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Speed of Light • How fast is the speed of light? – depends on material: – in vacuum, c = 3 x 108 m/s • What is this speed relative to? • What is the speed of sound relative to? – the ground? the air? the source? the receiver? (sound moved relative to the medium it was moving in: v = [B/ρ]1/2 ) Speed of Light From the E&M theory, c = 1/ o o where the sub 0’s indicate vacuum. So the speed must be relative to vacuum. Speed of Light So how do we determine any speed relative to vacuum? Idea: try to measure speed of light on earth, then we can see how fast the earth is moving through vacuum. [From the idea of relative speed, vresult = vin medium + vof medium] But speed of light is so large! How do we find a relatively small difference in speeds between vresult and vin medium ? Speed of Light The Michelson-Morley experiment used the Michelson interferometer in an attempt to find the speed of the earth through space. Let’s first review the Michelson interferometer: Michelson Interferometer Split a beam with a Half Mirror, the use mirrors to recombine the two beams. Mirror Half Mirror Light source Mirror Screen Michelson Interferometer If the red beam goes the same length as the blue beam, then the two beams will constructively interfere and a bright spot will appear on screen. Mirror Half Mirror Light source Mirror Screen Speed of Light But that assumed the apparatus was stationary. Since light should travel with respect to space (rather than with respect to the source or the receiver or the earth), we need to consider the whole apparatus as moving (with the earth & sun through space). So the light will have to travel DIFFERENT PATHS for the two beams: the up/down versus the left/right as we show next: Michelson Interferometer (c tup) = [L2 + (ve tup)2]1/2 (c tR) = L + ve tR Mirror ve tup L Light source ctup ctR ve tR Mirror L Half Mirror Screen Michelson Interferometer (c tdown) = [L2+(ve tdown)2]1/2 (c tL) = L - vetL Mirror ve tdown ctdown Half Mirror L ctL L Screen ve tL Mirror Speed of Light (c tup) = [L2 + (ve tup)2]1/2 ; (c tR) = L + ve tR (c tdown) = [L2 + (ve tdown)2]1/2 ; (c tL) = L - vetL Note that tup = tdown by symmetry, but that tR does NOT equal tL due to the opposite direction of the light. We now solve for tup-down = tud = tup + tdown and for tright-left = tRL = tright + tleft Speed of Light (c tup) = [L2 + (ve tup)2]1/2 (c tdown) = [L2+(ve tdown)2]1/2 tup-down = tud = tup + tdown = 2 tup so solving the first equation for tup gives: c2 tup2 = L2 + (ve tup)2 , or tup2(c2 – ve2) = L2 , or tup = [L2 /{c2 - ve2}]1/2 = L / {c2 - ve2}1/2 so tud = 2L/{c2-ve2}1/2 = 2L/c * [1/{1-(ve/c)2}1/2] . Speed of Light ctR = L + vetR and ctL = L - vetL , so tR = L / (c - ve) and tL = L / (c + ve) , so (getting a common denominator) tRL = tR + tL = L*[(c+ve) + (c-ve)] / (c2 - ve2) (and dividing numerator and denominator by c2) = 2Lc / (c2 - ve2) = 2L/c *[1 / {1 - (ve/c)2}]. Recall: tud = 2L/c * [1/{1-(ve/c)2}1/2]. Note that the two times are DIFFERENT. This should cause a difference in the interference pattern on the screen. Speed of Light tRL = 2L/c * [1 / {1 - (ve/c)2}]. tud = 2L/c * [1 / {1- (ve/c)2}1/2]. Note that if ve = 0, then the two times are the SAME. Let’s see about the time difference if ve > 0. Which of the two times is bigger? Speed of Light tRL = 2L/c * [1 / {1 - (ve/c)2}]. tud = 2L/c * [1 / {1 - (ve/c)2}1/2]. The ve/c is less than one, so the denominator {1 - (ve/c)2} is also < 1 . The square root of a fraction is bigger than the original fraction ( [¼]1/2 = ½ > ¼ ). Since we are dealing with the denominator, the one with the square root (which is bigger) will be the smaller overall quantity. So tud < tRL if ve > 0. Speed of Light tRL = 2L/c *[1 / {1 - (ve/c)2}]. tud = 2L/c * [1/ {1 - (ve/c)2}1/2]. t = tRL - tud = [2L/c]*{[1 / {1 - (ve/c)2}] - [1 / {1 - (ve/c)2}1/2]} use approximations that (for small x) 1/{1 - x} ≈ 1 + x and 1/{1 - x}1/2 ≈ 1 + x/2 so with x = (ve/c)2 we get: t = (2L/c)*(1/2)*(ve/c)2 = (L/c)*(ve/c)2. How big is this? Speed of Light t = (L/c)*(ve/c)2 . If we have L = 10 meters, and if ve = 3 x 104 m/s (due to earth rotating about the sun), THEN t = (10 m / 3 x 108 m/s) * [(3 x 104 m/s)/(3 x 108 m/s)]2 = 3.3 x 10-8 sec * (1 x 10-8) = 3.3 x 10-16 sec. Can we measure such a small time? Speed of Light t = 3.3 x 10-16 sec If we consider the merits of the Michelson Interferometer, we can see that we can detect a fraction of a wavelength, and so we can detect a fraction of a period. The period for visible light, say = 500 nm, is f = c = T, or T = /c = 5 x 10-7 m / 3 x 108 m/s = 1.6 x 10-15 sec. Speed of Light With t = 3.3 x 10-16 sec, and T = 1.6 x 10-15 sec we have t / T = 0.2 . Thus we should