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Physics 11Unit 9 – Special Relativity

1. Frame of reference• When an event happens, we can describe it in terms of the moment

and the place it occurs. To do this, we need a system called the frame of reference that consists of spatial coordinates (i.e., 𝑥, 𝑦, and 𝑧) and time.

Unit 9 - Special Relativity 2

• Every observer is at rest with respect to his own reference frame. But the descriptions of the same event by observers in different reference frames may be different. It is due to the relative motion. (We have discussed this in Unit 3.)


• In this unit, we shall study special relativity, a scientific theory proposed by Albert Einstein which is well-known even to non-scientific general audience. This theory focus on the motion of objects in an inertial reference frame.

• Inertial reference frame is a special type of reference frames in which the Newton’s law of inertia is always valid. That means, if the net force acting on a body is zero, the body is either at rest or moving at a constant speed.

• Let’s consider the following situation: An accelerating railroad car with a weight hanging inside it, suspended from the ceiling. There are two observers, one inside the car and one outside the car. What will they see about the weight?


• For the outside observer, she will see the suspended weight does not hang straight down because the tension needs to provide a horizontal component to give it an acceleration. This observer is in an inertial frame since the Newton’s law of inertia is valid to her.


• For the inside observer, however, the hanging weight still does not hang straight down but suspended by the cord at an angle 𝜃. To the on-board observer, the weight is at rest. In order for the weight to be displaced, there must be an additional fictitious force (or inertial force) acting on the weight!


• In the second example, an inertial force has to be introduced so that the observation can be explained. However, the force is just an artifact and is unreal!

• Since the Newton’s first law is invalid in this situation, the accelerating railroad car is not an inertial reference frame. In other words, it is a non-inertial reference frame.

• In general, all reference frames which are accelerating are non-inertial reference frames. (Strictly speaking, the Earth is also a non-inertial reference frame. Why?)


2. Postulates of special relativity• The theory of special relativity proposed by Albert Einstein was

developed based on two assumptions.

Assumption 1: The relativity postulate

• If two frames of reference move with constant velocity relative to each other (that means, inertial frames), then the laws of physics will be the same in both frames of reference.

Assumption 2: The speed-of-light postulate

• The speed of light in a vacuum is the same for any observer no mater what the velocity of the observer’s frame of reference is, and no matter what the velocity of the source of light is.


(1) The relativity postulate

• Please note that the special relativity theory only applies in the frames of reference which are not accelerating (or decelerating).

• There is no preference of reference frame used to describe an event as long as different reference frames are moving at constant velocity relative to one another.

• How to describe the following situation in which a car is moving 80 km/h forward?


• Regarding to the situation, there are two equivalent, and equally correct descriptions.

(1) The car is moving at 80 km/h forward relative to the road underneath.

Or

(2) The car is still and the road is moving 80 km/h backwardrelative to the car.

• The second one sounds weird, but is conceptually making sense!


• How do you describe the path of the ball in these two scenarios: (i) you are outside of the van; or (ii) you are inside the van?

• Can you predict the path correctly using the laws of mechanics?


• Indeed, if the same ball is thrown in the same way in these situations, you will observe exactly the same result.

• That means that the laws of physics are identical in these two reference frames.

• There is a deeper implication: You cannot tell if you are stationary or moving at a constant velocity!

• The problem can be resolved only if there exists an absolute reference frame which is truly stationary, but …

• the Earth is not, as it is orbiting around the Sun.

• the Sun is not, as it is rotating around the center of Galaxy.

• even the Galaxy is not, as it is moving in the Universe.

• So, in nature everything is moving relative to something else.


(2) The speed-of-light postulate

• In search of the mysterious aether, a substance believed to fill the empty space, Albert Michelson and Edward Morley performed the following experiment in 1887.


Michelson interferometer

• An animation demonstrating how the interferometer works.


• What they expected to see is that the light observed in January and July should travel at different speeds because of the motion of the Earth relative to the Sun.

• But it turned out that the speed of light remained the same whether the observe is moving toward the light source (i.e., in January) or away from the light source (i.e., in July)! No matter how the speed of light was measured, its value was constantly 3.0 × 108 m/s!

• How could that happen? It seems to have violated the principle of relative motion …


• The outcomes from the Michelson-Morley experiments and related work led to the following dilemma:

• (1) If the results of the Michelson-Morley experiment are correct, that is, the speed of light is constant regardless of the frames of reference in which it is measured, then the well accepted concept of relative motion must be wrong and has to be abandoned! At least, electromagnetic waves do not follow the principle of relative motion within the regime of classical physics.

• (2) If the principle of relative motion is true, then the speed of light will not be the same in all reference frames. It implies that all the discussions of electromagnetic waves (e.g. Maxwell) are wrong!

