Professor Ramamurti
Shankar: So,
what did we do the last time?
Here is what we were talking
about.
We were trying to describe an
event as seen by two observers.
One will be called S and
here is the x axis for
S;
the other is called S
prime.
S prime is sliding to
the right at velocity u.
Therefore, at a certain
time--Let me say S is me
and S prime is you.
When you pass me,
I'm sitting at the origin of my
coordinates, x = 0.
You're at the origin of your
moving coordinates,
x prime = 0,
and when we crossed each other,
that time we set as 0 in our
own clocks.
We synchronized them so that
the event, x = 0,
t = 0,
also had coordinate x
prime = 0,
t prime = 0.
When you and I crossed,
we pushed our stopwatches,
we synchronized our clocks,
our origins coincided,
and that's when it began.
So, after some time,
you are here.
This is your frame and this is
my frame.
If something happens here,
something could be anything.
I hope you understand what an
event is.
That's not a relativistic
notion.
That's the very old notion.
Something happens at some place;
firecracker goes off.
I ask you, "When did it happen?"
and "Where did it happen?"
and you are given the
coordinates.
So, everything that happens is
an event.
You saw Elvis,
I say, "Where did you see
Elvis?"
At the supermarket.
Okay.
When did you see Elvis?
I would also ask you what you
were smoking because this third
question, that's not an extra
coordinate but in this case I
would like,
well, I should tell you beyond
all the laughter that's a
serious issue.
Why is that not a coordinate?
Why not ask more and more
questions and call them all
coordinates?
The reason the other things are
not coordinates is very
important.
Why is time now suddenly a
coordinate?
Well, as you could have always
asked 100 years ago when did
something happen,
and the reason is that you will
see in the new relativistic
physics, x and t,
according to one person,
get mixed up into x
prime and t prime of the
other person.
The fact that space and time
coordinates can be combined to
give the new space and time
coordinates is why it‘s called
a coordinate.
I'll say more about it later.
Now, by the way,
I've posted some notes on this
topic for those of you who want
to have a second pass at it.
I also don't know how much the
textbook covers this topic
because I spend a lot of time on
this topic, not in proportion to
what's in the book.
So, you can look at those notes
which are posted today.
Before all the Einstein stuff,
if something occurred here,
you would say it's at a
distance x prime from
your origin.
I will say it's at a distance
x from my origin.
This difference between our
origins would be ut.
Therefore, before Einstein,
I would say x prime is
x - ut and you would say
x should be what you
think is a coordinate +
ut.
This is just going back and
forth from you to me and you can
already notice that to go back
and forth from you to me,
we just change the sign of the
velocity, because I'm going to
the right and you're going to
the left [Note:
Left and right mixed up here.].
That's why the formulas have
sign of u reversed.
What is the big change after
Einstein?
First is, I admit,
the possibility that maybe you
will think the time elapsed
since we synchronized our clocks
is not necessarily the same.
I leave that option open that
t and t prime are
not the same.
Second thing I do is--I have
already explained to you that
the velocity of light coming out
the same for different people is
very counterintuitive,
because if you are going to the
right, I expect you to get a
smaller speed and yet to get the
same answer.
So, we know that it's because
we don't agree anymore on clocks
and meter sticks being the same.
So in particular,
you will say,
"I don't buy your prediction
that I expect to get the answer
to be x - ut.
I'm going to put a fudge
factor, gamma,
which depends on the velocity
between you and me,
because your lengths are not my
lengths."
Likewise, I will tell you,
"I don't agree with your
expectation that I should get
x prime + ut
prime.
I don't believe your lengths
are really my lengths,
but I'm going to put the same
fudge factor,
gamma."
This is a very interesting
result.
If I think your meter sticks
are short so that I want to blow
up the answer with some amount;
you are allowed to say the same
about me.
It's one of the big paradoxes
in relativity that we can accuse
each other of using meter
sticks,
which are short,
and we'll go into a little bit
more how that's even possible.
But the postulates tell us that
if my fudge factor is gamma,
yours should also be the same
gamma.
So, gamma is what we don't know.
I showed you a trick to find
gamma.
I said, let's imagine that this
event here, which could've been
a firecracker,
is triggered by a light pulse
that was emitted when you and I
met here.
The light pulse goes racing and
sets off an explosion here.
That's the event we are talking
about.
That's not a generic event.
It's a particular event in
which x has to be
c times t because
that's the equation for a light
pulse, according to me.
Light goes at the velocity of
c, and the event has got
the same velocity,
according to you.
This event here should be
connected by a light pulse that
for x/t is c,
and x prime/t
prime is the same c.
Then, I said,
take these three equations,
multiply the left-hand side by
the right-hand side,
put in these numbers and
extract gamma,
and I will not repeat that
part, and gamma was this.
Once you've got this,
you can take this gamma and put
it back here.
Let's see what we get.
So I'm going to put what we get
here into this board.
You guys got all of that?
Okay.
So, I'm going to put it back
here, because this is just
background from last lecture.
What we have then is x
prime = x - ut divided by
this famous square root,
and if you go and solve for
t prime in the other
equation, it's a simple
algebraic manipulation,
so I don't want to waste my
time doing that.
You will find t prime
= t - ux/c^(2) divided by
the same square root.
These are priceless.
This is really all of
relativity, all the funny stuff
you hear about E =
Mc^(2), funny clocks,
the Twin Paradox.
Everything comes from this
equation.
And you people should be very
pleased that within,
what, six weeks of starting
Physics 200,
you have all the things you
need to understand these
equations, to understand where
they came from.
That's also the remarkable
thing about relativity.
A lot of the modern things in
physics require a lot of
mathematics.
In fact, if you want to take a
beam and you want to load it
with weight and you want to see
the stresses and strains,
the math in all that is 1,000
times more difficult than this
one, and yet the mathematics of
relativity is very,
very simple,
and accessible to all of you,
and there is nothing that I
know of that you don't know as
far as how to get these
equations.
This is what you do.
Everything follows from taking
these equations and analyzing
them, extracting the
consequences and that's really
more an issue of courage than
just intelligence because once
you've got these equations,
you will have to take them,
you'll have to follow them
where they take you.
That's what we're going to do.
The rest of the whole remaining
lecture, it's all about getting
information from these
equations.
Now, some of you are not used
to writing equations.
I've noticed,
when you see something like
this, it's not clear what is
being stated.
So, let me remind you one more
time.
u is a fixed number.
