# Lorentz Transformation Derivation

The Lorentz transformations in physics are defined as a one-parameter family of linear transformations. These transformations are named after Dutch physicist Hendrik Lorentz.

Following are the mathematical form of Lorentz transformations:

${t}’=\gamma (t-\frac{vx}{c^{2}})$ ${x}’=\gamma (x-vt)$ ${y}’=y$ ${z}’=z$

Where,

(t,x,y,z) and (t’,x’,y’,z’): coordinates of an event in two frames

v: velocity confined to x-direction

c: speed of light

$\gamma =(\sqrt{1-\frac{v^{2}}{c^{2}}})^{-1}$ : Lorentz factor

Frames of reference are divided into two groups inertial which includes relative motion with constant velocity and non-inertial which includes rotational motion with constant angular velocity and Lorentz transformations are used for inertial frames.

The Lorentz transformations are derived from Galilean transformation as it fails to explain why observers moving at different velocities measure different distance, a different order of events even after the same speed of light in all inertial reference frames.

## Lorentz transformation derivation

From Galilean transformation below which was studied for a beam of light, we can derive Lorentz transformations:

${x}’=a_{1}x+a_{2}t$ ${y}’=y$ ${z}’=z$ ${t}’=b_{1}x+b_{2}t$

The origin of the primed frame x’ = 0, with speed v in unprimed frame S. For the beam of light, let x = vt is the location at time t in unprimed frame S.

$∴ {x}’=0=a_{1}x+a_{2}t\rightarrow x=-\frac{a_{2}}{a_{1}}t=vt$

Where,

$\frac{a_{2}}{a_{1}}=-v$ ${x}’=a_{1}x+a_{2}t=a_{1}(x+\frac{a_{2}}{a_{1}}t)=a_{1}(x-vt)$ (rewriting the equation)

$a_{1}^{2}(x-vt^{2})+{y}’^{2}+{z}’^{2}-c^{2}(b_{1}x+b_{2}t)^{2}=x^{2}+y^{2}+z^{2}-c^{2}t^{2}$ $a_{1}^{2}x^{2}-2a_{1}^{2}xvt+a_{1}^{2}v^{2}t^{2}-c^{2}b_{1}^{2}x^{2}-2c^{2}b_{1}b_{2}xt-c^{2}b_{2}^{2}t^{2}=x^{2}-c^{2}t^{2}$ $(a_{1}^{2}-c^{2}b_{2}^{1})x^{2}=x^{2} or a_{1}^{2}-c^{2}b_{1}^{2}=1$ $(a_{1}^{2}v^{2}-c^{2}b_{2}^{2})t^{2}=-c^{2}t^{2} or c^{2}b_{2}^{2}-a_{1}^{2}v^{2}=c^{2}$ $(2a_{1}^{2}v+2b_{1}b_{2}c^{2})xt=0 or b_{1}b_{2}c^{2}=-a_{1}^{2}v$ $b_{1}^{2}c^{2}=a_{1}^{2}-1$ $b_{2}^{2}c^{2}=c^{2}+a_{1}^{2}v^{2}$ $b_{1}^{2}b_{2}^{2}c^{4}=(a_{1}^{2}-1)(c^{2}+a_{1}^{2}v^{2})=a_{1}^{4}v^{2}$ $a_{1}^{2}c^{2}-c^{2}+a^{4}v^{2}-a_{1}^{2}v^{2}=a_{1}^{4}v^{2}$ $a_{1}^{2}c^{2}-a_{1}^{2}v^{2}=c^{2}$ $a_{1}^{2}(c^{2}-v^{2})=c^{2}$ $a_{1}^{2}=\frac{c^{2}}{c^{2}-v^{2}}=\frac{1}{1-\frac{v^{2}}{c^{2}}}$ $a_{2}=-v\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$ $b_{1}^{2}c^{2}=\frac{1}{1-\frac{v^{2}}{c^{2}}}-1$ $b_{1}^{2}c^{2}=\frac{1-(1-\frac{v^{2}}{c^{2}})}{1-\frac{v^{2}}{c^{2}}}=\frac{\frac{v^{2}}{c^{2}}}{1-\frac{v^{2}}{c^{2}}}=\frac{v^{2}}{c^{2}}.\frac{1}{1-\frac{v^{2}}{c^{2}}}$ $b_{1}^{2}=\frac{v^{2}}{c^{4}}.\frac{1}{1-\frac{v^{2}}{c^{2}}}$ $b_{1}=-\frac{v}{c^{2}}.\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$ $b_{2}^{2}c^{2}=c^{2}+v^{2}.\frac{1}{1-\frac{v^{2}}{c^{2}}}=\frac{c^{2}(1-\frac{v^{2}}{c^{2}})+v^{2}}{1-\frac{v^{2}}{c^{2}}}=\frac{c^{2}-v^{2}+v^{2}}{1-\frac{v^{2}}{c^{2}}}=\frac{c^{2}}{1-\frac{v^{2}}{c^{2}}}$ $b_{2}^{2}=\frac{1}{1-\frac{v^{2}}{c^{2}}}$ $b_{2}=\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$ (which is similar to a1)

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

We can also write it as:

$a_{1}=\gamma$ $a_{2}=-\gamma v$ $b_{1}=-\frac{v}{c^{2}}\gamma$ $b_{2}=\gamma$

Following are the final form of Lorentz transformations:

$∴ {x}’=\gamma (x-vt)$ ${y}’=y$ ${z}’=z$ ${t}’=\gamma (t-\frac{v}{c^{2}}x)$

Thus, above is the Lorentz transformation derivation. To know more on other Physics related topics, stay tuned with BYJU’S.

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