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## Do GPS satellites move in the sky?

Following on from my blog “Is Tim Peake getting younger or older?” , a bit of fun to work out whether time was passing more slowly or more quickly for Tim Peake in the ISS than it is for us on the ground, it got me thinking about the global positioning system (GPS) that so many of us use on a daily basis. Whether it is using a SATNAV in our car, or a GPS-enabled watch to measure how far and fast we have run, or using maps on a smartphone, GPS must be one of the most-used satellite developments of the last few decades.

As I blogged about here, communication satellites need to be at a particular height above the Earth’s surface so that they orbit the Earth in the same time that it takes the Earth to rotate. In addition to their altitude, communication satellites can only orbit the Earth about the equator, no other orientation will allow the satellite to hover in the same place relative to a location on Earth.

But what about the satellites used in GPS? What kind of an orbit are they in?

## The GPS satellites’ orbits

It turns out that the GPS satellites are not in a geo-stationary orbit, but are in fact in an orbit which leads to their orbiting the Earth exactly twice in each sidereal day (for a definition of sidereal day see my blog here).

The GPS system consists of 31 satellites in orbit around the Earth

We can work out what radius from the Earth’s centre this needs to be by remembering that the speed of orbit is given by

$v = \sqrt { \frac{ GM }{ r } } \text{ (1) }$
where $v$ is the speed of orbit, $G$ is the universal gravitational constant, $M$ is the mass of the Earth and $r$ is the radius of orbit from the centre of the Earth (not from its surface).

A sideral day is 23 hours and 56 minutes, which in seconds is $8.6160 \times 10^{4}$ seconds. So, half a sidereal day is $4.308 \times 10^{4}$ seconds. We will call this the period $T$. The speed of orbit, $v$ is related to the period via the equation
$v = \frac{ 2 \pi r }{ T }$
where $r$ is the radius of the orbit, the same $r$ as in equation (1), and $2 \pi r$ is just the circumference of a circle. So, squaring Equation (1), we can write
$v^{2} = \frac{ GM }{ r } = \left( \frac{ 2 \pi r }{ T } \right)^{2}$
So, in terms of $r$ we can write
$r^{3} = \frac{ G M T^{2} }{ 4 \pi^{2} } \rightarrow r = \sqrt[3]{ \frac{ G M T^{2} }{ 4 \pi^{2} } }, \; \text{ so } r = 26.555 \times 10^{6} \text{ m}$
In terms of height above the Earth’s surface, we need to subtract off the radius of the Earth, so the altitude, which I will call $a_{gps}$, is going to be
$a_{gps} = 26.555 \times 10^{6} - 6.371 \times 10^{6} = 20.184 \times 10^{6} \text{ m } \text{ or } \boxed{ 20.2 \text{ thousand kilometeres} }$

## Why are GPS satellites in this kind of an orbit?

As I didn’t know what kind of an orbit GPS satellites were in before I wrote this blog, the next obvious question is – why are they in an orbit which is exactly half a sidereal day? It is clearly not coincidental! To answer this question, we need to first of all discuss how GPS works.

GPS locates your position by measuring the time a signal takes to get to your GPS device from at least four satellites. Your device can identify from which satellites it gets a signal, and the system knows precisely the position of these satellites. By measuring the time the signals take to you reach you from each of the satellites, it is able to calculate how far each one is from you, and then by using triangulation it can work our your location. There are currently 31 satellites in the system, so often there are more than four visible to your GPS device. The current 31 satellites have all been launched since 1997, the original suite of 38 satellites launched between 1978 and 1997 are no longer in operation.

As I mentioned in my blog about geostationary satellites, a satellite in a geostationary orbit can only orbit above the Earth’s equator. This would clearly be no good for a GPS system, as all the satellites would lie to the south of someone in e.g. Europe or North America. As I said above, there are currently 31 operational satellites; the 31 are divided into 6 orbital planes. If there were 30 satellites this would be 5 in each orbit. The orbits are inclined at $55^{\circ}$ to the Earth’s equator. Each orbit is separated from the other one by 4 hours (equivalent to $60^{\circ}$) in longitude.

As one can see approximately 6 hours in right ascension to both the east and west of one’s location, this means that there will be at least 3 of the orbits above the horizon, and sometimes more. If there were 5 satellites in each orbit this would mean that each one would pass a particular latitude 4 hours before the next one. So, at any particular time there should be some satellites further north than one’s location and some further south, as well as some further east and some further west. This configuration allows for the necessary triangulation to obtain one’s location.

The orbits are inclined at $55^{\circ}$ to the equator and separated by 4 hours (equivalent to $60^{\circ}$) in right ascension, as this diagram attempts to show

## Is the time-dilation effect due to SR or GR more important for these satellites?

We already showed in this blog that, for the International Space Station, the time-dilation due to Special Relativity (SR) has a greater effect on the passage of time than the time-dilation due to General Relativity (GR). What about for the GPS satellites?
The speed of orbit for the GPS satellites at a radius of $26.555 \times 10^{6}$ from the Earth’s centre is, using Equation (1),
$v = \sqrt{ \frac{ GM }{ r } } = 3.873 \times 10^{3} \text{ m/s}$
As we showed in my blog about Tim Peake, the speed of someone on the Earth’s surface relative to the centre of the Earth is $v_{se} = 463.35 \text{ m/s}$, so the relative speed between a GPS satellite and someone on the Earth’s surface is given by
$v = 3.873 \times 10^{3} - 463.35 = 3.410 \times 10^{3} \text{ m/s}$
Compare this to the value for the ISS, which was $7.4437 \times 10^{3}$, it is less than half the speed.

