Getting a sense for the way space time 
works in special relativity is very 
challenging and one of the ways that 
traditionally we 
Become accustomed to this and help wrap 
our minds around the weirdness, is to 
challenge ourselves with paradoxes. 
Almost invariably, they end up having to 
do with simultaneity because it's the 
relativity of what right now means that 
is really so current to everything that 
we are used to seeing. 
So, in this optional clip, I'm going to 
take advantage of my time and go through 
two of the most famous paradoxes in 
special relativity and discuss them in 
some detail and hopefully that will help 
clear things up 
and I suspect spur a lot of forum 
discussion because I'm not going to 
manage to clear everything up here. 
So let's start with a, a ladder paradox. 
Ladder paradox is if you want urban 
parking problem. 
The idea is you have a long ladder of 
length L and the short garage of length G 
but G is less than L and you want to fit 
the long ladder into the short garage. 
And the idea is very simple. 
You come running with a ladder at the 
garage at a very high speed. 
Then, therefore the long ladder is going 
to be Lorentz contracted to the point if 
v is high enough where it is shorter than 
g. 
Notice that by making v close enough to 
c, I can make the Lorentz contracted 
ladder as short as I want. 
And voila, the ladder fix in the garage. 
The ladder is fixed in the garage but 
note it only fixed in the garage while 
its moving very rapidly to the right. 
An obvious puzzle occurs because this may 
be true but relativity says I can look 
well, view the same situation from the 
point of view of somebody sitting on top 
of the ladder. 
And from the point of view of the ladder, 
the ladder is at rest, that's an inertial 
frame, the ladder is moving with a 
constant velocity to the right v. 
in the ladder frame the ladder itself is 
stationary, while the garage is moving at 
exactly the same speed v but to the left 
approaching the ladder. 
But from the ladder point of view, the 
ladder's length is L, its at rest. 
And near the hand, the garage is going to 
be low ends contracted. 
So, the garage is going to be even 
narrower than it really is. 
So, in the ladder frame we obtain the 
following picture and the two seem to be 
in conflict. 
In the picture on the left, the ladder is 
in the garage. 
In the picture on the right, there is no 
way the ladder could be in the garage 
because the garage is even shorter than 
its rest length which is as it is shorter 
than a ladder as we see at the top. 
You can make the puzzle sharper by 
saying, imagine that at the instant over 
here on the frame on the left where the 
ladder is inside the garage, I closed 
both doors. 
What happens then? 
Now I have a ladder in the garage. 
Well, now it's not ambiguous. 
The doors are closed. 
So, let's resolve this issue. 
there is, as always no real paradox. 
There's only confusion. 
And the confusion, as I said, is always 
to do with simultaneity. 
And, I'll solve this problem twice. 
The first time, algebraically, using 
Lorentz Transformation. 
And then, I'll demonstrate, graphically 
how this solution works and along the 
way, play with space time diagrams, and 
what we can or cannot read off of them. 
But I want to convince you that Lorentz 
Transformations are useful. 
So, to use Lorentz Transformations, what 
we do is, we have to be very careful that 
we're not talking about 
wishy-washy things, we want to be very 
specific about what are the things we're 
talking about. 
And the things we talk about are events, 
things that happened. 
And they happened a particular place, at 
a particular time. 
So what are the things we know? 
in order to use coordinates, I'm going to 
set up my usual space time diagram if you 
want, or my coordinates. 
I'm going to have the black coordinates 
presenting the garage because I'm 
thinking of the garage as at rest. 
And the latter is moving to the right and 
its coordinates as seen by the garage are 
t prime and x prime. 
I have to pick an origin for each of 
these. 
So I will pick the left end of the 
garage, remember the ladder is moving to 
the right. 
So the left end of the garage is going to 
be the origin for coordinates on the 
garage. 
So the garage stretches from x = 0 to x = 
G, so in the garages coordinates, 
the garage itself where this is x and t. 
The garage will be something like this. 
And this will be the point x = g. 
And the garage is stationary so it just 
sits there, 
the interior is the garage. 
I will pick the left end of the ladder to 
be x prime equals zero. 
And so, of course the left, the ladder is 
moving to the right, so if I set things 
up right and I am going to try to do that 
is in the standard way that I like the 
world line at the left end of the garage 
which is also because it is x prime 
equals zero. 
