Relativity: Physics unit 13 notes (FBISE class 11)

Introduction to relativity

Albert Einstein's theory of relativity transformed theoretical physics and astronomy during the 20th century. It superseded a 200-year-old theory of mechanics created by Isaac Newton. The ideas behind relativity can seem mysterious because they introduced new concepts such as four-dimensional spacetime, relativity of simultaneity, time dilation and length contraction.

The theory of relativity, given by Albert Einstein, includes two theories: special relativity and general relativity. Einstein published them in 1905 and 1915 respectively.

Special relativity (1905): Applies to all physical phenomena in the absence of gravity.
General relativity (1915): Explains the law of gravitation and its relation to the forces of nature.

Einstein explained the situations in which Newtonian physics might fail to deal successfully with phenomena, and in doing so he proposed revolutionary changes in human concepts of time, space and gravity.

Over the last century, many experiments have confirmed both special and general relativity. One of them is the observation that light from distant stars is deflected as it passes close to our Sun, which shows that gravity distorts or curves space. Instead of being an invisible force that attracts objects to one another, gravity is a curving or warping of space. The more massive an object, the more it warps the space around it, as shown in the cover picture of this unit (a massive sphere bending the grid of space around it, with the Earth following the curve).

Special relativity became an important and necessary tool in the new fields of nuclear physics, atomic physics and quantum mechanics. Nuclear power plants and nuclear weapons would be impossible without the knowledge that matter can be transformed into energy, so the theory has an enormous impact on the modern world. In this unit we discuss some important aspects of special relativity.

13.1Frame of referenceOften examined

Frame of reference: A coordinate system that can be used to determine the positions and velocities of objects within that frame.

For example, when a ball rolls in a street, you can say that the ball is moving because the frame of reference is the streets, whatever may be on the side of the street or the Earth itself. The origin, orientation and scale of a reference frame are specified by a set of geometric points whose positions are identified both mathematically and physically. All measurements of motion are made relative to a frame of reference, and different frames of reference can move relative to one another.

Why the frame must be named

Consider two persons A and B seated in a bus that is moving with a velocity $v$ , as shown in Figure 13.1. If we ask A about the velocity of B, A will say that B is at rest. If we ask the same question to a person C standing on the ground at the roadside, C will say that B is moving with velocity $v$ in the positive $X$ direction. So before specifying a velocity we have to specify the frame of reference.

Figure 13.1: two persons A and B are seated in a bus and the bus is moving with a velocity

v

. A sees B at rest; C sees B moving with velocity

v

.

The same person B has two different velocities because the two observers use two different frames.
Who is asked	Frame of reference	Velocity of B
A (sitting in the bus)	the bus	zero: B is at rest
C (standing on the road)	the ground	$v$ in the positive $X$ direction

Frames of reference are of two types: inertial and non-inertial.

Inertial reference frame

An inertial reference frame is a frame of reference in which Newton's law holds true: an object at rest remains at rest, and an object in motion remains in motion with a constant velocity on a straight line, unless acted upon by an external force.

Inertial frame of reference: A frame of reference that is at rest or moving with a uniform velocity along a straight line.

An inertial frame of reference does not accelerate. Any reference frame that moves with constant velocity relative to an inertial frame is itself inertial. The Earth can be considered an inertial reference frame for many experiments. The interior of a car moving along a road at constant velocity and the interior of a stationary house are examples of inertial reference frames.

Non-inertial reference frame

In a non-inertial frame of reference, objects experience acceleration even in the absence of applied forces, because the reference frame itself is accelerating relative to an inertial frame. Newton's laws do not hold in a non-inertial frame of reference.

Non-inertial frame of reference: A frame of reference that is accelerating.

An accelerating car is an example of a non-inertial reference frame.

Inertial and non-inertial frames compared.
	Inertial frame	Non-inertial frame
Motion of the frame	at rest, or moving with uniform velocity along a straight line	accelerating
Newton's laws	hold true	do not hold
Free object	stays at rest or keeps moving with constant velocity	seems to accelerate even with no applied force
Examples	a stationary house; the interior of a car moving at constant velocity; the Earth (for many experiments)	an accelerating car

13.2Special theory of relativityOften examined

In 1905, Albert Einstein introduced his famous theory of relativity. This theory deals with the structure of space-time. Einstein's special theory of relativity is based on two postulates.

1) The principle of relativity

Principle of relativity (first postulate): The laws of physics are the same for all observers in any inertial frame of reference.

As long as an object is moving in a straight line at a constant speed (that is, with zero acceleration), the laws of physics are the same for every observer. This implies that experiments performed in stationary and moving inertial frames of reference yield the same results. For example, it is impossible to determine experimentally whether an inertial reference frame is stationary or moving without observing it from an external frame of reference.

Consider two individuals, one standing in a stationary train and the other standing in a train moving at constant velocity $v$ , as shown in Figure 13.2. Both are observing the free fall of a ball under the action of gravity.

Figure 13.2: a ball falling freely under gravity in an inertial frame. In the stationary train and in the train moving at constant velocity, each person sees the ball fall along a vertical path.

Both individuals are in inertial frames of reference, where Newton's laws apply equally for both of them. Both persons observe the ball falling in a vertical path. If the trains are windowless, the passengers have no way of knowing whether their train is stationary or moving at constant velocity.

2) The principle of the constancy of the speed of light

Constancy of the speed of light (second postulate): The speed of light in vacuum is always constant for all observers, regardless of the relative velocity between the source of light and the observer measuring it.

No matter how fast an observer or the light source is moving, the measured speed of light always yields the same value:

c = 299792458 m s^{- 1}

This postulate has profound implications, particularly because it contradicts Newtonian mechanics. For instance, if you were travelling at $0.5 c$ in the same direction as a beam of light, Newtonian mechanics predicts that you would observe the speed of the light beam to be $0.5 c$ . According to special relativity, however, you would still observe the speed of light to be $c$ . This leads to a range of fascinating effects.

When a stationary person measures the speed of light emitted from a train at rest, the speed is simply $c$ . The speed of light remains $c$ even when the train is moving relative to the person, as shown in Figure 13.3 (a).

Figure 13.3 (a): the principle of the constancy of the speed of light. Whether the train is at rest (

v = 0

) or moving with velocity

v

, the person outside measures the light from the source on the train to travel at

c

.

The mirror thought experiment

A person on a train moving at the speed of light looks into a mirror placed at the front, as shown in Figure 13.3 (b). Will this person see a reflection of themselves in the mirror?

Figure 13.3 (b): a person on the train looks into the mirror placed in front. Light must travel from the person to the mirror and back to the eyes.

Reasoning step by step

Newtonian view: light moves at $c$ relative to the ground and the train, carrying the mirror, also moves at $c$ . The relative velocity between the light from the person and the mirror is zero.
So in Newtonian physics the light never reaches the mirror and the person would not see a reflection.
This contradicts the first postulate. A person in an inertial frame cannot tell whether the frame is stationary or moving at constant speed. If the reflection disappeared on a moving train, the person could tell that the train is not stationary, because a reflection would be visible if it were.
Special relativity asserts that the reflection is seen. The contradiction comes from Newtonian mechanics, and the thought experiment supports the constancy of the speed of light.
By the second postulate, light's speed is $c$ in every inertial frame. From the point of view of the person and the mirror in the moving train, the relative velocity of the light is still $c$ , so the light reaches the mirror and is reflected back to the person's eyes.

Therefore the person sees a reflection of themselves in the mirror.

Worked example: light from a moving spaceship

A spaceship moves at

0.6 c

relative to the ground and sends a pulse of light forward from its nose. (a) What speed would Newtonian mechanics predict for the pulse as measured from the ground? (b) What speed does the ground observer actually measure, and what does the astronaut measure?

A pulse leaves the nose of a ship moving at

0.6 c

. The ground observer measures the pulse at

c

(accent). The longer dashed line is the Newtonian sum

c + 0.6 c

, which is not what is measured.