This value of $v$ leads to a time dilation factor $\gamma$ in SR of
$\gamma = \frac{ 1 }{ \sqrt{ 0.9999999999} } \approx 1$
which means that the time dilation due to SR is negligible.
The time dilation due to GR is given by (see my blog here on how to calculate this)
$\left( 1 - \frac{ gh }{ c^{2} } \right) = (1 - 2.2 \times 10{-9}) = 0.9999999978$, or 22 parts in $10^{10}$. Compare this to the ISS, where it was about 1 part in $10^{11}$. Clearly the GR effect for GPS satellites is greater, by about a factor of 5, than it was for the ISS. But, conversely, the SR time-dilation effect has become negligible.

To conclude, the time dilation for GPS satellites is nearly entirely due to General Relativity, and not due to Special Relativity. Time is passing more quickly for the clocks on the GPS satellites than it is for us on Earth, the converse of what we found for the ISS, which is in a much lower orbit.

Because the timings required for GPS to work are so precise, the time dilation effect due to GR needs to be taken into account, and is one of the best pieces of evidence we have that time dilation in GR actually does happen.

## Is Tim Peake getting younger or older?

As anyone who hasn’t been living under a rock knows, the International Space Station (ISS) orbits the Earth with (typically? always?) six astronauts on board. It has been doing this for something like the last fifteen years. One of the astronauts currently on board is the Disunited Kingdom’s first Government-funded astronaut, Tim Peake.

The first British person to go into space was Helen Sharman, but she went into space in a privately funded arrangement with the Russian Space Programme in 1991. Other British-born astronauts have gone into space through having become naturalised Americans, and going into space with NASA. But, Tim Peake has gone to the ISS as part of ESA’s space programme, and his place is due to Britain’s contribution to ESA’s astronaut programme. So, he is the first UK Government-funded astronaut, which is why there has been so much fuss about it in these lands.

Official NASA portrait of British astronaut Timothy Peake. Photo Date: August 28, 2013.

Anyway, I digress. This blog is not about Tim Peake per se, or about the ISS really. I wanted this blog to be about whether Tim Peake is getting older or younger whilst in orbit. Of course everyone is getting older, including Tim Peake, as ‘time waits for no man’ as the saying goes. What I really mean is whether time is passing more or less quickly for Tim Peake (and the other astronauts) in the ISS compared to those of us on Earth.

As some of you might now, when an astronaut is in orbit he is in a weaker gravitational field, as the Earth’s gravitational field drops off with distance (actually as the square of the distance) from the centre of the Earth. Time will therefore pass more quickly for Tim Peake than for someone on the Earth’s surface due to this effect. This is time dilation due to gravity, a general relativity (GR) effect.

But, there is also another time dilation, the time dilation due to one’s motion relative to another observer, the time dilation in special relativity (SR). Because Tim Peake is in orbit, and hence moving relative to someone on the surface of the Earth, this means that time will appear to move more slowly for him as observed by someone on Earth. Interestingly (at least for me!), the SR effect works in the opposite sense to the GR effect.

Which effect is greater? And, how big is the effect?

## Time dilation due to SR – slowing it down for Tim Peake

As I showed in this blog, the time dilation due to SR can be calculated using the equation

$t^{\prime} = \gamma t \text{ where } \gamma = \frac{ 1 }{ \sqrt{ ( 1 -v^{2}/c^{2} )} }$

If he is in orbit at an altitude of 500km (I guessed at this amount, according to Wikipedia it is 400km, but it does not alter the argument which ensues) then his distance from the centre of the Earth (assuming a spherical Earth) is $6.371 \times 10^{6} + 500 = 6.3715 \times 10^{6}$ metres. The centripetal force keeping him in orbit is provided by the force of gravity, and in this blog I showed that the centripetal force $F_{c}$ is given by

$F_{c} = \frac{mv^{2} }{r}$

where $m$ is the mass of the object in orbit, $v$ is its velocity and $r$ is the radius of its orbit.This centripetal force is being provided by gravity, which we know is

$F_{g} = \frac{ GMm }{ r^{2} }$

where $G$ is the universal gravitational constant, and $M$ is the mass of the Earth. Putting these two equal to each other

$\frac{ mv^{2} }{ r } = \frac{ GMm }{ r^{2} } \rightarrow v^{2} = \frac{GM}{r}$

Putting in the values we have for the ISS, where $r=6.3715 \times 10^{6}, G=6.67 \times 10^{-11}$ and $M= 5.97237 \times 10^{24}$, we find that

$v^{2} = 6.2522 \times 10^{7} \rightarrow v = 7.907(067129) \times 10^{3} \text { m/s} = \boxed{ 7.907(067129) \text{ km/s} }$

But, this is the motion relative to the centre of the Earth. People on the surface of the Earth are also moving about the centre, as the Earth is spinning on its axis. But, we cannot calculate this speed as we have done above; people on the surface are not in orbit, but on the Earth’s surface. For something to stay e.g. 1 metre above the Earth’s surface in orbit it would have to move considerably quicker than the rotation rate of the Earth.

The Earth turns once every 24 hours, so for someone on the equator they are moving at

$v_{se} = \frac{ 2 \pi \times 6.3715 \times 10^{6} }{ 24 \times 3600 } = 463.348(5554) \text { m/s}$

where $v_{se}$ refers to the speed of someone on the surface of the Earth. Someone at other latitudes is moving less quickly, at the poles they are not moving at all relative to the centre of the Earth. The speed of someone on the surface will go as $v_{se} \cos (\theta)$ where $\theta$ is the latitude. This is why we launch satellites as close to the Earth’s equator as is feasible; we maximise $v_{se}$ and thus get the benefit of the speed of rotation of the Earth at the launch site to boost the rocket’s speed in an easterly direction.

