# Last class: Lorentz transformation Example: relativistic snake · Last class: Lorentz...

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Last class: Lorentz transformation (Relativistic version of Galileo transformation)

S

x

z

y

S' x'

z'

y'

v

(x,y,z,t)

(x',y',z',t')

)(

)(

2 xcvtt

zzyy

vtxx

−="

="

="

−="

γ

γ

Lorentz transformation (relativistic)

Example: relativistic snake

Snake’s proper length = 100cm V=0.6c – relativistic snake! – γ=5/4

Boy drops two hatchets 100cm apart in his reference frame. 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!

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)

€

t

€

t'R = γ tR −vxRc 2

$

% &

'

( ) =

540 − 0.6c ×100cm

c 2$

% &

'

( ) = −2.5ns

So the right hatchet fall before the snake’s head gets to it in both reference frames and the snake is unharmed

What is the distance between the places where the two

hatchets fall in snake’s frame? 1. From Lorentz transformations (TMZ):

€

x'R = γ(xR − vtR ) =54(100cm − 0) =125cm

2. From “relativistic common sense”: In snake’s frame: distance between the hatchets in the air: 100cm/γ; 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!!!

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’

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?

v = 0.5c

S’

u’=0.5c

S

€

ux' =Δx'Δt'

=γ Δx − vΔt( )

γ Δt − vΔx /c 2( )=

Δx − vΔtΔt − vΔx /c 2

=Δx /Δt − v

1− vΔx / Δtc 2( )=

ux − v1− uxv /c

2

€

uy' =Δy'Δt'

=Δy

γ Δt − vΔx /c 2( )=

Δy /Δtγ 1− vΔx / Δtc 2( )( )

=uy

γ 1− uxv /c2( )

€

" x = γ(x − vt)" y = y" z = z

" t = γ(t − vc 2

x)

Lorentz transformation

€

uz' =Δz'Δt'

=uz

γ 1− uxv /c2( )

Velocity transformation: Which coordinates are primed?

S x

z

y

S' x'

z'

y'

v (x,y,z,t)

(x',y',z',t')

u

Spacecraft Earth

u’ is what we were looking for! (i.e. velocity measured in S’)

Remember this? (from 2nd class)

mmmm 525)4()3( 22 ==+

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?

Time vs. space

Neither space nor time are invariant under Lorentz trans. May be some combination is.

Under Lorentz transformation we have: Time dilation (Time becomes “longer”) Length contraction (Distance becomes “shorter”)

May be if we take something like “time – length” we could get an invariant: “real” quantity independent of coordinate system

Spacetime interval

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:

The spacetime interval has the same value in all inertial reference frames! I.e. Δs2 is “invariant” under Lorentz transformations.

( ) ( ) ( ) ( ) ( )22222 zyxtcs ΔΔΔΔΔ −−−=

(Homework #3!)

21

21

21

21

tttzzzyyyxxx

−=Δ

−=Δ

−=Δ

−=ΔWith: Spacetime interval

Remember Lucy?

Lucy h

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:

tch Δ= or: 22)(0 htc −Δ=

Remember Ethel?

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

We can write:

Δy’ = h cΔt’

Δx’

€

0 = cΔ # t ( )2 − Δ # x ( )2 − h( )2And Lucy got

since 0=Δx

( ) ( ) ( ) ( ) ( ) 0,:,222220 =Δ=Δ−−−= ΔΔΔΔΔ= zhywithzyxtcsor: €

0 = cΔ # t ( )2 − Δ # x ( )2 − h( )2

Let’s define distance in a new way: that is invariant under

Lorentz transformation

€

Δs( )2 ≡ cΔt( )2 − Δx( )2 − Δy( )2 − Δz( )2

This distance is not in our usual space but in Spacetime

Spacetime

x

ct Here is an event in spacetime. Any light signal that passes through this event has the dashed world lines. These identify the ‘light cone’ of this event.

Spacetime

x

ct The blue area is the future of this event. The pink is its past.

Spacetime

x

ct Here is an event in spacetime.

The yellow area is the

“elsewhere” of the event. No physical signal can travel from the event to its elsewhere!

A

Spacetime

x

ct

A

B

(Δs)2 >0: Time-like events (A – D)

(Δs)2 <0: Space-like events (A – B)

(Δs)2 =0: Light-like events (A – C)

C D

(Δs)2 is invariant under Lorentz transformation.

€

Δs( )2 ≡ cΔt( )

2 − Δx( )2 − Δy( )

2 − Δz( )2

Example: Wavefront of a flash

x

y

z

t=0 t>0

(ct)2 - x2 - y2 - z2 = 0

Wavefront = Surface of a sphere with radius ct:

Spacetime interval for light-like event: (Δs)2 = 0

Einstein: 'c' is the same in all inertial systems. Therefore: (ct')2 - x'2 - y'2 - z'2 = 0 in all inertial systems! (Here we assumed that the origins of S and S' overlapped at t=0)

Remember Causality Paradox?

Causality: Event 1 (x,y,z,t) causes Event 2 (x’,y’,z’,t’)

If (x,y,z)=(x’,y’,z’), then it is only possible if Event 1 occurs BEFORE Event 2.

Both sets of coordinates are given in the SAME reference frame

Δs2>0: time-like events

These always occur in the same sequence, so these events can have a cause-and-effect relationship!

So causality is NOT violated in special

relativity