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V Galileo: The object would land at the base of the mast. Therefore, an observer on a ship, moving (with respect to land at velocity v) will observe the.

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Presentation on theme: "V Galileo: The object would land at the base of the mast. Therefore, an observer on a ship, moving (with respect to land at velocity v) will observe the."— Presentation transcript:

1 v Galileo: The object would land at the base of the mast. Therefore, an observer on a ship, moving (with respect to land at velocity v) will observe the ball to fall straight down. The time to fall, t: H = ½ gt 2 Galileo Aristotle Aristotle: If a man on top of a mast in a moving ship drops an object, it would fall toward the back of the ship. v

2 An observer on land would observe the fall falling in a parabola; however, it still lands near the base of the mast in time t: H = ½ gt 2 The different descriptions by the ship and land observers is that the land-observer says that the ball had initial horizontal velocity v, which it keeps as it falls. However, both observers agree completely on the physical description: F = ma = mg  a y = d 2 y/dt 2 = -g and d 2 x/dt 2 = 0. They just have different values for dx/dt: the observer on the ship says that dx/dt =0 while the observer on land says that dx/dt = v. They are both right! They are simply viewing the physics in different “reference frames”, that differ by a constant velocity v. v x y

3 Similarly, consider two people playing catch in a train car. Suppose A throws a ball to B at speed u. Both A and B will measure the speed to be u, even if the train is traveling at speed v.

4 However, if C is watching through a window from outside the train, he will say the ball is traveling at speed u + v. The are all correct; they are simply in different “reference frames”, that differ by a constant velocity v.

5 Galilean Relativity: The laws of physics are the same in all inertial frames of reference. At the time of Galileo, the only laws of physics that were known were mechanics. Reference frames which are traveling at constant velocity (i.e. uniform motion) with respect to distant stars: reference frames in which acceleration = 0 if there are no forces. That is: there is no experiment you can do that would tell you if you are moving with constant velocity (with respect to distant stars)  all inertial reference frames are equivalent  all uniform motion is relative  if you are not accelerating, you cannot tell that you are moving.

6 This principle is implicit in Newton’s laws of motion. Suppose Suppose two inertial reference frames, S and S’, where S’ has velocity v = v x x with respect to S. Suppose their origins coincide at t = t’= 0. Then Newton/Galileo would state: x’ = x – vt, y’ = y, z’ = z, and t’ = t Since t = t’: Then dx’/dt’ = dx/dt – v, dy’/dt’ = dy/dt, dz’/dt’ = dz/dt a x ’ = d 2 x’/dt’ 2 = d 2 x/dt 2 = a x a y ’ = d 2 y’/dt’ 2 = d 2 y/dt 2 = a y a z ’ = d 2 y’/dt’ 2 = d 2 z/dt 2 = a z Therefore F’ = ma’ = ma = F So both observers will not only agree on Newton’s Laws, but will agree on the value of F. 

7 Mid 19 th century “crisis”: Maxwell’s Equations, which seemed to perfectly describe electricity and magnetism are not consistent with Galileo’s “reference frame transformation” equations. This is most simply seen in the wave equation for an electromagnetic wave traveling in a vacuum:  2 E y /  t 2 = (  ) -1  2 E y /  x 2 = c 2  2 E y /  x 2 Problem: Maxwell’s Eqtns. do not specify a reference frame. For example for a light wave traveling along x direction, from Galileo’s transformation, one would expect that if S measured speed to be c, S’ would measure c-v. But Maxwell’s Eqtns. state they would both measure c !!

8 This is very different, for example, from sound, which travels at a speed 331 m/s in air – the reference frame is that in which the air is stationary. (i.e. if you are downwind from a bell when the wind speed = 10 m/s, you will measure the speed of the sound wave to be 341 m/s; if you are upwind, you will measure 321 m/s.)

9 In the late 19 th century, the assumption was that Galileo and Newton were correct and that Maxwell had missed something: that there was a material, called the ether (or aether), which carried electromagnetic waves (analogous to the air for sound waves). Since electromagnetic waves were so fast, and the ether had not yet been detected, it was assumed to have a very low density (i.e. to be very “light”) and ephemeral. Since light reaches us from the sun and stars, the ether was assumed to fill space. The hunt was on to detect the effects of the ether, i.e. to look for differences in the the speed of light traveling in different directions. These experiments were difficult, because c is so large. The most precise (and famous) of these experiments was the Michelson-Morley experiment (Case Western Reserve, Cleveland, published 1887), which tried to measure the effect on the speed of light by the earth’s motion [v earth = 3 x10 4 m/s = 10 -4 c] with respect to sun) through the ether.

10 Michelson Interferometer I  4E 0 2 cos 2 (  /2) = 4 I 0 cos 2 (  /2) Intensity at the detector (e.g. eye) will depend on the difference in phase (  ) of the two waves:  1 = 2  (x 1 / - ft 1 ),  2 = 2  (x 2 / - ft 2 ) where f = c/ is the frequency of the light. If x 1 = x 2,  = 2  c (t 2 -t 1 )/

11 For the arm parallel to the earth’s motion, the time to go between M 0 and M 2 and back: t 2 = L 2 /(c+v) + L 2 /(c-v) = 2L 2 c/(c 2 -v 2 ) For the perpendicular arm: t 1 = 2L 1 /(c 2 -v 2 ) 1/2

12 t 2 = 2L 2 c/(c 2 -v 2 ); t 1 = 2L 1 /(c 2 -v 2 ) 1/2 Therefore, the phase difference for the two paths  = 2  F(t 1 -t 2 )  0 if v  0, even if L 1 = L 2. e.g. MM experiment: v= v earth = 3 x 10 4 m/s, L 1 = L 2 = 11 m (large interferometer: mirrors configured for multiple passes), F = 6 x 10 14 Hz (green light)  t 1 -t 2 = 3.7 x 10 -16 s  = 0.22 radians  12.6 o Since it was not possible to set L 1 = L 2 with needed precision (|L 1 – L 2 | << ), after measuring the fringe pattern, the apparatus was rotated 90 o, which switches the two arms, so expected  (original) -  (rotated)  25.2 o. However, no change in the fringe pattern was observed. The experiment was repeated for different orientations and different times of year, but no change in fringe patterns was observed: observed  (original) -  (rotated) < 0.6 o

13 Michelson-Morley Experiment: Observation: The speed of light is equal in the two perpendicular arms of the interferometer. Possible Explanations: 1)The ether near the earth is dragged along by the earth, so earth is not moving with respect to its local ether. 1)The speed of light is c with respect to its source (like Galileo’s ball on ship), not with respect to an ether. 1)All moving objects held together by electromagnetic forces (i.e. everything) contract in their direction of motion: Lorentz 4) Maxwell was correct, the speed of light (in vacuum) is invariant (independent of its source and ether) and Galileo’s transformation equations (i.e. Newtonian physics) must be modified: Einstein (1905)  there is no ether, or need for one; EM waves can propagate through empty space (as per Maxwell) } contradicted by other experiments


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