That's your speed relative to
mine.
I see something happening,
I give it a pair of numbers,
xt, you give the same
event,
another pair of numbers,
x prime,
t prime,
and this is how they are
related.
So, I gave you an analogy,
but let me repeat the analogy.
If you're now talking about the
xy plane rather than the
xt plane,
that guard is not an event.
That guard is a coordinate of a
point, maybe where something is
sitting.
I give to it a pair of numbers,
x and y.
Now, you have a different
coordinate system,
rotated relative to mine,
by some angle θ.
To that same point,
same location,
you measure x prime up
to here and t prime from
there.
I'm sorry, x prime and
y prime are measured
along your axis,
and the formula there is
x prime = x cos θ + y
sin θ and y prime is
-x sin θ + y cos θ.
θ is the analog of the
velocity now.
If θ is 0,
you and I agree completely.
If θ is not 0,
your axis is rotated relative
to mine, and the same point has
two different numbers,
xy for me,
x prime/y prime
for you, and the relation
between them is this.
So, you plug in my x and
y, you can get your
x prime and y
prime.
Let me give a concrete example.
Let's take θ to
π/4 or 45°.
Then, as you know,
at 45, sine and cosine,
they're all the same.
One over root 2.
So, in that case,
x prime is x over
root 2 + y over root 2,
and y prime is -x
over root 2 + y over root
2.
That's a special case when
θ is 45°.
So, for every angle,
cosine and sine will reduce to
usual simple numbers that happen
to be the [one over]
square root of 2 here.
Then, it tells you my x
and y are related to your
x prime and y
prime in this manner,
and you can test it.
For example,
take a point with coordinates
(1,1).
(x,y) is 1,1.
If (x,y) is (1,1),
and my axis is related to yours
by 45°, (1,1) lies right there.
So, what I expect is your
y prime, there's
no y prime,
and the coordinate should be
entirely x prime.
If I put in (1,1) and indeed
you find if I put x = y =
1, y prime becomes 0.
And how about x
prime? It becomes 1+1
over root 2;
that means 2 over root 2 is the
square root of 2.
The square root of 2,
of course, would be the lengthy
measure this way because the
length of the vector is square
root of 2,
and the vector is entirely
along the x direction.
So, the length you will get for
this coordinate will be root 2.
So, come back to these
equations.
They're the same thing.
If you want,
I can write x prime as
x divided by this number
- u divided by this
number times t.
So, you can think of 1 over
this number here as the analog
of cos θ and
u over this number as the
analog of sine.
So, it's just that instead of
cosine and sine all being
upstairs, some of the numbers
are down, some of the numbers
are up.
But for a given u,
these are also numbers constant
depending on u,
and x prime is the
linear combination of x
and t,
and t prime is some
other linear combination of
x and t.
But I should warn you that it
is not an ordinary rotation.
In other words,
you cannot treat this as a
cosine and that as a sine.
They're not cosine or sine of
anything, because if they were,
this squared plus that squared
should add up to 1,
and they won't.
So, don't even try that.
This is not an ordinary
rotation but it is still what
you can call a linear
transformation.
Linear transformation means the
new numbers are related to the
first powers of the old numbers.
They are linear.
They don't involve t^(2)
and x^(2).
Okay, so that is the content of
the Lorentz transformation.
Everybody should understand
what they do by way of going
back and forth.
But now, you can go backwards.
You can say,
"Well, how do I write x
in terms of x prime and
t prime?"
There are two options open for
you.
One is, these are simultaneous
equations.
You've got to find a way to
solve for x and t
in terms of x prime and
t prime and all these
funny functions involving
u.
You treat them all as constants
and juggle them around,
multiply by this,
divide by that,
but you shouldn't do that
because you know what the answer
should be.
The answer should be the same
as what I got with the velocity
reversed.
So, x should be x
prime + ut prime or the
square root.
If I write the square root,
it means I don't feel like
putting what's in it.
It's the same old thing,
so you got tired of writing
that.
t prime will be t +
ux over c^(2) divided
by the square root.
So, one should be able to go
back and forth,
just like in this formula.
If you like,
I gave it to you as an example
long back as a homework problem.
You can write a new formula
that says x = x prime
cos θ - y sin θ and
y prime is equal to
something.
I mean, y is equal to
something something,
and the way to get that is to
either solve the equations or
realize that if I go from me to
you by a θ,
I go from you to me by -θ,
and all you have to do is
change sin θ to -sin
θ and leave the cos
θ alone.
Similarly here,
you change the sign of the
velocity u to get the
reverse transformations.
Okay, so that is the Lorentz
transformation.
Now, we are going to start
milking the transformation.
Everything is going to be
applying it to understand
various things.
The first step in getting the
mileage out of the Lorentz
transformation is,
take a pair of events.
I urge you, whenever you get a
problem with relativity,
to think in terms of events and
quite often think in terms of
pair of events.
Two events.
So, event one is,
let me give concrete events,
and most of the relativistic
examples involve some degree of
violence and so this one
involves a gun.
So, I take this gun.
I'm not going to point it at
any of you guys.
I fire the gun.
That's event 1.
The bullet hits the wall;
that's event 2.
You can take two events
connected by a bullet leaving me
and hitting the wall,
or you can take two events not
connected to anything,
okay.
You can take two unrelated
events.
It doesn't matter.
We're going to call them event
1 and event 2.
So, event 1 will have
coordinate x_1,
t_1,
right?
And according to you,
x_1 prime,
t_1 prime.
So, write the Lorentz
transformation that first
relates x_1
prime is x_1 -
ut_1 over the
square root.
Then right here,
t_1 prime is
t_1ux
_1 or c^(2)
divided by a square root.
Similarly, take the second
event and write the law for
that.
Well, it's the same thing with
the new numbers in it.
So, t_2 prime
is equal to t_2 -
ux_2 over
c^(2) divided by a square
root. Now,
take the difference of 2 - 1.
You can take 1 - 2 but it's
very common to define the
difference to be 2 - 1,
and call it
Δx_2 prime.
Δx prime is
x_2 prime
- x_1 prime.
All the Δs will be
defined to be 2nd – the 1st.
I'm telling you to take
x_1 prime,
subtract it from
x_2 prime on
the left-hand side and call it
Δ of x prime.
Δ of x prime is the
difference in the spatial
coordinates of the two events
according to you.
If you come to the right-hand
side, now this is something you
guys should be able to do in
your head.
