It’s Relatively Simple…

Nick Bremer, Erik Cox, Scott McKinney, Mike Miller, Logan Petersen, AJ Schmucker, Nick

Thull

IN REVIEW

• Einstein’s Postulates• Lorentz Frames• Minkowski Inner Product

Einstein’s Postulates

1. There exists a Lorentz frame for Spacetime.

2. A Lorentz transformation of a Lorentz frame gives a Lorentz frame.

1st Lorentz Postulate

– For stationary events, Physical Clock Time and Coordinate Time should agree.

– That is, we assume that stationary standard clocks measure coordinate time.

2nd Lorentz Postulate

• The velocity of light called c = 1.• Light always moves in straight lines with

unit velocity in a vacuum.• , time and spatial position• Note: Think of the light pulse as a

moving particle.

),( 0rvTTT

Minkowski Space(Geometry of Spacetime)

• The symmetric, non-degenerate bilinear form of the inner product has the properties

• <x,y>=<y,x>• <x1 + x2, y> = <x1, y> + <x2, y>• <cx,y> = c<x,y>• The inner product does not have to be positive definite, which

means the product of it with itself could be negative.• Non-degenerate meaning only the zero vector is orthogonal

to all other vectors• Spacetime has it’s own geometry described by the

Minkowski Inner Product.

Minkowski Inner Product• Defined on R4:• u = (u0,u1,u2,u3)• v = (v0,v1,v2,v3)• <u,v>:=u0v0- u1v1- u2v2- u3v3

• <•,•> also called the Lorentz Metric, the Minkowski metric, and the Metric Tensor

• M = R4 with Minkowski Inner Product• “•” represents the usual inner product (dot product)

in R3.• In this case you have an inner product that allows

negative length.

WorldLines

Vector Position Functions andWorldlines

• In Newtonian physics/calculus, moving particles are described by functions t r(t)

• r(t) = ( x(t) , y(t) , z(t) )

This curve gives the “history” of the particle.

Vector position functions andWorldlines cont…

• View this from R4 perspective t ( t, r(t) )

• In the above ‘t’ represents time and ‘r(t)’ represents the position.

• This can be thought of as a “curve in R4”, called the Worldline of the particle.

• A worldline (at +b) is given a non-Euclidian “geometry” in M by the Minkowski Metric.

A Brief Description of Relativistic Time Dilation

The Einstein – Langevin Clock

• Time is measured where the period between light emission and return is regarded as one unit.

Light source

Mirror

LCt 2 Where L = length of tube, and c = speed of light.

The Relativistic Time Dilation Factor

Let t’ = time of ½ pendulumLight Source

Mirror

L L = ct’

Consider a spaceship with an Einstein-Langevin Clock onboard. We will look at what happens after time t.

Time Relative to a Stationary Lorentz Frame

Light Source

Mirror

L

Ship has moved D = vt.

vt

ct

O

Z

Y

X

O’

Z’

Y’

X’

“e”

Observe:

(ct)2 = L2 + (vt)2

We’ll let t be the measure of ½ of the light pendulum according to the clock of the stationary Lorentz frame.

Relativistic Dilation Factor

• Solve for

22

22

22222

'

vc

ctt

vc

Lt

tvLtc

(ct)2 = L2 + (vt)2

We call

the relativistic dilation factor. This shows up in many equations in relativity theory. Too make things easier, we often set c = 1.

22)(

vc

cv

And recall: L = ct’

What Is Proper Time?Proper time is the elapsed time measured by a moving object.

However, this formula only applies for constant velocities.

Recall:

)1

1)( : Note(

')(1

'

2

2

vv

tvtv

tt

where represents the elapsed time of the moving object in proper time and t is the elapsed time of the moving object in coordinate time (using as a conversion factor).)(v

t

Solution:Take small time intervals on the worldline, each with approximately constant velocity.

)( ttx

)(tx vdtxd

ttxttx

)()(

t

Average Velocity

for small

2

1

2

1

22

011lim

t

t

t

ttdtvtv

1t

2ttvt 21'

With constant velocity, we can use:

As , summing the intervals together yields the following:

0t

Proper Time!

Proper Time

t

a

dttxt tv )(12

)(,

where t-a is the elapsed time measured on the stationary clock and the resulting integral is the elapsed time measured on the moving clock.

