V
Doppler effect:
There are two contributions to the Doppler effect: one due to the time dilation and the one due to
the movement of the source parallel to the line of sight.
The first effect will always be present, not matter how the source moves with re
Space-time displacement
Space and time form the space-time displacement 4-vector
X = (ct, x, y,z)
The coordinates of this 4-vector X transform following the Lorentzs transformations:
& x " = #x $ #%ct
(1) '
(ct " = #ct $ #%x
!
!
!
!
!
!
v
with " = .
c
The
Notes on the train and the tunnel paradox
Imagine a train and a tunnel that, when both a rest, have a length of L.
L
Now imagine that the train is moving toward the tunnel with a relative velocity of u.
According to an observer 0 at rest with r
Problem 2: Causality in the Klingon Empire
(a)
(b) We transform the three events from Kronos frame (unprimed) to
Federation negotiators frame (primed) using Lorentz transformations
x = (xct) and ct = (ctx). Here = 0.6, = 5/4. In Kronos
frame, coordinates
Thus
5
fY (y) = (y 1)3/2
2
with fY (y) = 0 otherwise.
For (2) we note that for z (1/2, 1)
1
Y
x
2
5
(y 1)3/2 dy
2
1/z
FZ (z) = P (1/Y z) = P
=
1/z
=
2
5
(y 1)3/2 dy
2
Applying the FTC, we have
fZ (z) =
5
2
1
1
z
3/2
1
z2
=
5
2z 2
1
1
z
3/2
with fZ (z) =
Answer
First note that E[Xin ] =
1
n+1 .
Then
100
.
3
2
E[Xi2 ] = 100E[X1 ] =
E[Y ] =
i
Also note, that as the Xi s are independent
Xi2
Var(Y ) = Var
1 1
5 9
2
= 100Var(X1 ) = 100
i
=
80
9
Then we note that
30 100/3
P (30 Y 35) = P
80/9
Y
35 100/3
80/9
Problem 3: Algebra
Our task is to show that the interval I = (ct)2 (x)2 (y)2
(z)2 between two spacetime events is invariant under Lorentz transformations. For brevity, we will drop the symbols (which is equivalent
to having the S and S frames synchronize
Problem 2: Mr. Tompkins and the word when
(a) Distance as measured in town frame is d = 5 km. So the distance
as seen by the moving traveler is contracted and is d which is given by
d = d/ = d
1
= 4 km
1 2
(1)
As seen by the traveler, the library gets cl
Homework 2 Solutions
Problem 1: The Train-Tunnel Paradox
In order to resolve this apparent paradox, we will examine the
three key events, both in the rest frame of the tunnel and in the rest
frame of the train. The setup of the problem makes it easier to
Problem 4: Distorted Meter Stick
(a) With = 4/5, = 5/3. In the frame S0 , we know the length
l0 = 100 cm and the angle 0 = 60 , so we can calculate x0 = 50 cm
and y0 = 86.6 cm. In the frame S, x is contracted (x = x0 / = 30 cm)
but y is not (y = y0 = 86.6
Physics 128, Modern Physics
Final, December ll, 2012 . Name
Question 1 (15 points- 5 points each)
Consider the following Lorentz transformation between coordinates in O and 0. Consider O and
O as having the same orientation for the x axis.
j
(c) Missile A gets launched at Cape Canaveral at time t = 3 AU/c
with an initial speed c. The missile deccelerates as it travels the xdirection through space (one dimensional), and at the position x 2.7
AU, it stops completely and stay at this position. T