Speed of Light How fast is the speed of light? –depends on material: – in vacuum, c = 3 x 10 8 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, 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, v result = v in medium + v of medium ] But speed of light is so large! How do we find a relatively small difference in speeds between v result and v in 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 Screen Light sourc e

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 Screen Light sourc e

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 t up ) = [L 2 + (v e t up ) 2 ] 1/2 ; (c t R ) = L + v e t R Mirror Half Mirror Screen Light sourc e L v e t up ct up L v e t R ct R

Michelson Interferometer (c t down ) = [L 2 +(v e t down ) 2 ] 1/2 ; (c t L ) = L - v e t L Mirror Half Mirror Screen L v e t down ct down L v e t L ct R

Speed of Light (c t up ) = [L 2 + (v e t up ) 2 ] 1/2 ; (c t R ) = L + v e t R (c t down ) = [L 2 +(v e t down ) 2 ] 1/2 ; (c t L ) = L - v e t L Note that t up = t down by symmetry, but that t R does NOT equal t L due to the opposite direction of the light. We now solve for t up-down = t ud and for t right-left = t RL.

Speed of Light (c t up ) = [L 2 + (v e t up ) 2 ] 1/2 (c t down ) = [L 2 +(v e t down ) 2 ] 1/2 t up-down = t ud = t up + t down = 2 t up so solving the first equation for t up gives: t up = [L 2 / {c 2 - v e 2 }] 1/2 = L / [c 2 - v e 2 ] 1/2 and t ud = 2L/[c 2 -v e 2 ] 1/2 = 2L/c * [1/{1-(v e /c) 2 } 1/2 ].

Speed of Light ct R = L + v e t R and ct L = L - v e t L, so t R = L / (c - v e ) and t L = L / (c + v e ), so t RL = t R + t L = L*[(c+v e ) + (c-v e )] / (c 2 - v e 2 ) = 2Lc / (c 2 - v e 2 ) = 2L/c *[1 / {1 - (v e /c) 2 }]. Recall t ud = 2L/c * [1/{1-(v e /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 t RL = 2L/c *[1 / {1 - (v e /c) 2 }]. t ud = 2L/c * [1 / {1- (v e /c) 2 } 1/2 ]. Note that if v e = 0, then the two times are the SAME. Let’s see about the time difference if v e > 0. Which of the two times is bigger?

Speed of Light t RL = 2L/c *[1 / {1 - (v e /c) 2 }]. t ud = 2L/c * [1 / {1 - (v e /c) 2 } 1/2 ]. The v e /c is less than one, so the denominator is also less than one. 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 will be the smaller overall quantity. So t RL > t ud if v e > 0.

Speed of Light t RL = 2L/c *[1 / {1 - (v e /c) 2 }]. t ud = 2L/c * [1/ {1 - (v e /c) 2 } 1/2 ].  t = t RL - t ud = [2L/c]*[1 / {1 - (v e /c) 2 }] * [1 - {1 - (v e /c) 2 } 1/2 ] use approximations that {1 - x} 1/2 = 1 - x/ and [1 / {1 - (v e /c) 2 }] ≈ 1 to get:  t = (2L/c)*(1/2)*(v e /c) 2 = (L/c)*(v e /c) 2. How big is this?

Speed of Light  t = (L/c)*(v e /c) 2. If we have L = 10 meters, and if v e = 3 x 10 4 m/s (due to earth rotating about the sun), THEN  t = (10 m / 3 x 10 8 m/s) * [(3 x 10 4 m/s)/(3 x 10 8 m/s)] 2 = 3.3 x sec * (1 x ) = 3.3 x sec. Can we measure such a small time?

Speed of Light  t = 3.3 x 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 m / 3 x 10 8 m/s = 1.6 x sec.

Speed of Light  t = 3.3 x sec, with T = 1.6 x sec Thus  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.

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 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? Up to now, we have more or less assumed, maybe without thinking about it, that there is one best system with one absolute time.

Inertial Reference Frames First a short discussion about inertial frames. Newton’s Second Law relates FORCES to ACCELERTIONS:  F = ma. Any frame in which Newton’s Second Law holds is called an inertial frame. There are frames that are not inertial. The earth’s surface is an example of a NON-INERTIAL FRAME.

Non-inertial Reference Frame 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 below the Low pressure system is moving faster and will overshoot. The air above the system is moving slower and will fall behind. L F F

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. 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 inerital 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 v earth = 0 ?

Transformation Equations 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 plane system. Caution: be extremely careful with knowing who is doing the measuring!

Galilean Transformation Eqs. A plane passes over a ground observer at time 0. 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 as seen by the ground observer.

Galilean Transformation Eqs. plane v xp plane = 0 at t = 0 (xp plane means the x position of the plane as measured by the plane observer) xp ground = 0 at t = 0 (xp ground means the x position of the plane as measured by the ground observer) xg plane = 0 (xg plane means the x position of the ground observer as measured by the plane observer) ground observer xg ground = 0 at t = 0 (xg ground means the x position of the ground observer as measured by the ground observer)

Galilean Transformation Eqs. (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: x of plane as seen by ground = xp G = v*t, and x of ground obs. as seen by plane = xg P = - v*t. If something happens (event 1) in the plane’s system, say x P1 in front of the plane at time t 1, then the ground observer will say that the event happened at: x G1 = x P1 + v*t 1.

Galilean Transformation Eqs. planev 1 xg plane = -v*txp plane = 0 at t > 0 ground xg ground = 0 at t > 0xp ground = v*t x P1 x G1 xp G

Galilean Transformation Eqs. x G = x P + v*t. This is the Gallilean Transformation Equation. The inverse relation is: x P = x G - v*t. [It is assumed in this system, that there is only one time, t, so t G = t P = t.] If something is moving in the plane’s system, then we have: dx G /dt = d(x P +vt)/dt, or v G = v P + v (where v is the speed of the plane itself).

Galilean Transformation Eqs. x G = x P + v*t, t G = t P, and v G = v P + v. The last 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 v P = c, then v G = c also. LTE’s: x 1 = (x 2 + v*t 2 ) / [1-(v/c) 2 ] 1/2 t 1 = (t 2 + v*x 2 /c 2 ) / [1-(v/c) 2 ] 1/2 Note that the only difference 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 if (v/c) is small, we get the Galilean Transformation Equations.