Note: t

a

dtat tv )(12

Note: T is Tau

Parameterize the Worldline Using Coordinate Time

Recall: Spacetime = E = the set of all possible events e

where E is modeled by ,,4M

4)(,)( txttz Parameterize using coordinate time:

)(,1,1 tvdtxd

dtdz

Differentiate:

Worldline: Eses )(

where and )(),()( sxstse s

Proper time represents a non-Euclidian arc-length. Proof: )(,1 tv

dtdz

Fact About Proper Time

)(,1,)(,1, tvtvdtdz

dtdz

)()(1 tvtv

2)(1 tv

Notice the similarities between the common arc-length formula and the expression above:

b

a

b

a

dtdtdz

dtdzdt

dtdy

dtdx ,

22

)(,, )(12

txtdtdtdtzd

dtzdt

a

t

a

tv

2)(1, tvdtdz

dtdz dtdt

dtzd

dtzd tv )(1

2,

This shows that Proper time is a non-euclidean arc-length.

t = b

t = a

x = x(t), y = y(t)

z(t) = ( x(t),y(t) )

For any curve, we can make a change of parameter using a function to get

)(tet )(tss ))(( tses

Since Proper Time is similar to arc-length, can the worldline be parameterized by proper time? Yes, provided the quantity called proper time is “independent of parameter”.

dtdz

dtdz

dsdt

dsdt

dtdz

dsdt

dtdz

dsdz

dsdz ,,,

2

dtdz

dtdz

dsdt

dsdz

dsdz ,,

dsdt

dtdz

dsdz

By Chain Rule

Differentiate z with respect to s:

Change parameter of the worldline:

dsdsdt

dtdz

dtdzds

dsdz

dsdz s

b

s

b

,,

Pick b so t(b) = a (i.e. if s = b, then t = a)

Make a substitution: )(stt dsdsdtdt

)))((),((,,)(

)(

stxstdtdtdz

dtdzdt

dtdz

dtdz t

a

st

bt

Thus showing proper time is independent of parameter.

Four-velocity

What is four-velocity?

4-dimensional, relativistic analog of traditional three-

velocity, represented by: ord

dz )(U

Four-velocity and Three-velocity

How is four-velocity related to traditional three-velocity?

Recall that we can parameterize worldlines by proper time:

)))((),(()( txtzz

)(v

dxd

ddt

ddzU

,)(

ddt

dtxdv

),(

)(),( v

dtxdv

Factor out

vv ,1)(

Note: shown later.)(vddt

Important Fact About Four-velocity

Four-velocity is a timelike unit vector

Proof: Show 1, d

dzddz

t

a

dttxt tv )(12

))(,(

Recall:

Differentiate using Fund. Thm. Calculus: )(1)(1

22

tvtvt

a

dtdtd

dtd

Using Chain Rule and inverses, we get:

)(,

11

)(12

v

dtdz

dtdzd

dt

tv

dtxd

dtdz

,1Recall:

dtdz

dtdzdt

xdddt

dtdz

ddz

,

1,1

dtdz

dtdzdt

xd

dtdz

dtdzdt

xdddz

ddz

,

1,1,,

1,1,

dtxd

dtxd

dtdz

dtdz

,1,,1

,

11

,1,,1

,1,,1

dtxd

dtxd

dtxd

dtxd

Thus showing and proving four-velocity

is a timelike unit vector.

1, d

dzddz

Classical Momentum

Conservation of Momentum

Recall the Classical Law of Conservation of Momentum

22112211 vmvmvmvm

It can be shown, however, that this Law is NOT Lorentz invariant for inelastic collisions.

Due to the principle of covariance, the laws of physics should hold for any Lorentz Frame.

Something must not be quite right!

So what do we do?

Answer: Would a 4-dimensional analog work?

Question: If the Classical Law isn’t quite right, how do we fix it?

What if we use 4-velocity[ ]in place of standard velocity?

)(U

Gearing up

Note that for small , || v

d

xddtxd

Recall: and

(space component of )

),()(

d

xdddtU

2||1 vdtxd

)(U

Relativistic Momentum

Our first analog of the Classical Law will look like:

22112211 UmUmUmUm

We’ll call this equation *. Let’s consider the time and space components of *.

Time Component

The time component of U(T) is simply: )(vddt

Which gives:

)()()()( 22112211 vmvmvmvm

As you may have noticed, this seems trivial since for small all it says is that . However, we will see that this will be a new law analogous to the Classical Law of Conservation of Energy.

v

2121 mmmm

Space Component

Now let’s examine the space part of *. Recall that:

dxdv

dtxd

21

dxd

dtxd

v

21

1

a.k.a. vv )(

21 vdtxd

(space component of ))(U

Space ComponentNow we can write the space component of * as:

222111222111 )()()()( vvmvvmvvmvvm

Notice that when is small,

and we get the Classical Law of Conservation of Momentum.

