Modern Physics Part 2: Special Relativity Last modified: 23/08/2018
Links Relative Velocity Fluffy and the Tennis Ball Fluffy and the Car Headlights Special Relativity Relative Velocity Example 1 Example 2 Example 3 Energy & Momentum Kinetic Energy Example The Classical Limit Example Massless Particles Photons Radiation Pressure Summary Special Relativity & Time (*) Time Simultaneity & Causality Time Dilation Length Contraction Example Relativistic Doppler Effect Example Starred (*) sections will not be included in exam.
Fluffy and the Tennis Ball Fluffy the cat is watching for birds. A car drives towards Fluffy with speed v car, and a cat-hating passenger in the car throws a tennis ball at her: v ball v car As we should remember: If the ball is thrown at a speed relative to the car of v ball,car, then the speed of the ball relative to Fluffy will be (assuming the car and ball are travelling in the same direction): v ball,fluffy = v car,fluffy + v ball,car The occupants of the car and Fluffy observe different speeds for the ball.
Fluffy and the Car Headlights Later, at night, Fluffy is again by the side of the road. Another car drives towards Fluffy with speed v car, and light from the car s headlights shines on Fluffy. As we know, this light can be considered to be a beam of photons. v car v photon Using the same logic as the previous tennis ball example, the speed of the photons seen by Fluffy would be expected to be: v photon,fluffy = v car,fluffy + v photon,car Again, the occupants of the car and Fluffy are predicted to observe different speeds for the light.
However this prediction is wrong! What actually happens is that the occupants of the car and Fluffy observe the same speed for the light from the headlights: i.e. v photon,fluffy = v photon,car. The familiar formula for relative velocity ( Galilean relativity ) doesn t work for large velocities (large compared to the speed of light). The resolution of this problem is found in Albert Einstein s theory of Special Relativity (1905, the same year as his explanation of the Photoelectric Effect), which connects observations between observers moving in different inertial reference frames. Inertial reference frame refers to a reference frame which is moving with constant velocity i.e. is not accelerating. Non-inertial reference frames are covered in Einstein s later theory of General Relativity (1916).
Special Relativity When developing Special Relativity, Einstein started from two postulates (assumptions): All observers in an inertial reference frame will measure the same speed for light: c = 299, 792, 458 m/s 3 10 8 m/s The Laws of Physics are the same for all observers in inertial reference frames. The resulting theory is very successful in describing the motion of fast-moving objects, but is a little complex mathematically. In this lecture, we will briefly discuss a few of the most important and interesting results of this theory, while avoiding most of the mathematical details.
Relative Velocity Einstein s theory predicts that the speed of an object C observed by B is related to the speeds of B and C seen by an observer A by the formula: v CB = v CA v BA 1 v CAv BA c 2 Note that as in the Galilean expression v AB = v BA. Setting v CA = c in the formula gives v CB = c for any value of v BA, consistent with Einstein s first postulate. When v CA, v BA c then the familiar Galilean relativity formula will give a good approximation: v CB = v CA v BA Because c = 3 10 8 m/s is so very large compared to most everyday speeds, this approximation is very often good enough.
Example A stationary observer A measures the speeds of two objects B and C to be 0.5c and 0.8c respectively (in the same direction): A rest B 0.5c C 0.8c Calculate the speed of C observed by B. We have v BA = 0.5c and v CA = 0.8c and need to calculate v CB. v CB = v CA v BA 0.8c 0.5c 1 v CAv BA v CB = c 1 0.5 0.8 = 0.3c 0.6 = 0.5c 2 Which is very different from the Galilean result: v CB = v CA v CB = 0.3c
Example A stationary observer A measures the speeds of two objects B and C to be 500 m/s and 800 m/s respectively (in the same direction): A rest B 500 m/s C 800 m/s Calculate the speed of C observed by B. Notice that both B and C are travelling faster than the speed of sound (340 m/s), which is very fast for normal objects. The method used is identical to the previous example: v CB = v CA v BA 1 v CAv BA v CB = c 2 800 500 1 500 800 c 2 = 300 1 4 105 c 2
You should be familiar with the binomial expansion: (1 + x) n = n k=0 ( ) n x n 1 + nx k for small values of x You may not know that this also applies for non-integer values of n. 300 v CB = 300 (1 + 4 ) 105 1 4 105 c 2 300.00000000133 m/s c 2 Which this time is extremely close to the Galilean result: v CB = v CA v CB = 300 m/s The difference between calculations will become even less for the lower speeds we saw in Semester 1 problems with cars, bicycles, helicopters etc.
