Chapter 2: The Special Theory of Relativity. A reference fram is inertial if Newton s laws are valid in that frame.

Transcription

1 Chapter 2: The Special Theory of Relativity What is a reference frame? A reference fram is inertial if Newton s laws are valid in that frame. If Newton s laws are valid in one reference frame, they are also valid in a reference frame moving at uniform velocity relative to the first frame: this is the Newtonian or Galilean principle of relativity. The Galilean transformation between two coordinate systems, K and K, where K is moving with velocity v in the x direction with respect to K is: x = x vt, y = y, z = z, t = t. Around 1802, there was evidence that light travelled as a wave. At that time, all known waves needed a medium in which to travel. So people assumed light needed a medium in which to travel, even though it appeared to travel in a vacuum in outer space. This medium was called the luminiferous ether or ether. The work of Maxwell showed that the speed of light in different media depends only on the electric and magnetic properties of matter. In a vacuum, v = c = 1/ µ 0 ɛ 0, where µ 0 and ɛ 0 are the permeability and permittivity of free space respectively. Reference frame K, K, K moving with respect to K with velocity v. An observer in K measures the speed of light to be c. Using the Galilean transformation, speed of light in reference frame K is c + v. But Maxwell s equations dont differentiate between the two reference frames. Physicists of the late ninteenth century proposed there was one preferred inertial reference frame in which the ether was at rest and the speed of light was c. The Michelson-Morely Experiment 1

2 Earth orbits the Sun at about 10 4 c - so find the effects of the Earth s motion through the ether. Fig. 2.2, p. 23 gives a diagram of the MM interferometer - a device to measure the phase difference between two waves. SOurce S passes through a glass plate A which is partially silvered: part of the light reflected, part is trasnmitted. Reflected light goes to mirror D and tramsitted light goes to mirror at C. Light is reflected at mirrors C, D and comes back to the partially silvered mirror A. B is added to make sure both beams of light pass through the same thickness of glass. INterference patterns observed due to difference in path lengths. Or because of the motion of the Earth through the ether, the fringe pattern should shift if the apparatus is rotated through 90 degrees such that arm AD becomes parallel to the motion of the Earth through the ether and arm AC is perpendicular to the motion. Let optical path lengths of AC and AD be l 1 and l 2 respectively. For constructive interference, 2(l 1 l 2 ) = nλ, where λ is he wavelength and n is an integer. When light travels from A to C, speed of light according to the Galilean transformation is c+v. On the return journey, from C to A, the speed of light is c v. The total time for the round trip journey to mirror M 1 is t 1 = l 1 c + v + l 1 c v = 2l 1 c ( 1 1 v 2 /c 2). Can show that the time of flight t 2 for the light to go to mirror M 2 and back is t 2 = 2l 2 1 c 1 v 2 /c 2, and the time difference between the two journeys is t = t 2 t 1 = 2 c ( l 2 1 v 2 /c 2) l 1 1 v 2 /c 2). Now rotate the apparatus by 90 degrees so that the ether passes along the length l 2 toward the mirror M 2. The time difference is now t = t 2 t 1 = 2 c ( l 2 1 v 2 /c l v 2 /c 2). 2

3 Look for a shift in the interference pattern when the apparatus is rotated by 90 degrees. The time difference is which can be simplified to t t = 2 c ( l 1 + l 2 1 v 2 /c l 1 + l v 2 /c 2, t t v2 (l 1 + l 2 ) c 3. Michelson s apparatus had l 1 l 2 = 1.2m. A time diffference of about s. For visible light, one wavelength amounts to T = 1/f = λ/c = s, so a time difference of s corresponds to about 0.04 fringes. Michelson deteced no fringe shift: a null result. Repeated with Morely. Many attempts to explain this null result. Appeared to be no single inertial reference frame in which the speed of light was c. Lorentz and Fitzgerald suggested the MM experiemtn could be understood if length is contracted by the factor 1 v 2 /c 2 in the direction of motion, v is the speed in the direction of travel. Now at is zero. Einstein s postulates: The principle of relativity: the laws of physics are the same in all inertial systems. Noo way to detect abolute motion and no preferred inertial system exists. Observers in all inertial systems measure the same value for the speed of light in a vacuum. Two events that are simulatneous in one reference frame (K) are not necessarily simulatneous in another reference frame (K ) moving with respect to the first frame. If β = v/c and γ = 1 then the Lorentz transformation equations are 1 v 2 /c2 x = γ(x βct), y = y, 3