Given

$v_{ship} = 0.6 c = 0.6 (3.00 \times 1 0^{8}) = 1.80 \times 1 0^{8} m s^{- 1}$ speed of the ship relative to the ground
$c = 3.00 \times 1 0^{8} m s^{- 1}$ speed of light

To find

(a) the Newtonian speed of the pulse
(b) the speed measured on the ground and in the ship

Newtonian mechanics simply adds the two speeds, as for a ball thrown forward from a moving train.
$c + v_{ship} = 3.00 \times 1 0^{8} + 1.80 \times 1 0^{8} = 4.80 \times 1 0^{8} m s^{- 1} = 1.6 c$
The second postulate says the speed of light has the same value in every inertial frame, whatever the motion of the source or the observer. So the ground observer measures the pulse at exactly $c$ .
$c = 3.00 \times 1 0^{8} m s^{- 1}$
The astronaut is also in an inertial frame, so the astronaut measures the same value.
$c = 3.00 \times 1 0^{8} m s^{- 1}$
Compare. The Newtonian answer $1.6 c$ is larger than $c$ and disagrees with the postulate, so it is the one that fails.

(a) 1.6 c (b) c = 4.80 \times 1 0^{8} m s^{- 1} (wrong, Newtonian) = 3.00 \times 1 0^{8} m s^{- 1} for both observers

13.3Consequences of the special theory of relativityOften examined

There are other surprising consequences of the special theory of relativity. They are summarised in the following five topics, each given its own section below.

Relativity of simultaneity: two events that are simultaneous for one observer may not be simultaneous for another observer in relative motion.
Mass-energy equivalence: mass and energy are equivalent, $E = m c^{2}$ .
Length contraction: objects appear shorter in the direction of their motion relative to the observer.
Time dilation: a moving clock runs more slowly than a clock at rest.
Mass variation: the mass of a moving object increases as its speed increases.

13.3Relativity of simultaneity

Relativity of simultaneity: Two events that are simultaneous for one observer may not be simultaneous for another observer in relative motion.

For example, consider a person standing next to a train track and compare the observations of a lightning storm made by this person with those of a person inside a moving train. The train is moving at nearly the speed of light.

Einstein imagined the following scenario, drawn in the figure below. The train is positioned at a point on the track equally between two trees. A bolt of lightning strikes both trees at the same time.

Lightning strikes two trees at the same time while the train passes the midpoint between them.

The two observers

The person standing beside the track is midway between the trees. The light from the two strikes reaches this person together, so this person receives the strike as simultaneous.
The person inside the train is moving toward one lightning bolt and away from the other. The light from the bolt ahead of the train has a shorter distance to cover, so it meets the person sooner.
So the person on the moving train sees the bolt ahead of the train first and the bolt behind the train later.

Einstein concluded that simultaneity is not absolute. Events that appear simultaneous to one observer may occur at different times for another observer. He realised that it is not the speed of light that changes but time itself that is relative. Time moves differently for objects in motion compared with those at rest. However, the speed of light remains constant and is observed to be the same by anyone, anywhere in the universe, regardless of their motion.

13.3Mass-energy equivalence

Energy and mass are equivalent and transmutable. This is expressed by the equation

E = m c^{2}

(13.1)

where $E$ is the energy, $m$ stands for the mass and $c$ for the speed of light.

Symbol	Meaning	SI unit
$E$	energy equivalent to the mass	$J$
$m$	mass	$kg$
$c$	speed of light in vacuum, $2.998 \times 1 0^{8} m s^{- 1}$	$m s^{- 1}$

Many scientists had observed that an object's mass increases with its velocity but never knew how to calculate it. This equation is the answer to their problem. It explains that the increase in the object's relativistic mass is equal to the kinetic energy divided by the square of the speed of light.

As the speed of light is very high, $c^{2}$ is enormous, so a tiny amount of mass is equivalent to a very large amount of energy. That is why atomic and hydrogen bombs are so powerful. The concept of mass defect in the atomic nucleus is also justified by the expression of mass-energy equivalence.

13.3Length contractionOften examined

Objects appear shorter in the direction of their motion relative to the observer.

Proper length ( $L_{0}$ ): The length of an object measured in its rest frame.
Length contraction: Observers in different reference frames that are in relative motion with the object always measure the length ( $L$ ) to be shorter than the proper length.

This phenomenon is described by the equation

L = L_{0} 1 - \frac{v ^{2}}{c ^{2}}

(13.2)

Symbol	Meaning
$L$	length measured by the observer who sees the object moving
$L_{0}$	proper length, measured in the rest frame of the object
$v$	speed of the object relative to the observer
$c$	speed of light

This effect, known as length contraction, occurs only in the direction of motion.

Length against speed

0.25

0.5

0.75

1

0.25

0.5

0.75

1

O

v / c

L / L_{0}

v = c

half the proper length

L = L_{0} 1 - v^{2} / c^{2}

The measured length

L

is a fraction of the proper length

L_{0}

. It is nearly

L_{0}

at everyday speeds and shrinks to zero as

v

approaches

c

.

Worked example: speed from a measured contraction

A spacecraft is

80 m

long when at rest. As it passes a station, the station measures its length as

48 m

. Find the speed of the spacecraft.

Given

$L_{0} = 80 m$ proper length, measured at rest with the craft
$L = 48 m$ length measured by the station

To find

$v$ as a fraction of $c$ and in $m s^{- 1}$

The rest length is the proper length $L_{0}$ . The station sees the shorter length $L$ .
$L = L_{0} 1 - \frac{v ^{2}}{c ^{2}}$
Divide both sides by $L_{0}$ to isolate the root.
$1 - \frac{v ^{2}}{c ^{2}} = \frac{L}{L _{0}} = \frac{48}{80} = 0.60$
Square both sides.
$1 - \frac{v ^{2}}{c ^{2}} = 0.36$
Solve for $v^{2} / c^{2}$ and take the root.
$\frac{v ^{2}}{c ^{2}} = 0.64 \Rightarrow \frac{v}{c} = 0.80$
Convert with $c = 3.00 \times 1 0^{8} m s^{- 1}$ .
$v = 0.80 c = 2.40 \times 1 0^{8} m s^{- 1}$
Check on the graph: the point $(0.80, 0.60)$ lies on the curve $L / L_{0} = 1 - v^{2} / c^{2}$ .

Reading the speed from the graph

0.25

0.5

0.75

1

0.25

0.5

0.75

1

O

v / c

L / L_{0}

0.60

0.80

(0.80, 0.60)

L / L_{0}

The measured fraction

L / L_{0} = 0.60

meets the curve at

v / c = 0.80

.

v = 0.80 c = 2.40 \times 1 0^{8} m s^{- 1}

13.3Time dilationOften examined

One of the key consequences of Einstein's special relativity is that time is experienced differently by moving objects. An object in motion undergoes time dilation, meaning that it experiences time more slowly compared with when it is at rest. Moving clocks are observed to tick more slowly than clocks that are stationary from the observer's perspective.

Proper time ( $t_{0}$ ): The time taken for an event to occur within its rest frame.
Time dilation: Observers in different reference frames in relative motion always measure the elapsed time ( $t$ ) to be longer than the proper time.

This is expressed by the equation

t = \frac{t _{0}}{1 - \frac{v ^{2}}{c ^{2}}}

(13.3)

Symbol	Meaning
$t$	time interval measured by the observer who sees the clock or event moving
$t_{0}$	proper time, measured in the rest frame of the event
$v$	speed of the event relative to the observer
$c$	speed of light

Measured time against speed

0.25

0.5

0.75

1

2

3

4

5

6

O

v / c

t / t_{0}

v = c

t = t_{0}

t = \frac{t _{0}}{1 - v ^{2} / c ^{2}}

The measured time

t

is always longer than the proper time

t_{0}

. It is almost equal to

t_{0}

at small

v

and grows without limit as

v

approaches

c

.

Worked example: a moving clock falls behind

A clock on a spaceship ticks once every

1.00 s

in the ship's own frame. The ship passes the Earth at

0.60 c

. (a) What interval between ticks do observers on Earth measure? (b) How many seconds does the ship's clock record while Earth clocks record one hour?