I want to subtract that from
that.
They share the same
denominator, so let me put the
denominator there.
In the numerator you will get
x_2 -
x_1,
which in my convention I will
call it Δx and the other
one will be uΔt.
What this tells you is that
differences in coordinates also
obey the same Lorentz
transformation.
Instead of saying the
coordinate of an event was x,
t and you get x
prime,
t prime from the Lorentz
transformation,
if you take a pair of events
and they are separated in space
by Δx and in time by
Δt according to me,
the separation according to
you, your Δx primes and
Δt primes are given by
similar formulas as the Lorentz
transformation;
put a Δ everywhere.
This just came from taking the
difference of two equations
applied to the two separate
events.
All right.
You can also do this backwards,
if you like.
If you want the differences
that I get in terms of yours and
yours in terms of mine,
you have to reverse the sign of
u.
So, I won't do that again.
Yep?
Student:
Shouldn't that be t?
Professor Ramamurti
Shankar: Here?
Student: Yes.
Professor Ramamurti
Shankar: This was the
formula for x in terms of
x prime and t
prime,
and t prime in terms of
t and x.
Student:
[inaudible]
Professor Ramamurti
Shankar: Here?
Professor Ramamurti
Shankar: Okay,
I've got to give it some
numbers.
This is a,
this is b,
this is c.
Where is the problem?
One of you [inaudible]
Professor Ramamurti
Shankar: Oh,
yes yes yes.
Of course.
Yes.
Thank you.
Okay, now a public apology is
forthcoming.
You wanted me to do this.
Student: Yeah.
Professor Ramamurti
Shankar: Yes?
Thank you very much.
Okay, do not hesitate to do
this, okay?
In fact, when I said I don't
know anymore than you do about
this, it looks like I know less
than you do about this.
So, the fact that I know this
doesn't mean I'm going to get it
right.
So, all of you who screwed up
in the Midterm,
remember there is hope for you.
Maybe you won't do well as
students but you can become a
professor here.
Apparently, it's all right for
us to get things wrong.
That's correct.
Very good.
Okay, so if you're following me
that well, I'm very happy now to
know you must be following what
I'm saying here.
So now, let us--Everything now
I keep telling you I'm building
it up in a big way but
everything really is going to
come from this version of the
Lorentz transformation for
differences.
Look, suppose you were smart
enough like Einstein and you did
use these equations for gamma.
What do you do next?
You write them down.
You can publish them and say
according to me,
this is a rule for
transformation,
but you cannot stop now.
You have to say what are the
implications of my equations
because these equations are
dramatic variations of Newtonian
laws.
For example,
let us do one thing.
Two events are separated by 10
meters according to me.
Δx prime,
if Δx is 10 meters,
then Δx prime is not 10
meters.
It is 10 minus something
divided by something.
So, that distance between two
events is changing.
That's not supposed to happen
just because you get into a
train.
Right?
If I hold my hands up and say I
caught a fish that big,
that distance should be the
same for me and the train or if
you look at me riding in the
train from outside the train you
should find the same distance.
We're saying it's not true.
Likewise, if I say two events
took place when I came to Yale,
I'm certain there I got a
degree four years later,
that's supposed to be true for
anybody.
But that's also not true.
If Δt is four years,
Δx is whatever you
like.
Maybe you never left New Haven,
so Δx is 0.
That's four years,
that's not four years.
These are all drastic
consequences and we've got to
explore the consequences.
The first thing I'm going to do
is to make sure that the
velocity transformations from
one frame to other will be the
requirements we set on it.
So, here's what I'm going to do.
Event 1: I fired the gun.
Okay.
Let me just call it Event 1,
I fire the gun.
Event 2: a bullet hits the wall.
So, the separation between
these two events is Δx
for me and the time it took the
bullet is Δt.
You can also see the bullet
from your train and you think
the distance between me and the
wall is something and the time
between the firing and hitting
the wall is something else.
What's the velocity of the
bullet according to you and me?
v velocity of bullet,
for me, would be
Δx/Δt.
Okay, now I should be a little
more careful with my rotation.
Usually Δ is used for
infinitesimal numbers,
especially when you're going to
define the velocity,
velocity is the limit of these
guys going to 0.
But it is the property of
Lorentz transformation that at
this level, when I say
differences in time and
differences in space,
they are not necessarily small.
Nowhere was I assuming that the
difference in x was a
small or the difference in
[inaudible]
was small [inaudible].
They could be events separated
by five light years,
you could still use them.
But at this stage,
don't think Δx has to
be infinitesimal.
It is simply a shorthand for
difference.
If I had all the time in the
world, I would write everything
as x_2 -
x_1,
but I'm using Δ as a
shorthand.
But when I'm going to find the
velocity of a bullet,
which could even be the
instantaneous velocity,
then, here I do want to take
the limit in which they go to 0
so it becomes dx/dt .
So, from now on,
for this purpose of velocity
calculation, you should take
them to be infinitesimal and
approaching 0,
but not in general.
These equations are valid but
arbitrarily big intervals in
time and space.
Now, w better be the
velocity for you.
That is, I am S and you
are S prime.
w is Δx prime
over Δt prime with
suitable limits.
Well, from these
transformational laws,
you can get them because here
is Δx prime and here is
Δt prime.
Let me divide this by this on
the left-hand side to get
w.
When I write Δx prime
over Δt prime,
you guys take the limits of
everything going to 0.
I don't feel like writing that.
In fact, it's true even without
the limit, but let's apply it in
the end to an instantaneous
velocity.
What happens on the right-hand
side?
Here is what Δx prime
is equal to.
Can you do this in your head?
If this Δx - uΔt,
this denominator cancels
between dividing this by this,
so it looks like--I just
divided this guy by this guy,
because that's when you divide
the left-hand side by the
left-hand side,
you have to divide the
right-hand side by the
right-hand side.
Now, you should have an idea of
what I'm planning to do.
What do we do next to this
expression here?
Yes?
Student:
[inaudible]
Professor Ramamurti
Shankar: A limit of what?
If we just take a limit of
Δt going to 0,
you're just going to get 0.
That's not the limit you want
to take.
You've got to bring velocity
into the picture.
Yes?
Student:
u over t?
Professor Ramamurti
Shankar: Pardon me?
Student:
u over the t?
Professor Ramamurti
Shankar: No.
For any value of u--
Student:
[inaudible]
Professor Ramamurti
Shankar: Yes.
