Special Theory Of Relativity Formula

Special Theory of Relativity Formula

Our understanding of space, time, and the nature of reality has been completely transformed by Albert Einstein’s 1905 proposal of the Special Theory of Relativity. It introduced profound concepts that fundamentally altered the way we perceive the universe. The Special Theory of Relativity has had profound implications for our understanding of the universe, from the behavior of particles at high speeds to the structure of space-time itself. It forms the foundation of modern physics and has led to advancements in fields such as cosmology, particle physics, and technology. In this post, student will learn in detail about special theory of relativity, its postulates and examples.

What is Special Theory of Relativity?

The Special Theory of Relativity is based on the principle of relativity, which states that the laws of physics are the same for all observers in uniform motion relative to each other. This principle implies that there is no privileged frame of reference in the universe.

Consider a scenario where two observers, Alice and Bob, are moving relative to each other in deep space. Alice is stationary relative to a distant star, while Bob is moving past her at a significant fraction of the speed of light (c). Alice observes that Bob’s clock appears to be ticking more slowly than her own clock. This is an example of time dilation, where time appears to pass more slowly for the moving observer (Bob) relative to the stationary observer (Alice).

Similarly, Bob observes that Alice’s ruler appears to be contracted in the direction of motion compared to his own ruler. This is an example of length contraction, where lengths appear shorter for objects in motion relative to an observer.

These examples illustrate how the Special Theory of Relativity fundamentally alters our understanding of space and time.

Constancy of the Speed of Light

One of the most fundamental postulates of the theory is that the speed of light in a vacuum is constant for all observers, regardless of their relative motion. This means that the speed of light (c) is the same for all observers, regardless of their motion relative to the source of light.

Time Dilation

According to the theory, time is not absolute but is instead relative to the observer’s frame of reference. Time dilation occurs when an observer moving relative to another observes that time appears to pass more slowly for the moving observer compared to the stationary observer. This effect becomes significant as objects approach the speed of light.

Length Contraction

Similarly, lengths contracted along the direction of motion relative to an observer. This means that objects in motion appear shorter in the direction of their motion than when they are at rest, as observed by a stationary observer.

Special Theory of Relativity Formula

The Special Theory of Relativity introduced by Albert Einstein in 1905 doesn’t have a single formula but rather a set of mathematical expressions derived from its fundamental postulates. However, some of the key formulas associated with the Special Theory of Relativity include:

Time Dilation Formula

The formula for time dilation describes how time appears to pass differently for observers in relative motion. It is given by:
\[ t’ = \frac{t}{\sqrt{1 – \frac{v^2}{c^2}}} \]
Where:
\( t’ \) is the dilated time observed by the moving observer.
\( t \) is the proper time experienced by the stationary observer.
\( v \) is the relative velocity between the two observers.
\( c \) is the speed of light in a vacuum.

Length Contraction Formula

The formula for length contraction describes how lengths appear contracted along the direction of motion for observers in relative motion. It is given by:
\[ L’ = L \sqrt{1 – \frac{v^2}{c^2}} \]
Where:
\( L’ \) is the contracted length observed by the moving observer.
\( L \) is the proper length experienced by the stationary observer.
\( v \) is the relative velocity between the two observers.
\( c \) is the speed of light in a vacuum.

Relativistic Energy-Momentum Relation

Einstein’s famous equation \( E = mc^2 \) is a consequence of the Special Theory of Relativity. It relates energy (\( E \)) to mass (\( m \)) and the speed of light (\( c \)). This equation signifies the equivalence of mass and energy and is a cornerstone of modern physics.

Solved Examples on Special Theory of Relativity Formula

Example 1: An astronaut travels in a spaceship at \( 0.8c \) (80% of the speed of light) relative to Earth. If the astronaut’s journey lasts for 5 years according to their own clocks (proper time), how much time has passed on Earth?

Solution:

Given:
\( v = 0.8c \)
\( t’ = 5 \) years (proper time)
\( c = 299,792,458 \) m/s (speed of light)

Using the time dilation formula:
\[ t = t’ \times \sqrt{1 – \frac{v^2}{c^2}} \]

Substitute the given values:
\[ t = 5 \times \sqrt{1 – \frac{(0.8c)^2}{c^2}} \]
\[ t = 5 \times \sqrt{1 – 0.64} \]
\[ t = 5 \times \sqrt{0.36} \]
\[ t ≈ 5 \times 0.6 \]
\[ t ≈ 3 \] years

According to Earth’s clocks, approximately 3 years have passed during the astronaut’s 5-year journey.

Example 2: A spaceship is observed to have a length of 100 meters when at rest. If the spaceship travels at \( 0.9c \) (90% of the speed of light), what is its observed length from Earth?

Solution:

Given:
\( L = 100 \) meters (proper length)
\( v = 0.9c \)
\( c = 299,792,458 \) m/s (speed of light)

Using the length contraction formula:
\[ L’ = L \times \sqrt{1 – \frac{v^2}{c^2}} \]

Substitute the given values:
\[ L’ = 100 \times \sqrt{1 – \frac{(0.9c)^2}{c^2}} \]
\[ L’ = 100 \times \sqrt{1 – 0.81} \]
\[ L’ = 100 \times \sqrt{0.19} \]
\[ L’ ≈ 100 \times 0.436 \]
\[ L’ ≈ 43.6 \] meters

The observed length of the spaceship from Earth is approximately 43.6 meters when it is traveling at \( 0.9c \).

Example 3: A particle with a rest mass of \( m_0 = 0.5 \) kg is accelerated to a velocity of \( v = 0.6c \) (60% of the speed of light). Calculate the relativistic kinetic energy and the total energy of the particle.

Solution:

Given:
Rest mass, \( m_0 = 0.5 \) kg
Velocity, \( v = 0.6c \)
Speed of light, \( c = 299,792,458 \) m/s

First, let’s calculate the relativistic kinetic energy using the formula:
\[ K = (\gamma – 1) \cdot m_0 \cdot c^2 \]
Where \( \gamma \) is the Lorentz factor, given by \( \gamma = \frac{1}{\sqrt{1 – \frac{v^2}{c^2}}} \).

1. Calculate \( \gamma \):
\[ \gamma = \frac{1}{\sqrt{1 – \frac{(0.6c)^2}{c^2}}} \]
\[ \gamma = \frac{1}{\sqrt{1 – 0.36}} \]
\[ \gamma = \frac{1}{\sqrt{0.64}} \]
\[ \gamma = \frac{1}{0.8} \]
\[ \gamma = 1.25 \]

2. Calculate relativistic kinetic energy (\( K \)):
\[ K = (1.25 – 1) \times 0.5 \times (299,792,458)^2 \]
\[ K = (0.25) \times 0.5 \times (8.98755179 \times 10^{16}) \]
\[ K = 4.493775895 \times 10^{16} \] Joules

Next, let’s calculate the total energy (\( E \)) of the particle using the relativistic energy-momentum relation:
\[ E = \gamma \cdot m_0 \cdot c^2 \]

3. Calculate total energy (\( E \)):
\[ E = 1.25 \times 0.5 \times (299,792,458)^2 \]
\[ E = 1.25 \times 0.5 \times (8.98755179 \times 10^{16}) \]
\[ E = 5.617219937 \times 10^{16} \] Joules

The relativistic kinetic energy of the particle is \( 4.493775895 \times 10^{16} \) Joules, and the total energy is \( 5.617219937 \times 10^{16} \) Joules.