The difference in speeds between the ISS and someone at the equator on the surface of Earth is therefore

$7.907(067129) \times 10^{3} - 463.348(5554) = \boxed { 7.443(718574) \times 10^{3} \text { m/s} }$

Referring back to my blog on time dilation in special relativity that I mentioned at the start of this section, this means that the time dilation factor $\gamma$, using this value of $v$, is

$\gamma = \frac{ 1 }{ \sqrt{(1 - (v/c)^{2})} } = \frac{ 1 }{ 0.9999999997 }$
(where $c$ is, of course, the speed of light).
This value of $\gamma$ is equal to unity to 3 parts in $10^{10}$, so it would require Tim Peake to orbit for about $3 \times 10^{9}$ seconds for the time dilation factor to amount to 1 second. $3 \times 10^{9}$ seconds is just over 96 years, let us say 100 years.

## The time dilation due to GR – speeding it up for Tim Peake

For GR, the time dilation works in the other sense, it will run more slowly for those of us on the Earth’s surface; we experience gravitational time dilation which is greater than that experienced by Tim Peake. In this blog here, I derived from the principle of equivalence the time dilation due to GR, and found

$\Delta T_{B} = \Delta T_{A} \left( 1 - \frac{ gh }{ c^{2} } \right)$

where, in this case, $\Delta T_{B}$ would be the rate of time passing on the Earth’s surface, $\Delta T_{A}$ the rate of time passing on the ISS,  $g = 9.81$ (the acceleration due to gravity at the Earth’s surface) and $h$ is the height of the orbit, which we have assumed (see above) to be 500 km = $500 \times 10^{3}$.

Plugging in these values we get that

$\frac{ \Delta T_{B} }{ \Delta T_{A} } = 1 - 5.45 \times 10^{-11}$

So, the GR effect is about one part in $10^{11}$ (100 billion). In six months, the number of seconds that Tim Peake will be in orbit is about $1.6 \times 10^{7}$ seconds, so a factor of about 10,000 less than for the GR effect to amount to 1 second. Tim Peake would need to be in orbit for about 5,000 years for the GR effect to amount to 1 second of difference!

## Conclusions

In conclusion, the SR effect on how quickly time is passing for Tim Peake is about 3 parts in 10 billion, in the sense that it passes more slowly for Tim Peake. The GR effect is even smaller, about one part in 100 billion, but in the sense that time is passing more quickly for him. The SR ‘time slowing down’ effect is greater than the GR time ‘passing more quickly effect’, by roughly a factor of 300.

Tim Peake is therefore actually ageing more slowly by being in orbit than if he were on Earth. But, he would need to orbit for nearly 100 years for this difference to amount to just 1 second! And, none of this of course takes into account the detrimental biological effects of being in orbit, which are probably not good to anyone’s longevity!

## Einstein’s general relativity centenary

There has been quite a bit of mention in the media this last week or so that it is 100 years since Albert Einstein published his ground-breaking theory of gravity – the general theory of relativity. Yet, there seems to be some confusion as to when this theory was first published, in some places you will see 1915, in others 1916. So, I thought I would try and clear up this confusion by explaining why both dates appear.

Albert Einstein in Berlin circa 1915/16 when his General Theory of Relativity was first published

## From equivalence to the field equations

Everyone knew that Einstein was working on a new theory of gravity. As I blogged about here, he had his insight into the equivalence between acceleration and gravity in 1907, and ever since then he had been developing his ideas to create a new theory of gravity.

He had come up with his principle of equivalence when he was asked in the autumn of 1907 to write a review article of his special theory of relativity (his 1905 theory) for Jahrbuch der Radioaktivitätthe (the Yearbook of Electronics and Radioactivity). That paper appeared in 1908 as Relativitätsprinzip und die aus demselben gezogenen Folgerungen (On the Relativity Principle and the Conclusions Drawn from It) (Jahrbuch der Radioaktivität, 4, 411–462).

In 1908 he got his first academic appointment, and did not return to thinking about a generalisation of special relativity until 1911. In 1911 he published a paper Einfluss der Schwerkraft auf die Ausbreitung des Lichtes (On the Influence of Gravitation on the Propagation of Light) (Annalen der Physik (ser. 4), 35, 898–908), in which he calculated for the first time the deflection of light produced by massive bodies. But, he also realised that, to properly develop his ideas of a new theory of gravity, he would need to learn some mathematics which was new to him. In 1912, he moved to Zurich to work at the ETH, his alma mater. He asked his friend Marcel Grossmann to help him learn this new mathematics, saying “You’ve got to help me or I’ll go crazy.”

Grossmann gave Einstein a book on non-Euclidean geometry. Euclidean geometry, the geometry of flat surfaces, is the geometry we learn in school. The geometry of curved surfaces, so-called Riemann geometry, had first been developed in the 1820s by German mathematician Carl Friedrich Gauss. By the 1850s another German mathematician, Bernhard Riemann developed this geometry of curved surfaces even further, and this was the Riemann geometry textbook which Grossmann gave to Einstein in 1912. Mastering this new mathematics proved very difficult for Einstein, but he knew that he needed to master it to be able to develop the equations for general relativity.