What you want to do now is to
divide the top and bottom of
this by Δt because you
got the velocity of the bullet
according to you.
I want to get the velocity of
the bullet according to me into
the picture.
That is Δx over
Δt, but this is not a
typical derivative in calculus,
okay?
You've got to divide everything
by Δt,
then you will get Δx
over Δt there - u
divided by 1 - u over
c^(2) times Δx
over Δt.
This is true even for finite
differences.
Now, take the limit.
Then, this becomes the velocity
of the bullet - u divided
by 1 - uv over
c^(2).
So, you've got to draw a box
around this guy.
This is another great formula.
So, I hope you know why I
divided by Δt.
I want to find the velocity of
the bullet according to you and
according to me.
For you, I took the distance
over time.
Well, I've got all these
distances and times in the top
and bottom.
I divide by the time so
everything turns into velocity.
So, this is the velocity of the
bullet according to me.
If you're going to the right,
at a speed u,
in the good old days,
what would I expect?
You've got to understand that.
My expectation in the old days
would be v - u,
right?
Whether it's going as speed
v, they're going as speed
u to the right.
So, you will see a diminished
speed by an amount equal to your
speed.
But now is the twist.
There's something in the
bottom, and the stuff in the
bottom is a number less than 1,
because of 1 - something.
We're going to jack up velocity.
Therefore, the velocity you
will actually measure is
somewhat bigger than what I
expected in the old days.
If you ever want to get back to
the good old days in any
relativistic calculation,
you should let the velocity of
light go to infinity.
Of course, it's not infinity.
What you really mean is
u/c and v/c are
negligible.
That means my velocity,
the bullet velocity,
they're all small compared to
the velocity of light;
we get back the answer from the
old days.
This is the answer from the new
days.
Now, let's find the beauty of
this result.
Let's get the reverse result.
Let us solve for the velocity I
get in terms of the velocity you
get.
Again, I think you realize you
can do the algebra,
but you guys should know that
to go from me back to you,
I should simply reverse the
sign of the relative velocity.
This is the backwards result.
In other words,
you're in a train and you fired
a bullet towards the front of
the train.
What speed do I attribute to
the bullet from the ground?
Well, I add the bullet speed to
the train speed.
That's what the numerator is.
The numerator says the answer
is somewhat less than that by
this number.
Yes?
Student:
[inaudible]
Professor Ramamurti
Shankar: Thank you very
much.
Yeah.
That's absolutely correct.
What I really need to do--quite
correct--it's not simply to
change u to -u,
but to change the velocities
that you were seeing and put
them in the place of velocities
that I was seeing.
That's correct.
This is what you want to get.
Now, let's look at the strength
of this result.
If someone tells you according
to relatively,
nothing can go faster than the
speed of light,
you can try to beat the system
as follows.
You can come to me and say,
"Can there be a gun whose
bullets go at three-fourths the
velocity of light?"
and I would say,
"yes," and you'd say,
"How about a train that goes at
three-fourths the velocity of
light?"
and I would say that seems to
be allowed.
Then you can say,
"Well, let me get into this
train at three-fourths the
velocity of light and fire a
bullet at three-fourths the
velocity,
then from the ground it should
appear to be going at 1.5 times
the velocity of light."
Well, that's the naive
expectation.
But if you do it now,
let's put w = ¾c
and u = ¾c,
the old answer is disaster;
it says 1.5c,
but the correct answer is that
plus (¾)^(2).
So, what is this guy?
This is 1.5c divided by
1 + 9/16.
You multiply everything by 16.
On the top you'll get 24,
the bottom will get 25 [times]
c.
You see, you can jack up the
velocity as seen by the ground
but it'll never be 1.5.
It'll still be less than the
velocity of light.
And the last thing you want to
check of this velocity addition
formula is why we started the
whole thing.
In other words,
suppose I see a pulse of light.
Let's go back now to this
formula.
w is v - u over 1
- uv over c^(2).
Rather than applying it to a
bullet or whatnot,
let's apply it to the light
pulse itself.
So, I saw a light pulse so I
say the object I saw,
which is what v stands
for, had a value
c.
And this is the pre-Einstein
expectation.
w should be c - u
because the pulse appears to
travel slower to a person moving
in the same direction as the
pulse.
But our new formula says it is
really that uc over
c^(2).
So, that canceled one part of
c, multiply top and
bottom by c,
and you will find it is
c.
So, the velocity of light will
always come out to be the same.
That is built into the formula
but it is a good thing to test.
So, what you find is,
velocities don't add in the
simple way.
If they did, you are in trouble.
You cannot get an upper limit
if they added in the simple way
because you can put a rocket
inside another rocket inside
another rocket and add up all
the speeds and even though each
one is less than 0,
the total could be whatever you
like.
But they don't add that way.
They add this way so that the
answer is either bigger than
what you think if v and
u are opposite,
or smaller than what you think
if the velocities in the top are
of the same sign.
No matter what you do,
the answer will always be less
than c.
Okay, so that is the first
conclusion from this.
Second conclusion,
which is again very staggering,
is that simultaneity is the
relative concept.
In other words,
if two events occur at the same
time for me, they don't occur at
the same time for you.
That is very surprising.
For example,
if you have twins born,
one in New-- sorry,
twins cannot be born in New
York and Los Angeles.
So, I've got to pick two kids
of different motherhood,
just happened to be born at the
same time.
When I say "time," we're not
talking about trivial three-hour
time difference between Los
Angeles and New York.
That's an artificial thing.
In the Einstein world,
you imagine,
we all have clocks that read
the same time.
So, by that measure,
the two kids are born at the
same time but if you watch that
from a moving train or a moving
rocket, you will disagree on
that.
You will, in fact,
say they are not simultaneous.
How does that come?
Well, it just comes from going
to this formula.
Δt prime is Δt -
uΔa x over c^(2)
divided by this number and all
I'm telling you is,
even though Δt is 0,
Δt prime is not 0.
Another shock.
Simultaneity is not absolute.
We used to think it's absolute.
In other words,
two events are occurring in Los
Angeles and New York.
They can be arranged to be
simultaneous for me,
living on the planet,
let's say,
but if you go in a rocket,
I expect you to agree they were
simultaneous.
I mean, how can they be
different?
Two things are happening right
now in different places,
it's got to be right now for
you.
But it's not.
It just comes from that formula.
Once again, if the velocity of
light is made much bigger than
everything in the problem,
simply to set it to infinity,
you will find Δt prime
is Δt.
