Chapter 12 Electrodynamics and Relativity Ether 12.1 The Special Theory of Relativity Properties of the ether: Since the light speed c is enormous, the ether had to be extremely rigid. So it did not impede the Ether: Since mechanical waves require a medium to motion of light. For a substance so crucial to electro- propagate, it was generally accepted that light also require a magnetism , it was embarrassingly elusive. Despite the medium. This medium, called the ether, was assumed to peculiar property just mentioned, no one could detect its pervade all mater and space in the universe. ghostly presence.
“Absolute” frame: The Maxwell’s equation was inferred Efforts to detect the ether: Michelson inspired by the that the speed of light should equal c only with respect to Maxwell took the problem of detecting the ether as a ether. This meant that the ether was a “preferred” or challenge. He developed his interferometer and used it to try “absolute” reference frame. to detect the earth’s motion relative to the ether. The result were not conclusive. ∂ 2 E 1 ∂ 2 E − = 0 ∂x2 c2 ∂t 2 1 2
The Michelson-Morley Experiment The Michelson-Morley Experiment (II) Michelson and Morley wanted to detect the speed of the Parallel: earth relative to the ether. If the earth were moving relative to L L (2L / c) the ether, there should be an “ether wind” blowing at the T = 0 + 0 = 0 same speed relative to the earth but in the opposite direction. 1 (c − v) (c + v) (1− v2 / c2 ) Michelson-Morley interferometer: Use light speed Perpendicular: variation to verify the existence of ether. 2L (2L / c) T = 0 = 0 2 (c2 − v2 )1/ 2 (1− v2 / c2 )1/ 2
L v2 ∆T = T −T ≅ 0 ( ) 1 2 c c2
3 4 The Michelson-Morley Experiment (III) The Two Postulates The two postulates in the theory of special relativity are: Using v=30 km/s, the expected shift was about 0.4 fringe. 1. The principle of relativity: All physical laws have the Even though they were able to detect shifts smaller than same form in all inertia frames. 1/20 of a fringe, they found nothing. 2. The universal speed of light: The speed of light in free space is the same in all inertial frames. It does not depend Possibilities: on the motion of the source or the observer. z The ether was dragged with the Earth. z No ether. Both postulates are restricted to inertial frames. This is why the theory is special. z Constant light speed. •The principle of relativity extends the concept of covariance from mechanics to all physical laws. •The constancy of the speed of light is difficult to accept at first. All the experimental consequences have confirmed its
5 correctness. 6
Some Preliminaries Some Preliminaries (II) Event: Event is something that occurs at a single point in A reference frame is assumed space at a single instant in time. to consist of many observers Observer: An observer is ether a person, or an automatic uniformly spread through the device, with a clock and a meter stick. Each observers can space. Each observer has a record events only in the immediate vicinity. meter stick and a clock to make measurements only in the Reference frame: A reference frame is a whole set of immediate vicinity. observers uniformly distributed in space. The frame in which an object is at rest is called its rest frame. To synchronize four equally spaced clocks, a signal is sent Synchronization of clocks: It is extremely important to out by clock A to trigger the other clocks---each of which has define precisely what is meant by the time in a given been set ahead by the amount of time it takes to travels from reference frame. This requires a careful procedure for the A to the given clock. synchronization of clocks.
7 8 Relativity of Simultaneity Relativity of Simultaneity How can we determine whether two events at different (another example) locations are simultaneous? Two events at different locations are simultaneous if an Two events that are simultaneous in one inertial observer midway between them receives the flashes at the system are not, in general, simultaneous in another. same instant. Relativity of Simultaneity: Spatially separated events that are simultaneous in one frame are not simultaneous in another, moving relative to the first.
9 10
Geometry of Relativity: Time Dilation Time Dilation (II) How does the relative motion of two frames affects the Now let us find the time interval recorded in the frame S, in measured time interval between two events? which the clock has velocity v. The time interval ∆t in frame S measured by two observers A and B at different positions. ∆∆tt ()cLv⋅=+⋅22 () 2 2L 220 τ = 0 c 2L 1 Tt=∆ =0 ⋅() c 1/− vc22 1 TT==γγ0 where A proper time, τ, is the time interval between two events as 1/− vc22 measured in the rest frame of a clock. In this frame both events occur at the same position. (Note: properÆown), Note that we have used c as the speed of light in both 11 frames---in accord with the second postulate. 12 Time Dilation (III) Time Dilation (IV) Since γ>1, the time interval T measured in frame S (by two
Another example: clocks) is greater than the proper time, T0, registed by the clock in its rest frame S’. The effect is called time dilation. h τ = (proper time) Two spatially separated clocks, A and B, record a greater c time interval between two events than the proper time recorded by a single clock that moves from A to B and is hvt22+∆() 1 h present at both events. ∆=tt ⇒∆= cc1/− vc22 1 ∆=t γγτ , where = 1 1/− vc22 γ = 1− v2 / c2 Moving clocks run slow.