Example A stationary observer A measures the speeds of two objects B and C to be 0.1c and 0.2c respectively (in the same direction): A rest B 0.1c C 0.2c Calculate the speed of C observed by B. In this case, v BA = 0.1c and v CA = 0.2c so: v CB = v CA v BA 0.2c 0.1c 1 v CAv BA v CB = c 1 0.2 0.1 = 0.102c 2 The Galilean result: v CB = v CA v CB = 0.1c differs by 2%, which is small, but not necessarily small enough to ignore.
From these examples we should notice a trend. For small speeds, the Relativistic and Classical calculations are only very slightly different. Using the Classical equations is generally easier mathematically, so is usually preferred. This applied to all of our calculations during Semester 1. For large speeds, the Relativistic and Classical calculations give dramatically different results. We must use Relativity. Classical Physics is inaccurate at these speeds. In between, for speeds starting around 0.1c, the difference is still small, but not small enough to ignore if we are doing very accurate calculations. If we are only interested in an approximate answer, we could get away with using Classical Physics, but Relativity becomes more and more important as speeds increase.
Energy & Momentum Another important equation, derived from Einstein s postulates, connects the energy E of an object and p, the magnitude of its momentum: E 2 = (pc) 2 + (m 0 c 2 ) 2 where the constant m 0 is called the rest mass of the object. When an object is stationary, it will have p = 0 and this formula gives the rest mass energy, E 0 : E 0 = m 0 c 2 The rest mass of an object is a form of potential energy, which can sometimes be converted into other forms of energy.
The energy of an object is connected to its rest mass and speed by the famous equation: E = mc 2 = γm 0 c 2 1 = 1 ( v )2m 0c 2 c Here m = γm 0 is known as the relativistic mass which INCREASES as 1 speed gets larger. The factor γ = appears often in relativity equations. 1 ( v c )2 Using this formula, together with the previous formula connecting E and p, we can find the relativistic definition of momentum: p = mv = γm 0 v When v c is small, γ 1 and the momentum is given by the familiar formula: p = m 0 v. Rest mass is the same as our traditional idea of mass.
Rest mass is a property of a particular type of particle, so two protons for instance travelling at different speeds will have the same value for m 0, but different values of m. What we have just seen for momentum will also happen with other physical quantities. Taking the limit γ 1 in relativistic equations will give our familiar equations for force, energy, torque etc. But as v c then γ and hence E. This means that the work needed to accelerate a particle closer and closer to the speed of light becomes larger and larger. An infinite (thus impossible) amount of work must be to be done to accelerate a mass to the speed of light. The universe has a speed limit. Nothing can travel faster than the speed of light!
Kinetic Energy The extra energy that a moving object has compared to the same object at rest is, of course, the kinetic energy of that object. Using our relativity equations: KE = E E 0 = (γ 1)m 0 c 2 For small values of v c, we can use the binomial expansion as seen earlier: γ = ( ( v ) ) 2 1 2 1 ( v ) 2 1 1 + c 2 c to determine a familiar formula for kinetic energy, valid when v c: ( KE = (γ 1)m 0 c 2 1 + 1 ( v ) ) 2 1 m 0 c 2 = 1 2 c 2 m 0v 2
Example An electron (rest mass = 9.1 10 31 kg) has an energy equal to four times its rest mass energy. For this electron, calculate the: (a) rest mass energy (in ev) (c) momentum (b) kinetic energy (in ev) (a) E 0 = m 0 c 2 = (9.1 10 31 )(3 10 8 ) 2 = 8.19 10 14 J = (8.19 10 14 )/(1.6 10 19 ) ev = 511 kev (b) kinetic energy = total energy - rest mass energy = 4E 0 E 0 = 3E 0 = 1.54 MeV (c) (pc) 2 = E 2 (m 0 c 2 ) 2 = (4m 0 c 2 ) 2 (m 0 c 2 ) 2 = 15(m 0 c 2 ) 2 p = 15m 0 c = 1.06 10 21 Note that using the Classical Physics formula p = 2m 0 KE gives a very different (and incorrect) value p = 6.69 10 22 Ns. Ns
What is the speed of this electron? E = 4m 0 c 2 = γm 0 c 2 γ = 4 = v c = 1 1 γ 2 = 1 1 4 2 = 15 4 v = 0.97c = 2.90 10 8 m/s = 0.97 1 1 ( v c )2 Or, since we have already found the momentum p, we can use the formula: p = γm 0 v v = p 15m0 c = = 0.97c γm 0 4m 0
The Classical Limit Taking γ 1 in relativistic equations is known as the Classical limit, as it gives our familiar Classical Physics equations. When will this be true? There is no exact line where Classical Physics stops and Relativity starts. As we saw earlier in our relative velocity examples, the two methods give very close results for low speeds, but the differences become larger with increasing speed. Remember Relativity is always correct, but using the Classical equations is usually much easier. How do we know if it is safe to use Classical Physics? The answer depends on what we know to begin with.