Given

$t_{0} = 1.00 s$ proper time: between two ticks at the same place on the ship
$v = 0.60 c$

To find

(a) $t$ , the interval measured on Earth
(b) the ship clock reading for $3600 s$ of Earth time

(a) Interval between ticks

The two ticks happen at the same place on the ship, so $t_{0} = 1.00 s$ is the proper time. Earth measures the longer time $t$ .
$t = \frac{t _{0}}{1 - v ^{2} / c ^{2}}$
Put in $v / c = 0.60$ .
$t = \frac{1.00}{1 - 0.36} = \frac{1.00}{0.80} = 1.25 s$

(b) Ship clock during one Earth hour

One Earth hour holds this many ticks of the ship clock, because each tick takes $1.25 s$ of Earth time.
$\frac{3600}{1.25} = 2880 ticks$
Each tick is $1.00 s$ on the ship, so the ship clock reads $2880 s$ , which is $48$ minutes.
$3600 s - 2880 s = 720 s = 12 minutes behind$

Where this lies on the curve

0.25

0.5

0.75

1

2

3

4

5

6

O

v / c

t / t_{0}

v = c

(0.60, 1.25)

t / t_{0}

At

v = 0.60 c

the Earth interval is

1.25

times the proper time.

t = 1.25 s; ship clock = 2880 s (12 minutes behind)

13.3Mass variationOften examined

The mass of a moving object increases as its velocity increases. This phenomenon is known as mass variation and is another expression of mass-energy equivalence. It is represented mathematically as

m = \dfrac{m_0}{\sqrt{1-\dfrac{v^{2}}{c^{2}}}}

(13.4)

Symbol	Meaning
m	relativistic mass (the mass of the moving particle)
m_0	rest mass of the particle
v	velocity of the particle relative to a stationary observer
c	speed of light

Mass against speed

The mass m is always more than the rest mass m_0. It rises steeply near v=c. The points show Example 13.1 (v=0.9c) and Example 13.2 (m=3m_0 at v=0.9428c).

This effect becomes noticeable only at relativistic speeds (speeds close to the speed of light). As an object is accelerated closer to the speed of light, its mass increases. The more massive it becomes, the more energy is required to achieve the same acceleration, making further acceleration more and more difficult. The energy that is put into attempting acceleration is instead converted into mass. The total energy of an object is its kinetic energy plus the energy embodied in its mass.

To accelerate even the smallest object to the speed of light would require an infinite amount of energy. Therefore material objects are restricted to speeds less than the speed of light.

Example 13.1

If a 0.5\,\mathrm{kg} body is moving at a speed equivalent to 90\,\% of the speed of light, what will be its mass in this situation?

Given

v = 90\,\%\ \text{of}\ c=0.9c
m_0 = 0.5\,\mathrm{kg} rest mass

To find

m

Use the mass variation relation (13.4).
m=\dfrac{m_0}{\sqrt{1-v^{2}/c^{2}}}
Put in the values. The c^{2} cancels inside the root.
m=\dfrac{0.5}{\sqrt{1-\dfrac{(0.9c)^{2}}{c^{2}}}}=\dfrac{0.5}{\sqrt{1-0.81}}
Simplify the root.
\sqrt{0.19}=0.4359
Divide.
m=\dfrac{0.5}{0.4359}=1.15\,\mathrm{kg}

m=1.15\,\mathrm{kg}

Example 13.2

At what speed would the mass of a proton be tripled? The rest mass of a proton is 1.673\times10^{-27}\,\mathrm{kg}.

Given

m_0 = 1.673\times10^{-27}\,\mathrm{kg} rest mass of a proton
m = 3m_0 tripled mass

To find

v

Use the relation (13.4).
m=\dfrac{m_0}{\sqrt{1-v^{2}/c^{2}}}
Put m=3m_0.
3m_0=\dfrac{m_0}{\sqrt{1-v^{2}/c^{2}}}
Cancel m_0. The rest mass drops out, so the answer is the same for any particle.
3=\dfrac{1}{\sqrt{1-v^{2}/c^{2}}}\ \Rightarrow\ \sqrt{1-\dfrac{v^{2}}{c^{2}}}=\dfrac13
Square both sides.
1-\dfrac{v^{2}}{c^{2}}=\dfrac19
Solve for v^{2}/c^{2}.
\dfrac{v^{2}}{c^{2}}=1-\dfrac19=\dfrac89
Solve for v.
v^{2}=\dfrac89\,c^{2}\ \Rightarrow\ v=\dfrac{\sqrt8}{3}\,c=0.9428\,c

v=0.9428\,c\approx2.83\times10^{8}\,\mathrm{m\,s^{-1}}

Assignment 13.1

Assignment 13.1 (1)

What is the speed of a rod relative to the observer, if the length of the rod is measured to be half of its proper length?

Given

L = \tfrac12 L_0 measured length
L_0 = L_0 proper length (rest frame)

To find

v

Write the length contraction equation (13.2).
L=L_0\sqrt{1-\dfrac{v^{2}}{c^{2}}}
Put L=L_0/2 and divide both sides by L_0.
\dfrac12=\sqrt{1-\dfrac{v^{2}}{c^{2}}}
Square both sides to remove the root.
\dfrac14=1-\dfrac{v^{2}}{c^{2}}
Solve for v^{2}/c^{2}.
\dfrac{v^{2}}{c^{2}}=1-\dfrac14=\dfrac34
Take the square root.
v=\dfrac{\sqrt3}{2}\,c=0.866\,c
Convert to metres per second with c=3.00\times10^{8}\,\mathrm{m\,s^{-1}}.
v=0.866\times3.00\times10^{8}\approx2.60\times10^{8}\,\mathrm{m\,s^{-1}}

v=\dfrac{\sqrt3}{2}\,c=0.866\,c

Assignment 13.1 (2)

The time period of a pendulum is measured to be 3\,\mathrm{s} in the inertial frame. What is the period when measured by an observer moving with a speed of 0.95c with respect to the pendulum?

Given

t_0 = 3\,\mathrm{s} period in the pendulum's own (inertial) frame
v = 0.95\,c speed of the observer relative to the pendulum

To find

t, the period measured by this observer

The period of 3\,\mathrm{s} is measured in the pendulum's own frame, so it is the proper time t_0. An observer moving at 0.95c measures the longer time t from Equation (13.3).
t=\dfrac{t_0}{\sqrt{1-v^{2}/c^{2}}}
Find v^{2}/c^{2}.
\dfrac{v^{2}}{c^{2}}=(0.95)^{2}=0.9025
Find the square root.
\sqrt{1-0.9025}=\sqrt{0.0975}=0.3122
Divide.
t=\dfrac{3}{0.3122}=9.61\,\mathrm{s}

t=9.61\,\mathrm{s}

Worked example: speed at which an electron's mass doubles

At what speed is the mass of an electron twice its rest mass? Also find the mass at that speed. Take m_0=9.11\times10^{-31}\,\mathrm{kg}.

Given

m = 2m_0
m_0 = 9.11\times10^{-31}\,\mathrm{kg} rest mass

To find

v
m

Start from the mass variation relation.
m=\dfrac{m_0}{\sqrt{1-v^{2}/c^{2}}}
Put m=2m_0. The m_0 cancels.
2=\dfrac{1}{\sqrt{1-v^{2}/c^{2}}}
Invert, then square both sides.
\sqrt{1-\dfrac{v^{2}}{c^{2}}}=\dfrac12\quad\Rightarrow\quad 1-\dfrac{v^{2}}{c^{2}}=\dfrac14
Solve for v/c.
\dfrac{v^{2}}{c^{2}}=\dfrac34\quad\Rightarrow\quad \dfrac vc=0.866
Convert to \mathrm{m\,s^{-1}}.
v=0.866(3.00\times10^{8})=2.60\times10^{8}\,\mathrm{m\,s^{-1}}
The mass is twice the rest mass.
m=2(9.11\times10^{-31})=1.82\times10^{-30}\,\mathrm{kg}

Mass doubles at 0.866c

The curve m/m_0 crosses 2 at v/c=0.866. Doubling the mass needs a speed far below c, but each further doubling needs a much smaller step in speed.

v=0.866c=2.60\times10^{8}\,\mathrm{m\,s^{-1}},\qquad m=1.82\times10^{-30}\,\mathrm{kg}

13.4Time as a fourth dimension in the spacetime model

Until the 20th century, it was thought that the three-dimensional geometry of the universe (its description in terms of locations, shapes, distances and directions) was distinct from time (the measurement of when events occur within the universe). According to Einstein, however, you need to describe the location of an object not only in three-dimensional space (length, width and height) but also in time. Hence time is the fourth dimension.

To know where you are, you have to know what time it is. The history of an object's location through time traces out a line or curve on a spacetime diagram. Each point in a spacetime diagram represents a unique position in space and time.

Spacetime: A mathematical model, as shown in Figure 13.4, that unites the three dimensions of space with the single dimension of time into a single four-dimensional continuum.
Spacetime diagram: A graphical representation of locations in space at various times.
Minkowski space: A geometric interpretation of special relativity that combines time and the three spatial dimensions of space into a four-dimensional continuum.

Figure 13.4: Minkowski space. The observer sits at the origin. The vertical axis is time; the future and past light cones lie above and below the flat 'hypersurface of the present', which stands for all of space at the present instant.