This is going to be my t prime access. 
This is, describes the motion, the 
straight world line, describes the motion 
of the left end of the ladder. 
And at some point, which I am going to 
call by convention, t equals zero and 
also t prime equal to zero. 
I will call that, pick as that event, 
event number one, or event number zero if 
you want. 
The instant when the. 
left end of the ladder enters the garage. 
So at this point, the ladder is all the 
way to the left of the garage. 
what about the right end of the ladder? 
Well, the right end of the ladder will be 
by con, by construction a distance l as 
seen in the primed coordinates to the 
right of the left end of the garage. 
So at coordinate x prime, of the right 
side of the garage is equal to l, and it 
too of course will be moving at speed v, 
at the same speed v. 
So its world line is going to be a 
parallel line here. 
this is the world line spiritually drawn, 
of the right end of the garage and as 
you've, as I've drawn it you can 
definitely see that at t equals zero, 
when the left end of the garage of the 
ladder enters the garage, the right end 
of the ladder has not yet exited the 
garage, and therefore at this point and 
in fact throughout the time that I am 
going to mark out here. 
The, this whole region is a region where 
the ladder is contained in the garage and 
then at this time, the right end emerges 
first and later the left end emerges. 
So the fact that I've drawn this to exist 
is the statement that while G is less 
than L G is bigger than the square root 
of 1 - v^2 / c^2 L. 
I have made the velocity high enough that 
the Lorentz Contracted Ladder does fit in 
the garage. 
Okay so this is event, the event that 
I'll call event number one and this here 
is event number two, event number two is 
the instant when the right end of the 
ladder is about emerge from the garage 
and that occurs of course at the position 
x^2 is equal to G as seen by the, 
by the garage frame and of course at that 
same, at that same instant, at that 
event, is the event where the world line 
of x prime equals L which is this blue 
line, the motion of the right side of the 
garage intersect x2 = g. 
So, that's event number two. 
The intersection of this green line with 
that blue line and I can compute it. 
In a moment I will. 
And then, the question of whether the 
ladder is inside the garage at any point 
or not, is the question of which event 
came first. 
If event number one, the left end of the 
ladder coming into the garage precedes 
event number two, the right end of the 
ladder emerging from the garage, then the 
ladder's in the garage. 
And in this picture it clearly is as seen 
in the frame of the garage. 
but the question is if on the other hand 
event number two precedes event number 
one, that means the right end of the 
ladder protrudes from the garage before 
the left end was in. 
The ladder was never in the garage. 
So let's figure out the order of these 
two events and again we will use Lorentz 
Transformations. 
So I've written out here the full set of 
Lorentz Transformations front and back 
forwards and backwards as you note they 
only differ by flipping the sign of'v'. 
And then. 
I am going to ask what is the time as 
seen by the garage frame at which this 
world line of the right side of the 
garage meets the world line of the right 
side of the ladder. 
Remember, this was the instant when the 
right side of the ladder pokes out of the 
right side of the garage, and so I look 
at this equation and I say x prime two is 
equal to L, so I say L over here. 
And that is going to be equal to'x', 'x' 
is'g' because it's at the right side of 
the garage and then minus'v' times't2' 
and then divided by the square root'1' 
minus'v' squared of over'c' squared. 
And okay from this I can solve, 
everything here is known, except for T2. 
So I multiply by the square root as I'm 
always doing, move the square root over 
to here. 
Notice I get the foreshortened length 
here, and what I find moving things 
around is that vt2 is G - L * the square 
root of 1 - v^2 / c^2 by construction we 
said that this is less than G so vt2 is 
positive because I was running fast 
enough. This number is small enough, the 
Lorentz constructed ladder fits into the 
garage. 
That means its right end will emerge from 
the garage after the left end entered. 
perfect that's what we set up in the 
garage's frame, the ladder fits into the 
garage. 
How does this look in the ladder's frame? 
Well and indeed there is this whole time 
as I said between the time the left end 
enters and the time that the right end 
exits that the ladder is completely 
contained in the garage. 
Good for us. 
