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.