Spacetime diagrams are useful in visualising and understanding relativistic effects, such as how different observers perceive where and when events occur. Spacetime diagrams can depict the geometry underlying phenomena like time dilation and length contraction without any mathematical equations.

Worked example: drawing objects on a spacetime diagram

On a spacetime diagram with position x along the horizontal axis and time t upward, sketch the history of three objects: A at rest, B moving at constant velocity, C speeding up. Explain each shape.

Spacetime diagram of three objects. Each point is one event (a position and a time). The history of an object is a line, or a curve when it accelerates (accent).

Choose the axes. Time is the vertical axis, space the horizontal one, and every point is an event: a position together with a time.
Object A is at rest. Its position does not change while time passes, so its history is a vertical line.
Object B covers equal distances in equal times. Its position grows steadily with time, so its history is a straight slanted line. A flatter line means more distance in the same time, that is a greater speed.
Object C covers more distance in each successive interval of time. The line therefore leans further and further over, which makes a curve.
Mark one point on B and read off its x and t from the axes. This shows why you must know the time as well as the place to say where an object is.

Formula sheet

Mass-energy equivalence

E=mc^{2}

E: energy; m: mass; c: speed of light (Eq. 13.1)

Unit: \mathrm{J}

Speed of light in vacuum

c=299\,792\,458\,\mathrm{m\,s^{-1}}

the same for all observers, whatever the motion of the source or observer

Unit: \mathrm{m\,s^{-1}}

Length contraction

L=L_0\sqrt{1-\dfrac{v^{2}}{c^{2}}}

L: length measured by the observer; L_0: proper length (rest frame); v: relative speed (Eq. 13.2)

Unit: \mathrm{m}

Time dilation

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

t: measured time; t_0: proper time (rest frame of the event); v: relative speed (Eq. 13.3)

Unit: \mathrm{s}

Mass variation

m=\dfrac{m_0}{\sqrt{1-\dfrac{v^{2}}{c^{2}}}}

m: relativistic mass; m_0: rest mass; v: speed relative to a stationary observer (Eq. 13.4)

Unit: \mathrm{kg}

Key terms

Frame of reference: A coordinate system that can be used to determine the positions and velocities of objects within that frame.
Inertial frame of reference: A frame of reference that is at rest or moving with a uniform velocity along a straight line; Newton's laws hold in it.
Non-inertial frame of reference: A frame of reference that is accelerating; Newton's laws do not hold in it.
Special relativity: Einstein's 1905 theory, which applies to all physical phenomena in the absence of gravity.
General relativity: Einstein's 1915 theory, which explains the law of gravitation and its relation to the forces of nature.
Principle of relativity: The laws of physics are the same for all observers in any inertial frame of reference.
Constancy of the speed of light: The speed of light in vacuum is constant for all observers, regardless of the relative velocity between the source of light and the observer measuring it; c=299\,792\,458\,\mathrm{m\,s^{-1}}.
Relativity of simultaneity: Two events that are simultaneous for one observer may not be simultaneous for another observer in relative motion.
Mass-energy equivalence: Mass and energy are equivalent and transmutable: E=mc^{2}.
Proper length: The length of an object measured in its rest frame, L_0.
Length contraction: Objects are measured shorter in the direction of their motion relative to the observer: L=L_0\sqrt{1-v^{2}/c^{2}}.
Proper time: The time taken for an event to occur within its rest frame, t_0.
Time dilation: A moving clock runs more slowly: the time measured by an observer in relative motion is longer, t=t_0/\sqrt{1-v^{2}/c^{2}}.
Rest mass: The mass m_0 of a particle measured when it is at rest.
Mass variation: The mass of a moving object increases as its velocity increases: m=m_0/\sqrt{1-v^{2}/c^{2}}.
Spacetime: A mathematical model that unites the three dimensions of space with the single dimension of time into a four-dimensional continuum.
Spacetime diagram: A graphical illustration of locations in space and times.
Minkowski space: A geometric interpretation of special relativity that combines time and the three spatial dimensions of space into a four-dimensional continuum.

Multiple choice questionsp. 255

The book prints no answer key for these multiple choice questions. Each key below is worked out from the unit's own teaching, and the reason is given with it.

1

Theory of relativity was formulated by:

AIsaac Newton
BStephen William Hawking
CAlbert EinsteinAnswer
DErnest Rutherford

Why

Albert Einstein published the special theory of relativity in 1905 and the general theory in 1915. Option A is the mechanics that relativity superseded, not the theory that replaced it. Options B and D are not associated with the theory of relativity.

2

A frame of reference is:

Aa graph plotted between distance and time.
Ba graph plotted between speed and time.
Cthe velocity of an object through empty space without regard to its surroundings.
Dan arbitrarily fixed point with respect to which motion of other points is measured.Answer

Why

A frame of reference is a coordinate system used to determine the positions and velocities of objects in that frame. Option D describes exactly this: a fixed point, or set of geometric points, with respect to which the motion of other points is measured.

Options A and B are graphs, not frames of reference. Option C describes a velocity with no reference at all, but all measurements of motion are made relative to a frame.

3

A pendulum has a time period of 5.0\,\mathrm{s} relative to the pendulum's frame of reference. What is the speed of the observer relative to the pendulum's frame of reference if the period measured by the observer is 10.0\,\mathrm{s}? (here, c is the speed of light).

A0.606c
B0.779c
C0.866cAnswer
D0.693c

Why

The pendulum is at rest in its own frame, so the period it measures is the proper time t_0=5.0\,\mathrm{s}. The observer measures the longer time t=10.0\,\mathrm{s}, so the time dilation equation applies.

t=\dfrac{t_0}{\sqrt{1-v^{2}/c^{2}}}\quad\Rightarrow\quad 10.0=\dfrac{5.0}{\sqrt{1-v^{2}/c^{2}}}

Solving gives \sqrt{1-v^{2}/c^{2}}=0.5, so 1-v^{2}/c^{2}=0.25, v^{2}/c^{2}=0.75 and v=0.866c. Check the wrong options: A gives a period of about 6.3\,\mathrm{s}, B gives about 8.0\,\mathrm{s} and D gives about 6.9\,\mathrm{s}. None gives 10.0\,\mathrm{s}.

4

An airliner traveling at 200\,\mathrm{m\,s^{-1}} emits light from the front of the plane. Which statement describes the speed of the light?

AIt travels at a speed of c+200\,\mathrm{m\,s^{-1}}.
BIt travels at a speed of c-200\,\mathrm{m\,s^{-1}}.
CIt travels at a speed c, like all light.Answer
DIt travels at a speed slightly less than c.

Why

By the second postulate, the speed of light in vacuum is c for every inertial observer, whatever the speed of its source. The light from the airliner therefore travels at c, like all light.

Options A and B add or subtract the plane's speed, as Newtonian velocity addition would do; this does not apply to light. Option D is wrong because light in vacuum does not travel slower than c.

5

An inertial frame of reference is the one:

Awhich is at rest.
Bwhich is moving with a uniform velocity along a straight line.
Cwhich has zero acceleration.
DAll of these.Answer

Why

An inertial frame is at rest or moves with uniform velocity along a straight line, so it has zero acceleration. Each of A, B and C is true of an inertial frame, so a single option is incomplete and the answer is D.

Short questionsp. 255–256

12025 Annual I, Set B, Q.2 (xii) OR3 marks

What is the difference between inertial and non-inertial frames of reference?

Solution

An inertial frame of reference is at rest or moves with a uniform velocity along a straight line, so it does not accelerate. Newton's laws hold in it: an object at rest stays at rest, and an object in motion keeps a constant velocity unless an external force acts on it. The Earth (for many experiments) and the inside of a car moving at constant velocity are inertial frames.

A non-inertial frame of reference is accelerating. Objects in it appear to accelerate even when no force acts on them, because the frame itself accelerates. Newton's laws do not hold in it, as in an accelerating car.

Property	Inertial frame	Non-inertial frame
Acceleration of the frame	Zero	Not zero
Newton's laws	Hold	Do not hold
Example	Car at constant velocity	Accelerating car

22026 Annual I, Set B, Q.3 OR2026 Annual I, Set D, Q.6 OR3 marks

State the postulates of special theory of relativity.