On the other hand, what happens if I look 
at the same thing from the point of view 
of the ladder, what I want to know is at 
what time did as measured by the ladder, 
did the right end of the ladder exit the 
garage, so I want't' prime at this same 
event. 
That turns out to be easier to extract 
from the inverse transformations. 
So again x is equal to g. 
So I put it g over here. 
Over here x prime is equal to l. 
And I want T prime, so again I can solve 
for T prime. 
And I find that V T prime again 
multiplying through the square root. 
I find that'vt' prime is one over, the 
square root of one minus'v' squared 
over'c' squared times'g' minus'l'. 
Now, G is less than L, this is a number 
less than one. 
T2 prime is negative. 
remember that T2, T prime = 0 and also, 
Tzero, = 0 was the time when the left end 
of the garage entered the garage. 
But here, the right end of the garage 
emerged before. 
So indeed, in the ladder's frame of 
reference, the ladder was never in the 
garage. 
You couldn't close both doors. 
Because, as far as the ladder is 
concerned, before the tail end of the 
ladder entered the garage, the front end 
of the ladder was protruding. 
So what happens you ask if the garage 
owner in his frame when the ladder is 
completely as he sees it contained in the 
garage, swiftly closes both doors. 
Well, lets follow that in a graphic 
representation which will give us a 
chance to play with space time diagrams 
and see some of their strengths and 
limitations. 
We could also do the calculation this way 
but I want to demonstrate a graphic 
method of solution. 
So hopefully. 
What we, what did we get from here? 
We got from here that the source of the 
paradox is the simultaneity issue in the 
Garage's frame. 
Indeed the ladder is contained in the 
garage. 
In the ladder's frame, this never happens 
and the reason is that simultaneity is 
relative, and because the ladder is 
moving, the relativity of simultaneity 
means that at the same time, that say, 
the left hand entered the garage, in the 
ladder's frame the right hand was already 
out, 
okay? So, what's our graphic 
representation? 
Well here's our graphic representation, 
this is already familiar. 
I was here careful to remind myself that 
I'm plotting CT, so the red line that 
goes off at 45 degrees is the world line 
of a light beam. 
You don't need a light beam here, but it 
helps me to draw the t prime, CT prime 
and x prime axis so that the light cone 
bisects them. 
And what I have here, these two black 
lines to represent the garage 
These two blue lines represent the world 
lines. 
X prime equal to zero is the T prime 
axis. 
And x prime equals to L is the bold line 
of the right side of the garage and so 
this point here is X prime equal to L 
and. 
Both sides of the garage are moving, of 
the ladder are moving. 
Both sides of the garage are stationary. 
I may have called the ladder a garage, 
but hopefully you understand what I am 
saying by now, and you're used to my 
ambiguity. 
A word on scales in these diagrams and a 
very important point. 
So, suppose that I were to draw a grid of 
lines so these are lines of constant t, 
these are lines of constant x and some 
units so these are, and, and I've picked 
the unit so they are compatible. 
In other words, that if in some units 
this is 1, 2, 3, 4 so my garage in this 
case is precisely four units long. 
Then CT is also 1, 2, 3, 4 and you can 
see that because these 
these things intersect the 45 degree. 
Light count. 
And so I've drawn these. 
How do I draw the point, t prime = to 
one? 
Where along the t prime axis is t prime = 
to one? 
and one way to think about it. 
One, naive way, is to just measure this 
same length along this axis. 
And because it is, 
The axis is, is, not vertical. 
That would lead you to some point. 
Around here and you'd maybe try to set 
that as t prime equals the one and that 
would be incorrect. 
The point is that the distance along this 
axis, in general, distances in this 
2-dimensional plot are completely 
meaningless. 
well, how do you measure distances? 
Well, the distance is this, between any 
two points. 
Is the square root of the difference in x 
squared, the horizontal difference, plus 
the vertical difference squared. 
That's, the Pythagorean Theorem which 
applies very well to the Euclidean 
surface of this 
A tablet upon which I am drawing it or to 
this Euclidean surface of this slide, but 
this is not a physical quantity. 
In particular, there is no reason why it 
isn't all related to x prime squared plus 
c^2 t prime squared. 
Its not a physically invariant object, 
Lorentz transformations change it and 
indeed this distance means nothing 
physical and reading it off the graph 
produces no useful information. 