• Which one is correct?


• Einstein attempted to resolve the dilemma by reconsidering the truemeaning of “time” and “space”.

• In Newtonian physics, time and space are absolute; they are irrelevant to the state and motion of the object. Time goes at a constant pace, and length is constant regardless of the observers.

• Einstein, however, disagreed with that. Let’s consider the following two cases.


• Case 1: Simultaneous events

• What would the people at C and D see respectively? Do the thunderbolts happen simultaneously or not?


• Case 2: Length measurement


3. Time dilation• A stunning consequence of the theory of special relativity is that time

passes at difference paces in different reference frames. Consider the following:


• The time interval measured by the astronaut is:

• But the time interval measured by the person on Earth is:

• These two time intervals are related by the time dilation formula:


∆𝑡0 = 2𝐷/𝑐

∆𝑡 =2𝐷

𝑐

1

1 −𝑣2

𝑐2

∆𝑡 =∆𝑡0

1 −𝑣2

𝑐2

• In this formula:

• ∆𝑡0: the proper time interval, which is the interval between two events as measured by an observer who is at rest with respect to the events and who views them as occurring at the same place.

• ∆𝑡: the dilated time interval, which is the interval measured by the observer who is in motion with respect to the events and who views them as occurring at different places.

• 𝑣: relative speed between the two observers.

• 𝑐: the speed of light in a vacuum


• Example: The spacecraft is moving past the Earth at a constant speed that is 0.92 times the speed of light. The astronaut measures the time interval between successive ticks of the spacecraft clock to be 1.0 s. What is the time interval that an Earth observer measures between ticks of the astronaut’s clock?

[2.6 s]


• Example: Alpha Centauri, a nearby star in our galaxy, is 4.3 light-years away. If a rocket leaves for Alpha Centauri and travels at a speed of 0.95 c relative to the Earth, by how much will the passengers have aged, according to their own clock, when they reach their destination? Assume that the Earth and Alpha Centauri are stationary with respect to one another.

[1.4 years]


• Example: Imagine your age is 30 and that you have a daughter who is 6 years old. You leave on a space trip in the year of 2017 and travel at a speed of 90% of the speed of light for a time of 5 years (as measured by you in the spaceship). When you return, how old will your daughter be?

[42 years old]


• Example: The average lifetime of a muon particle at rest is 2.2 × 10-6 s. A muon created in the upper atmosphere, thousands of meters above sea level, travels towards the earth at a speed of 0.998 c. Find, on the average, (a) how long a muon lives according to an observer on Earth, and (b) how far the muon travels before disintegrating.

(a) The observer on Earth measures the dilated lifetime. Therefore,


∆𝑡 =∆𝑡0

1 −𝑣2

𝑐2

=2.2 × 10−6

1 −0.998𝑐

𝑐

2= 35 × 10−6 s

(b) The distance traveled by the muon before it disintegrates is

• Because the distance that a muon has traveled is measured by the observer on Earth, the dilated lifetime should be used instead of the proper time which is measured in the reference frame of muon.

• This is a very strong evidence that verifies the time dilation. If its lifetime is 2.2 × 10-6 s, a muon particle could only fly

suggesting that it can never reach the Earth.


𝑥 = 𝑣∆𝑡 = 0.998 3.00 × 108 35 × 10−6 = 1.0 × 104 m

𝑥 = 𝑣∆𝑡 = 0.998 3.00 × 108 2.2 × 10−6 = 660 m

4. Length contraction

• Consider the example of the space travel from the Earth to Alpha Centauri (p.24) again. We have showed that the two observer recorded different times for the trip (due to time dilation!). Would they also measure different distances between the Earth and Alpha Centauri?

• From Earth’s observer:

• From the astronaut:


𝐿0 = 𝑣∆𝑡 = 0.95𝑐 4.5 years = 4.3 light − years

𝐿 = 𝑣∆𝑡0 = 0.95𝑐 1.4 years = 1.3 light − years

• The astronaut indeed found Alpha Centauri closer to the Earth than he expected at the beginning! It took him a shorter time and traveled a shorter distance to arrive the star.

• This effect is called the length contraction.

• Consider the following diagrams:


• For the Earth-based observer (part (a)), he measures the time of the trip to be ∆𝑡 and the distance to be 𝐿0. The relative speed of the spaceship is thus

• For the spaceship-based observer (part (b)), he measures the time of the trip to be ∆𝑡0, the distance to be 𝐿. The relative speed of the Earth to it is thus

• Since the relative speeds measured by them have to be the same,


𝑣 =𝐿0∆𝑡

𝑣 =𝐿

∆𝑡0

𝐿

∆𝑡0=𝐿0∆𝑡

• Substituting the time dilation formula into this expression, the following formula is obtained:

which after simplifying yields

• This is called the length contraction formula.


