# I m confused in 2 query 1- describe lorentz transformation equation and 2- derive the lorentz transformation between 2 inertial systems in which 1 is in motion is both ques is same or diff if same help me in solving this if different also help me plz dont post any links becaz in links ans is not properly and diagram r not there means i m not able to see diagrams plz help me

Dear Student,

The Lorentz transformations are just the equations giving the four coordinate of an event in one inertial frame in terms of the coordinates of the same event in another inertial frame.
Consider a fixed frame of reference S with Cartesian coordinates (x,y,z,t) and a moving frame of reference  w.r.t. S with velocity v in x-direction. The frame of reference coincide at .
Then,

Derivation of lorentz in my opinion is not required until higher grade under grad studies. It can be approached in various methods, one involves modifying galilean transformation equations, another is bit shorter but needs understanding of tensor arithmetic.

## An expanding sphere of light

Deriving 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 model 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

$b_{1}^{2} b_{2}^{2} c^{4} = ( a^{2} - 1 )( c^{2} + a_{1}^{2} v^{2} ) = a_{1}^{4} v^{2}$

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

$\boxed{ b_{1} = - \frac{ v }{ c^{2} } \cdot \frac{ 1 }{ (1 - v^{2}/c^{2} ) } }$

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.

Regards.

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