Download Announcements Today`s class I want to fly to a distant star that is

Announcements
• Reading for Monday: 2.1 – 2.5
Today’s class
• Velocity transformation
• Spacetime interval
• HW 3 is posted. Due next Wed. noon.
• Exam 1 soon. It will cover Chapters 1 & 2.
Start preparations now.
I want to fly to a distant star that is 200 lightyears away.
a) I can get there if I fly really close to the
speed of light
b) No way for me to get there before I die,
because it is not possible to go faster
than the speed of light.
Remember time dilation/length contraction!!!
I want to fly to a distant star that
is 200 light-years away.
a) If I fly REALLY close to the speed of light,
not only can I make it there, but noone on
Earth will notice that I was gone
b) When I come back from my trip I will not
know anyone
No matter how fast I fly, more than 200 years will go by on
Earth!!
Last class: Lorentz transformation
(Relativistic version of Galileo transformation)
y
y'
So we are already nearly done
with the transformation laws
(x,y,z,t)
S
x
z
(x',y',z',t')
S'
We now can convert:
locations:
x  x’ etc.
time:
t  t’
v
x'
Lorentz transformation
(relativistic)
z'
But we still have to figure out:
velocities:
u  u’
Lorentz transformation
v = 0.5c
x" = γ (x − vt)
y" = y
z" = z
v
t" = γ (t − 2 x)
c
u’=0.5c
S
€
S’
u is what we were looking for!
(i.e. velocity measured in S)
y
Suppose a spacecraft travels at speed v=0.5c relative to
the Earth. It launches a missile at speed 0.5c relative to
the spacecraft in its direction of motion. How fast is the
€ missile moving relative to Earth?
Δx'
γ ( Δx − vΔt )
Δx − vΔt
Δx /Δt − v
ux − v
ux' =
=
=
=
2 =
2
2
2
Δt' γ ( Δt − vΔx /c ) Δt − vΔx /c
1 − vΔx /( Δtc ) 1 − ux v /c
uy
Δy'
Δy
Δy /Δt
uy' =
=
=
=
2
Δt' γ ( Δt − vΔx /c ) γ 1 − vΔx /( Δtc 2 ) γ (1 − ux v /c 2 )
uz
Δz'
uz' =
=
Δt' γ (1 − ux v /c 2 )
(
€
Velocity transformation:
Which coordinates are primed?
)
y'
u
(x,y,z,t)
S
Earth
z
S'
x
z'
(x',y',z',t')
Spacecraft
v
x'
v = 0.5c
Relativistic transformations
u’=0.5c
S’
S
Suppose a spacecraft travels at speed v=0.5c relative to
the Earth. It launches a missile at speed 0.5c relative to
the spacecraft in its direction of motion. How fast is the
missile moving relative to Earth? (NOTE: Remember which
coordinates are the primed ones! And: Does your answer
make sense?)
a) 0.8 c
b) 0.5 c
c) c
Lucy
d) 0.25 c
The “object” could be light, too!
Suppose a spacecraft travels at speed v=0.5c relative to
the Earth. It shoots a beam of light out in its direction of
motion. How fast is the light moving relative to the
Earth? (Get your answer using the formula).
a) 1.5c
b) 0.5 c
c) c
d) d
e) e
e) 0
v
?
Lucy
v
?
... -3
-2
-1
0
George
1
2
3 ...
... -3
George has a set of synchronized clocks in reference frame
S, as shown. Lucy is moving to the right past George, and
has (naturally) her own set of synchronized clocks. Lucy
passes George at the event (0,0) in both frames. An
observer in George’s frame checks the clock marked ‘?’.
Compared to George’s clocks, this one reads
A) a slightly earlier time
B) a slightly later time
C) same time
-2
-1
0
1
George
2
3 ...
The event has coordinates (x = -3, t = 0) for George.
In Lucy’s frame, where the ? clock is, the time t’ is
, a positive quantity.
‘?’ = slightly later time
Time vs. space
In the reference frame of the snake.
Hatchets are 80 cm apart, but they fall 2.5 ns apart
In the reference frame of the boy:
Hatchets are 100 cm apart, but they fall
simultaneously
Spacetime interval
Neither space nor time are invariant under Lorentz trans.
May be some combination is
Remember this? (from 2nd class)
The distance between
the blue and the red ball
is:
If the two balls are not
moving relative to each
other, we found that the
distance between them
was “invariant” under
Galileo transformations...
…but not under Lorentz
transformations! (Length
contraction.)  need new
definition for distance?
Spacetime interval
Say we have two events: (x1,y1,z1,t1) and (x2,y2,z2,t2).
Define the spacetime interval (sort of the "distance")
between two events as:
With:
Spacetime interval
The spacetime interval has the same value in all inertial
reference frames! I.e. Δs2 is “invariant” under Lorentz
transformations.
(Homework #3!)
Remember Lucy?
Remember Ethel?
cΔt’
h
Lucy
Δy’ = h
Δx’
Event 1 – firecracker explodes
Event 2 – light reaches detector
Geometrical distance between events is cΔt’
Distance between x-coordinates is Δx’
and: (cΔt’)2 = (Δx’)2 + h2
And Lucy got
We can write:
0 = (cΔt #) 2 − (Δx#) 2
0 = (cΔt #) 2 − (Δx#) 2 − (h) 2
since
Event 1 – firecracker explodes
Event 2 – light reaches detector
Geometrical distance between events is h.
Time between the two events is Δt.
And we know that:
or:
or:
(Δs)2 ≡ (cΔt)2 − (Δx)2 − (Δy)
€
This
2
−
Spacetime
ct
Here is an event in
spacetime.
(Δz)2
distance is not in our usual space but in Spacetime
(h)2
€
€
Let’s define distance in a new
way: that is invariant under
Lorentz transformation
−
x
Any light signal that
passes through this event
has the dashed world
lines. These identify the
‘light cone’ of this event.
Spacetime
ct
Spacetime
ct
The blue area is the
future of this event.
x
The pink is its past.
Here is an event in
spacetime.
A
x
The yellow area is the
“elsewhere” of the
event. No physical
signal can travel from
the event to its
elsewhere!