If We Know Speed The value of the ratio v c can assist us. A commonly used guideline (not rule) is that it is OK to use Classical Physics, when v c 0.1. When v c = 0.1, γ = 1.005 and we would expect only a few percent difference between the calculations. For v c = 0.4, γ = 1.09 and we would expect differences of 10% or more in calculations, so Classical Physics really isn t good enough. 5 4 γ 3 2 1 Classical Physics OK? Must Use Relativity 0 0.2 0.4 0.6 0.8 1 v/c In between, the Classical equations will give an approximate answer correct to one or two figures, which may be good enough. For accurate calculations however, we would need to use the Relativistic equations.
If We Know Kinetic Energy What about if we know the kinetic energy of a particle and need to determine its speed? Do we use the Classical expression KE = 1 2 mv 2? Or the Relativistic formula KE = (γ 1)E 0? Let s rearrange the second formula: KE E 0 = γ 1 Using the γ values calculated on the previous page: v c = 0.1 KE E 0 = 0.005 and v c = 0.4 KE E 0 = 0.09 A suitable guideline would be that Classical Physics is OK when the kinetic energy is less than about 1% of the particle s rest mass, and relativity is definitely required when this ratio reaches about 10%.
Example In the previous lecture discussing the Photoelectric Effect, we calculated the maximum kinetic energy of a photoelectron produced by light of wavelength 450 nm on a sodium surface to be 0.4 ev. We then further calculated (using Classical Physics) the speed of such an electron to be 3.76 10 5 m/s. This an enormous speed! Surely we need to consider relativity? No - Classical Physics will be fine. We know this by comparing the Kinetic Energy, 0.4 ev with the rest mass energy of an electron, 511,000 ev (calculated earlier). We see that KE E 0, which tells us that non-relativistic calculations will be accurate enough. We expect γ for this electron to be very close to 1: v c = 3.76 105 3 10 8 = 0.00125 γ = 1.000001.
Massless Particles The equations of relativity allow for the possibility of particles being massless i.e. to have m 0 = 0. In this case the relationship between total energy and momentum becomes: E 2 = (pc) 2 + (m 0 c 2 ) 2 E = pc The formula E = mc 2 does not apply to massless particles. The equations of relativity further require that a massless particle will always travel at the speed of light c and at no other speed. Remember a massive particle (m 0 > 0) can travel at any speed v < c. (In Physics, massive very big )
Photons We are already familiar with a particle that always travels at the speed of light - the photon. So clearly the rest mass of a photon must be zero. As well as energy, photons have momentum: p = E c = hc/λ c = h λ What is the momentum of one of the photons from a DVD laser with λ = 650 nm? p = h λ = (6.63 10 34 )/(650 10 9 ) = 1.02 10 27 Ns
Radiation Pressure Since photons have momentum, they can apply a force to an object (remember Newton s 2 nd Law: F = dp dt ). Force on an absorbing surface Consider a single photon (with energy E γ and momentum p γ ) being absorbed by a surface that is initially at rest. Before and after the absorption we have to conserve momentum: Before After p γ rest p = p γ A beam of these photons will apply a force F to the surface: F = p ( ) ( ) ( ) no. of photons p P t = = p γ = P second photon c where P is the power of the light source. E γ
Force on an reflecting surface A perfect reflector will just change the direction of an incoming photon, leaving the wavelength unchanged. Again, momentum must be conserved during the reflection: Before After p γ rest p γ p = 2p γ The change in momentum of the surface is twice the change for the absorption case, and so the force on the surface is also two times larger: F = 2P c
Summary of Examinable Material A massive particle (i.e. rest mass, m 0 > 0) can travel at any speed v where v < c = 3 10 8 m/s. At speed v it will have energy: 1 ( v E = mc 2 = γm 0 c 2 = 1 and momentum: 1 p = γm 0 v = 1 ( v )2m 0v c This energy and momentum are connected by: E 2 = (pc) 2 + (m 0 c 2 ) 2 c )2m 0c 2