So, it is not true that I can measure off 
distances. 
They see and on the same scale as I did 
over there and thus, we figure out where 
the t prime equal to two save point is. 
So, how do I do it? 
Well, if I really want to plot where 
along the t prime axis, the point say t 
prime equals to 1 is, what I need to do 
is the following construction: remember 
there is something that is preserved in 
this weird and is physically meaningful 
in this low ends relativistic world, and 
that is that c^2 t^2 - x^2 is the same as 
c^2 sorry for the color violation here, 
minus x prime squared. 
And along the 
t prime axis well, that's the point where 
x prime = zero. 
So I can erase that. 
Furthermore, along the t prime axis, x = 
vt because this describes the motion of 
the left side of the ladder which is 
moving with speed v. 
So I get that c^2 - v^2. 
T squared is equal to equal to c squared 
t prime squared. 
I am going to eventually figure out 
exactly the time dilation, relativistic 
time dilation formula which is that t 
prime is t times square root of one minus 
v squared over c squared. 
What does that mean? 
That means that the time here, t prime at 
this point is less than one. 
If I draw the point t prime equal to one. 
It will be somewhere over here. 
And the point t'= two is somewhere over 
there. 
Notice that the scales of the two axis 
are different. 
This is confusing and troubling. 
It has to do with the fact that what is 
physically meaningful is this Lewenstein 
Distant, distance if you want this sort 
of weird distance with the minus sign 
rather than the 
Euclidean distance that were. 
Is the distance that we observe in the 
graph. 
And you can perform with negative value 
for the interval a similar calculation. 
And you will find that the scale of the 
x-axes also is extended. 
And in general in any space-time diagram 
the scale is smallest. 
And the Observers. 
For the observer that is addressed. 
In other words if I draw here the lines 
of constant x, the constant t prime and 
the lines of constant x prime then I find 
indeed that if I drew. 
X equals 1, 2, 3 and CT is 1, 2, 3. 
And I then want to draw the lines of x 
prime equal 1 2, 3. 
Then I need to draw them approximately 
here. 
And then x prime equals three would lie, 
lay, lie way over there. 
And likewise, t prime equals to 1 would 
be way up here, and 2 and 3 and so on. 
So the scales in this diagram are 
misleading. 
If you really want to lay scales on it, 
you need to use the invariant interval 
or, the appropriate hyperbolas to draw 
these distances. 
This is one of the reasons why doing 
calculations from these graphs can be 
confusing the thing that is always 
meaningful in these graphs is the order 
of things. 
In other words, what is true is that the 
entire region above the'x' axis is 
positive't' and that an event that is. 
further up than another is later as 
viewed by this observer and similarly the 
entire region above in this sense, the x 
prime axis is at +t prime and the region 
below it in this direction is at -t prime 
and the order of the events 
can be read off from this grid, 
you draw the grid and if you have some 
random event over here the infamous 
birthday party that we spoke of, over 
here. 
Then to read off its X and T we drop. 
Parallel to the x and t axis and we see 
that this happened at x a little larger 
than two and t getting close to four. 
If you want to plot this in x prime, t 
prime co-ordinates we. 
Draw lines parallel to the axis because 
remember this line is a line of constant 
T prime. 
It's parallel to the X prime axis, and 
this is a line of constant X prime. 
And we see that this happened at X prime 
a little less then one and T prime 
somewhere between two and three is when 
this birthday party happened. 
So this is how we read coordinates off of 
these space time diagrams, I hope that 
was helpful. 
And now I am going to erase all these 
grids which are just confusing. 
And note that I have marked here our 
favorite two events. 
Event number 1 is the instant t equal to 
zero when the left end of the ladder 
enters the garage and notice that as 
anticipated because I made this point 
which we know is L root 1 - v^2 over c^2 
less than G then there, the entire ladder 
at this point is enclosed in the garage 
and this goes on until the time of this 
event labelled two, event number two is 
the instant where the right end of the 
ladder starts protruding from the garage 
and so as I suggested there, there is 
this whole region here from t1 to t2 
where the entire ladder's inside the 
garage, 
okay? 
So, now we can ask what happens if I 
close the door. 
Well if I close the right door and oh, 
sorry and I forgot to note on the other 
hand we can see directly from this 
diagram that at event number two. 
