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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.

Riding on a beam of light

In this previous blog, I discussed how an experiment involving electrodynamics was not invariant under a Galilean transformation. Or, to put it another way, the laws of electrodynamics as stated would allow someone to determine whether they were at rest or moving, something which deeply troubled a young Albert Einstein. It is said that one of Einstein’s first “thought experiments” was to imagine himself travelling along on a beam of light. Light is the ultimate “free lunch”, the changing magnetic field produces a changing electric field which produces a changing magnetic field. It self-propogates at a speed of $3 \times 10^{8}$ metres per second in a vacuum.

Einstein realised that if he were travelling with the beam of light then, relative to him, the light would disappear as the electric and magnetic fields would be stationary relative to him. This worried him, as it suggested that one would be able to tell whether one was travelling or at rest, just by measuring the properties of light. Einstein realised, in an insight which possibly no one else was capable of, that the speed of light was fundamental to physics, and needed to always be constant. This led him to develop what we now call the special theory of relativity, most of which is expressed in a paper he published in 1905 called “On The Electrodynamics of moving bodies“.

Einstein’s special theory of relativity

Einstein’s Special Theory of Relativity is based on two very simple but far reaching principles

1. No experiment, mechanical or electrodynamical, can distinguish between being at rest or moving at a constant velocity.
2. That the speed of light in a vacuum, c, is constant to any observer, no matter how quickly the observer is moving.

From the second of these principles, with a simple thought experiment, we can derive the Lorentz transformations from first principles. These are the equations which allow us to translate from one frame of reference to another so that all the laws of Physics are invariant.

An expanding sphere of light

The thought experiment we will use to derive the Lorentz transformations from first principles is one of a flash of light originating at the origin of two frames of reference S and S’ which are moving relative to each other with a velocity $v$. We set up our experiment so that at time $t=0$ the origins of the two frames of reference are in the same place.

Two frames of reference S and S’ moving relative to each other with a velocity v have a flash of light originate at their respective origins at time t=0

The flash of light will expand as a sphere, moving with a velocity $c$ in both frames of reference, in accordance with Einstein’s 2nd principle of relativity. For reference frame S we can write that the square of the radius $r^{2}$ of the sphere is $x^{2} + y^{2} + z^{2} = c^{2}t^{2}$ so

$\boxed{ x^{2} + y^{2} + z^{2} - c^{2}t^{2}=0 } \qquad(1)$

For the reference frame S’ we can write that

$\boxed{ x^{\prime 2} + y^{\prime 2} + z^{\prime 2} - c^{2}t^{\prime 2} = 0 } \qquad(2)$

These two equations must be equal, as it is the same sphere of light and therefore the sphere must have the same radius in the two reference frames. Let us see if we can transform from one to the other using the Galilean transforms, which are

$\boxed {\begin{array}{lcl} x^{\prime} & = & x - vt \\ y^{\prime} & = & y \\ z^{\prime} & = & z \\ t^{\prime} & = & t \end{array} }$

$x^{\prime 2} + y^{\prime 2} + z^{\prime 2} -c^{2}t^{2} = (x-vt)^{2} + y^{2} + z^{2} - c^{2}t^{2}$

Expanding the brackets of the right hand side gives

$x^{2} - 2vtx + v^{2}t^{2} + y^{2} + z^{2} - c^{2}t^{2} \neq x^{2} + y^{2} + z^{2} - c^{2}t^{2}$

The left side of the equation should be equal to the right side, but the terms highlighted do not exist on the right hand side of the equation.

As we can see, the two expressions are not equal as the left hand side has the extra terms $-2vtx + v^{2}t^{2}$. This means that a Galilean transformations does not work. The extra terms involve a combination of $x$ and $t$, which suggests that both the equations linking $x$ and $x^{\prime}$ and $t$ and $t^{\prime}$ need to be modified, not just the equation for $x$ as is the case in the Galilean transformations.

Modifying the Galilean transformations

Let us assume that the transformations can be written as

$\boxed {\begin{array}{lcl} x^{\prime} & = & a_{1}x + a_{2}t \qquad(3) \\ y^{\prime} & = & y \\ z^{\prime} & = & z \\ t^{\prime} & = & b_{1}x + b_{2}t \qquad(4) \end{array} }$

We need to find the values of $a_{1}, a_{2}, b_{1}$ and $b_{2}$ which correctly transform the equations for the expanding sphere of light. We do this by substituting equations (3) and (4) into equation (2). Before we do this, we note that the origin of the primed frame $x^{\prime}=0$ is a point that moves with speed $v$ as seen in the unprimed frame S. Therefore its location in the unprimed frame S at time $t$ is just $x=vt$. So we can write equation (3) as

$x^{\prime} = 0 = a_{1}x + a_{2}t \rightarrow x = -\frac{a_{2}}{a_{1}} t = vt$

$\therefore \frac{ a_{2} }{ a_{1} } = -v$

Re-writing equation (3)

$x^{\prime} = a_{1}x + a_{2}t = a_{1}(x+\frac{ a_{2} }{ a_{1} } t) = a_{1}(x-vt)$

