RELATIVITYFIRST: CLASSICAL RELATIVIY

BASIC CONCEPTS:OBSERVER ─ maybe a human, but notnecessarily: an E.T., an amnimal, a ghost,a TV camera, or a XIX-century “wet plate” photographic camera may all bethough of as “observers”

BASIC CONCEPTS ─ continued:

A FRAME OF REFERENCE (or “reference frame”):a system in which the observer is located.

Examples: • Earth;• A specific object on Earth (e.g., a moving train or boat);• A spacecraft in outer space; • Another planet;• Our Galaxy;• A galaxy far, far away… and so on.

BASIC CONCEPTS ─ continued (2):

To describe the position of objects in a frame of reference,we introduce a coordinate system. In physics, we mostoften use a Cartesian system:

Just for refreshing your memory:Two-dimensional (2D) Cartesian system ─ examples:

BASIC CONCEPTS ─ continued (3):

Three-dimensional (3D) Cartesian system ─ examples (from Wikipedia):

Here it is explained how we determine the coordinates of a point Q: for findingthe x coordinate, we draw a plane parallelto the yz plane, passing through Q, and the x coordinate is where the plane intersects the x axis. And so on, for theother two coordinates.

Here is another example, also fromWikipedia, showing the “coordinate planes” ─ but don’t you think thereis something wrong with thePERSPECTIVE?!!!

BASIC CONCEPTS ─ continued (4):

Other important coordinate systems:Spherical (very important!): The position of a point is

expressed in terms of three coordinates (one linear and twoangular):

r ─ called the radius, is the distance between the point and the origin;

φ ─ called the azimuth angle,is the angle between the X axis and the projection of r on the xy plane;

θ ─ called the polar angle, orsometimes the zenith angle,is the angle between r and thethe z axis.

NOTE: φ for the azimuth and θ for the polar angle is the convention used in physics; mathematicians often use a reversed convention: θ for azimuth and φ for the polar angle.

Geographic coordinates:Note that they are similarto the spherical coordinates:

But instead of the polar angle θ,here we use latitude, the anglebetween the radius and the equatorial plane; in other words,Latitude = θ - 90º;

And instead of the azimuth, we usethe longitude. It’s essentially the same angle, but differently measured: azimuth changes from 0 to 360º, whereas the longitude from 0 to 180º to the East and to the West.

One more difference compared to the spherical system shownin the preceding slide: in geographic coordinates, we never specify the value of the

radius r . Can you tell why?

Cylindrical coordinates (not as often used as the spherical ones, but also important):

Here we use two linear coordinates:

The z-axis coordinate;

The ρ (rho) coordinate which is the distance between the origin andthe projection of thepoint on the xy plane;

And a single angular coordinate ─ namely,the azimuth angle φ.

RELATIVITY

Conventionally, we assume that the observer is locatedat the origin

Quotation from our texbook (K. Krane,Modern Physics):

“A theory of relativity is in effect a way for observers in differentframes of referenceto compare the resultsof their observations.”

INERTIAL FRAME

Suppose that an observer in her/his reference frame has thecapability of testing Newton’s Laws. If she/he finds that the Fisrst Law holds, such a frame is called an inertial frame.

Question: can you tell me what Newton’s First Law says?

Is it easy to find a real inertial frame? Well, no! A spacecraft in interstellar space, non-rotating and with all its rocket engines shut down, is something that comesto one’s mind… But you have to be a Star Trek character, otherwise it is not easy to find such an object.

Another question: if the observer finds that the Newton’s First Law holds, is it necessary to test whether the two other Laws hold?

INERTIAL FRAME, cont.Is our Mother Earth an inertial frame? NO! Because it’s rotating and orbiting the Sun. However, the deviations from the Newton’s Laws caused by these motions are really small, so in the first approximationwe can consider Earth as a “nearly inertial” frame.

Question: there are, however, some visible manifestations of thefact that Earth is not an inertial frame. Could you give some examples? Hint: if you go to the Oregon Coast, and you stay there for several hours ─ what do you see?Another hint: do you remember the “Foucalt Pendulum”?Another question: give some examples of reference frames thatare obviously non-inertial ─ such frames that you can tell right away: “Oh, this one is definitely not an inertial frame!”, without looking for subtle effects.

TRANSFORMATIONS (very important!)Transformation is a mathematical basis for comparing the physical situation as seen from different frames of reference.

Says the BLUE observer:“The red ball coordinates are (x,y) = (12, 5).”

Says the BLACK observer:“The red ball coordinates are (x’,y’) = (2, 3).”