While in the garage frame event number 
two happened after event number one and 
indeed for this whole time the ladder was 
inside the garage. 
In the ladder frame, event number two 
occurs before event number one, remember 
these are all the events that are 
simultaneous with the ladder entering the 
garage. 
Event number two corresponds. 
To some negative value of t prime down 
here and indeed it preceded as seen by 
the ladder event number one there as seen 
by the ladder the right end poked out of 
the garage before the left end got in and 
the ladder was never in the garage. 
So now you can ask, alright well, what 
happens if I close the door? 
Well if I close the door, then let me 
decide that the front door of the gara, 
the, the, the the door of the garage at g 
is going to be closed. 
What that means is that the right end of 
the ladder travels, at event number two 
it hits the garage door and from that 
moment on the left end, the right end of 
the ladder is at rest and moves along 
this green trajectory. 
Now. 
The way we're used to thinking about it, 
is a ladder is a ladder. 
If at this point in time. 
The ladder stopped, then the entire 
ladder stopped and if voila, we have the 
entire ladder contained in the garage as 
seen by the garage frame. 
But, this is confusing because in the 
ladder frame, remember, at when this 
event happened, in the ladder frame, the 
left end of the garage wasn't in the 
garage yet. 
The left end of the ladder wasn't in the 
garage yet. 
So what is this saying? 
Well, the answer to this is that we have 
made an unwarranted assumption here. 
Which is that when you stop the front end 
of the garage, of the ladder, you also 
stop the left end of the ladder. 
Remember, that at this point the right 
end of the ladder smashes into a wall, 
and it stops. 
Now, what is going to happen is the back 
end of the ladder just like our slinky, 
when I dropped it has not yet heard that 
the front end of the ladder has crashed 
into something. 
The left end of the ladder continues to 
come at speed v until somebody tells it. 
What tells it? 
Well, a sound wave, of course will be 
propagating down this ladder as the front 
end of it crumples, propagating back to 
tell the back end of the ladder, wait, 
there is an extra force, we need to stop. 
How fast does the sound wave travel? 
Well that depends on what the ladder is 
made of, but I know a limit. 
The sound wave will not travel back to 
the ladder in it's own frame, or in any 
other frame that's happily an invariant 
faster then the speed of light. 
So at this event when the front end of 
the ladder crashes into the door, you can 
imagine a sound wave propagating back up 
the ladder to the left, trying to tell 
the left end to stop. 
So, how fast can this signal propagate? 
Well, this is the fastest it can 
propagate. 
The fastest that the rear end of the 
ladder can hear, but the front has 
stopped, is the speed of light. 
So I've drawn a light speed, 45 degree 
propagation of the signal from the back 
of the ladder, from the front of the 
ladder to the back telling it to stop. 
And if this were the case. 
you notice that by the time this light 
being reaches the back end of the ladder, 
back end of the ladder is well within the 
garage because even in the ladder frame, 
we are now talking about a positive time. 
Right? 
And so, time has 
The, the, remember, the left end of the 
ladder entered the garage here. 
By now, the left end of the ladder is 
well in the garage. 
If you stop the front end of the ladder, 
the back end would keep coming. 
You would not, be able to stop it until 
it was in the garage. 
Now, what happens at the moment. 
To the actual ladder when the soundwave) 
propagates and reaches the back. 
Well the energies involved are such that 
most ladders would probably blow 
themselves up to smithereenes or crumble 
to atoms. 
But in any event by the time the rear end 
of the ladder appears that the front end 
has stopped. 
And maybe we have a very powerful ladder 
and it's elastic and it springs back out 
and re-expands by the time all that and 
crashes into the garage door. 
But the time, 
by the time all of that, 
any of that happens the ladder is well 
within the garage. 
And so the fact that you managed to close 
the doors. 
Is not an interruption when you close the 
doors. 
at this event, the ladder, had, the left 
end of the ladder was already in the 
garage, this door stopped the front end 
but the left hand kept, as seen by the 
ladder this left hand kept coming. 
So, there is no contradiction, there is a 
lot of kinetic energy to be lost and 
there might be a great explosion. 
And, of course, if the latter doesn't 
blow itself to smitherines, then of 
course there will be an elastic rebound. 