Now we substitute this expression and equation (4) into equation (2)

$a_{1}^{2}(x-vt)^{2} + y^{\prime 2} + z^{\prime 2} -c^{2}(b_{1}x+b_{2}t)^{2} = x^{2} + y^{2} + z^{2} -c^{2}t^{2}$

$a_{1}^{2} x^{2} -2a_{1}^{2} xvt + a_{1}^{2} v^{2} t^{2} - c^{2} b_{1}^{2} x^{2} - 2c^{2} b_{1} b_{2} xt -c^{2} b_{2}^{2} t^{2} = x^{2} - c^{2} t^{2}$

Equating coefficients:

$( a_{1}^{2} - c^{2}b_{1}^{2} ) x^{2} = x^{2} \rm{\;\; or \;\;} a_{1}^{2} - c^{2}b_{1}^{2} = 1 \qquad(5)$

$( a_{1}^{2} v^{2} - c^{2} b_{2}^{2} ) t^{2} = -c^{2} t^{2} \rm{\;\; or \;\;} c^{2} b_{2}^{2} -a_{1}^{2} v^{2} = c^{2} \qquad(6)$

$(2a_{1}^{2} v + 2b_{1} b_{2} c^{2} ) xt = 0 \rm{\;\; or \;\;} b_{1} b_{2} c^{2} = -a_{1}^{2}v \qquad(7)$

From equations (5) and (6) we can write

$b_{1}^{2} c^{2} = a_{1}^{2} - 1 \qquad(8)$

and

$b_{2}^{2} c^{2} = c^{2} + a_{1}^{2} v^{2} \qquad (9)$

Multiplying equations (8) and (9) and squaring equation (7) we get

$b_{1}^{2} b_{2}^{2} c^{4} = ( a_{1}^{2} - 1 )( c^{2} + a_{1}^{2} v^{2} ) = a_{1}^{4} v^{2}$

so

$a_{1}^{2} c^{2} - c^{2} + a^{4} v^{2} - a_{1}^{2} v^{2} = a_{1}^{4} v^{2}$

$a_{1}^{2} c^{2} - a_{1}^{2} v^{2} = c^{2}$

$a_{1}^{2} ( c^{2} - v^{2} ) = c^{2}$

$a_{1}^{2} = \frac{ c^{2} }{ c^{2} - v^{2} } = \frac{ 1 }{ 1 - v^{2}/c^{2} }$

so

$\boxed{ a_{1} = \frac{ 1 }{ \sqrt{ (1 - v^{2}/c^{2} ) } } }$

Thus we can write

$\boxed{ a_{2} = -v \cdot \frac{ 1 }{ \sqrt{ ( 1 - v^{2}/c^{2} ) } } }$

Using equation (8) we can write

$b_{1}^{2} c^{2} = \frac{ 1 }{ (1 - v^{2}/c^{2} ) } - 1$

$b_{1}^{2} c^{2} = \frac{ 1 - ( 1 - v^{2}/c^{2} ) }{ (1 - v^{2}/c^{2} ) } = \frac{ v^{2}/c^{2} }{ (1 - v^{2}/c^{2} ) } = \frac{ v^{2} }{ c^{2} } \cdot \frac { 1 }{ (1 - v^{2}/c^{2} ) }$

so $b_{1}^{2} = \frac{ v^{2} }{ c^{4} } \cdot \frac{ 1 }{ ( 1 - v^{2}/c^{2} ) }$

Taking the negative square root we can write

$\boxed{ b_{1} = - \frac{ v }{ c^{2} } \cdot \frac{ 1 }{\sqrt{ (1 - v^{2}/c^{2} ) }} }$

From equation (9) we can write

$b_{2}^{2} c^{2} = c^{2} + v^{2} \cdot \frac{ 1 }{ ( 1 - v^{2}/c^{2} ) } = \frac{ c^{2}( 1 - v^{2}/c^{2} ) + v^{2} }{ ( 1 - v^{2}/c^{2} ) } = \frac{ c^{2} - v^{2} + v^{2} }{ (1 - v^{2}/c^{2} ) } = \frac{ c^{2} }{ ( 1 - v^{2}/c^{2} ) }$

$b_{2}^{2} = \frac{ 1 }{ ( 1 - v^{2}/c^{2} ) }$

and so

$\boxed{ b_{2} = \frac{ 1 }{ \sqrt{ ( 1 - v^{2}/c^{2} ) } } }$

which is the same as $a_{1}$.

If we define

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

we can write

$a_{1} = \gamma, \;\;\; a_{2} = -\gamma v, \;\;\; b_{1} = -\frac{ v }{ c^{2} } \cdot \gamma \rm{\;\;\ and \;\;\;} b_{2} = \gamma$

Thus we can finally write our transformations as

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

These are known as the Lorentz transformations.

The Lorentz factor

The term $\gamma$ is know as the Lorentz factor.

The Lorentz factor $\gamma$ plotted against speed as a fraction of the speed of light.

As this plot shows, the Lorentz factor is essentially unity until the ratio $v/c$ (the ratio of the speed to the speed of light) becomes about half of the speed of light, or about $1.5 \times 10^{8}$ m/s. Given that even our fastest space ships only travel at a tiny fraction of the speed of light, it is not surprising that we have no direct experience of the weird effects that a Lorentz factor deviating significantly from one produce. Of course we see these effects in particle accelerators and cosmic ray showers, but human beings are a long way from attaining speeds where the Lorentz factor will deviate from unity.

In a future blog I will discuss some of these weird effects. They include time passing more slowly and distances shrinking. Very very weird; but very very real, they are shown to happen every day in our particle accelerators.

Galilean Relativity and Electrodynamics

Quite a few months ago now I derived the so-called Galilean transformations, which allow us to relate one frame of reference to another in the case of Galilean Relativity.

$\boxed {\begin{array}{lcl} x^{\prime} & = & x + vt \\ y^{\prime} & = & y \\ z^{\prime} & = & z \\ t^{\prime} & = & t \end{array} }$

It had been shown that for experiments involving mechanics, the Galilean transformations seemed to be valid. To put it another way, mechanical experiments were invariant under a Galiean transformation. However, with the development of electromagnetism in the 19th Century, it was thought that maybe results in electrodynamics would not be invariant under the Galilean transformation.

The electrostatic force between two charges

If we have two charges which are stationary, they experience a force between them which is given by Coulomb’s law.

$\vec{F}_{C} = \frac{ Q^{2} }{ 4\pi\epsilon_{0}\vec{r}^{2} }$ where $Q$ is the charge of each charge, $r$ is the distance between their centres, and $\epsilon_{0}$ is the permittivity of free space, which determines the strength of the force between two charges which have a charge of 1 Coulomb and are separated by 1 metre.

Coulomb’s law gives us the force between two charges. If the charges are the same sign the force is repulsive, if the charges are opposite in sign the force is attractive.

Moving charges produce a magnetic field

If charges are moving we have an electric current. An electric current produces a magnetic field. The strength of this field is given by Ampère’s law

$\oint \vec{B} \cdot d\vec{\l} = \mu_{0}I$ where $d\vec{l}$ is the length of the wire, $\vec{B}$ is the magnetic field, $\mu_{0}$ is the permeability of free space and $I$ is the current. So, if the two charges are moving, each will be surrounded by its own magnetic field.

A wire carrying a current produces a magnetic field as given by Ampère’s law.

The Lorentz force

If the two charges are moving and hence producing magnetic fields around each of them then there will be an additional force between the two charges due to the magnetic field each is producing. This force is called the Lorentz force and is given by the equation

$\vec{F}_{L} = Q\vec{v}\times\vec{B}$. If $r$ is the distance between the two wires, and they are carrying currents $I_{1}$ and $I_{2}$ respectively, and are separated by a distance $r$, we can write $B=\frac{\mu_{0}I}{2\pi r}$ which then gives us that the Lorentz force $F_{L} = \frac{ I_{1} \Delta L \mu_{0} I_{2} }{2 \pi r }$ and so the Lorentz force per unit length due to the magnetic field in the other wire that each wire feels is given by $\boxed{ \frac{ F_{L} }{\Delta L} = \frac{ \mu_{0} I_{1} I_{2} }{ 2 \pi r} }$. Writing the currents in terms of the rate of motion of the charges, we can write this as

$F_{L} = \frac{ \mu_{0} Q_{1} Q_{2} }{ 4\pi r^{2} } v^{2}$

The Lorentz force is the force on a wire due to the magnetic field produced in the other wire from the current flowing in it.

Putting it all together

Let us suppose the two charges are sitting on a table in a moving train. This would mean that someone on the train moving with the charges would measure a different force between the two charges (just the electrostatic force) compared to someone who was on the ground as the train went past (the electrostatic force plus the Lorentz force).

The force measured on one of the charges by the person on the train, for whom the charges are stationary, which we shall call $F$ will be

$F = \frac{ Q_{1}Q_{2} }{ 4 \pi \epsilon_{0}r^{2} }$.

The force measured on one of the charges by the person on the ground, for whom the charges are moving with a velocity $v$, which we shall call $F^{\prime}$ will be

$F^{\prime} = \frac{ Q_{1}Q_{2} }{4 \pi \epsilon_{0}r^{2} } + \frac{ \mu_{0} Q_{1} Q_{2} }{ 4\pi r^{2} } v^{2}$.

These two forces are clearly different, and so it would seem that the laws of Electrodymanics are not invariant under a Galilean transformation, or to put it another way that one would be able to measure the force between the two charges to see if one were at rest or moving with uniform motion because the forces differ in the two cases.

As I will explain in a future post, Einstein was not happy with this idea. He believed that no experiment, be it mechanical or electrodynamical, should be able to distinguish between a state of rest or of uniform motion. His solution to this problem, On the Electrodynamics of Moving Bodies, was published in 1905, and led to what we now call his Special Theory of Relativity. This theory revolutionised our whole understanding of space and time.