A Roller Coaster Ride through Relativity/Time Dilation

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Time Dilation[edit]

The simplest possible clock is a beam of light bouncing backwards and forwards between two parallel mirrors. If the mirrors are separated by a distance l, the time for each 'tick' (i.e. there and back) is t = 2l / c (Of course you remember that speed = distance over time and that time is distance over speed) It would be perfectly possible for an electronic circuit to count the number of ticks and provide a suitable display in hours, minutes and seconds. (I do not know if anyone has ever made such a clock. If the arm of the clock was 1 metre long, it would ‘tick’ at a rate of 6.7 GHz. This is not a lot faster than the clock speed of your average PC.)

Now suppose you are watching such a light clock passing by in a spaceship travelling at a sizeable fraction of the speed of light v. (Of course you can't see the light beam but you can imagine it!) The arm of the clock is at right angles to the direction of motion of the ship so you have no difficulty in verifying that the arm has a length l as the ship goes by. On the other hand, you can see (imagine!) the light beam moving in a diagonal line like this:

A simple light clock

Now because of the Fundamental Principle, you see the light beam moving at the normal speed of light. This means that it is going to take longer for it to travel the distance l and back again just like Albert on the river (albeit for a slightly different reason). In fact the effective velocity across the ship is \sqrt{c^2-v^2} and the time taken for the return trip is t={2l\over\sqrt{c^2-v^2}}

This means that the clocks on board the ship (and everything else as well) appear to me to run slow by a factor ofc\over\sqrt{c^2-v^2} or, as it is more usually written1\over\sqrt{1-v^2/c^2}. This factor is often called \gamma (the greek letter gamma) and is always greater than one. It rises to \infin as v gets closer and closer towards c.

We can summarise what we have derived so far as follows. If it takes T0 seconds to boil an egg, it will appear to me that the eggs in the spaceship take T seconds where:

T = \gamma T_0 = {1\over\sqrt{1-v^2/c^2}} T_0

Let's put some figures into the formula and see what we get:

v (as a % of c) \gamma
50 1.15
60 1.25
70 1.40
80 1.67
90 2.29
95 3.20
96 3.57
97 4.11
98 5.03
99 7.09

If we plot a graph of these figures we get this:

Graph of the relativistic gamma function

What this means is that if you travel at 50% of the speed of light, your clocks go 15% slower than clocks which are stationary. Travel at 80% of the speed of light and your clocks are 67% slower (ie at 3/5 of the rate of a stationary clock). Travel at 99% of the speed of light and every second of your time takes over 7 seconds of 'stationary clock' time!

That’s just ridiculous! You say. I mean, what would it be like to live in a world in which all the clocks went 7 times slower than normal? It would be a crazy world! Everything would appear to be slowed down! Cars would crawl along at 10 miles per hour. A game of football would take all day! A falling stone would appear to fall like a feather!

Hang on a minute, you are not thinking quite straight. If the clocks go 7 times slower, ...

Oh! I see what you mean: you interrupt. If the clocks go slower, stones will actually take 7 times fewer seconds to fall so everything will actually look as if it is happening faster! Is that it?

No, no, that's not right either! The point is that it is not just clocks which go slow - everything goes slow, you might say that time itself goes slow. When you watch a falling stone and time it with a stop watch, the stone falls slowly, the clock ticks slowly and your own mental processes think slowly as well. When the stone reaches the ground, the clock reads exactly what you expect it to read and the process seems to have taken exactly the same time as usual. In short, the whole process looks perfectly normal to you. In fact it doesn't just look perfectly normal to you, it is perfectly normal to you. You must remember that time is only dilated (ie stretched) from the point of view of someone else who is moving with respect to you. In your moving spaceship, everything looks normal to you. It is only me, on stationary Earth who sees your clocks going slow and your stones falling like feathers and your eggs which take ages to boil!

After a moment's thought you exclaim: That can't be right! and what's more, I can prove it!

Go on then.

Well you say that the clocks in the space ship appear to go slow because the spaceship is moving and you are stationary. But from my point of view in the spaceship, it would appear that it was the clocks on Earth which were going slow because, as you said yourself, all velocities are relative and you can't tell who is actually moving.

Bravo! You really are beginning to think like a relativist!

What do you mean, Bravo!? Haven't I just disproved your theory?

Well no, you haven't. What is wrong with both clocks appearing to go slow?

Surely that's obvious. All you have got to do is put the clocks side by side and see which one is going slow!

But how are you going to do that?

Well, as I go past you we will synchronise our clocks, and then a short while later we will see which clock has gone slow.

But by that time you will be hundreds of miles away.

Well, I'll send you a radio message with a time signal

But we will disagree about how long it takes the radio signal to get back to me.

Well - I will have to stop the spaceship and turn it round and..