11

1)(2

v

v v

Philosophy

It behooves us to stop and think here for a moment. It seems that the space component of * is very close to the Classical Law of Conservation of Momentum. Does this make sense?

It does. When moving from 3 dimensions to 4, we added time. If we look at the space part of our 4-dimensional analog, it seems reasonable to see things that were developed in 3-dimensional space.

So now what?

is very close to the Classical Law of Conservation of Momentum. If we could somehow redefine mass as some sort of rest mass times , then this equation would match the Classical Law.

222111222111 )()()()( vvmvvmvvmvvm

)(v

MassCan Mass Change?

If so, what happens if Mass is not constant?

How to Define Rest Mass

•Fix a Lorentz coordinatization

•Arbitrarily choose a particle at rest

•Define the rest mass of that particle to be one

In other words,

Equals the post collision mass and velocities

The initial mass and velocity

Than by solving for m1 and than looking as v1 approaches zero

as just the mass of an object having velocity zero

Rest mass is than defined

Relativistic Momentum also called 4- Momentum

or energy momentum

p = (Rest mass)(4-velocity)

The break down of p

The time component in P is related to the Newtonian concept of Kinetic energy.

While the space component in P is related to the Classical Law of Conservation of Momentum.

Substitution and Distribution

Recall,

Inspired by the tendency of Newtonian concepts to appear in spatial components of

relativistic momentum. We determine relativistic mass to equal

M=(rest mass)(time dilation factor)

Hence,

p can now be defined as a vector with relation to M.

Observations on Kinetic Energy

Background on Kinetic Energy

• Kinetic energy is the energy of motion

• Kinetic energy is a scalar quantity; it does not have a direction.

• The Kinetic energy of an object is completely described by magnitude alone.

221 vmKE

v = speed of the object

= mass of the objectm

Setup for Conversion to Newton

Newtonian Perspective versus the classical definition of momentum

vmp

dtpdF

:Newton’s 2nd Law of Motion Explains how an object will change velocity if it is pushed or pulled upon

Information for the Proof

The Kinetic Energy’s rate of change is particularly interesting:

vFvdtpdv

dtvdmvm

dtd

)21( 2

Claim:

Conversion to Newtonian KE

)21()21( 2 vvmdtdvm

dtd

Breakup the using the product rule for dot-products2v

)()21()21( vvdtdmvvm

dtd

Pullout the constant m21

Conversion (cont.)

)()21()()21(dtvdvv

dtvdmvv

dtdm

Distribute the dot-product

))(2)(21()()21(dtvdvm

dtvdvv

dtvdm

Combine the terms

dtvdmvm

dtvdv

)21)((2

Reduce

ConclusionRecall that , so we know that F

dtvdm

Fv

dtvdmv

vFKEdtd

)(

So…

Before we move on, this equation can be used to define Relativistic Energy as:

vFdtdE

Relativistic Energy

E = mc2

Working out the equation…• Let m = rest mass

UUm ,2

pp,

),(),,( pppp oo

UmmU ,

122 mmRecall: unit vector of four velocity: <U, U>

Recall: Four momentum p = mU

Continuing on…

pppo

2

),(),,( pppp oo

From the previous slide we had:

Recall: Relativistic Mass M = po

ppM 2

This is also the Minkowski Inner Product definition

Which gives us the following equation:

ppMm 22

Differentiating both sides with respect to …

ppMdtd

20

dtpdp

dtdMM

220

dtpdvM

dtdMM

220

vFMdt

dMM

220

By Product Rule for Dot Product

Recall: vMp

Recall: by Newton’s II

Fdtpd

t

Continuing on…

vFdt

dM 0

dtdE

dtdM

0

dtdM

dtdE

dtdt

dMdtdtdE

vFMdt

dMM

220

From the previous slide we had…

By factoring out 2M we get…

Recall: Relativistic Energy

vFdtdE

Almost there…

CME

dtdt

dMdtdtdE

From the previous slide…

If we consider smaller and smaller masses that get closer and closer to zero, the energy will also get smaller and smaller; therefore not allowing for any remainder C. So we will claim C is zero.

Which brings us to…

E = M!!!

proving the equation E = mc2 where

c (speed of light) is set to 1

Special Thanks

Thanks to Dr. Deckelman for his guidance on the creation of this presentation.