that was moving so BOTH twins should age in the same way. Some other people thought that the twins paradox is explained by the sheer acceleration/deceleration PROCESS, but it's not the case either. It's the different HISTORY of accelerations and decelerations that matters. The spaceship HAD to decelerate and accelerate to change the direction (to be able to come back to the starting point) and the Earth did not accelerate at all (except for its journey around the sun, but the solar system can be considered as an inertial system as a whole). Please notice that the accelerations and decelerations for the sake of the example were “almost instant”, but they were enough to determine which twin “was moving”. The scope of the time dilatation depends heavily on the speed of the spaceship and the length of the journey, NOT on the duration of the accelerations or decelerations alone. Below there is a fantastic visualization of the twin paradox (the Minkowski diagram of the twin paradox). The dots may be interpreted as the passing years (6 years for the travelling twin and definitely more years for the stationary twin – there are some dots missing for the stationary twin).

On a side note: I found also a fantastic visualization of an accelerating observer in special relativity:

The precise description of this “spacetime diagram” can be found here:
https://en.wikipedia.org/wiki/Minkowski_diagram

Here are some cool sites with different online calculators:
http://www.1728.org/reltivty.htm
http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/ltrans.html
http://nathangeffen.webfactional.com/spacetravel/spacetravel.php


UPDATE: Example II (“three observers”)

I analysed another example of how the special theory of relativity works. One of the crucial questions is: “What would happen if 2 spaceships (A and B) were going in opposite directions at very high speeds?” I added a special twist and placed the stationery observer (the observer on the Earth) between the spaceships, which makes my example much more interesting. To make the example easier (to makes some of the calculations similar to the previous example) I assumed that each spaceship is 2 light-years away from the Earth (so they are 4 light-years away from each other) and that the speed of each spaceship is 0.9c relative to the Earth:

Example II (“three observers”) – version 1 of the calculations.
Observer: the stationary observer (the observer on the Earth).
Reference point (for ALL the speeds involved): the Earth.
II.1.a. Distance of the journey of each of the spaceships: 2 light years.
II.1.b. Length of each of the spaceship: … (irrelevant value).
II.1.c. Time of the journey: 2 light-years / 0.9c = 2.2222 years (time on the Earth).
II.1.d. Mass of the spaceship: normal mass of the spaceship? (there’s no way to measure the mass of the spaceship from the Earth, but it's irrelevant – anyone outside the spaceship doesn't really care about its mass).

Example II (“three observers”) – version 2 of the calculations (the same calculations are true for each of the spaceships).
Observer: the observer on the spaceship A (one of the moving observers).
Reference point (for ALL the speeds involved): the Earth.
II.2.0.a. Speed relative to the spaceship B: 0.994475c (this speed can actually be measured by the observer on the spaceship A, but it is NOT relative to the Earth).
The moving observer sees not only the Earth, but he also sees the other moving spaceship almost exactly in the same line as he sees the Earth (not exactly in the same line, so it is not obscured by the Earth). According to the special theory of relativity the formula for the speed addition is this:

Please notice that if the speed “u” were 0 then the sum of the speeds (“s”) would be equal to the speed “v”. In our example v=0.9c and u=0.9c, so s=0.994475c. It means that the moving observer would measure this speed relative to the other spaceship. In other words: the moving observer sees the Earth “approaching” at the speed of 0.9c and next to it he sees the other spaceship “approaching” at the speed of 0.994475c.
II.2.0.b. Speed of the other spaceship relative to the Earth: 0.994475c – 0.9c = 0.094475c (the observer on the spaceship sees this speed this way).
II.2.0.c. CALCULATED (real) speed of the other spaceship relative to the Earth: 0.9c.
If the moving observer knows the special theory of relativity he can use the above formula for the speed addition and calculate the third speed “in reverse”. It means that from the formula he would know “v” (his speed relative to the Earth) and “s” (his speed relative to the other spaceship), so he would be able to calculate “u” (the real speed of the other spaceship relative to the Earth) as being equal to 0.9c. He can't observe it directly, but he can calculate it.
II.2.0.d. Relativistic factor for the speed of 0.9c is 2.294 (we know that from the previous example).
II.2.0.e. Relativistic factor for the speed of 0.994475c is 9.526 (you can calculate it using the formula from the previous example).
II.2.a.1. Distance of the journey to the Earth: 2 light years / 2.294 = 0.8718 light-years (we know that distance from the previous example – see the point I.2.a.).
II.2.a.2.a. Distance of the journey to the other spaceship NOT relative to the Earth: 4 light-years / 9.526 = 0.4199 light-years (distance NOT relative to the Earth but to the other spaceship).
II.2.a.2.b. Distance of the journey to the other spaceship relative to the Earth: 0.4199 light-years * 2.294 = 0.9633 light-years (distance relative to the Earth).
II.2.a.3 Distance between the other spaceship and the Earth: 0.9633 light-years – 0.8718 light-years = 0.0915 light-years (distance relative to the Earth).
II.2.b. Length of each of the spaceship: … (irrelevant value).
II.2.c.1. Time of the journey to the Earth: 0.8718 light-years / 0.9c = 0.9687 years (time on the spaceship – we know that time from the previous example – see the point I.2.c.).
II.2.c.2.a. Time of the journey to the other spaceship NOT relative to the Earth: 0.4199 light-years / 0.994475c = 0.4222 years (time on the spaceship NOT relative to the Earth but to the other spaceship).
II.2.c.2.b. Time of the journey to the other spaceship relative to the Earth: 0.4222 years * 2.294 = 0.9685 years (time on the spaceship relative to the Earth).
II.2.c.3. Time of the journey of the other spaceship to the Earth: 0.0915 light-years / 0.094475c = 0.9685 (time on the spaceship relative to the Earth).
All the times of the journeys (II.2.c.1., II.2.c.2.b. and II.2.c.3.) are the same! The values 0.9687 and 0.9685 are not exactly equal because I rounded all the numbers to 3 or 4 decimal places and several steps in calculations caused this minute difference (0.0002). Anyway, it means that each journey (the spaceship A to the Earth, the spaceship A to the spaceship B and the spaceship B to the Earth) will last equally long, so all the observers will meet at the Earth at the same time – exactly as it was calculated by the stationary observer (the observer on the Earth).
II.2.d.1. Mass of the spaceship relative to the Earth: normal mass of the spaceship * 2.294 = 2.294 times the normal mass of the spaceship (more than twice as “heavy”, but still able to accelerate further in a decent way).
II.2.d.2. Mass of the spaceship relative to the other spaceship: normal mass of the spaceship * 9.526 = 9.526 times the normal mass of the spaceship (almost 10 times as “heavy” – significantly more difficult to accelerate).
Please notice that the change of mass depends on the observer AND on the reference point. The mass of the spaceship AT THE SAME TIME/PLACE seems lower (when looking at the Earth) or bigger (when looking at the other spaceship). The mass is relative too! This example “proves” that it's not really a physical change in mass, but it’s more like “it seems that the spaceships mass is higher because we can't accelerate it as well as we did at lower speeds”. The higher the OBSERVED speed the bigger the “FEELING” that the mass of the spaceship is higher.

As you can see from the above calculations you have to be very careful to use the same observer AND the same reference point for ALL the final calculations! It is very easy to mess things up by joining calculations made for different observers OR with different reference points!

Example II (“three observers”) – time comparison.
II.3. One day/month/year on the spaceship was equal to 2.294 days/months/years on the Earth (2.2222 years on the Earth / 0.9687 years on the spaceship).

The calculations can be also made for a different reference point – for example we could assume that the spaceship B is stationary and both the other spaceship A and the Earth were moving (or instead of the Earth there could be just another spaceship). It would mean that the Earth is “running away” from the spaceship A (that the Earth is going in the same direction as the spaceship A). This time the speed of the spaceship A relative to the spaceship B (0.994475c) would be the speed “v” and the speed of the spaceship A relative to the Earth (0.9c) would be the speed “s” (the sum of speeds). The speed “u” (the speed of the Earth relative to the spaceship B) would be equal to -0.9c (the negative number means that the speed is not added, but subtracted from the speed 0.994475c).

Once again: this is exactly the same example, but with the REFERENCE POINT being the spaceship B (not the Earth). Let's calculate things from the POINT OF VIEW of the spaceship A.

