Special theory of relativity

The special theory of relativity, proposed by Albert Einstein in 1905, revolutionized our understanding of space and time. It reveals that time is not absolute, lengths can contract, and the speed of light remains constant for all observers — no matter how fast they are moving. At the heart of the theory lie two simple yet profound postulates that together overturn centuries of classical intuition and lay the foundation for modern physics.

In this page, you will explore the key concepts of special relativity, including inertial reference frames, the two postulates, and their remarkable consequences — such as time dilation, length contraction, mass-energy equivalence, and relativistic momentum and energy.

Inertial reference frame

We measure quantities such as displacement, velocity, and acceleration of an object relative to a reference called a reference frame, or frame of reference. We typically use an x-y-z coordinate system to define this reference frame.

For example, to find the velocity of a car moving along a road, we choose a fixed point described by the coordinate system and measure the car's displacement from that point. Dividing this displacement by the time taken gives us the velocity. Since the fixed point is on the ground, we say the ground is the reference frame.

Now consider a person running inside a train. If a passenger seated in the train measures the runner's velocity, they would choose a fixed point within the train for their measurement — making the train the reference frame.

A reference frame that is either at rest or moving at a constant velocity is called an inertial reference frame. In other words, an inertial reference frame is one that does not accelerate.

Postulates of special theory of relativity

There are two postulates of special theory of relativity. They are called postulate 1 and postulate 2.

Postulate 1: The laws of physics have the same form in all inertial reference frames.

Consider a car moving along a straight road, where the ground — which is at rest — serves as the reference frame. To study the car's motion, we apply the kinematic equations. Now consider a warship moving at a constant velocity across the ocean, with a car in motion on its landing platform. Since the ship moves at a constant velocity, it qualifies as an inertial reference frame, and an observer on board can apply the same kinematic equations to study the car's motion. This is precisely what the first postulate states: the laws of physics hold in the same form in every inertial reference frame, whether we are studying kinematics or dynamics.

This postulate also shows up in everyday life. Any activity you perform on the ground can be carried out identically inside an airplane or train moving at a constant velocity — with no observable difference. For instance, if you drop a ball while standing on the ground, it falls straight down and lands directly below the point of release. The same thing happens if you drop a ball while standing inside a train moving at a constant velocity: it falls straight down and lands directly below the release point, just as it would on the ground.

Postulate 2: The speed of light is a universal constant that does not change with the speed of the source or the observer.

Consider a cannon at rest that fires cannonballs, each leaving the barrel at some speed $v_b$. Now suppose the cannon begins moving in the same direction as the cannonballs at some speed $v_c$. Since both move in the same direction, their speeds add up, making the total speed of the cannonballs $v_b+v_c$. Conversely, if the cannon moves in the opposite direction, the speed of the cannonballs becomes $v_b-v_c$. This shows that the speed of the cannonballs depends on the speed of their source — the cannon.

This dependence on the source holds not only for objects like cannonballs, but also for waves such as sound. The observed speed also depends on the observer. For instance, when driving on a freeway, an oncoming car appears to approach faster than its actual speed, while a car moving away appears slower. In both cases, the observed speed shifts depending on the relative motion between the observer and the object.

Light, however, behaves differently. According to the second postulate, its speed is independent of both the source and the observer — it is always the same.

This postulate defies everyday intuition. The reason we don't notice this in daily life is that the speeds of ordinary objects are negligibly small compared to the speed of light, so relativistic effects go undetected. Nevertheless, the theory of relativity resolved several longstanding questions in electromagnetic theory that had previously gone unanswered.

Simultaneity

One of the key results of special relativity is that time is not an absolute quantity. For example, two events that are simultaneous for one observer may not be simultaneous for another. To understand this, consider the following thought experiment.

A train moves to the right at some velocity, with a passenger seated at its center. A second person stands on a platform outside. As the train approaches and the two observers momentarily align, a flash of light is emitted from the center of the train. The passenger inside the train observes how long it takes the light to reach the front and back ends. Since both ends are equidistant from the center, the light takes the same amount of time to reach each end. For the passenger, therefore, the two events — light arriving at the front and light arriving at the back — occur simultaneously.

Light from the center of a train reaching both ends at the same time for a passenger inside
Observer in the train. A flash of light is given off at t =0.

For the observer on the platform, however, the situation looks different. The back end of the train is moving toward the light, and the light is moving toward it — so the two close the distance between them more quickly. At the same time, the front end is moving away from the light, forcing the light to chase it. As a result, the light reaches the back end before it reaches the front. For the platform observer, then, the two events are not simultaneous.

