Special Relativity
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Number 149
Factsheet
Physics
Special Relativity is about the idea that there can be no absolute motion, you can only measure motion, length, mass and time dependent upon your position as an observer.
The idea of mass – energy equivalence also arises from Special Relativity theory.
Einstein’s Postulates of Special Relativity:
1. The Laws of Physics apply equally in all inertial frames of reference
2. The speed of light in a vacuum is measured as the same in all inertial frames
Evidence that supports the Theory of Special Relativity
The Michelson-Morley experiment shows that the speed of light is constant regardless of your apparent motion. It was this experiment that led to the development of the Theory of Special Relativity. What do Einstein’s Postulates of Special Relativity mean? 1. If you drop a ball, it falls, bounces and comes back up. If you are on a bus at constant speed, dropping the ball gives exactly the same result provided the bus is NOT ACCELERATING. Both are inertial frames of reference. Any experiment performed on one can be repeated on the other and the same result obtained. This means that any inertial frame of reference is essentially the same.
2. Ships A and B move away from the sun at constant speed. Even if A moves at ½c and B at ¾c, they are both inertial frames of reference and as such each will measure the speed of light with respect to their own ship as c (3 × 108 m s-1).
SUN
A B
½ c ¾c
Why?
From postulate 1 an observer on ship A could claim that A is stationary and that the Sun and ship B are moving away from A, so it is logical that it should measure the speed of light as c. Observers at Ship B and the Sun could both claim the same thing (that they are stationary) and so can also measure speed of light as c.
All have an equally valid claim and experiments show that they would indeed each measure speed of light to be the same.
Inertial Frames of Reference
These are any frame of reference with no external forces acting upon it or any of the objects inside it. If you place a ball on the floor of an aeroplane in level, uniform flight, the ball will remain still, obeying Newton’s Laws. The aeroplane is an inertial frame of reference.
However when it is accelerating the ball will roll backwards. For an observer on the aircraft, the ball appears not to be obeying Newton’s Laws of motion. The aeroplane cannot be an inertial frame of reference.
Time Dilation
Example Exam Question 1
(c = 3 × 108 m s-1)
(a) A train travels along a level straight track. Show that at a uniform speed of 30 m s-1 there is no significant effect of time dilation.
[2 marks] (b) As the train passes a point on the embankment a stop clock on
the train and on the platform are started simultaneously (i) An observer on the platform simultaneously reads the
platform clock 7 minutes later, and reads the clock on the train as 6 minutes. What is the new speed of the train as measured from the observer? [2 marks] (ii) What is the speed of the station as measured from the train?
[1 mark] (iii) An observer on the train can see both clocks. When the train clock reads 7 minutes, state what time the station clock
will read. [1 mark]
An observer watches another frame of reference moving at speed v with respect to him.
t0 is the time measured in the frame of reference (proper time)
t is the time measured by the observer v is the speed of the frame of reference
with respect to the observer c is the speed of light, (3 × 108 m s-1)
Exam Hint
:-• t0, the dilated time, is always smaller than t.
• when given a speed as a fraction of c, eg 0.5c, use 0.5c in the equation for ease. The two c’s will cancel out making the calculation simpler.
(
02 2)
t t 1 v c = −Example Exam Question 1 Answers
(a) You have to demonstrate that t ≈ t0.
A good approach is to show that t/t0≈ 1
(
02 2)
t t 1 v c = −substitute in the velocity and the speed of light and divide by t0
(
2 8 2) (
16)
0 t 1 1 t = 1 30 ( 3 10 )− × = 1 900 ( 9 10 )− ×(
14)
1 1 1 1 1 1 10− = ≈ = − × (As 1 - 1×10-14 = 0.99999999999999 ≈ 1 ) 0 0 t 1 t t t ≈ ⇒ ≈Hence relativistic effects are negligible as time is almost equal to dilated time.
b (i) For ease, convert times to seconds and decide which time is which:
the moving reference frame is the train, so the clock on the train which is measuring 6 minutes is t0. t is 7 minutes, the time as measured by the observer on the platform. t0 = 6 min = 360 s
t = 7 min = 420 s
You must rearrange the equation carefully. Practise this until you can get the equation shown below.
b (ii) From the inertial reference frame of the train, the station is moving backwards and the train is stationary. Therefore, the speed of the station as measured from the train is the same as the speed of the train as measured from the station. v = 1.13 × 108 m s-1
b (iii)6 minutes (360s)
As the station is viewed to be moving, then the situation from (b) (i) is reversed. If the station is moving with speed 1.1(3) × 108 ms-1 then it will experience time dilation just
as the train did when viewed from the platform.
