Results of Galilean Transformation equations can not be applied for the objects moving with a speed comparative to the speed of the light.

Therefore new transformations equations are derived by Lorentz for these objects and these are known as Lorentz transformation equations for space and time.

Let there are two inertial frames of references S and S’. S is the stationary frame of reference and S’ is the moving frame of reference. At time t=t’=0 that is in the start, they are at the same position that is Observers O and O’ coincides. After that S’ frame starts moving with a uniform velocity v along x axis.

Let an event happen at position P in the frame S’. The coordinate of the P will be x’ according to the observer in S’ and it will be x according to O in S.

The frame S’ has moved a distance “vt” in time t (refer figure).

What should be the relation between x and x’? As we can see from the figure that from frame S’

x’ α x – vt

or x’ = k (x – vt) (1)

where k is constant of proportionality that we will determine.

Similarly from frame S

x = k(x’ + vt’) (2)

Put equation (1) in (2)

x = k[k(x – vt) + vt’]

or x/k = kx – kvt + vt’

or vt’ = x/k – kx + kvt

or t’ = x/kv – kx + kvt

or t’ = kt – kx (1 – 1/k^{2})/v (3)

Similarly from frame S, time t will be

t = kt’ + kx’ (1 – 1/k^{2})/v (4)

(This equation can be derived by putting equation 2 in 1 and then solving.)

**Calculation of k**:

Let us suppose a flash of light is emitted from the common origin of S and S’ at time t=t’=0. From Einstein’s 2^{nd} second postulate, the flash of light travels with the velocity of light c and which remains same in both the frames.

After sometime, the position of the flash of the light as seen from observer O will be

x = ct

And as seen from O’ will be

x = ct’ (Here the form of Physics law is same that is position = (velocity)(time) from Einstein 1^{st} postulate)

Put these two values in equation (1) and (2) respectively, we get

ct’ = k (ct – vt) = kt (c –v)

and ct = kt’ (c + v)

Multiply above two equations

c^{2}tt’ = k^{2}tt’(c^{2} – v^{2})

or k^{2} = c^{2}/(c^{2} – v^{2})

or k^{2} = 1/ (1- v^{2}/c^{2}) (5)

or k = 1/√(1- v^{2}/c^{2}) (6)

The k is known as relativistic factor.

Substitute equation (6) in (1), we get

x’ = (x – vt)/(√1 – v^{2}/c^{2}) (7)

As it is assumed that frame S’ is moving only along x direction, therefore along y and z direction

y’ = y (8)

And z’ = z (9)

Equations 7-9 are known as Lorentz transformation equations for space.

**Let us derive Lorentz transformation equation for time:**

Cross-multiply equation (5)

1/k^{2} = 1 – v^{2}/c^{2}

Or 1 – 1/k^{2} = v^{2}/c^{2}

Put the above equation in equation (3)

t’ = kt – kx(v^{2}/c^{2})/v

or t’ = k (t – kxv/c^{2})

Put value of k from equation 5 in above equation, we get

t’ = (t – kxv/c^{2})/ (√1 – v^{2}/c^{2}) (10)

Equation (10) is **Lorentz transformation equation for time. **

Equations 7 -10 are known as **Lorentz transformation equations for space and time. These are again rewritten below:**

x’ = (x – vt)/(√1 – v^{2}/c^{2})

y’ = y

z’ = z

t’ = (t – xv/c^{2})/(√1 – v^{2}/c^{2})

If the frame is changed (that is from S), then the equations are known as **Lorentz inverse transformation equations for space and time. These are given as: **

x = (x’+ vt’)/(√1 – v^{2}/c^{2})

y = y’

z = z’

t = (t’ + x’v/c^{2})/(√1 – v^{2}/c^{2})

**Special case:**

If v <<< c

Then Lorentz equations will become Galilean by neglecting v^{2}/c^{2} or v/c^{2} wherever necessary as shown below:

x’ = x – vt

y’ = y

z’ = z

t’ = t

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