Light reaching the back of a moving train before the front, as seen by a platform observer
Observer on the platform. A flash of light is given off at t =0.

Time dilation

Let us consider a spaceship traveling past the earth at a speed $v$. In the spaceship, there is a laser, a mirror, a light receiver and a clock. Let the distance between the mirror and the receiver be d. Now, a person in the spaceship turns on the laser and the light from the laser hits the mirror and reflects back to the receiver.

Light traveling straight up to a mirror and back to a receiver inside a spaceship

For the observer on the spacecraft, the light travels a distance of 2d, before reaching the receiver. If $\Delta t_0$ is the time, the light takes to travel from the laser to the receiver, then

$\Delta t_0=\dfrac{2d}{c}$.

This is the time, the observer in the spaceship measures with his/her clock.

The same light following a longer diagonal path as seen by an observer on Earth

Now, consider, there is a person on the earth watching this experiment. But he/she sees a different path for the light as the spaceship and everything in there is moving. For him/her, the light travels the diagonal paths as shown in the figure above. It travels a distance $D$ from the laser to the mirror and another $D$ from the mirror to the receiver. So, the total distance traveled by the light is $2D$. If $\Delta t$ is the time taken by the light to travel from the laser to the mirror and to the receiver, then

$\Delta t=\dfrac{2D}{c}$

This is the time, the observer on the earth measures in his/her clock.

Considering, the distance traveled by the spaceship during the time $\Delta t$ as $2x$, we have

$2x=v \Delta t$

Next step is relating the two times, $\Delta t_0$ and $\Delta t$.

From the geometry of the figure (taking one of the right angled triangle), using the Pythagorean theorem, we can write

$d=\sqrt{D^2-x^2}$

Substituting $D$ and $x$ from the above equations, we get

$d=\sqrt{\dfrac{c^2\Delta t^2}{4}-\dfrac{v^2\Delta t^2}{4}}$

or

$d=\dfrac{\Delta t}{2}\sqrt{c^2-v^2}$

Substituting $d$ from the very first equation,

$\dfrac{c\Delta t_0}{2}=\dfrac{\Delta t}{2}\sqrt{c^2-v^2}$

Solving for $\Delta t$,

$\Delta t=\dfrac{\Delta t_0}{\sqrt{1-\dfrac{v^2}{c^2}}}$

To write the above equation in a compact form, we take,

$\dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}=\gamma$

Therefore,

$\Delta t=\gamma \Delta t_0$

When $v=0$, $\gamma = 1$.

For any non-zero velocity, $\gamma>1$, and $\Delta t>\Delta t_0$.

So the time interval between two events (here it is sending and receiving light) is greater for the observer on earth than for the observer on the moving spaceship. In other words, a moving clock runs slower than a clock that is at rest. This is known as time dilation.

Time dilation is noticeable only if the speed, $v$ is close to the speed of light. At the ordinary slow speeds, such as the speed of a car, or an airplane, the value of $\gamma$ is close to 1, so the time you measure in a car or in an airplane is the same as the time that you measure on the ground. So, time dilation is not noticeable at ordinary speeds.

Proper time

The time interval, $\Delta t_0$ is called proper time. When using the time dilation equation, you may be confused, which one is the proper time and which is not. Proper time is the time interval measured with a clock that is present in the place where both the events take place. In sending and receiving of light that we discussed above, sending is the first event and receiving is the second event. The clock in the spaceship was present in the place where both the events took place, so this clock measures the proper time.

Space travel

The kinematic equations that you learned in physics 1 come under the classical physics or non-relativity. Based on these equations, if you want to travel from Earth to reach a distant object, such as a star that is at a distance of 100 light years from the earth, it will take about 100 years to reach the star. But according to the theory of relativity, if we travel at a speed of $0.999c$, it will take less than 5 years. The only problem is to achieve such a high speed.

GPS navigation system and relativity

Global positioning system or GPS, uses radio signals and satellites to find the location on or near the earth. There is a network of about 30 satellites that revolve around the earth at an altitude of about 20200 km from the Earth's surface with an orbital period of about 12 hours. The satellites are placed in orbits such that at least four satellites are visible from any point on earth in a given time. Each satellite has an atomic clock with an accuracy of 1 nano second and with that they send out precisely timed radio signals. The GPS instrument that receives the signals compares the time difference between the signals and determines the position on the earth. Since the satellites are in motion, their clocks run slower than the Earth's bound clocks, according to the special theory of relativity. Calculation shows that a clock on a satellite falls behind a clock on the earth by 7 microseconds per day. This results in a time delay in receiving the signals from different satellites by the GPS, which leads to position inaccuracy. So, all the GPS systems take into account the time dilation and correct the time accordingly so that we can have an accurate GPS.