These equations were not ready until late 1915. Everyone knew Einstein was working on them, and in fact he was offered and accepted a job in Berlin in 1914 as Berlin wanted him on their staff when the new theory was published. The equations of general relativity were first presented on the 25th of November 1915, to the Prussian Academy of Sciences. The lecture Feldgleichungen der Gravitation (The Field Equations of Gravitation) was the fourth and last lecture that Einstein gave to the Prussian Academy on his new theory (Preussische Akademie der Wissenschaften, Sitzungsberichte, 1915 (part 2), 844–847), the previous three lectures, given on the 4th, 11th and 18th of November, had been leading up to this. But, in fact, Einstein did not have the field equations ready until the last few days before the fourth lecture!

The peer-reviewed paper of the theory (which also contains the field equations) did not appear until 1916 in volume 49 of Annalen der PhysikGrundlage der allgemeinen Relativitätstheorie (The Foundation of the General Theory of Relativity) Annalen der Physik (ser. 4), 49, 769–822. The paper was submitted by Einstein on the 20th of March 1916.

The beginning of Einstein’s first peer-reviewed paper on general relativity, which was received by Annalen der Physik on the 20th of March 1916

In a future blog, I will discuss Einstein’s field equations, but hopefully I have cleared up the confusion as to why some people refer to 1915 as the year of publication of the General Theory of Relativity, and some people choose 1916. Both are correct, which allows us to celebrate the centenary twice!

You can read more about Einstein’s development of the general theory of relativity in our book 10 Physicists Who Transformed Our Understanding of Reality. Order your copy here

## Time dilation in General Relativity

A student asked me last week if I could explain the difference between time dilation in Special Relativity (SR) and that in General Relativity (GR), so here is my attempt at doing so. Time dilation in SR comes about when something travels near the speed of light, and is due to the Lorentz transformations which ensure that experiments in any inertial frame are indistinguishable from each other.

I have already derived the Lorentz Transformations from first principles in this blog, and these equations are at the heart of SR, and show why time dilation occurs when one travels near the speed of light. In this blog here, I worked through some examples of time dilation in SR. But, what about time dilation in GR?

## How does time dilation come about in GR?

As I have already explained in this blog here, Einstein’s principle of equivalence tells us that whatever is true for acceleration is true for a gravitational field. So, to see how gravity affects time the easiest way is to consider how time would be affected in an accelerating rocket.

We will consider a rocket in empty space, away from any gravitational fields, which is accelerating with an acceleration $g$. We will have two people in the rocket, Alice and Bob. Alice is at the top end of the rocket, the nose end. Bob is at the bottom end of the rocket, where the tail is. Alice sends two pulses of light, one at time $t=0$, and the second one at a time $t= \Delta \tau_{A}$ later. They are received at the back of the rocket by Bob; the first pulse is received when the time is $t=t_{1}$, and the second one when the time is $t=t_{2} = t_{1} + \Delta \tau_{B}$, where $\Delta \tau_{B}$ is the time interval between flashes as measured by Bob.

This is illustrated in the figure below.

We can see how time dilation comes about in GR by considering a rocket accelerating in empty space with an acceleration $g$, and a light flashing from Alice at the front-end of the rocket and being received at the back-end by Bob.

We will set it up so that Bob’s position at time $t=0$ when the first flash is emitted by Alice is $z_{B}(0)=0$, and so his position at any other time is given by
$z_{B}(t) = \frac{1}{2}gt^{2} \text{ (Equ. 1) }$

(this just comes from Newton’s 2nd equation of motion $s=ut + \frac{1}{2}at^{2}$, see my blog here which derives those equations).

The position of Alice will just be Bob’s position plus the distance between them, which we will call $h$ (the height of the rocket), so
$z_{A}(t) = h + \frac{1}{2}gt^{2} \text{ (Equ. 2) }$

We will assume that the first pulse takes a time of $t=t_{1}$ to travel from Alice to Bob. The second pulse is emitted by Alice at a time $\Delta \tau_{A}$ after the first pulse, this is the time interval between each light pulse that Alice sends. This second pulse is received by Bob at a time of $t_{2} = t_{1} + \Delta \tau_{B}$, where $\Delta \tau_{B}$ is the time interval between pulses as measured by Bob using a clock next to him.

When the first pulse leaves Alice her position is $z_{A}(0)$, which from equation (2) is h, as she is at the top of the rocket. When Bob receives the pulse at time $t=t_{1}$ his position will be $z_{B}(t_{1})$ which, from equation (1) is $z_{B}(t_{1}) = \frac{1}{2} gt_{1}^{2}$. So, the distance travelled by the pulse is going to be
$z_{A}(0) - z_{B}(t_{1})= ct_{1} \text{ (Equ. 3) }$

as the speed of light is $c$ and it travels for $t_{1}$ seconds. Because the rocket is accelerating, the distance travelled by the second pulse will not be same (as it would be if the rocket were moving with a constant velocity). The distance travelled by the second pulse will be less, and is given by
$z_{A}(\Delta \tau_{A}) - z_{B}(t_{1} + \Delta \tau_{B}) = c(t_{1} + \Delta \tau_{B} - \Delta \tau_{A}) \text{ (Equ. 4) }$

We can use Equations (1) and (2), which give expressions for $z_{B} \text{ and } z_{A}$ as a function of $t$, to put in the values that $z_{A} \text{ and } z_{B}$ would have when $t = \Delta \tau_{A}$ for $z_{A}$ and $(t_{1} + \Delta \tau_{B})$ for $z_{B}$ respectively.