13 14
Example of Time Dilation Geometry of Relativity: Length Contraction Experimental evidence (muon decay): Consider a rod AB at rest in frame S, as shown below. The The reality of time dilation was verified in an experiment distance between its ends is its proper length L0: performed in 1941. The proper length, L0, of an object is the space interval between its ends measured in the rest frame of the object. Rest frame at ground: An elementary particle, the muon (µ), decays into other particle. The particle decay rate is N = N e−t /τ τ0 where = 2.2 µs is called the mean lifetime.
Moving frame at the upper atmosphere: Another source of generating muon is bombarded with cosmic ray protons. The muon generated with this method has the speed of v=0.995c. The mean lifetime is 10 times longer than their cousins that decay at rest in the laboratory. 15 16 Length Contraction (II) Length Contraction (III) Another example: An observer O’ in Frame S’, which moves at velocity v relative to frame S, can measure the rod’s length. By recording the interval b times at which O’ passes A and B. The measurements in the two frames are
cτ Frame S : L0 = ∆x = v∆t 1 LL00= ( : proper length) ⇒ L = L0 2 Frame S′: L = ∆x′ = v∆t′ γ Lvt+∆ Lvt −∆ ∆=tt12, ∆= 12cc LL L1 ∆=tt, ∆= ⇒∆=∆∆= ttt + 2 12cv−+ cv 12 c1/ − v22 c ccγγγ111 Lt=∆==γτ L 22220 Moving objects are shortened. 17 18
Effects of Length Contraction (I) Length Contraction Effects (II)
distortion rest frame
v = 0.0 c v = 0.5 c
muon frame v = 0.95 c v = 0.99 c
19 20 The Twin Paradox The Barn and Ladder Paradox Nothing in the theory of relativity catches the imagination more than the so-called twin paradox. Twin A stays on earth while twin B travels at high speed to a nearby start. When B returns, they both find that A has aged more than B. The paradox arises because of the apparent symmetry of before farmer’s view the situation: In B’s frame, it is A that leaves and returns, so one should also find that B has aged more than A. Who’s right? A > B ? What’s going on? B > A ladder’s view
21 22
The Lorentz Transformation The Addition of Velocity
The laws of electromagnetism are not covariant with respect ′ ′ ′ ′ to the Galiliean transformation. However with Lorentz dx =γγ (dx + vt ) = γdt (ux + v) transformation they are covariant. The space and time are vdx′ u′v dt = (dt′ + ) = γdt′(1+ x ) related shown as follows: c2 c2 Rest frame x′ = γ (x − vt) Taking the ratio of these equations we find u′ + v vx u = x ′ x 2 t = γ (t − 2 ) 1+ u′v c c x Moving frame x = γ (x′ + vt′) A extreme case c + v vx′ when u′x = c, we have ux = = c ′ 1+ cv c2 t = γ (t + 2 ) c 23 24 Covariant Vector, Contravariant Vector, The Structure of Spacetime: (i) Four-vectors γβ and Invariant Quantity v 0123γβ aaaaa= () xctxxxyxz≡===, , , , and β = the covariant vector (row): a µ 0123 c µ 0123 ≡−()aaaa xxx001=−(), a0 112 xxx=−() 1 the Lorentz transformations µ µ a 22 the contravariant vector (column): a a = xx= a2 33 3 xx= a xx00γγβ− 00 invariant quantity under Lorentz transformation xx11− 00 3 the Einstein 0 γβ γ µµv summation convention b ==Λ xxv 22∑ 1 xx0010 v=0 3 µµ µµ0123b ab= ab abµµ==− ab a a a a µ µ 330001 ∑ ()2 xx v=0 b 3 b the Lorentz transformation matrix 00 11 22 33 0 0 11 2 2 3 3 25 =−ab + ab + ab + ab =− ab + ab + ab + ab 26
The Invariant Interval Space-Time Diagrams (Minkowski Diagrams)
01 23 01 23 Two event A and B occurs at (xxxxA ,AAA , ,) and ( xxxx BBBB , , , )
