# Special relativity

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Special relativity is an extension of rational mechanics where velocity is limited to the speed of light c, a constant in the vacuum of matter and radiation.

## Speed limit

In classical mechanics, the absolute velocity is the sum of the velocity of the moving reference frame and the velocity relative to the moving reference frame. In special relativity one has to take into account the speed limit, e.g. the light speed. For colinear velocities, we get, as we shall show below :

$v_{x}= \frac{v + v'_{x}}{1+ \frac{v v'_{x}}{c^2}}$

where vx is the"absolute" and v'x the relative velocity. According to the relativity theory, all the velocities are relative; that's why the absolute velocity is replaced by the velocity in R, the observer's frame.

The relative velocity v'x is the velocity in the frame R' moving relatively to the observer in frame R. The frame R' moves at velocity v relatively to R. This formula gives a speed limit as may be seen by replacing v'x by c to get vx=c. For an infinite light speed one gets the Galilean addition of velocities:

$\left.v_{x}=v + v'_{x}\right.$

## Galilean transformation

In classical kinematics displacement is proportional to the velocity and time is independent of the velocity. When changing the reference frame, the total displacement x in reference frame R is the sum of the relative displacement x’ in R’ and of the displacement vt of R’ relative to R at a velocity v :

$x=x'+vt$

or, equivalently,

$x'=x-vt$.

This relation is linear when the velocity v is constant, that is when the frames R and R' are Galilean reference frames.

## Derivation of the Lorentz transformation

### General linear transformation

The more general relationship, with four constants α, β, γ and v is :

$x'=\gamma\left(x-vt\right)$
$t'=\beta\left(t+\alpha x\right)$

The Lorentz transformation becomes the Galilean one for β = γ = 1 et α = 0.

### Light invariance principle

The velocity of light is independent of the velocity of the source, as was shown by Michelson. We thus need to have x = ct if x’ = ct’. Replacing x and x' in these two equations, we have

$ct'=\gamma\left(c-v\right)t$
$t'=\beta\left(1+\alpha c\right)t$

Replacing t' from the second equation, the first one becomes

$c\beta\left(1+\alpha c\right)t=\gamma\left(c-v\right)t$

After simplification by t and dividing by cβ, one obtains :

$1+\alpha c=\frac{\gamma}{\beta}(1-\frac{v}{c})$

### Relativity principle

The relativity principle postulates that there are no preferred reference frames, at least for Galilean reference frames. The following derivation does not use the speed of light and allows therefore to separate it from the principle of relativity. The inverse transformation of

$x'=\gamma\left(x-vt\right)$
$t'=\beta\left(t+\alpha x\right)$

is :

$x=\frac{1}{1-\alpha v}\left(\frac{x'}{\gamma}-\frac{vt'}{\beta}\right)$
$t=\frac{1}{1-\alpha v}\left(\frac{t'}{\beta}-\frac{\alpha x'}{\gamma}\right)$

In accord with the principle of relativity, the expressions of x and t should be the same when permuting R and R' except for the sign of the velocity :

$x=\gamma\left(x'+vt'\right)$
$t=\left(t'+\alpha x'\right)$

and

$x=\frac{1}{1+\alpha v}\left(\frac{x'}{\gamma}+\frac{vt'}{\beta}\right)$
$t=\frac{1}{1+\alpha v}\left(\frac{t'}{\beta}-\frac{\alpha x'}{\gamma}\right)$

Identifying the preceding equations, we have the following identities, verified independently of x’ and t’ :

$x=\gamma\left(x'+vt'\right)=\frac{1}{1+\alpha v}\left(\frac{x'}{\gamma}+\frac{vt'}{\beta}\right)$
$t=\left(t'+\alpha x'\right)=\frac{1}{1+\alpha v}\left(\frac{t'}{\beta}-\frac{\alpha x'}{\gamma}\right)$

This gives the following equalities :

$\beta =\gamma=\frac{1}{\sqrt{1+\alpha v}}$

### The Lorentz transformation

Using the above relationship

$1+\alpha c=\frac{\gamma}{\beta}(1-\frac{v}{c})$

we get :

$\alpha =-\frac{v}{c^2}$

and, finally:

$\beta =\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$

We have now all the four coefficients needed for the Lorentz transformation which writes in two dimensions :

$x=\frac{x' + vt'}{ \sqrt[]{1 -\frac{v^2}{c^2}} }$
$t= \frac{t' + \frac{vx'}{c^2}}{ \sqrt[]{1 -\frac{v^2}{c^2}} }$