That goes back to Galilean
times or the old pre-Einstein
times.
That's why in relativity,
space and time form a new
space-time and time is called a
fourth component because if
c goes to infinity,
Δ prime is always
Δt and t prime is
always t.
If the coordinate never mixes
with anything else,
it doesn't deserve to be called
a coordinate,
whereas, it's the x and
y coordinates that mix
with each other when you do
rotations.
After Einstein,
the space and time coordinates
mix with each other to give you
new space and time coordinates
under Lorentz transformation,
which means,
when seen in a moving train.
That's why time is now elevated
to another dimension because it
transforms very much the way
x and y did.
The details of the
transformation are different.
We have sines and cosines
replaced by u and
u over the square root
but it's still mathematically
the way components of vectors
will transform.
Okay, so simultaneity is
relative.
So, you have to ask yourself,
"How did that happen?"
So, here's the famous example
that is given in all the books.
Maybe it was given by Einstein,
I'm not sure.
So, here's a train.
Let's say the train is at rest
and you are standing in the
middle and you want to arrange
for two events to be
simultaneous.
Oh by the way,
there is one catch.
If the two events occurred at
the same time and the same
place, Δx = 0,
Δt = 0;
then, the transformation will
tell you Δx prime is 0
and Δt prime is 0.
Because two events occurring at
the same time,
at the same place means,
for example,
my two hands came and clapped,
the two hands were at the same
time at the same place.
If someone said,
well, they were not at the same
place at the same time,
it means I didn't clap.
Two cars collide.
So, if two things occur at the
same time at the same place,
something happened that they
met in space-time.
You cannot find another
observer who says they did not
meet in space-time,
or they'll say the two cars did
not collide.
Even after relativity,
it is true that two events at
the same time and same place
occur the same time and same
place for all people.
Same time alone is not enough
and same place alone is not
enough.
You see that?
If two cars came and they
collided, they were at the same
time at the same place.
If the cars were at the same
time, here and there,
that's no accident.
If they were at the same place
because if this car went and two
minutes later that car went at
the same place,
nothing happens.
That's why you say this poor
guy was in the wrong place at
the wrong time.
You don't want to say this guy
was in the wrong place.
What does that mean?
You can go to the Battle of
Gettysburg site and stand now
and nothing happens to you
because it's the same place but
it's the wrong time.
Okay?
We all realize,
same place, same time,
is a congruence in space-time
and even after Einstein,
you cannot say something did
not happen.
Okay.
We leave that to politicians
please.
It did happen.
It still happened and the
equations have the property but
I've got several events
separated in space but at the
same time.
You're in the train.
Your job is to make two things
happen at the same time and
here's what you do.
You send a beam of light that
splits into a,
you know, here's a blown up
picture.
Beam of light comes and splits
like that and goes to the back
and front of the train.
You're in the middle of the
train, and you know that events
will be simultaneous for you.
The pulse travels on either
side and hits the two things at
the two ends and sets off two
explosions and you have done the
best you can to have
simultaneous events.
Okay now, I see you from the
ground.
So, all this happening in the
moving train,
you have every right to say
you're not moving but to
me--relative to me you are
moving.
That's a fact.
And I look at how well you did
with this.
So, you sent off these two
light pulses.
Then what happened?
This wall of the train is
moving away from the light
pulse.
This wall of the train,
the rear end,
is rushing to meet the light
pulse.
Now, the velocity of light is
the same for everybody so I can
think in terms of what happens.
If one wall is rushing to meet
the light pulse and one wall is
running away from the light
pulse,
I know very clearly the
firecracker in the back of the
train will go off first and in
the front of the train will go
off later.
That means that Δt will
not be 0.
In fact, you can see Δt
will be negative because the
event with the greater x
coordinate occurs later.
That's also an agreement of the
formula.
If Δt = 0,
then Δx prime will be
some negative number.
Now, why do we bring in the
light pulse?
We bring in the light pulse
because we can tell that they
were simply not simultaneous
because about the light we know
this.
Its velocity is the same for
everybody.
That's why all arguments in
relativity involve doing things
with light pulses or
communicating with a light pulse
because we know what light does.
It travels at one and the same
velocity for all people.
That's a postulate.
Therefore, we know you couldn't
have done any better in making
them simultaneous and I simply
disagree with you.
I'd say they just did not
happen at the same time,
and there's no question of who
is right.
Operationally,
for me, those two events were
separated in time by some amount
and for you they were not.
That's a great new idea that
things did not happen the same
time for all people,
and relativity tells you that
they didn't and tells you by how
much they were different.
Okay.
Then, I take the next surprise
here.
The next surprise has to do
with clocks.
The claim is that clocks will
not run at the same rate.
Okay, when we bought the
clocks, we compared them,
we got them from the same shop,
they were completely in synch
and you put one in your train
and you got into your train and
the claim is that I will find
that your clocks are running
slow.
Let me show that to you.
Again, if you want to do
anything with relativity,
at least learn this one thing.
Go back to Lorentz
transformation and think in
terms of events that are just
going to come out of the wash.
If you try something original
on your own, you people do all
kinds of stuff.
I've seen you going in a
cyclical circular or arguments
from which you cannot even come
out until somebody rescues you.
So don't do that.
So, I'm looking at a clock.
You've got to ask yourself,
how do I turn this issue of
time into a pair of events?
I have my clock,
so it goes tick-tock,
tick-tock.
I pick two events.
Event 1: a clock says tick,
and event 2:
clock says tock.
I tell you why I have tick and
tock because I want to have two
distant kind of events so that
we can talk about them.
So, this event is a clock that
I'm carrying with me.
The clock is with me.
So, let me put a clock at the
center of my coordinates system
or the origin.
It doesn't matter where it is.
Let me put it at the origin.
The space-time coordinate x
= 0, t = 0 is the
first tick of the clock.
How about the next tick?
The next tick,
this is x and t
coordinates, the tock of the
clock, let me call
τ_0 is the
time of the clock.
It's a time period;
how many seconds elapsed
between the tick and the tock.
The main question is,
"What is the location of the
second tick of the clock?"
Where does that happen?
Yes?
Student:
[inaudible]
Professor Ramamurti
Shankar: It doesn't move
with respect to me and I'm
talking about a clock that I'm
holding in my hand so if the
first event took place at x
= 0,
second one also takes place at
x = 0.
You understand?
I'm holding the clock.
