Galilean velocity transformation
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Galilean velocity transformation If an object has velocity u in frame S (note: velocities have a direction!), and if frame S’ is moving with velocity v along the positive x-axes of frame S, then the position of the object in S’ is: vt t x t x − = ) ( ) ( ' ... -3 -2 -1 0 1 2 3 ... ... -3 -2 -1 0 1 2 3 ... v u x x’ The velocity u’ of the object in frame S’ is therefore: A) u + v B) v - u C) u - v D) u E) -v
Transcript of Galilean velocity transformation
Galilean velocity transformation
If an object has velocity u in frame S (note: velocities have a direction!), and if frame S’ is moving with velocity v along the positive x-axes of frame S, then the position of the object in S’ is: vttxtx −= )()('
... -3 -2 -1 0 1 2 3 ...
... -3 -2 -1 0 1 2 3 ...
v
ux
x’
The velocity u’ of the object in frame S’ is therefore:
A) u + v B) v - u C) u - v D) u E) -v
Comparing inertial frames
... -3 -2 -1 0 1 2 3 ...
... -3 -2 -1 0 1 2 3 ...
v
Here are two inertial reference frames, moving relative to one another.
According to S, S’ is moving to the right, with v = 1 m/s.
x
x’
Comparing inertial frames
Here are two inertial reference frames, moving with respect to one another.
According to S’, S is moving to the left, with v = -1 m/s.
... -3 -2 -1 0 1 2 3 ...
... -3 -2 -1 0 1 2 3 ...
v
x
x’
But again: Both observers are right.
Important conclusion
Q: Who is right?a) Observer in Sb) Observer in S’c) Both are rightd) Neither is righte) It depends!Two observers in different reference frames can give a different description of the same physical fact (in this case, the relative velocity of the ‘other’ reference frame.) And they’re both right! The two frames are moving relative to each other.
Observer in S measures velocity of S’ to be +1 m/sObserver in S’ measures velocity of S to be -1 m/s
Comparing inertial frames
At time t = 0, the two frames coincide. A ball is at rest in frame S. Its position is
• x = 2 m in S• x’ = 2 m in S’
... -3 -2 -1 0 1 2 3 ...
... -3 -2 -1 0 1 2 3 ...
v
Comparing inertial frames
Frame S’ is moving to the right (relative to S) at v=1m/s. At time t = 3 sec, the position of the ball is
• x = 2 m in S• x’ = -1 m in S’
... -3 -2 -1 0 1 2 3 ...
... -3 -2 -1 0 1 2 3 ...
v
Important conclusion
• Where something is depends on when you check on it and on the movement of your own reference frame.
• Time and space are not independent quantities; they are related by rel. velocity.
• Definition: An event is a measurement of where something is and when it is there.
),,,( tzyx
Comparing inertial frames
At time t=0, the ball was at x = x’ = 2. At time t>0, the ball is still at x=2 in S but where is it in S’ at the time t>0?
a) x’ = x b) x’ = x + vt c) x’ = x-vt
... -3 -2 -1 0 1 2 3 ...
... -3 -2 -1 0 1 2 3 ...
v
Galilean position transformation
If S’ is moving with speed v in the positive x direction relative to S, then its coordinates in S’ are
ttzzyyvtxx
=′=′=′
−=′
Note: In Galilean relativity, time t=t’ is the same in both reference frames; why wouldn’t it be?!
... -3 -2 -1 0 1 2 3 ...
... -3 -2 -1 0 1 2 3 ...
v
Same thing as before, but now the ball is moving in S, too, with velocity u = –1 m/s.Is the ball faster or slower, as measured in Frame S’?
A – faster B – slower C – same speed
Galilean velocity transformation
u
Galilean velocity transformation
If an object has velocity u in frame S, and if frame S’ is moving with velocity v along the x-axes of frame S, then the position of object in S’ is:
vttxtx −= )()('
... -3 -2 -1 0 1 2 3 ...
... -3 -2 -1 0 1 2 3 ...
v
ux
x’
The velocity u’ of the object in frame S’ is therefore:
A) u + v B) u - v C) v - u D) u E) -v
Galilean velocity transformation
If an object has velocity u in frame S, and if frame S’ is moving with velocity v along the x-axes of frame S, then the position of object in S’ is:
vttxtx −= )()('
... -3 -2 -1 0 1 2 3 ...