𝐿

∆𝑡0=

𝐿𝑜∆𝑡0

1 −𝑣2

𝑐2

𝐿 = 𝐿0 1 −𝑣2

𝑐2

• The conventions of the length contraction formula are:

• 𝐿0 is called the proper length. It is the distance between two points as measured by an observer who is at rest with respect to them.

• 𝐿 is the length observed by an observer in relative motion with respect to the object.

• Please note that the effect of length contraction of objects is only observed along the direction of motion. Those dimensions that are perpendicular to the motion are not shortened.


• Example: An astronaut, using a meter stick that is at rest relative to a cylindrical spacecraft, measures the length and diameter of the spacecraft to be 82 m and 21 m, respectively. The spacecraft moves with a constant speed of 0.95 c relative to the Earth. What are the dimensions of the spacecraft, as measured by an observer on Earth?

[26 m by 21 m]


• Practice: Calculate the apparent length of a 100 m futuristic spaceship when it is travelling at the speeds given below.

• (a) 0.630 c

• (b) 0.866 c

• (c) 0.999 c


• Example: A rectangular painting measures 1.00 m tall and 1.50 m wide. It is hung on the side wall of a spaceship which is moving past the Earth at a speed of 0.90 c. (a) What are the dimensions of the picture according to the captain of the spaceship? (b) What are the dimensions as seen by an observer on the Earth?

[(a) 1.00 m by 1.50 m, (b) 1.00 m by 0.65 m]


• A challenging question: A meterstick, moving at a constant speed 𝑣, makes an angle of 30° with respect to the direction of motion. What is the value of 𝑣 so that the meterstick would make an angle of 45°with respect to a stationary observer?

[0.816 c]


• Some interesting highlights:

• (1) The term “proper” in proper time and proper length does not imply the idea of correctness in any sense. According to the relativity postulate, there is no preference of reference frames; in other words, there is no one reference frame more preferred to, or more correct than the others. The term “proper” only refers to the status of the observer relative to the objects of interest.

• (2) When two observers are moving at constant speed relative to each other, each measures the other person’s clock to run moreslowly, and each measures the other person’s length, along the person’s motion, to be contracted.


5. Relativistic momentum and mass variation

• In classical mechanics, the mass of any object is an unchanged quantity. That means, the mass of an object measured while it is at rest or it is moving remains constant.

• It implies that objects can be accelerated to whatever speeds, even beyond the speed of light as long as a sufficiently large force is applied. But we know it is not allowed within the framework of special relativity.

• The idea of the variation of mass has also be inspired by the experiments of measuring the mass-to-charge ratio of electrons by Walter Kaufmann in 1901, in which he observed that this ratio changed with the speed of electrons.


• When an object is moving at a speed much smaller than the speed of light, its momentum can be determined by the well-known formula:

• This relationship does not hold anymore, however, when the object is traveling at a speed comparable to the speed of light. In this situation, the definition of momentum has to be modified to the relativistic form:


𝑝 = 𝑚𝑣

𝑝 =𝑚𝑣

1 −𝑣2

𝑐2

• The two equations of momentum differ by the relativistic factor,

1 − 𝑣2/𝑐2, which appears also in the time dilation and length contraction formulas. This factor is always smaller than 1; therefore, the relativistic momentum is always greater than the non-relativistic momentum. Schematically,

• Since the relative speed is constant, the change of momentum implies an apparent change of mass:


𝑚 =𝑚0

1 −𝑣2

𝑐2

• Some key points regarding the variation of mass:

• (1) In the mass variation formula, 𝑚0 is called the rest mass or invariant mass, and 𝑚 is the relativistic mass.

• (2) The rest mass is the value of mass measured by an observer moving along with the object. Therefore, it corresponds to the Newtonian mass described in classical mechanics. The rest mass is the real mass of the object, measurable by the observers in allreference frames!

• (3) The relativistic mass of an object is the effective mass when it is moving at a speed 𝑣. It is not a true mass; indeed it is a quantity related to the energy possessed by the moving object.


𝑚 =𝐸

𝑐2(Einstein’s relation)

• (4) The mass variation formula suggests that the faster the object is moving, the larger the relativistic mass. But there is a limit; if the rest mass of the object is not zero, its speed must be smaller than 𝑐.


• Example: Let the mass of an electron be 𝑚, and the speed of light be 𝑐. Calculate what the momentum of an electron would be at each of the following speeds, according to the Newtonian equation for momentum. Repeat the calculations using the relativistic momentum. Compare your answers.