22 2 µ µµ lightlike (I == 0, ct d ) the displacement 4-vector: ∆≡xxxA −B
the interval between two events: I ≡∆ xxµ ∆µ =− ctd22 + 2
timelikeIctd<> 0 (22 2 ) spacelike Ictd>< 0 (22 2 ) lightlikeIctd== 0 (22 2 ) timelike (I <> 0, ct22 d 2 )
22 2 27 spacelike (Ictd>< 0, )? Occurs at the same time. 28 12.2 Relativistic Mechanics Proper Velocity 4-Vector (4-velocity) 12.2.1 Proper Time and Proper Velocity Proper time τ is invariant, whereas “ordinary” time t depends How to define the velocity? on the particular reference frame. µ µ Imagine you are on a flight to Moon, and the pilot announces Proper velocity has an enormous advantage over ordinary that the plane’s velocity relative to ground is 4/5c. velocity: it transforms simply. dl dx The numerator, dxµ , is a displacement 4-vector; u = the ordinary velocity η ≡ ηη dt d The denominator, d , is invariant. ηητ However, your watch runs slow due to time dilation. You ηγηβη001=−(), might be more interested in the distance covered per unit 112=−() proper time. ηγηβη More generally, µµv 22 η =Λvη dl 1 = τ η = =u the proper velocity 33 22 = dτ 1/− uc dct dt c η 0 ==c = 22 Which definition is more preferable/useful? 29 dd1/− uc 30 ττ
Mass-Energy Equivalence 12.2.2 Relativistic Energy and Momentum
Mass-energy equivalence: Since no internal process How to define the momentum? can even move the center of mass of a system, we can derive mass-energy equivalence. In classical mechanics Momentum is mass times velocity, but immediately a question arise: Should we use ordinary velocity or proper velocity? There is no priori reason to favor E EL one over the other. mv, x v t mu = ∆ = ∆ = 2 p ≡=mη the relativistic momentum c Mc 1/− uc22 − M∆x + mL = 0 = mmrelu rel : the relativistic mass
00 dt mc E 2 pm==η mc = = E = mc dc1/− uc22 mcτ2 where E ≡ relativistic energy 22 31 1/− uc 32 Kinetic Energy Conservation and Invariant Conserved quantity: same value before and after some How to define the kinetic energy? process. The relativistic kinetic energy is the total energy minus the Invariant quantity: same value in all inertia frame. rest energy: Mass is invariant, but not conserved. 2 mc 2 Charge is both conserved and invariant. EEEkin =−res = − mc 1/− uc22 Energy is conserved, but not invariant. 13uu24 Momentum is conserved, but not invariant. =+++−mc2 (1 ... 1) 28cc24 4 Velocity is neither conserved nor invariant. 132 mu =++mu 2 ... 28c Invariant: ppµ =−() p02 + ( pp ⋅ ) =− mc 22 µ 2 1 2 E Kmu= the classical defination of the kinetic energy ⇒−= pmc222 2 c2 33 34
12.2.3 Relativistic Kinematics Massless Particle: Photon Explore some applications of the conservation law to particle In classical mechanics there is no such thing as a massless decays and collisions. particle.
Example 12.7 Two lumps of clay, each In special relativity, p and E are still proportional to m. If u=c, of (rest) mass m, collide head-on at 3/5c. then the zero numerator is balanced by a zero in the They stick together. Question: what is denominator, leaving p and E indeterminate (zero over zero). the mass (M) of the composite lump? mu 0 p == 1/− uc22 0 When uc==⇒ and m0, mc2 0 Example 12.8 A pion at rest decays into E == a muon and a neutrino. Find the energy 1/− uc22 0 of the outgoing muon, in terms of the A massless particle could carry energy and momentum, two masses, mπ and mµ (assume mµ=0) provided it always travels at the speed of light.