4 z = z, t = γ(t βx/c), where the dashed quantities denote coordinates in frame K, moving in the x, x direction with velocity v with respect to frame K At time t = t = 0, the origins of the two coordinate systems are cooincident. These transformation enable you to findn out out the coordinates of the same event in two different inertial coordinate systems. The proper time is the temporal difference in the coordinates of two events occuring in a coordinate system in which the two events occur at the same spatial location. The proper length is the length of an object in a coordinate system in which the length is at rest. Let T 0 and L 0 be the proper time and proper length respectively measured in a reference frame K. How do these time and length intervals translate in a reference frame, K, that is moving with velocity v with respect to K. Then can use the Lorentz transformation equations to show that T = γt 0, where T is the time interval measured in a reference frame that is moving with velocity v with respect to K. γ = 1/ 1 v 2 /c 2. This is time dilation: moving clocks go slow. Also, L == L 0 /γ. Again using the Lorentz transformation equations can show that velocities in the reference frame K which is moving with velocity v in the x direction are related to velocities in the reference frame K by u x = u y = u z = u x v 1 (v/c 2 )u x, u y γ[1 (v/c 2 )u x ], u z γ[1 (v/c 2 )u x ]. 4

5 Experimental verification Muons are particles which decay exponentially according to N = N 0 exp( ln(2)t/t 1/2 ), where N, N 0 are the number of particles at time t, t = 0 respectively. At 2000m, you observe 1000 muons in a time interval T. The classically, the muons, travelling at 0.98c, take s to travel 2000m and from the decay equation you expect to see only 45 muons at ground level. In fact you observe 540 muons. This is because in the muons frame, the time interval is only s and hence you predict 538 muons to survive till ground level. Also another way of looking at this is to see that according to SR, the length 2000m only appears as 400m to the muons because of length contraction and again the muons travelling at 0.98c take s to make this trip. Also read up on Atomic Clock measurement on p. 45. Spacetime Now instead of describing an event as (x, y, z) need to describe it as (ct, x, y, z). The line connecting two events described in this way is a worldline. A spacetime diagram has ct on the y axis and a spatial coordinate, say x on the x axis. All physical prcesses have gradients in such a diagram given by c/v, ie. greater than 1. A light beam has a slope of 1. No physical process can have a slope less than 1 since it would then be travelling faster than the speed of light. This leads to the concept of past and future cones which give the locus of events that can influence the present and the locus of events that can be influenced by the present. Physical laws are formulated in terms of invariant quantities in these spacetime diagrams, notably the spacetime intervale between two events s 2 = x62 c 2 t 2. If s 2 = 0, the events have a lightlike separation. 5

6 If s 2 > 0, the events have a spacelike separation since the events cannot be causally connected. If s 2 < 0, the events have a timelike separation since the events can be causally connected. Doppler Effect A light source is moving toward a fixed receiver at velocity v. K is the reference frame in which the receiver is at rest. K is the receiver in which the source is at rest, that is K is moving to the receiver at velocity v. in frame K, the source emits n waves during a timne interval T. Then the length of the wave train is ct vt, and the wavelength is λ = (ct vt)/n, and the freequency is f = c/λ = cn/(ct vt). In its rest frame, the source emits n waves of frequency f 0 during the proper time T 0 such that n = f 0T 0. T, T 0 are related by T 0 = T/γ. Then can show that, f = 1 + β 1 β f 0. This is also the case when the source is fixed and the receiver approaches it with velocity v. Can also show that when the receiver and receiver are receeding from each other, 1 β f = f β Both cases can be summarized by 1 + β f = f 0, 1 β where we adopt the sign convention: +β when the source and receiver are appraoching each other and a β when they are receeding from each other. Relativistic Momentum Frank is at rest in a system K holding a ball of mass m. Mary holds a similar ball in a system K that is moving in the x direction with velocity v with respect to system K. 6

