## What isÂ Lorentz Transformation?

Lorentz transformation is the relationship between two different coordinate frames that move at a constant velocity and are relative to each other. The name of the transformation comes from a Dutch physicist Hendrik Lorentz.

**There are two frames of reference, which are:**

**Inertial Frames**– Motion with a constant velocity**Non-Inertial Frames**– Rotational motion with constant angular velocity, acceleration in curved paths

Lorentz transformation is only related to change in the inertial frames, usually in the context of special relativity. This transformation is a type of linear transformation in which mapping occurs between 2 modules that include vector spaces. In linear transformation, the operations of scalar multiplication and additions are preserved. This transformation has a number of instinctive features, such as the observer that is moving at different velocity may measure elapsed times, different distances and ordering of events but the condition that needs to be followed is that the speed of light should be same in all the inertial frames.

Lorentz transformation can also include rotation of space, a rotation which is free of this transformation is called Lorentz Boost. The space-time interval which occurs between any two events is preserved by this transformation.

## Lorentz Transformation Formula

Following are the mathematical form of Lorentz transformation:

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

Where,

- (t,x,y,z) ans (t’,x’,y’,z’) are the coordinates of an event in two frames
- v is the velocity confined to x-direction
- c is the speed of light

## Lorentz Transformation Equations

In the reference frame â€˜Fâ€™ which is stationary, the coordinates defined are x, y, z, t. In another reference frame Fâ€™ which moves at a velocity v which is relative to F and the observer defines coordinates in this moving reference frames as xâ€™ , yâ€™ , zâ€™ , tâ€™. In both the reference frames the coordinate axis are parallel and they remain mutually perpendicular. The relative motion is along the xxâ€™ axes. At t = tâ€™ = 0, the origins in both reference frames are same (x,y,z) = (xâ€™,yâ€™,zâ€™) = (0,0,0)

If the events x, y, z, t are recorded in reference frames F then in Fâ€™ these coordinates have the following value:

\(x^{‘}=\frac{x-vt}{\sqrt{1-\frac{v^{2}}{c^{2}}}}\)

\(t^{‘}=\frac{t-\frac{vx}{c^{2}}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}\)

\(y’=y\)

\(z’=z\)

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

Here lowercase gamma is the Lorentz factor

**Then the above equation becomes as follows and is known is Lorentz transformation:**

\(x’=\gamma (x-vt)\) |

\(t’=\gamma (t-\frac{vx}{c^{2}})\) |

yâ€™ = y |

zâ€™ = z |

**The above equations are collectively known as Galilean transformation.**

**Related Articles:**

## Lorentz Transformation Example

**Q1. Calculate the time interval of the signal using Lorentz transformation when a spacecraft S’ is crossed by another spacecraft S at a speed of c/2 on their way to alpha centauri. When spacecraft S crosses S’, captain of S’ sends a signal that lasts for 1.2s. Fine time interval of the single.Â **

**Ans:Â **Given:

Î”t’ =Â \(t_{2}^{‘}-t_{1}^{‘}\) = 1.2s

Î”x’ =Â \(x_{2}^{‘}-x_{1}^{‘}\) = 0

The time signal starts and ends at (x’,\(t_{1}^{‘}\))

Using first Lorentz transformation equation we get,

Î”t = t_{2} – t_{1}

Î”x = x_{2} – x_{1}

As the time interval of S’ is fixed, we can considerÂ Î”x’ = 0 and the time intervalÂ Î´t becomes:

\(\Delta t=\frac{\Delta t’}{\sqrt{1-\frac{v^{2}}{c^{2}}}}\)SubstitutingÂ Î”t’ = 1.2s we get

\(\Delta t=\frac{1.2s}{\sqrt{1-(\frac{1}{2})^{2}}}\)Î”t = 1.6s

### Physical Implications

The invariance of the speed of light is one of the critical requirements for Lorentz Transformation, this fact is also used in the derivation of this equation.

**There are 3 unintuitive but correct predictions of Lorentz transformations which are:**

- Length Contraction
- Time Dilation
- Relativity of Simultaneity