I've traveled forward in time.
It's τ_0
seconds later but the clock has
not gone anywhere so the two
events,
the two tickings of the clock,
are separated in space by 0 and
in time by
τ_0.
That means Δx =
0, Δt =
τ_0.
What do you get?
According to you,
the time difference between the
two ticks is Δt,
which is τ_0
- u etc.,
times 0 divided by this factor.
So, that means the time
difference between the two ticks
would be bigger.
If τ_0 is 1
second, if you divide 1 by [root
of]
1 - u^(2) over c^(2),
 you're going to find out
it is less.
There's more.
For example,
if this factor in the
denominator is .5,
then Δt would be 2
times τ_0. In
other words,
when your clock has taken 1
second to go from 1 tick to the
next, I will say according to me
the real time elapsed is 2
seconds.
It can be 10 seconds,
it can 100 seconds.
You can make Δt as big
as you like by letting you
approach c.
Yes?
Student:
[inaudible]
Professor Ramamurti
Shankar: Thank you.
I'm having this problem today
with the prime.
t prime is
τ_0 over this.
If you put
τ_0,
I put 1 second.
Question there, guys?
Student:
[inaudible]
Professor Ramamurti
Shankar: No,
u did not go to 0.
u is not 0.
uΔx over c^(2,)
it is the Δx that went
to 0.
Go the Lorentz transformation.
Δt prime is Δt -
uΔx over c^(2).
Δx = 0.
That's where I'm telling the
equation the clock is at rest
with respect to me.
That means that--
Student:
[inaudible]
Professor Ramamurti
Shankar: No,
no.
In this example,
let the unprimed observer be
the one holding the clock.
Then Δx is,
you know Δt is whatever
the time of the clock is.
1 second.
Δt prime is seen by
anybody else going at a speed
u relative to me.
You can also do it backwards by
taking the clock in the hands of
the S prime but you've
got to do a little more work.
It's a lot easier here because
I get to pump into the equation
two facts at one shot,
that the time between the two
ticks is 1 second and it's at
rest with respect to me,
which is why the Δx
vanished between the two ticks.
If I want to use your
equations, Δx prime
won't be 0 because the clock has
moved and you can do it.
I've, I think,
shown in my notes how to do
that but it's unnecessary.
This is the main result.
Now, here is the paradox.
You can go back to the
backwards equations.
Let's do this informally, okay.
Let's not write it again.
It will look like this.
Right.
You know that.
Let's take a clock you are
carrying.
It ticks off 1 second,
so between the tick and the
next tick, the time difference
is 1 second,
space difference is 0,
I'll find Δt is 1 over
a square root.
So, I will say your clock is
slow, you'll say my clock is
slow.
How is that possible?
How can it be that we accuse
each other of having--yes?
Student:
[inaudible]
Professor Ramamurti
Shankar: Oh,
you're absolutely right.
The predictions are in
agreement with that principle,
but I ask you to think about
the conflict you have,
yes?
Student:
[inaudible]
Professor Ramamurti
Shankar: Oh,
that's very good.
Student:
[inaudible]
Professor Ramamurti
Shankar: No,
that is correct.
Let me repeat what he said.
He said, if there are two
people, they were right next to
each other, they are at the same
height and they move apart and
each person looks at the other
person and finds the person
diminished in size because of
the distance,
and each person can tell the
other person you look small from
where I am.
In fact, even more paradoxical
is how in this world two people
can simultaneously look down on
each other.
You figure that out.
There are two people who
simultaneously have higher
opinion of themselves to the
other person and are the same
thing, they look down on each
other.
I have not found how many space
dimensions I should embed but
these people to make it that
possible, but this is certainly
correct.
A spatial resolution does give
the impression.
So, here's the answer that's
usually given to explain this to
you.
If I take a real clock like
this watch here and I ask you
why does it look slow to you
when I'm moving relative to you,
it's difficult because it's got
electronics and stuff and I
still don't know how to set the
clock on my VCR.
I'm not going to figure that
out.
So, for all of us guys who are
challenged, there is a clock
that's particularly simple,
and the clock works like this.
It has got two mirrors and a
light pulse just goes up and
down between the mirrors and
every time it completes a round
trip,
it sets off some detector and
it goes "click."
Now, this is very important to
mention that this is my x
coordinate and that's the
y coordinate and I did
not show you here there are
logical arguments and why,
if you and I are moving along
x, why our y
coordinates, if they agree in
the beginning,
continue to agree.
That's because if you say
something is two meters tall and
they pass each other,
you cannot have a disagreement
on the fact that heights are the
same.
But as in a linear dimension
there is room for discussion,
in the transverse dimension
there isn't.
So, anyway, I have this clock
that goes up and down.
Light pulse goes up and down.
It goes a certain distance,
L, in a vertical
direction, and 2L/c is
the time period of my clock.
That's the time for a round
trip.
You look at my clock and what
do you think is happening?
Remember, you are moving to the
right, relative to me,
so according to you my clock is
moving to the left,
and it looks like this.
Light beam is going on a zigzag
path, according to you.
You agree that's very clear.
For me, the pulse is going up
and down, for you it's going on
the hypotenuse.
You can imagine how with the
hypotenuse u^(2) and
c^(2) are going to come
from drawing the sides of the
triangle.
That's why the time would come
out longer.
Now, why do they like this
clock?
Because we know everything
about the operation of the
clock.
We know that this path is
longer than the straight up and
down path because the transfers'
coordinate is known to us.
Furthermore,
we know the velocity of light
is the same in all frames of
reference;
so, light going on a longer
path is simply going to take
longer.
So, I know your clock would
slow down.
Now, here is the beauty.
The beauty is that,
if you have a clock that's
going up and down,
straight up and down,
when I see it,
it's going to look like this.
So, it's as simple as saying
that if you had these two
clocks, what I think is going up
and down and you think it's a
zigzag this way,
and I look at your clock,
it's a zigzag the other way,
it's perfectly okay for me to
say to you,
your light pulse is going on a
hypotenuse and mine is going up
and down, and why you can say
that to me.
So, at least this light clock
explains to you why clocks will
appear to be slow.
Now, the question you can ask
is, what if I have some other
clock with gears and wheels and
teeth and whatnot?
How does that slow down?
The answer is,
we don't know exactly how to
explain that clock but I know
that if you carry a light clock
and another clock made of wheels
and gears,
they should run at the same
rate.