... -3 -2 -1 0 1 2 3 ...
v
ux
x’
( ) vuvdttdxvttx
dtd
dttxdu −=−=−=
′= )()()('
The velocity of the object in frame S’ is therefore:
Important conclusion
Two observers in different reference frames can give a different description of the same physical fact (in this case, the velocity of the ball.) And they’re both right!
So far we have talked about how observers from different reference frames describe the location (x) and the velocity (v) of objects. What other quantities did we talk about in fundamental physics?
F = m a
Dynamics
... -3 -2 -1 0 1 2 3 ...
F
F
F
S
... -3 -2 -1 0 1 2 3 ...
v
S’
DynamicsIn inertial frame S, we have (in x-direction, say)
How about in inertial frame S’? Well,
since you’re still applying the same forces, and
à no additional acceleration in an inertial frame.
maF =
adtduvu
dtd
dtdua ==−== )(''
FF' =
Inertial reference frames
Now, you’re playing pool on the train. The balls roll in straight lines on the table (assuming you put no English on them). In other words, the usual Newtonian law of inertia still holds. The frame as a whole is not accelerating.
V
Galilean relativity
The laws of mechanics(good old F = ma) are the same in any inertial frame of reference.
Einstein’s First Postulate of Relativity
The laws of physics (including electromagnetism) are the same in all inertial frames of reference.
Huston, we have a problem!!From fundamental Physics you might recall that Mr. Maxwell told us the speed of light ‘c’ would be:
smc /1000.31 8
00
×==µε
But Mr. Galileo just told us that c’ = c - v
Now, Mr. Einstein tells us that ε0 and µ0 are the same in any inertial frame of reference.
What gives??
Galilean velocity transformation
If an object has velocity u in frame S (note: velocities have a direction!), and if frame S’ is moving with velocity v along the positive x-axes of frame S, then the position of the object in S’ is: vttxtx −= )()('
... -3 -2 -1 0 1 2 3 ...
... -3 -2 -1 0 1 2 3 ...
v
ux
x’
The velocity u’ of the object in frame S’ is therefore:
A) u + v B) v - u C) u - v D) u E) -v
Today
• Maxwell vs. GalileoStrange things about the speed of light
• Is there a luminiferos ether? Let’s find out!
• InterferometersLight: the ultimate yardstick.
Mr. Maxwell told us, the speed of light ‘c’ is:
smc /1000.31 8
00
×==µε
Mr. Galileo told us that c’ = c – v
If the laws of physics are the same in all inertial frames then ε0 and µ0 (and c) have to be the same in all inertial frames.
So let’s make up the “luminiferous ether” to fix Galileo’s velocity transformation law (u’ = u - v)!
(We will have to check if there is a luminiferous ether!..)
we found a problem!!
Peculiar light-waves
• A sound wave propagates through air, with a velocity relative to the air (~330 m / sec)
• A water wave propagates through water, with a velocity relative to the water (1..100 m / sec)
• “The wave” propagates through a crowd in a stadium, with a velocity relative to the audience.
• An electromagnetic wave propagates through...
Answer (19th century physics): The “luminiferous ether.”
Ideas behind Einstein’s relativity
Is there an ether ?
(There where various other motivations for special relativity, but for simplicity we will focus here on the quest for detecting the ether.)
The ether
Suppose the earth moves through the ether fixed in space with speed v. A light wave traveling at speed c with respect to the ether is heading in the opposite direction. According to Galilean relativity, what is the magnitude of the speed of the light wave as viewed from the earth? (Assume the earth is not accelerating).a) |c| b) |c|+|v| c) |c|-|v| d) |v|-|c| e) something else
v c
Quiz on reading(closed book, no talking)
The Michelson-Morley Experiment tests if the speed of light in all inertial frames…
A – …is not the same in air and in vacuum
B – …is not the same in accelerating frames
C – …is not the same in all directions
D – …does not depend on the wavelength or color
E – …does not change when reflected by mirror.
Michelson and Morley…
…performed a famous*)
experiment that effectively measured the speed of light in different directions with respect to the “ether wind.”