Does it make any sense? 35 E ==pc hv photon 36 The Compton Effect (Example 12.9) The Compton Effect (ii) The derivation of Compotn’s scattering: Classically an For knownλ X ray frequency and final particle momentums electromagnetic wave carries moment given by p=E/c. We can furtherλ solveθ these two equations. Conservation of linear momentum: λ 2 2 2 ( p − p ′ cosλ ) + ( pλ′ sinθ ) = p
pppx : λλ=+′ cosθφ p cos 2 λ 2 ( p − p ′ ) + 2( p − pλ′ )m0c = p pp: 0=− sinθφ p sin hf h y ′ p = = λ 222 c λ Further solving these two equations, we obtain λλ λθθ (pp−+′′ cos ) ( p sin ) = p (1) ()ppmcpp−=− (1cos)θ Conservation of energy: λλ′′ λλ0 11 1 2 −=(1 − cosθ ) hf=+ hf′ K; K =− (γ 1) m0 c ppmcλθ ′ λλ0 h ()cpλλ−= cp′ K ⇒∆ =(1 − cos ) 2222 mc KKmccp+=2 0 0 h 22 = 0.00243 nm is called the Compton wavelength. (pp−+−′′ ) 2( ppmcp )0 = (2) mc λλ λλ 37 0 38
12.2.4 Relativistic Dynamics Work-Energy Theorem Newton’s laws The work-energy theorem (“the net work done on a particle Newton’s first law is built into the principle of relativity. equals the increase in its kinetic energy”) holds relativistically. dmpu Fp===, where mrel u Newton’s second law retains its validity in relativistic dt 1/− uc22 mechanics, provided we use the relativistic momentum. ddddmppl u Wd≡⋅=Fl ⋅= d l ⋅ dt =() ⋅ u dt dp mu ∫∫ ∫ ∫ 22 F = where pu==mrel dt dt dt dx 1/− uc dt 1/− uc22
d muu11 22 dm− udu ⋅=uuu22 1/ −uc − m 2 ⋅ Newton’s third law does not, in general, extend to the dt1/−−uc221/− u c dt 1/ uc 22 c dt relativistic domain due to the relativity of simultaneously. 11uduudu22 du =−+=33(122 )mu mu mu 22cdtcdt 22 dt Only in the case of contact interactions, where the two forces (1−−uc / )22 (1 uc / ) are applied at the same physical point, can the third law be dmcdE2 dE ==⇒==− WdtEE retained. 22 ∫ final initial 39 dt1/− uc dt dt 40 The Ordinary Force and The Minkowski Force Example 12.12: Hidden momentum The ordinary force: F is the derivative of momentum with As a model for a magnetic dipole m, consider a rectangular respect to ordinary time, transformation is ugly (both the loop of wire carrying a steady current. Picture the current as a numerator and denominator must be transformed). stream of noninteracting positive charges that move freely dp dp dp dt F yy y y within the wire. When a uniform electric field E is applied, the Fy ==βββ = = dtγγγ()(1)(1) dt−−−ccc dx uxx u charges accelerate in the left segment and deccelerate in the
dpzz dp dp z dt F z right one. Find the total momentum of all charges in the loop. Fz == = = dt()(1)(1) dt−−− dxβββ u u γγγcccxx Solution: 0 dpxxdp dp dE 0 −− The current is the same in all four dp() dp− dp F ==xxγβ =dt dt = dt c dt segments I=λu. x dt()(1)(1) dt−−− dx u u cccβββxx eN eN Il γ Iu==+− uNu so = The Minkowski force: K is the derivativeβ of momentum with ll+−±± e respect to proper time. β Relativistic momentum is µ Il ddtdppF µ dp pmNumNu=−γγ= γγ( −≠ ) m 0 K ≡= =, K ≡ +++−−−+−e dddtτ ττ1/− uc22 d 41 42