The rest mass energy of the particle is: E 0 = m 0 c 2 Any additional energy of the particle is its kinetic energy: E K = E E 0 = mc 2 m 0 c 2 = (γ 1)m 0 c 2 When γ 1, Classical Physics equations will be a very good approximation (and with easier maths!). This will be true (roughly) when v c 0.1 or E K E0 0.5% A massless particle (e.g. the photon), with m 0 = 0 can only travel at the speed of light c. With m 0 = 0, the above equation connecting energy and momentum becomes: E = pc
Time is the 4 th Dimension The Relativity equations we have seen so far have really just been slightly more complicated expressions for familiar quantities such as energy, momentum etc. with no major surprises. Things start to become odder when we also consider time. Classical Physics treats time as being separate to position co-ordinates. In relativity, time and position are more closely connected. Events are described by their co-ordinates in 4-dimensional space-time: (x, y, z, t) non-examinable We are familiar from Classical Physics with the idea that observers in different reference frames will measure different spatial co-ordinates (x, y, z) and hence different velocities etc. In relativity, these observers will also measure different times. This can have some surprising consequences, some of which we will now discuss briefly. The maths involved can be quite complex, so we will skip over most of the details, and none of this material will appear in the exam.
Simultaneity & Causality Two students A and B are standing on opposite sides of a brick wall. A third student C is watching and measures the times t A and t B when students A and B throw basketballs, and also the times t A and t B when the balls hit the wall. A t A t A t B t B B C non-examinable If student C observes t A = t B then the two balls are said to be thrown simultaneously. According to Classical Physics, C can move with any speed v, but will always see these two events to be simultaneous, since time and space are considered separate. More generally, Classical Physics predicts that in any moving reference frame, the observed order of events will always be the same. Ball A will always be seen to be thrown first or vice versa.
For large speeds, however, using Relativity, this is not true! Depending on C s speed and direction, he may see either t A < t B or t A > t B. The order of events is dependent on the reference frame. Fortunately, there is still some logic! Student C will always, in any reference frame, measure t A > t A (and also t B > t B). This is because these two events are causally connected. The ball can only hit the wall if it has been thrown first. Every observer will see events in this order. non-examinable If A and B throw the balls at random times, then the throwing of the balls are independent (non-causally connected) events and C s observation of the order in which the events occur can vary in different inertial frames. If the throwing of the balls are instead causally connected (maybe B only throws the ball after first seeing A s throw), then C s observation of the order will be the same in all reference frames.
Time Dilation Two students have identical stopwatches. Student A is at rest while student B is in a box moving at a speed v. Each student now measures the time taken for B s box to move between the two points X and Y - the ends of a long stick (that is hanging mysteriously in mid-air). rest A T A Y T B v B X non-examinable Classical Physics predicts that A s measurement T A, and B s T B should be equal (assuming no experimental error!). Relativity predicts (and experiments confirm) that the two measurements are not equal, but are connected by: T A = γt B (where again γ = 1 (1 ( v c )2 ) )
Recall that γ > 1 when v > 0, so we must have T B < T A. Student A observes Student B experience time running slower than A. This is often summarized by the phrase: Moving clocks run slow B does not notice anything different in terms of time and will in fact see A s clock run slower, because A is moving relative to B. non-examinable This phenomenon of different time measurements in moving reference frames is known as time dilation. Note: In this example B was moving toward A, but the same result is observed for motion away.