Solution

Einstein's special theory of relativity (1905) rests on two postulates:

Principle of relativity: the laws of physics are the same for all observers in any inertial frame of reference. No experiment can tell whether an inertial frame is at rest or moving at a constant velocity. Two people, one in a stationary train and one in a train moving at constant velocity, both see a ball fall vertically, so the passengers of a windowless train cannot tell whether their train is stationary or moving.
Constancy of the speed of light: the speed of light in vacuum, c=299\,792\,458\,\mathrm{m\,s^{-1}}, is the same for all observers, whatever the relative velocity of the source and the observer. A person moving at 0.5c in the same direction as a beam of light still measures its speed as c, not 0.5c.

The second postulate contradicts Newtonian velocity addition, which would give c-0.5c=0.5c. Keeping the measured speed at c leads to the consequences of relativity in section 13.3.

33 marks

Explain why it is impossible for a particle with mass to move faster than the speed of light.

Solution

The mass of a moving particle increases with its speed, according to mass variation:

m=\dfrac{m_0}{\sqrt{1-v^{2}/c^{2}}}

As v approaches c, the quantity 1-v^{2}/c^{2} approaches zero, so m grows without limit. A particle at 0.9c has m\approx2.3\,m_0, at 0.99c it has m\approx7.1\,m_0, and at 0.999c it has m\approx22\,m_0.

Accelerating a particle requires energy, and this energy goes into increasing its mass. Reaching c would therefore need an infinite amount of energy, which no real process can supply. Material objects are restricted to speeds less than c.

Also, for v>c the term under the square root becomes negative, so m would not be a real number.

43 marks

Imagine a train travelling at the speed of light. A person on this train looks into the mirror placed in front of him. Will this person see a reflection of himself in the mirror?

Solution

Yes, the person sees a reflection. Newtonian physics predicts no reflection: the relative velocity between the light and the mirror would be zero, so the light would never reach the mirror and no image would be formed.

This prediction contradicts the first postulate. A person in an inertial frame cannot tell whether the frame is stationary or moving at constant speed, but if no reflection were seen, the motion of the train would be detectable.

The second postulate settles the matter. The light travels at speed c relative to the person and to the mirror, so the light reaches the mirror, is reflected and returns to the person's eyes. He sees his reflection.

52025 Annual I, Set B, Q.2 (vi) OR3 marks

What happens to the density of an object as its speed increases?

Solution

The density increases. Density is \rho=m/V, and both the mass and the volume change with speed. Using mass variation and length contraction:

Mass increases with speed.
m=\dfrac{m_0}{\sqrt{1-v^{2}/c^{2}}}
Length contracts in the direction of motion, and the sides across the motion are unchanged, so the volume shrinks by the same factor.
V=V_0\sqrt{1-v^{2}/c^{2}}
Divide the mass by the volume.
\rho=\dfrac{m}{V}=\dfrac{m_0}{V_0\,(1-v^{2}/c^{2})}=\dfrac{\rho_0}{1-v^{2}/c^{2}}

Since 1-v^{2}/c^{2} is less than 1, the density is always greater than the rest density \rho_0. At 0.8c the factor is 1/0.36\approx2.8, so the density is about 2.8 times its rest value, and it grows without limit as v approaches c.

Note. The book gives no answer to this question. It is worked out here from the mass variation and length contraction relations of the unit.

63 marks

What is meant by the relativity of simultaneity?

Solution

The relativity of simultaneity means that two events that are simultaneous for one observer may not be simultaneous for another observer in relative motion.

Take a bolt of lightning that strikes two trees on either side of a train track, the train being exactly midway between them. A person standing beside the track sees both strikes at the same time. The person inside the moving train is moving towards one strike and away from the other, so he sees the bolt ahead of the train first and the bolt behind the train later. The two events are simultaneous for one observer but not for the other.

Einstein concluded that simultaneity is not absolute. It is time itself that is relative, not the speed of light, which is the same for everyone.

Comprehensive questionsp. 256

1

What is meant by frames of reference? Explain with the help of examples. Also discuss its types.

Solution

Frame of reference

A frame of reference is a coordinate system that can be used to determine the positions and velocities of objects within that frame. Its origin, orientation and scale are fixed by a set of geometric points whose positions are identified both mathematically and physically.

For example, when a ball rolls in a street, the ball is moving if the frame of reference is the street, a side of the street or the Earth itself. All measurements of motion are made relative to a frame, and different frames can move relative to one another.

The velocity of an object therefore has no meaning until the frame is specified. In Fig. 13.1, person A says that B is at rest, because they share the bus. Person C, standing on the road, says that B moves with velocity v in the positive x direction.

Figure 13.1: Two persons A and B seated in a bus moving with a velocity v. Person C stands on the road.

Types of frame of reference

There are two types of frame of reference.

Inertial frame of reference: a frame at rest, or moving with a uniform velocity along a straight line. It does not accelerate, and Newton's laws hold in it. The Earth can be treated as an inertial frame for many experiments, and so can the inside of a car moving along a road at constant velocity or a stationary house.
Non-inertial frame of reference: an accelerating frame. Objects in it accelerate even when no force acts on them, because the frame itself accelerates, so Newton's laws do not hold. An accelerating car is an example.

Any frame moving at constant velocity relative to an inertial frame is itself inertial.

2

Discuss the postulate of the special theory of relativity.

Solution

Einstein's special theory of relativity (1905) is based on two postulates. Each is explained below with the thought experiment that illustrates it.

First postulate: the principle of relativity

The laws of physics are the same for all observers in any inertial frame of reference. As long as an object moves in a straight line at constant speed, the laws of physics are the same for every observer. Experiments performed in a stationary frame and in a frame moving at constant velocity give the same results.

Consider two people, one standing in a stationary train and the other in a train moving at constant velocity v, both watching a ball fall freely under gravity (Fig. 13.2). Both see the ball fall vertically, and Newton's laws apply equally in both frames. If the trains are windowless, the passengers have no way of knowing whether their train is stationary or moving at constant velocity.

Second postulate: the constancy of the speed of light

The speed of light in vacuum is the same for all observers, regardless of the relative velocity between the source and the observer. Its value is c=299\,792\,458\,\mathrm{m\,s^{-1}}. A stationary observer measures the light from a moving train as c, and the light from a train at rest is also measured as c (Fig. 13.3 (a)).

This postulate contradicts Newtonian mechanics. A person moving at 0.5c in the same direction as a beam of light would, by Newtonian mechanics, measure its speed as 0.5c. Special relativity says the speed is still c.

Thought experiment: a mirror on a moving train

A person on a moving train looks into a mirror placed in front of him (Fig. 13.3 (b)). Newtonian physics predicts that no reflection is seen. The relative velocity between the light and the mirror would then be zero, so the light would never reach the mirror.

This outcome contradicts the first postulate. A person in an inertial frame cannot tell whether the frame is stationary or moving at constant speed, yet the absence of a reflection would reveal the motion of the train. Special relativity asserts that the reflection is seen. By the second postulate the light travels at c relative to the person and the mirror, so it reaches the mirror and returns to his eyes.

Figure 13.3 (b): A person on the train looks into the mirror placed at his front.

Together, the two postulates are the foundation of the special theory of relativity, and they lead to its consequences, discussed in section 13.3.

32025 Annual I, Set D, Q.6 OR

Explain the following consequences of the special theory of relativity: a) Relativity of simultaneity b) Length contraction c) Time dilation d) Mass variation e) Mass-energy equivalence.

Solution

(a) Relativity of simultaneity

Two events that are simultaneous for one observer may not be simultaneous for another observer in relative motion. Lightning strikes two trees placed on either side of a track, and the train is midway between them. The observer on the ground sees both strikes at the same time. The observer inside the moving train moves towards the front strike and away from the rear one, so the light from the front strike reaches him first. He sees the front bolt first and the rear bolt later.

Strikes A and B are simultaneous for an observer on the ground at M. The train moves towards B, so its observer sees B first.

Einstein concluded that simultaneity is not absolute. It is time itself that is relative, while the speed of light is the same for everyone.

(b) Length contraction

An object appears shorter in the direction of its motion relative to an observer. Its length measured in its own rest frame is the proper length L_0. An observer in a frame moving relative to the object measures a shorter length L:

L=L_0\sqrt{1-\dfrac{v^{2}}{c^{2}}}\qquad(13.2)

Contraction occurs only in the direction of motion. For example, at v=0.8c the factor is \sqrt{1-0.64}=0.6, so a rod of proper length 10\,\mathrm{m} appears 6\,\mathrm{m} long to the observer.