see AT LEAST a fringe shift of 0.2 , and probably more since we expect the sun to be moving through the galaxy, etc., and the above 0.2 shift is that due only to the earth rotating about the sun. Experimental Considerations Is it critical to have the two lengths exactly the same? No – having one length slightly different would not significantly affect the amount of the fringe shift. (To see this, replace L with L(1+) on one of the sides (with being very small), and see what affect this will have.) How do we determine which is the up-down side and which is the left-right side? If we are able to rotate the apparatus through 90o, we should see the fringe shift change from 0.2 shifts to zero and then forward to 0.2 shifts. Speed of Light Michelson & Morley determined that their apparatus was sensitive to about 0.01 shifts, and they expected NO LESS than 0.20 shifts. RESULT: Michelson & Morley detected about 0.01 shifts - a null result! Either the earth is stationary in the universe, (which we know it isn’t), or there is something wrong with our thinking! Speed of Light Up to now, we have more or less assumed, maybe without thinking about it, that there is one best system with one absolute time. Einstein had a different thought: Einstein said: If all the laws of physics apply in all inertial frames, why should the speed of light be different in all those inertial frames? Shouldn’t the speed of light also be the same in all inertial frames? Inertial Reference Frames First a short discussion about inertial frames. Newton’s Second Law relates physically identifiable FORCES to ACCELERATIONS: F = ma . Any frame in which Newton’s Second Law holds is called an inertial frame. There are frames that are not inertial. We’ll show 2 examples of NON-INERTIAL FRAMES: 1. an accelerating spaceship; 2. the earth’s surface. Example of a Non-inertial frame: Accelerating Spaceship Consider a spaceship far from any appreciable star or planet – so essentially no gravity. The spaceship is accelerating through space by firing its rocket engine. Now consider a person in the spaceship who is standing on a scale. Since the person is accelerating with the spaceship, and the scale is the only thing exerting a force on the person, Fscale = ma, where a is the acceleration of the spaceship. However, the person in the spaceship does not see himself/herself moving, much less accelerating. But the scale does read a force! What does the person think is causing that force? (remember there is no gravity force). scale engine Example of a Non-inertial frame: Accelerating Spaceship To be definite, let’s say the mass of the person is 100 kg, and the acceleration of the spaceship is 5 m/s2. The force the scale exerts (and so reads) will be 100 kg * 5 m/s2 = 500 Nt. The person now jumps up off of the scale by providing a bigger force on the scale, say 700 Nt. The acceleration of the person will now be 7 m/s2 during the time of the jump. The person will now be going faster than the spaceship and so move above the scale. But once the person leaves the scale, the person will no longer accelerate (no force applied), but the spaceship will continue to accelerate and so catch up and pass the person’s speed – the person appears to rise up and then fall down – just like there is real gravity in the spaceship. But there isn’t real gravity. This apparent gravity is a “strange” force from the person’s point of view. From our view (from the inertial frame), it is clear that there is no strange force. The spaceship frame in non-inertial. scale engine Another example of a Non-inertial frame: Surface of the earth Consider low pressure systems. Why does the wind circle the low pressure area instead of just filling it in? We can answer that question nicely if we consider the earth from space: Non-inertial Reference Frames Due to the rotating earth, the air South of the Low pressure system in the Northern Hemisphere is moving faster and will overshoot. The air North of the system is moving F slower and L F will fall behind. Non-inertial Reference Frames From the earth’s surface, it looks like the air from the South has run around to the East of the system, and the air from the North has run around to the West - a counter-clockwise rotation. L Non-inertial Reference Frames Thus from the earth’s surface, it looks like there is a strange force (called the Coriolis Force) acting. But from the space system, there is no need to invoke strange forces - it is clear that the result comes directly from known forces and Newton’s Second Law. Thus we can tell if our system is inerital or not - we simply see if we need “strange” forces. Inertial Reference Frames Any system that is moving with a constant speed relative to an inertial