Substituting from Equations (1) and (2) into Equation (3) we have
$z_{A}(0) = h, \; \; z_{B}(t_{1}) = \frac{1}{2}gt_{1}^{2}$

which makes equation (3) become
$h - \frac{1}{2}gt_{1}^{2} = ct_{1} \text{ (Equ. 5) }$

Doing the same kind of substitution into equation (4) we have
$z_{A}(\Delta \tau_{A}) = h + \frac{1}{2}g \left( \Delta \tau_{A} \right)^{2} \rightarrow h$

$z_{B}(t_{1} + \Delta \tau_{B}) = \frac{1}{2}g(t_{1} + \Delta \tau_{B})^{2} \rightarrow \frac{1}{2}gt_{1}^2 +gt_{1} \Delta \tau_{B}$

assuming that we can ignore terms in $(\Delta \tau_{A})^{2} \text{ and } (\Delta \tau_{B})^{2}$

Substituting these expressions into equation (4) gives
$h - \frac{1}{2}gt_{1}^{2} - gt_{1} \Delta \tau_{B} = c(t_{1} + \Delta \tau_{B} - \Delta \tau_{A}) \text{ (Equ. 6) }$

We now subtract equation (6) from (5) to give
$gt_{1} \Delta \tau_{B} = c \Delta \tau_{A} - c \Delta \tau_{B} \text{ Equ. (7) }$

Re-arranging equation (5) as $\frac{1}{2}gt_{1}^{2} +ct^{1} -h$ and using the quadratic formula to find $t_{1}$ we can write that
$t_{1} = \frac{ -c \pm \sqrt{ c^{2} + 2gh } }{ g } \rightarrow \frac{ -c + \sqrt{ c^{2} + 2gh } }{ g }$

(we can ignore the negative solution because the time is always positive). We will next use the binomial expansion to write
$\sqrt{ c^{2} + 2gh } \approx c ( 1 + \frac{gh}{ c^{2} } )$

(where we have ignored terms in $\left( \frac{2gh}{c^{2}} \right)^{2}$ and higher in the Binomial expansion), and so we can write for $t_{1}$
$t_{1} \approx \left( \frac{ -c + c \left( 1 + \frac{gh}{ c^{2} } \right) }{ g } \right) \rightarrow gt_{1} = c \left( \frac{gh}{ c^{2} } \right)$

Substituting this expression for $gt_{1}$ into equation (7) we now have
$c \left( \frac{gh}{ c^{2} } \right) \Delta \tau_{B} = c \Delta \tau_{A} - c \Delta \tau_{B}$

We can cancel the $c$ in each term and bringing the terms in $\Delta \tau_{B}$ onto one side and the term in $\Delta \tau_{A}$ on the other side we have
$\Delta \tau_{B} \left( 1 + \frac{ gh }{ c^{2} } \right) = \Delta \tau_{A}$

and so
$\Delta \tau_{B} = \frac{ \Delta \tau_{A} }{ \left( 1 + \frac{ gh }{ c^{2} } \right) }$

and using the binomial expansion for $(1 + gh/c^{2})^{-1}$ (and ignoring terms in $\left( \frac{ gh }{ c^{2} } \right)^{2}$ and higher), we can finally write
$\boxed{ \Delta \tau_{B} = \Delta \tau_{A} \left( 1 - \frac{ gh }{ c^{2} } \right) \text{ (Equ. 8) } }$

Because $\frac{gh}{c^{2}}$ is always positive, this means that $\Delta \tau_{B}$ is always less than $\Delta \tau_{A}$, or to put it another way the time interval as measured by Bob at the back-end of the rocket will always be less than the time interval measured by Alice where the light pulses were sent. This means that Bob will measure time to be going at a slower rate than Alice, Bob’s time will be dilated compared to Alice.

From the principle of equivalence, whatever is true for acceleration is true for gravity, so if we now imagine the rocket stationary on the Earth’s surface, with the top end in a weaker gravitational field than the bottom end, we can see that a gravitational field will also lead to pulses arriving at Bob being measured closer together than where they were emitted by Alice. So, gravity slows clocks down!

A very important difference between time dilation in SR and time dilation in GR is that the time dilation in GR is not symmetrical. In SR, both observers in their respective inertial frames think it is the other person’s clock which is running slow. In GR, both Alice and Bob will agree that it is Bob’s clock which is running slower than Alice’s clock.

In a future blog I will do some calculations on this effect in different situations, but as you can see from Equation (8), the size of the dilation depends on the acceleration $g$ and the difference in height between $A \text{ and } B$. I will also discuss whether it is time dilation due to GR or time dilation due to SR which affect the satellites which give us GPS the more, as both effects have to be taken into account to get the accuracy we seek in the GPS position.

## Einstein’s General Relativity – part 1 – the Principle of Equivalence

Next year, 2015, marks the centennial of Einstein’s theory of gravity, what we now call the General theory of Relativity (or just “General Relativity” – “GR”). It is widely recognised as one of the greatest achievements in science, and when Arthur Eddington validated one of its predictions in 1919 Einstein was catapulted to the status of an international star. It is often said that, whereas Einstein’s 1905 special theory of relativity (or “special relativity”) would have been thought of by someone else had Einstein not come up with it, general relativity was so far ahead of its time that we may still be waiting for it if it were not for Einstein’s unparalleled genius.

A portrait of Albert Einstein from around the period that he started developing his theory of gravity, General Relativity.

As it turns out, the development of Einstein’s new theory of gravity was not an easy one. Over the course of several blogs I will trace this tortuous path, which took the best part of ten years, mainly because he had to learn the mathematics of curved space and Tensor calculus to be able to express his ideas in equations. Today I will discuss the beginnings of GR, and in particular what we now call Einstein’s “principle of equivalence”, which he thought of in 1907.