Hidden momentum (relativistic effect) 12.3 Relativistic Electrodynamics The gain in energy (γmc2) is equal to the work done by the 12.3.2 How the Fields Transform electric force E. We have learned that one observer’s electric field is 2 IlEw another’s magnetic field. (γγ+−−=⇒= )mc eEw p 2 c What are the general transformation rules for electromagnetic Ilw is the magnetic dipole moment of the loop fields? as vectors m points into the page, and p is to the right, so Let’s start with “Charge invariant”. 1 pmE=×() Consider the simplest possible electric field. c2 A magnetic dipole in an electric field carries linear momentum, even though it is not moving. σ E = 0 yˆ σ This so-called hidden momentum is strictly relativistic, 0 E = yˆ ε0 ε and purely mechanical. 0
A more realistic model for a current-carrying wire can be found in the supplement. 43 44 See V. Hnizdo, Am. J. Phys. 65, 92 (1997). The Transformation of The Electric Field Example 12.13: The E-field of a moving point chagre. Are we sure that the field is still perpendicular A point charge q is at rest at the origin in system S0. Question: to the plates? What is the electric field of this same charge in system S, What if the field of a moving plane tilted, say, which moves to the right at speed v0 relative to S0? in the direction of motion? It doesn’t. Solution:
The total charge on each plate is invariant. 22 Qlwlwl==σσ00 0 where =− 1 vclww 0 / 0 and = 0 1 σσγσγ==⇒= EE⊥⊥ perpendicular 22000 00 components 1/− vc0
Qlwlw==σ 00 0 σ ,
whereσσ ll==00 and ww // // parallel =⇒=00 EE components 45 Very efficient as compared with Chap.10 Eq. 10.68. 46
The Transformation of The Magnetic Field The Transformation of The Magnetic Field Kx= ∓σ v ˆ γ vv 1 To derive the general rule we must ± 0 0 2 γεµ=+γγ(12 ) and cEEvB = ⇒ yyz = ( − ) start out in a system with both electric 000c v and magnetic fields.ε ⇒= BBzzγ () − E y σ c2 Eyz==− and Bvµ00σ 0
In a third system, S, traveling to the v is the velocity ⇒= Ezzyγ ()EvB + of SS relative to E = E right with speed v relative to S, the field 0 v x x ε ⇒= BBγ () + E would be σ yyc2 z Eyz==− and Bvµ0σ 0 vv+ 1 v ==0 , γ , σγσ = 1/+ vv c2 22 0 BB= 0 1/− vc x x E = E x x Eyyz=−γ ()EvB Ezzy=+γ ()EvB vv γσεγε0 γ σ 1 BB= BB=+γγ() EBBE =−() E ==( ) , where = x x yy22 zzz y y 22 cc 000 1/− vc γ 47 48 Two Special Cases 12.3.3 The Field Tensor E = E x x Eyyz=−γ ()EvB Ezzy=+γ ()EvB E and B certainly do not transform like the spatial parts of the vv two 4-vectors (4-velocity and 4-momentum). BBx = x BB=+γγ() EBBE =−() yycc22 zzz y What sort of an object is this, which has six components and 1. If B=0 in S, then transforms according previous results? vv B =−γγ()()EEyˆˆz Answer: Antisymmetric, second-rank tensor. cc22zz 1 00 01 02 03 ()vE tttt µµvv (symmetric tensor, =−2 × where vx= vˆ tt= c 10 11 12 13 10 distinct components) µv tttt t = tttt20 21 22 23 (antisymmetric tensor, ttµµvv=− 2. If E=0 in S, then 30 31 32 33 6 distinct components) tttt ˆˆ ˆˆ 01 02 03 E =−γ vB()()zyy − Bz =− vB zyy − Bz 0 ttt =×vB 01 12 13 µv −ttt0 where vx= vˆ t = −−tt02 120 t 23 49 50 03 13 23 −−−ttt0
The Tensor Transformation The Field Tensor and The Dual Tensor
γγβ− 00 01E 02Ey 03E 12 31 23 F ≡≡≡≡≡≡x , FFFBFBFB , z , , , . vvλ zyx aaµµλσ=Λλ 4-vector transformation − γβ γ 00 ccc vv Λ= tt=Λ Λ tensor transformation 0010 0///ExyzcEcEc λσ −−Ec/0 B B 0001 µv xzy F = the field tensor −−Ec/0 B B Work out the following transformation: yz x −−Ec/0 B B ttt01==−=+ 01 02γβ() t 02 t 12 γβ t 03 () t 03 t 31 zyx ttt23==+=− 23 31γβ() t 31 t 03 γβ t 12 () t 12 t 02 There was a different way of imbedding E and B in an By direct comparison, we can construct the field tensor F µv antisymmetric tensor.