Length Contraction A and B are both enthusiastic Physics students, and decide to use these time measurements to calculate the length, L of the stick. rest A T A Y Z T B v B X non-examinable Student A observes B move from X to Y at speed v in time T A, and calculates: L A = vt A Student B observes the stick move past at speed v in time T B, and calculates: L B = vt B Recalling the connection between the time measurements, we find: L A = vt A = v(γt B ) = γl B
Again, recalling that γ > 1 when v > 0, we must have L B < L A. Because Student A is at rest relative to the stick, L A is considered to be the correct, or proper length of the stick. Student B measures the length of the moving stick to be less than this. Moving lengths are measured to be shorter ( contracted ). non-examinable This length contraction only occurs in the direction of motion. Both students would measure the same distance between point Y and point Z on the ground for example.
Example If Students A and B measure times T A = 5 µs and T B = 4 µs respectively, (a) How fast is B travelling? (b) What do A and B measure the length of the stick to be? non-examinable (a) We know T A = γt B, therefore: γ = T A T B = 5 4 = 1.25, and v = c 1 1γ2 = c 1 1 (1.25) 2 0.6c (b) L A = vt A = 0.6c (5 10 6 ) = 900 m L B = vt B = 0.6c (4 10 6 ) = 720 m
Lorentz Transformation This changing of length and time is a little mind-bending. We are avoiding most of the maths involved, but the following example may help to give an idea of what s happening. non-examinable An observer measures the differences x and y between two positions A and B. y A x B y x A second observer in a rotated reference frame measures the differences x and y between the same two positions. y B y x y A x Clearly x > x and y < y, BUT both observers will agree on the distance between the points A and B: L AB = ( x) 2 + ( y) 2 = ( x ) 2 + ( y ) 2 θ x
In Special Relativity, changing to a moving reference frame is similar to a rotation in the x-t plane instead of x-y shown above. (This sort-of rotation is called a Lorentz transformation). Observers in two different inertial frames will measure different values for the intervals x and t between two events, as we ve already discussed. ALL observers will however agree on what is called the space-time separation of the events: ( x)2 (c t) 2 = ( x ) 2 (c t ) 2 non-examinable The different measurements of distance and time in relatively moving reference frames are essentially the different components of the same physical quantity. Our brains (and Classical Physics) treat space and time as separate, but the Universe does not!
Relativistic Doppler Effect Just as we have previously seen with sound, when light is emitted or detected by moving sources and receivers, there is a change in observed frequency due to the motion - the Doppler effect. At high speeds, this requires us to include the effects of Special Relativity. Assume a stationary source S emitting light of wavelength λ S. A receiver R is moving away from S at speed v. In S s reference frame, a wavefront of the light wave reaches the receiver R at time t = 0. The time taken for the next wavefront to reach R is T S : non-examinable S R S λ S vt S v ct S t = 0 t = T S The distance travelled by the light in time T S (in S s reference frame) is: ct S = λ S + vt S R v
Rearranging, we have: T S = λ S c v In the reference frame of R, because of time dilation, this time is measured to be: T R = T S γ non-examinable This time T R is the time measured by R between peaks of the light wave. i.e. the period of the wave. It is connected to the wavelength of light as seen by R: λ R = ct R = ct S γ = c λ S γ c v = (c2 v 2 ) 1/2 λ S = (c v) c + v c v λ S
In terms of frequency this can be rewritten: c v f R = c + v f S As we should expect, (and as seen with sound): f R < f S. If the receiver is instead moving towards the source, then this formula still applies with v v. Unlike the Doppler effect for sound, it doesn t matter which of R and S is moving, or even if they are both moving. The same formula applies, with v being the relative speed between R and S. non-examinable Relativistic Doppler Effect: f R = c + v c + v f c + S OR λ R = v c λ S + v with upper signs = motion away, and lower signs = motion towards
Example A spaceship is travelling at speed 0.2c towards a stationary student. If a passenger shines a green laser pointer (λ = 532 nm) at the student, what wavelength of light does the student measure? What would this measured wavelength be if the passenger continued to aim the laser at the student after the spaceship has passed by? Coming towards, we have λ S = 532 nm and v = 0.2c, so: non-examinable λ R = c v c 0.2c c + v λ S = c + 0.2c 532 = 0.8 532 = 434 nm (blue) 1.2 Moving away, the sign of v changes: c + v 1.2 λ R = c v λ S = 532 = 652 nm (red) 0.8 Even for non-visible wavelengths, light from an approaching source is often said to be blue-shifted and from a receding source, red-shifted.