## Einstein’s 1905 Special theory of Relativity

I have already blogged about Einstein’s ground-breaking Special theory of Relativity here. Just to recap, based on two assumptions

1. There is no experiment one can do to distinguish between one inertial (non-accelerating) frame of reference and another
2. The speed of light is constant in all inertial (non-accelerating) frames of reference

Einstein was able to show that these two postulates require that strange things happen to space and time when one travels an appreciable fraction of the speed of light. Lengths get shorter, and time passes more slowly. One of the other consequences of this theory is that Einstein predicted that no information can travel faster than the speed of light.

Einstein soon realised, after he had developed his theory, that Newton’s theory of gravity was in violation of special relativity because it violates both of the postulates on which special relativity is based. In Newton’s theory of gravity, the gravitational force between two objects acts instantaneously. So, according to Newton, if the Sun were to disappear, we would instantly notice its absence (the Earth would move in a straight line rather than continue in its orbit).

Secondly, you could have two inertial (non-accelerating) frames of reference in two different gravitational fields (e.g. one on the surface of the Earth and the other on the surface of the Moon), and a simple experiment like the swinging of a pendulum would yield a different result. This is because the force of gravity (which, along with the length of the pendulum’s string, determines its period of motion) would be different in the two places.

## Einstein’s “happiest thought”

In 1907 Einstein was still working in obscurity in the Patent Office in Bern. Although his special theory of relativity had been published two years before, it was yet to have received much attention. It wasn’t until 1908 that he would get his first academic appointment. In his largely boring patent clark job, Einstein had allowed his mind to wander just as he had done leading up to his miraculous year of 1905. This time, it was in pondering how he could fit Newton’s theory of gravity into his own special relativity. One day he had what he would later refer to as the “happiest thought of my life”. In a lecture on the origins of general relativity which he gave at Glasgow University in June 1933 (“The Origins of the General Theory of Relativity”), he expressed this 1907 thought as

If a person falls freely he will not feel his own weight

Very few of us have experienced free-fall, but most of us have been in a lift (elevator). Right at the start, when the lift starts moving, we temporarily feel heavier and our stomach may feel as if it is sinking. When we slow down at the top of the lift’s travel we temporarily experience the opposite, we feel lighter and our stomach may feel as if it is about to hit our diaphragm!

What Einstein realised is that, if a person were in a lift and the cable were to snap so that the lift fell freely towards the Earth, that person would feel weightless whilst the lift was falling. Their feet would come away from the floor of the lift, and if they took e.g. coins out of their pocket, those coins would not fall towards the floor of the lift but instead would appear to “float” next to the person.

Einstein realised in 1907 that being in a lift (elevator) which is falling freely would feel the same as being in empty space – you would feel weightless.

Einstein next illustrated his absolute genius – he went from this idea, which is fairly specific, to the much more general principle of equivalence – which states that:

there is no experiment you can do to distinguish between the effects of a uniform gravitional field and that of uniform acceleration

Einstein’s “happiest thought” led to his principle of equivalence, which simply states that being in a uniform gravitational field feels the same as accelerating in empty space. They cannot be distinguished from each other. The consequences of this idea are profound and far reaching.

## The first mention of what would become “General Relativity”

Einstein was under pressure from his German editor to write up a review of his principle of special relativity, and so in late 1907 he wrote an article entitled “Über das Relativitätsprinzip und die aus demselben gezogenen Folgerungen”
(On the Relativity Principle and the Conclusions Drawn from It) which appeared on the 4th of December 1907 in the journal Jahrbuch der Radioaktivität. In a section of this review article he included some ideas as to what would happen if he were to generalise his special theory of relativity to include the effects of gravity. He noted a few consequences (without going into the details as he had yet to work them out) – gravity would alter the speed of light and hence cause clocks to run more slowly (i.e. gravity would slow down time). He even postulated that generalising special relativity to include gravity may explain the drift in the perihelion of Mercury’s orbit, something which had been confusing astronomers for several decades.

## Gravity bends light

One of the more celebrated predictions of Einstein’s general theory of relativity is that gravity should bend light. As I mentioned above, in 1919 this was shown to be the case by England’s foremost theoretical astrophysicist of the day, Arthur Eddington. I will go into the details of what he measured in another blog in this series on general relativity, but to finish this part one I will explain how gravity bends light in Einstein’s model.

To understand how this happens, we have to go back to the principle of equivalence. Remember, this states that whatever is true inside a lift which is accelerating in empty space is also going to be true for a lift which is stationary in a uniform gravitational field.

Imagine that a beam of light enters the lift horizontally on the left hand side of the lift. Because the lift is accelerating, rather than follow a straight path across the lift, it will appear to follow a curve (actually a parabola), and it will exit at a lower point on the right hand side than where it entered (this is exactly the same kind of path as a ball would follow if it is projected horizontally from a platform e.g. 200m above the Earth’s surface).

Through the principle of equivalence, if a beam of light crossing an accelerating lift will follow a curve, so will a beam of light crossing a stationary lift which is in a gravitational field. So, gravity should bend light!

Light traversing a lift which is accelerating will appear to bend (in fact it will follow a parabolic path). Because of the principle of equivalence, light should be similarly affected by gravity.

As Einstein developed the mathematics of his general theory he was able to work out precisely how much a given gravitational field should bend light, and his predicted amount was found to be true for the Sun in a celebrated experiment in 1919 by Arthur Eddington.

In part two of this blog I will discuss some of the mathematical obstacles Einstein faced in bringing his general theory of relativity to fruition.

