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Subsections

1.2.1 Relativity parameters and useful formulas
1.2.2 Lorentz transformation
- 1.2.2.1 Space Transformation
- 1.2.2.2 Field Transformation
  - 1.2.2.2.1 Longitudinal field and pure longitudinal momentum

1.2 Basic notions of relativity

1.2.1 Relativity parameters and useful formulas

The definition of the relativistic momentum is expressed in function of the rest mass

(for an electron $m_e=9.10938\times 10^{-31}$ kg):

$\displaystyle \v{p} = \gamma m_0 \v{v}$

(1.3)

The relativistic factor $\gamma$ being:

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

(1.4)

The relative velocity $\beta$ being:

$\displaystyle \v{\beta} = \frac{\v{v}}{c}$

(1.5)

The energy of a particle of rest mass

travelling at $\v{v}$ is:

$\displaystyle \mathcal{E} = \gamma m_0 c^2$

(1.6)

The basic unit of energy used is the energy of a particule of charge

would gain while being accelerated between two conducting plates at a potential difference of one Volt: one electron volt (eV). By introducing the definitions of $\gamma$ , $\beta$ and

, we have the following expression for the energy:

$\displaystyle \mathcal{E}^2 = c^2p^2+m_0^2c^4$

(1.7)

$\displaystyle \v{p} = \gamma m_0 \v{v}= \gamma m_0 c \v{\beta} = \frac{\mathcal{E}}{c} \v{\beta}$

(1.8)

The latter expression justify the expression of the momentum in the unit eV/c, instead of the classic mecanics one: kg m/s. In the case of a conserved linear momentum, without external forces we can write:

$\displaystyle \v{v}=\frac{d\v{r}}{dt} = \frac{c^2 \v{p}}{\mathcal{E}}$

(1.9)

$\displaystyle \v{r} =$ cst $\displaystyle + \frac{c^2 \v{p}}{\mathcal{E}} t$

(1.10)

1.2.2 Lorentz transformation

1.2.2.1 Space Transformation

In the four dimensional space, we consider two reference frames $\mathcal{R}$ and $\mathcal{R'}$ with a relative velocity $\v{v}$ between them. The coordinates of a point in each frame are $X=\left(t,x,y,z \right)$ , and $X'=\left(t',x','y,z' \right)$ . The construction of Lorentz group transformation leads to the expression of any general Lorentz transformation:

$\displaystyle L$	$\displaystyle =$	$\displaystyle -\boldsymbol{\omega} \cdot \boldsymbol{S} - \boldsymbol{\zeta} \cdot \boldsymbol{K}$	(1.11)
$\displaystyle A$	$\displaystyle =$	$\displaystyle e^{L}$	(1.12)

where the coordinates from a frame to another are transformed by matricial multiplication by the Lorentz matrix

$\displaystyle X'=A\cdot X$

(1.13)

The 4 $\times$ 4 matrix

, has null diagonal term. Its other terms can be expressed by six parameters(degree of freedom) and each of them can be decomposed on a base of 6 canonical 4 $\times$ 4 matrices

and

, $i \in \{1 ; 2 ; 3\}$ , the first ones being antisymmetric contrary to the second ones that are symmetric(refer to [21] for the expression of these matrices). $\boldsymbol{\omega}$ and $\boldsymbol{\zeta}$ are 3 dimensional vectors that can be respectively intepreted in term of rotation and boost of the coordinate axis of the frame $\mathcal{R'}$ . The product $\boldsymbol{\omega} \cdot \boldsymbol{S}$ and $\boldsymbol{\zeta} \cdot \boldsymbol{K}$ are thus tensorial products so that the result is a 4 $\times$ 4 matrix.
For a boost without rotation ( $\boldsymbol{\omega}$ =0) in an arbitrary direction, the boost vector can be written in term of the relative velocity $\v{\beta}$ :

$\displaystyle \boldsymbol{\zeta} = \tanh^{-1}\left(\beta\right) \frac{\v{\beta}}{\beta}$

(1.14)

The lorentz transformation matrix reduces then:

$\displaystyle A_{\text{boost}}(\v{\beta}) =\left( \begin{array}{cccc} \gamma & ... ...beta_z}{\beta^2} & 1+ (\gamma-1)\frac{\beta_z^2}{\beta^2}\\ \end{array} \right)$

(1.15)

1.2.2.2 Field Transformation

Choosing the prime notation to refer to parameters in the moving frame, the fields in the laboratory frame are obtained using the following Lorentz transformation[21]:

$\displaystyle \v{E}$	$\displaystyle =$	$\displaystyle \gamma \left(\v{E}'-\v{\beta} \times \v{B}'\right) - \frac{\gamma^2}{\gamma+1} \left(\v{\beta} \cdot \v{E}' \right)\v{\beta}$	(1.16)
$\displaystyle \v{B}$	$\displaystyle =$	$\displaystyle \gamma \left(\v{B}'+\v{\beta} \times \v{E}'\right) - \frac{\gamma^2}{\gamma+1} \left(\v{\beta} \cdot \v{B}' \right)\v{\beta}$	(1.17)

Using the hypothesis that , the formulation reduces to:

$\displaystyle \v{E}$	$\displaystyle =$	$\displaystyle \gamma \v{E}' - \frac{\gamma^2}{\gamma+1} \left(\v{\beta} \cdot \v{E}' \right)\v{\beta}$	(1.18)
$\displaystyle \v{B}$	$\displaystyle =$	$\displaystyle \gamma \v{\beta} \times \v{E}'$	(1.19)

1.2.2.2.1 Longitudinal field and pure longitudinal momentum

It can be noticed that in the common situation where $\v{\beta}= \beta \v{e_z}$ , the field in the laboratory frame will be reduced to:

$\displaystyle E_z = \frac{\gamma +\gamma^2(1-\beta^2)}{\gamma +1} E'_z = E'_z$

(1.20)

Thus, the two fields are the same for any velocity in the hypothesis of pure longitudinal momentum.

Emmanuel Branlard
http://emmanuel.branlard.free.fr/