It will, this, the back end of the latter 
will now pop back to the right at high 
speed. 
Whoops. 
High speed but less than the speed of 
light crashing to the garage door and 
then whatever happens will happen 
but [COUGH] you can indeed enclose the 
ladder in the garage, at least in the 
instant. 
And the error that we were making when we 
were trying to compare the two frames and 
say you stock the ladder at the front was 
the assumption was that the ladder is a 
rigid object. 
One of the things we learned here is that 
the objects with a given shape and size. 
Do not make relativistic sense.An object 
whose pen has a size and the shape in a 
sense that when I move this side of it. 
And I drop it, the other side moves as 
well. 
This could not realistically be true, in 
fact, if you look very closely, it's not 
true. 
When I move this side, actually the pen 
bends a little. 
A sound wave goes travelling down the 
pen, and if it were a slinky with a 
slower sound wave, you'd be able to see 
it. 
And when the sound wave reaches here, 
then and only then, does the other end 
start rising. 
when you're dealing in realistic speeds, 
this little time delay makes all the 
difference. 
So thinking of an object as having a 
given size in realistic terms is a really 
bad idea. 
So hopefully we've learned some 
results of some of the ways to think 
about relativistic kinematics and 
hopefully you understand the resolution 
of the ladder diagram though as I said, 
it takes seeing the solution and then 
thinking about it a lot and arguing and I 
hope the forums will be lively. 
A perhaps even more famous paradox of 
special relativity is the twin paradox. 
And the twin paradox I'm going to 
demonstrate graphically. 
We won't do too many calculations there's 
not much to compute, 
so what's the story with the twin 
paradox? 
We have two twin brothers or sisters and 
they live until some age on Earth. 
And then, at some point, one of them gets 
into a spaceship traveling at high speed 
in some direction. 
close to relativistic speed, or traveling 
for a long time. 
And for a long time this twin travels off 
into space. 
So, he's going moving to the right. 
And then after have achieving, achieved a 
certain distance, he hops on another 
spaceship moving at the same speed but in 
the opposite direction. 
And, after a while returns to Earth and 
gets reunited with his twin. 
Now of course, when they left they were 
the same age. 
So the question is when twin number two 
returns from his voyage, who is younger? 
Are they the same age and the answer is 
no. 
They are not the same age and it's easy 
to make that calculation but then its 
puzzling to understand. 
Wait a minute, so let's first start with 
the question who is younger. 
Well. 
We have here the usual two events, event 
number zero and event number one and the 
Coordinates of these events, again I have 
my usual axis x prime, t prime t double 
prime here is the a new time axis which 
is the world line of the twin number two 
on his return but I'm going to start by 
dealing with this segment, it is sort of 
clear by symmetry that everything that 
happens along this segment happens 
symmetrically along this segment. 
And, so lets figure out what happened at 
event number one. 
So let's say that the twin on earth 
measured a time T1. 
And therefore twin on the, the twin that 
was traveling measured a time T prime 
one. 
Now where did this event occurred. 
Well this event occurred at the position. 
x1 which is equal to vt1 because the twin 
was moving with speed v, but it also 
occurred at position x prime one. 
And we know what x prime one is. 
X prime one is zero because in his own 
frame, the moving twin was addressed and 
therefore we could now set, the point is 
that we can now compute the invariant 
interval between here and here in the two 
different frames and set it to be the 
same. 
So, that tells us, and again I'm not 
going to maintain all of the colors, that 
c squared, t1 squared minus x1 squared. 
Has to be equal to'c' squared't1' prime 
squared minus'x1' prime squared. 
But we know some things, we know'x1' 
prime is zero, we know that'x1' is'vt1'. 
So, c squared t1 squared minus v squared, 
t1 squared is c squared t1 prime squared 
and we do the usual calculation and we 
find that t1 prime is t1 times the square 
root. 
Of one minus V squared over C squared. 
Or, written otherwise, t1 is t1 prime / 
by the square root of one - v^2 over c^2. 
So that, what we are saying is that this 
event appears to the stationary twin to 
have, happened at a later time than it 
appears to the moving twin. 
Why do I know it's a later time? 
Because I'm dividing by a number smaller 
than one. 