system is also an inertial system. This is true since acceleration is the CHANGE in velocity and does not depend on the amount of velocity. An example is that of riding in an airplane. As long as the ride is not bumpy and the plane is not accelerating (speeding up, slowing down, or turning), we do not realize that anything is different than on the ground! Relativity If all the laws of physics hold in all inertial frames, then why (according to Einstein) should the speed of light be different in different inertial frames? But how could the speed of light be the same in two frames that are moving with respect to one another? How could that even be considered to be possible? - But then, why did the Michelson-Morley experiment give vearth = 0 ? We’ll see in the Introduction to the 1st Relativity computer homework program how this works out. Transformation Equations But before we do that, we need to consider how to transform from one inertial system to another. Consider the classical reasoning (which fails to explain the Michelson-Morley experiment): Consider two inertial systems, call them the ground system and the airplane system. Caution: be extremely careful with knowing who is doing the measuring! Galilean Transformation Eqs. Classical Reasoning An airplane passes over a ground observer at time 0 sec. This means that the position of the plane according to the ground observer is 0 m at 0 sec. (It also means that the ground observer’s position according to the plane is 0 m at 0 sec.) We’ll consider that the plane is moving at some speed, v, to the right (+x direction) as seen by the ground observer. Galilean Transformation Eqs. plane v xpplane = 0 m at t = 0 s (xpplane means the x position of the plane as measured by the plane observer) xpground = 0 m at t = 0 s (xpground means the x position of the plane as measured by the ground observer) xgplane = 0 m at t = 0 s (xgplane means the x position of the ground observer as measured by the plane ground observer xgground = 0 m at t = 0 s observer) (xgground means the x position of the ground observer as measured by the ground observer) Galilean Transformation Eqs. Since the plane is moving to the right, this would mean that the ground observer is moving to the left as seen from the plane. Since the plane is moving at a constant speed, we can say: xof plane as seen by ground = xpg = v*t, and xof ground obs. as seen by plane = xgp = -v*t . If something happens (event 1) in the plane’s system, say x1p in front of the plane at time t1 , then the ground observer will say that the event happened at: x1g = x1p + v*t1 . Galilean Transformation Eqs. plane v xpplane = 0 m at t > 0 s xgplane = -v*t x1p xpg = v*t x1g ground xgground = 0 m at t > 0 s xpground = v*t 1 Galilean Transformation Eqs. xg = xp + v*t . This is the Gallilean Transformation Equation. The inverse relation is: xp = xg - v*t . If something is moving in the plane’s system, then we have: dxg/dt = d(xp+vt)/dt , or vg = vp + v and vp = vg - v (where v is the speed of the plane itself). Even though there are different position and velocity measurements in the two systems, it is assumed that there is only one time measurement, t, so tg = tp = t. Galilean Transformation Eqs. xg = xp + v*t, xp = xg - v*t, tg = tp, and vg = vp + v , and tp = tg, and vp = vg - v . The velocity equation was the basis for the prediction in the Michelson-Morley experiment which did not agree with the experiment’s results! A new set of transformation equations are necessary if we are to explain the results of the Michelson-Morley experiment. Lorentz Transformation Eqs. The derivation of these equations is done in PHYS 347. The requirement is that if ve = c, then vs = c also. I’m now using earth and spaceship (e,s) instead of ground and plane (g,p). An EVENT happens in system e at xe,te and in system s at xs,ts LTE’s: xe = (xs ± v*ts) / [1-(v/c)2]1/2 te = (ts ± v*xs/c2) / [1-(v/c)2]1/2 • Note that the only difference between the classical GTE and the LTE for the x equation is the inclusion of the denominator. • Note that the different frames will measure different times for the same event! • Note that as (v/c) → 0, the LTE equations become the Galilean Transformation Equations. Note on signs in LTE’s LTE’s: xe = (xs ± v*ts) / [1-(v/c)2]1/2 te = (ts ± v*xs/c2) / [1-(v/c)2]1/2 To determine the + or – signs in the LTE, simply ignore the denominator and consider the regular (Galilean) transformation for position: xa = xb ± vt If b is moving to the right with respect to a, then use the + sign since as time increases the position will become further to the right; if b is moving to the left with respect to a, then use the – sign. Use the same sign in the t equation as you do in the x equation.