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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!

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 and time travel

In this blog I derived, from first principles, the Lorentz transformations which are used in Einstein’s special theory of relativity to relate one frame of reference $S$ to another frame of reference $S^{\prime}$ which are moving relative to each other with a speed $v$.

$\boxed {\begin{array}{lcl} x^{\prime} & = & \gamma (x - vt) \\ y^{\prime} & = & y \\ z^{\prime} & = & z \\ t^{\prime} & = & \gamma ( t - \frac{ v }{ c^{2} }x ) \end{array} }$

So, these relate the length $x$ and time $t$ in two different reference frames which are moving relative to each other with a velocity $v$. One of the most intriguing and surprising consequences of Einstein’s special theory of relativity is that time is relative and not absolute. What this means in simple terms is that two observers in two reference frames $S$ and $S^{\prime}$ moving relative to each other with a velocity $v$ will measure time to be passing at different rates.

Time dilation

This phenomenon is known as time dilation. Let us consider our two reference frames $S$ and $S^{\prime}$. We will have a clock in frame $S^{\prime}$, which in that reference frame is stationary (e.g. a clock on a rocket ship, although the rocket ship is moving, the clock is stationary relative to the rocket ship).

Two successive events on the clock in $S^{\prime}$ are separated by a time interval $\Delta t^{\prime}$ which we are going to call the proper time $T_{0}$. The time interval in the other reference frame, $S$, is $\Delta t = T$. How does this compare to $T_{0}$?.

In the reference frame $S^{\prime}$ the clock is stationary, so we can say that the location of the clock in the x-dimension, $x^{\prime}$, does not change. That is, $\Delta x^{\prime} = 0$.

Using our equation which relates $t \; \text{and} \; t^{\prime}$ from above, we can write

$\begin{array}{lcl} \Delta t & = & \gamma (\Delta t^{\prime} + \frac{v}{c^{2}} \Delta x^{\prime}) \\ \Delta t & = & \gamma \Delta t^{\prime} \; \; (\text{as} \; \Delta x^{\prime} = 0 ) \\ \end{array}$

and so we can write

$\boxed {T = \gamma T_{0}}$

This means the time interval $T$ in frame $S$ will appear to be dilated by a factor of $\gamma$ compared to the proper time interval $T_{0}$.

A clock travelling at close to the speed of light will run more slowly compared to a stationary clock

Time dilation in Nature

We observe the effects of time dilation every day in Nature. Cosmic rays, high energy particles from space, strike molecules in our atmosphere and create particles from the high energy interactions (this is the same as happens in the Large Hadron Collider). One of the particles created in these reactions are muons, which decay very rapidly in about 2 microseconds second (2 millionths of a second). Given the distance between where they are created in the upper atmosphere and the Earth’s surface, they should not survive long enough to make it to the surface of the Earth. But they do. How? Because of time dilation, the muons are moving so quickly that $\gamma$ is appreciable more than 1, meaning that 2 microseconds in the muon’s frame of reference is much longer in our frame of reference. So, in the muon’s frame of reference it is indeed decaying in let us say 2 microseconds, but in our frame or reference it could survive for maybe a millisecond (thousandth of a second) or more, long enough to reach the surface of the Earth.

The symmetry of relativity

One aspect of relativity which confuses a lot of people is that it is symmetrical. Although an observer in frame $S$ will think that the clock in frame $S^{\prime}$ is ticking more slowly, if an observer in $S^{\prime}$ were to look at a clock which was at rest in frame $S$, that observer would think that the clock in frame $S$ is moving more slowly. Each would think that their clock is behaving normally, and it is the clock in the other’s reference frame which is showing the effects of time dilation.

If a twin sets off on a space trip where the rocket will travel close to the speed of light, then time dilation effects will come into play. This means that e.g. a 20-year old twin can set off on a space trip which for the twin who stays on Earth appears to last for 40 years, but because of time dilation effects maybe only 5 years will appear to pass for the twin on the rocket. Thus, the 60-year old twin who stayed on Earth will be greeted after 40 years by a 25-year old twin!!

In the example I have shown, 40 years for the twin who stays on Earth appears to pass as 5 years for the twin on the rocket. This means the time dilation factor is $40/5 = 8$, and as the time dilation factor is just the Lorentz factor $\gamma$, this means the rocket will need to travel at a speed of $99.2\%$ of the speed of light.

HANG ON!!! you say, what about the symmetry of relativity? Surely the twin in the rocket will think that the twin on Earth is aging more slowly, so why doesn’t he return to find the twin on Earth is only 25 and he is 60? Or maybe, because of the symmetry, they will both be 60 when the travelling twin returns?

No, what one has to realise is that there is no symmetry in this trip. In order for the travelling twin to leave the Earth and travel at close to the speed of light he has to speed up considerably. Also, in order to come back he has to slow down and reverse his direction, speeding up again once he’s turned his rocket around to come back to Earth. And, as he approaches Earth, he will have to slow down again. These large accelerations (changes in speed) which the travelling twin experiences break the symmetry, and so it really is the case that the travelling twin will return younger than the twin who has stayed on Earth. How much younger depends on how close to the speed of light the travelling twin travels.

Back to the future

Although it is possible therefore to “travel to the future”, as our twin in the example above does, what is not possible is to travel to the past. In order to do this one would need to travel faster than the speed of light, which Einstein’s theory does not allow. The results of neutrinos travelling faster than the speed of light, announced back in the Autumn of 2011, proved to be incorrect. One of the reasons that story caused so much interest is that travelling back in time has all kinds of problems associated with it, the movie “Back to the future” illustrated some of them. I will discuss time travel more in another blog.

Time for a photon

I will finish this blog with a question about photons (particles of light). Remember that Einstein’s theory of special relativity is based on the premise that light always travels at the same speed in a vacuum. The nearest star system beyond our Solar System is the Proxima Centauri system, which is 4.2 light years away. That means it takes light 4.2 years to travel from this system to us, which in terms of kilometres is 40 trillion kilometres ($4 \times 10^{13}$ kilometres!). Now you know why we use light years for such large distances.

So if light takes 4.2 years to travel the 40 trillion kilometres from Proxima Centauri to Earth, my question to you is

how long would it seem to take if you were a photon moving at the speed of light?

Answers on a postcard, or in the comment section below.