*) some say, the most successful failure…
How can we dothat, Morley?
You tell me,Michelson
Frame of reference
v
‘Ether’
-v -v
Ether ‘viewed’ in the laboratory on the earth:
Observer on the sun:
Ether in the laboratory framev vL
How can we measure the speed v of the ether?
If the ether would be a river, we could measure the speed of the water using a boat that travels at a known speed u’. (u’ is the relative velocity between the boat and the water.)
u'-v
If the boat travels the distance L within the time t, then we know v: L=(u’-v)t, therefore v = u’ – L/t
But: Very difficult with light! u’ = c à t ~ 10ns and v ~ 0.0001*c.à We would have to measure t with an absolute precision of ~0.0000000000001s and we have to know c very precisely!
Measuring only differences in c
-v -vL
B Lu’-v
u’+v
A
Compare the round-trip times tA and tB for paths A and B. This has the great benefit, that we do not have to measure the absolute times tA and tB (which are only a few ns) and we are less sensitive to uncertainties in the speed of light.
Michelson and Morley
-v -v
Mirrors
Light source
“Interferometer”Semi-transparentmirror
L
LDetector
The detector measures differences in the position of the maxima or minima of the light-waves of each of the two beams. (Yes, light is a wave!)
Intermezzo: Interferometers1881 Michelson invented a device now known as the ‘Michelson Interferometer.’ 1907 he received the Nobel prize for it!
We will see it in action in the famous Michelson-Morley experiment, which will lead us to the special relativity theory. à So the interferometer had a huge impact!!
Such interferometers are nowadays widely used for various precision measurements. State-of-the-art visible-light interferometers achieve resolutions of ~100pm! (X-ray interferometers are ~1pm).
(100pm = 1Å = diameter of a Hydrogen atom.)
The Michelson interferometer
Mirrors
Detector
Semi-transparentmirror
Light source
Light source
Electromagnetic waves
x
E
E-field (for a single color):
E(x,t) = E0 sin[ (2πx/λ ) ̶ ωt +φ]
λE0
φ
λ = 2πc/ω,
E
Bx
Wavelength λ of visible light is: λ ~ 350 nm … 750 nm.
ω = 2πf = 2π/T
Free physics simulations!http://phet.colorado.edu
Wave interference.jarRadio Waves.jar
Mirrors
Light source
EM-Waves in an interferometer
Light source
Semi-transparentmirror
L
L
Constructive interference
=+
Screen:
?
Esum(x,t) = ½ ·E0 sin(ωt+2πx/λ +φ) + ½ ·E0 sin(ωt+2πx/λ +φ) = ?
= E0 sin(ωt+2πx/λ +φ) = Elight source(x,t)
Light sourceLight source
L
L
Unequal arm lengths
ΔL/2ΔL
+ΔL/2
Destructive interference
=+
Screen:
?
Esum(x,t) = ½ ·E0 sin(ωt+2πx/λ +φ) + ½ ·E0 sin(ωt+2π(x+Δx)/λ +φ)
= ½ ·E0 sin(ωt+2πx/λ +φ) - ½ ·E0 sin(ωt+2πx/λ +φ) = 0
if Δx = λ / 2:sin(x+π) = - sin(x)
Light sourceLight source
Moving mirror: What do you see?
Screen
ΔL
ΔL
Intensity
λ/2
Light sourceLight source
Tilted mirror: What do you see?Fringes!
Screen
Application: Flatness measurement
Ultrahigh power laser(>3.5kW average and250MW peak power)
Michelsoninterferometer
Interference in daily life:
Do you want a bigger ‘interferometer’? There you go…
Gravitational wave detectors
Physics Review Letter, vol 116, 061102 (2016)
Do you remember this guy?The blue color originates from constructive interference of blue light and destructive interference of all other colors.
Summary for InterferometersMichelson interferometers allow us to measure tiny displacements. Displacements of less than 100 nm are made visible to the eye!
Interferometers find many applications in precision metrology such as for displacement, distance and mechanical stress measurements, as well as flatness measurements.
Interferometers have played an important role in physics:
Michelson-Morley experiment à special relativity
Testing general relativity: Gravitational wave detection
Global survey of groundwater (GRACE satellites)
End of our little “interferometer intermezzo”
