Download Speed of Light

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.