## Studying the Universe using gravitational waves

The European Space Agency announced last week (28th of November) that a space-based gravitational wave observatory will form one of its next two large science missions. This is exciting news for Cardiff University, as it has a very active gravitational waves research group headed up by Professor B S Sathyaprakash (known to everyone as “Sathya”). The Cardiff group, along with others in the Disunited Kingdom, played an important role in persuading ESA to make this observatory one of its next large science missions. Just over a year ago I blogged about some theoretical modelling of gravitational waves the Cardiff group had done.

The ESA plan is to launch a space-based gravitational wave observatory in 2034, which will have much more sensitivity than any current or even future ground-based gravitational wave observatory. Although NASA also had plans to launch a space-based gravitational wave observatory, there is currently none in existence, so this is really ground-breaking technology that ESA is announcing. From what I can understand, the current announcement by ESA is their commitment to the proposed LISA (Laser Interferometer Space Antenna) gravitational wave observatory. It would seem NASA has withdrawn their commitment to what was originally going to be a joint NASA/ESA mission.

ESA has chosen a space-based gravitational wave detector to be funded as one of its two key future missions. It should go into operation in 2034.

## What are gravitational waves?

Gravitational waves are ripples in the fabric of space. They were predicted by Einstein as part of his theory of general relativity, the best theory we currently have to describe gravity. In his theory, events which involve extreme gravitational forces (such as two neutron stars orbiting each other (or merging), or the creation of a black hole) will lead to the emission of these gravitational waves. As they spread out from the source at the speed of light, they literally deform space as they pass by, just as ripples deform the surface of a pond as they spread out from a dropped stone.

An artist’s impression of gravitational waves being produced as two black holes orbit each other.

## Current gravitational wave observatories

There are several current gravitational wave observatories, all ground-based. These include VIRGO (in Italy) and LIGO (Laser Interferometer Gravitational Wave Observatory) which is in the United States. LIGO is the most sensitive of the current generation of gravitational wave detectors. LIGO actually comprises three separate detectors; one in Livingston, Louisiana and two in Hanford, Washington State. Each of the three separate detectors consists of two long arms at right angles to each other, forming a letter “L”. The idea behind these detectors is that, if a gravitational wave were to pass the detector, each of the two arms would have its length changed differently by the deformation of space as the gravitational wave passes through. Thus the detectors work on the principle of an interferometer, looking for tiny changes in the relative length of the two arms. And, when I say tiny, I mean tiny. In a 4km arm they are looking for changes of the order of $10^{-18} \text{ m}$, or about one thousandth the size of a proton!

The principle of a gravitational wave detector. They are essentially “interferometers”, with two arms at right angles to each other. As the gravitational waves pass the detector, space will be deformed and alter the relative lengths of the two arms.

## LIGO (Laser Interferometer Gravitational Wave Detector)

Currently the most sensitive gravitational wave detector is LIGO. LIGO consists of three separate detectors, one in Livingston, Louisiana and two in Hanford, Washington State. The detector in Louisiana is shown below.

The LIGO detector in Livingstone, Louisiana. Each arm is 4km in length, and can detect changes in the relative length of the two arms of less than the size of a proton.

The detector in Louisiana, and one of the two detectors in Washington State, consist of two 4km long arms at right angles to each other. An event like the collapse of a 10 solar-mass star into a black hole is expected to produce a change in length in a 4km arm of about $10^{-18} \text{ m}$, which is about one thousandth the size of a proton. This is just at the limit of the detection capabilities of LIGO, which is why astrophysicists are wanting more sensitive detectors to be placed into space. The other detector in Washington State has arms which are 2km in length, but just as sensitive as the detector with 4km arms at frequencies above 200 Hz, due to a different design.

## LISA – Laser Interferometer Space Antenna

The ESA plans just announced will be based on the NASA/ESA plans for LISA, which have been on the drawing board for most of the last 10 years. ESA’s plan is to build two space-based interferometers, which will be in the form of equilateral triangles as this artist’s description shows.

An artist’s impression of LISA, the “Laser Interferometer Space Antenna”. Each interferometer will consist of an equilateral triangle, with each side 5 million km in length.

The plan for LISA is to have arms which are 5 million km long! Compare this to the 4km long arms of LIGO. The changes in the length of a 5 million km long arm would be roughly one million times more than for LIGO, so rather than $10^{-18} \text{ m it would be } 10^{-12} \text{ m}$, which should be well within the capabilities of the detectors. This means that less energetic events than the collapse of a 10-solar mass star into a black hole would be detectable by LISA. All kinds of astrophysical events which involve large changes in gravitational fields should be detectable by LISA, including the afore-mentioned creation of black holes, but also the merging of neutron stars, and even the merging of less massive stars.

But, possibly most exciting is the opportunity that gravitational waves provide to probe the very earliest moments after the Big Bang. With normal electromagnetic radiation (light, x-rays, infrared light etc.), we can only see as far back as about 300,000 years after the Big Bang. This is when the Cosmic Microwave Background Radiation was produced. Prior to this time, the Universe was opaque to EM radiation of any wavelength, because it was full of unbound electrons, and the photons would just scatter off of them and not get anywhere (see my blog here about the CMB). But, it was not opaque to gravitational waves, so they provide a way for us to see back beyond the CMB, and a unique way to learn about the conditions of the Universe in its earliest moments.

## The most distant galaxy yet discovered – 30 billion light years away!

A few weeks ago it was announced that a team had discovered what seems to be the most distant galaxy yet discovered. You can read the BBC story about it here, or if you like you can read the Nature science paper here to get as much detail as you could wish for. The galaxy, which has the catchy name z8_GND_5296, was discovered using the Hubble Space Telescope, with its distance being determined using the Keck 10m telescope on the summit of Mauna Kea.