Well, yeah, you say, of course. 
But on the other hand, I can draw this 
line. 
And say that is true, but that's because 
I am measuring the time at event one. 
I can invent an event number two here. 
Which is simultaneous with the, the, the 
moving twins decision to return and that 
event number two clearly occurs at a 
time. 
Earlier than T one. 
And if I compute I will find that in fact 
T two. 
Is exactly't1' prime times you can 
compute this. 
It's again these two happened at the same 
value of't' prime but one happened at 
the'x' equals zero and one happened at'x' 
prime equals zero. 
And the symmetry between these is of 
course the usual understanding that we 
have the stationary twin sees time 
dilation, sees the moving twin's clock 
running slow and the moving twin sees the 
stationary twin's clock running slow. 
So. 
The twin whose coordinates are black 
thinks that his brother is aging slowly. 
The twin whose coordinates are blue 
thinks his brother is aging slowly. 
So how is it that upon, but on the other 
hand, it is clear that since from at this 
point t1 is less, is clearly more than t1 
prime and since this is t1 prime is how 
much the moving twin aged, t1 is how much 
the stationary twin aged. 
Since the same thing is now repeated on 
their return, it's clear that if this is 
true, then 2t1 = 2t1 prime / that and 
upon their return the stationary twin 
will be older. 
How can that be? 
We just computed that on the one hand I, 
we I thought we had this symmetry 
relation. 
Relativity tells us that if the black 
twin sees the blue clock running slow so 
does the blue twin see the black clock 
running slow. 
What ruins relativity here? 
Well, when you think about it. 
It's pretty clear that these two twins 
are not related by Lorentz 
Transformation. 
They were related by Lorentz 
Transformation, as long as twin number 1, 
2 was moving to the right at this 
constant speed, v. 
But there is a big difference in their 
life histories. 
And the big difference occurs, of course, 
here, at this instant. 
Twin Number two jumped from one spaceship 
to the other. 
He was not inertial. 
At that instant twin number two was 
actually accelerating. 
Not only that but as I drew it, since his 
velocity changed instantaneously the rate 
of change of his velocity was infinite. 
He was experiencing an infinite 
acceleration. 
So that's not very realistic, but it 
makes the calculation simple. 
But this is why the. 
Life history of the two twins is 
different, it's different and it's not, 
you don't expect symmetry if blue thinks 
black clock runs slow then black thinks 
blue clock runs slow, the difference is 
that one of them accelerated and the one 
who accelerated will be younger and this 
is brought out by drawing some auxiliary 
world lines here, so what I've done is at 
this event, event number one. 
I took, I drew before the t double prime 
axis the world line of the moving twin as 
he returns. 
I can now draw the auxiliary light cone 
and therefore the x double prime axis. 
And here we see the resolution to the 
mystery because remember before I 
suggested that we introduce event number 
two over here which is. 
What is this? 
This event is from the point of view of 
the blue twin of the, the moving twin as 
long as he is moving away from earth, 
event number two is simultaneous with 
event number one. 
So if these guys were exchanging messages 
then indeed when he decided to change 
The, to, to jump ships. 
The moving twin was, would have been 
convinced that his brother was younger 
than he was. 
And in just as the stationary twin would 
have been convinced that his brother was 
younger than he was. 
Until that point, everybody's inertial, 
and life is symmetric. 
But at the moment that he jumps ships. 
What happens is that, now, simul-, what 
he considers, when he's moving to the 
left, to be simultaneous with this same 
event #one is actually event #three over 
here. 
Is line of simultaneity changed because 
you changed velocities instantaneously. 
So in particular over the instantaneous 
jump from ship to ship, his brother aged 
this much. 
All of this aging took place essentially 
instantaneously during the instant that 
the moving twin jumped chips. 
This is the resolution.This is where the 
symmetry is broken, indeed this segment. 
Is less. 
The time, the, the twin number, 
The twin on Earth aged during this 
segment is, indeed, less than the time 
that, the moving twin aged here. 
And the same bi-symmetry is true up 
there. 
But when you add this extra segment from 
jumping ships, it's unambiguous the twin 
on Earth is older. 
This is not a contradiction to 
relativity. 
There is a difference between them. 
You could measure the difference. 
One of them accelerated. 