They should run at the same
rate because if they ran at
different rates,
one slowed down and one didn't,
in comparing the two clocks you
can determine your velocity,
and that we know is impossible
by the postulate that you cannot
detect uniform velocity.
Therefore, if the light clock
does something,
all clocks must do the same
thing regardless of their
mechanism, and that includes
biological clocks.
So, if you have a clock which
is just yourself,
I look at you,
you know, over 15 or 20 years I
notice some changes.
You become taller,
then your hair turns white,
your teeth fall off.
That's a clock and that clock
should also slow down.
I don't care how your life
systems work but you are a clock
and you've got to slow down.
That's why we can make
predictions about what happened
to living systems even though
that's not our main business.
So, in particular,
if you take a clock,
which is made up of mechanical
parts and a human being,
and they travel at high speeds,
the aging of the human being
should slow down just like the
clock's ticking will slow it
down,
without knowing the reasons for
it.
So, this leads to a very famous
paradox called the Twin Paradox.
The Twin Paradox says that
you've got two guys,
twins.
Now, this is actually a valid
example.
They were born at the same
time, same place,
and then the one goes on a trip
at some speed u,
and goes around and comes back.
So, I think the person's been
gone for 20 years,
my twin.
So, let's say he was 20 when he
left, so I expect him to be 40
when he gets back.
But he'll come back younger
because as a clock,
he has slowed down.
So, what I think is 20 years,
this could be,
my time is 20 years,
but his time could be 10
because his factor downstairs is
.5.
So, he can come back being
younger than me.
Yes?
Student:
[inaudible]
Professor Ramamurti
Shankar: Oh,
but we've got to be careful.
So, his point had something to
do with acceleration,
but the real question is this.
See, if I'm on the ground and I
send my twin on the trip and we
meet, now the clocks are being
compared.
As long as the zigzag is going
this way and that zigzag is
going that way,
you can say what you like.
But what if the clocks are
brought head-to-head and
compared?
Can they both be slower than
the other?
We know that's impossible.
So, instead of clocks,
we use human beings and
dramatize the paradox and say,
"Who will be younger?"
Me or the twin who went on the
rocket?
Well, there can only be one
answer to that question,
and yet no matter what the twin
says, I have no reason to
believe that I moved.
You are the one who went on the
trip the opposite way so you've
got to be younger than me.
Yes?
Student:
[inaudible]
Professor Ramamurti
Shankar: Very good.
Okay.
I couldn't have said it better.
Let me repeat it.
The point is,
as long as you've got
oppositely moving clocks,
they can believe what they want
of each other,
but if you want the clocks to
be subject to a comparison,
then you can never compare
objects that are constantly
moving at opposite velocities.
So, somebody's got to turn
around and come back.
So, in the case of the twin,
he and I were abreast,
the twin got into the rocket to
set up the velocity difference
between him and me.
So, in the early stages,
he is the one undergoing
acceleration and not me.
Likewise, the twin went
somewhere and stopped and turned
around and came back.
The twin is suffering the
deceleration.
So, there are two periods at
least, during the voyage,
when the twin has no right to
claim that he is not moving,
that I am.
That's why we don't have the
same status.
Whereas for me,
the whole time Newton's laws
and the laws of inertia were
operating for me,
whereas for him and the rocket
during takeoff,
things started flying off and
during landing things started
flying off.
He cannot possibly claim he has
the same status as me.
That's why in that problem,
the relationship is not
symmetrical, and one person can
say I did not move because I was
always inertial,
the other will have,
I have to concede that he
moved.
Yes?
Student:
[inaudible]
Professor Ramamurti
Shankar: Yeah.
Then if you do it,
if you give them symmetrical
acceleration,
both leave like this and both
come back.
Then, both will agree on their
age but they will disagree with
the person on the ground.
Suppose there's some triplets.
You have triplets and two kids
are sent out that way,
third one stays back,
the third one that stays back
will be older than the others.
So, this is not science fiction
at all.
If you want to be alive for the
year 3000, you can do it.
You just have to get into a
rocket, at sufficient velocity
close to the speed of light,
so that this is 3000 years,
okay, and you figure out how
long you've got to live.
Maybe another 50 years.
Do the math,
you find the speed,
get into the rocket and go and
come back.
Now, this experiment is done
all the time with microscopic
particles, you know.
They are accelerated in
Fermilab, for example.
They go around in a ring and
just by virtue of their motion,
they live a very long time.
So, particles are supposed to
have a short lifetime which you
calculate in their own rest
frame,
live much longer because they
are moving, and one way to keep
them moving is to put them on an
accelerated ring and they live
for a very long time.
Yes?
Student:
[inaudible]
Professor Ramamurti
Shankar: No.
No, the rocket will think I'm
50 years older.
If you went on the rocket,
it's 50 years for you but for
people on the ground it's 3,000
years.
Student:
[inaudible]
Professor Ramamurti
Shankar: Yeah,
but if you don't want to age at
all, you can arrange it.
You want, what, five years?
Okay.
You just fixed the number.
You want to age only .5 years,
that's another number.
You'll never run out of numbers.
You can let u approach
c as much as you like and
you can live as long as you want
and you can come back whenever
you want.
Other issues like going
backwards in time--I'll come
back to that later.
There are problems with that.
We'll come to that,
but you can stop time as far as
we know, and we see it all the
time.
The muons produced in the upper
atmosphere, their lifetime is
something, 10 to - 6 seconds or
whatever.
Even if they travel at the
speed of light,
we see their time is not long
enough to cross the atmosphere
but they make it here.
How did they make it?
They make it because their
lifetime is computed in their
frame of reference.
They still think they lived
only their lifetime,
but I think they lived a much
longer time.
That's how they made it to here.
But how about from their point
of view?
They only lived a short time
and how did they go 100
kilometers?
That's the next thing I'm going
to tell you.
According to them,
the atmosphere is not 100
kilometers but maybe 8
kilometers long.
Now, that's the last point
which has to do with the length
contraction.
So length contraction says,
by the way, so thing to
remember what time dilatation
is: Every clock runs the fastest
in its own rest frame.
In any other frame of reference
in which a clock has any
velocity, it will appear to be
slower than the advertised time
period of that clock.
The last thing I want to do is
length contraction.
Length contraction says,
if you have a meter stick and
you and I bought them from the
same store,
once you're in this plane or
this rocket, I will claim your
meter stick is actually shorter
than a meter.
In fact, a shortening factor is
this.
This is the shortening factor.
