Let us today discuss and derive an interesting concept in relativity called the length contraction. You must be astonished to know that when an object moves with a speed comparative to the speed of light, its length contracted or more appropriate term is that it is apparent length contraction in relativity.

Let us start to discuss and derive why the length contraction occur in relativity:

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. The speed v is relativistic speed that is comparable to the speed of the light.

Let an object is placed in the frame S’. The coordinate of the initial point (A) of the object will be x1 (see the second line till A from S in figure) according to the observer in S and the coordinate of the final point will be will be x2 according to same observer.

The coordinate of the initial point (A) of the object will be x’1 (see the second last line till A from S’ in figure) according to the observer in S’ and the coordinate of the final point will be will be x’2 according to same observer.

Therefore the length of the object as seen by observer O’ in s’ will be

L’ = x’2 – x’1 (1)

The length L’ is called the proper length of the object. **Proper length** is defined as the length of the object measured by the observer which is in the same frame in which the object is placed.

The apparent length of the object from frame S at any time t will be

L = x2 – x1 (2)

As we have already derived and discussed Lorentz transformation equations for space and time and now use Lorentz transformation equations for space, that is

x’1 = (x1 – vt)/(√1 – v^{2}/c^{2}) (3)

x’ 2= (x2– vt)/(√1 – v^{2}/c^{2}) (4)

By putting equations (3) and (4) in equation (1) and solving, we get

L’ = (x2 – x1)/ (√1 – v^{2}/c^{2})

Substitute equation (2) in above equation,

L’ = L/(√1 – v^{2}/c^{2})

Or Apparent length that is the length from frame S will be

L = L’(√1 – v^{2}/c^{2}) (5)

This is the relation of the length contraction in relativity.

**Why it known as length contraction,** if we solve (√1 – v^{2}/c^{2}), then the value will always be in decimal (except when v = c).

If we multiply L’ with a decimal value, then L will be lesser than L’. This is the reason that it is called **length contraction**.

**For example let us discuss a numerical:** If an object is moving with speed 0.8c and its length is 1 meter, then what will be its apparent length?

Solution:

Given v = 0.8c

Proper length L’ = 1 m

To calculate L

As L = L’(√1 – v^{2}/c^{2}),

If we substitute the given values, we get

L = 0.6 m

Thus L < L’

**Special Case:**

If v <<< c, then v^{2}/c^{2 } will be negligible in L’(√1 – v^{2}/c^{2}) and it can be neglected

Then equation (5) becomes

L = L’

Similarly time dilation in relativity is derived and discussed in one of my articles.

Lorentz contraction

In a passenger car, a light source is placed on the floor. Two light rays (frequency is constant) are being sent to the two mirrors on the roof and return (only once : light paths form letter V). Number of waves existing on the light paths is invariant. This passenger car is moving.

How the above is visible to an observer on the ground ? This picture will be incompatible with constancy of light speed, Lorentz contraction and relativity of simultaneity.

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