We could put all the numbers in again but as we have done this calculation already to find the speed we know that the dilated time is 6 minutes (360s)
Notice this result with care!
The observer on the platform sees the train is moving. He reads the station clock as 7 minutes and the clock on the train as 6 minutes.
From the reference frame of the train:
The observer on the train sees the platform moving. When the train clock reads 7 minutes he sees the clock on the platform as 6 minutes.
This may seem to make no sense but it is the essence of relativity in that you get different results depending on your view point.
Exam
Hint:-• Some questions will involve algebraic manipulations so you should practice rearranging the formulae until you are comfortable with doing it.
• These calculations are quite complicated to enter correctly onto a calculator so ensure that you can get the answer shown in the example. Be careful to use brackets as shown so that you are square rooting the correct parts.
• “State…” indicates that no calculation is needed, so the answer should be obvious.
• These equations use ratios, which means you do not have to convert the units as was done in the above example. You could use minutes and get the same answer.
Length contractions (Lorentz contractions)
A ship is measured by the astronaut to be 100 m long.
100 m
As it flies past Earth at close to the speed of light it is measured to be less than 100 m long. The length has contracted.
If the astronaut tries to measure the length of the ship, the tape measure he uses is also moving and would contract by the same amount, so he would still measure it as being 100 m long. From the point of view of the astronaut, nothing about the ship has changed.
Notice that the ship does not shrink, it simply gets shorter in the direction it was travelling.
l0is the proper length, the length of the object when measured at rest (eg the length of the ship as measured by the astronaut).
l is the length as measured from a frame that is moving with respect to the object. (eg the Earth observer who is not moving alongside the ship and therefore is moving ‘with respect to it’.)
(
2 2)
0 l l= 1 v c−(
)
1 2 0 0 2 2 t t t v c 1 t 1 v c ⎛ ⎞ = ⇒ = ⎜⎝ − ⎟⎠ − 1 2 8 360 8 1 3 10 1 1.13 10 ms 420 − ⎛ ⎞ = × ⎜⎝ − ⎟⎠ = ×Physics Factsheet
149. Special Relativity
Reversing the situation
From the frame of the ship it can be said that the ship is stationary and the Earth is moving. From this view point the ship would see the Earth contract and look like a squashed ball in the direction it was apparently moving.
Reducing the distance
A very fast car travelling along a road will be seen to contract by an observer at the roadside.
v
For the driver, however, it is the road that is moving in relation to her. Whilst her car stays the same size, the road contracts.
v
The result is that at speeds approaching the speed of light the distance travelled by the car decreases as measured from the car.
Example Exam Question 2
A missile in a silo is measured to be 143cm long and 28 cm in diameter at its widest point. An on board computer will measure the length of the missile during the flight. (c = 3×108 ms-1)
(a) The missile is fired at 1×108 ms-1 with respect to its launch pad
(i) What is the length of the missile according to the on board instruments? [1 mark] (ii)What is the length of the missile as measured by an independent observer on the launch pad? [2 marks] (iii)What is the diameter of the missile as measured by this same observer? [1 mark] (b) The missile travels a distance of 1 km as measured from the
launch pad
(i) What is the distance travelled by the missile as measured by it’s on board computer? [2 marks] (ii) Show that the speed as measured by the launch pad observer is the same as the speed measured by the on board computer. [2 marks]
Example Exam Question 2 Answers
(a) (i) The length of the missile as measured at rest is 143 cm. As the computer and measuring equipment are at rest with the missile, then the computer reads the length as 143 cm. (ii)l is the length as measured from the launch pad (the
contracted length)
(Correct substitution of numbers) (correct answer & unit) (iii) The dimensions of the rocket only contract in the direction it is travelling. Therefore, the diameter remains unchanged at 28 cm
(b) (i) The proper length is l0, the distance travelled by the missile as measured from the launch pad. l is the distance as measured by the missile.