• (a) 0.10𝑐

• (b) 0.50𝑐

• (c) 0.87𝑐

• (d) 0.9999𝑐


• Example: The particle accelerator at Stanford University is 3 km long and accelerates electrons to a speed of 0.9999999997𝑐, which is very nearly equal to the speed of light. Find the magnitude of the relativistic momentum of an electron that emerges from the accelerator, and compare it with the non-relativistic value. The mass of electron is 9.11 × 10-31 kg.

[Rel: 1 × 10-17 kgm/s; ratio = 4 × 104]


6. Relativistic addition of velocities• The velocity of an object relative to an observer plays a key role in

special relativity. All the fundamental quantities (that is, time, length

and mass) are modified by the relativistic factor 1 − 𝑣2/𝑐2 when the object is moving at a speed comparable to the speed of light.

• In classical physics, calculating relative velocity is rather straightforward. Consider the following example:


• The velocity of the ball relative to the truck, 𝑣BT, is 8 m/s, and the velocity of the truck relative to the ground, 𝑣TG, is 15 m/s. What is the velocity of the ball relative to the stationary observer?

• Using the idea of relative motion, the velocity of the ball relative to the ground, and the stationary observer, is:

• This is the result obtained using the Newtonian mechanics, and it seems to make perfect sense. But it is wrong if relativity is taken into consideration!


𝑣BG = 𝑣BT + 𝑣TG

𝑣BG = 8 + 15 = 23 m/s

• For the case where the ball and the truck are moving along the same line, the special relativity gives the relative velocity:

• Imagine the truck is moving relative to the ground with a velocity of 𝑣TG = +0.8𝑐. A person riding on the truck throws a baseball at a velocity relative to the truck of 𝑣BT = +0.5𝑐. What is the velocity 𝑣BGof the baseball relative to a person standing on the ground?


𝑣BG =𝑣BT + 𝑣TG

1 +𝑣BT𝑣TG𝑐2

𝑣BG =𝑣BT + 𝑣TG

1 +𝑣BT𝑣TG𝑐2

=0.5𝑐 + 0.8𝑐

1 +(0.5𝑐)(0.8𝑐)

𝑐2

= 0.93𝑐

• For a general situation, the relative velocities are related by the velocity addition formula:

• The notations:

• 𝑣𝐴𝐵: The velocity of A relative to B

• 𝑣𝐴𝐶: The velocity of A relative to C

• 𝑣𝐶𝐵: The velocity of C relative to B

• When 𝑣𝐴𝐶 and 𝑣𝐶𝐵 are much smaller than the speed of light, this formula is reduced to the non-relativistic formula.


𝑣𝐴𝐵 =𝑣𝐴𝐶 + 𝑣𝐶𝐵

1 +𝑣𝐴𝐶𝑣𝐶𝐵𝑐2

• The addition formula allows us to calculate the relative velocities compatible to the speed-of-light postulate.

• For example:

• The speed of light relative to the ground-based observer is:


𝑣𝐿𝐺 =𝑣𝐿𝑇 + 𝑣𝑇𝐺

1 +𝑣𝐿𝑇𝑣𝑇𝐺𝑐2

=𝑐 + 𝑣𝑇𝐺

1 +𝑐𝑣𝑇𝐺𝑐2

=𝑐 𝑐 + 𝑣𝑇𝐺𝑐 + 𝑣𝑇𝐺

= 𝑐

• Example: Rocket A travels to the right and rocket B travels to the left, with velocities 0.8𝑐 and 0.6𝑐 respectively, relative to the Earth. What is the velocity of rocket A measured from rocket B?

• The speed of A relative to the Earth:

• The speed of B relative to the Earth:

• It is equivalent to say that the speed of the Earth relative to B is:

• Therefore, the speed of A relative to B is:


𝑣𝐴𝐸 = 0.8𝑐

𝑣𝐵𝐸 = −0.6𝑐

𝑣𝐸𝐵 = 0.6𝑐

𝑣𝐴𝐵 =𝑣𝐴𝐸 + 𝑣𝐸𝐵

1 + 𝑣𝐴𝐸𝑣𝐸𝐵/𝑐2=

0.8𝑐 + 0.6𝑐

1 + (0.8𝑐)(0.6𝑐)/𝑐2= 0.946𝑐

• Example: Consider a radioactive nucleus that moves with a constant speed of 0.5𝑐 relative to the laboratory. The nucleus decays and emits an electron with a speed of 0.9𝑐 relative to the nucleus along the direction of motion. Find the velocity of the electron in the laboratory frame.

[0.966𝑐]


• Example: System B is moving to the right relative to A with a velocity of 0.9𝑐. Observer B fires a rocket along the direction of this velocity with a speed of 0.8𝑐 relative to himself. What is the velocity of the rocket as measured by A, if the rocket moves (a) to the right; and (b) to the left?

[(a) 0.988𝑐; (b) 0.36𝑐]



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