speed of light how fast is the speed of light? –depends on material: – in vacuum, c = 3 x 10 8...

Speed of Light

• How fast is the speed of light?– depends on material:– in vacuum, c = 3 x 108m/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.

c o o = 1/

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

Mirror

Half Mirror

Screen

Lightsource


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

Mirror

Half Mirror

Screen

Lightsource

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:


(c tup) = [L2 + (ve tup)2]1/2 ; (c tR) = L + ve tR

Mirror

MirrorHalf Mirror

Screen

Lightsource

Lve tup

ctup

L ve tR

ctR


(c tdown) = [L2+(ve tdown)2]1/2 ; (c tL) = L - vetL

Mirror

MirrorHalf Mirror

Screen

L

ve tdown

ctdown

L

ve tLctR

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 and for

tright-left = tRL .

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:

tup = [L2 / {c2 - ve

2}]1/2 = L / [c2 - ve2]1/2

and

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

tRL = tR + tL = L*[(c+ve) + (c-ve)] / (c2 - ve

2)

= 2Lc / (c2 - ve

2) = 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 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 tRL > tud if ve > 0.

Speed of LighttRL = 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 {1 - x}1/2 = 1 - x/2 - ...

and [1 / {1 - (ve/c)2}] ≈ 1 to 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

t = 3.3 x 10-16 sec, with T = 1.6 x 10-15 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 LightEinstein 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.

LF

F


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


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 vearth = 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

vxpplane = 0 at t = 0

(xpplane means the x position of the plane as measured by the plane observer)

xpground = 0 at t = 0 (xpground means the x position of the plane as measured by the ground observer)

xgplane = 0 (xgplane means the x position of the ground observer as measured by the plane

observer) ground observerxgground = 0 at t = 0

(xgground 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: 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 xP1 in front of the plane at time t1 , then the ground observer will say that the event happened at: xG1 = xP1 + v*t1 .


plane v 1

xgplane = -v*t xpplane = 0 at t > 0

ground

xgground = 0 at t > 0 xpground = v*t

xP1

xG1

xpG


xG = xP + v*t . This is the Gallilean Transformation Equation.

The inverse relation is: xP = xG - v*t . [It is assumed in this system, that there is only one

time, t, so tG = tP = t.]

If something is moving in the plane’s system, then we have: dxG/dt = d(xP+vt)/dt , or

vG = vP + v (where v is the speed of the plane itself).


xG = xP + v*t, tG = tP, and vG = vP + 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 vP = c, then vG = c also.

LTE’s: x1 = (x2 + v*t2) / [1-(v/c)2]1/2

t1 = (t2 + v*x2/c2) / [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.

speed of light how fast is the speed of light? –depends on material: – in vacuum, c = 3 x 10 8...

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