Okay. 
So we've cleared up the paradox 
We understand that the twin that 
accelerated remained younger and there is 
not a violation of relativity because you 
could measure the difference between the 
two. 
One was inertial, one accelerating. 
Okay, but there is an unsatisfactory 
feeling to this resolution because what 
I've essentially done is hidden all my 
ignorance in this instant of infinite 
acceleration where instantaneously as 
seen by the moving twin is. 
Suddenly aged all the difference that 
makes up the, the, the, for the, for the, 
Lorentz, the time dilation that he 
otherwise would have seen. 
This infinity doesn't make sense. 
Can I do better? 
Well of course I can do better. 
What I really want to do, and in fact, 
the only physical thing to do, is you 
can't have infinite acceleration. 
You could apply a finite acceleration and 
you could round off this corner a little 
bit. 
And now I have a story without infinite 
acceleration. 
You can ask well what happens here? 
computing here is a little bit more 
difficult, because now the moving twin is 
not in an inertial frame. 
The way I cheated before was, that I, he 
jumped from one inertial frame to another 
inertial frame. 
And I ignored the acceleration that this 
implies, now I have to deal with this 
acceleration. 
And that requires a little more 
calculation then we have. 
But we can get a sense for what this 
implies, by imagining here's a trick that 
will cause him to accelerate. 
Remember I want him to be moving to the 
right and then accelerated to the left. 
And the way I'm going to do this is I'm 
going to imagine that 
Somewhere out here, I don't know. 
Little distance not exactly in the 
direction it is moving so it isn't 
crashing there. 
But not far from there lies a star a 
heavy massive object. 
And what I am going to imagine is that as 
this twin approaches, the star is 
accelerated and then goes into hyperbolic 
orbit and is deflected by the star so 
that he comes back to the earth. 
So basically I arrange the angles so that 
he overshoots the star. 
Gets captured affected by its gravity, 
loops around and comes back. 
Okay, now since he was going very fast I 
wanted this to be relativistic, he needs 
quite a bit of acceleration to flip the 
direction of that velocity, so he'll to 
get near the star or the star has to be 
very massive whichever way you want to 
think about it. 
And this gives us another point of view 
on what is going on here and why it is 
that in an unambiguous way, not the usual 
time dilation that is symmetric between 
two, 
Two inertial frames but in an asymmetric 
way honestly physically the moving twins 
clock was running slower than the 
stationary twins clock and it somehow has 
to do with the acceleration and the idea 
is imagine braking up the trajectory in 
sort of three pieces. 
Piece number one is the piece over here 
until he approaches the star. 
So, some at this point, we imagine we can 
ignore the gravity of the star and 
everything is inertial and indeed, 
The stationary twin thinks his brother's 
aging slower. 
The moving twin thinks the stationary 
twin's aging slower. 
And indeed the same thing obtains here. 
And all the symmetric Lorentz invariant. 
Lorentz transformation's back and forth 
describe what was going on here. 
And then there's the region in between 
these two events where there's this high 
acceleration. 
And now we have a sense of what is going 
on here. 
We know that during this period the blue 
clock runs slower than the black clock. 
But from our point of view its somewhat 
clear, the black clock is very far away 
from the star, the blue clock is running 
very near to the star. 
The blue clock runs slow during this 
period because of a gravitational red 
shift, and how much gravitational red 
shift? 
Well that depends how mass of the star in 
is, is and how close. 
And that is determined by how massive the 
star has to be and how close you have to 
get to it in order to deflect this 
velocity V change your velocity from plus 
V to minus V, zooming you back towards 
Earth. 
So at least without a calculation, you 
can do a calculation, but it's some 
intuitive level that is clear that indeed 
The accelerating phase of this moving 
twins trajectory is the part the phase of 
the journey during which unambiguously 
the moving twin ages less than the 
stationary twin in a way in which they 
both agree that this is not a question 
of. 
Whose clock is measuring it, but he the 
literally he is edging slower this is 
because we can now see the asymmetry. 
He is at the bottom of the gravitational 
level and his brother at the top. 
So, thinking about the gravitational time 
dilation is a very power tool and in 
particular, it gives us the way of 
reasoning our way through the twin 
paradox and understanding what there had 
been being going on.