Download Example: relativistic snake Solution: Boy is right What is the distance

Example: relativistic snake
Solution: Boy is right
Reason: Hatchets do not fall at the same
time in snake’s reference frame
Let’s assume that the left hatchet falls at
tL=tR=t’L=0 (S’-snake, S-boy)
Snake’s proper length = 100cm
V=0.6c – relativistic snake! – γ=5/4
t
Boy drops two hatchets 100cm apart in his reference frame.
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From boy’s reference frame:
Snake is moving – it is shorter – snake will be unharmed
From snake’s reference frame:
The hachets are moving – distance between them < my length
I will lose my head!
Both can’t be right!
What is the distance between
the places where the two
hatchets fall in snake’s frame?
1.  From Lorentz transformations (TMZ):
5
x'R = γ (x R − vt R ) = (100cm − 0) = 125cm
4
2. From “relativistic common sense”:
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$
vx ' 5 $
0.6c × 100cm '
t'R = γ & t R − 2R ) = &0 −
) = −2.5ns
%
(
c ( 4%
c2
So the right hatchet fall before the snake’s head gets to it
in both reference frames and the snake is unharmed
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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.
In snake’s frame: distance between the hatchets in the air:
100cm/γ=80cm; time delay between left and right: 2.5ns.
Distance traveled by snake during that time: 0.6c*2.5ns=45cm;
80+45=125cm; Same answer!"
Try to check your math this way. Easy to make a mistake!!!
Remember time dilation/length contraction!!!
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.
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
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!!
So we are already nearly done
with the transformation laws
We now can convert:
locations:
x ! x’ etc.
time:
t ! t’
But we still have to figure out:
velocities:
u ! u’
Lorentz transformation
v = 0.5c
x" = γ (x − vt)
u’=0.5c
y" = y
z" = z
Velocity transformation:
Which coordinates are primed?
S’
S
u’ is what we were looking for!
(i.e. velocity measured in S’)
v
t" = γ (t − 2 x)
c
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 )
(
€
y'
u
(x,y,z,t)
S
Earth
z
(x',y',z',t')
S'
x
z'
v
x'
Spacecraft
)
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€
v = 0.5c
Relativistic transformations
x" = γ ( x − vt )
vu −v
t " = γu("t =− 2 x ) 2
1c− uv / c
u−v
u' =
1 − uv / c 2
u’=0.5c
u=
S’
S
u=
u! + v
1 + u !v / c 2
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
d) 0.25 c
e) 0
u! + v
1 + u !v / c 2
u" =
u −v
1 − uv / c 2
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