Length contraction

Assume that there is a spacecraft travels from Earth to another planet, at a speed, $v$. Considering, $\ell_0$ is the distance between the Earth and the planet, the time to arrive the planet is

$\Delta t=\dfrac{\ell_0}{v}$.

This is the time required based on an observer on the Earth. Note that this time is not the proper time as the Earth bound clock is not present at the places where the events (the departure and the arrival) take place.

But for an observer on the spacecraft, the time is shorter due to the time dilation, as the moving clock runs slower. The time as measured by the observer on the spacecraft according to time dilation is

$\Delta t_0=\dfrac{\Delta t}{\gamma}$.

Substituting, $\Delta t$ from the previous equation,

$\Delta t_0=\dfrac{\ell_0}{v\gamma}$.

Now, calculating the travel distance, $\ell$ as observed by the person in the spaceship with the time $\Delta t_0$,

$\ell=v\Delta t_0$

Substituting, $\Delta t_0$,

$\ell=v\dfrac{\ell_0}{v\gamma}$

or

$\ell=\ell_0/\gamma$.

Since $\gamma\gt 1$, $\ell\lt\ell_0$.

The above equation tells us that the distance shortens with speed. This not only applies to distances but to the length of objects too. So, the above equation tells us that the length of a moving object is measured to be shorter than its length measured when the object is in rest. This is called length contraction.

It is important to note that the length contracts only in the direction of motion of the object. And, there will be no change in dimension of the object that is perpendicular to the direction of motion.

Proper length

The length $\ell_0$ is called proper length. Proper length is the length measured by an observer who is at rest relative to the object. For example, if a person measures the length of a table with a ruler, then that measurement is the proper length as there is no relative motion between the person and the table.

Time as a fourth dimension

You saw that at very high speeds called relativistic speeds, for an observer who is not moving with the object, length of the object shortens and the object gains time. Length is a dimension. Other dimensions are width and height and these can also change depends on the direction of motion of the object. So, time behaves like one of the dimension as it changes with speed like a dimension of an object. This brought to the idea of adding the time as the fourth dimension, and we have a four dimensional space-time.

Relativistic momentum

You have learned that the magnitude of the momentum of an object moving with velocity $v$ is

$p=mv$

This equation is valid only if the speed of the object is very small compared to the speed of light. When an object is moving at a relativistic speed, we need to use the following equation to find the momentum of the object:

$p=\gamma mv$.

Note that, if the speed of the object is much smaller compared to the speed of light, then the above equation reduces to the regular momentum equation. So, the relativistic momentum equation is valid for all the velocities.

The ultimate speed

Another important result of the special theory of relativity is that the speed of light is the ultimate speed. No object can travel with a speed equal or greater than the speed of light in vacuum. i.e., all objects must travel with a speed less than the speed of light.

Relativistic kinetic energy

You learned that the kinetic energy of an object moving at velocity $v$ is

$KE=\dfrac{1}{2}mv^2$

Like the momentum, this equation is also valid only at smaller speeds. At relativistic speeds, kinetic energy of the object is

$KE=(\gamma-1)mc^2$.

This kinetic energy equation is valid for all velocities.

Mass energy equivalence

We can rewrite the relativistic kinetic energy equation as

$KE=\gamma mc^2-mc^2$.

There are two terms on the right hand side, the second term is called the rest energy of the object as it is independent of the velocity of the object. If you add the rest energy to the kinetic energy of the object, you will get the total energy of the object (assuming the object has no potential energy).

i.e., the total energy of the object is

$E=KE + mc^2$

If the object is at rest, then $KE=0$ and we have

$E=mc^2$.

This is the famous Einstein’s equation.

Addition of relativistic velocities

When objects move towards or away from each other with relativistic speeds, i.e., at speed close to the speed of light, then the simple rule of addition or subtraction of velocities does not work. Assume that there is a spaceship moving away from the earth at a speed, $v$ relative to the earth. Now, if the spaceship fires a rocket at a speed, $u'$, with respect to the spaceship in the same direction as the spaceship's motion, then the speed of the rocket with the earth, $u$, according to the special theory of relativity is

$u=\dfrac{v+u'}{1+\dfrac{vu'}{c^2}}$.

If the rocket is moving opposite to the direction of the spaceship, then $u'$ should be replaced with $-u'$.