(c) Time dilation

A moving clock runs more slowly than clocks at rest relative to the observer. The time taken for an event within its rest frame is the proper time t_0. An observer in relative motion measures the longer time t:

t=\dfrac{t_0}{\sqrt{1-v^{2}/c^{2}}}\qquad(13.3)

A pendulum with a period of 5.0\,\mathrm{s} in its own frame is measured by an observer moving at 0.866c to have a period of 10.0\,\mathrm{s}, because the factor 1/\sqrt{1-0.75}=2 doubles the time.

(d) Mass variation

The mass of a moving object increases with its velocity. The relativistic mass m is related to the rest mass m_0 by

m=\dfrac{m_0}{\sqrt{1-v^{2}/c^{2}}}\qquad(13.4)

The effect is noticeable only at speeds close to c. A 0.5\,\mathrm{kg} body moving at 0.9c has a mass of 1.15\,\mathrm{kg} (Example 13.1). As the mass increases, more energy is needed for the same acceleration, so an object can never reach the speed of light.

(e) Mass-energy equivalence

Energy and mass are equivalent and can be transformed into each other:

E=mc^{2}\qquad(13.1)

The total energy of a moving object is its kinetic energy plus the energy embodied in its mass. The increase in relativistic mass is equal to the kinetic energy divided by c^{2}. Because c^{2} is very large, a tiny amount of mass is equivalent to a very large amount of energy. This is why atomic and hydrogen bombs are so powerful, and it is the basis of the mass defect in the atomic nucleus.

For example, an electron of rest mass 9.11\times10^{-31}\,\mathrm{kg} has an equivalent energy of 0.512\,\mathrm{MeV} (Numerical problem 1).

4

State and explain how time is considered the fourth dimension alongside the traditional three dimensions of space.

Solution

Why time is the fourth dimension

Until the 20th century, the three-dimensional geometry of the universe (locations, shapes, distances and directions) was thought to be distinct from time, which measures when events occur. Einstein showed that to describe an object you need not only its location in three-dimensional space (length, width and height) but also the time at which it is there. Hence time is the fourth dimension.

To know where you are, you need to know what time it is. The history of an object's location through time traces out a line or curve on a spacetime diagram, and each point of the diagram represents a unique position in space and time.

Spacetime model

Spacetime is a mathematical model that unites the three dimensions of space with the single dimension of time into one four-dimensional continuum. In 1908, Hermann Minkowski gave a geometric interpretation of special relativity in this form, now known as Minkowski space.

Figure 13.4: Minkowski space in simplified form. Time runs up the page. The light cones through the observer show the events that light from the observer can reach (future cone) and the events that can send light to the observer (past cone).

A spacetime diagram is a graphical representation of locations in space at various times. Spacetime diagrams can depict time dilation and length contraction without any mathematical equations.

Thus time is the fourth dimension: an event is located by three space coordinates and one time coordinate, and space and time together form a single four-dimensional continuum.

Numerical problemsp. 256

Take c=3.00\times10^{8}\,\mathrm{m\,s^{-1}} and 1\,\mathrm{MeV}=1.6\times10^{-13}\,\mathrm{J}. Lengths and times of a moving object are found from L=L_0\sqrt{1-v^{2}/c^{2}} and t=t_0/\sqrt{1-v^{2}/c^{2}}, where v/c is used as a ratio.

1

Calculate the equivalent energy of an electron with rest mass 9.11\times10^{-31}\,\mathrm{kg}.

Solution

Given

m_0 = 9.11\times10^{-31}\,\mathrm{kg}
c = 3.00\times10^{8}\,\mathrm{m\,s^{-1}}

To find

E in \mathrm{MeV}

Use Einstein's mass-energy equivalence.
E=m_0c^{2}
Substitute the values.
E=(9.11\times10^{-31})(3.00\times10^{8})^{2}=8.199\times10^{-14}\,\mathrm{J}
Convert joules to MeV, using 1\,\mathrm{MeV}=1.6\times10^{-13}\,\mathrm{J}.
E=\dfrac{8.199\times10^{-14}}{1.6\times10^{-13}}\,\mathrm{MeV}=0.512\,\mathrm{MeV}

E\approx0.512\,\mathrm{MeV}

Book answer. 0.512 MeV

22026 Annual I, Set D, Q.6 OR

The length of a spaceship is 100\,\mathrm{m}. What will be its length if the spaceship moves at a speed of 0.99c?

Solution

Given

L_0 = 100\,\mathrm{m}
v = 0.99c

To find

L

Length contraction applies in the direction of motion. Use the proper length L_0.
L=L_0\sqrt{1-\dfrac{v^{2}}{c^{2}}}
Substitute v/c=0.99.
L=100\sqrt{1-(0.99)^{2}}=100\sqrt{0.0199}
Evaluate.
L=100\times0.141=14.1\,\mathrm{m}\approx14\,\mathrm{m}

L\approx14\,\mathrm{m}

Book answer. 14 m

Note. The calculated length is 14.1 m, which the book rounds to 14 m. The two agree.

3

Particles called \pi-mesons are produced by accelerator beams. If these particles travel at 2.70\times10^{8}\,\mathrm{m\,s^{-1}} and live 2.60\times10^{-8}\,\mathrm{s} when at rest relative to an observer. How long do they live as viewed in the laboratory?

Solution

Given

t_0 = 2.60\times10^{-8}\,\mathrm{s}
v = 2.70\times10^{8}\,\mathrm{m\,s^{-1}}
c = 3.00\times10^{8}\,\mathrm{m\,s^{-1}}

To find

t, the lifetime in the laboratory

The lifetime at rest is the proper time t_0. The laboratory measures the longer time t, given by time dilation.
t=\dfrac{t_0}{\sqrt{1-v^{2}/c^{2}}}
Find the speed ratio. v/c=2.70/3.00=0.900, so v^{2}/c^{2}=0.810.
\sqrt{1-0.810}=\sqrt{0.190}=0.436
Substitute.
t=\dfrac{2.60\times10^{-8}}{0.436}=5.96\times10^{-8}\,\mathrm{s}

t\approx5.96\times10^{-8}\,\mathrm{s}

Book answer. 5.96\times10^{-8}\,\mathrm{s}

4

A neutral \pi-meson lives 1.40\times10^{-16}\,\mathrm{s} as measured in the laboratory, and 0.840\times10^{-16}\,\mathrm{s} when at rest relative to an observer. What is its velocity relative to the laboratory?

Solution

Given

t = 1.40\times10^{-16}\,\mathrm{s}
t_0 = 0.840\times10^{-16}\,\mathrm{s}

To find

v

The lifetime at rest is the proper time t_0, and the laboratory lifetime t is the longer one.
t=\dfrac{t_0}{\sqrt{1-v^{2}/c^{2}}}
Rearrange for the square root.
\sqrt{1-\dfrac{v^{2}}{c^{2}}}=\dfrac{t_0}{t}=\dfrac{0.840\times10^{-16}}{1.40\times10^{-16}}=0.600
Square both sides and solve for v^{2}/c^{2}.
1-\dfrac{v^{2}}{c^{2}}=0.360\quad\Rightarrow\quad\dfrac{v^{2}}{c^{2}}=0.640
Take the square root.
\dfrac{v}{c}=0.800\quad\Rightarrow\quad v=0.800\,c=2.40\times10^{8}\,\mathrm{m\,s^{-1}}

v=0.800\,c\approx2.40\times10^{8}\,\mathrm{m\,s^{-1}}

Book answer. 0.800\,c

FBISE 2025 and 2026 questions

Questions from the FBISE HSSC-I 2026 (1st Annual) and FBISE HSSC-I 2025 (1st Annual) papers that belong to this unit, answered the way an examiner expects.

2026 Annual I, Set B, MCQ 131 mark

Postulates of special theory of relativity applies to:

AOnly light
BOnly motion
CAll physical lawsAnswer
DOnly energy

Why

The postulates of special relativity are about all the laws of physics. The first says that the laws of physics are the same in every inertial frame, and the second adds that the speed of light is the same for all observers. Option A (only light) is the tempting one: the second postulate is about light, but the first covers every law, so the theory is not limited to light. Options B and D are likewise too narrow.

2026 Annual I, Set B, MCQ 141 mark

Correct Mass-energy relation is:

AE = mv
BE = mc^2Answer
CE = mg
DE = gh

Why

Einstein's mass-energy relation is E=mc^{2}: the energy equivalent of a mass m is the mass times the square of the speed of light. Options C (mg) and D (gh) look like weight and potential energy per unit mass, not relativity; option A (mv) is momentum, not energy.