= 0.9428 = 0.84 km (Correct substitution of numbers)
(correct answer & unit)
Exam
Hint:-We can check that this is an appropriate answer as we would expect this length to be smaller but as the speed is less than half the speed of light then the change is not too significant. (b) (ii)Speed measured from launch pad = 1×108 ms-1
speed measured
= distance measured by missile = l/t0
from missile time measured by missile
speed measured by missile
l0 /t is the distance as measured from the launch pad divided by the time as measured by the launch pad. This is clearly the speed as measured by the launch pad. Therefore, speed from launch pad = speed as measured by the rocket.
= speed measured from launch pad
Exam
Hint:-• Leaving the length as centimetres in aii) is fine just remember the answer will be in centimetres. If you want the answer in metres then use l0 = 1.43 m. Likewise with bi) which was left as km.
• As a check, l is always less than l0
If an observer measures the mass of an object moving relative to them, it will increase as the speed of the object increases: m0 is the mass of the object measured at
rest (proper mass) m is the relativistic mass
as v gets close to c then
Exam Hint:- as a check, m0 is always less than m
the closer to light speed an object gets, the closer its mass gets to being infinite. Therefore a near infinite force is required to get any more acceleration.
Relativistic mass increase
Particle accelerators and the speed of light
The Large Hadron Collider accelerates protons to 99.9999991% the speed of light. To reach 100% is impossible:
2 0 2 v l l 1 c = − =
(
)
(
)
2 8 2 8 1 10 143 1 134cm 3 10 ⎛ × ⎞ ⎜ − ⎟ = ⎜ × ⎟ ⎝ ⎠(
02 2)
m m 1 v c = −(
02 2)
m m 1 v c = −(
02 2)
(
0)
0 m m m m 0 1 1 1 c c = = = → ∞ − −(
)
(
)
2 8 2 0 2 8 2 1 10 v l l 1 1 1 c 3 10 ⎛ × ⎞ ⎜ ⎟ = − = ⎜ − ⎟ × ⎝ ⎠(
)
(
)
2 2 0 0 2 2 0 l 1 v c l l t t 1 v c t − = = = −How much of an effect does relativity really have?
Graph showing how the relativistic mass of a 1 kg object increases with speed. Speed given as fraction of speed of light
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
proportion of speed of light /c 0 1 2 3 4 5 6 relativistic mass /kg
You have to be travelling at 0.14c (4.2×107 m s-1) before you would notice a 1% increase in relativistic mass. The same is true for length
contraction and time dilation. At day to day speeds, the effect is unnoticeable.
This means that Newton’s Laws become unusable at relativistic speeds. For example: F = ma can only be used where the mass does not significantly increase due to relativity.
Mass Energy equivalence
The most common uses of this equation are: • Matter-antimatter annihilation
• Pair production
• Binding energy in nuclear fusion and fission
Mass, m, of a particle is equivalent to energy, E, depending on the speed of light squared. 1kg of mass is the same as 9×1016 J of energy!
It takes 9×1016 J of energy to produce 1 kg of mass.
E = mc2
When moving at relativistic speeds you must use the relativistic mass. The equation becomes:
(
)
2 0 2 2 m c E 1 v c = −Physics Factsheet
149. Special Relativity
Example Exam Question 3
(a) The relativistic mass of a particle is 10% higher than the rest mass. What is its velocity? [2 marks] (b) Show that the relativistic mass of a particle will always be larger than the rest mass. [2 marks] (c) Explain why the radius of a circular particle accelerator is always built slightly larger than is thought necessary by normal laws of circular motion. [2 marks]
Example Exam Question 3 Answers
a ) b ) as v < c, then v2/c2 < 1 (1- v2/c2) < 1 So
When m0 is divided by a number less than 1, the answer is always larger than m0. Therefore the relativistic mass will always be bigger.
