**Einstein's Principles of Relativity**

- All inertial frames are equivalent with respect to the laws of physics (Principle of Relativity)
- The speed of light in empty space is independent of the state of motion of its source (Constancy of the speed of light)

Assuming that there is no change of direction of motion, and we have a linear transformation, i.e.

(\Delta x)^\prime = (x_2 - x_1)^\prime = (x_2^\prime - x_1^\prime)

we can choose two frames of reference which are synchronised at the start of a light signal (i.e.

`x=x^\prime=0~,~t=t^\prime=0`

) and describe a light signal in these two frames, travelling in 1-dimension.Direction is preserved, hence:

\frac{\Delta x}{\Delta t} = \pm c \rightarrow \frac{\Delta x^\prime}{\Delta t^\prime} = \pm c

Combining these gives:

(\Delta x)^2 - c^2(\Delta t)^2 = 0

and, with the synchronisation of the two reference frames,

x^2 - c^2t^2 = x^{\prime 2} - c^2t^{\prime 2} = 0

for any interval,

(\Delta x)^2 - c^2(\Delta t)^2 = (\Delta x^\prime)^2 - c^2(\Delta t^\prime)^2

We have a linear transformation: If a body

*K*in reference frame

*S*moves with constant velocity

**u**, then

*K*in frame

*S'*moves with constant velocity

**u'**.

x^\prime = a_{11}x + a_{12}t + \mathrm{const}_1 ~~~ (A)

t^\prime = a_{21}x + a_{22}t + \mathrm{const}_2 ~~~ (B)

If

*S*and

*S'*are synchronised at some point

`t_0^\prime = t_0 = 0`

; `x_0^\prime = x_0 = 0`

then `\mathrm{const}_1 = \mathrm{const}_2 = 0`

.For 1D motion, the velocity

`\mathbf{u}=u\cdot \mathbf{e}_x`

. At the origin of *S'*(

*x'*= 0):

a_{11}x+a_{12}t = 0 \Rightarrow -\frac{a_{12}}{a_{11}} = \frac{x}{t} = u

Substituting back into (A):

x^\prime = a_{11}(x-ut) = a_{11}(u)(x-ut)~~~(A1)

i.e.

`a_{11}`

can be a function of *u*, since

*u*is constant for any given reference frame.

From the principle of Relativity,

*S*and

*S'*are equally valid descriptions of the object

*K*, and

*S*moves relative to

*S'*with velocity

*-u*.

\Rightarrow x = a_{11}(-u)(x^\prime - (-u)t^\prime)~~~(A2)

Invariance under time reflection gives:

t \rightarrow -t \Rightarrow u \rightarrow -u

with

*x*unchanged:

\begin{array}{cc}

x^\prime = a_{11}(u)x+a_{12}(u)t & x^{\prime\prime} = a_{11}(-u)x-a_{12}(-u)t \\

t^\prime = a_{21}(u)x+a_{22}(u)t & t^{\prime\prime} = a_{21}(-u)x-a_{22}(-u)t

\end{array}

\begin{array}{cc}

a_{11}(-u) = a_{11}(u) & a_{12}(-u)=-a_{12}(u) \\

a_{21}(-u)=-a_{21}(u) & a_{22}(-u)=a_{22}(u)

\end{array}

This implies that the coefficients of

`a_{11}~\mathrm{and}~a_{22}`

should be functions of `u^2`

.For a light signal,

`x=ct~;~x^\prime=ct^\prime`

from Einstein's 2nd principle. Substituting these into (A1) and (A2) gives:

x^\prime = a_{11}(u^2)x\left(1-\frac{u}{c}\right)

x = a_{11}(u^2)x^\prime\left(1+\frac{u}{c}\right)

Transforming from

*S*to

*S'*and back again should give the identity transformation, i.e.

1 = a_{11}(u^2)a_{11}(u^2)\left(1-\frac{u}{c}\right)\left(1+\frac{u}{c}\right)

\Rightarrow a_{11}^2(u^2) = \frac{1}{1-\frac{u^2}{c^2}}

\Rightarrow a_{11} = \gamma = \frac{1}{\sqrt{1-\frac{u^2}{c^2}}} = \frac{1}{\sqrt{1-\beta^2}}

With

`\beta =\frac{u}{c}`

. These are the Lorentz `\beta, \gamma`

factors.From

`-\frac{a_{12}}{a_{11}} = \frac{x}{t}=u`

, we get

a_{12} = \frac{-u}{\sqrt{1-\beta^2}}

Hence

`x^\prime = \gamma(x-ut)`

.For

`a_{21}, a_{22}`

, take (A2) and solve for *t'*:

x = a_{11}(x^\prime+ut\prime) \Rightarrow t^\prime = \frac{1}{u}\left(\frac{x}{a_{11}}-x^\prime\right)

t^\prime = \frac{1}{u}\left[\frac{x}{\gamma}-\gamma(x-ut)\right] = \frac{1}{u}\left[\left(\frac{1}{\gamma}-\gamma\right)x+\gamma ut\right]

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

Hence:

t^\prime = \frac{1}{u}(\gamma u t - \gamma\beta^2x) = \gamma(t-\frac{ux}{c^2})

a_{22} = \gamma ~;~ a_{21}=-\frac{u\gamma}{c^2}

Thus, the 1-dimensional Lorentz transformations between two inertial frames moving with a relative velocity

*u*are:

x^\prime = \frac{x-ut}{\sqrt{1-\frac{u^2}{c^2}}}

t^\prime = \frac{t-\frac{ux}{c^2}}{\sqrt{1-\frac{u^2}{c^2}}}

The inverse transformations are:

x = \frac{x^\prime+ut^\prime}{\sqrt{1-\frac{u^2}{c^2}}}

t = \frac{t^\prime+\frac{ux^\prime}{c^2}}{\sqrt{1-\frac{u^2}{c^2}}}

The derivation of the Lorentz transformations for position and time is very clear and mathematically elegant. If I can suggest you something: when you introduce the transformations that map the coordinates of the frame S' to the ones in the frame S'', a more explicit explanation of what you are introducing, and a stress on the fact that such a transformation must satisfy x=x'', would be ideal.

ReplyDeleteSo, how does the treatment of accelerating frames using Special Relativity lead to hyperbolic geometries?

You don't treat accelerating frames, in SR. You treat accelerated motion as a series of instantaneous inertial frames. I'll write it up another day, perhaps!

ReplyDelete