2025 Annual I, Set B, MCQ 151 mark

When an object moves with a very high speed its length ______ in the direction of its motion:

AContractsAnswer
BExpands
CRemains same
DBecomes zero

Why

By length contraction L=L_0\sqrt{1-v^{2}/c^{2}}, an object moving at very high speed is measured shorter along its direction of motion. Option B is wrong because lengths never expand in this effect (the proper length L_0 is the largest). Option D would need v=c, which a body with mass cannot reach.

2025 Annual I, Set B, MCQ 161 mark

Relativistic mass is always:

ALess than rest mass
BGreater than rest massAnswer
CEqual to rest mass
DZero

Why

From m=m_0/\sqrt{1-v^{2}/c^{2}} the root is less than 1 for any moving body, so the relativistic mass m is greater than the rest mass m_0 (it rises without limit as v\to c). Option C (equal) is the tempting one: it holds only at v=0, or when the speed is so small that the effect cannot be noticed.

2025 Annual I, Set B, Q.2 (vi) OR3 marks

What happens to the density of an object as it moves in a direction with very high speed relative to speed of light? Justify.

Solution

The density of an object increases as its speed approaches the speed of light. Density is \rho=m/V, and both the mass and the volume change with speed.

Mass increases with speed (mass variation).
m=\dfrac{m_0}{\sqrt{1-v^{2}/c^{2}}}
Length contracts along the direction of motion, and the sides across the motion are unchanged, so the volume shrinks by the same factor.
V=V_0\sqrt{1-v^{2}/c^{2}}
Divide the mass by the volume.
\rho=\dfrac{m}{V}=\dfrac{m_0}{V_0\,(1-v^{2}/c^{2})}=\dfrac{\rho_0}{1-v^{2}/c^{2}}

Since 1-v^{2}/c^{2} is less than 1, \rho>\rho_0. At 0.8c the factor is 1/0.36\approx2.8, so the density is about 2.8 times its rest value, and it grows without limit as v\to c.

2025 Annual I, Set B, Q.2 (viii) OR3 marks

At what speed would the mass of proton be doubled? The rest mass of proton is 1.673\times10^{-27}\,\mathrm{kg}.

Solution

Given

m_0 = 1.673\times10^{-27}\,\mathrm{kg} rest mass of a proton
m = 2m_0 doubled mass

To find

v

Use the mass variation relation.
m=\dfrac{m_0}{\sqrt{1-v^{2}/c^{2}}}
Put m=2m_0. The rest mass cancels, so the answer is the same for any particle.
2=\dfrac{1}{\sqrt{1-v^{2}/c^{2}}}\ \Rightarrow\ \sqrt{1-\dfrac{v^{2}}{c^{2}}}=\dfrac12
Square both sides.
1-\dfrac{v^{2}}{c^{2}}=\dfrac14
Solve for v^{2}/c^{2}.
\dfrac{v^{2}}{c^{2}}=\dfrac34
Take the square root and use c=3.00\times10^{8}\,\mathrm{m\,s^{-1}}.
v=\dfrac{\sqrt3}{2}\,c=0.866\,c=0.866\times3.00\times10^{8}\approx2.60\times10^{8}\,\mathrm{m\,s^{-1}}

v=0.866\,c\approx2.60\times10^{8}\,\mathrm{m\,s^{-1}}

2025 Annual I, Set B, Q.2 (xii) OR3 marks

What is the difference between inertial and non-inertial frame of reference?

Solution

An inertial frame of reference is at rest or moves with a uniform velocity along a straight line, so it does not accelerate. Newton's laws hold in it: a body at rest stays at rest, and a body in motion keeps a constant velocity unless an external force acts. The Earth (for many experiments) and the inside of a car moving at constant velocity are inertial frames.

A non-inertial frame of reference is accelerating. Bodies in it appear to accelerate even when no force acts on them, because the frame itself accelerates, so Newton's laws do not hold in it. An accelerating car is an example.

Property	Inertial frame	Non-inertial frame
Acceleration of the frame	Zero	Not zero
Newton's laws	Hold	Do not hold
Example	Car at constant velocity	Accelerating car

2026 Annual I, Set B, Q.3 OR2+2+2 = 6 marks

a. State postulates of special theory of relativity. b. Briefly discuss time dilation and length contraction. c. At what speed would the mass of the proton be tripled? The rest mass of a proton is 1.673\times10^{-27}\,\mathrm{kg}.

Solution

(a) Postulates of special relativity (2 marks)

Einstein's special theory of relativity (1905) rests on two postulates.

Principle of relativity: the laws of physics are the same for all observers in any inertial frame of reference (a frame at rest or moving with constant velocity). No experiment can tell whether such a frame is at rest or moving uniformly. For example, a ball falls vertically for a person in a stationary train and for a person in a train moving at constant velocity.
Constancy of the speed of light: the speed of light in vacuum, c=299\,792\,458\,\mathrm{m\,s^{-1}}, is the same for all observers, whatever the relative velocity of the source and the observer. An observer moving at 0.5c along a light beam still measures its speed as c, not 0.5c.

(b) Time dilation and length contraction (2 marks)

Time dilation. A moving clock runs slower than a clock at rest relative to the observer. If t_0 is the proper time (the time between two events measured by one clock at rest at the events) and t is the time measured by an observer who sees that clock moving with speed v, then

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

Since \sqrt{1-v^{2}/c^{2}}<1, t>t_0: the moving clock ticks more slowly. For example, a pendulum of period 5.0\,\mathrm{s} in its own frame has a period of 10.0\,\mathrm{s} for an observer moving at 0.866c, since 1/\sqrt{1-0.75}=2. The effect is noticeable only at speeds close to c.

Length contraction. An object is measured to be shorter in the direction of its motion relative to the observer. If L_0 is the proper length (measured in the rest frame of the object) and L the length measured by an observer for whom the object moves with speed v, then

L=L_0\sqrt{1-\dfrac{v^{2}}{c^{2}}}

Since the root is less than 1, L<L_0. The contraction is only along the direction of motion; the width and height are unchanged. At v=0.8c the factor is 0.6, so a 10\,\mathrm{m} rod is measured as 6\,\mathrm{m}.

The three effects against v/c: the time and mass factor 1/\sqrt{1-v^{2}/c^{2}} rises without limit as v\to c, while the length factor \sqrt{1-v^{2}/c^{2}} falls to zero. They are noticeable only close to c.

(c) Speed at which the mass of a proton is tripled (2 marks)

Given

m_0 = 1.673\times10^{-27}\,\mathrm{kg} rest mass of a proton
m = 3m_0 tripled mass

To find

v

Use the mass variation relation.
m=\dfrac{m_0}{\sqrt{1-v^{2}/c^{2}}}
Put m=3m_0 and cancel m_0 (the rest mass drops out).
3=\dfrac{1}{\sqrt{1-v^{2}/c^{2}}}\ \Rightarrow\ \sqrt{1-\dfrac{v^{2}}{c^{2}}}=\dfrac13
Square both sides.
1-\dfrac{v^{2}}{c^{2}}=\dfrac19
Solve for v^{2}/c^{2}.
\dfrac{v^{2}}{c^{2}}=\dfrac89
Take the square root.
v=\dfrac{\sqrt8}{3}\,c=0.943\,c\approx2.83\times10^{8}\,\mathrm{m\,s^{-1}}

v=0.943\,c\approx2.83\times10^{8}\,\mathrm{m\,s^{-1}}

Note. Part (a) is short question 2 of the book, and part (c) is Example 13.2 with the mass tripled.

2026 Annual I, Set D, MCQ 141 mark

The rest energy of mass 2.00g is:

A1.80\times10^{18}\,J
B180\times10^{12}\,J
C1.80\times10^{14}\,JAnswer
D1.80\times10^{16}\,J

Why

Rest energy E=m_0c^{2} with m_0=2.00\,\mathrm{g}=2.00\times10^{-3}\,\mathrm{kg} and c=3.00\times10^{8}\,\mathrm{m\,s^{-1}}:

E=(2.00\times10^{-3})(3.00\times10^{8})^{2}=1.80\times10^{14}\,\mathrm{J}

This is option C. The wrong powers in A (10^{18}) and D (10^{16}) come from forgetting to convert grams to kilograms or from mis-squaring c.

Note. Option B, 180\times10^{12}\,\mathrm{J}, equals 1.80\times10^{14}\,\mathrm{J} as well, so options B and C have the same value as printed. C is the answer in standard form.

