# A Roller Coaster Ride through Relativity/Gravitational Time Dilation

## Gravitational Time Dilation

Suppose you are playing with your red laser pen in the accelerated rocket but instead of shining it across the rocket, you shine it straight towards the front of the rocket. What, if anything, will happen to the light then?

I know the answer to that one! It will just go straight forwards at the speed of light!

Well done! You have obviously learned your lessons well. But something happens, all the same.

What​?

Its colour changes. You see, by the time the light has travelled a distance h up the rocket (relative to you in the rocket, that is), the rocket has increased its speed by δv = ah/c. What this means is that, in effect, the receiver at the front of the rocket is continually moving away from the source with speed ah/c. Now if we restrict ourselves to small heights and small accelerations (ie if ah << c) this produces a Doppler shift equal to:

${\displaystyle \lambda =\lambda _{0}(1+ah/c^{2})}$

Now we know that the effects of gravity are exactly the same as the effects of acceleration so we must conclude that, when light climbs up through a gravitational field, its wavelength increases according to the formula:

${\displaystyle \lambda =\lambda _{0}(1+gh/c^{2})}$

where g is the gravitational field strength (assumed uniform over the distance h)

I am sure you know that the gravitational potential energy of a mass m lifted a height h in a gravitational field of strength g is mgh. You may also know that gravitational potential (${\displaystyle \phi }$) is defined as the gravitational potential energy per unit mass. It follows that when you move up a distance h through a gravitational field g, the change in gravitational potential ${\displaystyle \delta \phi }$ is equal to

${\displaystyle \delta \phi ={mgh \over m}=gh}$

This means that we can write our equation for the Doppler shift more generally like this:

${\displaystyle \lambda =\lambda _{0}(1+\delta \phi /c^{2})}$

The greater the gravitational potential difference which the light has to climb, the greater the Doppler shift which is produced. On the other hand, because of the c2 term, the shift is very small indeed. The gravitational potential at the surface of a star of radius R and mass M is equal to:

${\displaystyle \phi =-{GM \over R}}$

(The minus sign is necessary because the gravitational potential is defined to be zero at an infinite distance from the star. At the surface it is less than zero. For the proof of this formula see Appendix H)

In rising from the star, a beam of light passes up through a change in gravitational potential equal to:

${\displaystyle \delta \phi ={GM \over R}}$

so the formula for the gravitational red shift experienced by a beam of light emitted from the surface of a star of mass M and radius R is:

${\displaystyle \lambda =\lambda _{0}\left(1+{GM \over Rc^{2}}\right)}$

For the Sun, M = 2 x 1030 kg, R = 7 x 108 m so the expression GM/Rc2 computes to 2 x 10−6. A Doppler shift of less than one part in a million is very small and corresponds to a recession speed of only 600 ms−1 - much less than the Doppler shift in wavelength induced by the thermal motion of atoms on the surface of the Sun, which makes the effect difficult to detect. Nevertheless, the effect is real and, in the case of more massive stars, it should not be ignored. What is more important at the moment is the implication this analysis has on the rate of passage of time at the two ends of the rocket.

Suppose that instead of shining a laser pen towards the front of the rocket, you use it to send timing signals to a fellow astronaut situated at the front. If, by your watch, you send timing signals every T0 seconds, these will be Doppler shifted just like the light beam and your friend will receive timing signals every T seconds where:

${\displaystyle T=T_{0}(1+\delta \phi /c^{2})}$

What this means is that the clock at the back of the (accelerating) rocket goes slower than the clock at the front! It also means that a clock at the bottom of a well goes slower than a clock at the top! What is more, this effect has actually been demonstrated! Unbelievable though it may seem, the effect was observed at Havard university in 1960 over a distance of 22.5 m. (if you are wondering why such a strange distance it is because that is the height of the stair well at the Jefferson Physical Laboratory!) A quick calculation shows that the Doppler shift over this distance is equal to 2.5 x 10−15. What the Havard scientists, Pound and Rebka, did was to build a gamma ray source and a gamma ray detector so finely tuned that, if the frequency of the source changed by only this amount, the detector would reject the signal. The source and the detector were placed together at the bottom of the shaft and matched exactly. Then the detector was taken upstairs and, as expected, the signal was rejected. To confirm the effect absolutely, the source was raised at a steady speed towards the detector. The blue doppler shift caused by the speed was just sufficient to counteract the red shift caused by the gravity field and the detector sprang into life again. What was this speed? 2 x 10−15 times the speed of light, of course. You want me to calculate it for you? All right. It works out to be 2.7 millimetres per hour!

The effect is also measurable using atomic clocks in a high flying aircraft. You may recall that the velocity time dilation effect of a jet plane flying at 200 ms−1 for 10 hours was 8 ns slow. Now suppose that the plane was flying at a height of 10,000 m during this time. The change in gravitational potential between this height and the ground is approximately gh = 10 x 10,000 = 100,000 J kg−1 so the clock in the plane will run faster than the clock on the ground by a factor of gh/c2. For a period of 10 hours, this works out to be 40 ns, so the effect of gravitational time dilation in this case is actually larger, and in the opposite sense, to the effect of velocity time dilation.

The derivation of the gravitational time dilation formula presented so far has two disadvantages. Firstly, I have used an approximate formula for the Doppler shift only applicable for small values of h and a. More seriously though, it ignores the second order effects of special relativity due to the fact that the increase in speed of the rocket during the time t (whose time t anyway?) is not exactly equal to at. In order to generate a formula which is applicable to situations in which ah is large, we need to look at another way of producing artificial gravity.

Science fiction writers are fond of describing large futuristic space stations as a large spoked wheel, spinning slowly on its axis.

