Theory of relativity/Special relativity

Resource type: this resource contains a lecture or lecture notes.

Special relativity is an extension of classical mechanics (Galilean relativity) that deals with the consequences of the speed of light being the same for all observers.

Articles

• space, time, and the Lorentz transform
• momentum
• energy
• E = mc²
• spacetime diagrams and vectors

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 a frame designated with a prime(') moving with respect to another unprimed frame along the x,x' axis with a relative velocity of $v$ between the frames, the x,x' component of the velocity of a test particle is related between the frames according to:

$u_{x}={\frac {v+u'_{x}}{1+{\frac {vu'_{x}}{c^{2}}}}}$

where $u_{x}$ is the x component of the test particle's velocity with respect to the unprimed frame, and $u'_{x}$ the x' component of the test particle's velocity with respect to the primed frame. According to the relativity theory, all the velocities are relative.

This formula gives a speed limit as may be seen by replacing $u'_{x}$ by c to get $u_{x}=c$.

For an infinite light speed one gets the Galilean addition of velocities:

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

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. 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 :

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

Velocity addition law

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:

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

Minkowski metric

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

$\left.d\sigma ^{2}=dx^{2}+dy^{2}\right.$

extended to three dimesions of space this becomes

$\left.d\sigma ^{2}=dx^{2}+dy^{2}+dz^{2}\right.$

The path a real massive particle takes through spacetime called its worldline as experienced by the particle will actually be timelike and a generalized version of the Pythagorean theorem to four dimensional flat spacetime yields the description of that interval as an invariant calculable from any frame by

$ds^{2}=dc\tau ^{2}=dct^{2}-\left(dx^{2}+dy^{2}+dz^{2}\right)=\left(1-{\frac {u^{2}}{c^{2}}}\right)dct^{2}$

where u is the velocity of the test particle relative to the ct,x,y,z frame. The differential form of the Lorentz transformation writes:

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

$dct=\gamma \left(dct'+{\frac {vdx'}{c}}\right)$

For example, consider for a moment just x and t. 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:

$ds^{2}=dct^{2}-dx^{2}$

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

Let us develop and simplify:

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

$ds^{2}={\frac {1}{1-{\frac {v^{2}}{c^{2}}}}}\left(1-{\frac {v^{2}}{c^{2}}}\right)\left(dct'^{2}-dx'^{2}\right)$

$ds^{2}=dct'^{2}-dx'^{2}$

The Minkowski metric is conserved in the Lorentz transformation.

Though the term "the metric" a term loosely used for both the "line element" and the "metric tensor" interchangeably. To be clear, the "line element" of special relativity is

$ds^{2}=dct^{2}-\left(dx^{2}+dy^{2}+dz^{2}\right)=\eta _{\mu \nu }dx^{\mu }dx^{\nu }$

and the metric tensor of special relativity is the set of coefficient elements read off of that line element $\eta _{\mu \nu }$

Taking

$x^{0}=ct$,

$x^{1}=x$,

$x^{2}=y$,

$x^{3}=z$,

here for special relativity the metric elements are

$\eta _{00}=1$

$\eta _{11}=\eta _{22}=\eta _{33}=-1$

All other elements are 0.

Written as a matrix this is

$||\eta _{\mu \nu }||={\begin{bmatrix}1&0&0&0\\0&-1&0&0\\0&0&-1&0\\0&0&0&-1\end{bmatrix}}$

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:

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

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 described above. The variation dT of the kinetic energy being equal to the work of the applied force f for a displacement dx, we have:

$dT=fdx=fvdt={\frac {d\left(\gamma mv\right)}{dt}}vdt=mvd\left(\gamma v\right)$

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

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

The variation of the kinetic energy becomes $dT=mc^{2}d\gamma$. Integrating this equation, one obtains:

$T=\left.\gamma mc^{2}-T_{0}\right.$

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

$T_{0}=\left.mc^{2}\right.$

The kinetic energy is:

$T=\left(\gamma -1\right)mc^{2}$

equal to the difference between mass-energy $mc^{2}$ and relativistic energy $\gamma mc^{2}$.

Total relativistic energy

The relativistic energy is the time element of the momentum four vector

$E_{R}=p^{0}c$

and for a test particle with mass is given by

$E_{R}=\gamma mc^{2}$

But this energy does not include the potential energy it may be associated with in interacting with other matter. In order to develop the Klein-Gordon equation with the inclusion of a potential one defines a canonical momentum four-vector which is the sum of the momentum four-vector with its interacting charge times the Lorentz gauge vector potential it experiences

$P^{\lambda }=p^{\lambda }+{\frac {q}{c}}\phi ^{\lambda }$

The total relativistic energy then is the sum of the relativistic energy and the potential energy due to the vector potential.

$E_{Tot}=E_{R}+q\phi ^{0}$

The length of the momentum four-vector yields its mass as an invariant

$m^{2}c^{2}=\eta _{\mu \nu }p^{\mu }p^{\nu }$

replacing the momentum 4-vector with the difference between the canonical momentum and the vector potential from above into this equation and operating the result on the wave function yields the Klein-Gordon equation for the inclusion of a vector potential.

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

See also

• Topic:Special relativity
• Extended special theory of relativity
• Four-force