The inverse Lorentz transformation writes :

$x'= \frac{x - vt}{ \sqrt[]{1 -\frac{v^2}{c^2}} }$
$t'=\frac{t - \frac{vx}{c^2}}{ \sqrt[]{1 -\frac{v^2}{c^2}} }$

The true base of special relativity is the Lorentz transformation generalizing that of Galieo at velocities near that of light. The Lorentz transformation expresses the transformation of space and time, both depending on the relative velocity between the observer's and relative frames R and R'. Another demonstration may be found in Einstein's book [1]. The direct Lorentz transformation is, in two dimensions:

$x= \gamma\left(x' + vt'\right)$
$t= \gamma(t' + \frac{vx'}{c^2})$

The inverse Lorentz transformation is:

$x'= \gamma\left(x - vt\right)$
$t'=\gamma\left(t - \frac{vx}{c^2}\right)$

We have then four equations to be used as needed, using the Lorentz factor :

$\gamma=\frac{1}{ \sqrt[]{1 -\frac{v^2}{c^2}} }$

The Lorentz transformation remains valid in differential form for a constant velocity :

$dx= \gamma\left(dx' + vdt'\right) = \gamma\left(v'_{x} + v\right)dt'$
$dt=\gamma\left(dt' + \frac{vdx'}{c^2}\right)=\gamma\left( 1+ \frac{vv'_{x}}{c^2}\right)dt'$

From these two formulas we get the formula at the top of this page:

$v_{x}= \frac{dx}{dt}= \frac{v + v'_{x}}{1+ \frac{v v'_{x}}{c^2}}$

## Minkowski metric

The euclidean space is characterised by the validity of the Pythagoran theorem which may be written as a two-dimensional metric:

$\left. ds^2= dx^2 + dy^2\right.$

With y=ict, one obtains the Minkowski metric representing the pseudo-euclidean space of special relativity:

$ds^2=-c^2 d \tau^2=dx^2+dy^2=dx^2-c^2dt^2=-c^2\left(1- \frac{v^2}{c^2}\right)dt^2$

where v is the velocity of the frame R' relative to R. The differential form of the Lorentz transformation writes:

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

Replacing x and t as a function of x' and t' with the Lorentz transformation , one obtains the same Minkowski metric except for the primes:

$ds^2=dx^2-c^2dt^2=\gamma^2\left(dx' + v dt'\right)^2 -c^2\gamma^2\left(dt' + \frac{vdx'}{c^2}\right)^2 =\gamma^2\left[\left(dx' + v dt'\right)^2 -c^2\left(dt' + \frac{vdx'}{c^2}\right)^2\right]$

Let us develop and simplify:

$ds^2=\gamma^2\left(dx'^2 + v^2 dt'^2 +2vdx'dt' -c^2dt'^2 - \frac{v^2dx'^2}{c^2} -2vdt'dx' \right)= \frac{1}{1- \frac{v^2}{c^2}}\left(1- \frac{v^2}{c^2}\right)\left(dx'^2 -c^2dt'^2\right)=dx'^2-c^2dt'^2$

The Minkowski metric is conserved in the Lorentz transformation.

## Time dilation

Let us consider a clock in its rest frame R' moving at a velocity v relative to a frame R where is an observer. The clock rate is Δt’ at rest, in its proper frame R' and Δt viewed from R. Since the clock is at rest in R', its position is constant in R', say x'=0. To apply the Lorentz transformation, we have to choose the right equation among the four of the direct and reciprocal Lorentz transformation. We choose the one containing Δt’, Δt and x':

$\Delta t= \gamma\left(\Delta t' + \frac{vx'}{c^2}\right)= \frac{\Delta t'}{ \sqrt[]{1 -\frac{v^2}{c^2}} }$

The time interval between two beats appears larger on a moving clock than on a clock at rest. One says that time is dilated or that the clock is running slow. The time of the moving clock does not flow any more when the clock moves at light speed, but only for the distant observer, at rest. A high speed particle of limited lifetime like a meson coming from outer space, will have an apparently much larger lifetime when viewed from the Earth but its proper lifetime remains unchanged.

Let us consider now that an observer places himself in the moving frame R' and looks at a clock placed in the rest frame R. We shall have the same formula, but with t and t' reversed. Indeed the movement is relative; there is no absolute movement but a symmetry between both Galilean frames.