In fact, what astronomers measure is not the distance of a distant galaxy, but its redshift, which astronomers denote with the letter $z$. Redshift is the movement of the spectral lines of a galaxy to longer wavelengths due to the expansion of the Universe, the expansion discovered by Edwin Hubble in 1929. The redshift of this newly discovered galaxy has been found by Keck to be $z=7.51$, beating the previous record of $z=7.21$. But how do astronomers translate this into a distance?

## The cosmological definition of redshift

It turns out that measuring distances in astronomy is one of the most difficult things to do for several reasons. Not only are there very few direct ways to measure the distance to an object, after all we can hardly lay down a measuring tape between us and the stars and galaxies! But, to make it even worse, there also are various definitions of distance! In a future blog I will talk about the most direct ways we have to measure distance, but how we translate from these measurements to a distance also depend on the geometry of the Universe, which Einstein showed in his General Theory of Relativity is determined by the effects of gravity.

The geometry of the Universe is determined by its average density $\Omega$, and how this relates to something called the “critical density” $\Omega_{0}$, which is the dividing line between whether the Universe will carry on expanding forever, or stop expanding and start to collapse. If average density $\Omega > \Omega_{0}$ the Universe will stop expanding and collapse. If $\Omega < \Omega_{0}$ the Universe will carry on expanding forever, and if the average density $\Omega = \Omega_{0}$ the Universe is on the dividing line between the two, and is said to have a flat geometry. Without going into the details here, most cosmologists believe that we live in a Universe where $\Omega = \Omega_{0}$, that is a flat Universe.

The preferred method for measuring large distances “directly” is to use something called a Type Ia Supernova, I will blog about this method again in a future blog. But, we can only see Type Ia supernovae out to distances corresponding to a redshift of about $z=1$. The galaxy in this story is much further away than this, $z=7.51$. So, to calculate its distance we have to use a model for the expansion of the Universe, and something called Hubble’s law.

The measured redshift of a galaxy (or any object) is just given by

$z = \frac{ \lambda - \lambda_{0} }{ \lambda_{0} } \text{ (Eq. 1) }$

where $\lambda$ is the observed wavelength and $\lambda_{0}$ would be the wavelength of a spectral line (usually for a galaxy it is a line called the Lyman-alpha line) in the laboratory.

As long as the redshift is much less than 1, we can then write that

$z=\frac{ v }{ c } \text{ (Eq. 2) }$

where $v$ is the recession velocity of the galaxy and $c$ is the speed of light. In the case of $z$ not being less than 1, we need to modify this equation to the relativistic version, so we write

$1 + z = \sqrt{ \frac{ 1+ v/c }{ 1 - v/c } } \text{ (Eq. 3) }$

In our case, $z=7.51$, so we need to use this relativistic formula, and when we do we get that the recession velocity of the galaxy is $97\% \text{ of c }$, the speed of light.

Re-arranging equation 1 we can write $1 + z = \frac{ \lambda }{ \lambda_{0} }$. In principle, the distance and redshift are just related via the Hubble law

$v = H_{0} d \text{ (Eq. 4) }$,

where $v$ is the recession velocity of the galaxy, $H_{0}$ is the Hubble constant, and $d$ is the distance of the galaxy.

Things get a lot more complicated, however, when we take into account General Relativity, and its effects on the curvature of space, and even the definition of distance in an expanding Universe. I will return to this in a future blog, but here I will just quote the answer one gets if one inputs a redshift of $z=7.51$ into a “distance calculator” where we specify the value of Hubble’s constant to be $H_{0} = 72 \text{ km/s/Mpc }$ and we have a flat Universe ($\Omega=1$) with a value of $\Omega_{M}=0.25$ (the relative density of the Universe in the form of matter) and $\Omega_{vac} = 0.75$ (the relative density of the Universe in the form of dark energy).

Putting these values in gives a co-moving radial distance to the galaxy of $9103 Mpc \text{ or } 29.7 \text{ billion light years}$. (I will define what “co-moving radial distance” is in a future blog, but it is the distance quoted in this story, and is the measurement of distance which is closest to what we think of as “distance”).

The redshift also gives a time when the galaxy was formed, with $z=0$ being the present. We find that it was formed some 13.1 billion years ago, when the Universe was only about 700,000 years old.

## A galaxy 30 billion light years away??

Going back to the “co-moving radial distance”, I said it is about 30 billion light years. A light year is, of course, the distance light travels in one year. So how can a galaxy be 30 billion light years away, implying the light has taken 30 billion years to reach us, if the Universe is only 13.7 billion years old?? This sounds like a contradiction. The solution to this apparent contradiction is that the Universe has expanded since the light left the galaxy. This is what causes the redshift. In fact, the size of the Universe now compared to the size of the Universe when the light left the galaxy is simply given by

$1 + z = \frac{ a_{now} }{ a_{then} }$

where $a$ is known as the scale factor of the Universe, or its relative size. For $z=7.51$ we have $a_{now} = (1 + 7.51)\times a_{then} = 8.51 a_{then}$, so the Universe is 8.51 times bigger now than when light left the galaxy (this is what causes the redshift, it is the expansion of space, not that the galaxy is moving through space with a speed of 97% of the speed of light). It is the fact that the Universe is over 8 times bigger now than when the light left the galaxy which allows its distance measured in light years to be more than a distance of 13.7 billion light years that one would naively think was the maximum possible! So, there is no contradiction when one thinks about things correctly.