(
2 2)
0 m 110% 1 1.1 m =100% = = 1 v c− 2 1 v c 1 1.1 = −(
02 2)
m m 1 v c = −(
1 v c2 2)
1 − <Practice Questions
For all questions c = 3×108 ms-11. (a) Explain what is meant by: (i) proper time, (ii) proper length? [2 marks] (b) (i) What was the purpose of the Michelson Morley experiment? [2 marks] (ii) What were the implications of the result? [3 marks] (c) State the two postulates of special relativity. [2 marks] (d) A moving bus can, in theory, be taken as an inertial frame of reference, but why in practice is it not? [2 marks] (e) In a perfectly smooth and fully enclosed lift moving at uniform speed, what experiment could be performed within the lift to prove that it is moving? [1 marks] 2. A spaceship passes Earth at 0.7c and flashes a beam of light for 3 µs. (a) What is the duration of the flash as measured from the Earth? [2 marks] (b) What distance does the ship travel in that time as measured from: (i) the ship (ii) the Earth [2 marks] (c) If the ship has mass 1.34×106 on the launch pad, what is its
mass as measured from Earth while it is in flight? [2 marks] 3. (a) Is it possible for a particle accelerator to accelerate a proton
to the speed of light? [1 mark] (b) Explain your answer. [1 mark] (c) Suggest how is it possible for photons (light particles) to travel at the speed of light? [1 marks] (d) (i) What is the rest energy of a proton (mp = 1.67×10-27)?
[2 marks] (ii) What is the relativistic energy of the proton travelling at 200,000,000 m s-1? [2 marks]
(e) The accelerator is 1.2km long. What is the relativistic distance travelled by the protons at this speed? [2 marks]
Exam Hint
:-Practisethis type of question for time and length too. (c) to move an object in a circle a centripetal force of F = mv
2 r
is needed. The relativistic mass of the particle will increase as it moves faster, so m would be larger than expected. As r ∝ m, the radius of the circle travelled by the particles would increase as the mass increased.
Answers
1. (a) (i) The time for an event as measured by someone at rest to the event (eg the time for a car to complete a journey as measured inside the car).
(ii) The length of an object when measured at rest to it. (b) (i) To prove the existence of the ether, measure the
(absolute) speed of the Earth through the ether. (ii) There is no ether (or it cannot be detected), there is no
‘absolute speed’ (speed can only be given relative to other objects), the speed of light is measured as the same with respect to an observer regardless of how that observer is moving.
(c) 1. The Laws of Physics apply equally in all inertial frames of reference.
2. The speed of light in a vacuum is measured as the same in all inertial frames
(d) Bumps in the road make it bounce (accelerate up and down). The bus will also change speed during the journey. (e) None
2 (a) v = 0.7c, t0 = 3ì s, t = ? (correct numbers identified & substituted)
(b) distance = speed × time (ì = ×10-6)
(i) = (0.7 × 3×108) × 3×10-6) = 630 m
(ii) = (0.7 × 3×108) × 4.2×10-6) = 882 m
(c)m0=1.34×106, m=?, v=0.7c (correct numbers substitute) = 1.2×108 m s-1
3. (a) No
(b) As v approaches c, the mass → ∞. It is not possible to get a force big enough for the required acceleration
(c) they have zero rest mass
(d) (i) E = mc2 = 1.67×10-27× (3×108)2 = 1.50×10-10J
(ii)
(substitution) (answer and unit) (e) l0=1.2 km, l=? (and substitution)
= 1.2 × 0.7454 = 0.89 km
Acknowledgements:
This Physics Factsheet was researched and written by Kieron Nixon
The Curriculum Press,Bank House, 105 King Street,Wellington, Shropshire, TF1 1NU Physics Factsheets may be copied free of charge by teaching staff or students, provided that their school is a registered subscriber.
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