Aha! You are going to STOP the spaceship! As soon as you do that, the motion of the ship stops being uniform and we must examine very carefully what happens when the motion of the clock changes. The slowing down of the ship introduces an asymmetry into the situation which means that when you bring the clocks back together, the clock that was in the spaceship will indeed be found to be slow compared to the clock which remained back on Earth.

Are you sure?

Sure I'm sure. Believe it or not, this experiment has actually been done using extremely accurate atomic clocks and the results confirm Einstein's theories exactly. Mind you, the effect is incredibly small. Suppose you synchronise two clocks on the ground and then fly one of them round in a jet plane at a speed of 200 ms-1 for say 10 hours. What will the expected time difference due to special relativity be? (I must say special relativity, because in the real experiment the effects of general relativity have to be taken into account as well.) The first thing we have got to do is to work out γ for a speed of 200 ms-1. If you try doing this on a calculator you will run into a problem. The velocity of light is very large. It is in fact 300,000 km every second or 3 x 108 ms-1. This means that and if you try subtracting this from 1 an ordinary calculator will just give you an answer of 1 because the number is far too small and the calculator does not have enough digits.

If you know something about indices you will know that the formula

\gamma = 1 \over\sqrt{1-v^2/c^2}

can also be written as

\gamma=(1-v^2/c^2)^{-1/2}

Now there is an exceedingly useful little theorem called the Binomial Theorem which says that when x is much less than one

(1+x)^n=(1+nx)

If we apply this theorem to our formula, we arrive at a much simpler expression for \gamma (but remember this is only true when v is a lot smaller than c)

If you try calculating \gamma using this formula with a calculator you will still run into difficulty when you add the 1 but you should still be able to confirm that for v = 200 ms-1 \gamma = 1.00000000000022

What this means is that for every second as measured by the moving clock takes 1.00000000000022 s as measured by the stationary clock. In 10 hours, the difference amounts to 10 x 60 x 60 x 0.00000000000022s which equals 0.000000008 s or 8 ns easily within the capabilities of an atomic clock.

Note that, because the moving clock runs slower than the stationary clock, when the two clocks are compared after the trip, the moving clock will read less than the stationary clock by a factor of \gamma.

Does this mean that if I travel to a distant star and back, everybody will be older than me when I return?

You bet it does. Suppose in some future century you choose to visit Alpha Centauri, 4 light years away, travelling in a Thomson Astro-cruiser at 80% of the speed of light. To your twin brother left behind on Earth the journey will take (4 / 0.8 =) 5 years out and 5 years back - ie 10 years in all. From his point of view though (and his point of view is more special than yours because it is he who remains 'stationary' all the time) your clocks run 1.67 times more slowly, so you age by only 10/1.67 = 6 years.

So how long does the journey actually take? 6 years or 10?

Both. It takes 6 of your years and 10 of his! You can't really say which time it actually takes. Both points of view are equally valid. On the other hand, every pair of events has what is known as a proper time interval between them, this being the time as measured by a clock which travels between the two events at a constant speed (or as in the case of two events which occur at the same place, is stationary).

The journey under consideration involves three events:

Event A : departure from Earth

Event B : turn-around at Alpha Centauri

Event C : return to Earth

The proper time interval between A and B is the time as measured by your clock ie 3 years. Likewise the proper time interval between B and C is also 3 years. But the proper time interval between A and C is not 6 years, it is 10 years - the time as measured by your stay-at-home twin. You can see that proper times do not necessarily add up. You could if you like define journey time to be the amount of time the voyager experiences, which is the sum of all the proper times for each section of the journey. In this case the journey time is 6 years and the proper time is 10 years, but by moving around fast enough you can make the journey time as small as you like. If you went to Alpha Centauri and back at 99% of the speed of light you could do the trip in just over a year (while your twin brother aged about 8 years); at 99.9% of the speed of light the trip would take less than 5 months. Here are some more examples:

speed

(% of c)

proper time

=distance / speed

journey time

=proper time / γ

50% 16.0 years 14 years
80% 10.0 years 6 years
90% 8.9 years 4 years
99% 8.1 years 1 year
99.9% 8.0 years 131 days
99.99% 8.0 years 41 days
99.999% 8.0 years 13 days

Wow! Could you really get to Alpha Centauri and back in a fortnight's holiday? That's really cool!

Well, you could spend two weeks of your time getting there and back, but your boss back on Earth would be hopping mad, since he will have had to wait eight years for you to return.

Also, the formula we have been using assumes you can accelerate a rocket all the way to 99.999% of the speed of light in no time at all. If you did that, everybody inside would be reduced to jelly! But it might be possible to accelerate a rocket at a more moderate acceleration for a long period of time and gradually build up a large enough speed. Let's look at this possibility.