Who is right? Both are right, of course, because each observer gives the coordinates in her own reference frame. But ask the BLUEObserver: “What are the coordinates in the BLACK observer frame?”

SimpleExample:

TRANSFORMATIONS (cont.)The blue observer answers: “I cannot know ─ unless you give methe transformation formula”.

Let’s intro-duce vec-

tor R thatsimply des-cribes the position ofthe (x’y’)frame’s origin In the (x,y)coordinates.Note that here:

R= (10,2)

!Correct! )3,2()25,1012(

);2,10( );5,12(

' so ,' : thatnote Also,

Rr

Rr

RrrrRr

Tran

sform

ation

Fo

rmu

la:

Galilean Transformation (GT)(extremely important!!!)

GT may seem trivial at the first glance, but it is really a veryimportant stuff. We will explain the GT in a step-by-step manner.

First, think of the following situation: our friend, the “black”observer is again watching the ball, but the ball is now movingwith constant speed in the x direction. There is a stopwatch showing the elapse of time, and there is an “invisible magic pen” that at every full second marks the x coordinate of the ball, as seen by the observer:

x

Q: What represents the speed of the ball V in this plot?

A: Yes! You are right! The distance between two red marks(i.e., the distance traveled in 1 s) is the graphic representation of V !

Galilean transformation – cont.Now, the situation in the “black observer’s” frame is watchedby the “blue” observer in her frame.

Q1: Are the ball position coordinates the same in the “blue observer’s” frame as in the “black observer’s” frame?

A1: Of courseNOT the same!

Q2: Is the ball speed seen by both observers the same? A2: Yes, the speed is the same!

Galilean Transformation, cont. Now, the situation in the frame of the “black” observer is the same.She sees exactly the same ball motion as before!But her entire frame is now moving with constant speed In the x direction.

The position of the ball ● in the blue frame at every full second ismarked by a red “tick” ; and the position of the moving frame’s originIs marked by a black “tick”.

The distance between two black marks _|__|_ now represents the

velocity of the black frame relative to the blue frame (“translational

velocity”); call it U .

And the distance between two red marks on the blue axis, ,

now represents the ball’s velocity in the blue frame; call it V .

Call the ball’s velocity in the black frame as V´

Galilean Transf.,Cont.

' :Obviously VUV

Galilean Trans.(more formally)

Consider two frames, xyz and x’y’z’, movingrelatively to each other with a constant velocity u.For simplicity, we will draw only two axes, x & y.

As far as the choice of x,y,z axes is concerned:

One has total freedom in choosing their orientation! (The Constitution

of Physics, 1st Amendment ☻).

So, for convenience, let’s choose the orientation of the x and x’ axesalong the direction of the relative frame motion.

Also, let’s choose the time coordinate such that at t = 0 the originsof the two frames are at the same point:

y, y’

x, x’’

y y’

x x’’

t = 0 After sometime t > 0:

u·t

Galilean Transformation (more formally, continued)

x, x’

y, y’ y y’

x x’

●●

A point

t = 0: After a while (t > 0):

zz

yy

xx

'

'

'

zz

yy

tuxx

'

'

'

u·t

Galilean Transformation (more formally, continued 2)

Position vector:

),,( zyxr

.,, where

),,(,,

dt

dzv

dt

dyv

dt

dxv

vvvdt

dz

dt

dy

dt

dx

dt

rdv

zyx

zyx

Velocity vector and its components:

xyz reference frame:

Galilean Transformation (more formally, continued 3)

Position vector:

),,(

)',','('

zytux

zyxr

. ; :and

)(' :So

.'

,'

,'

where

),,('

,'

,''

'

''

'

'''

'''

zzyy

x

x

zyx

zyx

vvvv

uvdt

dtu

dt

dxdt

tuxd

dt

dxv

dt

dzv

dt

dyv

dt

dxv

vvvdt

dz

dt

dy

dt

dx

dt

rdv

Velocity vector and its components:

x’y’z’ reference frame:

Galilean Transformation (more formally, continued 4)What about the acceleration in both systems? Well, it can be readilyshown that all three components of the acceleration vectors in thexyz and x’y’z’ systems are equal:

xyz system:

),,(,,2

2

zyxzyx aaa

dt

dv

dt

dv

dt

dv

dt

vd

dt

rd

dt

d

dt

rda

x’y’z’ system:

.0 hence and const. because ),,(

,,,,)(

),,(,,''' '''

'''

2

2

dt

duuaaa

dt

dv

dt

dv

dt

du

dt

dv

dt

dv

dt

dv

dt

uvd

aaadt

dv

dt

dv

dt

dv

dt

vd

dt

rd

dt

d

dt

rda

zyx

zyxzyx

zyxzyx

Galilean Transformation ─ summary:

tuxx '

yy '

zz '

uvv xx '

yy vv '

zz vv 'zz aa '

yy aa '

xx aa '

And, of course*: tt ' ─ i.e., time is the same in both systems

This is the full set of equations expressing the Galilean Transformation

* Why “of course”? Could you give me a good definition of time?