2026 Annual I, Set D, MCQ 151 mark

According to special theory of relativity time dilation means:

ATime increases for moving observerAnswer
BTime decreases for moving observer
CTime becomes zero
DTime is constant for moving observer

Why

Time dilation means that time between two events is stretched for an object in motion: the moving clock ticks more slowly, so the time interval t=t_0/\sqrt{1-v^{2}/c^{2}} is greater than the proper time t_0. The paper's option A, 'time increases for moving observer', states this as the lengthening of the time interval. Option B is the tempting wrong one: it confuses the slowing of the clock with a shorter interval. Option C (time becomes zero) would need v\to\infty, and D (constant) denies the effect.

Note. The wording of the options is loose: strictly, an observer in relative motion measures a longer interval (t>t_0) on the moving clock. Option A is the intended answer.

2025 Annual I, Set D, MCQ 151 mark

Relativistic mass is always:

ALess than rest mass
BGreater than rest massAnswer
CEqual to rest mass
DZero

Why

From m=m_0/\sqrt{1-v^{2}/c^{2}} the root is less than 1 for any moving body, so the relativistic mass m is greater than the rest mass m_0 (it rises without limit as v\to c). Option C (equal) is the tempting one: it holds only at v=0, or when the speed is so small that the effect cannot be noticed.

2025 Annual I, Set D, MCQ 161 mark

Inertial frame of reference is a frame in which ______ law of motion holds.

A1stAnswer
B2nd
C3rd
D4th

Why

An inertial frame is one that is at rest or moves with uniform velocity along a straight line, and in it Newton's first law (law of inertia) holds: a body at rest stays at rest and a body in motion keeps its constant velocity unless a force acts. In a non-inertial (accelerating) frame this fails, since bodies seem to accelerate with no force. Options B and C are tempting but Newton's second and third laws are not what define the frame; D is not a law of motion.

2026 Annual I, Set D, Q.6 OR2+1+3 = 6 marks

a. State postulates of special theory of relativity. b. Briefly discuss length contraction as a consequence of special relativity. c. The length of spaceship is 100\,\mathrm{m} long. What will be its length if the spaceship moves with a speed of 0.99c?

Solution

(a) Postulates of special relativity (2 marks)

Einstein's special theory of relativity (1905) rests on two postulates.

Principle of relativity: the laws of physics are the same for all observers in any inertial frame of reference (a frame at rest or moving with constant velocity). No experiment can tell whether such a frame is at rest or moving uniformly. For example, a ball falls vertically for a person in a stationary train and for a person in a train moving at constant velocity.
Constancy of the speed of light: the speed of light in vacuum, c=299\,792\,458\,\mathrm{m\,s^{-1}}, is the same for all observers, whatever the relative velocity of the source and the observer. An observer moving at 0.5c along a light beam still measures its speed as c, not 0.5c.

(b) Length contraction (1 mark)

Length contraction. An object is measured to be shorter in the direction of its motion relative to the observer. If L_0 is the proper length (measured in the rest frame of the object) and L the length measured by an observer for whom the object moves with speed v, then

L=L_0\sqrt{1-\dfrac{v^{2}}{c^{2}}}

Since the root is less than 1, L<L_0. The contraction is only along the direction of motion; the width and height are unchanged. At v=0.8c the factor is 0.6, so a 10\,\mathrm{m} rod is measured as 6\,\mathrm{m}.

(c) Length of the spaceship at 0.99c (3 marks)

Given

L_0 = 100\,\mathrm{m} proper length
v = 0.99c

To find

L

Length contraction applies along the direction of motion. Use the proper length L_0.
L=L_0\sqrt{1-\dfrac{v^{2}}{c^{2}}}
Substitute v/c=0.99.
L=100\sqrt{1-(0.99)^{2}}=100\sqrt{0.0199}
Evaluate.
L=100\times0.141=14.1\,\mathrm{m}\approx14\,\mathrm{m}

L\approx14\,\mathrm{m}

Note. Part (c) is numerical problem 2 of the book. The marks printed are 2+1+3.

2025 Annual I, Set D, Q.6 OR2x3 = 6 marks

Discuss the following consequences of special theory of relativity: i) Time dilation ii) Length contraction iii) Mass increase.

Solution

Special relativity predicts three effects at speeds close to c. Each gets 2 marks: the formula, the meaning and an example.

(i) Time dilation

Time dilation. A moving clock runs slower than a clock at rest relative to the observer. If t_0 is the proper time (the time between two events measured by one clock at rest at the events) and t is the time measured by an observer who sees that clock moving with speed v, then

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

Since \sqrt{1-v^{2}/c^{2}}<1, t>t_0: the moving clock ticks more slowly. For example, a pendulum of period 5.0\,\mathrm{s} in its own frame has a period of 10.0\,\mathrm{s} for an observer moving at 0.866c, since 1/\sqrt{1-0.75}=2. The effect is noticeable only at speeds close to c.

(ii) Length contraction

Length contraction. An object is measured to be shorter in the direction of its motion relative to the observer. If L_0 is the proper length (measured in the rest frame of the object) and L the length measured by an observer for whom the object moves with speed v, then

L=L_0\sqrt{1-\dfrac{v^{2}}{c^{2}}}

Since the root is less than 1, L<L_0. The contraction is only along the direction of motion; the width and height are unchanged. At v=0.8c the factor is 0.6, so a 10\,\mathrm{m} rod is measured as 6\,\mathrm{m}.

(iii) Mass increase

Mass variation. The mass of a moving object increases with its speed. If m_0 is the rest mass and m the relativistic mass at speed v,

m=\dfrac{m_0}{\sqrt{1-v^{2}/c^{2}}}

Since the root is less than 1, m>m_0. A 0.5\,\mathrm{kg} body moving at 0.9c has mass 0.5/\sqrt{1-0.81}=1.15\,\mathrm{kg}. The effect is noticeable only close to c. As v\to c the mass grows without limit, so more and more energy is needed for the same acceleration, and an object with mass can never reach the speed of light: the energy put in goes into mass instead. This is another expression of mass-energy equivalence, E=mc^{2}.

The three effects against v/c: the time and mass factor 1/\sqrt{1-v^{2}/c^{2}} rises without limit as v\to c, while the length factor \sqrt{1-v^{2}/c^{2}} falls to zero. They are noticeable only close to c.

Exam focus

Discuss time dilation, length contraction and mass increase (formula, meaning, example). (long)
25D Q.6 OR asked all three (2 marks each); 2026 Set B Q.3 OR (b) asked time dilation and length contraction, and 2026 Set D Q.6 OR (b) asked length contraction. Same as comprehensive question 3.
Go to answer
State the postulates of special relativity with an example for each. (short)
2026 Set B Q.3 OR (a) and 2026 Set D Q.6 OR (a), 2 marks each; short question 2 and comprehensive question 2 of the book.
Go to answer
Speed at which a particle's mass is doubled or tripled (m=m_0/\sqrt{1-v^{2}/c^{2}}). (numerical)
2025 Set B Q.2 (viii) OR (doubled) and 2026 Set B Q.3 OR (c) (tripled); both use the proton, and the rest mass cancels. Example 13.2 of the book.
Go to answer
Length of a spaceship moving at 0.99c (L=L_0\sqrt{1-v^{2}/c^{2}}). (numerical)
2026 Set D Q.6 OR (c), 3 marks; numerical problem 2 of the book. Square and substitute, then take the root.
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Difference between inertial and non-inertial frames of reference. (short)
2025 Set B Q.2 (xii) OR, and 2025 Set D MCQ 16 (Newton's first law holds in an inertial frame); short question 1 of the book.
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What happens to the density of a body at very high speed? Justify. (short)
2025 Set B Q.2 (vi) OR; short question 5 of the book. Combine mass variation and length contraction.
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Mass-energy equivalence E=mc^{2}: formula and rest energy of a given mass. (mcq)
2026 Set B MCQ 14 and 2026 Set D MCQ 14 (rest energy of 2.00\,\mathrm{g}); numerical problem 1 of the book.
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One-line MCQs: length contracts, relativistic mass exceeds rest mass, time dilation, postulates. (mcq)
2025 Set B MCQs 15 and 16, 2025 Set D MCQ 15, 2026 Set B MCQ 13 and 2026 Set D MCQ 15. Learn the direction of each effect.
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Relativity of simultaneity, and time as the fourth dimension of spacetime. (long)
Not asked in the 2025 or 2026 papers, but short question 6 and comprehensive question 4 of the book and two SLOs cover them.
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