7 Frank throws his ball along his y axis and Mary throws her ball with exactly the same speed along her negative y axis. The two balls collide in a perfectly elastic collision and each of them catches their own ball as it rebounds. Each measures the speed of his/her ball to be u 0 before and after the collision. The according to Frank, for the ball he has thrown, u Fx =, u Fy = u 0. If the momentum is p = mv, the momentum of the ball thrown by Frank is entirely in the y direction: p Fy = mu 0. The collision is perfectly elastic and the ball returns to Frabnk along the y axos. The change of momentum is p Fy = 2mu 0. If the conservation of momentum is valid, need to measure the speed of Mary s ball as viewed by Frank. We obtain, u Mx = v. Then u My = u 0 sqrt1 v 2 c 2. Before the collision, for Mary s ball, as viewed by Frank, p Mx = mv, and p My = mu 0 1 v 2 /c 2. After the collision, as viewed by Frank, p MX = mv, and p My = mu 0 1 v 2 /c 2, so that the change in momentum of Mary s ball as viewed by Frank is p M = p My = 2mu 0 1 v 2 /c 2. Thus conservation of momentum as viewed by Frank is not valid because p F + p M does not equal to zero. This is the case even if we use the velocity transformation equations from SR. So change the definition of momentum in order to make the principle of conservation of linear momentum valid. Define p = Γ(u)mu, where Γ(u) = 1 1 u 2 /c 2, and u is the speed of the particle, not the relative velocity between the inertial reference frames. Now linear momentum is conserved. Relativistic Energy 7

8 If we change the definition of momentum, since force is the rate of change of momentum and since kinetic energy is related to work done, which is force times distance, we need to change the definition of kinetic energy from the classical 1 2 mv2. Then since F = dp dt = d dt (γmu) = d d ( mu 1 u 2 /c 2. Kinetic energy is the work done on a particle by a net force. Thus W 12 = 2 1 F.ds = K 2 K 1. If the particle starts from rest, K 1 = 0. Then the work W and kinetic energy K are d W = K = dt (γmu).udt. After some manipulation by parts, can show K = mc 2 (γ 1). This is the relativistic definition of energy. Can show that for small speeds, this definition is close to 1 2 mu2. Total Energy and Rest Energy From before, which is K = mc 2 (γ 1) = γmc 2 mc 2, γmc 2 = K + mc 2. The term mc 2 is the rest energy, denoted by E 0, E 0 = mc 2. The total energy of the particle is the sum of the kinetic energy and rest energy, E = γmc 2 = K + mc 2 = K + E 0. This is the equivalence of mass and energy - even when a particle is at rest, it still has energy as long as it has mass. Mass and energy are no longer separate quantities and we have the conservation of mass-energy or conservation of energy where now we have to include rest mass energy in the equations. 8

9 Energy and Momentum Recall p = γmu. Then p 2 c 2 = γ 2 m 2 u 2 c 2 = γ 2 m 2 c 4 (u 2 /c 2 ) = γ 2 m 2 c 4 β 2. Then using the equation of β 2, we find p 2 c 2 = γ 2 m 2 c 4 m 2 c 4, and so Rearranging, or p 2 c 2 = E 2 E 2 0. E 2 = p 2 c 2 + E 2 0, E 2 = p 2 c 2 + m 2 c 4. This relates the total energy of a particle to its momentum. For a photon, this equation implies that the total energy is E = pc. It has zero mass. Can show that massless particles travel at the speed of light. 9