In fact, way back,
in the derivation of the laws
of transformation,
the fudge factor I calculated,
taking your lengths into my
lengths, is precisely connected
with this factor.
If you say the length to
something, I will tell you that
the length is actually less than
that because your meter sticks
are short.
But let me prove that to you.
So, let's take a rod that's
moving.
It's moving at a speed u;
you're carrying the rod and I
want to find its length.
What should I do?
That's the question, okay?
So, the rod is moving at speed
u.
I want to find its length.
So, Δx prime is Δx
- uΔt over the square root.
I'm going to make a pair of
events to find the length of the
rod.
Event 1, remember this rod is
zooming past me,
Event 1 is when this front end
of the rod hits a certain
marking on my graduated x
axis and event 2 is when the
back end of the train hits my
origin,
and the distance between those
two is the length of the rod,
provided one condition.
How do you find the length of a
moving rod?
You find this end and that end
at the same time.
Otherwise you will screw up,
right?
If you find this end now,
go on a lunch break and come
back and the road is over there.
And I said that's the rear end
of the rod, and you get back to
be the length;
that's not how you find the
length of a moving object.
You find the length by
measuring the coordinates of the
two ends at the same time.
Consequently,
at the same time means
Δt must be 0;
Δx is the length
according to you.
According to the person going
with the rod,
the two events,
since they take place at the
two ends of the rod,
are separated by the length of
the rod because the rod,
according to you,
is not going anywhere.
Suppose you have the rod.
It's not moving relative to you.
Something happens at one end,
something happens with the
other end.
What's the spatial distance
between them?
It's the length of the rod.
That's the meaning of the
length of the rod.
Therefore, if you
cross-multiply,
you get this result,
L is
L_0 times--;So,
a rod will appear longest in
its rest frame and to anybody
else it will appear shorter.
Clocks appear fastest in the
rest frame;
rods appear longest in the rest
frame.
For anybody to whom a rod is
moving, the rod appears short.
So, here's another paradox then.
I make a hole in my x
axis.
I make the hole to be half a
meter long.
This is half a meter.
You and I bought a meter stick
but you are moving at a velocity
where this number has become
half.
u over c is 1
over root 2.
u^(2) over c^(2)
is whatever.
Look, it's whatever it takes to
make this factor half.
So, I expect your meter stick
to have shrunk to half its
length.
So, I make a hole on the table,
half a meter long,
the rod goes by,
I think it would fall into the
hole.
That's my expectation.
You come and say,
my meter stick is a meter
stick.
The hole that you dug that you
think is half a meter long
actually is a quarter meter
long.
So, there's no way my meter
stick is going through that hole
in the table because it's four
times as long as the hole in the
table.
Do you understand that?
Normally, you make a meter
stick, you put it on top of a
hole of the same length,
it will fall down.
But if it's a moving meter
stick, which has contracted
according to me from one meter
to half a meter,
half a meter hole is enough for
it to fall.
But from your vantage point,
my half a meter hole looks like
a quarter of a meter hole.
Your meter stick still looks
like a meter.
The question is,
when this experiment occurs,
will it fall or not?
Hey, these are all paradoxes.
If you invented the theory,
you have got to defend these
paradoxes.
You cannot get two answers to
one question.
That's a logical issue.
So, the answer to these
problems is, if you know one
point of view in which it is
correct, that's the right
answer.
Then, the other person has to
figure out what happened.
So, in my point of view,
if I made a half a meter hole
and your rod has shrunk to half
the length, it's going to fall.
It will fall.
The question is,
how does the other person
reconcile himself to the fact
that here is,
according to that person,
the hole is a quarter meter
long and the rod is one meter
long?
Well, I should--for it to fall,
you've got to give it a little
tilt, okay, so that it can
actually fall in.
I give it a little here.
How can this object fall into
that?
The way other person explains
it is, he said first,
this end went in here.
Other end was sticking way out.
Sometime later,
the second end went in here.
That's how we got to the hole.
Namely, the two ends did not go
at the same time.
That way, it's certainly
possible for a one-meter stick
to go through a
quarter-centimeter hole in the
table.
I mean, give the rod a little
tilt so it can fall into the
hole.
The tip enters first;
the tail is way outside but
after a while the tail goes in
and in that manner;
the rod goes in.
So, they can reconcile this
paradox by saying the two ends
did not go at the same time.
Also, if you ask the other
person who thinks something is
two meters long and the other
person thinks it's half a meter
long,
how did they reconcile it?
I say, I measure the two ends
at the same time but you will
tell me, "You did not measure
the ends of my meter stick at
the same time.
You measured one end and you
goofed off and you came back and
measured the second end.
By the time the rod had slipped
to the right,
you measured that end and you
goofed off.
You waited until it came there
and then you measured the other
end.
That's where you got half
instead of one."
And nobody is right or wrong.
I did measure the two ends
simultaneously according to me
but you don't have to agree they
were simultaneous.
So, it's by going soft on what
simultaneity means that you are
able to reconcile this.
Because in Newtonian days,
simultaneity is absolute;
length is absolute.
In relativity,
simultaneity is relative and
length is also relative because
the operational way to find the
length of a body will not
satisfy all observers.
If I find the length of your
moving meter stick,
the two ends at the same time,
you will say I didn't measure
them at the same time and the
formulas relative will allow
that.
That's how we live with the
fact that we accuse each other
of having relatively short meter
sticks.
I blame you on your
measurement, but that blame is
not a genuine blame because you
can dig a hole equal to the
reduced length and things will
be falling.
So, you have every right to say
things are shorter,
and I will say I didn't shrink;
I fell in because my nose went
in first and then a little later
the tail went.
There are various paradoxes.
I'll just leave you with one.
You have a garage.
Your parents built you a
garage, 12 meters long.
You bought a car 14 meters long.
Can you park it?
Well, if you go at sufficiently
high velocities,
so the 14 gets shrunk to 12,
there will be a brief instant
in which both ends of your car
are inside the garage.
Of course, you will have to
smash through the back end of
the garage but you will
maintain,
"Yes, I smashed into that but
my car was in the garage for
some time" and the parent will
say,
no, the front end went in and
smashed the rear.
The back hadn't even come in.
A little later the back came.
So, everyone will agree you
broke the garage,
you broke the car,
and the lengths will contract
relatively to each other and the
phenomena is that according to
you,
there was a time the whole car
was in the garage,
that according to the parents
it was not.
 