## Length contraction

Now consider a ruler at rest in a frame R' moving at a constant velocity v relative to a frame R where is an observer. The length at rest of this ruler is Δx' for an observer in R'. The ruler appears to have a length Δx for the observer in R. In order to measure the length of the ruler, the latter has to take an instantaneous picture of the two rulers, for example at time t=0, with Δt=0. He obtains then their lengths Δx and Δx’. He will use the equation of the Lorentz transformation where Δx, Δx’ and Δt appear:

$\Delta x' =\gamma\left(\Delta x - v \Delta t\right)=\frac{\Delta x}{ \sqrt[]{1 -\frac{v^2}{c^2}} }$

This formula has the same form as for the time, except that the primes are on the left side. For this reason, lengths contract instead of dilating for the time. Then one writes usually:

$\Delta x =\Delta x' \sqrt[]{1 -\frac{v^2}{c^2}}$

## Acceleration transformation

### Lorentz method

In classical kinematics, accelerations do not depend on the velocity of the Galilean frame since the velocity of the frame being constant, its derivative, the acceleration, is zero. In special relativity, due to both time dilation and length contraction, the change of Galilean frame changes acceleration.

Let be dvx/dt and dv'x/dt the accelerations of a particle of abscissas x and x' in the frames R of the observer and R', moving. Since the acceleration is the second derivative of space relative to time and if the frames R and R' are approximately Galilean, the Lorentz factor γ is a constant to the fourth order, as pointed out by Einstein (the slowly accelerated electron)[2] , we have:

$\frac{dv_{x}}{dt}= \frac{d^2x}{dt^2}=\frac{d^2 \left(\frac{x'}{\gamma }\right)}{d\left(\gamma t'\right)^2}= \gamma^{-3} \frac{d^2x'}{dt'^2}=\gamma^{-3} \frac{dv_{x'}}{dt'}=\left(\sqrt[]{1 -\frac{v^2}{c^2}}\right)^{3} \frac{dv_{x'}}{dt'}$

Let us bind the accelerated particle to its frame R'. We then have v=vx; the frames are thus no more Galilean. At low speeds, we are in the newtonian domain where γ≈1; the accelerations are practically equal in R et R'. For a velocity near the speed of light, the variation dv/dt of the velocity is small and, then, the acceleration, as viewed from R, is low. In both cases the frames are approximately Galilean. It is only at intermediate velocities that γ may be large with a variation of the velocity not negligible. This approximation seems to be valid according to the few available experimental data.

We may write v'=vx' :

$\gamma^{3}\frac{dv}{dt}= \frac{dv'}{dt'}$

Using an identity due, as it seems, to Lorentz [3]

(H. A. Lorentz, The theory of electrons and its applications to the phenomena of light and radiant heat, Courier Dover Publications, 2003), we have

$d\left(\gamma v\right)=\gamma dv+v d\gamma=\frac{1}{ \sqrt[]{1 -\frac{v^2}{c^2}} } \left(\ 1+ \frac{v^2}{ 1 -\frac{v^2}{c^2}} \right)dv=\gamma^{3}dv$

one obtains then:

$\frac{d\left(\gamma v\right)}{dt}=\frac{dv'}{dt'}$

### General method

Starting with the Lorentz Transform for velocity in the S' frame, as well as the transform for time in the S' frame we are able to derive the equation for acceleration in Relativistic circumstances.

$u'=\frac{u-v}{1-\frac{uv}{c^{2}}}$

and

$t'=\gamma(t-\frac{vx}{c^{2}})$

Taking the differential of both, and remembering to use the quotient rule, we get:

$du'=\frac{du(1-\frac{uv}{c^{2}})-(u-v)(-\frac{vdu}{c^{2}})}{(1-\frac{uv}{c^{2}})^{2}}$

and

$dt'=\gamma(dt-\frac{vdx}{c^{2}})$

Now we divide these two equations and divide top and bottom by dt:

$\frac{du'}{dt'}=\frac{\frac{1}{dt}}{\frac{1}{dt}}\frac{du(1-\frac{uv}{c^{2}})+(u-v)(\frac{vdu}{c^{2}})}{\gamma(dt-\frac{vdx}{c^{2}})(1-\frac{uv}{c^{2}})^{2}}=\frac{\frac{du}{dt}(1-\frac{uv}{c^{2}})+\frac{du}{dt}(u-v)(\frac{v}{c^{2}})}{\gamma(\frac{dt}{dt}-\frac{dx}{dt}\frac{v}{c^{2}})(1-\frac{uv}{c^{2}})^{2}}=\frac{\frac{du}{dt}[1-\frac{uv}{c^{2}}+\frac{uv}{c^{2}}-\frac{v^{2}}{c^{2}}]}{\gamma(1-\frac{uv}{c^{2}})^{3}}$

