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How to add velocities in special relativity

As I mentioned in this blogpost, in special relativity any observer will measure the speed of light in a vacuum to be $c$, irrespective of whether the observer is moving towards or away from the source of light. We can think of the speed of light as a cosmic speed limit, nothing can travel faster than it.

But, let us suppose that we have two reference frames $S$ and $S^{\prime}$ moving relative to each other with a speed of $v=0.9c$, 90% of the speed of light. Surely, if someone in frame $S^{\prime}$ fires a high-speed bullet at a speed of $u^{\prime}= 0.6c$, an observer in frame $S$ will think that the bullet is moving away from him at a speed of $u = v + u^{\prime} = 0.9c + 0.6c = 1.5c$, which seemingly violates the comic speed limit.

What have we done wrong?

We cannot simply add velocities, as we would do in Newtonian mechanics. In special relativity we have to use the Lorentz transformations to add velocities. How do we do this? Let us remind ourselves that the Lorentz transformations can be written as

The Lorentz transformations to go either from reference frame $S \text{ to } S^{\prime}$, or to go from $S^{\prime} \text{ to } S$.

Calculating a velocity in two different reference frames

To calculate the velocity $u$ of some object moving with a velocity $u^{\prime}$ in reference frame $S^{\prime}$ we need to use these Lorentz transformations.

We start off by writing

$x = \gamma \left( x^{\prime} + vt^{\prime} \right) \text{ (1) }$

and

$t = \gamma \left( t^{\prime} + \frac{x^{\prime}v}{c^{2} } \right) \text{ (2) }$

We will now take the derivative of each term, so we have

$dx = \gamma \left( dx^{\prime} + vdt^{\prime} \right)$

and

$dt = \gamma \left( dt^{\prime} + \frac{dx^{\prime}v}{c^{2} } \right)$

We can now write $dx/dt = u$ (the velocity of the object as seen in frame $S$) as

$\frac{dx}{dt} = \frac{ \gamma \left( dx^{\prime} + vdt^{\prime} \right) }{ \gamma \left( dt^{\prime} + \frac{dx^{\prime}v}{c^{2} } \right) }$

The $\gamma$ terms cancel, and dividing each term on the right hand side by $dt^{\prime}$ gives

$\frac{dx}{dt} = \frac{ \left( dx^{\prime}/dt^{\prime} + vdt^{\prime}/dt^{\prime} \right) }{ \left( dt^{\prime}/dt^{\prime} + \frac{dx^{\prime}v}{c^{2} dt^{\prime} } \right) } = \frac{ \left( u^{\prime} + v \right) }{ \left( 1 + \frac{ u^{\prime} v}{c^{2} } \right) }$

$\boxed{ u = \frac{ \left( u^{\prime} + v \right) }{ \left( 1 + \frac{ u^{\prime} v}{c^{2} } \right) } }$

where $u^{\prime}$ was the velocity of the object in reference frame $S^{\prime}$.

Going back to our example of $v = 0.9c$ and $u^{\prime} = 0.6c$, we can see that the velocity $u$ as measured by an observer in reference frame $S$ will be

$u = \frac{ 0.6c + 0.9c }{ \left( 1 + \frac{ (0.6c \times 0.9c) }{ c^{2} } \right) } = \frac{ 1.5c }{ 1 + 0.54 } = \frac{ 1.5c }{ 1.54} = \boxed {0.974c}$, not $1.5c$ as we naively calculated.

The constancy of the speed of light

What happens if a person in reference frame $S^{\prime}$ shines a light in the same direction as $S^{\prime}$ is moving away from $S$? In this case, $u^{\prime}=1.0c$. Putting this into our equation for $u$ we get

$u = \frac{ 0.6c + 1.0c }{ \left( 1 + \frac{ (0.6c \times 1.0c) }{ c^{2} } \right) } = \frac{ 1.6c }{ 1 + 0.6 } = \frac{ 1.6c }{ 1.6} = 1.0c$

So they both agree that the light is moving away from them with the same speed $c$!

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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.

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