The 1g rocket problem[edit]

Suppose we construct a rocket which can accelerate with a continuous acceleration of 1 g (10 ms-2). We set out from Earth, accelerating at 1 g until we are half way to the star we want to visit; turn round and decelerate at the same rate; have a look at the star for a while; accelerate back again for half the journey; turn round and decelerate all the way home. (Life on board a rocket like this would be just like life on Earth, because the acceleration would give the effect of artificial gravity.)

The formula for the time dilation effect in such an accelerated system is this:

{aT \over c} = \sinh \left({aT' \over c} \right)

where T' is the proper time of the journey (ie the time as experienced by the space traveller) and T is the (longer) time experienced by those who stay at home; a is the acceleration of the ship and c is of course the velocity of light.

If we work in units of years and light years, the velocity of light c is, of course, 1 light-year per year.

By an extraordinary coincidence, the acceleration due to gravity at the Earth's surface g (10 ms-2) works out to be almost exactly 1 light-year per year2. Check out the working in the box below.

1 year = 365 × 60 × 60 = 3.15 ×107 s
Light travels at 3.00 × 108 ms-1 so
1 light-year (ly) = 9.46 × 1015 m
The speed of light is, of course, 1 ly/year (ly y-1)
The acceleration due to gravity at the Earth's surface is 9.8 ms-2
This is equal to 9.8 × (3,15 × 107)2 /9,46 × 1015
= 1.03 light years per square year (ly y-2)

So putting a = g = 1 ly y-2 and c = 1 ly y-1 our formula reduces to just

T = \sinh(T')

where T (and T’) is in years.

(For the proof of this and other interesting formulae connected with 1 g accelerated rockets see the Appendix A at the end of the book.)

When working out the time dilation effect we must remember that we must do the calculation for each quarter of the journey separately.

Here is a table which tells you how many years will pass on Earth during voyages of different lengths. Owing to the exponential nature of the sinh function, the figures rocket up dramatically and you can see that in principle it would be possible for a human being living now to return to Earth within 60 years to see what it will be like in 6½ million years time!

total time on ship (y) ¼ time on ship (y) ¼ time at home (y) total time at home (y)
6 1.5 2.4 9.4
10 2.5 6 24
20 4 74 297
40 10 11,000 44.000
60 15 1,600,000 6500000

That’s incredible! Could you really travel into the future just by building a fast enough rocket?

Unfortunately, two practical difficulties stand in the way of this dream. Firstly, it will be necessary to build a rocket motor that can sustain an acceleration of 1g for many decades. Sadly, no known or even theoretically possible propulsion system comes anywhere near this requirement. Secondly, a spacecraft travelling at nearly the velocity of light would probably be destroyed by all the microscopic interstellar dust particles slamming into it at nearly the speed of light. As we shall see later, the kinetic energy of a particle the size of a grain of sand travelling at a speed at which \gamma = 1,000,000 is equal to that of a 6 megaton bomb!

I see what you mean. In any case – I still don’t really believe it

Well, you are not alone alone. For many years in the 1950's, respected scientists were still discussing the famous Twins Paradox and even now the internet is littered with postings purporting to show that the effect is neither possible nor logical. Believe me, though. It is.

At the top of the second hill[edit]

The roller coaster has now nearly reached the summit of the second hill and as it slows down momentarily you get a chance to have a quick look round. Over on the right the hands of the big clock appear to be moving normally again but somehow, in the minute it has taken you to negotiate the first big dip, the clock, which appeared to you to run slow when we started the descent, has inexplicably skipped a couple of minutes ahead.

That's funny. I thought that clock was going slow. How come my watch seems to have lost a few seconds?

Yes, that is curious isn't it? It is another example of the twin's paradox. While we were careering down the slope, it seemed to us that everyone else's clocks were going slow. But of course, to everyone else, it was our clocks which were going slow. The difference only became apparent when we slowed down to a crawl again.

Tell me again – why is it that it is our clocks which ended up being slower than theirs and not the other way around?

Because it was we whose speed changed. An object which remains at rest or travels at uniform speed is in what is called an inertial frame. But our frame of reference was not inertial because we accelerated down the hill and slowed down as we went up again. The people who stay in an inertial frame always turn out to be older then the people who accelerate and decelerate.

Yes - I think I get it.

Have a look at this.

Looking down you can see one of the children playing shove-halfpenny give his coin a tremendous whack. All the other children's coins were too big to go down the crack but there is something different about this one. It doesn't seem round, it seems oval and, incredibly, it drops right through the crack!

How did that happen? I hear you cry. The answer is that:

Bizarre consequence number 2
Moving objects shrink along their direction of motion.

Hold on to that Principle! I shout. You are going to need it again! The roller coaster tips violently forward and we are plunging downwards again...

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