A little more simplification yields the end product:

$\frac{du'}{dt'}=\frac{\frac{du}{dt}[\gamma^{-2}]}{\gamma(1-\frac{uv}{c^{2}})^{3}}=\frac{\frac{du}{dt}}{\gamma^{3}(1-\frac{uv}{c^{2}})^{3}}$

Our Goal:

$a'=\frac{a}{\gamma^{3}(1-\frac{uv}{c^{2}})^{3}}$

## Relativistic Newton's Second Law of Motion

Let us multiply both sides of the acceleration transformation equation by the constant rest mass m0:

$\frac{d\left(m_{0}\gamma v\right)}{dt}=\frac{d\left(m_{0}v'\right)}{dt'}=F'$

In the frame R' where the velocity of the particle is low (in fact zero), one may apply Newton's Second Law of Motion. The right side represents the force F' in the frame R'. If one admits that the force does not depend one frame since it is applied to the particle, we have F=F' and, then

$F=\frac{d\left(m_{r}v\right)}{dt}$

where mr is the relativistic mass, appearing to the distant observer, varying in function of the velocity: $m_{r}=\frac{m_{0}}{ \sqrt[]{1 -\frac{v^2}{c^2}} }$

## Kinetic energy

In a frame moving at velocity v relative to the observer, contrarily to the Galilean transformation, the Lorentz transformation gives an acceleration depending on the relative speeds of the referentials, even Galileans (we limit ourselves to the case where velocity and acceleration are colinear). In order to produce the acceleration a = dv/dt, it is necessary to apply a force, defined by the relativistic Newton's Second Law of Motion which is a derivative relative to time of the momentum mrv. The variation dT of the kinetic energy being equal to the work of the applied force F for a displacement dx, we have:

$dT=Fdx=Fvdt=\frac{d\left(m_{r}v\right)}{dt}v dt=m_{0}vd\left(\gamma v\right)$

Let us use an identity similar to that of Lorentz above:

$vd\left(\gamma v\right)=v\gamma dv+v^2 d\gamma=\frac{v}{ \sqrt[]{1 -\frac{v^2}{c^2}} } \left(\ 1+ \frac{v^2}{ 1 -\frac{v^2}{c^2}} \right)dv=c^2d\gamma$

The variation of the kinetic energy becomes dT=m0dγ. Integrating this equation, one obtains:

$T=\left. m_{0}\gamma c^2\right.$

The kinetic energy should be zero when the velocity v is zero, e.g. when γ=1. The integration constant is thus:

$T=\left. m_{0}c^2\right.$

The kinetic energy is:

$T=\left(m_{r}-m_{0}\right) c^2$

equal to the difference between rest mass m0 and relativistic mass mr multiplied by the universal factor $c^2$. These two masses have indices in order to avoid any confusion.

## Total relativistic energy E=mc²

The total relativistic energy E = mc² must not be confused with the total classical mechanical energy. The sum of the kinetic energy and the potential energy remains a constant value without being an absolute value.

Drivers know that the distance they may travel is proportional to their volume or mass of gas with a coefficient K depending on its heat content. It may be assumed that there is a maximum value of energy of any type contained in a given mass. The maximum energy avalable in a given mass is obtained when all the mass is converted into energy (radiative,thermal, mechanical, electrical…). A higher energy content is impossible, because there is no matter anymore. The problem is to find the universal coefficient K. Let us apply relativity. The maximum energy available is then Er = Kmr in the observer's frame and E0=Km0 in its proper frame of the object of mass m0. The difference between these two energies is

$T=\left(Km_{r}-Km_{0}\right)$

is due uniquely to the velocity, the relativistic mass depending only on the rest mass and the relative velocity between the object and the observer. The application of the Lorentz transformation, of Newton's law and of the definition of energy has shown in the preceding paragraph that the relativistic kinetic energy is:

$T=\left(m_{r}-m_{0}\right)c^2$

Identifying these last equations, one finds

$\left .K=c^2\right.$

The total relativistic energy is then:

$\boldsymbol {\left .E=mc^2\right.}$

where m is the mass, static or dynamic, depending if the relative velocity is small or comparable to the speed of light.

## Conclusion

We have derived the most celebrated equation of the twentieth century from first principles:

Linear transformation of space and time

Light invariance principle

Relativity principle

Newton's second law of motion