Time for a practical example involving Galilean Transformation!

Consider a “well-behaved” river in which the current has the samespeed u everywhere, and the width L is everywhere the same,

L

There is a buoy moored in the middle of the river, and we will use itas the origin of the xy reference frame. And there is a raft with no engine or any other propulsion system floating in the middle of the current (so it floats downriver with speed u relative to the banks).The raft will be the origin of our x’y’ reference frame.

GT practical example, continued:

Suppose that there is a swimmer (●) sitting at the river bank. When the raft passes, he starts swimming straight towards it. The current carries downstream the raft and the swimmer alike.So the swimmer always stays on the y’ axis.

But for a person watching from the shore it looks that he swims along a slanting direction ─ we can see that clearly if we mark theposition of the swimmer with red circles (o) at regular intervals.

Suppose that another swimmer also takes off from the shore when the raft passes him ─ however, she wants to swim across and reach the other bank at a point exactlyopposite the take-off point:

She has to compensate for the current speed ─ otherwise,the current would carry her downstream, like the first swimmer.

Here it is shown how the trajectory of the second swimmerlooks in the “raft frame”. For an observer sitting on the raftthe water does not move – but the buoy is moving in thebackward direction:

In the “buoy frame” the swimmer’s trajectory is parallel to the y axis (it was shown in the preceding slide, and here you can also see that the distance between the swimmer and the y axisdoes not change). But for the “raft observer” the swimmer’s path is inclined in the “up-river” direction.

Suppose that in still water (e.g., in a lake) bothswimmers swim with the same speed c .

Now, we want to know: how much time it takes each swimmerto make a full “round-trip” to the other bank and back?

With Swimmer #1 it is not a big challenge to find the time. In the “raft frame”, her velocity component in the y’ directionis vy’ = c , and the distance to be swum each way is L.

Hence, the total time needed for making a round-trip is:

c

Lt

2

That was easy. But for some reasons that will soon become clear, it is much more interesting for us to find the “round-triptime” of the other swimmer.

Swimmer #2: In the “buoy frame” Swimmer #2 swims parallel to the y axis. The total distance he covers in a round-trip is 2L , so that the “round-trip time” is:

yv

Lt

2

where vy is his speed component in the y direction. The problemis, however, that we don’t know its value!

How to find it? Here the Galilean Transformation comes in handy.In the xy system (the “buoy frame”) the swimmer’s velocity component are:

0) :know weall( ? and 0 yyx vvvSo, we apply the Galilean Transformation:

unknown) (still ? and '' yyx vvuv

It’s not yet enough to calculate the time. But we also know….

Swimmer #2, continued:

),( '',yx vvv

, cv

cvvv yx 2'2', )()(

…that in still water the swimmer speed is . And, note, in thex’y’ frame (i.e., the “raft frame”) the water is “still”.

We can write the velocity vector in the x’y’ frame as:

And we know that its magnitude (i.e., the vector’s “length”) is:

Which can also be written in terms of the components:

But we know that vx’ = -u, so we can write:

2

222'2'2 1 )(

c

ucucvcvu yy

Swimmer #2 cont.:We can now write the expression for the round-trip time inelegant form:

22 /1

122

cuc

L

v

Lt

y

Upstream-downstream swimming time:

Another thing we want to know is how much it would take the swimmers to swim the same distance L (measured along the rivershore) upstream and then back downstream.

It’s a less challenging task than before. It is straightforward thatfor an observer sitting at the river bank the swimmers’ speed during the upstream and downstream trip is, respectively:

ucvucv xx dwnstr.upstr. )( and ;)(

Upstream-downstream swimming time, cont.:

The total time needed for making an upstream-downstream trip isthen:

)/1(

12

)/1(

2

))((

)()(

22222

22

dwnstr.upstr.tot.

cuc

L

cuc

Lc

uc

ucucL

ucuc

ucLucLuc

L

uc

Lttt

)/1(

12 ;

/1

1222dwn.-up22across cuc

Lt

cuc

Lt

Let’s summarize: