Life in the fast lane:the kinematics of Star Trek

Last week we learned that length and time are not absolute, but instead, depend on the frame of reference of the observer.

We derived the relations:

2

21

'

cv

TT

2

21

cv

LL p

Using only geometrical arguments and applying Pythagorean's theorem.

time dilation

length contraction

dcT/2 222 2/2/ cTdvT

2/2 cTd in red frame

in blue frame

Time always runs slower when measured by an observer moving with respect to the clock.

The length of an object is always shorter when viewed by an observer who is moving with respect to the object.

To an observer at rest in one frame, the light in the other frame travels further because the other frame is moving with respect to the observer. Since c is constant in all frames, to each observer time appears to be moving more slowly and length appears to be contracted in a reference frame moving with respect to the observer.

Everything is slowed/contracted by afactor of:

in a frame moving with respect to the observer.

2

21

1

cv

x

y

Let’s look at this another way: thinking Newtonianly…

Professor Hoffman rolls a cat toy across the floor towards her cat.

x

y

The same cat, the same cat toy, different (arbitrary) choice of

coordinate systems.

By Pythagorean’s theorem, the toy rolls:

22 yxd

Easier visualization, easier calculation—Professor Hoffman is happy (although her cat probably doesn’t care as long as he gets his

toy.)

This is an example of rotating your coordinate axes in space.

Coordinate systems

Time, the fourth dimension?

“Spacetime”

x

ct

In the previous slide, the two space dimensions were shown to be interchangeable. A similar relationship can be used to express the relationship between space and time in relativity.

x=ct

Light propagating in one dimension in a spacetime coordinate system as viewed from a frame S. The distance traveled is equal to the speed of light times the time elapsed.

The red and blue spaceships viewed in spacetime coordinates.

The red and blue spaceships are at x=0, t=0 when oneemits a pulse of light.

At that instant one of the spaceships starts to move away from the other with velocity v.

The lightcone- the distance light has traveled since x=0, t=0 as a function of time.

Who perceives him/herself to be at the center of the lightcone? A passenger on the red spaceship or a

passenger on the blue spaceship?

The red spaceship’s reality. The blue spaceship’s reality.

They are both at the center of the light cone!

This can be achieved by rotating their coordinate axes as Professor Hoffman did when playing with her cat, except one thing…notice that one set of axes is not orthogonal! You need to add a Lorentz boost.

The velocity of a particle in spacetime coordinates is:

slope

c

tc

xc

t

xvelocityparticle

The speed of light is absolute, therefore the axes must be drawn to preserve the speed of light!

x

ct

a light pulse is sent out

x=ct

ctx tcx

in frame S

in frame S’

x’

ct’ and x’ axes must be symmetrically placed with respect to the light pulse!

ct’

the enterprise is traveling in a reference frame with v<c w/r to S

ct ct’

xx’ Vtx

Newtonianly:

Vtxx

tt

x

ctx=ctct’

x’

ctc

Vxx x

c

Vcttc

The axes aren’t perpendicular but are scaled by some factor. They must be symmetric w/r to the light pulse. What is this factor?

Hendrik Lorentz: a guy who was almost as smart as Einstein

Maxwell’s highly successful equations…

Contain a constant velocity!

In the 19th century, these equations were thought to hold only in the luminiferous ether!

Lorentz wanted to make these equations covariant (the same in all reference frames). He came up with the Lorentz transformation:

2

21

1

cv

Recognize this?

Gauss’ law

no magnetic monopoles

Faraday’s law

Ampere’s law

Galilean transformation from S to S’:

vtxx

Guess that the relativistic version has a similar form but differs by some dimensionless factor G:

S S’v

x x’

)( vtxGx

(we know that this must reduce to the Galilean transformation as v/c ->0)

The transformation from S’ to S must have the same form:

tvxx )( tvxGx

From first postulate of relativity-laws of physics must have the same form in S and S’

)( vtxGx )( tvxGx substitute into

solve for t’, you get:

v

xG

tGt 11' 2

Now we need an expression for the velocity dx’/dt’ in the moving frame:

Take derivatives of: )( vtxGx

v

xG

tGt 11' 2

)(' vdtdxGdx

v

dx

GdtGdt )1

1('

2

)/)(11(1 2 VuG

vu

td

xdu

From second postulate of relativity- the speed of light must be the same for an observer in S and S’

where u=dx/dt

u=dx/dt=c u’=dx’/dt’=c

)/)(11(1 2 VuG

vu

td

xdu

Plug this into:)/)(11(1 2 VcG

vcc

and get:

Solve for G:2

21

1

cv

G

The Lorentz transformations!S S’

v

x x’

The transformation:

To transform from S’ back to S:

ct ct’x=ct

x

x’

Spacetime is invariant!

True, observers will differ on the length of objects or the time events occurred, however, they will always agree on how far apart two events are in spacetime.

E1

E2

To find the space and time coordinates of an event in a specific frame, draw lines from the event parallel to the axis of that frame.

2122

12222 xxttcxtcs

2122

12222 xxttcxtcs

22 ss !

ct

x

x=ctx=-ctA B

C

O

Could an event at O cause A?

Yes, because a “messenger” at O would not have to travel at a speed greater than the speed of light to get there.

Could an event at O cause B?

A light signal sent from O could reach B.

Could an event at O cause C?

No, the spacetime distance between O and C is greater than could be covered by light. It would require time travel.

xtc

xtc

xtc

here,now

where light that is here now may have been in the past

where light that is here now may go in the future

Warp speed revisited Can we achieve warp speed?

It’s beginning to become clear that we cannot.

The Lorentz factor approaches 1 asymptotically, but never exceeds it:

1

We will gain more insight into why this happens when we talk about relativistic mass next time.

One more thought about the kinematics of Star Trek: a worked example

The enterprise is speeding along at some “realistic” velocity (less than the speed of light). Let’s pretend that these stars emit light of a single wavelength. Will a passenger on the enterprise perceive the star to be the same color as a passenger on a Klingon ship that is hovering near the star? (Hint, think about the waves in the still versus the moving stream from the last lecture.)