Example II (“three observers”) – version 3 of the calculations.
Observer: the observer on the spaceship A.
Reference point (for ALL the speeds involved): the spaceship B.
II.3.0.a. Speed relative to the Earth: 0.9c (this speed can actually be measured by the observer on the spaceship A, but it is NOT relative to the spaceship B).
II.3.0.b. Speed of the Earth relative to the spaceship B: 0.994475c – 0.9c = 0.094475c (the observer on the spaceship A sees this speed this way).
II.3.0.c. CALCULATED (real) speed of the Earth relative to the spaceship B: -0.9c (the “-“ sign determines only the direction of the movement – the same direction as the spaceship A). The observer on the spaceship A can't observe it directly, but he can calculate it.
II.3.0.d. Relativistic factor for the speed of -0.9c is 2.294 (we know that from the example I – the direction of the movement is not important).
II.3.0.e. Relativistic factor for the speed of 0.994475c is 9.526 (we know that from the example II).
II.3.a.1. Distance of the journey to the spaceship B: 4 light-years / 9.526 = 0.4199 light-years (relative to the spaceship B).
II.3.a.2.a. Distance of the journey to the Earth NOT relative to the spaceship B: 2 light-years / 2.294 = 0.8718 light-years (distance NOT relative to the spaceship B but to the Earth).
II.3.a.2.b. Distance of the journey to the Earth relative to the spaceship B: 0.8718 light-years / 2.294 = 0.3800 light-years (distance relative to the spaceship B).
II.3.a.3 Distance between the Earth and the spaceship B: light-years 0.4199 – 0.3800 light-years = 0.0399 light-years (distance relative to the spaceship B).
II.3.b. Length of each of the spaceship: … (irrelevant value).
II.3.c.1. Time of the journey to the spaceship B: 0.4199 light-years / 0.994475c = 0.4222 years (time on the spaceship A).
II.3.c.2.a. Time of the journey to the Earth NOT relative to the spaceship B: 0.8718 light-years / 0.9c = 0.9687 years (time on the spaceship A NOT relative to the spaceship B but to the Earth).
II.3.c.2.b. Time of the journey to the Earth relative to the spaceship B: 0.9687 years / 2.294 = 0.4223 years (time on the spaceship A relative to the spaceship B).
II.3.c.3. Time of the journey of the Earth to the spaceship B: 0.0399 light-years / 0.094475c = 0.4223 (time on Earth relative to the spaceship B).
All the times of the journeys (II.3.c.1., II.3.c.2.b. and II.3.c.3.) are the same! The values 0.4222 and 0.4223 are not exactly equal because I rounded all the numbers to 3 or 4 decimal places and several steps in calculations caused this minute difference (0.0001). Anyway, it means that each journey (the spaceship A to the Earth, the spaceship A to the spaceship B and the Earth to the spaceship B) will last equally long, so all the observers will be at the Earth at the same time – exactly as it was calculated with a different reference point (the Earth in the previous 2 variations of the calculations).
II.3.d.1. Mass of the spaceship relative to the Earth: normal mass of the spaceship * 2.294 = 2.294 times the normal mass of the spaceship (more than twice as “heavy”, but still able to accelerate further in a decent way).
II.3.d.2. Mass of the spaceship relative to the other spaceship: normal mass of the spaceship * 9.526 = 9.526 times the normal mass of the spaceship (almost 10 times as “heavy” – significantly more difficult to accelerate).

Example II (“three observers”) – time comparison #2 (different values for different reference points).
Before we can compare anything we have to calculate time of journey from the point of view of the spaceship B:
II.3.0. Time of the journey of the spaceship A: 4 light-years / 0.994475c = 4.0222 years (time on the spaceship B).
II.3. One day/month/year on the spaceship A was equal to 9.527 days/months/years on the spaceship B (4.0222 years on the spaceship B / 0.4222 years on the spaceship A).

We could do the calculations for the “moving Earth” because in the above example NEITHER observer accelerated, so ANY reference point was equally good. However, if we analysed a “triplets” example (all the observers are triplets – one stays on the Earth and the other 2 triplets depart the Earth in opposite directions, travel at the same speed the same distance and then come back to Earth also travelling at the same speed) then there could be only 1 correct reference point – the Earth. Only the Earth was NOT accelerating or decelerating – the other 2 triplets were accelerating and decelerating to change their direction (to be able to come back to the starting point). The Minkowski diagram of the triplets paradox would look like this:

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