A centrifugal space station

As the station spins, astronauts at the periphery subjectively feel an outward force, popularly (and correctly) known as the centrifugal force which feels to them just like gravity. From the point of view of an observer in a shuttle craft waiting to dock in the centre of the station, we can see that what is in fact happening is that the metal structure of the station is exerting an inward force on the astronauts which is giving them an inwards centripetal acceleration causing them to move in a circle.

Correction. I am not allowed to say that that is what is in fact happening. General Relativity tells us that the astronauts point of view is just as valid as ours. So what does the universe really look like from the point of view of the astronauts on board the space station? To them, the station is, of course, stationary. But the station is generating a rather strange form of gravity which is zero at the centre of the station and which increases steadily as you move towards the rim. Moreover, the force of gravity is outwards, not inwards.

Now you know that the centripetal acceleration of a mass m rotating round in a circle of radius r with angular speed ω is equal to 2 so the strength of gravity at the rim will be: g = 2 (where R is the radius of the station) and the gravitational potential difference between the centre and the rim will be given by:

${\displaystyle \Delta \phi =\int _{0}^{R}r\omega ^{2}\,dr={\tfrac {1}{2}}R^{2}\omega ^{2}}$

What about clocks placed at the centre and at the rim? From our point of view outside the station, we can see that the clock at the rim is going more slowly than the clock at the centre because the clock at the rim is moving while the clock at the centre is not. In fact we can write down the relation between the time as measured by the centre clock Tcentre in terms of the time measured by the clock at the rim Trim as follows using the special relativity time dilation formula:

${\displaystyle T_{centre}={1 \over {\sqrt {1-v^{2}/c^{2}}}}\,T_{rim}={1 \over {\sqrt {1-R^{2}\omega ^{2}/c^{2}}}}\,T_{rim}}$

More generally, by putting ${\displaystyle R^{2}\omega ^{2}=2\Delta \phi }$ we can write down the general formula for the relativistic time dilation between two points where a gravitational potential difference exists as follows:

${\displaystyle T={1 \over {\sqrt {1-2\Delta \phi /c^{2}}}}\,T_{0}}$

(If you compare this formula with the similar formula we derived earlier on this page you will see that the latter is simply the binomial expansion of this one approximated to the second term.)

Now the astronauts on board the space station will observe exactly the same time dilation effect as we do - it is just that they will explain it differently.

(It is very important to note that, like velocity time dilation, gravitational time dilation is a relative phenomena which depends not on the local strength of gravity, but on the difference in gravitational potential between two points. At the very centre of the Earth, for example, there is no gravity - but clocks will still run more slowly there than they do on the surface.)

So much for clocks - what about rulers? This is where things start to get difficult. Suppose that, from our stationary birds eye view in the shuttle, we watch one of the astronauts measuring out the radius of the space station with a metre ruler. Since the ruler is always at right angles to its direction of motion, we can see that it remains the same length and his measurement of the radius agrees with ours, namely R.

But what happens when he measures the circumference? We see that his ruler is contracted because of the speed and so, instead of getting the expected value of 2πR, he ends up with an answer that is bigger than this! What is he to make of this? Does gravity alter the fundamental mathematical constants like π? No, I think that is going too far. What we can say is that gravity distorts the structure of space in such a way that, in a non-uniform gravitational field, the circumference of a circle is no longer equal to 2π times the radius. In fact this is such an important conclusion that it deserves a box of its own.

Bizarre consequence number 19
In a non-uniform gravitational field, space is non-Euclidean; the circumference of a unit circle is not equal to 2π and the angles of a triangle no longer add up to 180º.

So what happens to rulers in a gravitational field? Do rulers shrink or what?

Well I find it difficult to give a straight answer to that question. It is true that if I measure the radius and the circumference of a large star with a metre ruler (!) I will find that the circumference is less than 2π times the radius. (The reason why it is less not more is that real gravity is an attractive force, unlike the centrifugal effect inside the space station which is an outward force.) This would appear to suggest that as you take the ruler down to deeper and deeper depths inside the star - ie to lower and lower gravitational potentials - the ruler shrinks. And because gravitational potential is a scalar quantity, the ruler has to shrink in all directions, not just parallel or perpendicular to the local gravitational field. However, if a ruler shrinks in all directions, what sense does it make to say that it shrinks at all? The truth is that the effect of gravity is far more subtle than just a question of shrinking rulers. No, gravity actually distorts the space in which the rulers exist. But to go any further along that road requires the use of mathematical concepts far beyond my competence.

## Oblivion

Wow! That was a bit weird you shout as the roller coaster shoots up to the top of the next hill. Where do we go now? you ask.

I don't have time to tell you about the last bizarre consequence because the roller coaster track ahead of us goes horizontally for a few meters and then - just stops!

As the train reaches the end of the track it tilts over 90° and, for an instant we are suspended over the most terrifying sight. Below us is a deep, interminable shaft. Along the sides of the shaft are electric lights at regular intervals but as we look down the shaft, the lights in the distance get redder and redder until they fade from view completely. At the bottom of the shaft - no, there is no bottom - there is just an aching blackness.

Suddenly we are in free fall; the lights are flashing past us faster and faster; as we look up, the shining blue sky which we left behind is turning first violet, then ultra violet, and now begins to bathe us in X-rays and even g-rays; and then the aching blackness which we saw from above appears to grow in size until it is all around us, even above us. The bright patch of gamma-rays which was our home on Earth is vanishing fast and suddenly, it is gone. We are completely alone in the darkness with nothing to tell us where we are or how fast we are travelling except that we begin to experience a very unpleasant sensation of being stretched and squashed at the same time. Rapidly the stretching and squashing increases to the point where our bodies are dismembered and pulled into long thin strands, but we are past pain - we have passed into oblivion . . .