0 BBBxyz Ex = Ex E =−γ ()EvB E =+γ ()EvB yyzzzy µv −−BEcEcxzy0// G = the dual tensor vv −−BEcyz/0 Ec x / BB= BB=+γγ() EBBE =−() x x yy22 zzz y cc −−BEcEczy//0 x 51 52 12.3.4 Electrodynamics in Tensor Notation Maxwell’s Equations in Tensor Notation (i) Reformulate the laws of electrodynamics (Maxwell’s Maxwell’s equations can be written in the following forms. equations and the Lorentz force law) in relativistic notation. µv ∂F Gauss’s law = µ J u How the sources of the fields, ρ and J, transform? ∂xv 0 Ampere’s law with Maxwell’s correction QQ ρρρ== and Ju , where = (the proper charge density) µv 00 01 02 03 VV0 ∂∂∂∂∂FFFFF 0 0 µµ==+++=0 0 J V ∂∂∂∂∂xxxxxv 0123 ρρ== γρ0 , where VucV =− 122 / (length contraction) 00 0 1 ∂E ∂E ∂E ρ V (x ++y z ) =µρc ⇒∇⋅= E Ju== u =( u ) =ηη , where =u (proper velocity) cx∂∂∂ y z 0 ε ργρ00 ργ 0 ρ γ 0 ∂∂∂∂∂FFFFF1v 10111213 The current density 4-vector: JcJJJµ = (ρ , , , ) µµ==+++=1 J 1 x yz ∂∂∂∂∂xxxxxv 01230 11∂E ∂B ∂B ∂E Conservation of charge: −+−=→−+∇×=x z y J (BJ ) ( ) ∂J ∂J ∂J 3 ∂J i ct22∂∂∂ y z00x ct ∂ xx ∇⋅J =x +y +z = ∂ρ ∑ i µ ∇⋅J =− ∂∂∂xyzi=0 ∂ x ∂J 1 ∂∂EE 0 = 0 +µ =2 and 3 ( − +∇×BJ ) =µµεµ ⇒∇× B − = J ∂t ∂∂ρρ()cJ ∂ ∂x µ 2 yz,0, yz 000 −=−=− 53 ct∂∂ t 54 ∂∂tctx() ∂0 µµ
Maxwell’s Equations in Tensor Notation (ii) The Minkouski Force and Relativistic Potentials Maxwell’s equations can be written in the following forms. The Minkowski force on a charge q is given by ∂Gµv Faraday’s law 11 = 0 KEuBF=+×=q[( )] ∂xv Gauss’s law for magnetic field 1/−−uc22 1/ uc 22 ∂∂∂∂∂GGGGG0v 00010203 The electric and magnetic fields can be expressed in terms of µ ==+++=0 0 ∂∂∂∂∂xxxxxv 0123 a scalar potential and a vector potential. ∂B ∂B ∂B ∂A (x ++y z ) = 0 ⇒∇⋅= B 0 E =−∇V − BA=∇× ∂∂∂xyz ∂t 1v 10111213 ∂∂∂∂∂GGGGG µ µ ==+++=1 0 AVcAAA= ( ,xyz , , ) 4-vector potential ∂∂∂∂∂xxxxxv 0123 µ v µ 111∂B ∂E ∂E ∂B v ∂∂AA x z y F =− the definition of the field tensor −−+ =→+∇×= 0 (E )x 0 µ ct∂∂ cy cz ∂ ∂ t ∂∂xxv ∂Aµ ∂∂BB = 0 the Lorentz gauge +=µ 2 and 3 ( +∇×EE ) = 0 ⇒∇×+ = 0 µ ∂∂ttyz, ∂x 55 56 39.3 Covariance Homework of Chap.12 Covariance: The laws of mechanics are covariance---they retain their form---with respective to Galiliean transformation. Newton’s second low, F=ma, in one frame has the same form, F’=ma’, in another. However, the Maxwell’s equations does not satisfy this requirement when applying the Galiliean Prob. 3, 4, 6, 25, 30, 33, 38, 46 transformation. x′ = x − vt; t′ = t Three Problems: 1. The force between the charge depends on the frame of reference employed. 2. Maxwell’s equations are valid in only one special frame with the Galiliean transformation. 3. The applied electromagnetism law will change with
57 reference frame. 58
The Relativistic Doppler Effect The Relativistic Doppler Effect (II) In the classical Doppler effect for sound waves, the In order to obtain the Doppler effect, we have to calculate observed frequency depends differently on the velocities of the time interval measured in two frames. Note that the time the source and the observer. The underlying reason is that dilation is assumed in the following calculation. for sound there is a medium (the air) that serves as an “absolute” reference frame. d v (a) Source at rest, observer moves (b) Source moves, observer at rest T = ∆t + = γ (1+ )T0 (velocity modulation) (wavelength modulation) c c v v c + v v′ v ± vo T = T f ′ = = f f ′ = = fo 0 o λ′ v ± v c − v λo v s Longitudinal Transverse In contrast, for light there is no absolute frame: The relativistic Doppler effect for light depends only on the c − v 1 f = f f = f relative velocity between the source and the observer. c + v o γ o 59 60