Special Theory of Relativity Formula

In the special theory of relativity, the traditional concept of absolute universal time is replaced with the concept of time that is based on reference frame and physical position. Also known as special relativity, it is a physical theory that describes the interaction between space and time. It states that the length, time, momentum, and energy are all affected by the velocity of one reference frame relative to another. It describes the relativity of motion, specifically the velocity of anything travelling faster than the speed of light. For instance, an object travelling in space near the speed of light will have a different length, time, momentum, and energy than an object travelling on the ground.

Special Theory of Relativity Formula

γ = 1 / √(1 – (v/c)²)

where,
γ is the relativistic factor,
v is the velocity of object,
c is the speed of light with a constant value of 3 × 10⁸ m/s.

Do Check,

• Time Dilation Formula
• Length Contraction Formula

Solved  Examples on Special Theory of Relativity Formula

Example 1: A particle has a velocity of 0.54c, where c is the speed of light. Calculate the value of the relativistic factor γ.

Solution:

We have,

v = 0.54c

Using the formula we get,

γ = 1 / √(1 – (v/c)²)

  = 1 / √(1 – (0.54c/c)²)

  = 1 / √(1 – (0.54)²)

  = 1/√0.7084

  = 1/0.84166

  = 1.1881

Example 2: A particle has a velocity of 0.26c, where c is the speed of light. Calculate the value of the relativistic factor γ.

Solution:

We have,

v = 0.26c

Using the formula we get,

γ = 1 / √(1 – (v/c)²)

   = 1 / √(1 – (0.26c/c)²)

   = 1 / √(1 – (0.26)²)

   = 1/√0.9324

   = 1/0.96

   = 1.042

Example 3: A particle has a velocity of 0.78c, where c is the speed of light. Calculate the value of the relativistic factor γ.

Solution:

We have,

v = 0.78c

Using the formula we get,

γ = 1 / √(1 – (v/c)²)

   = 1 / √(1 – (0.78c/c)²)

   = 1 / √(1 – (0.78)²)

   = 1/√0.3916

   = 1/0.625

   = 1.6

Example 4: A particle has a velocity of 0.86c, where c is the speed of light. Calculate the value of the relativistic factor γ.

Solution:

We have,

v = 0.86c

Using the formula we get,

γ = 1 / √(1 – (v/c)²)

   = 1 / √(1 – (0.86c/c)²)

   = 1 / √(1 – (0.86)²)

   = 1/√0.2604

   = 1/0.51

   = 1.96

Example 5: Calculate the velocity of a particle if its relativistic factor γ is 1.5. 

Solution:

We have,

γ = 1.5

Using the formula we get,

γ = 1 / √(1 – (v/c)²)

√(1 – (v/c)²) = 1/γ

1 – (v/c)² = 1/(1.5)²

1 – (v/c)² = 0.444

(v/c)² = 1 – 0.444 = 0.556

(v/c) = √(0.556)

v/c = 0.7456

v = c × 0.7456 = 3.0 × 10⁸ × 0.7456

v = 2.237 × 10⁸ m/s

Example 6: Calculate the velocity of a particle if its relativistic factor γ is 2.

Solution:

We have,

γ = 2

Using the formula we get,

γ = 1 / √(1 – (v/c)²)

√(1 – (v/c)²) = 1/γ

1 – (v/c)² = 1/2²

(v/c)² = 0.75

v/c = √(0.75)

v = c × 0.886 = 3 × 10⁸ × 0.886

v = 2.59 × 10⁸ m/s

Example 7: Calculate the velocity of a particle if its relativistic factor γ is 1.3.

Solution:

We have,

γ = 1.3

Using the formula we get,

γ = 1 / √(1 – (v/c)²)

√(1 – (v/c)²) = 1/γ

1 – (v/c)² = 1/(1.3)²

(v/c)² = 0.743

v/c = √0.408 = 0.693

v = 0.693 × 3 × 10⁸

v = 2.079 × 10⁸ m/s